Chemistry the Core Concepts Part I 2e - Janusa 2010

Michael A. JanusaGlenn V. Lo

ChemistryThe Core ConceptsPart I

2nd Edition

Michael A. JanusaGlenn V. Lo

CHEMISTRYThe Core ConCepTs

parT I2nd Edition

Kona Publishing and Media GroupHigher Education DivisionCharlotte, North Carolina

Cover Design and Typesetting: diacriTech

Copyright © 2010 by Kona Publishing and Media Group

All rights reserved. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of Kona Publishing & Media Group.

ISBN: 978-1-935987-17-8Library of Congress Control Number:

ABOUT THIS E-TEXT

This textbook is an attempt to provide a rigorous but learner-friendly introduction to Chemistry. This textbook should serve the needs of science and non-science majors alike.

1. the very first concepts are defined in simple terms that even someone with a minimal back-ground in science can understand

2. the order of presentation of topics is meticulously chosen so that sufficient background has been provided by the time each new concept is introduced

3. examples are immediately provided to illustrate and/or apply each new concept4. “test yourself ” questions provide additional reinforcement; links to videos explaining the

answers are provided5. “study resources” provide study tips and links to useful and interesting websites6. additional study questions provided in the companion website (i-assign.com) can be assigned

as computer-graded homework7. useful physics and math background/review, often relegated to appendices in typical text-

books, are integrated into the development of the topics

For alternative sequencing of topics (for a typical first-semester course), it is useful to organize the chapters into the following learning units:

Unit 1: Introductory Concepts (Ch. 1–8)Unit 2: Atomic and Molecular Structure and Properties (Ch. 9–16)Unit 3: Counting Atoms, Molecules, and Ions (Ch. 17–20)Unit 4: Reactions in Aqueous Solutions (Ch. 21)Unit 5: Molarity and solution stoichiometry (Ch. 22)Unit 6: Ideal gas behavior and gas stoichiometry (Ch. 23–25)Unit 7: Real Gases, Liquids, and Solids (Ch. 26)Unit 8: Redox Reactions (Ch. 27) Unit 9: Solutions (Ch. 28)Unit 10: Thermochemistry (Ch. 29)

iii

iv About This E-Text

Units 2, 3, and 4 may be covered in any order after Unit 1. Units 3 and 4 must be covered before Unit 5. Unit 3 must be covered before Unit 6 or 8. Units 2 and 6 must be covered before Unit 7. Units 5 and 6 must be covered before Unit 9.

The companion website, i-Assign.com, is a web-based homework delivery and collection system, as well as a test bank authoring system and test generator. The automation afforded by i-Assign reduces a teacher’s workload while providing the following benefits.

1. By setting periodic deadlines for graded homework, teachers are able to encourage students to study on a regular basis and focus on a manageable amount of material.

2. Teachers can set up practice homework to give students more opportunities for mastery.3. Homework and test individualization minimizes cheating.4. Teachers can add their own questions.5. Multiple versions of a test, which can be printed or delivered online, can be easily generated.

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TABLE OF CONTENTS

Chapter 1: What is Chemistry? . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1

Chapter 2: Measurements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7

Chapter 3: Atoms, Ions, and Molecules . . . . . . . . . . . . . . . . . . . . . .29

Chapter 4: Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .43

Chapter 5: Classification of Matter . . . . . . . . . . . . . . . . . . . . . . . . . .77

Chapter 6: The Periodic Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . .89

Chapter 7: Compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

Chapter 8: Chemical Changes . . . . . . . . . . . . . . . . . . . . . . . . . . . .117

Chapter 9: History of Atomic Theory . . . . . . . . . . . . . . . . . . . . . . .137

Chapter 10: Quantum Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . .149

Chapter 11: Atomic Orbitals . . . . . . . . . . . . . . . . . . . . . . . . . . . . .165

Chapter 12: Electron Configuration . . . . . . . . . . . . . . . . . . . . . . . .181

Chapter 13: Periodic Trends in Atomic Properties . . . . . . . . . . . . . .193

Chapter 14: Molecular Structure . . . . . . . . . . . . . . . . . . . . . . . . . .205

Chapter 15: Quantum Mechanical Description of Molecules . . . . . . .235

Chapter 16: Intermolecular Forces . . . . . . . . . . . . . . . . . . . . . . . .245

Chapter 17: The Mole Concept . . . . . . . . . . . . . . . . . . . . . . . . . . .257

Chapter 18: Molar Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .263

Chapter 19: Elemental Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . .273

Chapter 20: Stoichiometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .281

Chapter 21: Reactions in Aqueous Solutions . . . . . . . . . . . . . . . . . .293

Chapter 22: Molarity and Solution Stoichiometry . . . . . . . . . . . . . . .313

Chapter 23: Ideal Gas Behavior . . . . . . . . . . . . . . . . . . . . . . . . . . .325

Chapter 24: Counting Gas Particles . . . . . . . . . . . . . . . . . . . . . . . .339

Chapter 25: Kinetic Molecular Theory . . . . . . . . . . . . . . . . . . . . . .355

Chapter 26: Real Gases, Liquids, and Solids . . . . . . . . . . . . . . . . . .371

Chapter 27: Redox Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . .405

Chapter 28: Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .427

Chapter 29: Thermochemistry . . . . . . . . . . . . . . . . . . . . . . . . . . .439

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CHAPTER

What is Chemistry?

1.1 ScienceChemistry is a science, so we begin by defining science. Science is a study that seeks to understand something based on empirical (or factual) information. Major ideas in science are consistent with unbiased and verifiable facts; they are not mere opinions. A more thorough way of defining science is to describe what scientists do. Scientists:

• conduct research to acquire knowledge or solve problems.• summarize knowledge into laws and theories.• disseminate knowledge.

1.1.1 Scientific ResearchScientists engage in research to acquire knowledge. In doing so, they follow a procedure called the Scientific Method. It involves the following:

• Formulating the question or problem.• Formulating a hypothesis.• Conducting experiments to test a hypothesis.• Analyzing experimental results and making a conclusion.

A hypothesis is a tentative answer to a question or problem and is defined as an educated guess. What makes it educated? A scientist tries to find out as much of what is known or related to the problem before making a hypothesis. This is done by reading scientific publications (the scientific literature), making observations, and relating the observations to what is already known from experience. Observations or data are factual information based on what is sensed (seen, touched, heard, tasted, smelled) or measured.

A hypothesis must be testable. It is tested by doing an experiment. Observations made or data collected in an experiment should be reproducible. Performing replicate measurements (multiple trials) is standard procedure in scientific work. Things that could affect the outcome of an experiment are called variables. A common type of research attempts to establish relationships between two variables. In this case, a series of

1

2 Chemistry: The Core Concepts

experiments is performed where one variable (the independent variable) is deliberately chosen as another variable is measured (dependent variable). When doing this type of research, it is important to try to identify all the other possible independent variables and hold all of those variables constant.

The analysis of experimental results should attempt to determine if they support the hypothesis. The experimental results may be inconsistent with the hypothesis and lead one to a conclusion that the hypothesis needs to be rejected or revised. If the hypothesis is revised, further experiments obviously need to be done.

▼ Example 1

A patient comes into a doctor’s office with an illness. The first question that a doctor might try to answer is “what is the illness”.(a) What would the doctor have to do to formulate a

hypothesis?(b) Suppose the doctor’s hypothesis is that the illness is some-

thing due to a certain type of bacteria. What experiment could be done to test the hypothesis?

Answers:(a) The doctor could examine the symptoms (patient’s tempera-ture, rashes, etc.) and ask the patient about other symptoms, previous illnesses, medication the patient may be currently taking, etc. The doctor could then relate the information to what he has learned from medical school and his experiences with previous patients. The doctor could also refer to medical publications.(b) There may be established laboratory procedures for detecting the presence of the bacteria in a blood or urine sample. If the test turns out to be negative, then the doctor will have to come up with another hypothesis.

1.1.2 Disseminating KnowledgeSharing of information is important for the advancement of knowledge. Once a scientist reaches a conclusion about a novel problem that others could find useful, it is important that the information be disseminated. One way this is done is by writing a paper that describes the work. The paper should not only present the results, but should also describe the procedures in sufficient detail so that other scientists can duplicate the experiments and verify the results. The paper is then submitted for publication in scientific journals. Reputable journals use a peer review process; articles submitted for publication are sent to other scientists to examine for reliability.

Chapter 1: What is Chemistry? 3

1.1.3 Laws and TheoriesOrganizing and summarizing information are important for the advancement of knowl-edge. Scientists do these by formulating laws and theories.

A pattern tends to emerge when numerous related observations are examined. A statement that summarizes the pattern is called a law (of nature). For example, the state-ment “heat naturally flows from a hot object to a colder one” is a law of nature. For us to be able to say that this is a law, we must have made this observation numerous times. Laws generally have no known exceptions at the time they are formulated. Over time, excep-tions might be found, but it is customary to still refer to the original statement as a law. Oftentimes, the statement of a law specifies exceptions; such a law is called a limiting law. A law can be written in the form of a mathematical equation; an example is the so-called ideal gas law, PV = nRT. The ideal gas law is an example of a limiting law; it is only valid as P approaches zero.

The word theory is often mistakenly used when referring to a hypothesis. This may be fine in layman’s terms, but not in science. In science, theory and hypothesis are not the same. A theory is a set of fundamental ideas that is consistent with everything (observations and laws) that is known about a subject. These fundamental ideas are called postulates. The postulates of a theory are assumptions; they cannot be proven to be true but they are accepted for as long as there are no known exceptions. Laws with no known exceptions could be among of the postulates of a theory. A theory becomes widely accepted (or well-established) if there is overwhelming evidence to support it. All one needs to disprove a theory is one fact that contradicts one of the postulates of the theory. However, should such an exception be found, it is typically not necessary to discard a theory; rather, one or more postulates is/are modified in order to account for the excep-tion. The more a theory has been modified, the more reliable it becomes. So what is the difference between a hypothesis and a theory? A hypothesis is an educated attempt to explain or predict something. A good way to come up with an educated explanation or prediction is to examine the known facts in light of a well-established theory. If you do that, you could say that your hypothesis is a theoretical explanation or a theoretical prediction but it is not the theory.

1.2 Chemistry and Related DisciplinesChemistry is an example of a natural science. In a natural science, the “thing” that we seek to understand is nature itself. Other examples of natural sciences are physics, biology, geology, and astronomy.

Physics and Chemistry, the physical sciences, are both defined as the study of matter and its properties. Matter is the stuff that the universe is made of; anything that has mass


and occupies space. The term property refers to characteristics of matter, such as color, taste, melting point, boiling point, etc. Physics focuses on the mathematical description of the most basic property of matter, motion. Physicists precisely define concepts such as force, energy, heat, light, and electric charge to summarize everything that is known about motion into a few fundamental equations. Chemistry, on the other hand, focuses on transformations of matter; we call these transformations chemical changes or chemical reactions. Chemists try to understand matter in terms of the idea that all matter is made up of tiny particles called atoms, and in terms of theories about how these particles behave. We could say that chemists study matter from an atomic perspective. Chemists study macroscopic or bulk systems (things we can see) and interpret the results from a microscopic standpoint. In Chemistry, the word microscopic refers to atomic sizes, which are much smaller than cells in living things. You could say that what a chemist calls microscopic, a biologist would call sub-microscopic.

Why study Chemistry? By understanding atoms, we are better able to make matter behave the way we want it to. A strong motivation for studying science is that it allows us to develop technology that would make our lives better; the word technology refers to the practical application of science. Here are some examples of practical applications of Chemistry: how to make a drug more potent and have fewer side effects, how to make something taste more like sugar but less fattening, how to make a battery produce more current over a shorter period of time, how to make touch-sensitive computer screens, how to make computer chips process information faster, how to counteract the effect of a poison, how to make less expensive, sturdier, lighter building materials, etc.

Most modern scientific work is interdisciplinary or multidisciplinary. A discipline is a field of study. When something is interdisciplinary, it involves two or more fields. For example, Mathematics and Physics are used extensively in Chemistry to characterize the behavior of matter (in bulk samples or at the atomic level); work involving these may be classified as Physical Chemistry or Chemical Physics. Chemistry and Physics are the most basic of all the sciences. Both are used extensively in the other sciences, such as Biology (which deals with life) and Geology (which deals with the earth). Much of the progress in modern biology is due to (and relies on) an understanding of Chemistry. Living things are made up of atoms! Biochemists and molecular biologists are primarily concerned with studies of things associated with life from an atomic perspective (or chemical per-spective); they deal with molecules such as proteins, carbohydrates, lipids, nucleic acids, which are found in living things. The word molecule refers to two or more atoms that are bound together.

Chapter 1: What is Chemistry? 5

TEST YOURSELF

For answers and other study resources, see: http://i-assign.com/ebook/answers/chapter1.htm.

1. Consider the following statements: “A drop of alcohol on a lab bench disappears after 5 minutes. The drop disappeared because molecules of alcohol moved apart and spread into the entire room.” These statements, respectively, are:

A. hypothesis and theory B. observation and hypothesis C. observation and theory D. law and theory

2. Consider the following statement: “Matter expands when heated.” This statement is an example of:

A. a law B. a theory C. a hypothesis D. an observation

3. Which of these statements is/are true? 1. It is good to do research on a problem or question before formulating a hypothesis. 2. It is not possible to prove a theory. A. 1 only B. 2 only C. both D. neither

http://i-assign.com/ebook/answers/chapter1.htm

7

2CHAPTER

Measurements

A measurement is a quantitative comparison. It usually (not always!) has two parts: a number and a unit. The unit tells us what the basis is for the comparison. For example, when we say that something has a mass of 5 kg, we mean that its mass is 5 times that of the metallic cylinder stored in a vault at the BIPM (Bureau International des Poids et Mesures, www.bipm.org, or International Bureau of Weights and Measures) in Sevres, France. The cylinder in BIPM is defined to have a mass of 1 kg and is an example of what is called a standard reference. By reference, we mean that it is the basis for comparison. By standard, we mean that people (governments) agree to use it as the reference; as a matter of convenience and necessity people have to agree on a reference. In this chapter, we will discuss how to assess the reliability of measurements, how to properly report measure-ments as well as values calculated from measurements, and how to convert units.

2.1 Reliability of MeasurementsThere are two characteristics of a measurement that we use to assess its reliability: accu-racy and precision. Although often used interchangeably, accuracy and precision are two different things. Accuracy refers to the “correctness” of a measured value, while precision refers to its reproducibility or consistency. Precision is necessary in order to be accurate; it is difficult to be correct if you are inconsistent. However, good precision does not neces-sarily lead to high accuracy; it is possible to be consistently wrong (precisely inaccurate).

2.1.1 AccuracyWe can determine the accuracy of a measured value by comparing it with the expected (correct or true) value. To assess the accuracy of a measured value, we calculate either the absolute error:

Absolute Error = Measured value − Correct value or the relative error:

Relative Error = Absolute Error / Correct ValueRelative error is often expressed as a percentage; in this case it is referred to as the percent error. The smaller the absolute or relative error is, the more accurate the measurement.

http://www.bipm.org


▼ Example 1

On a laboratory exam, students were asked to determine the amount of ascorbic acid in an “unknown” tablet. One student deter-mined the amount to be 225.0 mg. The correct value is 250.0 mg. Calculate the absolute error and the relative error.Answer:Absolute Error = 225.0 mg − 250.0 mg = −25.0 mgRelative Error = −25.0 mg / 250.0 mg = −0.100, or −10.0%

A negative error means that the measured value is less than the correct value. Similarly, a positive error means that the measured value is larger than the correct value.

Why should we bother measuring something if we already know what it is? In a teaching laboratory, it would be a way for the teacher to check if a student has learned the skills pertinent to the course (as illustrated in the preceding example). In general, the reason is to check if the instrument or method that we would be using for subsequent measurements is reliable. Checking if an instrument is giving correct values and adjust-ing it so that it gives correct values, or figuring out how to calculate the correct values, is called calibration or standardization. We do this using standard reference materials and/or standard operating procedures. For something to be considered a “standard” its validity should ultimately be traceable to an authorized agency; in the United States, this agency is the NIST (National Institutes of Standards and Technology, www.nist.gov). In cases where a standard is not defined, the “correct” value is assumed to be one that is predicted by a well-established theory (the theoretical value), or one previously reported in a peer-reviewed publication (the literature value).

We can tell there is a problem with the accuracy of a measurement if we repeat it several times and find that the errors are “one-sided” (that is, consistently positive or consistently negative). This type of error is called a systematic or determinate error and is avoidable. It could, for example, be due to an improper calibration of the instrument; to avoid it, we re-calibrate. It could be due to a shortcoming of the method used; to avoid it, we modify the method.

2.1.2 PrecisionIn general, it is not possible to obtain exactly the same result every time we repeat a mea-surement regardless of how carefully we try to replicate the procedure. This is due to natural fluctuations in the property being measured and in the response of measuring instrument to the property. Errors associated with these natural fluctuations are called indeterminate or random errors. Random errors cannot be avoided, but can be minimized

http://www.nist.gov

Chapter 2: Measurements 9

by doing multiple trials. If errors are truly random, positive and negative errors tend to be equally probable. Thus, it is good practice to perform multiple trials and take the average as the measured value; errors in the individual trial values will tend to cancel each other out when we calculate the average. The simplest way of assessing the precision the measured value is to calculate the range, the difference between the highest and lowest trial values:

Range = highest value − lowest valueThe smaller the range is, the more precise the measured value.

▼ Example 2

A student determined the time it takes for rock to fall from the top of a building to the ground below. Based on the following results, calculate the range.Trial 1: 3.12 s Trial 2: 3.15 s Trial 3: 3.20 s Trial 4: 3.22 s

Answer: 0.10 sThe range is the difference between the highest and lowest trial values. In this case, the highest value is 3.22 s and the lowest value is 3.12 s. Therefore:Range = 3.22 s − 3.12 s = 0.10 s

2.2 Reporting Measured ValuesMeasurements should be reported in a manner that informs the reader about its precision. The best way is to specify the uncertainty along with the measured value. If we report a measurement as “525.26 ± 0.21 g,” we are saying that our best estimate is 525.26 g, but if the measurement were repeated, we are likely to get a value that can be up to 0.21 g smaller, or up to 0.21 g larger. The uncertainty is “± 0.21 g,” which is read as “plus-or-minus 0.21 g.”

The uncertainty should be rounded off to the first nonzero digit; one additional digit may be kept if the first nonzero digit is less than 3. The measured value should be reported so that the last decimal place is the same as that of the uncertainty.

▼ Example 3

The length of an object is measured several times and the average was found to be 24.3828 mm, with an uncertainty of ± 0.0286 mm. How should the measured value and uncertainty be reported?


Answer: 24.38 ± 0.03 mm, or 24.383 ± 0.029 mmThe first nonzero digit in the uncertainty is the 2 in the hun-dredth’s place. Therefore, uncertainty should be rounded off to the hundredth’s place: ± 0.0286 → ± 0.03. The average should also be rounded off the hundredth’s place: 24.3828 → 24.38.Since the first nonzero digit in the uncertainty is less than 3, we may keep an extra digit: ± 0.0286 → ± 0.029. If we write our uncertainty to the thousandth’s place, then we also write our average to the thousandth’s place: 24.3828 → 24.383.

How do we determine the uncertainty? The simplest way is to take half of the range.

▼ Example 4

The percentage of calcium in a tablet was determined four times, and the following results were obtained:Trial 1: 27.530% Trial 2: 20.894% Trial 3: 29.227% Trial 4: 21.187%

How should the average be reported?Answer: 25%We can calculate the average by adding up the values from the four trials and dividing by the number of trials (4). If we do this, our calculator should give us 24.71%The range is the difference between the highest and lowest value.Range = 29.227% − 20.894% = 8.333%

The uncertainty is approximately equal to half the range:Uncertainty = ±8.333/2 = ± 4.167 → ± 4 (round off to the first nonzero digit)

Therefore, the level of uncertainty is in the ones place; the average should be reported to the nearest unit: 24.71 → 25.

A more sophisticated method of determining uncertainty is to calculate the standard error of the mean or standard deviation of the mean. The formula for standard deviation of the mean is:

x x

N N

ii=1

N

− < >( )−( )

∑ 2

1


In this expression:• N is the number of trials.• xi is the measured value for the ith trial.• <x>is the average or mean of all the xi values.• xi−<x> is called the deviation of xi from the mean.

Basically, what we do is square each deviation, add up all the squared deviations, divide the sum by N(N−1), then take the square root of the result. If we calculate the standard error of the mean for the data in the preceding example, we find it to be 2.1. Thus, the uncertainty is ± 2.1% or ± 2%. The estimated uncertainty based on the range is generally larger than the stan-dard error of the mean; we could say that it is a more conservative estimate of the uncertainty.

We do not always have to quote the uncertainty when reporting measurements. If we do leave out the uncertainty we should still give the reader an indication of what the magnitude of the uncertainty is. The last digit in the reported value should be in the same decimal place as the first nonzero digit of the uncertainty. We say that it is the last significant digit of the measurement. Because of the uncertainty, any digit after this digit is meaningless and should not be reported.

▼ Example 5

How should 25.03 ± 0.27 mg be reported if the uncertainty is left out?Answer: 25.0 mgThe magnitude of the uncertainty is in tenth’s place since the first nonzero digit of the uncertainty (± 0.27) is in the tenth’s place. So, if we were to leave out the uncertainty, we report the measured value to the nearest tenth. We say that this last digit (in this case, in tenths place) is the last significant digit; given the magnitude of the uncertainty, there is no good reason for us to report additional digits.

How should we report values for each trial? If the measuring instrument has a digi-tal readout, the fluctuations in the readout should give us an estimate of the uncertainty. If the reading does not seem to fluctuate, it suggests that our instrument is not sensitive enough to detect the natural fluctuations in whatever it is we are trying to measure; in this case, we use all the digits shown on the readout and we say that the instrument’s sensitiv-ity limits the precision. If the instrument has an analog scale, it is customary to take the uncertainty as 1/10 of the difference between adjacent marks. The last digit in the reported value, the last significant digit, should be in the same decimal place as the first nonzero digit of the estimated uncertainty.


▼ Example 6

Adjacent marks on a meter stick are 1 mm apart. How should readings from this scale be reported?Answer: We estimate the uncertainty to be 1/10 of 1 mm, or 0.1 mm. Therefore, we should report readings from this scale to the nearest tenth of a millimeter.

2.3 Significant FiguresThe proper way to report a measurement is to stop at the last significant digit. We can define significant digits or significant figures as all the digits in a properly reported mea-surement except for powers of ten and zeros that are solely intended to locate a decimal point. The last significant digit gives us information about the precision of the measure-ment. The significant figures in a reported measurement are determined as follows:

Rule 1. If the number is written in scientific notation, all the digits in the coefficient are significant. (A number is said to be written in scientific notation if it is written as a num-ber between 1 and 10 times a power of 10.)

Rule 2. If the number is not written in scientific notation, the first nonzero digit is the first significant digit and all digits to the right of the first nonzero digit are significant. EXCEPTION: if the number does not have a decimal point, trailing zeros may or may not be significant; unless otherwise specified, it is customary to assume that they are not significant.

▼ Example 7

How many significant digits are in the following:(a) 1.50 × 10−3 kg (b) 2.835 × 104 mm (c) 0.00250 m (d) 73.00 km (e) 385 mL (f) 3800 mg

Answers:(a) 1.50 × 10−3 kg has 3 significant digits (the 1, 5, and 0 in

1.50); rule 1(b) 2.835 × 104 mm has 4 significant digits (the 2, 8, 3 and 5

in 2.835); rule 1(c) 0.00250 m has 3 significant digits (2, 5, and 0); rule 2(d) 73.00 km has 4 significant digits (7, 3, 0, and 0); rule 2(e) 385 mL has 3 significant digits (3, 8, and 5); rule 2(f) 3800 mg can have 2, 3, or 4 significant digits; rule 2

exception


The 3 and 8 in 3800 are significant. The trailing zeros may or may not be significant because there is no decimal point. Unless otherwise specified, assume that they are not significant; there-fore 3800 mg has only 2 significant digits. This is an example of a measurement that is not properly reported but is, unfortunately, commonly encountered. To avoid ambiguity, it is best to write a number like this in scientific notation:• Write 3800 as 3.8 × 103 if the trailing zeros are not

significant.• Write 3800 as 3.80 × 103 if the first zero is significant and

the second is not.• Write 3800 as 3.800 × 103 if both zeros are significant.

2.4 Calculations Involving Measurements

2.4.1 Dealing with UnitsWhen performing calculations involving measurements, units are treated like any algebraic quantity. When a unit is attached to a number, multiplication is implied. As a consequence:

• numerical values of measurements can only be added or subtracted if they have the same unit.

• when using measured quantities in multiplication, division, or exponentiation, the units must be manipulated accordingly.

▼ Example 8

Calculate: 5 m + 2 m = ?Answer: 7 mSince both measurements have the same units, we can add the numerical values: 5 m + 2 m = (5 + 2) m, or 7 mThe unit (m) can be factored out just like x can be factored out of an algebraic expression like (5x + 2x): 5x + 2x = (5 + 2)x, or 7x.


▼ Example 9

Calculate: (2 cm)(3 cm) = ?Answer: 6 cm2

The unit (cm) is treated just like x in the algebraic expression “(2x)(3x)”:

(2x)(3x) = 6x2

▼ Example 10

Calculate: (6 m) / (2 min) = ?Answer: 3 m/min, or 3 m min−1

Dividing by min is the same as multiplying by the reciprocal of min, which is the same as min raised to the −1 power (min−1).

2.4.2 Significant Digits in Calculated ValuesThe result of a calculation should be written in a manner that reflects the level of uncer-tainty in the numbers used in the calculation.

Rule for Addition and/or Subtraction. Determine the least precise term, the term with the largest uncertainty. Express the answer to as many decimal places as the least precise term.

▼ Example 11

Calculate: 25.0 g + 1.003 g = ?Answer: 26.0 gThe uncertainty in 25.0 g is at least ±0.1 g.The uncertainty in 1.003 is at least ±0.001 g.The term 25.0 g is less precise than 1.003 g; its uncertainty is larger.Therefore, the answer needs to be expressed to the nearest tenth, since the less precise term (25.0) is expressed to the nearest tenth.The sum is (26.003 g) should be rounded off to 26.0 g.


▼ Example 12

Calculate: 4.2 × 10−3 m + 4.4 × 10−4 m − 3.50 × 10−5 m = ?Answer: 4.6 × 10−3 mThe best thing to do in this case is to first make the powers of 10 the same. Choose the largest power (−3).• 4.4 × 10−4 → 0.44 × 10−3 [Math review: move decimal point

in coefficient one place to the left while adding one to the exponent. Note that −4 + 1 = −3.]

• 3.5 × 10−5 → 0.0350 × 10−3

Then add and/or subtract the coefficients:4.2 × 10−3 + 0.44 × 10−3 − 0.0350 × 10−3 = (4.2 + 0.44 − 0.0350) × 10−3 = (4.605) × 10−3

Here, we round off the coefficient to the nearest tenth since 4.2 has the largest uncertainty (at least ±0.1). Therefore, the answer is 4.6 × 10−3 m

Rule for Multiplication and/or Division. Determine the least precise term, the term with the fewest number of significant digits. Express result to as many significant digits as the least precise term.

▼ Example 13

Calculate: (16.00 g) / (2.00 mL) = ?Answer: 8.00 g/mLThe term 16.00 g has 4 significant digits.The term 2.00 mL has 3 significant digits.The term 2.00 mL has fewer significant digits (3) and is consid-ered as less precise. Therefore, the answer should have 3 signifi-cant digits, the same as the less precise term (2.00 mL).

Multi-step Calculation. Apply the appropriate rule for each step, but keep an extra digit in the intermediate steps in order to avoid build-up of round off errors. It would be helpful to mark the last significant digit for the intermediate result so that you can easily keep track of it.


▼ Example 14

Calculate: (25.1)(1.0) + 2.5Answer: 28Step 1. Multiply (25.1)(1.0)• 25.1 has three significant digits• 1.0 has two significant digits• Therefore: (25.1)(1.0) = 25.1; should only have 2 significant

digits (5 is the last significant digit; the “1” after that is not significant, but we keep it since we will use the result for the next step)

Step 2. Add: 25.1 + 2.5• 25.1 has an uncertainty of at least ±1 because 5 is the last

significant digit here; we are just keeping the extra digit to avoid round-off error

• 2.5 has an uncertainty of at least ±0.1• The less precise term has an uncertainty of at least ±1• 25.1 + 2.5 = 27.6 → 28; the answer should have an

uncertainty of at least ±1

Calculations Involving Exact Numbers. Exact numbers do not have any uncertainty. They are, therefore, considered to be perfectly precise and have an “infinite” number of significant figures for as long as they are not rounded off. They can never be the least precise term in a calculation. Exact numbers include small counts (as in 13 pencils) and numbers that are defined to be exact. In the context of the definition “1 inch = 2.54 cm”, the numbers 1 and 2.54 are exact and are considered as having an infinite number of significant digits.

▼ Example 15

Consider the following formula where 1.8 and 32 are exact: F = 1.8C + 32.

Calculate F if C = 20.0.Answer: 68.0First we multiply 1.8 × 20.0• 20.0 has 3 significant figures• 1.8 has an infinite number of significant figures• 20.0 × 1.8 = 36.0, should have 3 significant figures

Then we add 32 to the result:• 36.0 has an uncertainty of at least ±0.1• 32 has zero uncertainty since it is exact


• 36.0 is the less precise term with an uncertainty of at least ±0.1

• the answer should have an uncertainty of at least ±0.1:36.0 + 32 = 68.0

Rounding Off Precisely Known Numbers. There are quantities that are known to a much greater precision than others. We could round these off to fewer digits, but the rounded value must still be more precise than the least precise term in the calculation.

▼ Example 16

The speed of light, c, in the formula E=mc2, is defined to be exactly 2.99792458 x 108 m/s. If, in a calculation, m = 10.0 kg, to what extent can we round off the value of c?Answer: For multiplication, the less precise term is the term with the fewer number of significant figures. In this calculation m, which is 10.0 kg, is the less precise term with only 3 significant figures. The value of c should have at least 4 significant figures: 2.998 × 108 m/s is acceptable.

Significant Figures Involving Logarithms. When we calculate the logarithm of another number, it is useful to write it in scientific notation in order to determine the number of significant digits needed for the result. The number of decimal places in the result should be the same as the number of significant digits in the coefficient of the original number.

▼ Example 17

Express log(2.7 × 104) to the correct number of significant figures.Answer: 4.43If we take the logarithm of this number, we get:log(2.7 × 104) = log(2.7) + log(104)

= 0.431 + 4 = 4.43

The number 2.7 has two significant digits; the logarithm of 2.7 should be rounded to 0.43 (also two significant digits).


The integer part of the answer (4) essentially comes from the power of 10, which is an exact number; it has an infinite number of significant digits; we could have written it as 4.0000000...

Similarly, if we reverse the process, the number of significant digits in the antilog should be the same as the number of decimal places (after the decimal point) in the original number.

▼ Example 18

Express 10−2.699 to the correct number of significant figures.

Answer: 2.00 × 10−3

There are 3 decimal places in the original number (−2.699). The antilog should have 3 significant digits.

2.5 SI UnitsNumerous units have been defined over the years. Different countries have adopted differ-ent sets of standards. In the English system, still widely used in the US and UK, the units used include inches, feet, yard and miles for length, pounds and ounces for mass, etc. Nowadays, however, the only system that is acceptable for scientific work and for interna-tional trade is the International System of Units, which is generally referred to as SI (from the French name: Le Systeme International d’Unites).

The SI system defines seven fundamental quantities and a base unit for each of these fundamental quantities. The fundamental quantities are: length, mass, time, temperature, amount of substance, electric current, and luminous intensity. The corresponding base units are meter (m), kilogram (kg), second (s), kelvin (K), mole (mol), ampere (A), and candela (cd). The abbreviations for the units, indicated in parentheses in the preceding sentence, are also well defined. The official abbreviation for second is s (lower case), not “sec.” The temperature unit is “Kelvin” (K), not “degrees Kelvin” (°K).

Units for quantities that can be thought of as obtainable from measurements of two or more fundamental quantities are called derived units. For example, speed is a ratio of distance traveled (a measurement of length) to the elapsed time (a measurement of time). Therefore, the unit for speed — m/s or m s−1 (meters per second) — is a derived unit. Some derived units have special names. For example, the unit for force, kg m s−2, is called a Newton (official abbreviation: N).

One reason why numerous units have been invented is convenience. It is more con-venient to express the height of a person as 6 ft rather than, say, 0.0011 miles. We do not


lose this convenience when using SI because the system also defines official prefixes that can be used to scale a unit up or down. For example, the SI prefix centi (officiall abbrevia-tion: c) means 10−2, or one one-hundredth. Therefore, 1.5 × 10−2 m is equivalent to 1.5 cm. Similarly, 1,500,000 g (or 1.5 × 106 g) is equivalent to 1.5 Mg; M (upper case) is the official abbreviation for the prefix mega, which means 106 (or one million). The most commonly used SI prefixes are:

Prefix: Giga, Mega, Kilo, Hecto, Deca, Deci, Centi, Milli, Micro, Nano, PicoOfficial abbreviation: G, M, k, h, da, d, c, m, μ (Greek letter mu), n, pMeaning: 109, 106, 103, 102, 101, 10−1, 10−2, 10−3, 10−6, 10−9, 10−12

The prefix micro (μ) is also commonly abbreviated using mc. For example, you’ll find microgram abbreviated as mcg in vitamin bottles; however, μg is the official abbreviation.

Some commonly used units that were defined before the SI system was adopted have been officially redefined in terms of SI units. A Liter (L) is now defined as exactly 0.001 cubic meters (1L = 10−3 m3). A milliliter (mL) is one cubic centimeter (1 mL = 1 cm3). One teaspoon is now 5 mL or 5 cm3; one tablespoon is now 15 mL. One inch is now defined to be exactly 2.54 cm.

2.6 Unit Conversions

2.6.1 Algebraic SubstitutionQuantities given in one unit can be expressed in another unit if we know equivalent quantities for both units. Simply treat units like any algebraic term and make appropriate substitutions.

In general, numbers used in defining equivalent quantities are defined to be exact. Therefore, converting measurements from one unit to another should not affect the num-ber of significant figures. In other words, the number of significant figures in the answer should be the same as in the original measurement.

▼ Example 19

Convert 3.5 mm to m.

Answer: 3.5 × 10−3 mSince 1 mm = 10−3 m, we simply replace mm in the 3.5 mm by 10−3 m. 3.5 mm = 3.5 (10−3 m), or 3.5 × 10−3 m


▼ Example 20

Convert 9.8 × 10−4 m to mm

Answer: 9.8 × 10−1 mmSince 1 mm = 10−3 m, (1/10−3) mm = 1 m, or 103 mm = 1 mReplace m by 103 mm: 9.8 × 10−4 m = 9.8 × 10−4 (103 mm), or 9.8 × 10−1 mm

▼ Example 21

The gas constant R is 0.08206 L atm mol−1 K−1. Express R in m3 Pa mol−1 K−1.Given: 1 L = 10−3 m3, 1 atm = 1.01325 × 105 Pa

Answer: 8.314 m3 Pa mol−1 K−1

R = 0.08206 L atm mol−1 K−1 = 0.08206 (10−3 m3) (1.01325 × 105 Pa) mol−1 K−1

= 8.314 m3 Pa mol−1 K−1

▼ Example 22

A house has a floor area of 1.500 × 103 ft2. What is the floor area in in2?Given: 1 ft = 12 in

Answer: 2.160 × 105 in2

Area = 1.500 × 103 ft2 = 1.500 × 103 (12 in) 2 = 2.160 × 105 in2

▼ Example 23

Convert 658 nm to μm.Answer: 0.658 μmWe are not likely to find the relationship between nm and μm listed anywhere. However, we know that the SI prefix nano (n) means 10−9 and the SI prefix micro (μ) means 10−6. Therefore, we know that:• 1 nm = 10−9 m• 1 μm = 10−6 m, or 106 μm = 1 m


These relationships suggest a strategy for doing the needed con-version. First, we convert nm to m, then convert m to μm.658 nm = 658 (10−9 m) = 658 × 10−9 m= 658 × 10−9 (106 μm) = 658 × 10−3 μm, or 0.658 μm

2.6.2 Factor Label MethodA popular method for converting units involves the use of conversion factors. This method is often referred to as “factor label” or “dimensional analysis.” A conversion factor is just a ratio of equivalent amounts in two different units. For example, since

1 mm = 10−3 m,we can set up two possible conversion factors:

110

1013

3mmm

or mmm−

−

We multiply whatever it is we want to convert by one of these so that the original unit cancels out.

▼ Example 24

Convert 3.5 mm to m.Answer: 3.5 × 10−3 mWe multiply it 3.5 mm by the conversion factor that has mm in the denominator.

3 5101

3 5 103

3. .mmm

mmm× = ×

−−

▼ Example 25

Convert 2.5 g/mm to g/m.Answer: 2.5 × 103 g/mWe multiply 2.5 g/mm by the conversion factor that has mm in the numerator because mm is in the denominator of the original measurement.

2 51

102 5 103

3. .g

mmmm

mgm

× = ×−−


The justification for the factor label method is that the conversion factors are ratios of equivalent quantities and are therefore equal to one. Multiplying anything by one does not change its value. Quantities expressed in different units are also directly proportional. When two quantities are directly proportional, their ratio is constant. Put another way: if one is zero, so is the other; when one changes, the other changes by the same factor. For example, the ratio of (length in meters) to (length in millimeters) is a constant. Therefore:

length in mlength in mm

mmm

= 101

3−

So, if we know the length in millimeters, we can solve for the length in meters by multiply-ing both sides of the equation by the length in millimeters:

length in m length in mm mmm

= ( )×101

3−

2.6.3 Temperature ConversionsOne exception to the unit conversion methods described above involves temperature readings. Readings on the Celsius, Fahrenheit, and Kelvin scales are, by definition, not directly proportional. We can easily tell that this is the case since a zero reading on one scale does not correspond to a zero reading on the other scales. In fact, 0°C is equivalent to 32°F and 273.15K. In these cases, it is best to just use the defining formulas. Let TF, TC, and TK be the temperature readings in Fahrenheit, Celsius and Kelvin. If we know TC, then:

T FC

T F

TKC

T K

F C

K C

= °°

⎛⎝⎜

⎞⎠⎟

+ °

=°

⎛

⎝⎜

⎞

⎠⎟ +

1 81

32

11

273 15

.

.

In other words, multiply the Celsius reading by 1.8, then add 32 to get the Fahrenheit reading, and add 273.15 to the Celsius reading to get the Kelvin temperature. Similarly, if we know TF or TK, then we can solve for TC.

T T FF

C

TT K

KC

CF

CK

= °°

°⎛⎝⎜

⎞⎠⎟

=

°⎛⎝⎜

⎞⎠⎟

−

−

321 81

273 1511

.

.


T T FF

C

TT K

KC

CF

CK

= °°

°⎛⎝⎜

⎞⎠⎟

=

°⎛⎝⎜

⎞⎠⎟

−

−

321 81

273 1511

.

.

Note: All the numbers (1, 1.8, 32, and 273.15) in these formulas are exact numbers by definition.

▼ Example 26

What are the Fahrenheit and Kelvin readings equivalent to 25.000°C?

Answer: 77.000°F and 298.15K

TF

CC F F

TKC

F

K

= °°

⎛⎝⎜

⎞⎠⎟

°( ) + ° = °

=°

⎛⎝⎜

⎞⎠⎟

1 81

25 000 32 77 000

11

25

.. .

.0000 273 15 298 150°( ) + =C K K. .

While temperature readings on the different scales are not directly proportional, temperature differences or changes on these scales are. Based on the definitions above, a one degree change on the Celsius scale is equivalent to a 1.8 degree change on the Fahrenheit scale and also equivalent to a one unit change on the Kelvin scale. Let ΔTF, ΔTC, and ΔTK be the corresponding changes in temperature on the three scales. Then:

ΔΔΔΔ

TT

KC

TT

FC

K

C

F

C

=°

= °°

111 81.

In other words, we can use algebraic substitution or the factor label method when dealing with temperature changes or temperature differences.


▼ Example 27

If the temperature goes up from 25°C to 34°C, by how much did the temperature change in Kelvin?Answer: 9K

Δ ΔT TKC

C CKC

CKC

KK C= ×°

= ° ° ×°

= ° ×°

=11

34 2511

911

9( ) ( )−

Here’s another way of solving this problem. We convert 34°C and 25°C to Kelvin first, then take the difference:

ΔT CKC

K CKC

KK = ° ×°

+⎛⎝⎜

⎞⎠⎟

° ×°

+⎛⎝⎜

⎞⎠⎟

3411

273 15 2511

273 15. .−

Note that when we take the difference, the 273.15K terms cancel out.

▼ Example 28

If the temperature changes from 50.0°F to 62.0°F, by how much did the temperature change in degrees Celsius.

Answer: 6.67°C

Δ ΔT TCF

F F)CF

FCFC F= × °

°= ° − ° × °

°= ° × °

°=1

1 862 0 50 0

11 8

12 01

1 8.( . .

..

.66 67. °C

Δ ΔT TCF

F F)CF

FCFC F= × °

°= ° − ° × °

°= ° × °

°=1

1 862 0 50 0

11 8

12 01

1 8.( . .

..

.66 67. °C

Here is a longer way of solving this problem. We could convert the Fahrenheit readings to Celsius first, then take the difference:

ΔTF F

FC

F FFC = ° °

°°

⎛⎝⎜

⎞⎠⎟

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

° °°

62 0 321 81

50 0 321 8

..

..

− − −

11°⎛⎝⎜

⎞⎠⎟

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥C

Since the two fractions above have the same denominator, we can combine the numerator and get

ΔTF F F F

FC

FC = ° ° ° °°

°⎛⎝⎜

⎞⎠⎟

= °( . ) ( . ).

( . .62 0 32 50 0 32

1 81

62 0 50 0− − − − °° × °

°= ° × °

°= °F

CF

FCF

C).

..

.1

1 812 0

11 8

6 67


ΔTF F F F

FC

FC = ° ° ° °°

°⎛⎝⎜

⎞⎠⎟

= °( . ) ( . ).

( . .62 0 32 50 0 32

1 81

62 0 50 0− − − − °° × °

°= ° × °

°= °F

CF

FCF

C).

..

.1

1 812 0

11 8

6 67

which is equivalent to the expression we have earlier. Note that the 32°F terms in the numerator cancel out.

2.7 Unitless QuantitiesNot all measurements have a unit. The basis of comparison for those that do not have units is implied in the definition of the quantity being measured. Some of these quantities are defined as ratios of two quantities that have the same unit. When we calculate the ratio, the units cancel out. For example, specific gravity is a unitless quantity. It is defined as the ratio of the density of an object to the density of a reference material (such as water). Another example is transmittance. It is defined as the ratio of the intensity of light that passes through a solution to the intensity of light that that passes through a reference solution.

Other unitless quantities are defined as logarithms of unitless quantities. Absorbance is the negative of the logarithm of transmittance. pH is the negative of the logarithm of the activity of hydrogen ions; activity is a unitless quantity.


TEST YOURSELF


1. Which of the following has the fewest number of significant figures? A. 10.50 mg B. 20700 mL C. 0.003100 g D. 1.500×10−3 m

2. The amount of acid present in a sample was measured several times and the average value was determined to be 3601.2 mg, with an uncertainty of ±30 mg. To how many significant figures should the average be reported?

A. 2 B. 3 C. 4 D. 5

3. Five replicate measurements of the calcium content of a sample yielded the following results: 5.042%, 5.131%, 5.434%, 4.852%, 4.728%.

To how many significant digits should the average be reported? A. 1 B. 2 C. 3 D. 4

4. What is the proper way of reporting the temperature shown on the thermometer scale (Celsius) below?

A. 30 °C B. 30.0 °C C. 30.00 °C D. 3.0×101 °C

29 30 31

5. To how many significant figures should the result of the following calculation be reported: 2500.0 + 25 g?

A. 2 B. 3 C. 4 D. 5

6. To how many significant figures should the result of the following calculation be reported?

2500 mL + 25 mL? A. 2 B. 3 C. 4 D. 5



7. How should the result of the following calculation be reported? 80.00 g / 10.0 mL ? A. 8 g/mL B. 8.0 g/mL C. 8.00 g/mL D. 8.000 g/mL

8. How should the result of the following calculation be reported? 0.800 m / 0.010 s A. 80 m/s B. 80.0 m/s C. 8.0×101 m/s D. 8×101 m/s

9. How should the result of the following calculation be reported? 0.0072 + (0.0050)(39.0)? A. 0.20 B. 0.21 C. 0.202 D. 0.2022

10. How should the result of the following calculation be reported, assuming that 1.8 and 32 are exact numbers: 32 + (1.8)(2.37)?

A. 36 B. 36.3 C. 36.27 D. 36.266

11. The volume of a sphere of radius r is given by the formula:

V r= 43

3π

The radius of a sphere is measured to be 5.00 cm. What is the proper way to report the volume? Note: π = 3.141592653589793238462643383279...

A. 520 cm3 B. 523 cm3 C. 524 cm3

12. Which of the following is equivalent to 101.9 MHz? A. 1.019×106 Hz B. 1.019×108 Hz, C. 1.019×104 Hz D. 1.019×10−5 Hz

13. Which of the following is not equivalent to 10−3 mg? A. 10−6 g B. 1μg C. 10−4 cg D. 106 ng

14. Given: 1 inch is exactly equivalent to 2.54 cm. A measurement of 7.00 cm is equivalent to how many inches?

A. 2.76 in B. 3 in C. 17.8 in D. 20 in

15. Given: 1 inch is exactly equivalent to 2.54 cm, and 12 inches is exactly 1 foot. A measurement of 7.0 cm is equivalent to how many feet?


16. A measurement of 2.50 mm is equivalent to ? A. 2.50×106 nm B. 2.50×10−9 nm C. 2.50×10−6 nm D. 2.50×109 nm

17. A car is traveling at a speed of 40.0 km hr−1. How fast is the car moving in km min−1? A. 2.40×103 km min−1 B. 6.67×10−1 km min−1 C. 1.50 km min−1

18. The density of air is 1.29×10−3 g/L. What is the density of air in mg/mL? A. 1.29×10−3 mg/mL B. 1.29 mg/mL C. 1.29×10−6 mg/mL

19. The area of a circle is found to be 2.4 cm2. What is the area in m2? A. 2.4×102 m2 B. 2.4×10−2 m2 C. 2.4×10−4 m2 D. 5.8×104 m2

20. What temperature reading on the Fahrenheit scale corresponds to 20.0°C? A. 20.0°F B. 68.0°F C. 36.0°F D. −6.67°F

21. What temperature reading on the Celsius scale corresponds to 20.0°F? A. 20.0°C B. 68.0°C C. 36.0°C D. −6.67°C

22. At what temperature are the Celsius and Fahrenheit readings equal? A. −40° B. 32° C. 100° D. 212°

23. What is the Kelvin temperature equivalent to 25.00°C? A. 298K B. 298.15K C. −248.15K

24. What is the Kelvin temperature equivalent to 50.00°F? A. 323.15K B. 283.15K C. 263.15K D. 395.15K

25. If the temperature increases by 20.0 degrees Celsius, by how much did it increase in degrees Fahrenheit?

A. 36.0° B. 68.0° C. 6.67° D. 20.0°

26. During a chemical reaction, the temperature of a mixture dropped by 10.0°C. By how much did the temperature drop in Kelvin?

A. 10.0K B. 283.15K C. 263.15K

29

3CHAPTER

Atoms, Ions, and MoleculesChemistry is about understanding the behavior of matter in terms of atoms. Matter is what makes up our universe. Matter is made up of atoms, which are too small to see. In this chapter, we examine the parts of an atom and discuss what typically happens to these parts of the atom.

3.1 Subatomic ParticlesAn atom is made up of even smaller particles: protons, neutrons, and electrons. Protons and neutrons are in the nucleus, a very tiny spot in the middle of an atom. Protons and neutrons are collectively called nucleons.

Electrons are much smaller than protons and neutrons and they are found around the nucleus. Electrons move so fast that it is convenient to imagine electrons as a cloud around the nucleus.

An atom is mostly empty space. The typical diameter of a nucleus is 10−14 – 10−13 m, while the diameter of the rest of the atom (the “electron cloud”) is typically around 10−10 m. The diameter of each individual electron is estimated to be less than 10−18 m. To imagine the relative magnitudes of these numbers, imagine you are in a giant world where a typical atomic nucleus is 1 cm instead of about 10−13 m. In this magnified world, the diameter of a typical electron cloud would be at least 1000 cm and each individual electron would be a tiny speck. Put another way, if you put two atoms “side by side,” the distance between the nuclei would be at least 1000 times larger than the diameter of the nuclei themselves. Between the nuclei, where electrons are roaming about, is mostly empty space, which would be at least 1,000,000,000 times larger than the space actually occupied by the nuclei.

▼ Example 1

A tennis ball has a diameter of 6.5 cm. If the nuclei of two neighboring atoms were the size of tennis balls, how far away from each other would the nuclei be?Answer: About 1000 times compared to the diameter: 1000 × 6.5 cm = 6.5 × 103 cm, or 65 m.


Note: protons and neutrons are made up of even smaller particles called quarks. In the so-called “standard model”, all matter in our universe is said to be made up of quarks and leptons; there are 6 kinds of quarks and 6 kinds of leptons. An electron is a lepton. In Chemistry, we are generally not concerned about quarks and leptons (other than the electron).

3.2 Atoms and ElementsIn general, samples of matter that we encounter in nature can be separated into two or more components with different properties. (The word “property” refers to characteristics of matter such as color, melting point, etc.) However, there are samples that cannot be separated into components with different properties. These samples are called elements. We explain why these samples cannot be separated into distinct components by assuming that an element is made up of only one kind of atom.

There are a little over a hundred known elements to date; less than one hundred are naturally-occurring. Newly discovered elements are generally created in the laboratory and are not stable. Differences in properties of elements are due to differences in their atoms.

3.3 The Atomic NumberAll atoms of the same element have the same number of protons; atoms of different ele-ments have different number of protons. An element’s atomic number is defined as the number of protons in the each of the atoms of that element. Elements are listed in a chart called a periodic table (see Figure 1) in increasing atomic number (left-to-right, top-to- bottom). As a matter of convenience, elements are assigned symbols that are derived from their names. The first letter of the symbol is capitalized and the second letter, if there is one, must be written in lower case. For example, the symbol for oxygen is O. For magne-sium, the symbol is Mg.

▼ Example 2

What is the symbol for an element whose atoms have 10 protons in their nucleus?Answer: If the atoms of an element have 10 protons in their nucleus, then this element has an atomic number of 10. Referring to a periodic table, we find that the element with atomic number of 10 is neon (Ne).

Chapter 3: Atoms, Ions, and Molecules 31

Figure 1. Periodic Table


▼ Example 3

How many protons are found in the nucleus of a magnesium (Mg) atom?Answer: Referring to a periodic table, we find that the atomic number of magnesium is 12. Therefore, any atom of magnesium has 12 protons.

3.4 IsotopesAll the atoms of an element have the same number of protons. But they may have different number of neutrons. Atoms of the same element that have different numbers of neutrons are called isotopes. To make a distinction between isotopes, we specify the total number of nucleons (protons and neutrons) next to the symbol for the element. This number is called the mass number and is written either a superscript to the left of the symbol or following a dash after the symbol for the element.

▼ Example 4

How many protons and neutrons are found in the nucleus of an atom of O-16 or 16O?Answer: 8 protons and 8 neutronsThe number 16 in O-16 or 16O is the mass number, which is the total number of protons and neutrons. Referring to a periodic table, we find that the atomic number of oxygen is 8; this means that any oxygen atom has 8 protons. Therefore, every atom of O-16 must have 8 neutrons in its nucleus.

In general, naturally-occurring samples of matter are not isotopically pure. For example, a typical sample containing oxygen atoms will be found to have:

• 99.757% O-16,• 0.038% O-17, and• 0.205% O-18.

These percentages are called the natural abundance of the oxygen isotopes. What these mean is that out of every 100,000 O atoms, we would expect 99,757 to be O-16, 38 would be O-17, and 205 would be O-18. A modern instrument that is used to separate isotopes and measure isotopic abundance is a mass spectrometer.


3.5 Mass of Protons, Neutrons, and ElectronsNot only are electrons very small in size, they also contribute very little to the mass of an atom. The currently accepted masses of protons, neutrons, and electrons are given below:

• Mass of proton: 1.6726 × 10−27 kg or 1.0073 u• Mass of neutron: 1.6749 × 10−27 kg or 1.0087 u• Mass of electron: 0.00091 × 10−27 kg or 0.0005486 u

The unit u is an abbreviation for atomic mass unit, which is also frequently abbreviated as amu; this unit is also called the Dalton (D). We can see from the numbers given above that a proton’s mass is about 2000 times that of an electron (1836 times to be more precise). On the other hand, a neutron and a proton have almost the same mass. To the nearest tenth of an atomic mass unit, we can say that the mass of proton or neutron is 1.0 u, while that of an elec-tron is 0.0 u. This is why the number of electrons is ignored in the defining the mass number of an atom. To the nearest atomic mass unit, the mass of an atom is equal to its mass number.

▼ Example 5

Which of these has the largest total mass: 1 proton, 3 electrons, or 2 neutrons?Answer: 2 neutrons• The mass of one proton is 1.0 u.• Three electrons have a total mass of 3 × 0.0 u, or 0.0 u.• Two neutrons have a total mass of 2 × 1.0 u, or 2.0 u.

3.6 Mass of AtomsThe mass of a C-12 atom is defined to be exactly 12 u. The masses of electron, proton, and neutron given in the preceding section are based on this definition. If we add up the masses of the 6 protons, 6 electrons, and 6 neutrons found in a C-12 atom, we find that the total does not add up to exactly 12 u. We cannot get the most precise value for the mass of an atom by simply adding the precise masses of protons, neutrons, and electrons that make it up. The actual mass of an atom is slightly different and needs to be looked up. The discrep-ancy is called the mass defect (m) and is related to the binding energy (E), the energy needed to separate all protons and neutrons from one another, according to Einstein’s equation:

E = mc2.In this equation c is the speed of light, 2.998 × 108 m/s.


▼ Example 6

What is the most precisely known mass of an atom of 18O?Answer: 17.9991603 uWe have to look up this number. Note that the sum of the individual masses of 8 protons and 10 neutrons alone should be larger than 18 u, but an actual O-18 atom has a mass slightly less than 18 u.

Most elements have two or more naturally-occurring isotopes. The masses of specific isotopes are typically not listed in simple periodic tables; what is typically listed is the average mass of atoms in a naturally-occurring sample. The numerical value of the aver-age mass (in u) is called the atomic weight. For example, the atomic weight of Mg is often listed as 24.305. This means that, in a naturally-occurring sample of Mg, the average mass of the atoms is 24.305 u; there is not a single Mg atom that actually has a mass of 24.305 u.

If we want information about specific isotopes of an element, we need to look it up from more extensive data sources. For example, we can look up the following information about naturally-occurring isotopes of Mg from webelements.com:

(http://webelements.com/magnesium/isotopes.html)• Mg-24, mass = 23.99 u, natural abundance = 78.99%.• Mg-25, mass = 24.99 u, natural abundance = 10.00%.• Mg-26, mass = 25.98 u, natural abundance = 11.01%.

A natural abundance of 78.99% for Mg-24 means that about 79 out of every 100 atoms have a mass of 23.99 u. The average mass of can be calculated from the information above using the formula:

Average Mass = Sum of (relative abundance × mass)In other words,

Average Mass of Mg = (78.99%)(23.99 u) + (10.00%)(24.99 u) + (11.01%)(25.98 u) = 24.31 u

3.7 Electrical ChargeElectrons and protons are said to be electrically-charged, while neutrons are electrically-neutral. The unit for electric charge is the Coulomb (official abbreviation: C). Electrons carry the smallest magnitude of charge known, which is −1.602 × 10−19 C. This amount of charge is usually assigned a value of “−1” (atomic units). Protons have a charge of +1.602 × 10−19 C, or “+1” (atomic units). Neutrons are not charged (charge = 0). All atoms are electrically neutral. This means that they have equal numbers of protons and electrons.

http://webelements.com/magnesium/isotopes.html


▼ Example 7

An atom has 10 protons. How many electrons does it have?Answer: 10 electronsAn atom is electrically neutral. It has equal numbers of protons and electrons. If an atom has 10 protons, it also has 10 electrons.

▼ Example 8

What is the nuclear charge of a magnesium atom?Answer: +12 (atomic units), or (12)(+1.602 × 10−19 C)By nuclear charge, we mean the total charge of the nucleus. The charge of the nucleus is due to the protons. Referring to a periodic table, we find that the atomic number of magnesium is 12. Therefore, any magnesium atom has 12 protons in its nu-cleus. Since each proton carries a charge of +1 atomic unit, the charge of the nucleus is +12. In Coulombs, the charge is (12)(+1.602 × 10−19 C)

Particles with like charges are known to repel each other, while those with unlike charges attract each other. Therefore, electrons repel each other, and protons repel each other, but a proton and an electron attract each other. The attraction and repulsion forces due to charge are called Coulombic or electrostatic forces. But if protons repel each other, why are they clustered very close together in the nucleus? A much stronger force, which works only at very short distances, called the strong force (or strong nuclear force), holds the protons and neutrons together in the nucleus. The attractions due to the strong force counteract the repulsive Coulombic forces among the protons in the nucleus.

3.8 RadioactivityNeutrons are needed to stabilize a nucleus that has more than one proton. If there are too many or too few neutrons, the atom would be unstable. The nuclei of unstable atoms tend to break up over time; the break up of a nucleus is called nuclear fission. These unstable atoms are said to be radioactive.

It is tempting to think that there would be no naturally-occurring radioisotopes since they are unstable. In fact, there are naturally-occurring radioisotopes. These are the ones that have extremely long half-lives. The half-life of a radioactive atom is the time it takes


for half of the atoms to decay (transform into something else). For example, U-238 has a half-life of 4.46 billion years. This means that half of U-238 that existed 4.46 billion years ago are still here. Less stable radioisotopes can be created in the laboratory by bombarding atoms with energetic (very fast moving) particles, such as neutrons from a nuclear reactor, or charged particles accelerated within a cyclotron.

3.9 Formation of Ions and MoleculesMost naturally-occurring atoms have a stable nucleus (non-radioactive). For most every-day phenomena, the nuclei of atoms remain intact. However, atoms may:

• lose electrons to other atoms.• gain electrons from other atoms.• share electrons with other atoms.

These lead to the formation of ions and molecules.

3.10 Cations and AnionsEvery atom is electrically neutral; each one has an equal number of protons and electrons. An ion has an unequal number of electrons and protons. We define the charge of an ion as the number of protons minus the number of electrons.

Charge of ion = (Number of protons) − (Number of electrons)Note that the charge of an atom, which has an equal number of protons and electrons, is zero.

An ion with more protons is positively charged and is called a cation (pronounced as cat eye on). An ion with more electrons is negatively charged and is called an anion (pronounced as an eye on).

▼ Example 9

What is the charge of an ion with 12 protons and 10 electrons? Is it a cation or anion?Answer: +2, cationThe charge of the ion is:Charge = (Number of protons) − (Number of electrons) = 12 − 10 = +2.

Since the charge is positive, the ion is a cation.


3.11 Monatomic IonsWhen an atom loses or gains one or more electrons, the result is a monatomic ion. When writing symbols or formulas of ions, the charge must be indicated as a superscript (upper right-hand corner). The standard way of indicating charges is to put the magnitude in front of the algebraic sign. If the magnitude of the charge is 1, the number is omitted.

▼ Example 10

What type of ion (cation or anion) is formed when a magnesium atom loses two electrons? Write the symbol for the ion.Answer: cation, Mg2+

Loss of two electrons means that there will be two fewer electrons (compared to protons).Charge = (Number of protons) minus (Number of electrons) = +2

Loss of electron(s) leads to the formation of a positively charged ion, a cation. Since the charge is +2, the symbol for the ion is Mg2+. Note that the 2 is written in front of the + sign. In very old textbooks, you may see the symbol written as Mg+2 or Mg++.

▼ Example 11

What type of ion (cation or anion) is one that has 7 protons and 10 electrons? Write the symbol for the ion.Answer: anion, N3−

The charge is (Number of Protons) minus (Number of Electrons) = 7 − 10 = −3. Since the charge is negative, the ion is an anion. Referring to a periodic table, we find that an atom with 7 protons is a nitrogen atom. So an ion with 7 protons is derived from a nitrogen atom. The symbol for the ion is N3−. In old textbooks, you may see this written as N−3 or N − − −.

In the examples above, we are assuming that the ion being described is monatomic; that is, derived from just one atom. It is possible to have ions with more than one atom; these are called polyatomic ions.


3.12 MoleculesA molecule is an electrically-neutral group of two or more atoms that share electrons. We say that the atoms in a molecule are held together by covalent bonding.

A formula is used to describe the composition of a molecule. In a formula, we list the symbols of the elements that make up the molecule and use subscripts to indicate how many atoms of each element are in the molecule. Unwritten subscripts are implied to be 1. To eliminate possible confusion, symbols of elements must be written properly: the first letter must be capitalized; any succeeding letter must be in lower case.

▼ Example 12

About twenty percent of air molecules are O2 molecules. How many atoms are in a molecule of O2?Answer: 2 atomsThe formula for the molecule (O2) has a subscript of 2 for O. This means that there are two O atoms in one molecule of O2.

▼ Example 13

Which represents a molecule: Co or CO?Answer: CO a molecule consisting of two atoms — one carbon atom (symbol: C) and one oxygen atom (symbol: O). Co is the symbol for the element cobalt; it represents one atom.

▼ Example 14

How many atoms are in a molecule of acetic acid, CH3COOH?Answer: 8 atomsThe formula CH3COOH implies that the molecule contains two C atoms, two O atoms, and four H atoms. Note that if there is no subscript indicated, the subscript is implied to be 1. In the formula above, The subscripts shown for H are 3 and 1, for a total of 4. The formula of this molecule could also be written as C2H4O2. It is typically written as CH3COOH to reflect the way the atoms are connected.


▼ Example 15

How many electrons and protons are in a molecule of ammonia (NH3)?Answer: 10 electrons, 10 protonsA molecule of NH3 consists of one nitrogen atom and three hydrogen atoms. Referring to the periodic table, we find that each N atom has 7 protons; each H atom has 1 proton. Therefore, a molecule of NH3 consists of 10 protons (7 in the N atom and 1 in each of the 3 H atoms). Since a molecule is electrically-neutral, it has an equal number of protons and electrons. Therefore, one molecule also has 10 electrons.

3.13 Polyatomic IonsA polyatomic ion is an electrically-charged group of two or more atoms that share electrons. The atoms sharing electrons are said to be held together by covalent bonding. You can think of a polyatomic ion as a “charged molecule”, just as a monatomic ion is a “charged atom.”

The formula of a polyatomic ion lists the symbols of the elements that make up the ion. In the formula, we use subscripts to indicate how many atoms of each element are in one ion, and a superscript to indicate the charge.

▼ Example 16

The polyatomic ion called nitrate has a formula of NO3−. How

many protons and electrons does a nitrate ion have?Answer: 31 protons, 32 electronsReferring to a periodic table, we find that each N atom has 7 protons and each O atom has 8 protons. Therefore, a nitrate ion (NO3

−) has 31 protons, 7 in the N atom and 8 in each of the three O atoms: 7 + 3(8) = 31.Since the NO3

− ion has a charge of −1, it has one more electron compared to protons; therefore, the ion has 31+1 (or 32) electrons.


▼ Example 17

The polyatomic ion called mercury(I) or mercurous has a formula of Hg2

2+. How many protons and electrons does a mercurous ion have?Answer: 160 protons, 158 electronsReferring to a periodic table, we find that a mercury atom has 80 protons. Therefore, a mercurous ion (Hg2

2+)has 2×80, or 160 protons. The +2 charge means that the ion has two fewer electrons compared to protons. Therefore, the number of electrons is 160−2, or 158. To check, recall that charge is equal to (Number of Protons) minus (Number of electrons): 160 − 158 = +2.


TEST YOURSELF


1. Which of the following subatomic particles are found in the nucleus of an atom? A. protons and neutrons B. protons and electrons C. neutrons and electrons D. protons, neutrons, and electrons

2. The symbol 13C or C-13 represents an atom with… A. 13 protons B. 13 electrons C. 13 neutrons D. 13 nucleons

3. Which of the following has the largest mass number? A. 56Mn B. 56Fe C. 56Cr D. they all have the same mass number

4. How many neutrons does an O-18 atom have? A. 8 B. 10 C. 16 D. 18

5. Which of the following has the largest total mass? A. 2 protons + 4 electrons B. 2 neutrons + 2 protons C. 6 electrons D. they all have the same mass

6. The mass of one atom of Mg-26 is closest to: A. 24.3 u B. 24.0 u C. 26.0 u D. 12.0 u

7. Which of the following is false? A. An atom’s mass is concentrated in the nucleus B. The atom’s nucleus is positively charged C. The number of electrons in an atom is equal to the number of protons D. The number of protons in an atom is equal to the number of neutrons

8. A magnesium atom becomes a magnesium ion (Mg2+) by… A. gaining two electrons B. losing two electrons C. gaining two protons D. losing two protons



9. A nitrogen atom becomes a nitride ion (N3−) by… A. gaining 3 electrons B. losing 3 electrons C. gaining 3 protons D. losing 3 protons

10. How many protons and electrons are in a nitride ion (N3−)? A. 7, 10 B. 7, 4 C. 10, 7 D. 4, 7

11. How many protons and electrons are in a magnesium ion (Mg2+)? A. 10, 12 B. 12, 10 C. 24, 22 D. 22, 24

12. Which of the following describes a cation? A. an aluminum ion (Al3+) B. an oxygen atom that has gained 2 electrons C. an ion with 17 protons and 18 electrons D. an ion with more neutrons than protons

13. Which of the following describes an anion? A. a sodium ion (Na+) B. a calcium atom that has lost 2 electrons C. an ion with 8 protons and 10 electrons

14. How many hydrogen atoms are in 5 molecules of acetic acid (CH3COOH)? A. 3 B. 4 C. 15 D. 20

15. How many electrons are in a sulfur trioxide (SO3) molecule and a sulfite ion (SO32−)?

A. 40, 40 B. 40, 42 C. 38, 40 D. 78, 80

16. In a fictitious world, element X has only two isotopes: 77X with a mass of 77.0 u and 78X, with a mass of 78.0 u. The relative abundance of the heavier isotope is 30.0%. The average mass of the X atoms is...

A. 77 u B. 77.3 u C. 77.5 u D. 78 u

17. In a fictitious world, element Z has only two isotopes: 35Z with a mass of 35.00 u and 37Z, with a mass of 37.00 u. What is the relative abundance of the lighter isotope if the average mass is 35.8 u?

A. 40.0% B. 50.0% C. 60.0%

43

4CHAPTER

Energy

In this chapter, we examine the concept of energy. Energy is an important property of matter. Understanding this property of matter requires an understanding of how atoms and molecules move and why they move the way they do. We start by reviewing how physicists describe and explain motion.

4.1 Describing MotionMotion is quantitatively described using the concepts of displacement, distance, speed, velocity, and acceleration. To facilitate understanding of these concepts, we will restrict our discussion to one-dimensional motion (that is, motion in a straight line).

4.1.1 The Displacement VectorIn order to locate anything, we need to first define a point of reference; we call this the origin. Then, we specify two things:

• how far it is from the origin, and• which direction we need to take in order to get there from the origin.

If the movement of an object is restricted to a straight line, it is convenient to:• select a point on the line as the origin. At the origin, the location is defined

to be zero.• assign positive values for locations on one side of the origin and negative values

for locations on the other side. If d is the distance from the origin, then d is the location if it is on one side of the origin and -d is the location if it is on the other side.

It is common to use the letter x to represent location. Movement towards one side is then referred to as movement towards the +x direction and movement towards the other side is referred to as movement towards the −x direction. If the line of motion can be described as being “upward and downward” or “northbound and southbound,” the letter y or z is commonly used.


The difference in the location of an object at two instances is called its displacement. Suppose it is located at x1 and x2 at times t1 and t2, respectively. The displacement is defined as:

Δx x x2 1= −

When the Greek letter delta (Δ) is written in front a symbol, it means “change in whatever the symbol stands for.” Thus, Δx means “change in x”; it is calculated by subtracting the initial value (x1 in this case) from the final value (which in this case is x2).

▼ Example 1

An object moves 3.0 m to the left. What is the displacement if we take locations right of the origin as positive and locations left of the origin as negative?Answer: −3.0 m Since the final location is to the left of the original location, we expect x2 to be less than x1. Therefore, x2 minus x1 would be negative:

Δx x x 3 m2 1= =− − .0

We say that the object moved 3 meters in the −x direction.

Displacement is an example of a vector quantity, something that has magnitude and direction. For one-dimensional motion, we specify the direction of vectors using positive and negative algebraic signs.

4.1.2 DistanceThe distance that an object has traveled is equal to the absolute value of the displacement if it has not changed direction.

d = |Δx |If it changed direction, then we add up the distances traveled for each leg of the trip to get the total distance traveled. Distance is always a positive number.

▼ Example 2

An object moves 2.0 m to the right, then 3.0 m to the left. What are the displacements and distance traveled for each leg. Calculate the total displacement and distance traveled.Answer:Assuming right corresponds to the +x direction, the displacement and distance traveled for the first leg are:

Δx 2 m d 2 m= + =. , .0 0

Chapter 4: Energy 45

For the second leg, the displacement and distance traveled are:Δx 3 m d 3 m= =− . , .0 0Δx 3 m d 3 m= =− . , .0 0

The total displacement is:Δx 2 m 3 m 1 m= (+ ) + ( ) =. . .0 0 0− −

but the total distance traveled is:d 2 m 3 m 5 m= ( ) + ( ) =. . .0 0 0

We can see that the object ends up being 1.0 m to the left of its original location; the displacement is −1.0 m. However, it has traveled a total distance of 5.0 m.

4.1.3 SpeedSpeed refers to how fast an object is moving. The average speed of an object is calculated by dividing the distance traveled by the amount of time that has elapsed:

s dt

=Δ

▼ Example 3

Suppose an object travels 10.0 m in 2.0 min. Calculate its average speed?

Answer: 5.0 m/min or 5.0 m min−1

We calculate the average speed by dividing the distance by the elapsed time.

sdt

mm= = =

Δ10 02 0

5 0 1.. min

. min−

We keep only two significant digits in the answer since the less precise term in the calculation, 2.0 min, has only two significant digits.

In the preceding example, we say that the average speed is 5.0 meters per minute. It does not necessarily mean that the object traveled 5.0 meters every minute. It could have traveled 8.0 m during the first minute, then 2.0 m during the second minute. This would still amount to a total distance traveled of 10.0 m in 2.0 min. In fact, there are an infinite number of possibilities. The actual speed at a specific point in time is called the instanta-neous speed (or simply the speed). If you were traveling in a car, your instantaneous speed would be what your speedometer says at any given instant; it will most likely be changing throughout your entire trip.


If the speed of an object does not change, then we say that it is moving at constant speed and the average speed will also be equal to this constant speed.

4.1.4 VelocityVelocity, v, is a vector, just like displacement. Its magnitude is the speed. For one- dimensional motion, we use positive values (v = s) if the object is moving in whatever we choose as the positive direction. We use negative values (v = –s) if it is moving in the opposite direction.

▼ Example 4

An object is moving with a velocity of −5.0 m/s. What is its speed?Answer: s = |−5.0 m/s| = 5.0 m/sSpeed, by definition, is always a positive number. The negative sign simply means that the object is moving in whatever was chosen as the negative direction.

4.1.5 AccelerationAn object is said to be accelerating if it changes velocity. The average acceleration is defined as:

a vt

= ΔΔ

The SI unit for acceleration is meters per second per second (m/s/s) or meters per second squared (m/s2 or m s−2).

▼ Example 5

True or False. An object moving at constant speed is not accelerating.Answer: False.Velocity could change if the object changes direction, even if it does not change speed.

As illustrated in the following examples,• the acceleration and velocity vectors point in the same direction (have the same

algebraic sign) if an object speeds up without changing direction.• the acceleration and velocity vectors point in opposite directions (have opposite

signs) if the object slows down or reverses direction.


▼ Example 6

An object moving on a straight line speeds up from 4.0 m s−1 to 8.0 m s−1 in 4.0 s. Calculate the average acceleration.

Answer: +1.0 m s−2

It is customary to take the original direction of motion as positive. Therefore, let us take the initial velocity as positive:

v1 = +4.0 m s−1

Since the object is not changing direction, the final velocity is also positive:

v2 = +8.0 m s−1

The average acceleration is:

avt

v vt

m s m ss

m s= = = = +ΔΔ Δ

2 1 28 0 4 04 0

1 0− − −( . ) ( . )

..

/ /

which turns out to be positive as well. This means that, on average, the object’s speed increased by 1.0 m s−1 every second. In 4.0 seconds, the speed went up by 4.0 m s−1 (from 4.0 m s−1 to 8.0 m s−1). The acceleration and velocity vectors are both positive; both vectors are pointing in the same direc-tion when an object is speeding up. Had we taken the direction of motion as negative, we would have gotten a negative acceleration as well (−1.0 m s−2).

▼ Example 7

An object moving in a straight line slows down from −4.0 m s−1 to −2.0 m s−1 in 2.0 s. Calculate the average acceleration.

Answer: +1.0 m s−2

The acceleration is:

avt

v vt

m s m ss

m s= = = = +ΔΔ Δ

2 1 22 0 4 02 0

1 0− − − − −( . ) ( . )

..

/ /

which is positive. Note that when the velocity becomes less negative, it means that it is slowing down while moving in the negative direction. Any time an object slows down in a given direction, we should get opposite algebraic signs for acceleration and velocity.


▼ Example 8

A molecule moving at 500.0 m s−1 towards a wall, collides with the wall, then moves in the opposite direction at a constant speed of 500.0 m s−1 during a time interval of 0.10 s. Calculate the average acceleration during this time interval?

Answer: −1.0 × 104 m s−2

Let us take the initial velocity as positive:v1 = +500.0 m s−1

Since the final velocity is in the opposite direction, we assign it a negative value:

v2 = −500.0 m s−1

Therefore, the average acceleration is:

avt

v vt

m s m ss

= = =ΔΔ Δ

2 1 500 0 500 00 10

− − −( . ) ( . ).

/ /

= − × −1 0 104 2. m s

▼ Example 9

True or False. A negative acceleration means that an object is slowing down.Answer: False.It is true only if the original direction of motion is taken as posi-tive. If the original direction of motion is taken as negative, then a negative acceleration means that the velocity is becoming even more negative and the object is speeding up in that direction.

4.2 Explaining MotionNewton’s Laws of Motion gives us an explanation for why things move the way they do. Newton’s 1st law basically says that the natural tendency for objects is to move at con-stant velocity. Newton’s 2nd law says that objects accelerate if there is an unbalanced force (a push or a pull) acting on it. Newton’s 3rd law says that if one object exerts a force on another, then it also experiences an equal and opposite force from that other object. The third law is often stated as “for every action, there is an equal and opposite reaction.”

Force is a vector and the sum of all the forces acting on an object is called the net force. Newton’s 2nd law tells us how to calculate force. It says that the net force acting on


an object (F) is directly proportional to the observed acceleration (a) of the object; the proportionality constant is the mass of the object (m). Mathematically, we say that

F = m aThe unit for force is Newton, N, which is equivalent to 1 kg m s−2. This equation means that, if an object is experiencing acceleration, the net force acting on it must be nonzero. The greater the acceleration, the greater the net force. It also means that a greater force is needed cause the same acceleration for a larger mass.

▼ Example 10

A force of 5 N pushes a object to the right. A force of 7 N pulls it to the left. What is the net force on the object?Answer: 2 N (to the left)If we assume right as the +x direction, then the net force is the sum of (+5 N) and (−7 N):

F = (+5 N) + (−7 N) = −2 NA negative net force means that it is directed to the left, opposite of what we assumed as the positive direction.

▼ Example 11

Which of the following will experience greater acceleration if sub-jected to the same net force?A. a proton B. an electronAnswer: BWe can rearrange F = ma to: a = F/mFor the same force (same numerator), a larger mass (larger denominator) will give us a smaller acceleration. A smaller mass will give us a larger acceleration. A proton’s mass is about 2000 times larger than that of an electron.

Since mass is always positive, the force vector points in the same direction as the accel-eration vector. If an object is speeding up as it is moving in a particular direction, then it must be experiencing a net pull or push in that direction. If it is slowing down, then it must be experiencing a net push or pull in the opposite direction. If it reverses direc-tion, then it must be experiencing a net push or pull in the direction opposite the original direction. Therefore, a positive acceleration implies a positive net force; a negative accel-eration implies a negative net force.


▼ Example 12

An object heading east slows down. What is the direction of the net force vector?Answer: West.If it slows down as it is heads east, something must be pulling or pushing it westward. If we take the eastward direction as positive, then the velocity vector is positive, but the acceleration vector is negative (because acceleration and velocity have opposite signs for something that is slowing down), and the force vector is also negative (which means it is pointing west).

▼ Example 13

The speed of a 10.0 kg object in “free fall” increases by 9.8 m/s every second. Calculate the net force on the object.Answer: 98 N (downward)Let us assume that the downward direction is negative. Since the object is increasing speed as it moves downward, its acceleration is also negative. Therefore:

a 9 8 m s 2= − −.The acceleration must be due to a net force that is also negative (pointing downward):

F = m a = (10.0 kg)(−9.8 m s−2) = −98 N

4.2.1 WeightThe value 9.8 m s−2 in the preceding example is observed for all falling objects near the surface of the earth, in the absence of any air around the object. It is called the accel-eration due to gravity, and is represented by the letter g. This is due to the force that the earth exerts on the object; the magnitude of this force is called the object’s weight (on earth):

W = m g

The reason the acceleration would be less than 9.8 m s−2 in the presence of air is that, as the object falls, it collides with air molecules which exert an upward force on the object.


▼ Example 14

What is the weight of a 0.50 kg book? How much force does the book exert on the earth?Answer: 4.9 N (upward)Near the surface of the earth, the weight is:

W = m g = (0.50 kg)(9.8 m s−2) = 4.9 N.If we take the downward direction as negative, then we say that the force on the book due to gravity is −4.9 N. By Newton’s third law, the book is also pulling on the earth with a force of +4.9 N.

4.2.2 Normal ForceConsider a 0.50 kg book, at rest, on a table. Since the book is not moving, its velocity is constant and equal to zero. Since velocity is constant, acceleration is also equal to zero (and also constant). Since acceleration is zero, the net force is also zero. But we saw in an earlier example that the earth does exert a downward force on the book and we calculated it to be −4.9 N. Therefore, there must be an upward force on the book that is +4.9 N in order for the net force to be equal to zero. Where does this upward force come from? It comes from the interactions of the atoms on the table’s surface with the atoms of the book. Once these atoms get too close, they start to repel each other. It is customary to just lump all of these forces together and call the net upward force as the normal force.

4.2.3 FrictionConsider a book, at rest, on a table. Suppose we attempt to push it sideways by applying a force of, say, 10.0 N, but it does not move. Since the book is not moving, its velocity is con-stant and equal to zero. Since velocity is constant, acceleration is also equal to zero (and also constant). Since acceleration is zero, the net force is also zero! Since we are exerting a 10.0 N force on the book, there must be another force with a magnitude of 10.0 N acting on the book in the opposite direction. We call this force friction. Just like the normal force, it comes from the interactions of the atoms on the table’s surface with the atoms of the book.

4.2.4 Fundamental ForcesAll forces in nature can be classified into just four fundamental types: electromagnetic, strong, weak, and gravitational. Strong and weak forces are significant only between particles at very, very short distances from each other (such as protons and neutrons in


the nucleus of an atom). In Chemistry, we are mostly concerned with gravitational and electromagnetic forces. Interactions among atoms and molecules primarily involve electro-magnetic forces.

▼ Example 15

What type of fundamental force is mainly responsible for friction?Answer: electromagneticFriction is due to interactions among atoms. Forces among atoms are primarily electromagnetic in nature.

4.2.4.1 Gravitational ForceNumerous observations of the way objects interact suggest that any pair of particles exerts a force on each other that depends on their masses and how far they are from each other, according to the equation:

FG m m

rgrav = 1 22

where G is a constant (6.67 × 10−11 N m2 kg−2), m1 and m2 are the masses of the particles and r is the distance between them. This force is called gravitational force.

▼ Example 16

Consider the earth (with mass m1 = 6.0 × 1024 kg) and an object (with mass m2) as two particles separated from each other by a distance equal to the radius of the earth (r = 6.4 × 106 m). Explain why objects near the surface of the earth experience an acceleration of 9.8 m/s2.Answer:We can calculate the gravitational force of attraction between the earth and an object near the surface of the earth:

FN m kg kg m

mm sgrav = =

( . )( . )( . )

( .6 67 10 6 0 10

6 4 109 8

11 2 2 242

6 2

× ××

− −−22

2)m

How near is “near”? Even if an object is, say, 40 km (or 0.04 × 106 m) above the surface, the denominator in the preceding calculation will still be the same (within two significant digits).

4.2.4.2 Electromagnetic forcesIt is known that rubbing two objects together could lead to attractions between them that are stronger than can be accounted for by gravitational forces. The two objects are said to


be (electrically) charged. Charged objects are known to either attract or repel each other, suggesting that there are two kinds of charge; we classify them as positive or negative. Objects with like charges (both positive or both negative) repel each other; objects with unlike charges attract each other. The observed force is called electrostatic or Coulombic force. By assigning a negative charge for each electron and a positive charge for each proton, we can account for observations pertaining to electrostatic interactions using Coulomb’s law. Coulomb’s law says that the magnitude of the force between two objects depends on the magnitude of the charges and the square of the distance (r) between them. If q1 and q2 are the magnitudes of the charges, then

Fk q q

rcoul = 1 22

where k is a constant equal to 9.00 × 109 N m2 C−2. The unit for charge is called the Coulomb (C). The charge of an electron is −1.602 × 10−19 C, and the charge of a proton is +1.602 × 10−19 C. The charge of anything is equal to the algebraic sum of the charges of all the electrons and protons in it. The transfer of electrons from one object to another when they are rubbed together is what causes them to be electrically charged and explains why they attract each other. Rubbing two objects together makes one positively charged and the other negatively charged.

▼ Example 17

The most probable distance of the electron from the nucleus of a hydrogen atom is 52.9 pm. Compare the gravitational and C oulombic forces at this distance. Given: mass of electron is 9.11 × 10−31 kg, mass of proton = 1.67 × 10−27 kg, 1 pm = 10−12 m.Answer: Plugging in the masses, charges and distance in the appropriate formulas, we find that the magnitude of the Coulombic force is about 1039 times larger.

FN m kg kg kg

grav =( . )( . ) ( . )

( .6 67 10 9 11 10 1 67 10

52 9

11 2 2 31 27× × ×− − − −

×××−

−

103 63 1012 2

47

mN

).=

FN m kg kg kg

grav =( . )( . ) ( . )

( .6 67 10 9 11 10 1 67 10

52 9

11 2 2 31 27× × ×− − − −

×××−

−

103 63 1012 2

47

mN

).=

FN m C C C

coul =( . )( . ) ( . )

( .9 00 10 1 602 10 1 602 10

52 9 10

9 2 2 19 19× × ××

− − −

−−−

12 288 25 10

mN

).= ×

FN m C C C

coul =( . )( . ) ( . )

( .9 00 10 1 602 10 1 602 10

52 9 10

9 2 2 19 19× × ××

− − −

−−−

12 288 25 10

mN

).= ×


In addition to electrostatic forces, charged particles are also affected by magnetic forces. A magnetic force is produced by a moving charged particle. Some particles such as electrons and protons also have an intrinsic magnetism; by this, we mean they produce a magnetic force by being what they are.

4.3 EnergyThe concept of energy gives us an alternative way of describing motion. Oftentimes, it is more convenient to use.

4.3.1 Kinetic EnergyIf an object of mass m has a velocity v, then its kinetic energy is defined as:

K mv= 12

2

The unit for kinetic energy is Joule (J), which is defined to be kg m2/s2 or kg m2 s−2.

▼ Example 18

Which has more kinetic energy?A. 2.00 kg mass moving at 2.00 m s−1

B. 2.00 kg mass moving at −2.00 m s−1

Answer: SameA and B have the same mass but different velocities. But kinetic energy depends on the square of the magnitude of velocity, not the velocity itself. In other words, it depends only on speed, not direction. Kinetic energy is a scalar, not a vector. In this particular case, A and B have the same speed (2.00 m/s), even if they are moving in opposite directions.

▼ Example 19

What is the kinetic energy of a nitrogen molecule (mass = 4.65 × 10−26 kg) moving at a speed of 5.00 × 102 m s−1?

Answer: 5.81 × 10−21 J

K = (1/2) m v2 = (1/2)(4.65 ´ 10−26 kg) (5.00 ´ 102 m s−1)2

= − −5.81 kg m s , or 5.81 J× ×10 1021 2 2 21


4.3.2 Work and the Work-Energy TheoremWork is defined as the application of a force over a distance. If a force f is applied while an object undergoes a displacement Δx, then the work done (w) is:

w f x= Δ

▼ Example 20

What is the work done due to gravity when a 1.00 kg object falls a distance of 0.50 m?Answer: 4.9 JThe force of gravity on the object is pointing downward. Let us take the downward direction as the positive direction. So, the acceleration is g.

f m a 1 kg 9 8 m s 9 8 N2= = (m)(g) = ( )( ) =. . .00 −

The displacement is also positive since the object moved downward:

Δx = +0.50 mTherefore, work is: w f x 9 8 N 5 m 4 9 J= = = +Δ ( . )( . ) .0 0

The unit for work is Joule, just like kinetic energy. In fact, it can be shown that the work associated with the net force on the object is equal to the change in its kinetic energy:

ΔK w=

This is known as the work-energy theorem. Thus, work can also be defined as energy that is transferred due to the application of force over a distance. If f and Δx have opposite signs, w is negative; this means that application of a force against the direction of motion slows down an object or decreases its kinetic energy (ΔK = w < 0). If f and Δx have the same sign, w is positive; this means that application of a force in the same direction as the direction of motion speeds up an object or increases its kinetic energy (ΔK = w > 0).

▼ Example 21

By how much does the kinetic energy of a 1.00 kg object change when it falls a distance of 0.50 m? What is the speed at this point if the object starts from rest?


Answer: increase by 4.9 J, 3.1 m s−1

In the preceding example, we calculated the work done to be +4.9 J. Thus, according to the work-energy theorem, the kinetic energy will increase by 4.9 J. We expect the object to speed up since the force (due to gravity) is downward and the object is moving downward.Since the object starts from rest, the kinetic energy after it has fallen 0.50 m is 4.9 J. Using the formula for kinetic energy:

K mv=12

2

we can solve for the speed:

vK

mkg m s

kgm s= = =

2 2 4 91 00

3 12 2( . )

.. /

−

4.3.3 Potential EnergyAny pair of particles is said to have a potential energy (V) due to the force they exert on each other, which depends only on the distance between them. Thus, potential energy is defined as energy due to position. For any set of particles, the total energy (E) is defined as the sum of the kinetic energies of all the particles and the potential energies for every pair of particles.

▼ Example 22

Using Ki and Vij to represent the kinetic energy of particle “i” and the potential energy due to the interaction of particles “i” and “j”, write the expression for the total energy of three particles (“1”, “2”, and “3”).Answer: E K K K V V V1 2 3 12 13 23= + + + + +The total energy is the sum of:• K1, the kinetic energy of particle “1”• K2, the kinetic energy of particle “2”• K3, the kinetic energy of particle “3”• V12, the potential energy of interaction of particles “1”

and “2”• V13, the potential energy of interaction of particles “1”

and “3”, and• V23, the potential energy of interaction of particles “2”

and “3”


Potential energy is defined such that, for any set of particles, the total energy is constant if there are no external forces affecting the particles. We could say that if we leave a set of particles alone, then its total energy is conserved; the potential and kinetic energies may change as the particles move, but the total energy remains the same. This means that a drop in potential energy implies an increase in the kinetic energy (faster movements) and vice versa.

▼ Example 23

Consider a ball and the earth as two particles; ignore the rest of the universe. As the ball is dropped from a certain height, what happens to the potential energy?Answer: The potential energy decreases as the ball falls.We know that the ball will speed up as it falls. Therefore, its kinetic energy increases as it falls. Since the total energy must remain constant, the potential energy must be decreasing.

▼ Example 24

Describe what happens to the potential energy as an electron moves away from a proton?Answer: it increasesSince the electron and proton have opposite charges, they attract each other. The electron will be slowing down as it moves away from the proton. Its kinetic energy will be decreasing. Therefore, the potential energy will be increasing.

The value of potential energy does not really mean anything. What we are really interested in are changes in the value of the potential energy as particles move. The change in potential energy gives us information about the change in kinetic energy (whether a particle has sped up or slowed down). As a matter of convenience, we need to establish a “reference” so that we can have values for potential energy that we can use for calculations. For a pair of charged particles, a commonly used reference is the case where the particles are at infinite distance from each other; if we define the potential energy to be zero at this reference distance, then we can calculate the potential energy at any distance (r) using the following equation:

V kq q

r= 1 2


where q1 and q2 are the charges and k is the same constant used in Coulomb’s law (k = 9.00 × 109 N m2 C−2). Examining the equation, we can see that as r approaches infinity (∞), the denominator becomes very large, making the potential energy (V) approach zero. As the two particles approach each other, the denominator (r) approaches zero and V rises towards +∞ if the particles have like charges (q1 and q2 both positive or both negative); V drops towards –∞ if the particles have unlike charges.

▼ Example 25

One of the two curves shown below corresponds to the potential energy between two electrons. Which one?

Answer: the blue curveTwo electrons will repel each other. As they move farther apart, they will be speeding up. This means they will be gaining kinetic energy. Increasing kinetic energy means decreasing potential energy. The blue curve shows a potential energy that decreases as the distance (r) increases. It eventually levels off since the repul-sion diminishes when the electrons are very far apart; at very large distances the speed of separation will level off.


▼ Example 26

Shown below (red curve) is the potential energy for a proton and an electron (as in a hydrogen atom). If the green line repre-sents the total energy, describe the two particles when they are separated at distances corresponding to points A, B, and C on the graph.

Answer:Based on Newton’s 2nd law and Coulomb’s law:• Of the three points shown, point A corresponds to the case

where the particles are closest to each other (about 100 pm apart). Note: 1 pm = 10−12 m.– If the particles happen to be approaching each other, the

potential energy will be drastically dropping, which means that they will be speeding up their approach until they crash into each other. The kinetic energy, the vertical difference between the green curve (total energy) and red curve (potential energy) will be getting larger as r gets smaller.

– If they happen to be moving away from each other, the potential energy will be increasing (red curve rising as r increases), and the particles will be slowing down (kinetic energy decreasing) until their separation corresponds to the distance at point B; this distance is around 420 pm.

• If the particles are separated by about 420 pm (which cor-responds to point B), the potential and total energy would be equal (the red and green lines cross at point B). This means the kinetic energy is zero at this instant; the particles are motionless (momentarily); all the energy is potential. However, since there is an attractive force between the particles, they will start moving towards each other and speed up until they crash into each other.


• At distances larger than that at point B, as in point C (1000 pm), the kinetic energy is negative! Why? At these distances, the green curve is below the red curve; this means that the total energy is less than the potential. By definition, kinetic energy cannot be negative. K = (1/2)mv2. Mass (m) is always positive and the square of the speed (v2) is always positive. The part of the chart below the (red) potential energy curve is said to be a “forbidden region.” In other words, with a total energy as indicated by the green line, the particles can only separate as far as about 420 pm; from there, they move towards each other.

Note: The prediction we made for points A and B was troubling to scientists towards the end of the 19th century. It suggests that anytime an electron approaches a positive nucleus, it will eventually crash into it, which would mean that the atom as we now know it cannot possibly be stable long enough for any of us to exist. This, and other “anomalous” phenomena discovered around that time were eventually resolved with the development of modern physics. The physics covered in this chapter, which was developed by Isaac Newton and others prior to the 20th century, is now referred to as classical physics. Among the strange predictions of modern physics is that there is a finite probability of the particles being in the forbidden region mentioned above. Thus, the said region is generally referred to as a “classically-forbidden” region.

▼ Example 27

Shown below (in red) is the potential energy for a proton and an electron (as in a hydrogen atom). If the green line (at 1 × 10−18 J) represents the total energy, describe what happens when the par-ticles are separated by 1000 pm.


Answer:The blue point corresponds to the situation where the total energy is 1 × 10−18 J and the particles are separated by 1000 pm. According the classical physics:• If they happen to be approaching each other, the distance (r)

would be decreasing, the potential energy (red curve) would be decreasing, the kinetic energy would be increasing, the particles would be speeding up towards each other and even-tually crash into each other.

• If they happen to be moving apart, the distance (r) would be increasing, the potential energy (red curve) would be increas-ing but eventually level off at zero, the kinetic energy would be decreasing and eventually level off at 1 × 10−18 J, the parti-cles would be slowing down but eventually settle into a con-stant rate of separation and continue moving apart forever.

4.4 Bond EnergyWe now apply the concepts to potential and kinetic energy to understand bonding of atoms. Since the bulk of an atom’s mass is concentrated in the nucleus, the movement of the nucleus is taken as the that of the atom.

Consider the potential energy plots shown in Figure 1 for two protons. The red curve represents the how the potential energy (V), due to electrostatic repulsion, depends on the distance (r) between them.

The potential energy drops towards zero as they separate (r approaches + ∞) and rises as the protons approach each other (r approaches zero). The rising potential energy as two protons approach each other means that they will be slowing down (losing kinetic energy), which is what we expect since like charges repel each other.

Pairs of charged particles are hardly ever by themselves in the universe. For example, the two protons in an H2 molecule are surrounded by two electrons, which have attrac-tive interactions with the protons (nuclei). When we take the effect of electrons into account, the behavior of the two protons in H2 is better described by the blue curve in Figure 1. This blue curve is called the molecular potential energy curve of H2; the distance between the nuclei (r) is taken as the distance between the two H atoms. V(r) is zero at infinite distance (very large r). But as the nuclei approach each other, they actually gain kinetic energy due to the presence of electrons; the potential energy dips below zero until they reach a distance of about 74 pm. As they get even closer, the potential energy starts to rise sharply; this means that they start to slow down as the electrons are no longer able


to alleviate the repulsion between the nuclei. The total energy of an average H2 molecule at room temperature is shown on the graph by the green line; this line is flat (horizontal) since the total energy is constant if the particles are left alone. If we zoom in on this section (see Figure 2), we can conclude that the two protons in the average H2 molecule are “stuck” together. With only as much total energy as indicated by the green line in Figure 2, the nuclei can only get as close as around 70 pm and cannot move farther than about 80 pm from each other. The nuclei will be oscillating between these two distances. This “bonding” of two positive nuclei (that would otherwise repel each other) is made possible by the sharing of electrons by the two atoms and is called covalent bonding. The distance where the minimum in V(r) occurs is called the bond length. A common way of

Figure 1. Potential Energy of Two Protons


describing the “size” of a hydrogen atom is to imagine it as a sphere with a radius equal to half of this bond length. At the closest possible distance, we could say that the two atoms have “collided.”

▼ Example 28

Consider the potential energy curve for H2 shown in red below. For which of the two energy levels indicated do we expect the two atoms to be able to move farther from each other?A. the level indicated by the blue line B. the level indicated by the

green line

Figure 2. Potential Energy and Average Total Energy at Room Temperature of H2


Answer: AThe distances where the total energy is all potential (no kinetic; where red curve intersects with blue or green line) represent the closest and farthest that the atoms can be from each other. For the energy level indicated by the green line, the allowed dis-tance is between about 50 pm and 130 pm. For the energy level indicated by the blue line, the atoms can be as close as about 40 pm, but can be as far away from each other as possible. The blue line does not intersect the red curve at large distances since the red curve (the potential energy) approaches zero as r approaches infinity and the blue line is above zero.Note that since the typical potential energy curve rises very sharply at short distances, giving two atoms more energy would have more impact on their ability to separate than on their ability to approach each other.

The “total” energy” referred to in the preceding discussion is called the relative or internal energy” of the molecule; the kinetic energy is due to motion of the nuclei that we could describe as vibration and rotation and the potential energy accounts for the total energy of the electrons and the repulsion between the nuclei. The true total energy of the


molecule is the sum of the relative energy and translational energy, the kinetic energy associated with the movement of the molecule as a whole.

The shape of the potential energy curve for H2 is typical for any pair of atoms, mol-ecules, or oppositely-charged ions. Because of its shape, the potential energy curve is often referred to as a potential well. The extent to which a pair of particles can separate from each other depends on how the total energy compares to the “depth” of the potential well. Pairs of oppositely charged ions and pairs of atoms in a stable molecule tend to have much deeper wells that pairs of molecules.

How can two bonded atoms be separated? They will have to gain energy. This can hap-pen if the molecule collides with a particle (a “third body”) that has a high kinetic energy (fast moving particle). If the amount of kinetic energy that can be transferred from the third body is at least equal to the potential well depth, then it will be sufficient to separate any pair of atoms (in the well) that it encounters. Thus, the well depth is called the bond dissociation energy. If this amount of energy is transferred to a pair of bonded atoms, the relative kinetic energy of the separated atoms will be the same. We could say that the net result is the conversion of kinetic energy (of a third body) into potential energy of the separated atoms.

▼ Example 29

The average N2 molecule at temperature T (in Kelvin) has a transla-tional energy of about (3/2)kBT, where kB = 1.38 × 10−23 J K−1. The dissociation energy (De) for a pair of N atoms is 1.6 × 10−18 J. Estimate the temperature at which the average N2 molecule has enough translational energy to be able to break up another N2 molecule into two N atoms by collision?Answer: About 77000 KThe average translational energy must be at least equal to the dissociation energy. So:

32

k T De≥

or:

TDk

JJ K

Ke

B

≥ =−

− −

23

21 6 1031 38 10

7 7 1018

23 14=

( . )( . )

.×

××

How can two, initially separated, atoms end up strongly bonded to each other? The potential well must be deep and they need lose kinetic energy to a third body so that they end up being “trapped” in the potential well. The net result is the conversion of the poten-tial energy of the two separated atoms into kinetic energy of a third body; the potential energy of the resulting molecule is lower.


▼ Example 30

Consider the potential energy curve shown in red below for H2. For a pair of H atoms with total energy indicated by the blue line, at which of these distances can collision with a third body cause bond formation, leading to the formation of H2 molecule with total energy indicated by the green line? A. 100 pm B. 200 pm

Answer: A.At 200 pm separation, the green line (total energy) is below the red line (potential). The total energy cannot be less than the potential energy; it’s “classically forbidden.” Bond formation is possible, but the energy of the resulting molecule will be higher than that indicated by the green line. At 100 pm separation, bond formation is possible and the energy of the resulting molecule can be anything higher than the potential energy at 100 pm, which is below the green line.

4.5 Energy TransferChemistry deals with changes that matter undergoes. These changes are generally accom-panied by a transfer of energy. Energy that is transferred can be classified as either heat or work. We will precisely define these terms in this section.


4.5.1 Thermal Energy, Temperature, HeatThe terms thermal energy, temperature, and heat are used interchangeably in common everyday conversation. The distinctions among these terms are best understood from an atomic perspective.

The thermal energy of a sample of matter is defined as the sum of the kinetic energies of atoms in a sample. Thermal energy is an example of an extensive or extrinsic property of matter; an extensive property is directly proportional to the amount of sample. If we imagine duplicating a sample, the sample and the duplicate together would have twice as much thermal energy as the original sample.

Temperature often defined as the degree of “hotness” or “coldness” gives us an indi-cation of the average kinetic energy of the atoms in a system. When the temperature of a sample increases, it means that the average kinetic energy of the atoms has increased. Temperature is an example of an intensive or intrinsic property; it is independent of the amount of sample. If we imagine duplicating a sample, the sample and the duplicate together would still have same temperature (not twice as hot!); the atoms in both sample and duplicate have the same average kinetic energy.

Based on experience, we know that if we put something hot in contact with something cold; the temperature of the hot object would drop and the temperature of the cold object would increase. We interpret this observation as the result of a transfer of energy from the hot object to the cold one; the energy transferred is called heat.

To summarize, thermal energy is energy that a sample of matter has, whereas heat is energy that is transferred into or out of a sample of matter when we put it in contact with something that has a different temperature.

▼ Example 31

Consider the following:Sample A: 5.0 g Cu at 50°CSample B: 10.0 g of Cu at 50°C

Which sample has atoms with higher average kinetic energy? Which sample has more thermal energy? Will heat flow between samples A and B?Answer: The atoms in both samples have the same average kinetic energy since the temperature is the same. However, sample B has more thermal (or total kinetic) energy since it has more atoms. There will be no heat flow since the temperature is the same.


▼ Example 32

Consider the following:Sample A: 5.0 g Cu at 50°CSample B: 5.0 g of Cu at 100°C

Which sample has atoms with higher average kinetic energy? Which sample has more thermal energy? Will heat flow between samples A and B?Answer: The atoms in sample B have a higher average kinetic energy since the temperature of sample B is higher. Since the number of atoms is the same for both samples, we can also con-clude that Sample B has a higher thermal (or total kinetic) energy. Heat flows from a hotter object to a colder one; heat will from sample B to sample A if they are put in contact.

4.5.2 Heat and WorkWe can make the temperature of an object change without putting it in contact with some-thing warmer or colder. Stirring water, for example, can cause its temperature to increase. This is because heat flow is not the only way to transfer energy. Energy that is transferred any other way besides heat is classified as work.

At the atomic level, the energy transfer associated with heat can be thought of as being due to random collisions of atoms. On the other hand, energy transfer associated with work, can be thought of as resulting from collisions due to some kind of organized motion of atoms (“mechanical work”) or electrons (“electrical work”).

▼ Example 33

Which of the following ways of increasing the temperature of a nail is classified as heating? A. Putting it on a hot stove B. Hitting it with a hammer C. Passing electricity through itAnswer: A.Heating is energy transfer due to a temperature difference; random collisions between atoms of the stove and atoms of the nails cause a redistribution of thermal energy. The collisions of the atoms of the nail and hammer result from an orchestrated, synchronized motion of the atoms of the hammer; the energy


transfer is classified as mechanical work. Electricity is just a flow of high-energy electrons. The temperature increase resulting from passing electricity through the nail is due to the organized move-ment of high-energy electrons through the nail. The collisions between electrons and the atoms of the nail lead to a transfer of energy; the energy transfer is classified as electrical work.

4.5.3 Energy Transfer and Temperature ChangeA temperature change does not necessarily mean that energy was transferred into or out of a sample. Similarly, transfer of energy into or out of a sample does not necessarily lead to a temperature change. The reason for these is that temperature gives an indication of the average thermal (kinetic) energy only. The atoms within the sample itself could move around so that there is a net conversion of potential energy into kinetic energy or vice versa. These movements are associated with what are known as phase changes and chemical changes. In the next section, we will ignore phase and chemical changes. We will deal with phase and chemical changes in other chapters.

4.5.4 Heat Measurement4.5.4.1 Units for HeatThe original unit for heat, the calorie, is defined as the amount of heat needed to raise the temperature of one gram of water by 1°C. James Joule’s experiment showed that this tem-perature change can also be accomplished by doing work. Joule used gravity to pull down a weight that turns a paddle to stir water in a bucket. He showed that heat and work are, in fact, just two different ways of transferring the same thing (energy). In terms of the SI unit for energy (which is named in honor of Joule), Joule’s experiments showed that 1 calorie is equivalent to about 4.18 J. The modern definition of a calorie as used in Chemistry, or the “thermochemical calorie,” is:

1 cal = 4.184 J (exactly)

4.5.4.2 Symbol and Sign Convention for HeatIn order to keep track of energy transfers, it is convenient to assign positive values to heat if it flows into a sample and negative values if it flows out of a sample. The commonly used symbol for heat is q.


▼ Example 34

A hot spoon on a kitchen counter cools to room temperature. What algebraic sign do we assign to q?Answer: negativeAs the hot spoon cools down, heat flows out of it. If we were interested in q for the spoon’s surroundings, we would assign that a positive value since heat that flowed out of the spoon went into the surroundings.

4.5.4.3 Heat and Heat CapacityWe can determine how much heat flowed into or out of a sample by simply measuring the temperature change, provided we know its heat capacity. Heat capacity (C) is the amount of heat flow that would cause the temperature of sample to change by one degree Celsius (or one Kelvin). The unit for heat capacity is Joule per Kelvin (J/K or J K−1) or Joule per degree Celsius (J /°C or J °C−1). If the temperature of a sample changes by an amount equal to ΔT, then the heat flow (q) is:

q C T= Δ

This equation tells us that, for a given sample, a larger change in temperature involves a larger heat flow. An increase in temperature means heat flow into the sample (q > 0; positive). A drop in temperature (ΔT < 0; negative) means heat flow out of the sample (q < 0; negative).

▼ Example 35

The heat capacity of a sample is 20.9 J/K. Calculate the heat flow if the temperature of the sample drops from 50.0°C to 40.0°C.Answer: −209 JThe change in temperature is the difference between the final and initial temperature:

ΔT T T 4 C 5 C 1 C or 1 Kfinal initialo o o= = =− − − −0 0 0 0 0 0 0 0. . . . .

Remember that a change of one degree Celsius is equivalent to a change of one Kelvin. We do not add 273.15 because we are dealing with temperature changes. Note that ΔT is negative; the temperature dropped.The amount of heat flow is: q C T 2 9 J/K)( 10.0K) 209 J= = − = −Δ ( 0. ,

q C T 2 9 J/K)( 10.0K) 209 J= = − = −Δ ( 0. , which means that 209 Joules of heat flowed out of the sample.


For homogeneous samples (samples that have uniform properties throughout), we may find values of heat capacities tabulated on a per gram basis. The heat capacity per gram is called specific heat capacity, or simply specific heat. The unit for specific heat is Joule per Kelvin per gram (J K−1 g−1) or Joule per degree Celsius per gram (J C−1 g−1). If the specific heat of a sample is c, then the actual heat capacity is just c times the mass (m) of the sample.

C mc=In other words, we can also write q as:

q mc T= Δ

▼ Example 36

The specific heat of lead is 0.128 J K−1 g−1. What is the heat capacity of a 10.0 g sample of lead? How much heat would raise the temperature of this sample from 25.0°C to 30.0°C.Answer: 1.28 J K−1, 0.64 JThe heat capacity is:

C mc 1 g 128 J K g 128 J K1 1 1= = ( )( ) =0 0 0. . .− − −

The temperature change is:ΔT 3 C 25 C 5 C or 5 Ko o o= =0 0 0 0 0. . . .−

and the heat flow is:

q C T or mc T 128 J K 5 K 64 J1= = ( )( ) =Δ Δ . . .− 0 0

4.5.4.4 Determining Heat CapacityTo determine the heat capacity of a sample, we mix it in a calorimeter with another sample that has a known heat capacity and is at a different temperature. A calorimeter is a thermally-insulated container, one that does not allow heat to flow in or out. Therefore, in a calorimeter, any heat flow out of one sample has to flow into the other sample. If q1 is the heat flow for a sample with unknown heat capacity C1, and q2 is the heat flow for a sample with known heat capacity (C2), then:

q q1 2+ = 0or:

C T C T1 1 2 2Δ Δ+ = 0


The temperature changes, ΔT1 and ΔT2, are measured experimentally, so we can solve for the unknown heat capacity:

CqT

qT

C TT1

1

1

2

1

2 2

1

= =−

=−

Δ ΔΔ

Δ

We can calculate its specific heat by dividing the heat capacity (C1) by its mass (m1):

cCm1

1

1

=

▼ Example 37

A piece of metal at 5.0°C is mixed with 100.0 g of water at 20.0°C in a calorimeter. The final temperature of the mixture is 15.0°C. Calculate the heat capacity of the metal. What additional information would you need to determine the specific heat of the metal?[Note: the specific heat of water is 1.000 cal K−1 g−1, or 4.184 J K−1 g−1.]

Answer: 2.1 × 102 J K−1, mass of metalThe heat capacity of the metal is:

CC T

T12 2

1

=− Δ

Δ

where:C2 is the heat capacity of water

= ( )( ) =1 g 4 184 J K g 418 4 J K1 1 100 0. . .− − −

ΔT2 is the temperature change for the water

= ( ) =15 C 2 C 5 C or 5 Ko o o. . . , .0 0 0 0 0− − −

ΔT1 is the temperature change for the metal

= −( ) =15 C 5 C 1 C or 1 Ko o o. . . , .0 0 0 0 0 0

Therefore,

CC T

TJ K K

KJ

K12 2

1

1418 4 5 010 0

209210 0

= = =− − − − −−Δ

Δ( . )( . )

.( )

.

= =209 2 1 101 2 1J K J K− −. ×

Note that the final answer should only have two significant digits. The −2092 J in the intermediate result is the amount of heat


flow for the water (q1); it is negative since heat flowed out of the water; its temperature decreased. The negative of (−2092 J) is +2092 J, the amount of heat flow for the metal; it is positive since heat flowed into the metal; its temperature increased.

If we divide C1 by the mass of the metal, we can determine its specific heat.

Modern techniques for determining heat capacity use electrical work to change the temperature of the sample. If welec is the amount of energy transferred to the sample and its temperature is found to change by an amount ΔT, then the heat capacity is:

Cw

Telec=

Δ


TEST YOURSELF


1. An object is slowing down as it moves to the +x direction. The acceleration of the object is…

A. positive B. negative C. zero

2. An object is speeding up as it moves in the −x direction. The net force on the object is…

A. positive B. negative C. zero

3. When subjected to the same amount of force, which of these atoms would experience greater acceleration?

A. H B. He

4. When an atom moving in the +x direction collides with a wall and reverses direction, which of the following is true?

A. the acceleration of the atom is negative B. the force on the atom in positive C. the force on the wall is negative

5. At typical atomic distances, the predominant fundamental force between charged particles is…

A. strong B. weak C. electromagnetic D. gravitational

6. Which has greater kinetic energy? A. A He atom moving at 300 m s−1

B. A Ne atom moving at 300 m s−1

C. neither; they have the same kinetic energy

7. As two protons approach each other, their potential energy… A. increases B. decreases C. remains the same

8. The formation of a bond between two atoms involves a conversion of: A. potential energy to kinetic energy B. kinetic energy into potential energy



9. The bond dissociation energy of a diatomic molecule is 4.9 × 10−19 J. If one of these molecules, with a relative kinetic energy of 1.2 × 10−19 J, were to gain 5.3 × 10−19 J of energy from a third body, calculate the relative kinetic energy of the separated atoms. Sketch the molecular potential energy curve.

10. Passing electricity through a wire dipped in water causes the temperature of both the wire and water to increase. The temperature change in the wire is due to _______ and the temperature change in the water is due to _______.

A. heat, heat B. heat, work C. work, heat D. work, work

11. A pot of water at room temperature is placed on a hot stove. If 5000 calories of heat is transferred, q for the pot of water is ______ and q for the stove is ______.

A. +5000 calories, +5000 calories B. −5000 calories, −5000 calories C. +5000 calories, −5000 calories D. −5000 calories, +5000 calories

12. How much heat is needed to raise the temperature of a system from 20.0°C to 25.0°C if the heat capacity of the system is 8.00×102 J K−1?

A. 1.60 × 102 J, B. 2.22 × 105 J C. 4.0 × 103 J

13. What is the specific heat of a substance if a 30.0 g sample has a heat capacity of 3.00 J/K?

A. 0.100 J °C−1 g−1 B. 90.0 J g K−1 C. 10.0 J−1 K g

14. What is the final temperature of an object, initially at 25.0°C, if 5.00 × 103 J of heat flows out of it? Assume that the heat capacity is 2.50 × 103 J K−1.

A. −250.0°C B. −2.00°C C. 23.0°C D. 27.0°C

15. It takes 8.4 × 103 J of heat to raise the temperature of a system from 25.0°C to 35.0°C. What is the heat capacity of the system?

A. 8.4 × 102 J K−1 B. 8.4 × 104 J/°C, C. 2.9 × 101 J/K

16. What is the specific heat of a substance if the temperature of a 35.0 g sample increases from 25.0°C to 27.0°C with the absorption of 14.0 J of heat?

A. 0.400 J g−1 B. 7.0 J K−1 C. 0.200 J K−1 g−1

77

5CHAPTER

Classification of MatterMatter can be classified according to its physical state, or according to its composition. In this chapter, we will examine the atomic interpretation of the differences among the categories.

5.1 Physical States of MatterThe physical state of a sample of matter can be described as solid, liquid, or gaseous. Intuitively, we know the distinctions between the three physical states. For example, we can easily tell that ice, wood and rock are solids, water, wine and gasoline are liquids, and that air, helium (in balloons) and neon (inside neon lamps) are gases.

We can formally define solid, liquid, and gas by describing two of their properties: shape and volume.

• Solids have definite shape and volume.• Liquids have definite volume, but the shape follows the shape of the bottom of the

container.• Gases do not have definite volume or shape. They entirely fill up whatever container

they are put into. They follow the entire shape and volume of the container.

Liquids and gases are also known as fluids; they have an indefinite shape because of their tendency to flow. When we say that solids and liquids have “definite volume,” we mean that it would require a great deal of effort to compress them (compared to gases) and that, unlike gases, they do not just spread out and fill up whatever container we put them in.

5.2 Phase ChangesA change in physical state is also called a phase change. When we heat a solid, it eventually turns into a liquid. This process is called melting or fusion. Further heating of the liquid eventually converts it to a gas. This process is called evaporation or vaporization. Sometimes, a solid can change directly into a gas; this process is called


sublimation. All of these processes are said to be endothermic; heat is absorbed from the surroundings when they occur. The opposite of these processes are exothermic; heat is released to the surroundings as they occur. The opposite of melting is freezing. The opposite of vaporization is condensation or liquefaction. The opposite of sublimation is deposition.

▼ Example 1

What terms are used to describe the following changes?A. a mothball “disappearing”B. ice turning into liquid waterC. water droplets forming on the outside of a cold glass of

waterD. ice “shrinking” in a frost-free freezer E. sweat “disappearing” from your skin

Answers:A. sublimation B. melting C. condensation D. sublimation E. vaporization

▼ Example 2

Explain why you feel cold when you come in drenched from the rain?Answer: Vaporization is an endothermic process. As the water vaporizes, water molecules draw heat from your body.

▼ Example 3

Explain why exposure to steam is more dangerous than exposure to boiling water?Answer: As boiling water cools (when it comes in contact with your body), heat is transferred to your body. With steam, more heat is transferred to your body because the condensa-tion of steam to boiling water is exothermic. In fact, it can be shown that the condensation alone generates nine times more heat.

Chapter 5: Classification of Matter 79

5.3 Atomic Interpretation of Physical States and Phase Changes

All matter is made up of tiny particles called atoms, ions, and molecules. Differences in the characteristics of physical states can be explained with the notion that these particles:

• are always moving,• move more or less independently of each other at very, very large distances,• repel each other at very, very short distances (mainly due to repulsions of positively

charged nuclei),• attract each other at intermediate distances,• gain kinetic energy (move faster) when heated, and• are better able to move away from each other when they gain kinetic energy.

When we talk about distances between atoms being large or short, we are referring to distances compared to the size of the atoms. A distance of 0.1 to 1.0 nm would be considered “intermediate,” with the strongest attraction occurring at a distance of 0.2–0.5 nm between neighboring nuclei. At shorter distances (less than about 0.1 nm) repulsion between nuclei (the internuclear repulsion) rises very sharply. A distance of 3 nm would be considered “very large”; 1 nm is 10−9 m.

Explaining shape. Unlike liquids and gases, solids have definite shape because the particles that make up the solid do not have much freedom of movement. The amount of energy they have is just enough for them to vibrate about more or less fixed positions. As solids are heated, the particles gain energy. The increase in energy allows them to move faster and increase their range of motion as they are better able to overcome the attractive forces among them. The motion eventually becomes less restricted as the solid turns into a liquid, and eventually into a gas. The motion is least restricted among particles in the gas phase. They are able to spread out as far from one another as possible. Thus, a gas fills up whatever container you put it in.

Explaining compressibility. In solids and liquids, the particles are already as close to each other as possible (0.1–0.2 nm); pushing them even closer together will require a great amount of force to overcome the strong internuclear repulsions that would occur at the short distances. In the case of gases, the particles are very far apart. They can be pushed closer together without much resistance from the internuclear repulsions.


▼ Example 4

A 24.8 L sample of air at room temperature contains approxi-mately 6.0 × 1023 molecules. Estimate the average distance between molecules.Answer: 3.5 nmTo calculate the average distance, imagine spreading out the molecules so that they are evenly spaced. We do this by imagining a 24.8L box and dividing it into 6.0 × 1023 cubic compartments, with one molecule in each compartment. Thus, the volume for each compartment is

Vm

m= ××

= ×24 8 106 0 10

4 13 103 3

2326 3.

..

−−

Note that 24.8 L is equivalent to 24.8 × 10−3 m3. The volume of a cube is equal to a3, if a is the length of the side. So, to get the length of the side of each cubic compartment, we take the cube root of the volume.

a V m m= = ×( ) = ×− −13 26 3

13 94 13 10 3 5 10. .

or 3.5 nm. This is the center-to-center distance between adjacent cubes and our estimated average distance between the nuclei of neighboring air molecules. For comparison, if we imagine an N2 molecule, which comprises about 80% of air molecules, as a sphere, it would have a radius of about 0.16 nm. This is about 20 times smaller than 3.5 nm.

When a substance is made up of molecules, heating will generally first lead to the spreading out of the molecules. A much higher temperature would be required to separate the molecules into atoms. Typical attractions between molecules are weaker than attrac-tions between neighboring atoms within a molecule. For example, the energy needed to separate an H atom from a CH4 molecule is about 330 times larger than the energy needed to separate two CH4 molecules from each other.


▼ Example 5

In the figure below, which process best represents the vaporiza-tion of liquid nitrogen (N2)?

Answer: B.The process illustrated in B occurs at 77K (or −196°C); at room temperature nitrogen exists as a gas and is made up of N2 mol-ecules. In order to break up the N2 molecules into N atoms, we would need to heat our sample up to about 70000K.

5.4 Elements, Compounds and MixturesSamples of matter that we encounter in nature can generally be separated into two or more distinct components. We consider two components to be distinct if they have different sets of properties. However, once we start separating matter into its components, we reach a point where we have a sample that can no longer be separated into distinct components. We say that what we have at that point is an element. So far, a little over a hundred different elements have been identified; less than 100 are naturally-occurring.

If there are only a few elements, why is there so much variety in nature? The answer is that elements can combine in many, many different ways to form compounds. A com-pound is made up of two or more elements, but its properties do not resemble those of the elements that make it up. Each compound has a unique set of properties that differentiates it from other compounds. Collectively, we refer to elements and compounds as substances. Much of the matter we find in nature is a mixture of substances. A sample of matter con-taining only one substance is called pure substance. Pure substances are rarely found in nature; they are produced in the laboratory by separating mixtures.


▼ Example 6

Under ordinary conditions, hydrogen and oxygen are gases; a mixture of hydrogen and oxygen is also a gas. Water, on the other hand, is a compound of hydrogen and oxygen; under ordinary con-ditions, it is a liquid. Hydrogen peroxide is another compound of hydrogen and oxygen; its set of properties is different compared to water, hydrogen, or oxygen.

How can we tell if a sample is a mixture or a pure substance? A pure substance is usu-ally homogeneous; by this, we mean that the properties are uniform throughout the entire sample. The only exceptions are at certain conditions where it is possible for more than one physical state of the substance to exist. At the melting point of a substance, for exam-ple, you will see both solid and liquid state; in this case, you have a heterogeneous sample of the pure substance.

Most heterogeneous samples are mixtures, but a mixture can also be homogeneous; such a mixture is called a solution. How can we tell if a homogeneous sample of matter is a mixture or a pure substance? The traditional method is to examine the behavior of the system when it is heated or cooled. While a pure substance is melting, its tempera-ture remains constant; in practice we say it is constant if it does not change by more than 1–2°C. We say that a pure substance has a sharp melting point (or freezing point), while mixtures have a wide melting (or freezing) range. Similarly, a pure substance has a sharp boiling point, while mixtures have a wide boiling range.

Heterogeneous mixtures can be classified as coarse mixtures, suspensions, and col-loids. Coarse mixtures are the most commonly encountered and are relatively easy to recognize; an example of a coarse mixture is a salad. In a suspension, very fine particles are evenly dispersed in a liquid or gas; examples: dust and air, orange juice with pulp. The dispersed particles in a suspension are visible to the naked eye and settle on standing. Colloids or colloidal dispersions are similar to suspensions; very fine particles are also evenly dispersed but they are too small for the unaided eye to actually discern; examples: fog, milk, lump-free gravy. The dispersed particles in a colloid are detectable by shining a bright light on the sample; the scattering of light by colloidal particles is called the Tyndall effect. Unlike in suspensions, the dispersed particles in a colloid do not settle on standing.

5.5 The Law of Definite CompositionThe Law of Definite Composition, also known as the Law of Constant Composition or the Law of Definite Proportions, is a statement of the general observation that if a pure sub-stance can be broken into two or more elements, the proportion of the elements obtained is


well defined, regardless of how much substance was used or where the substance came from. Similarly, if the process were reversed, the same proportion of elements would be used up to make the substance. A substance that can be broken into two or more elements is called a compound. Therefore, the law essentially says that a compound has a definite composition.

▼ Example 7

Example: the percentage of Cl by mass in 10.0 g of a compound is 25.0%. What is the percentage of Cl by mass in 20.0 g of this compound?Answer: 25.0%It does not matter how much of the compound we have. The percentage of any element in the sample is a constant.

▼ Example 8

A 16-g sample of a compound contains 12 g C and 4 g H. What is the mass of a sample of this compound that contains 24 g C? What is the mass of H in this sample?Answer: 32 g, 8 gThe law of definite composition says that, since:

massmass

gg

C

total

= =1216

0 75.

for the 16 g sample, then the ratio is also 0.75 for any sample of this compound. In other words, for the other sample:

massmass

gmass

C

total total

= =240 75.

Solving for masstotal, we get 32 g. Note that 12 is 75% of 16, and 24 is 75% of 32. The percentage of C in both samples is the same.We can figure out the mass of H by subtracting the mass of C from the total mass:

32 g − 24 g = 8 g.Alternatively, we can use the fact that the mass-to-mass ratio of C to H is constant. For the 16 g sample, the ratio is:

massmass

gg

C

H

= =124

3


Therefore, for the sample that contains 24 g C, the ratio is also equal to 3.

massmass

gmass

C

H H

= =243

Solving for massH, we get 8 g.

5.6 Atomic Interpretation of Elements, Compounds, and Mixtures

A pure sample of an element is made up of only one kind of atom. Separating it into two or more parts will give us parts that are made up of the same kind of atoms. Thus, the parts will have the same properties.

Compounds and mixtures are made up of at least two elements; therefore, they are made up of at least two kinds of atoms. Unlike a mixture, a compound has a definite composition because the atoms of the different elements in any sample of the compound are in a well-defined (fixed) ratio.

▼ Example 9

Which of the figures below best illustrate the atomic composition of a pure element, a pure compound, and a mixture?

Answer: Figure A best represents a sample of a pure element;it shows only one kind of atom (red circles). Figure C best repre-sents a sample of a pure compound; it shows two kinds of atoms, but the atoms are organized in a well-defined manner; it is made up of only one kind of molecule. Any “sample” you draw from Figure C will give you one blue square for every red circle. Figure B best represents a mixture of two elements. If you were to take a random “sample” of atoms from Figure B, you would not always get the same blue-to-red ratio.


▼ Example 10

Consider the pictorial representation of the molecular composi-tion of a mixture. Assuming filled circles represent C atoms and unfilled squares represent O atoms, we can say that the figure represents a mixture of: A. an element and two compoundsB. two elements and a compoundC. three compoundsD. two elements

Answer: AThere are three substances in the mixture. One element (O2) and two compounds (CO and CO2).


TEST YOURSELF


1. For which of the following are the atoms, on average, farthest from one another? A. air B. gasoline C. copper wire

2. Water is made up of molecules. For which physical state is the motion of the water molecules most restricted?

A. solid (ice) B. liquid C. gas (steam)

3. What physical state of matter is being described by the following phrase: “particles are as close to one another as possible and unable to move”

A. solid B. liquid C. gas D. none of these

4. Consider the pictorial representation shown below for a change in physical state at the molecular level.

The picture best represents which of these changes: A. evaporation B. melting C. sublimation D. condensation

5. Which of the pictorial representations best accounts for the conversion of water from liquid to steam?



6. What type of matter cannot be separated into two or more distinct components? A. element B. compound C. mixture D. pure substance

7. A naturally-occurring sample of matter most likely consists of… A. a single element B. a single compound C. a mixture D. pure substance

8. A sample of matter is heated and found to start melting at 75.0°C. While melting, the sample temperature continues to rise to about 85.0°C. This sample of matter is most likely…

A. an element B. a compound C. a mixture

9. Samples A, B, and C appear to be identical. Each sample is found to melt sharply at 122°C. However, a mixture of sample A and B is found to melt from 115–120°C, while a mixture of sample A and C is found to melt sharply at 122°C. Which of the following is true?

A. Samples A and B are composed of the same substance B. Samples A and C are composed of the different substances C. A mixture of B and C will have a sharp melting point D. None of the above

10. If a medicine bottle has a “shake well before using” label, it most likely contains…… A. a solution B. a colloid C. a suspension

11. Which of the following is a coarse mixture? A. air B. 18K gold C. gasoline D. salad

12. A 25.0 gram sample of a compound is found to be 24% chlorine by mass. What is the percentage of chlorine in a 50.0 g sample of the compound?

A. 24% B. 48% C. 12% D. It depends where the sample came from

13. A 25.0 gram sample of a compound is found to be 24% chlorine by mass. How much chlorine is in a 50.0 g sample of the compound?

A. 24 g B. 48 g C. 12 g


14. A 17.0 g sample of a compound contains 14.0 g nitrogen and 3.0 g hydrogen. How much hydrogen does a 34.0 g sample of the compound contain?

A. 3.0 g B. 6.0 g C. 14.0 g D. 28.0 g

15. A 32.0 g sample of a compound contains 16.0 g oxygen, 12.0 g carbon, and 4.0 g hydrogen. How much oxygen is present in a sample of this compound that contains 36.0 g carbon?

A. 12.0 g B. 5.33 g C. 48.0 g D. 36.0 g

16. A 7.0 g sample of a compound contains 3.0 g of element “X” and 4.0 g of element “Y”. What is the mass ratio of element “X” to element “Y” in a 70.0 g sample of this compound?

A. 0.75 B. 1.33 C. 7.5 D. 13.3

17. Consider the pictorial representation shown below for the molecular composition of a mixture. The figure represents a mixture of…

A. an element and two compounds B. two elements and a compound C. three compounds

89

6CHAPTER

The Periodic Table

The periodic table is the most important resource for anyone studying Chemistry. The elements are arranged in the periodic table in a manner that facilitates understanding of the patterns in their properties.

6.1 Classifying the ElementsIn a typical periodic table, elements are organized into 18 vertical columns with two additional (horizontal) rows below the 18 columns. Elements that belong to same vertical column are said to belong to the same family or group. Elements in the same horizontal row are said to belong to the same period.

▼ Example 1

Which of the following belongs to period 3: Na, Sc, or In?Answer: NaThe term Period refers to a (horizontal) row. Period 3, the third horizontal row, is highlighted in green below. Na is the first element in period 3. Sc is in the fourth period. In is in the fifth period.


The representative elements. In the traditional American periodic table, columns 1, 2, and 13 through 18 are labeled using the numbers 1 through 8 (Arabic or Roman) fol-lowed by the letter A. In other words, column 1 is group IA or 1A, column 2 is group IIA or 2A, column 13 is group IIIA, column 14 is group IVA, etc.. Sometimes, column 18 or group VIIIA is labeled as group 0 (zero). These “A” groups are known as the main groups or representative groups, and they are indicated in Figure 1 by the red shading.

In the traditional American periodic table, columns 3 through 12 are labeled using IIIB (or 3B), IVB (or 4B), VB, VIB, VIIB, VIII (or VIIIB, or 8B; next 3 columns), IB, and IIB. The elements in these “B” groups are called the transition elements or transition metals and they are indicated in Figure 1 by the blue shading.

The two rows at the bottom are known as the inner transition elements (lanthanides and actinides); they are indicated in Figure 1 by the yellow shading. In the extended- or long-form of the periodic table, these rows are inserted between group IIA and group IIIB in rows 6 and 7. In other words, the lanthanides actually belong to period 6 (or row 6) and the actinides belong to period 7.

The representative elements Ga, In, Sn, Tl, Pb, and Bi are also known as the post-transition metals. These are shown in green in Figure 2.

Some groups have special names; see Figure 3. Elements in group IA, except hydrogen, are called the alkali metals. Elements in group IIB are called the alkaline earth metals. The transition elements are also known as the transition metals. Elements in group VIIIA (column 18) are called the noble gases. Elements in group VIIA (column 17) are known as the halogens. Elements in group VIA (column 16) are known as the chalcogens.

Figure 1. Representative, Transition and Inner Transition Elements

Chapter 6: The Periodic Table 91

Figure 2. The Post-transition Metals

Figure 3. Groups with Special Names: Alkali Metals (1A), Alkaline Earth Metals (2A), Chalcogens (6A), Halogens (7A), and Noble Gases (8A)


6.2 Properties of ElementsPatterns in properties of elements can be correlated with their location in the periodic table.

6.2.1 Metallic CharacterElements on the left side of the periodic table are observed to be metallic in pure form; see Figure 4. Metals are shiny, malleable, ductile, good electrical conductors, and good heat conductors. Something that is malleable can be hammered into thin sheets. Something that is ductile can be drawn out into a thin wire. Elements on the right side of the period are classified as nonmetals. Elements in between (B, Si, Ge, As, Sb, Te and Po) are known as the metalloids.

▼ Example 2

Which element is a good electrical conductor, Cu, Ge, Kr?Answer: Cu Cu is a metal. Ge is a metalloid. Kr is a nonmetal. Metals are the good electrical conductors.

Figure 4. Metals, Nonmetals and Metalloids

6.2.2 Physical StatesMost naturally-occurring elements are solids under ordinary conditions. The exceptions (see Figure 5):

• Liquids: mercury and bromine. On a warm summer day, outside an air-conditioned laboratory, the following elements may also be found as


liquids: gallium (melts at 29.76°C or 85.57°F) and caesium (melts at 28.44°C or 83.19°F).

• Gases: hydrogen, nitrogen, oxygen, fluorine, chlorine, and the noble gases. Except for hydrogen, we find these elements on the rightmost side of the periodic table.

Figure 5. Physical States of Elements

6.2.3 Tendency to Form Monatomic IonsThe charges of naturally-occurring monatomic ions can be correlated with the periodic table. The correlation is summarized in Figure 6. Ions derived from atoms of metallic elements tend to be positively charged. Ions derived from atoms of nonmetallic elements tend to be negatively charged.

Monatomic ions derived from elements belonging to group IA except H are always found in nature to have a charge of +1. Monatomic ions from Group IIA elements are always found in nature to have a charge of +2. The monatomic ion formed from alumi-num is always found in nature to have a charge of +3. For transition and post transition metals, there are typically more than one type of cation found in nature. The most com-mon charge of monatomic ions derived from transition metals is +2. Monatomic ions derived from atoms of zinc, cadmium, and mercury are always found in nature with a charge of +2.

Monatomic ions derived from nonmetallic elements in group VIIA (column 17, the halogens) are called halides and are always found to have a charge of −1. Monatomic ions derived from elements of group VIA (column 16) are always found to have a charge of −2. For group VA (column 15), the observed charge is always −3.


Monatomic ions derived from hydrogen are always observed to have a charge of either +1 or −1.

What about the other elements? The noble gases (group VIIIA), metalloids and nearby nonmetals tend not to form monatomic ions.

Monatomic cations share the same name as the atoms they are derived from. For example, K+, the ion derived from a potassium atom, is called potassium ion. However, names of monatomic cations derived transition and post-transition metals must specify the charge of the ion in Roman numeral, as in cobalt(III) for Co3+. Note that there is no space between the last letter in cobalt and the opening parenthesis. For monatomic anions, the name differs from that of the parent atom; the name of the anion ends in ide. For example, O, the neutral atom is oxygen, but the negatively charged O2− is oxide.

▼ Example 3

Which of the following is called magnesium ion?A. Mg+ B. Mg− C. Mg2+ D. Mg3+

Answer: CAll the choices are ions of magnesium, but the only one that we call magnesium ion is Mg2+ because it is the only one that we can find in nature. The other ions can be made in laboratory, but would be very short-lived.

Figure 6. Naturally-occurring Monatomic Ions


▼ Example 4

What is the formula of lead(IV) ion?Answer: Pb4+

IV is the Roman numeral for 4. The symbol for lead is Pb.

▼ Example 5

What is the formula of nitride ion?Answer: N3−

The ide ending in name suggests that the ion is negatively charged. The nitr at the beginning of the name suggests that this ion is derived from a nitrogen atom (N). The only naturally-occurring monatomic ion derived from a nitrogen atom has a charge of −3.

6.2.4 Covalent BondingWhen two atoms share electrons, they are said to be held together by a covalent bond. In contrast, ions with opposite charges are attracted to each other and are said to be held together by an ionic bond.

Since atoms of metallic elements tend to form cations and atoms of nonmetallic elements tend to form anions, we can say that ionic bonding is likely to occur between a metal and a nonmetal. Between two atoms of nonmetallic elements, electron sharing or covalent bonding is more likely to occur. Indeed, the pure forms of some nonmetallic elements are found in nature as collections of molecules. For example, oxygen can exist as a collection of O2 molecules or O3 molecules (ozone). O2 is the form of oxygen that makes up about 20% of the air that we breathe. The elements that, in pure form, naturally occur as a collection of molecules under ordinary conditions are listed below:

• Diatomic: H2, N2, F2, Cl2, Br2, I2• Polyatomic: P4, S8

When two atoms are sharing electrons, the sharing is generally unequal; this means that the electrons, on average, spend more time closer to one of the two atoms. We say that the electrons are polarized towards one of the atoms, that the bonding is polar covalent. Equal sharing, pure covalent bonding, happens only between atoms of the same element. Ionic bonding can be thought of as an extreme case of polar covalent bonding. It is best to say that the bond between a metal and nonmetal is likely (not definitely) ionic. In the periodic table, metallic and nonmetallic characteristics increase as you move away from


the metalloids. Thus, ionic bonding is more likely to happen between a metal on the far bottom left corner of the periodic table and a nonmetal on the far top right side (exclud-ing the noble gases); the most ionic bond would occur between Fr and F. We tend to get into the “gray areas” when dealing with metal-nonmetal pairs closer to the metalloid line. In fact, a transition metal atom can form a covalent bond with a nonmetal or another transition metal atom; an entire subfield of chemistry (coordination chemistry) deals with this topic.

Three common polyatomic ions where a metal atom shares electrons with nonmetal atoms are permanganate (MnO4

−), chromate (CrO42−), and dichromate (Cr2O7

2−).When mixed with water, transition and post-transition metal ions are, in fact, sharing

electrons with water molecules. Copper(II) ions in water, for example, are more accurately represented by the formula Cu(H2O)4

2+ rather than Cu2+. The Cu2+ ion is surrounded by four H2O molecules such that electrons are shared between the O (of the H2O mol-ecules) and Cu. The sharing of electrons in this case is called coordinate covalent bonding. Cu(H2O)4

2+ is an example of what is called a complex ion.

6.2.5 Metallic BondingBetween a pair of metal atoms, covalent bonding is more likely to occur. However, among a large number of metal atoms, electrons tend to be very loosely held and it is convenient to visualize the nuclei as being immersed in a “sea” of electrons; the bonding among the atoms in this case is called metallic bonding. This explains why metals are good electrical conductors.

6.2.6 Noble GasesThe noble gases are found in nature as collections of neutral atoms. Under extreme labora-tory conditions, atoms of noble gases can be made to form molecules, but they have no natural tendency to form molecules or ions.


TEST YOURSELF


1. Which of the following elements is a transition metal? A. silicon B. lead C. sodium D. chromium

2. Which of the following elements is a post-transition metal? A. silicon B. lead C. sodium D. chromium

3. Which of the following elements is a metal that belongs to a main group or representative group?

A. magnesium B. zinc C. sulfur D. uranium

4. All of the following are true, except… A. Sodium is an alkali metal B. Neon is a noble gas C. Calcium is an alkaline earth metal D. Oxygen is a halogen

5. Which of the following is a metalloid? A. potassium B. silicon C. iron D. helium

6. Which of the following elements is the best electrical conductor? A. carbon B. germanium C. manganese D. iodine

7. Which of the following naturally-occurring ions is paired with its correct symbol? A. potassium ion, K2+ B. sulfide, S3−

C. barium ion, Ba2+ D. iodide, I+

8. Which of the following pairs of atoms is most likely to form an ionic bond? A. C and H B. Ca and H C. Hg and Hg

9. What type of bond is more likely between a carbon atom and an oxygen atom? A. ionic B. covalent C. metallic



10. Which of these naturally occurs as a monatomic gas under ordinary conditions? A. helium B. sodium C. oxygen D. nitrogen

11. Which of these elements naturally occurs as a diatomic gas under ordinary conditions?

A. bromine B. iodine C. mercury D. chlorine

12. Which of these elements is not a gas under ordinary conditions? A. bromine B. helium C. hydrogen D. oxygen

13. What type of bond holds the atoms together in the polyatomic ions chromate (CrO4

2−) and nitrate (NO3−)?

A. ionic, ionic B. ionic, covalent C. covalent, ionic D. covalent, covalent

99

7CHAPTER

Compounds

In this chapter, we will learn how to interpret names and formulas of compounds.

7.1 Classifying Common CompoundsCompounds can be classified as either ionic or molecular. Ionic compounds are made up of ions held together by ionic bonds. Molecular compounds are made up of molecules. Within molecules, atoms sharing electrons are said to be covalently-bonded. Note that electron sharing (or covalent bonding) can also be present in ionic compounds if at least one of ions is polyatomic; in a polyatomic ion, the atoms are held together by covalent bonds. Ionic bond refers to the attraction between neighboring ions of opposite charges.

Shown in Figure 1 is a representation of how the ions are arranged in a regular pattern in barium chloride (BaCl2); this regular arrangement is called a crystal lattice. The green spheres represent chloride ions (Cl−) and the orange spheres represent barium ions (Ba2+). This figure clearly illustrates why we say that there are no BaCl2 “molecules”; we cannot tell which Cl− ions go with which Ba2+ ion. All we know is that the count-to-count ratio

Figure 1. Crystal Lattice of BaCl2

from http://www.molsci.ucla.edu/pub/explorations.html#Crystalline Solids.

http://www.molsci.ucla.edu/pub/explorations.html#Crystalline


7.1.1 Classifying Compounds Based on Name or FormulaMost compounds can be easily classified as ionic or molecular based on the name or the formula. Typically:

• names of ionic compounds either start with the name of a metallic element, or with ammonium.

• formulas of ionic compounds either start with the symbol of a metallic element, or with “NH4” (which may be enclosed in parentheses); the cation in these cases is the polyatomic ion called ammonium, NH4

+.

▼ Example 1

Classify the following compounds as ionic or molecular: sodium sulfide, carbon disulfide, ammonium sulfideAnswer: ionic, molecular, ionic• Sodium is a metallic element. The cation in this compound is

sodium ion (Na+).• Carbon is a nonmetallic element.• Ammonium sulfide is ionic; the cation is ammonium (NH4

+).

Figure 2. Crystal Lattice of CO

from http://cst-www.nrl.navy.mil/lattice/struk/b21.html (public domain)

of barium-to-chloride ions is 1-to-2. So, we refer to one barium and two chloride ions together as one formula unit; not one molecule.

Solid samples of molecular substances can also be crystalline, as illustrated in Figure 2 for carbon monoxide, but we can clearly distinguish which atoms “go together” as part of one molecule.

http://cst-www.nrl.navy.mil/lattice/struk/b21.html

Chapter 7: Compounds 101

▼ Example 2

Classify the following as ionic or molecular: Mg(OH)2, CH3OH, NH4ClAnswer: Mg is the symbol for magnesium, a metallic element. Therefore, Mg(OH)2 is an ionic compound; the cation is mag-nesium ion (Mg2+). C is the symbol for carbon, a nonmetallic element. Therefore, CH3OH is a molecular compound. N is a nonmetallic element but it is part of an “NH4” grouping in NH4Cl; this compound is an ionic compound; the cation is ammonium ion (NH4

+).

7.1.2 Classifying Compounds Based on PropertiesIonic compounds typically have very high melting points. This means that most ionic compounds are solids at room temperature. If a compound is a liquid or gas at room temperature, it is most likely to be molecular. However, if it is a solid, it could be ionic or molecular. For molecular compounds, in general, the larger the molecules, the higher the melting point, and the higher the boiling point.

▼ Example 3

Which of these compounds is most likely to be gaseous at room temperature: NaCl, CH4, or C8H18?Answer: CH4,NaCl is an ionic compound; we expect it to have a very high melt-ing point and it does. CH4 and C8H18 are molecular compounds. The smaller the molecules of a compound, the more likely we will find the substance as a gas at room temperature. CH4, meth-ane, is the primary component of natural gas. It is a gas at room temperature. C8H18 is octane, a component of gasoline; it is a liquid at room temperature. When something is a gas at room temperature, it means that its boiling point is much colder than room temperature.

Another way we can tell if a compound is ionic is if it conducts electricity when melted.


▼ Example 4

Which of these is not a good electrical conductor?A. liquid NaCl, B. solid Cu, C. solid NaNO3

Answer: solid NaNO3NaCl is an ionic compound; we expect ionic compounds to be good electrical conductors if melted. Cu is a metallic element; metals are good electrical conductors. NaNO3 is an ionic compound; it would be a good electrical conductor if melted, but not in solid form.

Ionic compounds are strong electrolytes. Pure water is a very poor conductor of electric-ity, but dissolving an ionic compound in water produces a mixture that is a good electrical conductor.

Molecular compounds are, in general, nonelectrolytes. However, there are molecular compounds that once dissolved in water, can break up into ions, or react with water and pro-duce ions in the process; this process is called ionization. Most of these compounds are weak electrolytes because the production of ions occurs only to a small extent. In fact, there are only six known molecular compounds that are completely ionized in water and are, therefore, strong electrolytes; these are the six strong acids: HCl (hydrochloric acid), HBr (hydrobromic acid), HI (hydriodic acid), HNO3 (nitric acid), H2SO4 (sulfuric acid), and HClO4 (perchloric acid). Acid molecules break up in water producing hydrogen ions (H+). We say that the acid is ionized or dissociated in water. Except for the six strong acids, acids are weak electrolytes.

▼ Example 5

What are the ions produced when hydrochloric acid (HCl) is dissolved in water?Answer: H+ and Cl−. When a hydrogen ion (which has a +1 charge) breaks off from an HCl molecule (which is neutral), what is left behind must have a −1 charge.

▼ Example 6

Is hydrofluoric acid (HF) a strong or a weak electrolyte?Answer: weak. HF is not one of the six known strong acids. Most of the HF molecules will remain intact when HF is dissolved in water; a very small fraction of the HF molecules will be dissociated into H+ and F− ions.


7.2 Formula UnitsThe formula of a compound tells us the composition of one formula unit of the com-pound. For molecular compounds, the formula unit is the molecule. For ionic compounds the formula unit is the smallest collection of ions with a total charge of zero.

▼ Example 7

How many H atoms are in one formula unit of propanoic acid (CH3CH2COOH)?Answer: 6The subscripts of hydrogen in CH3CH2COOH are 3, 2, and 1. Note that unwritten subscripts are implied to be 1. Adding up the sub-scripts, we get 3 + 2 + 1 = 6. Therefore, there are 6 hydrogen atoms in one formula unit (or one molecule) of propanoic acid.

▼ Example 8

Explain why the formula of a compound made up of calcium ions (Ca2+) and nitride ions (N3−) is written as Ca3N2?Answer: One formula unit of an ionic compound must have a total charge of zero. The simplest set of calcium (Ca2+) and nitride (N3−) ions with a total charge of zero is: three calcium ions and two nitride ions:Charge of three calcium ions: (3)(+2) = +6Charge of two nitride ions: (2)(−3) = −6Total charge = 0

Therefore, the formula of calcium nitride is Ca3N2. Note that charges of the ions are not indicated when writing the formula of an ionic compound.

7.3 Names and Formulas of Common IonsThe first step in reading or writing formulas of common compounds is to classify the com-pound as either ionic or molecular. Once we recognize the formula to be that of an ionic compound, reading it is just a matter of naming the cation, then the anion. It is, therefore, important to be familiar with the names and formulas of common ions.

7.3.1 Monatomic IonsNames and formulas of naturally-occurring monatomic ions are discussed in section 6.2.3.


7.3.2 Polyatomic IonsNames of polyatomic ions are trivial. By trivial, we mean that we just have to know them; there is nothing to figure out. It is like saying that a person’s name is, say, John. Table 1 lists the polyatomic ions that are commonly encountered in compounds.

Table 1. Common Polyatomic IonsPolyatomic cations:

• mercury(I) or mercurous: Hg22+

• ammonium: NH4+

Polyatomic anions with names ending in ate:borate: BO3

3−

carbonate: CO32−

bicarbonate or hydrogen carbonate: HCO3−

nitrate: NO3−

chlorate: ClO3−

bromate: BrO3−

iodate: IO3−

phosphate: PO43−

hydrogen phosphate: HPO42−

dihydrogen phosphate: H2PO4−

arsenate: AsO43−

selenate: SeO42−

sulfate: SO42−

bisulfate or hydrogen sulfate: HSO4−

perchlorate: ClO4−

perbromate: BrO4−

periodate: IO4−

permanganate: MnO4−

chromate: CrO42−

dichromate: Cr2O72−

molybdate: MoO42−

acetate: C2H3O2− or CH3COO−

persulfate (or peroxydisulfate): S2O82−

thiosulfate: S2O32−

dithionate: S2O62−

cyanate: OCN−

thiocyanate: SCN−

Polyatomic anions with names ending in ite have one less oxygen that those ending in ate, and have the same charge.

Example: nitrate is NO3−, nitrite is NO2

−

Exception: dithionate is S2O62−, dithionite is S2O4

2−

Polyatomic ions with names ending in ite but beginning with hypo have one less O and the same charge compared to those without the hypo prefix. Example: chlorite is ClO2

− and hypochlorite is ClO−.Polyatomic anions with names ending in ide.

hydroxide: OH−

cyanide: CN−

peroxide: O22−

superoxide: O2−

carbide: C22−

persulfide: S22−

azide: N3− (note: this is different from nitride, N3−, which is monatomic)


7.4 Names and Formulas of Binary Compounds

Binary compounds are compounds made up of only two elements.

▼ Example 9

Which of the following is a binary compound? CH3OH, O2, HClAnswer: Among the substances listed, only HCl is made up of two elements: hydrogen and chlorine. CH3OH is a ternary compound; it’s made up of three elements (C, H, and O). O2 is not even a compound.

It is easy to recognize names of binary compounds. It consists of two words. We should see the name of an element in the first word, and the stem of the name of an element in the second word. The second word should end in ide.

▼ Example 10

Which of the following are binary compounds? ethyl acetate, ammonium chloride, sodium phosphate, magnesium chloride, carbon tetrachloride, disulfur difluorideAnswer: Among the compounds listed, only the last three (magnesium chloride, carbon tetrachloride, and disulfur difluoride) are binary. Note that there may be a prefix (such as di and tetra) preceding the name of the element.

7.4.1 Reading Formulas of Binary CompoundsGiven the formula of a binary compound, how do we determine its name? First, we need to classify the compound as ionic or molecular. We can tell it is ionic if the first element is metallic. Next, we apply the following rules:

For binary, ionic compounds, the name consists of two words. No prefixes are used regardless of the subscripts.

• First word: name of the cation, which is just the name of the metallic element. If the metallic element is a transition or post-transition metal, specify its charge in Roman numerals inside parentheses.


• Second word: name of the anion, which is just the stem of the name of the nonmetallic element, followed by ide.

Exceptions:• If the second element is N and its subscript is 3, the anion is azide (N3

−), which is a polyatomic ion with a charge of −1 instead of nitride (N3−), which is monatomic with a charge of −3.

• If the second element is O and its subscript is 2, the anion could be two oxide ions (O2−), or one peroxide ion (O2

2−), or one superoxide ion (O2−). Figure out which

one using the rule that the formula unit must be neutral.• If the second element is S and its subscript is 2, the anion could be two sulfide ions

(S2−), or one persulfide ion (S22−). Figure out which one using the rule that the

formula unit must be neutral.

▼ Example 11

What are the cation and anion in FeS? What is the name of the compound?Answer: Fe2+ is the cation, S2− is the anion. The name is iron(II) sulfide.We know the compound is ionic because Fe is a metal. We know the cation is monatomic and derived from Fe. Since Fe is a transi-tion metal, we have to figure out its charge based on the charge and subscript of the anion.S is sulfur and the only naturally-occurring monatomic ion from sulfur has a charge of −2 and is called sulfide; the anion is S2−.We figure out that the charge of iron here is +2, because the total charge of the ions must add up to zero. So the name of the cation is iron(II).

▼ Example 12

What are the cation and anion in Na2O2? What is the name of the compound?Answer: Na+ is the cation, O2

2− is the anion. The name is sodium peroxide.We know the compound is ionic because Na is a metal. The only monatomic ion derived from Na, which belongs to group IA, has a charge of +1. Therefore, the cation is Na+, and there are two of this in Na2O2. The total positive charge is 2(+1), or +2.


We need to figure out if the “O2” part of Na2O2 represents one polyatomic ion or two monatomic ions. Based on Table 1, the pos-sible polyatomic ions are:peroxide (O2

2−) and superoxide (O2−)

The only possible monatomic ion is oxide, the only known naturally-occurring monatomic ion from oxygen is O2− (oxide). One peroxide has total charge of −2. One superoxide has a total charge of −1. Two oxides have a total charge of 2(−2), or −4.Since the total positive charge is +2, the total negative charge must be −2. Therefore, the anion must be peroxide.One formula unit of this compound consists of two Na+ ions and one O2

2− ion.

For binary, molecular compounds, the name consists of two words.• First word consists of: prefix corresponding to the subscript of the first element,

then the name of the first element.• Second word consists of: prefix corresponding to the subscript of the second ele-

ment, then the stem of the name of the second element, then ide.

Prefixes: 2=di, 3=tri, 4=tetra, 5=penta, 6=hexa, 7=hepta, 8=octa, 9=nona, 10=deca. A subscript of one is not written in formulas; no prefix is used when the subscript is one.

▼ Example 13

P4O10 is a binary molecular compound. What is its name?Answer: tetraphosphorus decoxide. Note: the “a” is dropped from a prefix ending in “a” if it is followed by “o”. Here, we used “dec” instead of “deca.”

7.4.2 Writing Formulas of Binary CompoundsGiven the name of a binary compound, how do we determine its formula? First, we need to determine if it is ionic or molecular. Then, we apply the following rules:

For a Binary Molecular Compound:• write the symbols of the elements mentioned in the name, then• write subscripts corresponding to their prefix.


▼ Example 14

Carbon tetrachloride is a binary molecular compound. What is its formula?Answer: CCl4.The subscript for carbon is implied to be 1 (no prefix). The sub-script for chlorine is 4 since tetra means 4.

For a Binary Ionic Compound:• based on the charges of the cation and anion, figure out the smallest possible sub-

scripts so that (cation subscript)(cation charge) + (anion subscript)(anion charge) = 0,

then• write the symbol for each element and indicate the appropriate subscript.

▼ Example 15

Iron(III) sulfide is an ionic compound. What is its formula?Answer: iron(III) is Fe3+. Sulfide is S2−. So, one formula unit of this compound must have 2 cations and 3 anions. 2(+3) + 3(−2) = 0. The formula is: Fe2S3.

7.4.3 Binary Compounds with Trivial NamesSome compounds are more commonly referred to by their trivial names. You should be familiar with these:

• H2O, water• H2O2, hydrogen peroxide• CO, carbon monoxide• CH4, methane• SiH4, silane• GeH4, germane• NH3, ammonia• PH3, phosphine• AsH3, arsine• NO, nitric oxide• N2O, nitrous oxide• N2H4, hydrazine


7.5 Names and Formulas of Compounds Containing Polyatomic Ions

Given the formula of a compound containing one or two polyatomic ions, how do we determine its name? For any ionic compound, we simply name the cation, then the anion.

▼ Example 16

What is the name of Fe(NO3)3?Answer: Iron(III) nitrate.The cation is Fex+; we need to figure out what x is since Fe is a transition metal.The anion must be NO3

−, which is called nitrate (see Table 1) and has a charge of −1. Since there are three anions in Fe(NO3)3, each with a −1 charge, the total negative charge is 3(−1), or −3. Therefore, the total positive charge must be +3; x = 3. The iron in this compound must have a +3 charge. So the cation is iron(III).

Given the name of a compound containing one or more polyatomic ions, how do we determine its formula?

• Write the formulas of the ions in the order they are mentioned; omit the charges, and write subscripts, so that

(cation subscript)(cation charge) + (anion subscript)(anion charge) = 0.• If the subscript of a polyatomic ion is larger than 1, enclose the entire formula

for the ion in parentheses, then write the subscript after the closing parenthesis.

▼ Example 17

What is the formula of ammonium sulfate?Answer: (NH4)2SO4Ammonium is NH4

+, sulfate is SO42−. So one formula unit

of this compound should have two cations and one anion. 2(+1) + 1(−2) = 0. The simplest combination of ions with a total charge of zero is two ammonium ions and one sulfate ion. Write a subscript of 2 for ammonium, and no subscript for sulfate. The formula is (NH4)2SO4.


7.6 Names and Formulas of AcidsSome molecular compounds produce H+ ions when dissolved in water. These compounds are called acids and they are often named as though they are ionic; the cation, in this case, is hydrogen (H+). When writing formulas for these compounds, H is usually written first, with a subscript corresponding to the number of H+ ions produced when one molecule of the compound goes into water. The rest of the formula corresponds to the formula of the anion formed when the molecule dissociates (breaks up) in water; the magnitude of the charge of the anion corresponds to the number of H+ ions that can be obtained from one molecule of the acid.

▼ Example 18

Give two possible names for H2S.Answer: dihydrogen sulfide, hydrogen sulfideBased on the rule for naming binary molecular compound, it is called dihydrogen sulfide. However, because it is an acid, we can also name it as if it is ionic, and simply refer to it as hydrogen sulfide. The name of the cation is hydrogen (H+) and the name of the anion formed when two H+ ions break off from H2S is S2−; the anion is called sulfide.

Another way acids are named is based on the name of the anion produced when the H+ ions break off from the acid molecules. Anions have names ending in ide, ite, or ate.

If the name of anion ends in ide, the acid name starts with hydro followed by the stem of the name of the anion, followed by ic acid.

▼ Example 19

Give two possible names for HCl.Answer: hydrogen chloride, hydrochloric acidHCl is a binary molecular compound called hydrogen chloride. The ion produced with H+ breaks off from HCl is chloride (Cl−). Therefore, the name is hydrochloric acid.

▼ Example 20

Hydrogen cyanide (HCN) is also known as ______ acid.Answer: hydrocyanic


The anion formed when H+ breaks off from the molecule is cyanide (CN−). The other name for hydrogen cyanide is hydrocyanic acid.

If the name of the anion ends in ate, the acid name starts with the stem of the name of the anion, followed by ic acid.

▼ Example 21

Give two possible names for HNO3.Answer: hydrogen nitrate, nitric acidThe anion formed when H+ breaks off from HNO3 is nitrate (NO3

−). Therefore, the name is hydrogen nitrate or nitric acid.

If the name of the anion ends in ite, the acid name starts with the stem of the name of the anion, followed by ous acid.

▼ Example 22

Give two possible names for HNO2.Answer: hydrogen nitrite, nitrous acidThe anion formed when H+ breaks off from HNO2 is nitrite (NO2

−). Therefore, the name is hydrogen nitrite or nitrous acid.

7.7 Salts and HydratesIf the anion of an ionic compound is not oxide or hydroxide, it is called a salt. If the anion is hydroxide, the compound is called a base; if the anion is oxide, the compound is called a base anhydride.

▼ Example 23

Which of the following are salts? NaCl, CaCO3, NaOH, CaO, HCl, CH4

Answer: NaCl and CaCO3HCl and CH4 are not ionic compounds.NaOH and CaO are ionic, but their anions are hydroxide (OH−) and oxide (O2−).


Salt hydrates are compounds that produce water (H2O) and salt when heated;the salt produced is called the anhydrous salt. The formula of the hydrate is written as follows:

• write the formula of the salt.• write a large dot.• write the ratio of number of water molecules produced to the number of formula.

units of salt produced.• write H2O.

▼ Example 24

When Epsom salt (a hydrate) is heated, 14 water molecules are produced for every 2 formula units of anhydrous MgSO4 pro-duced. What is the formula of the hydrate?Answer: MgSO4 • 7H2OThe ratio of water molecules to anhydrous formula units is 14-to-2; 14/2 = 7.

To name a hydrate, simply name the salt, then specify the extent of hydration using the following prefixes:

1 = mono, 2 = di, 3 = tri, 4 = tetra, 5 = penta, 6 = hexa, 7 = hepta, 8 = octa, 9 = nona, 10 = deca, 1/2 = hemi, 3/2 = sesqui

▼ Example 25

Name this compound: CaSO4 • (1/2) H2O.Answer: calcium sulfate hemihydrate

▼ Example 26

Write the formula for sodium carbonate decahydrate.Answer: Na2CO3 • 10 H2O


TEST YOURSELF


1. Which of the following is an ionic compound? A. ethyl acetate B. sulfuric acid C. sodium hydrogen carbonate D. hydrazine

2. Which of the following is a molecular compound? A. phosphorus trichloride B. ammonium chloride C. lithium aluminum hydride D. sodium dihydrogen phosphate

3. Which of the following is a molecular compound? A. Na2S B. NH3 C. (NH4)2SO4 D. N2

4. Which of the following is a solid with a very high melting point and conducts electricity when melted?

A. CaCl2 B. CCl4 C. H2O

5. Which of the following is a strong acid? A. H2SO4 B. HC2H3O2 C. HCOOH D. HNO2

6. Which of the following is a strong electrolyte? A. CH3OH B. KOH C. HF D. NH3

7. Which of the following is a strong electrolyte? A. HCl B. C6H12O6 C. H2S D. NH3

8. A molecule of fructose consists of six carbon atoms, twelve hydrogen atoms, and six oxygen atoms. The formula of fructose is…

A. C6H12O6 B. CHO C. CH2O

9. A molecule of acetic acid consists of four hydrogen atoms, two carbon atoms, and two oxygen atoms. Which of these cannot possibly be the formula of acetic acid?

A. H4C2O2 B. H2CO C. HC2H3O2 D. CH3COOH



10. Which of these is not the formula of the compound represented by the figure below, if blue spheres represent carbon atoms, white spheres represent hydrogen atoms and red sphere represents an oxygen atom?

A. CHO B. C2H6O C. C2H5OH D. CH3CH2OH

11. What is the formula of an ionic compound consisting of Ca2+and Cl− ions? A. CaCl B. Ca2Cl C. CaCl2 D. Ca2Cl2

12. What is the formula of an ionic compound consisting of Na+and CO32− ions?

A. NaCO3 B. Na2(CO3)2 C. NaCO D. Na2CO3

13. What is the formula of an ionic compound consisting of Ca2+and CO32− ions?

A. CaCO3 B. Ca2(CO3)2 C. CaCO D. Ca2CO3

14. What is the formula of an ionic compound consisting of Mg2+and PO43− ions?

A. Mg2PO3 B. Mg3(PO4)2 C. Mg3PO2 D. MgPO4

15. Which of the following is the correct name for PCl5? A. phosphorus chloride B. phosphorus chlorine C. phosphorus pentachlorine D. phosphorus pentachloride

16. Which of the following is the correct name for MgCl2? A. magnesium chloride B. magnesium dichloride C. magnesium chlorine D. magnesium dichlorine


17. Which of the following is the correct name for Na2S? A. sodium sulfide B. disodium sulfide C. sodium sulfite D. sodium sulfate

18. Which of the following is correct name for CuCl2? A. copper(II) chloride B. copper(I) chloride C. copper chloride(II) D. copper(I) dichloride

19. Which of the following is the correct formula for dinitrogen tetroxide? A. NO B. N2O3 C. N2O4 D. NO2

20. Which of the following is the formula for iron(III) sulfide? A. FeS B. Fe3S C. Fe2S3 D. Fe3S2

21. Which of the following is the formula for cobalt(II) chloride? A. CoCl B. CoCl2 C. Co2Cl D. Co2Cl2

22. Which of the following is the correct formula for sodium phosphate? A. NaP B. NaPO3 C. Na3PO4 D. Na3(PO4)2

23. Which of the following is the correct formula for ammonium sulfide? A. (NH4)2S B. NH4SO3 C. (NH3)2SO4

24. Which of the following is the correct formula for iron(III) sulfate? A. Fe3 S B. FeSO4 C. Fe2(SO3)3 D. Fe2(SO4)3

25. Which of the following is the correct formula for potassium hydrogen phosphate? A. KHPO4 B. K2HPO4 C. KH2PO4

26. Which of the following is the correct name for K2CO3? A. dipotassium carbon trioxide B. potassium carbide C. dipotassium carbonate D. potassium carbonate

27. Which of the following is the correct name for FeSO4? A. iron(I) sulfate B. iron(II) sulfate C. iron(III) sulfite D. iron sulfite


28. Which of the following is the correct name for BaO2? A. barium oxide B. barium peroxide C. barium dioxide D. barium(IV) oxide

29. Which of the following is sulfurous acid? A. H2SO4 B. H2SO3 C. H2S

30. Which of the following is another name for hydrogen cyanide (HCN)? A. hydrocyanic acid B. cyanic acid C. cyanous acid

31. Which of the following is a salt? A. Mg(OH)2 B. HgO C. Mg(NO3)2

32. Which of the following is a base? A. Ca(OH)2 B. CaCl2 C. H2S

33. How many O atoms are in one formula unit of MgSO4 . 7H2O? A. 4 B. 5 C. 7 D. 11

34. When heated, a salt hydrate yields 6 H2O molecules for every 4 formula units of K2CO3. What is the name of the compound?

A. potassium carbonate tetrahydrate B. potassium carbonate hexahydrate C. potassium carbonate hemihydrate D. potassium carbonate sesquihydrate

117

8CHAPTER

Chemical Changes

The term chemical change, or chemical reaction, refers to the transformation of one or more substances into others. We say that a chemical change has occurred if there is a change in composition. If a change does not involve a change in composition, then it is called a physical change.

8.1 Reactants and ProductsWe refer to the substances involved in a chemical change as reactants and products. Reactants are transformed into products. Therefore, as a chemical change occurs, the amount of reactants will be decreasing and the amount of products will be increasing.

▼ Example 1

Suppose a drop of water added to a white solid results in the formation of a blue solid. Identify the reactants and products.Answer: the white solid and water are reactants; the blue solid is a product.This reaction is something that we would actually observe if we were to add a drop of water to copper sulfate (CuSO4), which is a white solid. The reaction produces a compound called copper sulfate pentahydrate (CuSO4

. 5H2O), which is a blue solid.

▼ Example 2

During a chemical change, the amount of substance X increases from 5.0 g to 6.0 g. Is X a reactant or a product?Answer: productSince the amount of X increased, we call it a product of the reaction.


8.2 Atomic Interpretation of Chemical ChangeAt the atomic level, a chemical change involves a reorganization of atoms into different formula units. For physical changes, the formula units remain intact; they simply move apart (as in vaporization of a liquid or breaking of a rock) or move closer together (as in condensation of a gas into a liquid).

▼ Example 3

Consider the following pictorial representation for a change at the atomic level.

Is the change chemical or physical?Answer: chemicalOne of the atoms represented by unfilled circles appears to have broken off from each of the two molecules shown on the left.

▼ Example 4

Consider the following pictorial representation for a chemical change at the atomic level.What are the reactants? Products? Which molecules are not involved in the reaction?

Chapter 8: Chemical Changes 119

Answer: H2 is a reactant; we have less of it after the reaction. O2 is also reactant. H2O is a product; we have more of it afterthe reaction. N2 is neither a reactant nor a product. We have the same amount before and after.

8.3 Chemical EquationsA chemical equation is a symbolic representation for what happens in a chemical change. In a chemical equation:

• the formulas of the reactants (separated by + signs) are written first,• an arrow pointing to the right is then drawn,• the formulas of the products (separated by + signs) are written next, and• numbers, called coefficients, are written in front of the formulas of reactants and

products; unwritten coefficients are implied to be 1.

It is preferable, but not required, to indicate the physical states of the reactants and products by using the following labels: (s) for solid, (l) for liquid, (g) for gaseous, and (aq) for aqueous solution; something is said to be in aqueous solution if it is dissolved in water.

▼ Example 5

Consider the chemical equation for the reaction of calcium carbonate, CaCO3, with hydrochloric acid, HCl(aq).CaCO3(s) + 2 HCl(aq) ➝ CaCl2(aq) + H2O(l) + CO2(g)

Identify the reactants and products, as well as their physical states. What are the coefficients?Answers:• The reactants are solid CaCO3 and HCl in aqueous solution;

they are written first.• The products are CaCl2 in aqueous solution, liquid H2O, and

gaseous CO2; they are written to the right of the arrow.• The coefficient of HCl is 2. The coefficients of CaCO3, CaCl2,

H2O, and CO2 are all 1.

The purpose of coefficients is to balance the chemical equation. Balancing a chemical equation means adjusting the coefficients so that the number of atoms of each element is the same on both sides of the arrow. This is done because no atoms should be created or destroyed during a chemical change. To count the number of atoms, we simply multiply the coefficient by the subscript.


▼ Example 6

Consider the chemical equation for the reaction of calcium carbonate, CaCO3, with hydrochloric acid, HCl(aq).CaCO3(s) + 2 HCl(aq) ➝ CaCl2(aq) + H2O(l) + CO2(g)

How many atoms of Cl atoms are shown on either side of the arrow?Answer: 2The Cl atoms are found in HCl on the left and CaCl2 on the right:CaCO3(s) + 2 HCl(aq) ➝ CaCl2(aq) + H2O(l) + CO2(g)

The coefficient of HCl is 2; the subscript of Cl in HCl is implied to be 1: 2 × 1 = 2The coefficient of CaCl2 is implied to be 1; the subscript of Cl in CaCl2 is 2: 1 × 2 = 2

There are an infinite number of sets of coefficients that we can use to balance a chemical equation. We can multiply all of the numbers in a valid set of coefficients by the same factor and come up with another set of coefficients that will also give us a balanced equation.

▼ Example 7

Consider the balanced chemical equation for the reaction of calcium carbonate, CaCO3, with hydrochloric acid, HCl(aq):CaCO3(s) + 2 HCl(aq) ➝ CaCl2(aq) + H2O(l) + CO2(g)

If the coefficient of HCl were changed to 6, what other coefficients would be needed to keep the equation balanced?Answer: The other coefficients need to be changed to 3, 3, 3, and 3.The coefficients shown are (1, 2, 1, 1, and 1). Changing the coefficient of HCl to 6 means multiplying what it is right now by 3: 2 × 3 = 6. To keep the equation balanced, all other coefficients must be multiplied by 3.

It is customary, but not necessary, to use the smallest set of whole number coefficients.How can we tell if we have the smallest set of whole number coefficients?• Find the greatest common factor (GCF) of all the coefficients. If the GCF is 1, then

we have the smallest set of whole number coefficients.How can we tell that the GCF is 1?• Divide all the coefficients by the smallest. If one of the resulting numbers is not a

whole number, then the GCF is already 1.


▼ Example 8

Which of the following sets of coefficients can be reduced to a simpler set of whole numbers?A. 1, 3, 3, 2 B. 5, 2, 3, 4 C. 3, 6, 6

Answer: C• For set A, (1, 3, 3, 2), the smallest whole number is 1.

The greatest common factor is 1.• For set B, (5, 2, 3, 4), the smallest whole number is 2.

Dividing all the numbers by this gives us a set (2.5, 1, 1.5, 2) that includes non-integers. This means that the greatest common factor for set B is already 1.

• For set C, (3, 6, 6), the smallest whole number is 3. Dividing all the numbers by this gives us a simpler set of whole numbers, (1, 2, 2).

Coefficients in chemical equations can be fractional, as in:

1/2 CaCO3(s) + HCl(aq) → 1/2 CaCl2(aq) + 1/2 H2O(l) + 1/2 CO2(g)

How could we have half of a molecule? We cannot. However, coefficients do not necessarily mean individual counts; in practice, they are often taken as group counts. Because atoms and molecules are so small, it is convenient to count them in groups called moles. A mole of molecules is about 602 billion trillion molecules. In the example above, 1/2 CO2 would be interpreted not as half of a CO2 molecule, but as half a mole of CO2 molecules (or 301 billion trillion CO2 molecules).

If, in balancing an equation, we end up using fractional coefficients and later decide to change the coefficients to whole numbers, all we have to do is multiply each of the fractional coefficients by the least common multiple (LCM) of the denominators.

▼ Example 9

Rewrite the following chemical equation using whole number coefficients:1/2 N2 + 3/2 H2 ➝ NH3

Answer: N2 + 3 H2 ➝ 2 NH3The coefficients are (1/2, 3/2, and 1) or (1/2, 3/2, and 1/1). The denominators are 2, 2, and 1. The least common multiple of these numbers is 2. Multiplying all the coefficients by 2 gives us (2 × 1/2, 2 × 3/2, and 2 × 1) or (1, 3, and 2).


8.4 Strategies for Balancing Chemical Equations

Balancing chemical equations means adjusting the coefficients after correct formulas of reactants and products have been written. In other words, subscripts should never be changed when balancing a chemical equation. When balancing an equation, it is helpful to:

• first try to balance elements that appear only once on either side.• balance by adjusting coefficients (in the deficient side) upward.• recheck previously balanced elements after every adjustment.

▼ Example 10

Balance: CH4 + O2 ➝ CO2 + H2O,Answer: CH4 + 2 O2 ➝ CO2 + 2 H2OIt is better to try to balance C and H before O since O appears twice on the right side.Balance C. We can see that the coefficients of CH4 and CO2 must be equal since these are the only formulas containing C and the subscript of C in both cases is the same. C is already balanced as is; at this point, the coefficients of CH4 and CO2 are both 1.1 CH4 + O2 ➝ 1 CO2 + H2O

Balance H. There are 4 H atoms in CH4, but only 2 in H2O. The coefficients of CH4 and H2O must be in a 1-to-2 ratio. There should be twice as many H2O molecules compared to CH4 in order to have the same number of H atoms on both sides. Balance H by adding another H2O to the right side (increasing the coefficient of H2O up to 2) since there are fewer H atoms on the right.1 CH4 + O2 ➝ 1 CO2 + 2 H2O

Since 1 × 4 = 2 × 2, H is now balanced.Recheck C. C is still balanced.Balance O.• Left side: O atoms are found in O2, which has an implied

coefficient of 1. 1 × 2 = 2 1 CH4 + 1 O2 ➝ CO2 + 2 H2O• Right side: O atoms are found in CO2 and H2O. (1 × 2) +

(2 × 1) = 4 1 CH4 + O2 ➝ 1 CO2 + 2 H2O1


• We need 2 more O atoms on the left side. We do this by adding another O2, by changing the coefficient of O2 from 1 to 2.

1 CH4 + 2 O2 ➝ 1 CO2 + 2 H2O1

Now, O is balanced: 2 × 2 = (1 × 2) + (2 × 1)Recheck C and H. Both are still balanced.

Balancing an equation can be facilitated with simple algebra:• Assign variables names to the coefficients.• Set up algebraic equations that must be satisfied to balance each element.• Assign a number to one of the variables, then solve for the others.

▼ Example 11

Use algebra to balance the following reaction: CH4 + O2 ➝ CO2 + H2O

Solution:Let the coefficients be w, x, y, and z:

w CH4 + x O2 ➝ y CO2 + z H2O• Eq. 1, To balance C: (w)(1) = (y)(1)• Eq. 2, To balance H: (w)(4) = (z)(2)• Eq. 3, To balance O: (x)(2) = (y)(2) + (z)(1)

In Eq. 3, (x)(2) is the number of O atoms on the left, (y)(2) + (z)(1) is the number on the right. On the left, O is found in O2 which has a coefficient of x. On the right, O is found in CO2, which has a coef-ficient of y and H2O, which has a coefficient of z.We can assign any number to one of these variables and solve for the others. Let w = 1. Then:• Substituting w = 1 into Eq. 1 gives us: (1)(1) = (y)(1), or y = 1• Substituting w = 1 into Eq. 2 gives us: (1)(4) = (z)(2); solving

for z, we get z = 2• Substituting y = 1 and z = 2 into Eq. 3 gives us: (x)(2) = (1)(2) + (2)(1); solving for x, we get x = 2

Thus: w = 1, x = 2, y = 1, and z = 2.If we started with w = 2, we would have gotten x = 4, y = 2, and z = 4.We could have assigned a value to y or z first; we would still be able to figure out all the other coefficients. However, starting with a value for x, which is essentially trying to balance O first, will give us three equations in three unknowns (w, y, and z), which will be more complicated to solve.


8.5 Signs of a Chemical ChangeEach substance has a unique set of properties that makes it different from other substances. If there is a change in the properties as a result of putting two or more substances in contact, or subjecting a sample of matter to heat, light or electricity, then it is very likely that a chem-ical change has occurred. If we cannot explain a change in terms of unchanged molecules or ions simply moving around, then it is very likely that we are dealing with a chemical change.

Some signs that a chemical change has most likely occurred are unusual changes in color, odor, taste or burning (a flame is a hot glowing gaseous product of a combustion reaction). Some signs that a chemical change may have occurred include: production of heat, absorption of heat, production of gas bubbles in a liquid, and formation of a solid when two liquids are mixed (precipitation).

A lack of visible change does not imply that a chemical change has not occurred. For example, sodium hydroxide dissolved in water, NaOH(aq), is a colorless liquid. Acetic acid in water, CH3COOH(aq), is also a colorless liquid. If we mix the two liquids, we still get a colorless liquid; we will not be able to observe a visible change. But there is, in fact, a chemical change. NaOH(aq) is bitter, CH3COOH(aq) is sour, but the resulting mixture is salty. CH3COOH(aq) smells like vinegar while NaOH(aq) is odorless. The resulting mix-ture could be odorless or could still smell a little bit like vinegar (if there are acetic acid molecules left unreacted). If we were to measure the temperature of the liquids before and after mixing, we would also find that the temperature of the resulting mixture is higher; this reaction generates heat; we say it is an exothermic reaction.

Production of heat generally accompanies most physical and chemical changes. Therefore, generation of heat alone does not imply that a chemical change has occurred. Example: steam condensing into a liquid generates a lot of heat, but it is just a physical change. Production of heat and light together, as in the burning of paper, involves a chemical change; these are called combustion reactions. However, the production of heat and light by an incandescent light bulb does not involve a chemical change.

Production of gas bubbles in a liquid could occur by stirring, shaking, or heating the liquid. It can also occur by pulling a vacuum above the liquid. In these cases, the change involved is just a physical change. However, if bubble formation results simply by mixing a liquid with something else (no stirring, heating, shaking or vacuuming involved), then the gas bubbles are very likely to be the product of a chemical change. If a chemical reac-tion produces a gas that is not very soluble in the liquid, the gas will be observed leaving the water as bubbles (“fizz”). A familiar example is carbon dioxide, CO2, which fizzes out when baking soda is mixed with vinegar. The fizz observed when opening a can of carbon-ated drink does not involve a chemical change; the can is pressurized; opening it lowers the pressure and lowers the solubility of carbon dioxide.


Precipitation usually indicates a chemical change. However, it can happen if you mix different liquids such that a substance dissolved in one of the liquids suddenly becomes insoluble; in this case, the precipitation is not considered a chemical change.

A change in color, odor or taste does not necessarily imply that a chemical change has occurred. For example, our eyes perceive a mixing of yellow and blue as green. So, mixing a yellow liquid with a blue liquid to produce a green liquid does not necessarily mean that a chemical change has occurred. But if mixing a yellow liquid with a blue liquid produces a red liquid, then we can be confident that a chemical change has occurred. If you mix something sweet and something sour and your taste buds senses a “sweet and sour” taste, it does not necessarily mean that a chemical change has occurred as a result of the mixing. But if you sense a bitter taste instead, then a chemical change has occurred as a result of the mixing.

▼ Example 12

Which of these most likely describes a chemical change?A. Food rotting,B. Gasoline burning,C. Stain fading when treated with bleachD. Water boiling,E. Ice melting,F. A chalk cut into two pieces,G. Green leaves turning brown in autumn,H. Sugar dissolved in water

Answers: A, B, C, G

8.6 Simple Types of ReactionsThe simplest types of reactions can be classified as combination, decomposition, displacement, metathesis, and combustion.

8.6.1 Combination ReactionsCombination reactions, also known as composition or synthesis reactions, involve two or more reactants and only one product. Commonly observed combination reactions include the following:

• combination of two elements to form a compound.• combination of ammonia (NH3) with an acid to form an ammonium salt.• combination of a metal oxide with water to form a metal hydroxide.• combination of a nonmetal oxide with water to form an acid.• combination of a metal hydroxide with carbon dioxide to form a metal bicarbonate.


▼ Example 13

Write the chemical equation for the reaction of aluminum metal with liquid bromine.Answer: 2 Al(s) + 3 Br2(l) ➝ 2 AlBr3(s)Aluminum metal is Al(s). Bromine is made up of diatomic mol-ecules (Br2) under ordinary conditions. The expected product of a metal and nonmetal is an ionic compound. The metallic element is expected to form a cation; the only naturally-occurring mona-tomic ion from Al has a charge of +3; the only naturally-occurring monatomic ion from Br has a charge of –1. Thus, the product is AlBr3. Since ionic compounds generally have high melting points, we expect AlBr3 to be a solid under ordinary conditions.

▼ Example 14

Write the chemical equation for reaction of LiOH with CO2 gas.Answer: LiOH(s) + CO2(g) ➝ LiHCO3(s)Metal hydroxides are known to react with CO2 to form metal bicarbonate. Thus, we predict the product of lithium hydroxide and CO2 to be lithium bicarbonate (LiHCO3). LiOH and LiHCO3 are both ionic compounds and are expected to be solids under ordinary conditions. This is the reaction used in the space shuttle to prevent the build up of CO2 in the air that astronauts breathe; we exhale CO2.

▼ Example 15

Write the chemical equation for the reaction of SO2(g) (from a volcano) with rain.Answer: SO2(g) + H2O(l) ➝ H2SO3(aq)SO2 is a nonmetal oxide. Nonmetal oxides react with water to form acids. H2SO3 is sulfurous acid.

▼ Example 16

Write the chemical equation for the reaction of ammonia gas (NH3) with hydrogen chloride gas (HCl).Answer: NH3(g) + HCl(g) ➝ NH4Cl(s)


The product is an ammonium salt, ammonium chloride. Since ammonium chloride is an ionic compound, we expect it to be a solid under ordinary conditions.

▼ Example 17

Write the chemical equation for the reaction of calcium oxide with water.Answer: CaO(s) + H2O(l) ➝ Ca(OH)2(s)The product of a metal oxide and water is a metal hydroxide. This reaction is highly exothermic and is used for on-the-spot food warming in self-heating cans.

8.6.2 Decomposition ReactionsA decomposition is the opposite of a combination reaction. It involves only one reactant, and two or more products. Commonly observed decomposition reactions occur when the following compounds are heated (thermal decomposition):

• metal carbonates; the products are metal oxide and CO2(g).• metal bicarbonates; the products are metal hydroxide and CO2(g).• metal hydroxides; the products are metal oxide and H2O(g).• acids; the products are nonmetal oxide and H2O(g).• metal chlorates; the products are metal chloride and O2(g).• peroxides; the products are oxide and O2(g).• binary compounds; the products are the elements.

▼ Example 18

Write the chemical equation for the decomposition of calcium carbonate.Answer: CaCO3(s) ➝ CaO(s) + CO2(g)This is the reaction that occurs when limestone is heated in a furnace in a process called calcination. The product, calcium oxide (CaO), is called quicklime and is used to make cement.


▼ Example 19

Write the chemical equation for the decomposition of sodium azide (NaN3).Answer: 2 NaN3(s) ➝ 2 Na(s) + 3 N2(g)This is the reaction that occurs in a car’s airbag. The N2 gas produced by the reaction causes the airbag to inflate.

▼ Example 20

Write the chemical equation for the thermal decomposition of KClO3.

Answer: 2 KClO3(s) ➝ 2 KCl(s) + 3 O2(g)This is a commonly used procedure for generating O2 in introductory chemistry laboratory courses.

8.6.3 Displacement ReactionsA displacement reaction, also known as single replacement, involves the replacement of an element in a compound by another element. A metallic element may displace another metallic element from an ionic compound, or hydrogen from acid, depending on the conditions. Most of these reactions involve compounds in aqueous solution.

The term activity series refers to a list that we can use to predict which metallic element is more likely to displace another. A subset of this series that includes the most common metallic elements is:

[K Ca Na] [Mg Al Mn Zn Cr] [Fe Ni Sn Pb] H [Cu Bi Sb] [Hg Ag Pt Au]The metal listed first can displace one that is listed later. The elements in the first

group (K, Ca, and Na) are so reactive that they can readily displace H2 from cold water; the other product is an aqueous solution of the metal hydroxide. The second group readily displaces H2 from steam; they react slowly with room temperature water. The first, second, and third groups are able to displace H2 from acids.

▼ Example 21

Write the chemical equation for the reaction of zinc metal with copper(II) chloride.Answer: Zn(s) + CuCl2(aq) ➝ ZnCl2(aq) + Cu(s)The expected products are zinc chloride and copper metal. Naturally occurring zinc ion is always observed to have a


+2 charge and naturally occuring chloride ion is always observed with a –1 charge. Therefore, the formula of the zinc chloride product is ZnCl2.

▼ Example 22

Write the chemical equation for the reaction that occurs when potassium (K) is mixed with an aqueous solution of NaCl.Answer: 2 K(s) + 2 H2O(l) ➝ 2 KOH(aq) + H2(g)The H2O comes from the aqueous solution. Even if K were to displace Na from NaCl, the Na produced will immediately react with H2O since Na is a very reactive metal. In fact, formation of Na is not observed when this reaction is carried out. What is observed is the formation of bubbles, which indicates the forma-tion of a gaseous product that is insoluble in water. We can test the gas and verify that it is, in fact, H2.

A displacement reaction can also involve displacement of a nonmetal by another nonmetal. The commonly observed reactions involve halogens; among the halogens, F is found to be the most active, followed by Cl, Br, and I.

▼ Example 23

Write the chemical equation for the reaction of chlorine gas with an aqueous solution of potassium bromide.Answer: Cl2(g) + 2 KBr(aq) ➝ 2 KCl(aq) + Br2(l)Remember that the elemental form of the halogens is diatomic.

8.6.4 Metathesis ReactionsMetathesis reactions are also known as double decomposition or double replacement. A metathesis reaction usually involves two compounds that produce ions in water. The reaction can be thought of as an exchanging of partners. The positive ion from one com-pound combines with the negative ion from the other compound. Commonly observed metathesis reactions include:

• precipitation reactions. In a precipitation reaction, the mixing of two aqueous solutions leads to the formation of a product that is insoluble in water. As a result, the insoluble product comes out of solution and makes the mixture appear cloudy.


Eventually, this insoluble product, called the precipitate, settles to the bottom of the container. In the chemical equation the physical state of the precipitate is indicated by (s); alternatively, a downward pointing arrow ( ) is drawn next to the formula.

• acid-base neutralization reactions. This type of reaction occurs when an acid reacts with a metal hydroxide or metal oxide. The products are water (H2O) and a salt. Precipitation may occur at the same time if the salt happens to be insoluble in water.

▼ Example 24

Write the chemical equation for the precipitation of lead(II) iodide when aqueous solutions of sodium iodide and lead(II) nitrate are mixed.Answer: 2 NaI(aq) + Pb(NO3)2(aq) ➝ PbI2(s) + 2 NaNO3(aq)

or 2 NaI(aq) + Pb(NO3)2(aq) ➝ PbI2 + 2 NaNO3(aq)

▼ Example 25

Write the chemical equation for the neutralization of sulfuric acid by sodium hydroxide in aqueous solution.Answer: H2SO4(aq) + 2 NaOH(aq) ➝ 2 H2O + Na2SO4(aq)

Bubble formation is generally observed for metathesis reactions that lead to the formation of H2CO3 or H2SO3. The reason for this is that these compounds tend to decompose to H2O and CO2 or SO2. CO2 and SO2 are gaseous in pure form and are not very soluble in water. Bubble formation and a sharp odor are also observed for metathesis reactions that lead to the formation of H2S. H2S is gaseous in pure form and is not very soluble in water.

▼ Example 26

Write the chemical equations to explain the bubbles observed when baking soda, NaHCO3(s), reacts with vinegar, HC2H3O2(aq).Answer: The reaction of baking soda (or sodium bicarbonate) with acetic acid (in vinegar) is a metathesis reaction that leads to the formation of carbonic acid (H2CO3): NaHCO3(s) + HC2H3O2(aq) ➝ NaC2H3O2(aq) + H2CO3(aq)

→

➝


Carbonic acid then decomposes to H2O and CO2.H2CO3(aq) ➝ H2O(l) + CO2(g)

CO2 is not very soluble in water and it comes out of solution as gas bubbles.

8.6.5 CombustionCombustion, or burning, involves a reaction of a material with oxygen (O2), accompanied by the production of heat and light. The products of combustion are the oxides of the elements present in the material.

▼ Example 27

Write the chemical equation for the combustion of propane gas (C3H8).

Answer: C3H8(g) + 5 O2(g) ➝ 3 CO2(g) + 4 H2O(g)Note that the products are the oxide of carbon (carbon dioxide, CO2) and the oxide of hydrogen (water, H2O). If there is insufficient O2 available, carbon monoxide (CO) may also be formed. If the compound contains nitrogen, N2 is formed unless a very large excess of oxygen is used (in which case, oxides of nitrogen are formed such as NO and NO2). If the compound contains sulfur, SO2 and SO3 are also produced.


TEST YOURSELF


1. A mixture contains 25.0 g of substance X and 10.0 g of substance Y. A few minutes later, due to a chemical change, it is found that the mixture now contains 30.0 g of substance X and 5.0 g of substance Y. For the reaction described here...

A. X is the reactant and Y is the product B. Y is the reactant and X is the product C. Both X and Y are reactants D. Both X and Y are products

2. Consider the pictorial representation shown below for a process at the atomic/ molecular level. Which process involves a chemical change?

A. 1 only B. 2 only C. both D. neither

3. Consider the pictorial representation shown below for a process at the atomic/ molecular level. Which process does not involve a chemical change?



4. Which pictorial representation best accounts for the conversion of water from liquid to steam?

5. Consider the pictorial representation of a chemical change shown below. Which of the following is not a product of the reaction?

A. C2H2 B. H2O C. CO2

6. Consider the chemical equation for the hydrogenation of acetylene: C2H2 + 2H2 → C2H6

Which of these is the formula of a reactant in this reaction? A. C2H2 B. 2H2 C. C2H6

7. Consider the following observations of a chemical change: A small amount of orange-red solid is heated in a test tube. A little explosion occurred and a green powder was obtained. The mass of the green solid was found to be less than that of the original orange-red solid. Which of the following chemical equations is consistent with the observations above?

A. Mg(s) + O2(g) → 2 MgO(s) B. (NH4)2Cr2O7(s) → N2(g) + Cr2O3(s) + 4 H2O(g) C. 2 HgO(s) → 2 Hg(l) + O2(g) D. Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)


8. Which of the following is not a balanced chemical equation? A. 2 H2 + O2 → 2 H2O B. 4 H2 + 2 O2 → 4 H2O C. Both A and B D. None of the above

9. Which of the following is the balanced chemical equation for the following reaction?

Mg + O2 → MgO A. Mg + O → MgO B. Mg + O2 → MgO + O C. Mg2 + O2 → 2 MgO D. 2 Mg + O2 → 2 MgO

10. Consider the following chemical equation where coefficients are represented by the letters w, x, y, and z: w Al2O3 + x H2SO4 → y Al2(SO4)3 + z H2O

If w = 1, then which of these is true? A. y = 2 B. z = 3 C. x = 6

11. Consider the combustion of propane: C3H8 + O2 → CO2 + H2O (unbalanced) Using the smallest set of whole numbers, the sum of the coefficients in the balanced

chemical equation is: A. 7 B. 9 C. 12 D. 13

12. Consider the pictorial representation of a chemical change shown below:

Using the smallest set of whole numbers, the coefficient of C2H2 in the balanced chemical equation for this reaction is...

A. 1 B. 2 C. 3 D. 4


13. Consider the pictorial representation of a chemical change shown below:

Using the smallest set of whole numbers, the coefficient of NH3 in the balanced chemical equation for this reaction is...

A. 2 B. 3 C. 4 D. 6

14. Which of the following observations is a definite sign that a chemical change has occurred?

A. Mixing a blue-colored liquid with a yellow-colored liquid, yields a green-colored mixture

B. Mixing a blue-colored liquid with a colorless liquid yields a yellow-colored liquid C. Neither D. Both

15. Which of the following observations implies that a chemical change has definitely not occurred?

A. Mixing two colorless liquids yields a colorless liquid B. A piece of paper, held next to a flame, catches fire and leaves a black residue C. Both D. Neither

16. Which of the following is a correct chemical equation for a combination reaction between a metallic element and nonmetallic element?

A. Mg(s) + N2(g) → MgN2(s) B. Mg(s) + O2(g) → MgO2(s) C. Mg(s) + N(g) → MgN D. None of these


17. Which of the following is a decomposition reaction? A. 2 CO(g) + O2(g) → 2 CO2(g) B. Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) C. Ba(OH)2(aq) + H2SO4(aq) → 2 H2O(l) + BaSO4(s) D. 2 H2O2(aq) → 2 H2O(l) + O2(g)

18. What is the salt produced by the neutralization reaction of magnesium hydroxide, Mg(OH)2, and hydrochloric acid, HCl?

A. MgCl B. MgCl2 C. HOH D. none of these

19. What is chemical equation for the neutralization of stomach acid, HCl, by milk of magnesia, Mg(OH)2?

A. HCl + Mg(OH)2 → HMg + Cl(OH)2

B. 2 HCl + Mg(OH)2 → 2 HOH + MgCl2

C. HCl + Mg(OH)2 → H2O + MgCl D. HCl + Mg(OH)2 → H2O + MgHCl

20. What is the chemical equation for the neutralization of nitric acid, HNO3, by calcium oxide (CaO)?

A. HNO3 + CaO → HO + CaNO3

B. 2 HNO3 + CaO → H2O + Ca(NO3)2

C. HNO3 + CaO → NO4 + CaH D. 2 HNO3 + CaO → H2O + CaNO3

21. Which of the following is more likely to occur? A. Cu(s) + MgCl2(aq) → Mg(s) + CuCl2(aq) B. Cu(s) + AgNO3(aq) → Cu(NO3)2(aq) + Ag(s)

22. Silver phosphate is insoluble in water. All sodium salts are soluble in water. Given this information, write the chemical equation for the precipitation reaction that occurs when aqueous solutions of sodium phosphate and silver nitrate are mixed.

23. Write the chemical equation for the combustion of urea, CO(NH2)2.

137

9CHAPTER

History of Atomic TheoryIn this chapter, we examine how our notion about atoms has developed over time.

9.1 AtomismThe word atom literally means uncut (a = not, tomos = cut). The philosophy of atomism, espoused by the Greek philosopher Democritus (460 BC-370 BC), asserts that the world is made up of two things: atoms and the void. In other words, matter of made up of tiny, indestructible particles called atoms and in between atoms, there is nothing. However, atomism did not flourish, as it was contrary to the ideas supported by more influential people of that time (Plato, Aristotle). Furthermore, the Greek philosophers were thinkers, not workers. They did not have any experimental support for their ideas.

The scientific method as we know it, the use of empirical (experimental) data to support ideas, was not developed until centuries later. Publications regarding the “proper way” to do science date as far back as the early 1600s. Before that, alchemists (early chemists) developed laboratory techniques that were eventually used for scientific work. The alchemists’ goals were to turn lead into gold, cure diseases with herbs, develop the elixir of life (panacea) and develop a universal solvent that can dissolve anything (“alkahest”). Alchemists were considered scientists but, these days, their approach to achieving their goals is not what we would consider as “scientific”. Although alchemists did not advance our knowledge about atoms, they promoted observation and experimen-tation and served as bridge between Greek philosophers’ “pure thought” and modern scientific method.

9.2 Dalton’s Atomic TheoryJohn Dalton elevated the idea of atomism to a scientific theory in 1803. In Dalton’s atomic theory, the following assumptions (postulates) are made:

• matter consists of small indivisible particles called atoms.• an element consists of identical atoms (same mass).


• atoms of different elements have different masses.• atoms of two or more elements combine in small whole-number ratios to form

compounds.• a chemical change involves the rearrangement of atoms.

Extensive work by several scientists provided the empirical support for Dalton’s ideas. Among the most notable are the works of Antoine Lavoisier (the “father” of Modern Chemistry) and Joseph Proust.

9.2.1 Law of Conservation of MassLavoisier’s work led to the formulation of the Law of Conservation of Mass, which says that the amount of matter is constant even during a chemical change. Dalton’s assumption that atoms are indestructible and merely rearrange during chemical changes explains this general observation.

9.2.2 Law of Definite ProportionsProust’s work led to the formulation of the Law of Definite Proportions, or Law of Definite Composition, which says that compounds (substances made up of more than one element) have a well-defined composition. By this, we mean that the mass-to-mass ratio of any two elements in the compound is the same for any sample of the compound. It also means that the percentage of any element in any sample of the compound is well-defined (constant).

The Law of Definite Proportions is explained by Dalton’s assumptions that all atoms of an element have the same mass and that the count-to-count ratios of atoms from different elements in a compound are well-defined small whole number ratios.

▼ Example 1

Suppose a compound is made up of elements X and Y, such that the count-to-count ratio of X to Y is 1:2. Also suppose that each X atom has a mass of x units, and each Y atom has a mass of y units. Show that the mass-to-mass ratio of X to Y is constant.Answer:Imagine a sample containing n X atoms. If the count-to-count ratio of X-to-Y is 1:2, then this sample must have 2n Y atoms:

countcount

n2n

12

X

Y

= =

Chapter 9: History of Atomic Theory 139

If each X atom has a mass of x units and each Y atom has a mass of y units, then the mass-to-mass ratio would then be:

massmass

nx2ny

x2y

X

Y

= =

Since x and y are constants, x/2y is also a constant; the mass-to-mass ratio of X to Y is, therefore, constant regardless of what n is.

▼ Example 2

In the preceding example, show that the percentage of X (by mass) in the compound is also constant.Answer:The ratio of the mass of X to the total mass is

massmass mass

n xn x 2 n y

nxn x y

xx y

X

X Y+=

+=

+=

+( )2 2

which is also a constant since x and y are constants. If we multiply this value by 100, we get the percentage of X in the compound (by mass), which is also a constant.

9.2.3 Law of Multiple ProportionsDalton himself formulated the Law of Multiple Proportions, which says that if two elements can form more than one compound, the ratio of their mass ratios can be reduced to a ratio of small whole numbers.

▼ Example 3

Verify that the ratio of C-to-H mass ratios for CH4 and C2H4 is a small whole-number ratio.Answer:Assume that the mass of each C atom is c and that the mass of each H atom is h. The C-to-H mass ratio in CH4 is:

massmass

c4 h

C

H

=


while the C-to-H mass ratio in C2H4 is:

massmass

2c4 h

C

H

=

The ratio of mass ratios is:

which simplifies to 1/2, a ratio of two whole numbers.

▼ Example 4

A compound of elements X and Y is found to be 85.71% X by mass. According to the Law of Multiple Proportions, which of the following is a more plausible mass percent of X for another compound of X and Y? A. 80.00%, B. 84.32%Answer: AFirst, we determine the X-to-Y mass ratios.• For the known compound, assuming a 100 g sample: mass of X = 85.71% of 100g = 85.71 g mass of Y = 100g − 85.71 g = 14.29 g X-to-Y mass ratio: 85.71/14.29 = 5.998• For choice A, assuming a 100 g sample: mass of X = 80.00% of 100 g = 80.00 g mass of Y = 100 g − 80.00 g = 20.00 g X-to-Y mass ratio: 80.00/20.00 = 4.000• For choice B, assuming a 100 g sample: mass of X = 84.32% of 100 g = 84.32 g mass of Y = 100 g − 84.32 g = 15.68 g X-to-Y mass ratio: 84.32/15.68 = 5.378

Then, we calculate the ratio of mass ratios.• For the known compound and the sample in choice A: 5.998/4.000 = 1.500, which is 3/2 (a small whole

number ratio)• For the known compound and the sample in choice B: 5.998/5.378 = 1.115; the closest possible ratio of

small whole numbers is 10/9 (which is equal to 1.111)

c4 h

2 c4 h

⎛

⎝⎜

⎞

⎠⎟

⎛

⎝⎜

⎞

⎠⎟c

4 h

2 c4 h

⎛

⎝⎜

⎞

⎠⎟

⎛

⎝⎜

⎞

⎠⎟

c4 h

2 c4 h

⎛

⎝⎜

⎞

⎠⎟

⎛

⎝⎜

⎞

⎠⎟2 c

4 h

c4 h

2 c4 h

⎛

⎝⎜

⎞

⎠⎟

⎛

⎝⎜

⎞

⎠⎟c

4 h

2 c4 h

⎛

⎝⎜

⎞

⎠⎟

⎛

⎝⎜

⎞

⎠⎟


How can we express a number, x, as a ratio of whole numbers? If x is a whole number, then it already is; it is a ratio of that whole number to 1. If it is not a whole number, then we look for a whole number (n) such that nx is a whole number; the whole number ratio is nx/n.

▼ Example 5

Express 1.199 as a ratio of whole numbers.Answer: 6/5We just need to find n, such that (n)(1.199) is a whole number.Try 2. 2(1.199) = 2.398, not a whole number.Try 3. 3(1.199) = 3.597, not a whole number.Try 4. 4(1.199) = 4.796, not a whole number.Try 5. 5(1.199) = 5.995, close enough; round off to 6.

The ratio is 6/5.

9.3 The Discovery of the ElectronTheories evolve as we discover new information. Some of Dalton’s assumptions about atoms are no longer accepted. Atoms are, in fact, divisible. We now know that they are made up of even smaller particles called protons, neutrons, and electrons. The discovery of the electron is credited to Joseph John Thomson, although our experience with it dates back to the time of Thales of Miletus (600 BC), where it was found that rubbing amber against fur makes them attract each other. We now interpret this phenomenon in terms of electrons being transferred from fur to amber. The fur becomes positively-charged and the amber becomes negatively-charged. In 1600, William Gilbert discovered that amber was not the only material that can attract other materials when rubbed; he coined the word electric (from elektron, the Greek word for amber) and referred to materials that attract each other when rubbed as electrically charged. In 1629, Niccolo Cabeo discovered that electrically charged materials can also repel each other. This led to the notion of two kinds of charge; like charged materials repel each other, materials with unlike charge attract each other. In 1733, Charles du Fay proposed two kinds of charge: resinous and vitreous. By his definition, vitreous charges are transferred to a glass rod when it is rubbed against silk, and resinous charges are transferred to amber when it is rubbed against fur. We now know that what du Fay called resinous charges are negatively charged electrons. We also now know that when glass is rubbed against silk, it becomes positively charged, not because positively charged particles (vitreous charge) are transferred to the glass, but because (negatively-charged) electrons move out of the glass. The idea of just one type of charged particle being transferred when materials are rubbed is credited to Benjamin Franklin (1752), although he assumed that the charge that was transferred was positive.


Measurement of charges is attributed to Charles Augustin de Coulomb who formulated (in 1785) the law that now bears his name: the force between two charged particles is directly proportional to the product of the magnitudes of the charge and inversely proportional to the square of the distance between them. The SI unit for electrical charge is Coulomb (C).

J. J. Thomson used cathode ray tubes (CRT) in his experiments (1897). Inside a CRT, a piece of metal (the cathode) is subjected to high voltage. As a result, a beam of particles, which Thomson called corpuscles, is ejected from the metal. The CRT beams behaved the same way (when subjected to a magnet or to electrically charged plates) regardless of the nature of material used as cathode. This means that whatever these particles are, they are found in all atoms. We now know that the particles in the beam are electrons; the name “electron” was coined by G. Johnstone Stoney. In J. J. Thomson’s CRT experiments, the charge-to-mass ratio for the electron, but not the mass itself, was determined.

Figure 1. Cathode Ray Tube

Robert Millikan’s oil-drop experiment (1909) established the charge of the electron. In his experiments, Millikan found that when he charged oil droplets, the charges were all multiples of −1.60 × 10−19 C. This suggested that −1.60 × 10−19 C is the smallest possible charge and this is the charge of each electron.

▼ Example 6

If we were to repeat Millikan’s experiment, which of the following values for the charge of an oil droplet are we least likely to observe?A. −1.122 × 10−18 C B. −1.553 × 10−18 C C. −1.759 × 10−18 C

from http://commons.wikimedia.org/wiki/File:JJ_Thomson_exp2.jpg (public domain)

http://commons.wikimedia.org/wiki/File:JJ_Thomson_exp2.jpg


Answer: BThe values we would expect to get should be very close to an inte-ger multiple of the charge of an electron, which −1.60×10−19 C. To check if the given choices are integer multiples, we divide each one by −1.60×10−19 C.• A. (–1.122 × 10−18 C)/(−1.60 × 10–19 C) = 7.01 → 7• B. (−1.553 × 10−18 C)/(−1.60 × 10−19 C) = 9.70• C. (−1.759 × 10−18 C)/(−1.60 × 10−19 C) = 10.99 → 11

Among the choices given, choice B is farthest from being an inte-ger multiple.

Based on J. J. Thomson’s measured charge-to-mass ratio and Millikan’s measured charge, the mass of the electron was found to be 9.11 × 10−31 kg, much smaller than the mass of the lightest known atom (hydrogen).

With the discovery of the electron, the plum pudding model was proposed for the atom. Instead of the structureless, indivisible spheres that Dalton envisioned, atoms are now proposed to be positively charged spheres with negatively charged electrons embed-ded in them, just like plums in a pudding.

Figure 2. The Plum Pudding Model

from http://commons.wikimedia.org/wiki/File:Plum_pudding_atom.svg (public domain)

http://commons.wikimedia.org/wiki/File:Plum_pudding_atom.svg


9.4 The Discovery of the NucleusThe work of Ernest Rutherford provided the next major advancement in our knowledge about atoms. In his experiments (1909), Rutherford bombarded a piece of gold foil by alpha particles. Alpha particles are high speed particles produced by radioactive materials; an alpha particle is essentially a helium nucleus (2 protons and 2 neutrons). If the plum pudding model were correct, then the mass of an atom would be spread evenly throughout the space occupied by the atom and all of the alpha particles would be expected to pass through with very minor deflection (like bullets through a tissue paper). As it turned out, most (not all) of the alpha particles passed through the gold foil with very minor deflection. A very small fraction of the particles were either deflected at very large angles or bounced back (“backscat-tered”). This led to the proposition of the currently accepted nuclear model for the atom, where:

• the mass of atom is concentrated in a very tiny spot (the nucleus),• the nucleus is positively charged,• electrons move around the nucleus, and• the space where electrons move about is much, much larger than the space

occupied by the nucleus (the diameter of the region where electrons move is at least 1000 times larger than the diameter of the nucleus).

The nuclear model explains Rutherford’s experimental results. Most of the alpha particles passed through because the gold foil is mostly empty space; the regions occupied by the nuclei are very, very small compared to the spaces between them. The very small fraction of alpha particles that pass near the nuclei would be deflected at large angles. The alpha particles that are backscattered are the ones that collide head-on with the nuclei.

9.5 The Discovery of the NeutronThe discovery of the neutron is credited to Sir James Chadwick. The discovery was made in 1932, years after Rutherford discovered the nucleus. Neutrons were more difficult to detect because they are electrically neutral. Electrically charged particles like electrons and protons are easier to detect because they respond to magnets and electrically charged plates.

Chadwick bombarded beryllium atoms with alpha particles. He showed that the resulting radiation was a beam of neutral particles that have a mass of one amu.


TEST YOURSELF


1. Assume circles and squares in the figures below represent atoms. According Dalton’s atomic theory, which of the figures below best represents a sample of a compound?

2. Assume that circles and squares in the figure below represent atoms. According to Dalton’s theory, which of the changes indicated is possible?

3. Consider the pictorial representation shown below for the molecular composition of a mixture. Using Dalton’s atomic theory, the figure represents a mixture of…

A. an element and two compounds B. two elements and a compound C. three compounds



4. If the mass of atom X is 10 times that of atom Y, Dalton would explain that an observed X-to-Y mass ratio of 5:1 in a compound means that the X-to-Y atom count ratio in the compound is....

A. 1:1 B. 1:2 C. 2:1 D. 2:3

5. Shown below is a diagram illustrating Rutherford’s alpha scattering experiment. Alpha particles are represented by red circles. Which of the three labeled alpha particles (A, B, or C) will have the most typical trajectory?

6. Which of these is true? A. Dalton assumed that atoms are made up of electrons, protons, and

neutrons B. J.J. Thomson determined the charge and mass of electrons C. In Rutherford’s alpha scattering experiment, all of the alpha particles passed

through the gold foil D. The alpha particles used in Rutherford’s experiment are positively charged

particles.


7. Match the pictorial representations showing four neighboring atoms to the scientist most closely associated with the atomic model.

A. 1-Dalton, 2-J.J.Thomson, 3-Rutherford B. 1-J.J.Thomson, 2-Dalton, 3-Rutherford C. 1-J.J.Thomson, 2-Rutherford, 3-Dalton D. 1-Rutherford, 2-J.J.Thomson, 3-Dalton

8. If we were to repeat Millikan’s oil drop experiment, which of the following charge are we least likely to observe for an oil drop?

A. 1.0 × 10–19 C B. 1.6 × 10–19 C C. 4.8 × 10–19 C

149

CHAPTER

Quantum Theory

Towards the end of the 19th century, experiments involving light interacting with atoms and molecules showed “anomalous” results that could not be explained by classical phys-ics. Classical physics refers to the physics developed by Isaac Newton, James Maxwell, and others prior to the 20th century. Describing motion using Newton’s laws of motion is called Newtonian mechanics or classical mechanics. Classical principles that describe electricity and magnetism are based on the work of James Maxwell.

Quantum mechanics and Relativity were developed to address failures of classical phys-ics in describing the world of the very small and the world of the very fast, respectively. Quantum mechanics and relativity are considered as the two major branches of modern physics. The currently accepted model of the atom is best described by quantum mechanics.

10.1 Electromagnetic RadiationLight is also known as electromagnetic radiation. Much of the behavior of light can be explained by thinking of it as a wave. However, towards the end of the 19th century, scien-tists observed phenomena that could not be explained by the wave model. In these cases, we have to think of light as a stream of particles, which we now call photons.

10.1.1 The Wave ModelEarlier studies dealing with light have led scientists to the conclusion that light is an elec-tromagnetic wave. Let us try to understand what these two words mean.

A wave is a periodic (repeating) disturbance that transfers energy from one place to another. Imagine dropping a rock into a body of water. A water wave is produced (ripples travel outward from where the rock fell) and an object floating elsewhere will bob up and down as the wave passes through it. We say that the energy from the rock has been trans-mitted to the object.

The term electromagnetic means that the disturbance:• is due to oscillation of charged particles.• exerts a force on any charged particle that is in its way.

10


The things that are “waving” in an electromagnetic wave are electric and magnetic fields. The plot in Figure 1 illustrates how the electric field strength (E) associated with the wave varies vs. distance. Perpendicular to each electric field wave, is a magnetic field wave.

The value of the electric field strength at a given location refers to the strength of the force that a particle (with a charge of 1 Coulomb) would “feel” at that location; positive and negative values indicate the direction of the force. The SI unit for the electric field strength is Newton per Coulomb, N C−1. The electric field is strongest at the crests (maxi-mum E) and troughs (minimum E); the magnitude of the electric field strength at these locations is shown in Figure 1 as Eo, and is called the amplitude of the wave. Locations where the magnitude of the electric field is zero are called nodes. The wavelength, repre-sented by the Greek letter lambda (λ) in Figure 1, refers to the distance between two con-secutive crests. This is also the distance between two consecutive troughs, or the distance between two nodes that are separated by one node.

Physical Interpretation of Wavelength. The wavelength of an electromagnetic radiation is related to what we perceive as the color of the light. The term visible spectrum refers to the range of possible wavelengths that we can see, 400 nm to 800 nm. The symbol nm stands for nanometer, or 10−9 m. The mnemonic ROYGBIV (Red, Orange, Yellow,...) relates wavelength to color. Red light is on the long wavelength end of the visible spectrum (630-800 nm), whereas violet is on the short wavelength end (about 400 nm).

▼ Example 1

Which color of light corresponds to longer wavelength: red or blue?Answer: Red

Figure 1. Electric Component of Electromagnetic Wave

Most electromagnetic radiation is invisible to us. The term electromagnetic spectrum refers to the entire range of possible wavelengths. Electromagnetic radiation is classified according to wavelength (shortest to longest) into gamma rays, x-rays, ultraviolet, visible, infrared, microwave, and radiowave.

Chapter 10: Quantum Theory 151

▼ Example 2

Which type of electromagnetic radiation corresponds to longer wavelength: microwave or x-ray?Answer: microwave

Physical Interpretation of Frequency. Electromagnetic waves travel at a speed of 2.998 × 108 m/s. For most calculations, we can round this off to 3.00 × 108 m/s. We refer to this as the speed of light and, typically, represent this by the letter c. When the wave has traveled a distance equal to its wavelength, we say that it has completed one cycle. The time it took for this to happen is called the period (T). The number of cycles completed per second is called the frequency of the wave and is just equal to 1/T; frequency is com-monly represented by the Greek letter nu (ν). Since speed is distance traveled per unit time, the speed of the wave is the distance traveled in one cycle (λ) divided by the time for one cycle. Therefore:

cT T

= = ⎛⎝⎜

⎞⎠⎟

=λ λ λν1

We can rearrange this equation to:

νλ

= c

and see that the shorter the wavelength, the higher the frequency. Therefore, frequency is also associated with what we perceive as color. The unit for frequency is reciprocal second (s−1) or Hertz (Hz).

▼ Example 3

Calculate the frequency of green light that has a wavelength of 500.0 nm.Answer: 6.00 × 1014 s−1, or 6.00 × 1014 HzWe simply plug in the speed of light and wavelength into the formula that relates frequency, speed of light, and wavelength. To avoid problems with units, express all quantities in base SI units. Thus, we plug in 500.0 nm as 500.0 × 10−9 m.

νλ

= = ××

= ×−

−−c m s

ms

2 998 10500 0 10

6 00 108 1

9114.

..


▼ Example 4

Which color of light corresponds to higher frequency, red or blue?Answer: blueBlue has shorter wavelength, higher frequency.

Physical Interpretation of Amplitude. The amplitude of the wave, Eo, is related to what we perceive as the intensity of light. Intensity, S, is defined as the energy transported per unit time, per unit area, and is given by the equation:

S EnergyTime Area

PowerArea

c E=⋅

= = 12 0 0

2ε

where, εo (“epsilon naught”, 8.854 × 10−12 C2 N−1 m−2) is the constant known as the permittivity of vacuum and c is the speed of light. The unit for intensity is Joule per second per square meter (J s−1 m−2). A Joule per second is a unit for power, the energy delivered per unit time, and is also called a Watt (W). So, the unit for intensity can also be written as W m−2.

▼ Example 5

A 5.00 mW laser beam has a cross sectional area of 2.00 mm2. What is the intensity of the beam? Calculate the amplitude of electric field component of the electromagnetic wave.Answer: S = 2.50 × 103 W m−2, Eo = 1.37 × 103 N C−1

Intensity is power per unit area; we are given the power (5.00 mW) and the area (2.00 mm2). It is best to express these quantities in terms of unprefixed units when doing calculations. 1 mW = 10−3 W and 1 mm = 10−3 m.

SPowerArea

W

mWm= = ×

( )= ×

−

−

−5 00 10

2 00 102 50 10

3

3 23 2.

..

Rearranging the equation that relates intensity to amplitude, we can solve for the amplitude as

ESc

Wm

C N m m s0

3 2

12 2 1 2 8 1

2 2 2 50 10

8 854 10 2 998 100

= =×( )

×( ) ×

−

− − − −ε

.

. .(( )= × −1 37 103 1. N C


Here is how the units for the electric field strength works out to N C−1.

W m

C N m s

J s

C N m s

N m

C N m

N C N N C N C

−

− − −

−

− − −

− − − −

= =

= = =( )

2

2 1 2 1

1

2 1 1 2 1

2 1 2 2 −−1

Note that a Joule (J) is equivalent to a Newton meter (N m).

10.1.2 The Particulate ModelTowards the end of the 19th century, experiments dealing with light showed behaviors that are inconsistent with the notion of light as an electromagnetic wave. These experi-ments deal with the following phenomena: blackbody radiation, photoelectric effect, atomic and molecular spectra.

In all of these, the “anomaly” is resolved by assuming that the energies of light and matter are quantized. Thus, the theory that was eventually developed, based on these phenomena, came to be known as quantum theory. When we say something is quantized, it means that its values are restricted; it cannot have just any value. In the case of electromagnetic radiation, it turns out that, in these “anomalous” cases, we have to think of the energy as being carried by a stream of particles, now called photons. The energy carried by each photon is given by:

E h hcphoton = =ν

λ

where h is a very small number called Planck’s constant. We say that light energy is quantized because it is only available in multiples of hν. The value of Planck’s constant is 6.626 × 10−34 J s. Max Planck discovered this constant in trying to explain blackbody radiation and it turned out to be a universal constant; the same constant appears in the equations that were successfully used to explain the other phenomena listed above.

▼ Example 6

Which has more energetic photons: radiowave or x-ray?Answer: x-raysX-rays have very short wavelengths, very high frequency, very energetic photons. This is the reason why x-rays are considered dangerous. One should not have too much exposure to x-rays. X-ray photons are energetic enough that they can “knock off”


electrons from atoms in your body, which could lead to all sorts of chemical changes. X-rays are called ionizing radiation. Magnetic resonance imaging (MRI, the other major imaging technique used in medicine) uses radiowaves, which are on the other end of the elec-tromagnetic spectrum. Radiowaves have very low energy photons.

▼ Example 7

Calculate the energy of each photon of red light with a wavelength of 632 nm.Answer: 3.14 × 10−19 J

Ehc J s ms

mphoton = =×( ) ×( )

×= ×

−

−λ

−6.626 10 2.998 10

632 103.14 1

34 8 1

9 00−19J

Ehc J s ms

mphoton = =×( ) ×( )

×= ×

−

−λ

−6.626 10 2.998 10

632 103.14 1

34 8 1

9 00−19J

Note that we replaced nm by 10−9 m.

Because h is a very small number, photon energies in Joules are also very small num-bers. It is often convenient to express photon energies in terms of a smaller energy unit, the electron volt (eV): 1 eV = 1.602 × 10−19 J.

▼ Example 8

A photon has an energy of 3.14 × 10−19 J. Express this value in eV.Answer: 1.96 eV

3 14 101

1 602 101 9619

19..

.× ××

⎛

⎝⎜

⎞

⎠⎟ =−

−JeV

JeV

▼ Example 9

The frequency of radiation used in a microwave oven is 2.45 GHz. If a 600.0 W oven is operated for 60.0 s, how many photons are generated.Answer: 2.22 × 1028 photons1 GHz = 109 Hz or 109 s−1. So, the energy of one photon is

E h J s s Jphoton = = ×( ) ×( ) = ×− − −ν 6 626 10 2 45 10 1 62 1034 9 1 24. . .


A Watt (W) is a unit of power, which is energy per unit time: 1 W = 1 J s−1. Therefore, if we multiply the wattage by the time in seconds, we get the total energy generated by the oven:

Etotal = (600.0 J s−1) (60.0 s) = 3.60 × 104 J

If we have N photons and multiply it by the energy of each photon, we get the total energy. In other words:

Etotal = N Ephoton

or:

NE

EJJ

total

photon

= = ××

= ×−3 60 10

1 62 102 22 10

4

2428.

..

10.2 The Photoelectric EffectThe photoelectric effect is one of the phenomena that could not be explained with the notion that light is a wave. A photoelectron is an electron ejected from a metallic surface when the surface is exposed to light. It was found that:

• there is a minimum frequency (νo) of light needed to cause electrons to be ejected; the threshold frequency depends on the nature of the metallic surface.

• the kinetic energy of the electrons ejected depends on the frequency, but not on the intensity of the light; this was puzzling because the intensity of light is the amount of energy it delivers per unit time per unit area.

• if electrons are ejected, more of them are ejected if the light is made more intense.

Albert Einstein explained the observations as follows. The kinetic energy of the ejected electron is the difference between the energy of the photon and the energy needed to dislodge the electron from the metal. If the individual photons do not have sufficient energy to dislodge electrons, no photoelectrons will be observed regardless of how intense the radiation is. If the photon energies are sufficient, then a higher intensity would mean more photons and, consequently, more photoelectrons produced. But the kinetic energy of the electrons depend only on the photon energy (hence, on the frequency, not on the intensity). Based on the data, Einstein deduced that

• the photon energy is hν, where h is the same constant that Planck earlier discovered when trying to explain blackbody radiation.

• the minimum photon energy needed to dislodge the electron, called the work function of the metal, is hνo.


▼ Example 10

The work function for Al is 4.08 eV. Will photoelectrons be ejected if it is exposed to ultraviolet radiation of frequency 300.0 nm? If so, what would be the kinetic energy of the photoelectrons?Answer: Yes, the kinetic energy of the photoelectron would be 0.05 eVThe photon energy for 300.0 nm is:

Ehc J s m s

mphoton = =×( ) ( )

×

=

− −

−

×

λ

6 626 10 2 998 10

300 0 10

6 62

34 8 1

9

. .

.

. 22 10 19× − J

which corresponds to 4.134 eV. The energy of the photoelectron would be the difference between the photon energy and the energy needed to remove the electron:KE of photoelectron = 4.134 eV − 4.08 eV = 0.05 eV

10.3 Atomic and Molecular SpectraWhen light is passed through a prism, the prism causes the waves corresponding to different wavelengths to bend at different angles. If the light comes from a source where there is a very high concentration of particles (atoms, molecules or ions), we tend to get a continuous band of colors, the familiar rainbow of colors. If we pass that light through a sample that has a very low concentration of particles before we pass it through the prism, we are likely to see a pattern of dark lines interspersed within what would otherwise be a continuous band of colors; see Figure 2. We say that light waves (with wavelengths corresponding to the missing colors) were absorbed by the particles in the sample. Thus, the observed pattern is known as the absorption spectrum of the sample. Figure 2 shows the absorption spectrum of hydrogen atoms.

If we were to examine light from a source where the concentration of particles is very low, we are not likely going to get a continuous band of colors. If we project the light that passes through the prism onto a screen, we would see a pattern of bright colored lines with dark regions in between. The pattern depends on what atoms or molecules are


producing the light, and is called the atom or molecule’s emission spectrum. The emission spectrum from hydrogen atoms is shown in Figure 3.

Each atom or molecule has a characteristic spectrum that can serve as its “fingerprint” for identification purposes. Figure 4 shows the absorption spectrum of sodium. Spectra from atoms tend to have sharp lines; thus, they are often referred to as line spectra.

How come atoms and molecules do not absorb or emit a continuous band of colors? The explanation provided by Niels Bohr is as follows:

• Energies of atoms and molecules are quantized. They cannot have just any amount of energy.

• Transitions between allowed energy levels can occur when a photon is absorbed or released.

• The energy of photon must be equal to the difference in energy between two allowed levels (this is known as the Bohr frequency condition). Since the energies are restricted, this means that the energy differences are also restricted. Thus, not every photon energy can be observed, which means that not every wavelength or frequency can be observed.

ΔE E h cphoton= = ⎛

⎝⎜⎞⎠⎟λ

Figure 2. Absorption Spectrum of Hydrogen

Figure 3. Emission Spectrum of Hydrogen

Figure 4. Absorption Spectrum of Sodium


▼ Example 11

Theory predicts that the two lowest allowed energy levels of hydro-gen are −13.6 eV and −3.40 eV, respectively. What is the energy of the photon that will be absorbed by a hydrogen atom for it to be excited from −13.6 eV to −3.40 eV? What wavelength of light does this correspond to?Answer: 121.6 nm

Ephoton = ΔE = (−3.40 eV) − (−13.6 eV) = 10.2 eVWe rearrange:

E hc

photon = ⎛⎝⎜

⎞⎠⎟λ

to solve for λ:

λ = =×( ) ×( )

×( ) =− −

−h cE

J s m s

J

6 626 10 2 998 10

10 2 1 602 101 2

34 8 1

19

. .

. .. 22 10 7× − m

or 122 nm. Note that we have substituted 1.602 × 10−19 J for eV so that our answer comes out in meters (m). This spectral line is observed experimentally, and is known as the Lyman alpha line.

10.4 Matter WavesLight exhibits wave-like behavior, as well as particle-like behavior. The term wave-particle duality is used to describe this dual nature of light. Nature tends to be symmetrical, so it makes sense to expect that what we normally consider as particles (like electrons) would also exhibit wavelike behavior. In the 1920s, Louis de Broglie suggested that matter, which we normally think of as being made of particles, may also exhibit wave-like behavior. In 1925, Davisson and Germer provided experimental support for this idea: diffraction patterns were observed for a beam of electrons. Diffraction is a phenomenon associated with waves.

Figure 5 illustrates diffraction of light, which happens when light is passed through a narrow slit. Instead of showing up only around where the slit is located, the light spreads out. It spreads out more the narrower we make the slit. We also observe dark spots in between bright spots. This can only be explained if we think of light as a wave; see Figure 6. As waves emerge from the barrier (through the slit), they bend. As a result some waves will have farther to travel than others before they hit the wall on the other side. At certain


Figure 5. Single Slit Diffraction

Figure 6. Explaining Single Slit Diffraction

from http://en.wikipedia.org/wiki/File:Diffraction1.png GNU Free Documentation License

from http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslitd.html#c1

http://en.wikipedia.org/wiki/File:Diffraction1.png

http://en.wikipedia.org/wiki/File:Diffraction1.png


angles, the shifting of the waves results in what is known as destructive interference and we see a dark spot on the wall where this occurs.

It is now generally accepted that wave characteristics apply to all matter. The so-called de Broglie wavelength (λ) associated with a particle is inversely proportional to the momentum of the particle. Momentum is the product of mass (m) and speed (v).

λ = =hp

hm v

If a higher value for one quantity implies a lower value for another, the two quantities are said to be inversely proportional. When two quantities are inversely proportional, their product is a constant, which is called the proportionality constant. In the case we have here, the product of wavelength and momentum is Planck’s constant (h). Because of this, wave behavior is only discernible for very small particles (very small masses). Diffraction, for example, is observed when the particle encounters obstacles that are separated by a distance comparable to its de Broglie wavelength. For large masses, the de Broglie wave-length tends to be very, very small to be detectable.

▼ Example 12

Calculate the de Broglie wavelength of a 1.00 kg object moving at 1.00 m/s.Answer: 6.63 × 10−34 m

λ = = ×

( ) = ×−

−−h

m vJ s

kg m sm

6 626 10

1 00 1 006 63 10

34

134.

( . ) ..

This is very, very, very small. This is much smaller than the diam-eter of a typical nucleus (10−13 − 10−14 m), or even an electron (10−18 m). It is undetectable. We expect, in this case, that classi-cal physics would adequately describe the particle.Note: here is how the unit in the answer works out to meters (m).

Jskg m s

kg m s skg m s

kg m m skg m s

m2

−

−

− −= = =−

1 1 1

2 1

▼ Example 13

Calculate the de Broglie wavelength of an electron (mass = 9.11 × 10−31 kg) moving at a speed of 1.00 × 106 m s−1.


Answer: 7.27 × 10−10 m

λ = = ××( ) ×( ) = ×

−

− −−h

m vJ s

kg m s

6 626 10

9 11 10 1 00 107 27 10

34

31 6 1

10.

. .. mm

This wavelength is comparable to x-rays and to the distance between the nuclei of adjacent atoms in a solid, which is typically about 1-2 Angstroms; one Angstrom is 10−10 m. We expect the electrons here to exhibit wavelike behavior similar to x-rays.

10.5 The Uncertainty PrincipleThe Uncertainty Principle is a consequence of wave-particle duality. It says that it is impossible to precisely determine the location and velocity of particles at the same time. The uncertainties are inconsequential in our macroscopic world, but tend to be significant in the world of the very small. For example, to see anything, you need light. Light bounc-ing off an object travels to the detector (such as your eyes), which tells you where the object is located. If we were to try to locate an electron, we would need to bounce a pho-ton off it. Since the photon behaves like a particle, the act of locating an electron affects the electron’s motion. The impact is insignificant for macroscopic objects.

We could try to precisely locate electrons by passing them through a narrow slit. The narrower we make the slit, the more we know about the location of the electrons that pass through the slit. However, we find that the narrower we make the slit, the less we know about the direction of motion as the electrons emerge from the slit; this manifested by the spreading out of the observed diffraction pattern. This means we lose information about the velocities of the electrons by attempting to more precisely locate them.

Wave-particle duality and the uncertainty principle resolve the problem regarding the stability of the nuclear atom. Classical mechanics predicts that the electron will crash into the nucleus and this will happen so fast that the nuclear structure suggested by Rutherford’s experiment could not be stable. The uncertainty principle suggests that the closer an electron gets to the nucleus, the less uncertain we would be about its location, the more uncertain we would be about its velocity (how fast it is moving and where it is headed), just like the electron beam that goes through a narrow slit spreads out more the narrower we make the slit.

10.6 Quantum MechanicsBy the 1920s, it became evident that classical physics is just an approximation. This approximation happens to work well in the macroscopic world we deal with everyday.


However, a more general theory is needed in order to account for observations we make on the world of the very small. Thus, quantum mechanics was invented. Quantum Mechanics, as formulated by Erwin Schrodinger, postulates (assumes) that a mathemati-cal function (called the wavefunction) is associated with particles and that all accessible information about the system can be obtained from this wavefunction. The Schrodinger equation, which is also a postulate, is an example of what mathematicians call a differential equation; when we solve it, we get a function (the wavefunction). Quantum Mechanics is consistent with Heisenberg’s Uncertainty Principle. It can be described as a probabilistic theory, as opposed to one that is deterministic. In Quantum Mechanics, we do not try to locate particles; we only try to calculate probabilities or likelihood of finding particles in whatever region we are interested in.

For the macroscopic world, Quantum Mechanics makes essentially the same predic-tions as classical physics (although it would be more cumbersome to use it in this case since classical physics would work just fine). Otherwise, it would be unacceptable. The requirement that Quantum Mechanics should be consistent with classical physics where classical physics is known to be valid is called the Bohr Correspondence Principle.


TEST YOURSELF


1. Illustrated below is a snapshot of the electric field component of an electromagnetic wave as it travels through space. At which of the locations indicated would a charged particle experience the strongest force?

2. Which of the following wavelengths is associated with the electromagnetic radiation that humans can see?

A. 500 nm B. 300 nm C. 0.0005 cm

3. Which of the following regions of the electromagnetic spectrum has the longest wavelengths?

A. x-rays B. visible C. microwave D. radiowave

4. What is the frequency of the electromagnetic wave with a wavelength of 250 nm? A. 1.2 × 106 Hz B. 1.2 × 1015 Hz C. 6.0 × 1014 Hz

5. What is the energy of a photon of ultraviolet radiation with a wavelength of 250 nm? A. 5.0 eV B. 2.5 eV C. 4.0 × 10−2 eV

6. Which of the following regions of the electromagnetic spectrum has the most energetic photons?

A. x-rays B. visible C. microwave D. radiowave

7. Which of the following characteristics of electromagnetic radiation is not associated with what we perceive as color?

A. wavelength B. frequency C. wave speed D. photon energy



8. How many photons are produced by a 5.0 mW green laser beam (λ = 500.0 nm) in 20.0 s?

9. If a metallic surface with a work function of 2.3 eV is exposed to light, electrons will not be ejected if the photon energy is

A. 2.1 eV B. 2.5 eV

10. If electrons are not ejected from a metallic surface when exposed to blue light, we expect that exposure to red light would

A. cause electrons to be ejected B. not cause electrons to be ejected

11. If a metallic surface with a work function of 2.5 eV is exposed to light with photon energy of 2.8 eV, the maximum kinetic energy of the electrons is...

A. 0.3 eV B. 5.3 eV

12. An atom loses 2.0 eV of energy by emitting light. The energy of the photon released is...

A. 1.0 eV B. 2.0 eV C. 4.0 eV

13. The three lowest allowed energy levels of a particle are 0.5 eV, 1.5 eV, and 2.5 eV. If the particle happens to have an energy of 0.5 eV, it will not absorb a photon with energy of...

A. 0.5 eV B. 1.0 eV C. 2.0 eV

14. The lowest allowed energy levels of a particle are: a, 4a, 9a, and 16a. Which of these cannot possibly be the energy of a photon emitted by a particle with energy of 9a?

A. 5a B. 7a C. 8a

15. Theory predicts that the allowed energies of hydrogen are given by the formula:E = (−13.6 eV) / n2, where n = 1, 2, 3, …

which of these photon energies cannot be absorbed by a hydrogen atom that has an energy of −13.6 eV?

A. 5.1 eV B. 12.1 eV C. 13.6 eV

16. The average speed of N2 molecules at 298K is 475 m s−1. Calculate the de Broglie wavelength of the average N2 molecule (mass = 4.65 × 10−26 kg). Would you expect a beam of N2 molecules moving at 475 m s−1 to be diffracted when it encounters atoms that are about 10−10 m apart?

165

11CHAPTER

Atomic Orbitals

In this chapter, we examine the quantum mechanical description of electrons in atoms and monatomic ions.

11.1 Wavefunctions and OrbitalsOne of the postulates of modern quantum theory is that anything that can be known about an atom or molecule can be obtained from a mathematical function, called a wavefunction, which is customarily represented by the Greek letter psi (ψ). This, of course, assumes that Mother Nature gives us access to this information. The wavefunction is determined by solving what is known as the Schrodinger equation.

▼ Example 1

True or False. If we know ψ for an electron in an atom, then at any given time, we can use ψ to precisely determine where it is, how fast it is moving, and in which direction it is headed.Answer: FalseExperimental evidence suggests that it is not possible to precisely determine, at the same time, both velocity and location of an electron. This is known as Heisenberg’s Uncertainty Principle. This is the reason why the notion of electrons moving around in well-defined orbits around the nucleus is no longer acceptable.

An orbital is a wavefunction that gives us information about an electron. Among the information we can obtain from an orbital is a description of the region around the nucleus where we are most likely to find the electron. Thus, it is customary to refer to an electron as being “in an orbital” or as “occupying an orbital”.


11.2 Quantum NumbersFor each atom, ion, or molecule, quantum theory gives us an infinite set of orbitals that we can use to describe the electrons. It is convenient to refer to these orbitals by a set of numbers (called quantum numbers) that is unique for each orbital. Quantum numbers, which arise when we solve the Schrodinger equation, are restricted numbers; they cannot have just any value. For orbitals describing electrons in atoms and monatomic ions, the quantum numbers and their allowed values are as follows:

• n, the principal quantum number, can only be a positive integer: n = 1, 2, 3, …• l (“ell”), the orbital angular momentum quantum number (also known as azimuthal

quantum number or orbital quantum number) can only be a non-negative integer that is less than n: l = 0, 1, …, n − 1

• m (or ml, “em sub ell”), the magnetic quantum number can only be an integer from −l to +l.

▼ Example 2

Which of the following is an allowed value for the quantum number n?A. 0 B. 0.5 C. 2 D. −3

Answer: C. 2The number 0.5 is not allowed because it is not an integer. The numbers 0 and −3 are integers but they are not positive.

▼ Example 3

Which of the following is an allowed value for the quantum num-ber l if n = 2?A. −1 B. 1.2 C. 0 D. 2

Answer: C. 0If n = 2, the only allowed values of l are 0 and 1.

▼ Example 4

Which of the following is an allowed value for the quantum num-ber m if n = 3 and l = 2?A. −3 B. 1.5 C. 2 D. 3

Answer: C. 2If l = 2, then, the only allowed values of m are −2, −1, 0, 1, and 2.

Chapter 11: Atomic Orbitals 167

11.3 Shells and SubshellsOrbitals can be classified into shells or levels. All orbitals that have the same principal quantum number (n) are said to belong to the same shell or level.

Orbitals can also be classified into subshells or sublevels. All orbitals in the same shell that have the same orbital angular momentum quantum number (l) are said to belong to the same sublevel or subshell.

Subshells are often specified by writing a code letter for l next to the value of n. This is known as the spectroscopic notation. The code letters are:

• s if l = 0• p if l = 1• d if l = 2• f if l = 3• g if l = 4

▼ Example 5

What is spectroscopic notation for the subshell with n = 2 and l = 1?Answer: 2pThe number 2 is the value of n. The letter p is the code letter for l = 1. We call this subshell the “2p subshell.”

▼ Example 6

How many orbitals belong to a d subshell?Answer: 5The letter d is the code letter for l = 2. For l = 2, the allowed values of m are −2, −1, 0, 1, and 2. Each of these corresponds to one orbital.

In general, the number of orbitals that belong to the same subshell is equal to 2l + 1:• If l = 0, 2(0) + 1 = 1; there is only one orbital in an s subshell.• If l = 1, 2(1) + 1 = 3; there are three orbitals in a p subshell.• If l = 2, 2(2) + 1 = 5; there are five orbitals in a d subshell.• If l = 3, 2(3) + 1 = 7; there are seven orbitals in an f subshell.

We can also say that s orbitals come in sets of 1, p orbitals come in sets of 3, d orbitals come in sets of 5, etc. Each orbital in a subshell has a unique m quantum number.


We can use spectroscopic notation to refer to a subshell or to any one of the orbitals in that subshell. So, when we say “2p orbital,” we mean any one of the three orbitals in the 2p subshell. If we want, we can specify the m, as in 2p+1; this would be the orbital in the 2p subshell that has m = +1.

▼ Example 7

How many orbitals belong to the n = 2 shell? List them.Answer: 4The possible sets of quantum numbers for n = 2 are:

n l m

2 0 0

2 1 −1

2 1 0

2 1 −1

One of the orbitals is in the 2s subshell (n = 2, l = 0, m = 0). This orbital is called the 2s orbital.Three of the orbitals are in the 2p subshell (n = 2, l = 1). The m quantum number distinguishes the orbitals. These orbitals are called 2p+1, 2p0 and 2p−1.

In general, the nth shell has n2 orbitals:• For n = 1, n2 = 1; there is only one orbital (n = 1, l = 0, m = 0).• For n = 2, n2 = 4; there are 4 orbitals; 1 in the 2s, 3 in the 2p; 1 + 3 = 4.• For n = 3, n2 = 9; there are 9 orbitals: 1 in the 3s, 3 in the 3p, 5 in the 3d; 1 + 3 + 5 = 9.

11.4 Information from OrbitalsThe value of ψ itself, which could be a complex number, has no physical significance. To obtain information about the electron, we need to perform mathematical operations on ψ, depending on what information we need.

11.4.1 Probability of Finding ElectronThe square of the absolute value of ψ is assumed to be a probability density function; this assumption is called the Born Postulate. From this, we can calculate the probability of finding the electron in any particular region.


From ψ, we can obtain the so-called radial distribution function or surface density function, S(r), which tells us how the probability of finding the electron depends on the distance (r) from the nucleus, as illustrated in Figure 1 for the 1s and 2s orbitals of hydro-gen. The value of r where the radial distribution reaches a maximum value corresponds to the most probable distance of the electron from the nucleus. Examining Figure 1 carefully, we find that if the electron is described by the 1s orbital, its most probable distance from the nucleus is 52.9 pm. If the electron is in the 2s orbital, its most probable distance is farther away.

The most probable distance of the 1s electron of H, 52.9 pm, is called the Bohr radius. It is the same distance that Bohr, in an earlier model, predicted to be the first allowed orbit for the electron in a hydrogen atom. Bohr’s model is no longer accepted because it assumes well-defined circular orbits for electrons, which is inconsistent with the Heisen-berg uncertainty principle. With modern quantum theory, we now say that the Bohr radius is just the most probable distance of the electron from the nucleus for an electron in the 1s orbital of hydrogen; the electron can actually be anywhere from very close to the nucleus to very far away.

We can deduce from Figure 1, that on average, the electron would be closer to the nucleus if it is in the 1s orbital than if it is in the 2s orbital. We can actually calculate the aver-age distance, <r>, of an electron from the nucleus if it is the only electron in the atom or ion:

r o= + −+( )⎡

⎣⎢

⎤

⎦⎥

⎡

⎣⎢⎢

⎤

⎦⎥⎥

nl ln

aZ

221 1

21

1

Figure 1. Radial Distribution Function for H (1s and 2s)


where ao is the Bohr radius (52.9 pm), and Z is the nuclear charge (the number of protons in the nucleus). For atoms with more than one electron, we can estimate <r> by replac-ing the nuclear charge Z by an effective nuclear charge Zeff , a smaller number, since the electron is not going to feel the “full force” of the nucleus due to the presence of other electrons.

▼ Example 8

For which of these orbitals is the distance of the electron, on average, farthest from the nucleus?A. 2s of H B. 2p of H C. 2s of He+

Answer: A• If we plug in n, l, and Z into the formula for <r>, we find that• For H, 2s: Z = 1, n = 2, l = 0: <r> = 6 ao• For H, 2p: Z = 1, n = 2, l = 1: <r> = 5 ao• For He+, 2s: Z = 2, n = 2, l = 1: <r> = 3 ao

Comparing 2s and 2p in H, we find that the electron will (on aver-age) be farther from the nucleus in the 2s orbital.Comparing 2s in He+ and 2s in H, we find that the electron will (on average) be farther from the nucleus in H. In He+, there are 2 protons in the nucleus pulling the electron in; in H, there is only one proton.

In general, the average and most probable distance depends on Z, n and l:• For the same Z and l, the larger the n, the larger the average distance.• For the same Z and n, the larger the l, the smaller the average distance.• For the same n and l, the larger the Z, the smaller the average distance.

Comparisons of most probable distances will show the same pattern.We can use ψ to generate plots that would help us visualize the probability of finding

the electron. In Figure 2, the probability of finding the electron assigned to a 3p0 orbital is represented by an electron cloud (or charge cloud); regions where the cloud is thicker represent regions where the probability of finding the electron is higher.

A more commonly used representation is called a boundary surface plot; see Figure 3. A 3-dimensional surface is drawn such that the probability of finding the electron (being described by the orbital) is low outside the surface. The surface is chosen so that |ψ|2 is


constant anywhere on the surface. The algebraic sign (+ or −) of the function is oftentimes indicated, or color-coded. The shape of the boundary surface resembles the shape of the electron cloud and depends on the quantum numbers l and m. The s orbitals are spheri-cally symmetrical. The p orbitals can be described as consisting of two lobes. The size of the surface depends on the quantum number n; the larger the n, the farther the surface extends from center (where the nucleus is located).

Evaluation of orbitals with nonzero m generally yields complex numbers. It is common practice to combine orbitals with the same |m| so that an equivalent pair of orbitals that yield real numbers are obtained. For example, the 2p+1 and 2p−1 orbitals can be combined so that we do not need to deal with complex numbers; the resulting orbitals are called 2px and 2py. The p orbitals shown in Figure 3 are labeled px, py and pz, instead of p+1, p−1 and p0. The 2pz is, in fact, another name for 2p0. Similarly, for l = 2, combinations of d+1 and d−1 orbitals give rise to dxz and dyz orbitals; combinations of d+2 and d−2 orbitals give rise to dx2 − y2 and dxy orbitals; the dz2 orbital is identical to the d0 orbital (l = 2, m = 0).

Figure 2. Electron Cloud Representation of the 3p0

from http://www.jce.divched.org/JCEDLib/LivTexts/pChem/JCE2005p1880_2LTXT/QuantumStates/Bookfolder/L25OrbitalShapes.htm

http://www.jce.divched.org/JCEDLib/LivTexts/pChem/JCE2005p1880_2LTXT/QuantumStates/Bookfolder/L25OrbitalShapes.htm

http://www.jce.divched.org/JCEDLib/LivTexts/pChem/JCE2005p1880_2LTXT/QuantumStates/Bookfolder/L25OrbitalShapes.htm


Figure 3. Boundary Surface Plots of Real Orbitals

from http://www.chm.davidson.edu/ronutt/che115/AO.htm

11.4.2 Orbital EnergiesAn orbital gives us information about the energy of an electron. For hydrogen and ions with only one electron (such as He+), the formula for orbital energy predicted by modern quantum theory is:

E R ZnH= − ⎛

⎝⎜⎞⎠⎟

2

where RH is called the Rydberg constant (13.6 eV or 2.18 × 10−18 J), n is the principal quantum number, and Z is the nuclear charge (in atomic units; the number of protons in the nucleus).

▼ Example 9

In a hydrogen atom, which orbital describes an electron with higher energy: 3s or 3p?Answer. In both cases, n = 3. So the orbital energy is the same.

http://www.chm.davidson.edu/ronutt/che115/AO.htm


The potential energy is defined to be zero when the electron is at infinite distance from the nucleus. Therefore, a negative value for the energy means that the electron must gain energy from somewhere in order for it to escape from the atom. The minimum energy required is called the ionization energy and it is simply −E. In other words, the ionization energy (IE) is:

IE E R ZnH= − = ⎛

⎝⎜⎞⎠⎟

2

If the energy available is more than the minimum required, the excess energy becomes kinetic energy of the freed electron. An experimental technique known as photoelectron spectroscopy allows us to verify predictions of quantum theory; patterns in the measured kinetic energies of the electrons removed from atoms and molecules give us information about the orbital energies.

▼ Example 10

What is the minimum energy needed to remove an electron from the 2s orbital of H? If this electron were to encounter a photon with energy of 5.0 eV, what would be its kinetic energy once it leaves the atom?Answer: 3.4 eV, 1.6 eVThe energy of the electron in the 2s orbital of H is:

E RZn

eV eVH= − ⎛⎝⎜

⎞⎠⎟

= − ( ) ⎛⎝⎜

⎞⎠⎟

= −2 2

13 612

3 40. .

since Z = 1 and n = 2.This means it has to gain at least 3.40 eV to be able to leave the atom.The photon energy is 5.0 eV: 5.0 eV − 3.4 eV = 1.6 eV.The extra 1.6 eV of energy becomes kinetic energy of the freed electron.

▼ Example 11

From which orbital would it be hardest to remove an electron?A. 1s of H B. 1s of He+ C. 2s of He+

Answer: B• For 1s of H, Z = 1, n = 1: E = −13.6 eV (1/1)2 = −13.6 eV IE = 13.6 eV


• For 1s of He+, Z = 2, n = 1: E = −13.6 eV (2/1)2 = −54.4 eV, IE = +54.4 eV

• For 2s of He+, Z = 2, n = 2: E = −13.6 eV (2/2)2 = −13.6 eV, IE = +13.6 eV

From orbital energies, we can calculate the energy of photons that can be absorbed or emitted. The energy of the photon must be equal to the difference between allowed ener-gies; this is called the Bohr frequency condition.

E Ephoton = Δ

We can also calculate the wavelength of the light (λ) since:

Eh c

photon =λ

▼ Example 12

What is the energy of the photon released when an electron in the 3p orbital of H loses energy and ends up in the 2s orbital? What is the color of the light produced?Answer: 1.89 eV or 3.03 × 10−19 J, red lightThe energy of the photon must be equal to the difference in the orbital energies for 3p and 2s.

E E E R Rphoton 3p 2s H H= − = − ⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

− − ⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=13

12

2 2

−− −⎛⎝⎜

⎞⎠⎟

= = × −R eV JH

19

14

1 89 3 03 10 19. .

E E E R Rphoton 3p 2s H H= − = − ⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

− − ⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=13

12

2 2

−− −⎛⎝⎜

⎞⎠⎟

= = × −R eV JH

19

14

1 89 3 03 10 19. .

where we used 13.6 eV or 2.18 × 10−18 J for the Rydberg con-stant (RH). To calculate the wavelength, it is preferable to use the value of the photon energy in the SI unit (J).

λ = =×( ) ×( )

×=

− −

−

hcE

J s m s

Jphoton

6 626 10 2 998 10

3 03 106 56

34 8 1

19

. .

.. ×× =−10 6567m nm

λ = =×( ) ×( )

×=

− −

−

hcE

J s m s

Jphoton

6 626 10 2 998 10

3 03 106 56

34 8 1

19

. .

.. ×× =−10 6567m nm

For atoms with more than one electron, calculating the actual energy of the orbitals is complicated; we can approximate by replacing the nuclear charge Z by an effective nuclear charge (Zeff). The effective nuclear charge will depend on the quantum number l. Therefore the orbital energy will depend on both quantum numbers n and l.


11.4.3 Magnetic PropertiesAny type of curved motion of a charged particle generates a magnetic field. Thus, the motion of electron around the nucleus generates a magnetic field. The magnetic quantum number, m, gives us information about the magnetism generated by the electron.

Imagine an electron moving around in a circle as illustrated in Figure 4. A particle that is moving in a circle is said to have an angular momentum vector that is pointing perpendicu-lar to the plane of the circle; for the motion indicated in Figure 4, the red arrow indicates the direction of the angular momentum vector. If the particle happens to be an electron, then it is said to have a magnetic moment vector that is pointing in the direction opposite that of the angular momentum vector. In other words, if an electron were moving as shown in Figure 4, we can imagine that there is a tiny invisible bar magnet in the middle of the circle and the red arrow would be pointing to the south pole of that magnet.

The magnetic moment, μ, a measure of the strength of the magnetism, depends on the area of the circle and how fast the charged particle is moving; it is quantized and the allowed values depend on the orbital quantum number l:

= +( )l l B1 μμ

where μB is a constant called the Bohr magneton. However, because of the Uncertainty Principle, the direction of the magnetic moment vector is not well defined. In other words, we do not exactly know where the red arrow in Figure 4 would be pointing. We do know that the angle it makes with the z-axis depends the quantum number m. Therefore, for a given l, we say that there are 2l + 1 ways that the magnetism (due to the orbital motion of the electron) can be oriented, as illustrated in Figure 5 for l = 2. The five possible orientations correspond to the five possible values of the magnetic quantum number, m, for l = 2. Each of the arrows shown in Figure 5 represents a possible orientation of the

Figure 4. Angular Momentum of Particle Moving in a Circle

From http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magmom.html#c1


north-south line of the magnetism. The arrow should not be taken as stationary; it could be anywhere on the cone that is superimposed on it in the sketch. We say that it is precess-ing about the z-axis on this cone of uncertainty.

▼ Example 13

How many ways can the magnetism due to the orbital motion of an electron in a 2p subshell be oriented?Answer: 3Since p is the code letter for l = 1, there are three possible values for m for an electron in the 2p subshell: m = −1, 0, or +1. These correspond to the three 2p orbitals that could be used to describe the electron, and these correspond to the three different ways that the magnetism due to the orbital motion can be oriented.

11.5 Electron SpinThe magnetic properties of atoms suggest that electrons have an intrinsic magnetism, in addition to the magnetism that is generated by their motion around the nucleus. By intrin-sic, we mean that they are magnetic just by being electrons. Since magnetism is associated with a circular type of motion of a charged particle, it is useful to imagine the intrinsic magnetism of an electron as resulting from a spinning motion. The motion of an elec-tron around the nucleus is analogous to that of a planet revolving around the sun, while electron spin is analogous to the rotation of the planet around its own axis. To be consis-tent with experimental results, we have to assume that the magnetism due to electron spin

Figure 5. Five Possible Orientations of Magnetic Moment for l = 2

adapted from http://physchem.ox.ac.uk/~hill/tutorials/qm2_tutorial/am/index.html


can be oriented in only two possible ways. An electron in these two possible cases is also referred to as having a spin quantum number, ms, of either +1/2 (spin up) or −1/2 (spin down). Thus, an electron in an atom is completely described by specifying four quantum numbers (n, l, m, ms). The quantum numbers n and l tell us the shell and subshell, the quantum number m tells us which orbital, and ms tells us the spin.

▼ Example 14

How many ways can the magnetism due to spin of an electron in a 3d orbital be oriented?Answer: 2For any electron, there are only two ways that the magnetism due to spin can be oriented. We refer to these ways as “spin up” (ms = +1/2) and “spin down” (ms = −1/2).

11.6 Pauli’s Exclusion PrincipleThere is a limit to the number of electrons that we can assign to an orbital, and that limit is 2. If we do assign two electrons to an orbital, one must be “spin up”, the other must be “spin down.” This is known as Pauli’s Exclusion Principle. Another way of stating Pauli’s Exclusion Principle is to say that “no two electrons in an atom can have the same set of quantum numbers.” If two electrons have the same n, l, and m (same orbital), they must have opposite spins (one must have ms = +1/2, the other must have ms = −1/2)

▼ Example 15

How many electrons can be assigned to a 4p orbital?Answer: 2For any orbital, the limit is 2; one electron has ms = +1/2, the other has ms = −1/2.

▼ Example 16

How many electrons can be assigned to 4p orbitals?Answer: 6The 4p orbitals come in sets of 3, and a maximum of 2 electrons can be assigned per orbital. All six will have n = 4, l = 1. They will differ in m and ms values; the six possible (m, ms) values:(1, 1/2), (0, 1/2), (−1, 1/2)


and (1, −1/2), (0, −1/2), (−1, −1/2).

The first three are “spin up”; the other three are “spin down.”

▼ Example 17

How many electrons can be assigned to a 4p subshell?Answer: 6A 4p subshell consists of three 4p orbitals. Maximum “occupancy” is 2 electrons per orbital; 2 × 3 = 6.

▼ Example 18

How many electrons can be assigned to the n = 3 shell?Answer: 18In the n = 3 shell, there are 3 subshells (3s, 3p, 3d).The 3s subshell consists of 1 orbital.The 3p subshell consists of 3 orbitals.The 3d subshell consists of 5 orbitals.The total number of orbitals is 1 + 3 + 5, or 9.Maximum occupancy is 2 electrons per orbital; 2 × 9 = 18In general, since there are n2 orbitals in the nth shell, the number of electrons that can be assigned to a shell is 2n2.


TEST YOURSELF

For answers and other study resources, see: http://i-assign.com/ebook/answers/chapter11.htm

1. Which of the following are allowed values of the quantum numbers n and l, respectively?

A. 1.5 and 0 B. 3 and 3 C. 2 and 1 D. 0 and 1

2. Which of the following is not an allowed value of the magnetic quantum number (m) for an electron in a 3d orbital?

A. 3 B. 2 C. 0 D. −1

3. What is the value of the orbital quantum number for an electron in a 3p orbital? A. 0 B. 1 C. 2 D. 3

4. Which of the following is not a valid name for an atomic orbital? A. 1s B. 2s C. 2d D. 3p

5. How many ways can the intrinsic magnetism of an electron in a 4d orbital be oriented?

A. 2 B. 3 C. 5 D. 10

6. What is the maximum number of electrons that can be assigned to a 2p orbital of an atom?

A. 2 B. 3 C. 6 D. 10

7. How many sets of quantum numbers can an electron in a 4p subshell have? A. 2 B. 3 C. 6 D. 10

8. Which of the following can accommodate the most number of electrons? A. a 4f orbital B. the 3d subshell C. the n=2 shell D. a 3d orbital

9. How many ways can the magnetism generated by the orbital motion of an electron in a 3p orbital be oriented?

A. 1 B. 3 C. 5 D. 7



10. Calculate the wavelength required to remove an electron from the 1s orbital of Li2+ if the electron is to have a kinetic energy of 2.0 eV.

11. Calculate the energy needed to excite an electron from the 1s orbital to the 2p orbital of H.

12. Atomic orbitals are mathematical functions that allow us to determine all of the following except …

A. the location and velocity of an electron at any given time B. the probability of finding an electron in any given region C. how an atom would behave in a magnetic field D. the energy needed to remove an electron from an atom

181

12CHAPTER

Electron Configuration

12.1 Ground and Excited ConfigurationsThe term electron configuration refers to a listing of the number of electrons assigned to subshells of an atom. The number of electrons assigned to each subshell is indicated as a superscript next to the name of the subshell; unwritten numbers are implied to be 1. If no electron is assigned to a subshell, the name of the subshell is omitted.

▼ Example 1

What does the electron configuration 1s2 2s2 2p1 represent? Answer: The configuration 1s2 2s2 2p1 refers to assignment of 5 electrons to three different subshells: 2 electrons are assigned to the 1s subshell, 2 electrons are assigned to the 2s subshell and one electron is assigned to the 2p subshell. This electron configuration can also be written as 1s2 2s2 2p.

▼ Example 2

Which atom or ion can have an electron configuration of 1s2 2s2 2p1? Answer: Strictly speaking, any monatomic species that has 5 electrons, such as B, C+, N2+, Be−, Li2−.

The term ground state configuration refers to the most stable (lowest energy) way of assigning the electrons. All other ways are higher energy or excited state configurations. Excited state configurations are unstable and can be formed when an atom absorbs light or collides with other particles. Excited atoms eventually revert to their ground state configura-tion either by giving off light or as a result of collisions with other particles. If ground or excited is not specified, we assume that the term electron configuration refers to the ground state.


▼ Example 3

Which of the following is an excited electron configuration of H?A. 1s1 B. 1s2 C. 2p D. none of these

Answer: C. 2p• The configuration 1s1 means one electron assigned to the

1s orbital (subshell). This is the ground state of H because the electron has the lowest energy if assigned to the 1s orbital.

• The configuration 1s2 implies two electrons. H has only one electron.

• The configuration 2p means one electron assigned to the 2p subshell; we can write this as 2p1. This would be an excited state since an electron will not have its lowest possible energy in the 2p subshell.

▼ Example 4

Write ground and excited electron configurations for He.Answer: A helium atom has 2 electrons. Therefore, we need to list how these 2 electrons are to be assigned.• The ground state configuration has both electrons

assigned to the lowest energy subshell, which is the 1s subshell: 1s2

• Any other way of assigning the two electrons would be an excited state. Examples of excited state of He: 1s12s1, 1s12p1, 2s2, 2s12p1, etc.

12.2 Orbital Energies and the Aufbau Principle

One may be tempted to think that the ground state configuration is obtained by simply assigning all electrons to the 1s subshell. That would be incorrect because there is a limit to the number of electrons that can be assigned to an orbital and there is only one orbital in the 1s subshell. Pauli’s Exclusion Principle says that no more than two electrons can be assigned to an orbital.

Chapter 12: Electron Configuration 183

▼ Example 5

When writing electron configurations, what is the maximum number of electrons that can be assigned to the 2p subshell?Answer: 6There are three orbitals in the 2p subshell. The maximum number of electrons that can be assigned to any orbital is 2: 3 × 2 = 6.

For atoms with more than one electron, the ground state configuration is obtained by using the Aufbau principle: assign electrons to lowest energy orbitals first. Whereas orbital energy depends only the principal quantum number (n) for H and ions with only one electron, this is not the case for atoms or ions with more than one electron, where it also depends on the orbital angular momentum quantum number (l). The reason for this is that in atoms with more than one electron, each electron will be shielded from the nuclear charge by the other electrons, and the extent of shielding is different for electrons in different subshells. In H and atoms and ions with only one electron, there is no shielding; the orbital energy is the same for all subshells (within the same shell). It turns out that, in a typical atom:

• Orbitals with higher (n + l) have higher energies.• For orbitals with the same (n + l), the orbitals with higher n have higher energies.

▼ Example 6

In a typical atom, arrange the following orbitals in order of increasing energy: 3p, 4s, and 3d. Answer: 3p < 4s < 3dCompare (n + l ) values:• 3p, n = 3, l = 1, n + l = 4 • 4s, n = 4, l = 0, n + l = 4 • 3d, n = 3, l = 2, n + l = 5

Therefore, among these, 3d has the highest energy. For 3p and 4s, which have the same (n + l ) value, 4s has a higher energy because it has the higher n.

We can use the periodic table as a memory aid for the typical order of filling of orbit-als, as shown in Figure 1. The order is: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p.


Keep these in mind when using the memory aid:• The s block (shown in green) spans 2 columns. We can assign a maximum of

2 electrons to an s subshell since an s subshell consists of only one orbital.• The p block (shown in purple) spans 6 columns. We can assign a maximum of 6

electrons to a p subshell since there are 3 orbitals in a p subshell and each orbital can be assigned to a maximum of 2 electrons: 3 × 2 = 6.

• The d block (shown in blue) spans 10 columns. We can assign a maximum of 10 electrons to a d subshell since there are 5 orbitals in a d subshell and each orbital can be assigned to a maximum of 2 electrons: 5 × 2 = 10.

• The f blocks (shown in red) span 14 columns. We can assign a maximum of 14 electrons to an f subshell, up to 2 for each of the 7 orbitals in the subshell.

▼ Example 7

Write the ground state electron configuration for Scandium (Sc).Answer: 1s2 2s2 2p6 3s2 3p6 4s2 3d1

First, we determine the atomic number of Sc. We can refer to a periodic table and find that it is 21. The number of electrons in an atom is equal to the number of protons, which is equal to the atomic number. Therefore, we need to assign 21 electrons to orbitals following the Aufbau principle. The ground state electron configuration of Sc is:1s2 2s2 2p6 3s2 3p6 4s2 3d1

Figure 1. Using The Periodic Table to Remember Typical Order of Orbital Filling


The numbers in the superscripts should add up to 21. Note that 2p6 does not mean that we are assigning 6 electrons to a 2p orbital. What it means is that we are assigning 6 electrons to the 2p subshell, which consists of 3 orbitals.

▼ Example 8

Write an excited state electron configuration for Scandium (Sc).Answer: one of many possible answers is

1s2 2s2 2p6 3s2 3p6 4s2 4p1

To write an excited state configuration, all we have to do is not follow the Aufbau principle. But we must still follow Pauli’s principle (no more than 2 electrons per orbital). In the answer given here, we assign the last electron to a higher energy orbital (in the 4p subshell) instead of the 3d subshell.

12.3 Patterns in Electron Configuration and the Periodic Table

Consider the electron configurations of Ne and Na. Neon (Ne) has 10 electrons and its electron configuration is:

1s2 2s2 2p6

Sodium (Na) has 11 electrons, one more than Ne. Its electron configuration is 1s2 2s2 2p6 3s1

which is oftentimes abbreviated as: [Ne] 3s1

The symbol [Ne] is an abbreviation for 1s22s22p6 and refers to what we call a noble gas core or, more specifically, the neon core. It is very convenient to write electron configurations for atoms this way: specify the noble gas core, then write the configuration for the remain-ing electrons. A list of ground state electron configurations written this way can be found at the NIST Atomic Spectral Database, and is partially reproduced in Table 1. Note that [He] is the noble gas core for all elements in the second row of the periodic table (elements 3 through 10); [Ne] is the noble gas core for the third row (elements 11 through 18); [Ar] is the noble gas core for the fourth row, etc.


Table 1. Electron Configurations of First 92 Elements from NIST Atomic Spectral Database http://physics.nist.gov/ PhysRefData/DFTdata/configuration.html

1 H 1s1 32 Ge [Ar] 3d10 4s2 4p2 63 Eu [Xe] 4f 7 6s2

2 He 1s2 33 As [Ar] 3d10 4s2 4p3 64 Gd [Xe] 4f 7 5d1 6s2

3 Li [He] 2s1 34 Se [Ar] 3d10 4s2 4p4 65 Tb [Xe] 4f 9 6s2

4 Be [He] 2s2 35 Br [Ar] 3d10 4s2 4p5 66 Dy [Xe] 4f 10 6s2

5 B [He] 2s2 2p1 36 Kr [Ar] 3d10 4s2 4p6 67 Ho [Xe] 4f 11 6s2

6 C [He] 2s2 2p2 37 Rb [Kr] 5s1 68 Er [Xe] 4f 12 6s2

7 N [He] 2s2 2p3 38 Sr [Kr] 5s2 69 Tm [Xe] 4f 13 6s2

8 O [He] 2s2 2p4 39 Y [Kr] 4d1 5s2 70 Yb [Xe] 4f 14 6s2

9 F [He] 2s2 2p5 40 Zr [Kr] 4d2 5s2 71 Lu [Xe] 4f 14 5d1 6s2

10 Ne [He] 2s2 2p6 41 Nb [Kr] 4d4 5s1 72 Hf [Xe] 4f 14 5d2 6s2

11 Na [Ne] 3s1 42 Mo [Kr] 4d5 5s1 73 Ta [Xe] 4f 14 5d3 6s2

12 Mg [Ne] 3s2 43 Tc [Kr] 4d5 5s2 74 W [Xe] 4f 14 5d4 6s2

13 Al [Ne] 3s2 3p1 44 Ru [Kr] 4d7 5s1 75 Re [Xe] 4f 14 5d5 6s2

14 Si [Ne] 3s2 3p2 45 Rh [Kr] 4d8 5s1 76 Os [Xe] 4f 14 5d6 6s2

15 P [Ne] 3s2 3p3 46 Pd [Kr] 4d10 77 Ir [Xe] 4f 14 5d7 6s2

16 S [Ne]3s2 3p4 47 Ag [Kr] 4d10 5s1 78 Pt [Xe] 4f 14 5d9 6s1

17 Cl [Ne] 3s2 3p5 48 Cd [Kr] 4d10 5s2 79 Au [Xe] 4f 14 5d10 6s1

18 Ar [Ne] 3s2 3p6 49 In [Kr] 4d10 5s2 5p1 80 Hg [Xe] 4f 14 5d10 6s2

19 K [Ar] 4s1 50 Sn [Kr] 4d10 5s2 5p2 81 Tl [Xe] 4f 14 5d10 6s2 6p1

20 Ca [Ar] 4s2 51 Sb [Kr] 4d10 5s2 5p3 82 Pb [Xe] 4f 14 5d10 6s2 6p2

21 Sc [Ar] 3d1 4s2 52 Te [Kr] 4d10 5s2 5p4 83 Bi [Xe] 4f 14 5d10 6s2 6p3

22 Ti [Ar] 3d2 4s2 53 I [Kr] 4d10 5s2 5p5 84 Po [Xe] 4f 14 5d10 6s2 6p4

23 V [Ar] 3d3 4s2 54 Xe [Kr] 4d10 5s2 5p6 85 At [Xe] 4f 14 5d10 6s2 6p5

24 Cr [Ar] 3d5 4s1 55 Cs [Xe] 6s1 86 Rn [Xe] 4f 14 5d10 6s2 6p6

25 Mn [Ar] 3d5 4s2 56 Ba [Xe] 6s2 87 Fr [Rn] 7s1

26 Fe [Ar] 3d6 4s2 57 La [Xe] 5d1 6s2 88 Ra [Rn] 7s2

27 Co [Ar] 3d7 4s2 58 Ce [Xe] 4f 1 5d1 6s2 89 Ac [Rn] 6d1 7s2

28 Ni [Ar] 3d8 4s2 59 Pr [Xe] 4f 3 6s2 90 Th [Rn] 6d2 7s2

29 Cu [Ar] 3d10 4s1 60 Nd [Xe] 4f 4 6s2 91 Pa [Rn] 5f 2 6d1 7s2

30 Zn [Ar] 3d10 4s2 61 Pm [Xe] 4f 5 6s2 92 U [Rn] 5f 3 6d1 7s2

31 Ga [Ar] 3d10 4s2 4p1 62 Sm [Xe] 4f 6 6s2

http://physics.nist.gov/�PhysRefData/DFTdata/configuration.html




The term valence means outermost. This is an important term to know since it is very frequently used to describe patterns in electron configuration. What patterns do we observe when we write electron configurations?

All atoms in the same (horizontal) row of the periodic table have the same valence shell. To be more specific, the valence shell of all atoms in row n is the nth shell.

▼ Example 9

In which row of the periodic table would you find an atom with an electron configuration of [Ar] 4s2 3d2? Answer: fourth row; the valence shell is the fourth shell since the 4s orbital belongs to the fourth shell; the 3d orbital belongs to an inner shell. The atom is Ti.If we examine Table 1, we find Ti listed with an electron configuration of [Ar] 3d2 4s2. The 3d2 is written before the 4s2. There is nothing wrong with writing the configuration either way. This is like saying “there are two people in room A and two people in room B.” It would be equivalent to saying “there are two people in room B and two people in room A.”

All atoms in the same group (or vertical column) of the periodic table have the same valence configuration type. If the nth shell is the valence shell, then the valence configura-tions are:

• Group IA. ns1, Examples: Li: 2s1, Na: 3s1 • Group IIA. ns2, Examples: Mg: 3s2, Ca: 4s2 • Group IIIA. ns2np1, Examples: B: 2s22p1 , Ga: 4s23d104p1 • Group IVA. ns2np2, Examples: C: 2s22p2, Ge: 4s23d104p2 • Group VA. ns2np3, Examples: P: 3s23p3, Sb: 5s24d10 5p3 • Group VIA. ns2np4, Examples: O: 2s22p4, Se: 4s23d104p4 • Group VIIA. ns2np5, Examples: Cl: 3s23p5, I: 5s24d105p5 • Group VIIIA. ns2np6, Examples: Ne: 2s22p6 (except He: 1s2) • Transition Metals. ns2(n−1)dx, where x = 1 to 10. Example: Sc: [Ar] 4s23d1

Examine the valence configurations above. Why do you think groups IA and IIA are collectively referred to as the “s block” elements? Why do you think groups IIIA through VIIIA (columns 13 through 18) are called the “p block” elements? Why do you think the transition metals are called the “d block” elements?


▼ Example 10

Which atom has an electron configuration of 1s22s2…etc…4p5?Answer: Bromine (Br). The valence shell is 4. Therefore, the element belongs to period 4 (horizontal row #4). The configuration ends in 4p5, so this element belongs to the fifth column of the p blockcolumn 17 (group VIIA). The configuration given is for an atom of bromine (Br).

For transition metals, the valence configuration is ns2(n−1)dx. Note that (2 + x) gives us the column where the element is located in the periodic table:

• for column 3 (group IIIB): x = 1, or 2 + x = 3 • for column 4 (group IVB): x = 2, or 2 + x = 4• etc.• for column 12 (group IIB): x = 10, or 2 + x = 12

However, there are exceptions; see Table 1. Notable exceptions are Cr, Ni, Cu, Nb, Ru, Rh, Pd, and Pt. A commonly used justification for these exceptions is that electrons are “shifted” in order to have half-filled (d5) or fully-filled d subshells (d10); there seems to be some stability attained with half-filled or fully-filled subshells. But this justification is tenu-ous since the exceptions are not consistent. Consider Ni, Pd, and Pt which all belong to the same column. The electron configurations are:

• For Ni, the electron configuration is [Ar] 4s2 3d8 (no shifting!)• For Pd, the electron configuration is [Kr] 4d10 , instead of [Kr] 5s2 4d8 ; two elec-

trons shifted; “stability of completely filled subshell” justification • For Pt, the electron configuration is [Xe] 6s1 4f14 5d9 (only one electron shifted!)

Characterizing atoms with more than one electron using orbitals is, strictly speaking, an approximation. A much more accurate theoretical treatment is beyond the scope of an introductory Chemistry textbook.

12.4 Orbital DiagramsAn orbital diagram gives more detailed information than an electron configuration. In an orbital diagram, electrons are represented by arrows. Boxes or blanks are used to represent orbitals. Upward-pointing arrows represent electrons with +1/2 spin; downward-pointing arrows represent electrons with −1/2 spin. Figure 2 illustrates one of many possible orbital diagrams that can be drawn for an oxygen atom, which has 8 electrons. Note that Pauli’s principle must be followed: no more than 2 electrons per orbital; if there are two electrons in an orbital, one must be spin-up, the other spin-down.


▼ Example 11

Which atom is represented by the following orbital diagram?

Answer: OxygenThe diagram shows 8 arrows. Each arrow represents an electron.

The orbital diagrams shown for oxygen in the preceding example and in Figure 2 are just two of many (in this case, 15) allowed ways of distributing the electrons. These repre-sent two of the most stable, lowest energy distributions for the ground state configuration of oxygen. To get these distributions, we fill the orbitals in the highest occupied subshell singly, with electrons of the same spin, before putting a second electron (of opposite spin) in any of the orbitals. When we do this, we are following what is known as Hund’s Rule of Maximum Multiplicity. The justification for Hund’s rule is spin correlation; electrons with the same spin repel each other less than electrons with opposite spins.

Figure 2. Orbital Diagram for an Atom with 8 Electrons


▼ Example 12

Is there anything wrong with the orbital diagram shown below?

Answer: No, there is nothing wrong with it. It would be one of many possible orbital diagrams for an excited configuration of an oxygen atom. An oxygen atom has 8 electrons; thus 8 arrows are drawn. Pauli’s Exclusion principle is not violated; none of the orbitals have more than two electrons and the electrons have opposite spins in orbitals that have two electrons. The configuration in this case is 1s2 2s2 2p3 3s1.

12.5 Paramagnetism Magnetic properties of materials are due to unpaired electrons. Materials made of atoms, molecules, or ions that have one or more unpaired electrons are said to be paramagnetic. An oxygen atom, as shown in section 12.4, has unpaired electrons. Oxygen atoms are like tiny magnets. They will be attracted to other magnets. Whenever the electron configuration of an atom has partially filled subshells, the atom is paramagnetic. Atoms that are not paramagnetic are said to be diamagnetic.

▼ Example 13

Explain why N atoms are paramagnetic.Answer: The ground configuration for a nitrogen atom is 1s2 2s2 2p3. The 3 electrons in the 2p subshell are spread out over the three 2p orbitals, with parallel spins.


▼ Example 14

Explain why Be atoms are not paramagnetic.Answer: The ground configuration for Be is 1s2 2s2. All the subshells are completely filled. All the electrons are paired up; the magnetism due to their spins cancel each other out.


TEST YOURSELF

For answers and other study resources, see:http://i-assign.com/ebook/answers/chapter12.htm.

1. Which of the following is not a valid electron configuration for a lithium atom, which has 3 electrons?

A. 1s3 B. 1s2 2s1 C. 1s2 2p1 D. 1s1 2p1 3s1

2. In a typical atom, which of these orbitals is filled last? A. 3p B. 3d C. 4s D. 4p

3. Which of the following is the ground state electron configuration of Sulfur? A. 1s2 2s2 2p6 2d6 B. 1s2 2s2 2p6 3s2 3p4

C. 1s2 2s2 2p7 3s1 3p3 D. 1s2 2s2 2p6 3s2 3p6

4. Which of the following is the ground state electron configuration of Al? A. [Ne] 3s2 3p1 B. [He] 2s2 2p6 3s2 3p1

C. [Mg] 3p1 D. all of the above

5. What is the ground state electron configuration for zirconium? A. [Kr]5s25p2 B. [Kr]5s25d2 C. [Kr]5s24d2

6. How many electrons are unpaired in the ground state of Oxygen? A. 0 B. 1 C. 2 D. 3

7. Which of the following atoms is not paramagnetic? A. Na B. Mg C. C D. Ti


193

13CHAPTER

Periodic Trends in Atomic PropertiesThe way the periodic table is organized allows us to easily predict trends in properties of atoms. These trends can be explained using electron configurations. In this chapter, we will examine ionization energy, atomic size, and electronegativity. We will explain the observed patterns in the charges of naturally occurring ions.

13.1 Ionization EnergyThe first ionization energy of an atom refers to the minimum energy needed to remove an electron from a neutral atom in the gas phase.

▼ Example 1

Write the chemical equation for the first ionization energy of Na.Answer: Na(g) → Na+(g) + e−

Note that the physical states of Na atoms and Na+ ions are both gaseous.

The second ionization energy of an atom refers to the energy needed to remove a second electron. The energies needed to remove additional electrons are called the third ionization energy, fourth ionization energy, fifth ionization energy, etc.

▼ Example 2

What is the term for the energy associated with the following process?Na+(g) → Na2+(g) + e−

Answer: the second ionization energy of Na atomThis is called the second ionization energy of Na atom because Na+ represents an atom that has already lost one electron. We can also call it the “ionization energy of the sodium ion (Na+).”


Whenever the term ionization energy is used, it is customary to assume that it is referring to the first ionization energy. It is also customary to assume that the electron being removed is in the outermost shell of the atom. Higher energies would be needed to remove electrons from inner shells; in fact, measurements of the energies (using an exper-imental technique called photoelectron spectroscopy) needed to remove electrons from inner shells give us experimental validation of the concepts of sublevels (or subshells).

▼ Example 3

The first ionization energy of Na is the minimum energy needed to remove the electron from which orbital of Na?Answer? 3sThe first ionization energy is the minimum energy needed to remove an electron from an atom. The minimum would corre-spond to the value for the easiest electron to remove and that would be the valence (outermost) electron. For, Na, the electron configuration of Na is 1s2 2s2 2p6 3s1, or [Ar] 3s1. Thus, the easiest electron to remove is in the 3s orbital.

A good compilation of ionization energy data can be found at webelements.com. Since the typical values in Joules are very small, ionization energies are often given in kilojoules per mole (kJ mol−1). One mole of atoms is a collection of 6.022 × 1023 atoms, or about 602 billion trillion atoms. In some tabulations, eV is used; 1 eV = 1.602 × 10−19 J.

▼ Example 4

Experimental first ionization energy of Na is reported as 495.8 kJ mol−1. What is the minimum energy needed to remove an electron from the 3s orbital of one sodium atom?Answer: 8.233 × 10−19 J, or 5.139 eVThe given value is for a mole of atoms, or for 6.022 × 1023 atoms. Therefore, we divide this by 6.022 × 1023 to get the value for just one atom:

495 8 106 022 10

8 233 10

8 233 101

3 1

23 119

19

..

.

.

××

= ×

× ×

−

−−

−

J molmol

J

JeVV

JeV

1 602 105 13919..

×=−

495 8 106 022 10

8 233 10

8 233 101

3 1

23 119

19

..

.

.

××

= ×

× ×

−

−−

−

J molmol

J

JeVV

JeV

1 602 105 13919..

×=−

Chapter 13: Periodic Trends in Atomic Properties 195

In Figure 1, we can see that the general trends in ionization energy are:• Across a period (horizontal row): ionization energy increases.• Going down a group (vertical column): ionization energy

decreases.

A good way of summarizing the trend in ionization energy is to say that it is increas-ing towards the top right corner of the periodic table. Helium has the highest ionization energy. However, there are exceptions. Among the main group elements, the most no-table breaks in the trend across a period are between group IIA and IIIA (columns 2 and 13), and from group VA to VIA (columns 15 and 16). There is also a break in the trend through the transition and inner transition metals.

Figure 1. First Ionization Energies

0

5

10

15

20

25

30

H He Li Be B C N O F

Ne

Na M Al

Si P S Cl

Ar K

Ca

Sc Ti V Cr

Mn

Fe

Co Ni

Cu

Zn

Ga

Ge

As

Se Br

Kr

Firs

t Ion

izat

ion

Ener

gy, e

V

Period 1 Period 2 Period 3 Period 4


▼ Example 5

In any given period, which element has the highest ionization energy?Answer: the noble gasIt would be most difficult to remove an electron from a noble gas atom. Electrons would be easiest to remove from alkali metals (group IA).

▼ Example 6

From which atom is it harder to remove an electron: Na or K?Answer: NaNa and K are in the same column (group IA). The trend in ionization energy is decreasing going down a column. Na is above K in group IA. Therefore, the ionization energy is larger for Na. It would be harder to remove an electron from Na.

13.2 Trends in Atomic SizeThe trend in atomic size is opposite that of the ionization energy. The general trend is

• Across a period (horizontal row): size decreases• Going down a group (vertical column): size increases

▼ Example 7

Arrange the following in order of increase atomic size: Na, K, Cl?Answer: Cl < Na < K. Explanation: Cl is found to the right of Na in row 3. So Cl is smaller than Na. K is below Na in group IA. So K is larger than Na.


13.3 Explaining Trends in First Ionization Energy and Atomic Size

The trend of increasing ionization energy can be explained by examining the trends in the factors that determine how hard it is to remove an atom’s valence electron. Valence means “outermost.”

Across a period, the number of protons in the nucleus increases. In other words, the nuclear charge increases as we go from left to right. This leads to a stronger inward pull on the outermost electrons. Therefore, this trend would tend to make it harder to remove an electron from the outermost shell.

Across a period, the number of valence electrons also increases. Since electrons repel each other, this trend would tend to make it easier to remove an electron from the outer-most shell. We can think of an outer electron as being partly shielded or screened by other electrons from the pull of the nucleus.

Therefore, we can say that an outer electron feels an effective nuclear charge that pulls it in toward the nucleus; the stronger this effective nuclear charge, the more difficult it is to remove an electron, and the higher the ionization energy. Since the push from an additional electron is partly directed toward the nucleus while the pull from an additional proton is always directed that way, the effective nuclear charge generally increases as we go across a row (from left to right).

In general, breaks in the trend among the main group elements occur when going from an atom with a filled subshell (such as those in Group IIA; valence configuration: s2) or from an atom with a half-filled subshell (such as those in Group VA; valence configura-tion: p3) to the atom in the next group.

• Valence electrons in group IIA are in the s subshell, whereas the valence electron in group IIIA is in the p subshell. Electrons in the valence s orbitals of a group IIA atom are able to penetrate closer to the nucleus and are able to achieve a higher effective nuclear charge, compared to the electron in the valence p orbital of the neighboring group IIIA atom.

• Electrons in half-filled subshells all have the same (parallel) spins; adding one more electron would mean adding an electron with opposite spin, which would involve a significantly larger increase in electron-electron repulsions. Electrons with parallel spins repel each other less compared to electrons with opposite spins.

The break in the trend through the transition metals has to do with additional elec-trons going into a d subshell, which is in an inner shell; the valence configuration remains the same. For example, the electron configuration of Sc is [Ar]4s23d1, whereas Ti has an electron configuration of [Ar]4s23d2.


▼ Example 8

For which pair of atoms does a break occur in the ionization energy trend across a period: Mg and Al, Si and P, S and ClAnswer: Mg and AlAmong main group elements, breaks occur between group IIA and IIIA and between group VA and VIA.

The trend within a group is primarily due to the size of the valence shells. Going down a group, the effective nuclear charge would be similar, but the valence shells get larger; the farther a valence electron is from the nucleus, the easier it is to remove from the atom. This follows from Coulomb’s law: the attraction between the valence electron and the nucleus is inversely proportional to the square of the distance between them.

The explanation for the trends in atomic size is similar to that of ionization energy. In general, the higher the ionization energy, the more strongly a valence electron is held, the closer it would be to the nucleus, the smaller the atom.

13.4 Explaining Ion-formation PatternsIt is more difficult to remove a second electron after the first one is gone. Similarly, it is harder to take out a third one after two have been taken out. In other words, the trend in successive ionization energies is increasing. The explanation is simple: fewer electrons means lower electron-electron repulsion and a higher effective nuclear charge for the remaining electrons, making them harder to remove.

A drastic increase in ionization energy occurs once all the valence electrons have been taken out. This is because the remaining electrons (the core electrons) are significantly much closer to the nucleus than the valence electrons.

▼ Example 9

Predict where the drastic increase occurs in the successive ionization energies of Be.Answer: Between the second and third ionization energies. Be has an electron configuration of [He]2s2; it has two valence electrons (in the 2s orbital); after the removal of the two valence electrons, we’re left with the noble gas core and it becomes drastically more difficult to remove the next one. We can look up the successive ionization energies of Be from webelements.com and find: 899.5, 1757.1, 14848.7, and 21006.6 kJ/mol. Note that the second is less than 2 times larger compared to the first and the fourth is less than 2 times larger compared to the third,


but the third is more than 8 times larger compared to the second. A similar pattern will be observed for all members of group IIA; this explains why naturally occurring ions from this group have a charge of +2. Similarly, atoms of group IA tend to form ions with a +1 charge. In general, atoms in groups IA and IIA form cations that are isoelectronic with the noble gas in the preceding row.

▼ Example 10

Explain why naturally-occurring monatomic ions of fluorine (the fluoride ions) have a −1 charge.Answer:A fluorine atom, with an electron configuration of [He]2s22p5, has two ways of achieving a noble-gas-like configuration:• losing 7 electrons (which requires a great deal of energy;

does not happen; we do not find F7+ ions in nature).• gaining one electron, forming fluoride ion (F−) which is

isoelectronic with a neon atom. The electron configuration of F− is [He]2s22p6, just like Ne.

Why is F2− not observed in nature? If it were formed, its electron configuration would be [Ne]3s1 and the valence electron is not likely to stay long enough for us to observe it. In general atoms of nonmetallic elements tend to gain enough electrons to become isoelectronic with the noble gas in the same row.

We can also explain why the most commonly observed charge of transition metal ions is +2, and why transition metal ions with charges other than +2 are observed. Most transition metal atoms have two electrons in the valence shell; in general, these electrons are easier to remove than the d electrons in an inner shell.

▼ Example 11

Explain the formation of Fe2+ and Fe3+.Answer:Fe has an electron configuration of [Ar] 4s2 3d6. The valence shell is the 4th shell; the two valence electrons are in the 4s orbital. These electrons are easier to remove compared to the 3d electrons.The electron configuration of Fe2+ is [Ar] 3d6, not [Ar] 4s2 3d4.Because the electron configuration of transition metal ions with a charge of +2 is not noble-gas-like, it is not drastically more difficult


to remove another electron or two; therefore charges of +3 and +4 could also observed.The electron configuration of Fe3+ is [Ar] 3d5.

13.5 Sizes and Ionization Energies of Ions vs. Atoms

Cations are smaller than the atoms that they came from; their ionization energy is higher. Fewer electrons in the valence shell, but the same number of protons in the nucleus, means lower electron-electron repulsion, and a higher effective nuclear charge. The naturally-occurring cations of the main group elements have lost all electrons in the valence shell of the original atom, leaving behind the core, which is obviously smaller and has an extremely high ionization energy.

▼ Example 12

Compare size and ionization energy of the following: Mg, Mg+, Mg2+?Answer: Mg has the largest size; Mg2+ has the smallest size. Mg has the lowest ionization energy; Mg2+ has the highest ionization energy.

Anions are larger than the atom that they come from; their ionization energy is lower. More electrons in the valence shell, but the same number of protons in the nucleus, means more electron-electron repulsion, and a smaller effective nuclear charge. Once the valence shell is filled, additional electrons will need to go to an even larger shell.

▼ Example 13

Which has the largest size: F, F−, F2−?Answer: F2−

▼ Example 14

Which has a higher ionization energy: F2− or Na?Answer: NaIt would be harder to remove an electron from Na. Both F2− and Na have the same electron configuration. But Na has a higher


nuclear charge (+11 due to 11 protons in the nucleus) compared to F2− (+9 due to 9 protons in the nucleus). Since the electron configurations are the same; the screening of the valence elec-tron is the same. Therefore, the effective nuclear charge felt by the valence electron of Na would be higher than that of F2−. This explains why F2− is not found in nature; it cannot hold on to its valence electron. Even Na has a hard time holding on to its electron; it is highly reactive.

▼ Example 15

Compare the size and ionization energy of Mg2+ and Ne.Answer: Mg2+ has a higher ionization energy and is smaller in size.Mg2+ is isoelectronic with Neon. So the screening of the valence electrons is the same. However, Mg2+ has more protons compared to Ne (12 vs. 10). Therefore, the effective nuclear charge is higher for Mg2+. It would be harder to remove an electron from Mg2+.

13.6 ElectronegativityElectronegativity refers to the ability of an atom to attract electrons that it is sharing with another atom. The trend in electronegativity generally follows that of ionization energy and the reason for the trend is pretty much the same. Across a period (horizontal row), electronegativity increases from left to right, and within a group (vertical column), elec-tronegativity decreases. We can also say that the trend in electronegativity is increasing towards the upper right-hand corner of the periodic table. Thus, fluorine (F) is the most electronegative element. Note that, based on the trend, Ne and He would be expected to be more electronegative than F. However, in nature, He, Ne and the other noble gases are not found sharing electrons with other atoms; they are usually ignored when discussing electronegativities.

▼ Example 16

Which is more electronegative: Na or N?Answer: NNonmetals are more electronegative than metals. Nonmetals are found on the upper right-hand corner of the periodic table.


TEST YOURSELF


1. The ionization energy of bromine atoms is the energy associated with which of the following?

A. Br(g) → Br+(g) + e− B. Br2(l) → 2 Br+(g) + 2e−

C. Br(g) + e− → Br−(g) D. Br2(l) + 2e− → 2 Br−(aq)

2. Which of the following has the highest ionization energy in the period where it belongs?

A. Ne B. Na C. Br D. Mn

3. Which of the following has the highest ionization energy? A. Na B. Rb C. N

4. Which of the following has the smallest atomic radius? A. O B. S C. Li

5. The increasing trend in ionization energy across a period, from left to right, is due to…

A. Increasing repulsions among electrons in the valence shell B. Increasing effective nuclear charge C. Increasing number of core electrons D. Increasing size of the valence shell

6. The decreasing trend in ionization energy going down a group is, is due to… A. Increasing repulsions among electrons in the valence shell B. Increasing effective nuclear charge C. Decreasing number of core electrons D. Increasing size of valence shell



7. Which atom’s first five ionization energies best resembles the pattern shown in the graph?

A. Na B. Mg C. Al D. K

8. What is the electron configuration of naturally occurring aluminum ion? A. [Ne] 3s2 3p1 B. [Ne] 3s2 C. [Ne] 3s1 D. [Ne]

9. All of the following have the same electron configuration except…? A. Ne B. Na+ C. O− D. O2−

10. What is the electron configuration of Fe3+? A. [Ar] 4s2 3d6 B. [Ar] 4s2 3d3 C. [Ar] 4s1 3d4 D. [Ar] 3d5

11. Which of the following has the largest size? A. K B. K+ C. Se D. Li

12. Which of the following has the largest size? A. Cl B. Cl− C. F D. F−

13. Which of the following has the highest ionization energy? A. Ne B. Mg2+ C. Mg D. F

14. Which of the following has the highest electronegativity? A. Na B. Br C. F

15. Which type of element has the lowest electronegativity? A. a metal B. a nonmetal C. a metalloid

205

14CHAPTER

Molecular Structure

In this chapter, we examine patterns in the way atoms form covalent bonds and learn how to predict the shapes of molecules and polyatomic ions.

14.1 The Octet/Duet RuleThe chemical properties of atoms primarily depend on their valence configuration. When we talk about chemical properties of atoms, we are referring to their tendency to lose, gain or share electrons. The pattern that emerges when one examines the chemical properties of atoms in the main groups is that they seem to have a tendency to be noble-gas-like; this means losing, gaining, or sharing electrons so that they end up being surrounded by the same number of valence electrons as noble gas atoms. Noble gas atoms, except helium, have 8 valence electrons; a helium atom has two valence electrons. Thus, this tendency to noble-gas-like is referred to as the octet/duet rule.

Atoms belonging to the same family or group (vertical column) have similar chemical properties because they have the same valence configuration. Columns in the traditional peri-odic table are numbered such that the number corresponds to the number of valence electrons.

▼ Example 1

C belongs to group IVA (column 14). How many valence electrons does a carbon atom have?Answer: C has four valence electrons.IV is the Roman numeral for four. The electron configuration of a carbon atom is 1s2 2s2 2p2. The valence electrons are the electrons in the 2s and 2p subshells, which are in the 2nd or n = 2 shell.

▼ Example 2

Which of the following has the same number of valence electrons as a C atom: N or Si?


Answer: Si.C and Si belong to the same group (IVA). N and C belong to the same period (horizontal row 2). It is atoms in the same group that have the same number of valence electrons.

▼ Example 3

What is the electron configuration of magnesium ion?Answer: 1s2 2s2 2p6

• The electron configuration of Mg atom is 1s2 2s2 2p6 3s2 or [Ne] 3s2

• For Mg2+ ion, the electron configuration is like that of a Neon atom: 1s2 2s2 2p6

• The valence shell of the magnesium ion is the second shell, and there are eight electrons in that shell. We say that by losing two electrons, a magnesium atom acquires an octet for its valence configuration.

▼ Example 4

What is the electron configuration of hydride ion?Answer: 1s2

• The hydride ion is the negative ion derived from a hydrogen atom.

• The electron configuration of H is 1s1

• The noble gas closest to H in the periodic table is He, which has two electrons and an electron configuration of 1s2

• We predict the hydride ion to be isoelectronic with He; we say that when a hydrogen atom becomes a hydride ion, it acquires a duet for its valence configuration.

14.2 Lewis SymbolsA Lewis symbol is a visual representation of the valence configuration of a main group atom or monatomic ion. Main group elements or representative elements belong to the “A groups” in the traditional periodic table (columns 1, 2, and 13–18). Dots are drawn around the symbol for the atom; each dot represents a valence electron.

Chapter 14: Molecular Structure 207

▼ Example 5

Draw the Lewis symbol for a carbon atom.Answer: since C has four valence electrons, we draw four dots around the symbol for carbon.

For ions, dots are added or taken away depending on the charge, square brackets are drawn, and the charge is indicated in the upper right hand corner.

▼ Example 6

Draw the Lewis symbols for magnesium ion.Answer: Magnesium belongs to group IIA. Therefore, a Mg atom has two valence electrons; its Lewis symbol has two dots. But naturally-occurring magnesium ion has a charge of +2; this means two fewer electrons compared to a magnesium atom. The Lewis symbol for magnesium ion has no dots.

▼ Example 7

Draw the Lewis symbols for fluoride ion.Answer: Fluorine belongs to group VIIA. Therefore, a fluorine atom has 7 valence electrons and its Lewis symbol has seven dots. Naturally-occurring fluoride ion has a charge of –1; this means one additional electron compared to a fluorine atom. Therefore, the Lewis symbol for fluoride ion has 8 dots.

By following the octet/duet rule, we can say that the atoms end up as ions with a noble-gas-like configuration, or as being isoelectronic with a noble gas.


14.3 Lewis StructuresA Lewis structure is a visual representation of how the atoms in a molecule or polyatomic ion are sharing valence electrons. In a Lewis structure,

• each atom is represented by its symbol.• all valence electrons must be represented by dots or lines.• each dot represents one valence electron.• a line is equivalent to two dots.• lines or dots drawn between symbols of two atoms represent valence electrons

shared by the two atoms.

For polyatomic ions, the Lewis structure must be enclosed in square brackets and the charge of the ion must be indicated as a superscript in the upper right-hand corner.

▼ Example 8

How many valence electrons are shared in the Lewis structure shown below? How many are not shared?

Answer: 16 shared, 8 not sharedThe total number of valence electrons shown in the structure above is 24 (16 shared; 8 unshared). There are 8 lines drawn between symbols; each of these 8 lines represents 2 shared electrons; 8 × 2 = 16. The total number of valence electrons shown (24) should be equal to the total number of valence elec-trons from the individual atoms that make up the structure. Note that Nitrogen belongs to group VA; a nitrogen atom has 5 valence electrons. Each H atom has only one electron. Carbon belongs to group IVA; each C atom has 4 valence electrons. Oxygen belongs to group VIA; each O atom has 6 valence electrons.• From N: 2 × 5 = 10• From H: 4 × 1 = 4• From C: 1 × 4 = 4• From O: 1 × 6 = 6

10 + 4 + 4 + 6 = 24


In a Lewis structure, each line or pair of dots shared between two atoms is called a bonding pair. A pair of atoms can have up to three bonding pairs between them.

• If only one bonding pair is shown between two atoms, we say that the atoms have a single bond between them.

• If there are two bonding pairs between these two atoms, we say that they have a double bond.

• If there are three bonding pairs between these two atoms, we say that they have a triple bond.

Each pair of unshared electrons is called a lone pair.

▼ Example 9

How many single bonds, double bonds, triple bonds, and lone pairs are shown in the structure below.

Answer:• Single bonds: 6 (four N-H and two C-N)• Double bonds: 1 (between C and O)• Lone pairs: 4 (two on the O, one on each of the N)

For polyatomic ions, the total number of valence electrons shown in the Lewis structure should account for the charge. A negative charge means additional electrons; a positive charge means fewer electrons.

▼ Example 10

How many valence electrons should be shown in the Lewis struc-ture for SO4

2–?Answer: 32 electrons• The S atom contributes 6 valence electrons since S belongs

to group VIA.• Each of the four O atoms also contributes 6 valence electrons

since O belongs to group VIA.• Total number of valence electrons from one S and four O

atoms is: (1)(6) + (4)(6) = 30.• To this, we need to add two more electrons since the ion has

a charge of –2.30 + 2 = 32.


▼ Example 11

How many valence electrons should be shown in the Lewis structure for NH4

+?Answer: 8 electrons• The N atom contributes 5 electrons since N belongs to

group VA.• Each H atom contributes one electron.• Total number of valence electrons from one N and 4 H:

(5) + (4)(1) = 9.• We need to subtract one electron because the ion has a

charge of +19 − 1 = 8.

When atoms share electrons, they also appear to have a tendency to acquire a noble-gas-like configuration. Atoms tend to share electrons such that each one appears to be surrounded by an octet (or a duet in the case of hydrogen).

▼ Example 12

Describe how the octet/duet rule is illustrated in the Lewis structure below.

Answer:• The O atom is surrounded an octet (8 valence electrons):

four electrons in the two lone pairs and four electrons in the double bond.

• The C atom is also surrounded by an octet: four electrons in the double bond, and two electrons in each of the two single bonds.

• Each N atom is surrounded by an octet: six in the three single bonds and two in the lone pair.

• Each H atom has a duet: two electrons in each of the single bonds. H cannot exceed a duet, it can bond to only one atom. We say that hydrogen is always a terminal atom in a molecule or polyatomic ion; we only find H atoms at the edges of a molecule or polyatomic ion.


14.4 Strategies for Drawing Lewis StructuresHere is a systematic procedure for drawing a structure that follows the octet/duet rule.

Step 1. Start by connecting all the atoms with single bonds. Attach the H atoms last since they should be terminal atoms; H atoms can only be involved in one single bond. Add lone pairs to complete the octet.

Step 2. Count all the valence electrons in the structure; compare with the expected (correct) total number of valence electrons.

• If the number of valence electrons shown in the structure is correct, you are done.

• If the number of valence electrons shown in the structure is larger than the correct number: remove one lone pair each from neighboring atoms and add a bonding pair between the two atoms.

• Repeat until you get the correct number of valence electrons.

▼ Example 13

Draw the Lewis structure for the carbon monoxide molecule, CO.Answer:Step 1. Start by drawing a tentative structure with the two atoms connected by a single bond; then draw lone pairs in order to have an octet around each atom. In order for both atoms to have an octet, we need three lone pairs on each one.

Step 2. Count the number of valence electrons and compare with the correct number.• The structure as drawn shows 14 valence electrons.• The correct number should be 10. C belongs to group IVA;

C will contribute 4 valence electrons. O belongs to group VIA; O will contribute 6 valence electrons.

• We have only 10 valence electrons to distribute between the two atoms, but we are showing 14 in the tentative structure. Therefore, we remove a lone pair each from the neighboring atoms add a bonding pair between the two atoms. If we do this, we get:


• We now have 12 valence electrons. Again, we know this structure is incorrect because CO only has 10. So, we take out two more lone pairs (one each from neighboring atoms) and add another bonding pair between them.

If we were to explore all the possible ways of drawing a valid Lewis structure for diatomic molecules or ions, we find that there are only five unique ways that it can be done. These five ways are shown in Figure 1.

Figure 1. Lewis Structures for Diatomic Species

Note that each way has a unique number of valence electrons: 2, 8, 14, 12, and 10 for structures 1 through 5, respectively.

▼ Example 14

Which of the following diatomic species does not have a single bond?H2, HF, F2, O2, O2

2–

Answer: O2 ; it has a double bond• O2 has 12 valence electrons; each O atom in O2 has 6

valence electrons: 6 × 2 = 12. Therefore, the Lewis structure of O2 should resemble Structure 4 in Figure 1, which has a double bond.

• H can only form a single bond. Therefore, H2 and HF are expected to have a single bond. We expect H2, with only two


valence electrons, to look like structure 1 in Figure 1. We expect HF to look like structure 2 in Figure 1. H contributes one valence electron and F contributes 7; only structure 2 above has 8 valence electrons.

• The F atoms in F2 also have a single bond between them; each F atom in F2 contributes 7 valence electrons; 2 × 7 = 14; only structure 3 in Figure 1 shows 14 valence electrons.

• The O atoms in peroxide ion, O22–, also have a single

bond between them; each O atom contributes 6 electrons and the –2 charge means two additional electrons: 2 × 6 + 2 = 14.

▼ Example 15

Draw the Lewis structure for hydronium ion (H3O+).

Answer: +1Since each H can only have a single bond, the Lewis structure must be one where each H is sharing a pair of electrons with O. In order for the O atom to have an octet, we should draw one lone pair on it.

We then verify if we have the correct number of valence elec-trons shown. The three single bonds and one lone pair account for 8 valence electrons. H3O+ should have 8;one from each of the H atoms, 6 from the O atom, minus one because the + charge indicates one electron lost: 3(1) + 6 − 1 = 8. Finally, we draw square brackets around the structure and indicate the + charge as a super-script.

14.5 Isomeric StructuresWhen we have more than two atoms in a molecule or polyatomic ion, there is usually more than one way of connecting the atoms. Each of the unique ways is called an isomer; each one represents a different molecule.


▼ Example 16

Draw the two possible isomers of a molecule made up of one H, one N, and one C atom.Answer:The atoms can be connected as H-C-N or C-N-H. Since H cannot have more than one single bond, C-H-N is not allowed.

It should be easy to verify that both structures satisfy the octet/duet rule and have the correct number of valence electrons. These structures represent two different molecules, two different compounds. The first one is called hydrogen cyanide, the other is called hydrogen isocyanide. These two molecules are called isomers. Note that connecting the atoms as H-N-C is the same as C-N-H; similarly the H-C-N connectivity is the same as N-C-H.

14.6 Resonance StructuresIt is possible to have more than one way of distributing valence electrons for a given mol-ecule or ion. Each of these unique ways is called a resonance structure. Unlike isomers, the connectivity in resonance structures is the same; resonance structures represent the same molecule. The existence of two or more resonance structures generally indicates that none of the structures adequately represents reality. Reality is probably a hybrid of the reso-nance structures, although some structures may be considered “more significant” (that is, resemble reality better than the others). When drawing resonance structures, we draw double-headed arrows between them to indicate resonance.

▼ Example 17

Draw resonance structures for a OCS, assuming that the C is between O and S.Answer:The total number of valence electrons in OCS is 16: 6 from O, 4 from C, and 6 from S. If we explore all the possibilities, we find that any triatomic molecule or ion with 16 valence


electrons will have the resonance structures similar to the ones shown below:

14.7 Formal ChargesNot every possible isomeric structure is stable enough to be observed in nature. We can use formal charges to determine which isomeric structures are more likely to be observed. If more than one resonance structure can be drawn for a given molecule, we can also use formal charges to determine which structure is more significant. In general,

• structures with small formal charges (0, +1, –1) are more likely to be stable isomers or are more significant resonance structures that those that have larger formal charges (such +2, –2).

• structures where neighboring atoms have formal charges of the same sign are less likely to be stable or significant.

• if neighboring atoms have opposite formal charges, we would expect the more electronegative atom to have the negative formal charge and the less electronegative atom to have the positive formal charge.

To calculate the formal charge of an atom in a structure, use the formula: F.C. = F – A,

where• F is the number of valence electrons in the Free atom (neutral, not bonded to

another atom), and• A is the apparent number of valence electrons that the atom owns in the structure.

To determine A, assume the atom owns half of electrons shared with other atoms and all of the electrons in its lone pairs.

In other words,A = B + 2L,

where B is the number of bonding pairs and L is the number of lone pairs.


▼ Example 18

Determine the formal charge of S in the structure below.

Answer:• S belongs to group VIA, so a free sulfur atom has 6 valence

electrons. F = 6• In the structure, S appears to own 5 electrons. It has three

bonding pairs (in the triple bond) and one lone pair. B + 2L = 5. Put another way, it appears to own 2 in the lone pair, plus half of the six electrons in the triple bond (3). Therefore:

A = 2 + 3 = 5.• The formal charge of S is: F.C. = F – A = 6 – 5 = +1 Compared to a free S atom, the S atom in the structure

appears to have lost one electron.

How do we come up with structures that have low formal charges? In general, if a free atom has x valence electrons, it would have a formal charge of zero if it has (8-x) bonding pairs.

▼ Example 19

In what ways can a C atom have a formal charge of zero?Answer:A carbon atom has 4 valence electrons. It needs 4 more to acquire an octet: 8 − 4 = 4. Therefore, a C atom gets a formal charge of zero when it has four bonding pairs, as in• four single bonds.• two single bonds and one double bond.• one single bond and one triple bond, or• two double bonds.

Because of this, carbon is said to be tetravalent; the Greek prefix tetra means four. A carbon atom by itself, has 4 valence elec-trons (it belongs to group IVA); it just needs four more electrons to achieve an octet and be noble-gas-like. Carbon atoms are found in molecules of organic compounds. The study of organic compounds is very important. Living things are primarily made of organic compounds.


▼ Example 20

In what ways can an O atom have a formal charge of zero?Answer:An O atom has 6 valence electrons. It needs two more to acquire an octet: 8 – 6 = 2. Therefore, a O atom gets a formal charge of zero when it has two bonding pairs (and two lone pairs), as in• two single bonds and two lone pairs.• one double bond and two lone pairs.

As illustrated in the preceding examples, C has a formal charge of zero in structures where it has four bonding pairs, while O has a formal charge of zero in structures where it has two bonding pairs. For nitrogen atoms, the typical pattern is three bonding pairs (three single bonds as in NH3, a double bond and a single bond as in HONO, or a triple bond as in HCN); in these cases, the formal charge of N is zero. Halogens (F, Cl, Br, and I) tend to form just one single bond. The free halogen atoms have 7 valence electrons; by forming one single bond, they achieve an octet and have a formal charge of zero.

Another good strategy for coming up with structures with reasonable formal charges is to surround a less electronegative atom with more electronegative atoms. In these structures, the more electronegative atoms are more likely to get a negative formal charge (as they should) and the less electronegative atom is more likely to get a positive formal charge.

14.8 Exceptions to the Octet/Duet RuleThe octet/duet rule is not a strict rule.

Atoms can have less than an octet. Notable examples B and Be. Experimental evidence suggests that the B atom in BCl3 is best represented by a structure where the B is sur-rounded by only six valence electrons (three bonding pairs). Similarly, Be is BeH2 is best represented by a structure where Be is surrounded by only four electrons (two bonding pairs). One characteristic of these molecules is that they tend to react so that the central atom, which does not have an octet, ends up with an octet by bonding with a molecule or ion with a lone pair.

There are molecules that have an odd number of electrons. Notable examples are NO and NO2. In these cases, it is impossible to have all electrons paired and it would not be possible to follow the octet rule. Molecules with unpaired electrons are called radicals and are experimentally detectable using techniques that take advantage of the fact that these molecules are paramagnetic.


Atoms in period 2 cannot have more than an octet, but atoms beyond period 2 can have more than an octet (“expanded octets”). Recall that s, p, and d subshells can accommodate 2, 6, and 10 electrons. The valence shell of atoms in period 2 cannot accommodate more than an octet because they only have s and p subshells. However, beyond period 2, the valence shell can accommodate more than 8 electrons due to the availability of d subshells.

▼ Example 21

Which atom cannot have more than an octet: C or Si?Answer: C. Carbon belongs to period 2. Si belongs to period 3 and has d subshells in its valence shell that allows it to accommodate more than an octet.

The use of expanded octets allows us to avoid large formal charges as illustrated in the following example.

▼ Example 22

Which of the two resonance structures shown below for sulfate, SO4

2–, is less significant?

Answer: structure AThe formal charge of S in structure A is +2 (not good); in struc-ture B, its formal charge is zero.It should be noted that structure B is one of six equally significant Lewis structures; another one would be:


and another one would be:

Reality would best be represented by a hybrid of all six structures; a single bond between S and O is generally found to be longer than a double bond. In sulfate, we would expect each of the S-to-O bonds to be of the same length; we also expect each one to be longer than a typical S-to-O single bond but shorter than a typical S-to-O double bond.

14.9 Molecular GeometryThe subject of molecular geometry deals with shapes of molecules. A Lewis structure is a two-dimensional representation of a molecule or ion; its intent is to show how the valence electrons of atoms are distributed in the molecule. A Lewis structure does not necessarily have to depict the actual shape of the molecule or ion. But we can use it to deduce what the actual shape is.

Since atoms are always in motion, when we talk about shapes of molecules or poly-atomic ions, we mean the shape based on the average location of the atoms.

14.9.1 VSEPR TheoryTo determine the shape, we use a simple theory called VSEPR, which stands for Valence Shell Electron Pair Repulsion. The basic idea behind VSEPR is that regions of high elec-tron density around each atom in the molecule tend to be oriented as far away from one another as possible. These regions of high electron density are often referred to as electron domains or electron groups. The justification for the idea is simple: electrons repel each other. Once we know how the electron domains around an atom are oriented, it becomes easy to imagine the shape of the molecule.

The first step in figuring out the shape of a molecule is to determine the steric num-bers of atoms that are bonded to at least two other atoms.

14.9.2 Steric NumberThe term steric number refers to the number of electron groups around an atom. Once we have determined the steric number, we can imagine how these groups are oriented relative to each other.


To determine steric number,• a single bond is counted as one group.• a double bond is counted as one group.• a triple bond is counted as one group, and• a lone pair is counted as one group.

▼ Example 23

What are the steric numbers of C and N in the structure below?

Answer: The steric number of C is 3; for each of the two N atoms the steric number is 4.• There are 3 groups of electrons “sticking out” of the carbon

atom: two single bonds and one double bond.• There are four groups of electrons sticking out of each of the

nitrogen atoms: 3 single bonds and 1 lone pair.

14.9.3 Predicting Bond Angles and Shapes Based on Steric Number

Once we have determined the steric number of an atom, we can predict the measure of the angle formed by the electron groups sticking out of that atom.

Steric Number = 2. If there are only two groups of electrons sticking out of an atom, the two groups will be as far away from each other by being oriented 180° apart. The three bonding patterns where this occurs are illustrated in Figure 2. Since the two electron groups form a straight angle with the atom at the vertex, we say that the electron domain geometry around the atom is linear, or that the atom is a linear center.

Figure 2. Bonding Patterns for Steric Number=2


The shape of a triatomic molecule or ion is described as linear if the (nuclei of the) three atoms fall on a straight line; this happens if the steric number of the central atom is 2.

▼ Example 24

Based on the Lewis structures of H2O and CO2, which molecule is linear: H2O or CO2?

Answer: CO2• The steric number for C is 2, the two double bonds will be

oriented 180o apart. Therefore, the three atoms CO2 will be on a straight line.

• In the case of H2O, the steric number of O is 4. There are four groups of electrons sticking out of the O atom: two single bonds and two lone pairs. The single bonds will not be oriented 180° apart.

Steric Number = 3. Three groups of electrons will be as far from each other as possible if they are oriented 120° apart. Typical bonding patterns are shown in Figure 3. Since the three groups sticking out of the atom are on the same plane and are pointing towards the corners of a triangle, we say that the electron domain geometry around the atom is trigonal planar. Or we say that the atom is a trigonal planar center. Strictly speak-ing, the angles differ slightly from 120° when the groups are not identical. Lone pairs repel other groups more strongly than a single bond. A double bond repels other groups more strongly than a lone pair. Therefore, except for the case where all three groups are single bonds, the angles indicated in Figure 3 are approximate.

Figure 3. Bonding Patterns for Steric Number = 3


▼ Example 25

Consider the Lewis structure for formaldehyde.

Which angle would be larger: HCO or HCH?Answer: The HCO angle will be larger.• The C atom is a trigonal planar center; its steric number is 3• Because the steric number of C is 3, the groups will be

oriented about 120° apart.• Because the C=O bond is a double bond, it is “bulkier” and

will “push” the single bonds closer to each other. The actual angles are: 121.27° for HCO and 117.46° for HCH. The 3-D model shown below resembles reality better than the Lewis structure above.

from http://commons.wikimedia.org/wiki/File:Formaldehyde-3D-balls-A.png (public domain)

▼ Example 26

Describe the OCO bond angles in carbonate ion based on the Lewis structure shown below.

Answer: all OCO angles are exactly 120°.

http://commons.wikimedia.org/wiki/File:Formaldehyde-3D-balls-A.png

http://commons.wikimedia.org/wiki/File:Formaldehyde-3D-balls-A.png


• The C atom is a trigonal planar center• All three resonance structures have one C=O double bond

and two C-O single bonds, suggesting that they are all equally significant. All three C-to-O bonds are, therefore, equivalent.

• The geometry of this ion is said to be of the type AX3. The A stands for the central atom and the X3 stands for three atoms bonded to A.

• The model below better resembles reality than the Lewis structure.

from http://en.wikibooks.org/w/index.php?title=File:Trigonal-3D-balls.png&filetimestamp=20061212171212 (public domain)

▼ Example 27

Describe the geometry of nitrite ion based on the Lewis structure shown below.

Answer:• The N atom is a trigonal planar center since its steric

number is 3; there are 3 groups of electrons sticking out of the N. The ONO bond angle is, therefore, expected to be very close to 120°.

• The ion can be described as bent. Any time there are only three atoms in a molecule or ion, the shape is either linear or bent. It is linear if the bond angle is 180°; otherwise, it is bent.

• The geometry of this ion is said to be of the type AX2E. The A stands for the central atom, the X2 stands for the two atoms bonded to A, and E stands for one lone pair.

http://en.wikibooks.org/w/index.php?title=File:Trigonal-3D-balls.png&filetimestamp=20061212171212

http://en.wikibooks.org/w/index.php?title=File:Trigonal-3D-balls.png&filetimestamp=20061212171212


• The model below better resembles reality than the Lewis structure:

from http://en.wikibooks.org/w/index.php?title=File:Bent-3D-balls.png&filetimestamp=20061212175343 (public domain)

Steric Number = 4. Four groups of electrons sticking out of an atom will be as far from each other as possible if they are oriented such that each one is pointing towards the corners of a regular tetrahedron. A regular tetrahedron is a perfectly symmetrical four-sided solid (tetra = four, hedron = side). To visualize a regular tetrahedron, imagine carv-ing it out of a cube as shown in Figure 4. The corners indicated by the green circles are the corners of a tetrahedron; the solid carved out, which can be described as four equi lateral triangles forming the sides of a pyramid, is a regular tetrahedron.

Figure 4. Carving Out A Regular Tetrahedron From A Cube

Another way to imagine a tetrahedron is to imagine three oranges forming a triangle on a table; put a fourth orange on top of the center of the three oranges; the four oranges would be the corners of a tetrahedron.

To imagine an atom with a steric number of 4, imagine it in the middle of the cube in Figure 4; the four groups of electrons will be directed towards the green circles. If there is an atom at the other end of each group, the molecule would look like Figure 5. The shape of the molecule is said to be tetrahedral and type AX4.

http://en.wikibooks.org/w/index.php?title=File:Bent-3D-balls.png&filetimestamp=20061212175343

http://en.wikibooks.org/w/index.php?title=File:Bent-3D-balls.png&filetimestamp=20061212175343


The angle between any pair of electron groups sticking out of a tetrahedral center is very close to 109.5°. There may be slight distortions if the groups are not all equivalent.

If one of the groups sticking out of a tetrahedral center happens to be a lone pair, as in Figure 6, the shape of the molecule or ion is called triangular pyramidal or trigonal pyra-midal. The bond angles are close to, but slightly less than 109.5°, because the lone pair will repel the single bonds more strongly and push them a little closer together. The geometry of the molecule is said to be AX3E; three atoms are bonded to the central atom, which has one lone pair .

If two of the groups sticking out of a tetrahedral center happen to be lone pairs, as in Figure 7, then the shape of the molecule is called bent; there are only three atoms and they are not on a straight line. The bond angle is close to, but slightly less than, 109.5° because of the lone pair. The geometry of the molecule is said to be of type AX2E2. Two atoms are bonded to the central atom, which has two lone pairs.

Figure 5. Model Of A Tetrahedral Molecule; Type AX4

from http://upload.wikimedia.org/wikipedia/commons/5/58/Methane-3D-balls.png (public domain)

Figure 6. Triangular Pyramid; type AX3E

from http://en.wikibooks.org/w/index.php?title=File:Ammonia-3D-balls.png&filetimestamp=20070515220035 (public domain)

http://upload.wikimedia.org/wikipedia/commons/5/58/Methane-3D-balls.png

http://en.wikibooks.org/w/index.php?title=File:Ammonia-3D-balls.png&filetimestamp=20070515220035




▼ Example 28

Describe the molecular geometry of H2S, NF3, and CCl4 based on their Lewis structures.

Answer:• The central atom (S, N, and C) has a steric number of 4 and

is a tetrahedral center in all three cases. The bond angles are expected to be close to 109.5°.

• Due to the lone pair, the shape of NF3 is called trigonal pyra-midal or triangular pyramidal.

• Due to the two lone pairs, the shape of H2S is called bent.

▼ Example 29

Consider the Lewis structure shown below. Which bond angles are closest to 109.5°?

Answer: The angles that have N at the vertex, HNH and HNC.

Figure 7. Bent Molecule; Type AX2E2

from http://en.wikibooks.org/w/index.php?title=File:Water-3D-balls.png&filetimestamp=20070425083824 (public domain)

http://en.wikibooks.org/w/index.php?title=File:Water-3D-balls.png&filetimestamp


• The steric number of both N atoms in this molecule is 4. There are four groups of electrons sticking out of each N atom. Therefore, these groups will be oriented about 109.5° apart.

• The steric number of C is 3; it is a trigonal planar center.

Steric Number = 5. If there are five groups of electrons sticking out of an atom, as illustrated in Figure 8, we say that the atom is a trigonal bipyramidal center. If there is an atom at the other end of each of these five groups, the shape of the molecule or ion itself is called trigonal bipyramidal; the geometry is said to be type AX5.

Figure 8. Trigonal Bipyramidal Geometry; Type AX5

To visualize this geometry, imagine the central atom in the middle of the earth. Three of the groups will be sticking out towards corners of an equilateral triangle along the equa-tor; hence these three groups, which would be 120° apart are called the equatorial groups. The two other groups are called the axial groups; imagine one is sticking up towards the north pole and the other group sticking out towards the south pole. Each axial group is perpendicular to each of the equatorial groups.

If one or more of the groups sticking out of trigonal bipyramidal center is a lone pair, they take up equatorial positions because these positions are, on average, farther from the other groups than the axial positions. The shapes of the molecules for types AX4E, AX3E2, and AX2E2 are shown in Figure 9.

from http://en.wikibooks.org/w/index.php?title=File:Trigonal-bipyramidal-3D-balls.png&filetimestamp=20061212170609 (public domain)

http://en.wikibooks.org/w/index.php?title=File:Trigonal-bipyramidal-3D-balls.png&filetimestamp=20061212170609

http://en.wikibooks.org/w/index.php?title=File:Trigonal-bipyramidal-3D-balls.png&filetimestamp=20061212170609


▼ Example 30

What is the shape of the SF4 molecule?Answer: see-saw• The molecule has 34 valence electrons (6 from S; 7 from

each of the four F atoms).• The S atom is surrounded by four F atoms; each F atom

atom is surrounded by an octet; this accounts for only 32 valence electrons (4 × 8 = 32).

• The remaining two electrons are assigned to the S as a lone pair.

• Therefore, the molecular geometry is type AX4E. The mole-cule has 4 S-F bonds and one lone pair on the central atom.

AX4E; “see-saw” shape (example: SF4)from http://en.wikibooks.org/w/index.php?title=File:Seesaw-3D-balls.png&filetimestamp= 20070430211139 (public domain)

AX3E2; “T-shape” (example: BrF3)from http://en.wikibooks.org/w/index.php?title=File:T-shaped-3D-balls.png&filetimestamp= 20070430211156 (public domain)

AX2E3: shape is linear (example: ICl2–)

from http://en.wikibooks.org/w/index.php?title=File:Linear-3D-balls.png&filetimestamp= 20061212175353 (public domain)

Figure 9. Trigonal Bipyramidal Geometries, With Lone Pairs

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Steric Number = 6. Six groups of electrons sticking out of an atom stay as far apart as possible by taking mutually perpendicular orientations, as illustrated in Figure 10. If there is an atom at the end of each group (example: SF6), the shape of the molecule resembles an 8-sided solid with an equilateral triangle on each side. Thus, the shape is called octahedral (octa = 8, hedron = side). An atom with a steric number of 6 is called an octahedral center.

Figure 10. Octahedral Geometry; Type AX6

For an octahedral center, as illustrated in Figure 11• if one of the six groups happens to be a lone pair (type AX5E), the shape of the

molecule is called square pyramidal (example: BrF5).• if two of the six groups are lone pairs (type AX4E2), the shape of the molecule is

square planar (example: ClF4–).

• if three of the six groups are lone pairs (type AX3E3), the shape of the molecule is T-shape.

• if four of the six groups are lone pairs (type AX2E4), the shape of the molecule is linear.

from http://en.wikibooks.org/w/index.php?title=File:Octahedral-3D-balls.png&filetimestamp =20061212175404 (public domain)

http://en.wikibooks.org/w/index.php?title=File:Octahedral-3D-balls.png&filetimestamp


AX4E2; square planarfrom http://en.wikibooks.org/w/index.php?title=File:Square-planar-3D-balls.png&filetimestamp=20061212170229 (public domain)

AX3E3; “T-shape”from http://en.wikibooks.org/w/index.php?title=File:T-shaped-3D-balls.png&filetimestamp=20070430211156 (public domain)

AX2E4: linearfrom http://en.wikibooks.org/w/index.php?title=File:Linear-3D-balls.png&filetimestamp=20061212175353 (public domain)

AX5E; square pyramidalfrom http://en.wikibooks.org/w/index.php?title=File:Square-pyramidal-3D-balls.png&filetimestamp=20070430211148 (public domain)

Figure 11. Octahedral Geometries With Lone Pairs

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TEST YOURSELF


1. How many electrons are shown in the Lewis symbol for a nitrogen atom? A. 5 B. 7 C. 8 D. 14

2. How many dots are shown in the Lewis symbol for a nitride ion? A. 5 B. 7 C. 8 D. 14

3. How many electrons are shared between C and N in the Lewis structure shown below?

A. 3 B. 4 C. 6 D. 8

4. How many electrons are shared in the structure of formaldehyde shown below? A. 4 B. 6 C. 8 D. 12

5. Which of the structures shown is valid for HOCl?

6. Which of the following has a double bond? A. HCl B. O2

2– C. N2 D. PF

7. How many valence electrons must be shown in the Lewis structure of carbonate ion (CO3

2–)? A. 20 B. 22 C. 24 D. 30 E. 32



8. Consider the following partially-drawn structure of 2-butene (C4H8) where only the C atoms are shown:

C—C——C—C When this structure is completed following the octet/duet rule, how many hydrogen

atoms are bonded to the first carbon? A. 1 B. 2 C. 3 D. 4

9. Which of the structures shown below are resonance structures?

A. Structures I and II B. Structures II and IV C. Structures I and III D. Structures II and III

10. Which of the Lewis structures shown below for azide ion (N3−) has a nitrogen atom

with a formal charge of –2?

A. Structure I only B. Structure II only C. Both structures D. neither structure

11. Which of the Lewis structures shown below has zero formal charges for all the atoms?

12. Based on formal charges, which of the Lewis structures shown below is the most likely structure of CH4O?


13. Based on formal charges, which of the Lewis structures shown below is the most likely structure of HNO2?

14. Which of the following atoms must have an octet when sharing electrons with other atoms?

A. N B. B C. P D. none of these

15. Which of the following atoms can have more than an octet of valence electrons when it shares electrons with other atoms?

A. S B. N C. C D. H

16. Which of the structures below represent a linear molecule?

A. 1 and 2 B. 2 and 3 C. 3 and 4 D. 1 and 4

17. Which molecule has the smallest bond angles? A. CH4 B. NH3 C. H2O

18. For the molecule represented by the Lewis structure below,

the N —C—N bond angle is closest to... A. 180° B. 109.5° C. 120°


19. Which of the following can be described as having a “see-saw” shape? A. ICl4

– B. ICl4+ C. ICl5

20. Draw a structure for phosphate ion (PO43−) where the phosphorus atom is

surrounded by four oxygen atoms and has a zero formal charge.

235

15CHAPTER

Quantum Mechanical Description of MoleculesQuantum mechanics can be applied to molecules in much the same way it is applied to atoms. The wavefunctions that describe electrons in a molecule are called molecular orbitals. In this chapter, we examine two popular methods of constructing molecular orbitals.

15.1 The Born Oppenheimer ApproximationIn determining molecular wavefunctions, we use the Born-Oppenheimer approxima-tion. In this approximation, the geometry of the molecule is first defined; the locations of the nuclei are fixed. Then, we solve the Schrodinger equation with this restriction (that the locations of the nuclei are fixed, not moving). The energies we get are called the electronic energies. For each possible geometry, we get a set of molecular orbitals and electronic energies. By examining the set of molecular orbitals for the most stable geometry (the equilibrium geometry), we can explain and/or predict properties of the molecule.

How is the equilibrium geometry determined? Consider a diatomic molecule. The geometry of a diatomic molecule is defined by specifying the internuclear distance (r), the distance between the nuclei. At a series of r values, we solve the Schrodinger equation; we get the allowed electronic energies for each r value. By adding the electronic energy to the potential energy of repulsion between the nuclei, we get a molecular potential energy function. Since we get a set of allowed energies for each r, we get a set of molecular potential energy functions for the molecule; of these, the one with lowest values corresponds to what is called the ground electronic state of the molecule. For diatomic molecules, the molecular potential energy for the ground electronic state resembles the blue curve in Figure 1. The equilibrium geometry corresponds to the distance where the molecular potential energy is at a minimum.

The potential energy curve in Figure 1 can be interpreted as follows. Imagine two positively charged nuclei (initially at infinite distance from each other) approaching each other. Since like charges repel, we would expect them to slow down, which means a decrease in kinetic energy and an increase in potential energy as they approach each


other. However, this is not what happens because the repulsions are alleviated by the presence of electrons between the nuclei. Electrons are negatively charged and exert attractive forces on the positively charged nuclei. As illustrated in Figure 1; the potential energy typically drops first, then increases once the nuclei get too close to each other. At very short distances, the repulsions between the nuclei can no longer be alleviated by the presence of the electrons and we see a sharp increase in potential energy. A balance between repulsive and attractive forces occurs at the equilibrium geometry.

To facilitate interpretation of molecular orbitals, they are calculated in terms of the atom-ic orbitals (of the atoms in the molecule). Two popular ways of doing this are the LCAO-MO (linear combination of atomic orbitals - molecular orbital) and VB (valence bond) methods.

15.2 Valence Bond MethodThe valence bond (VB) method accounts for bonding between two atoms in terms of overlapping atomic orbitals. What this means is that the functions associated with the atomic orbitals of bonded atoms are combined so that the resulting function (the molecular orbital) predicts a high probability of finding the electron between the two atoms. For two atoms linked by a single bond, valence bond theory proposes a “head on” overlap, called a sigma bond. Figure 2 illustrates what happens when 1s orbitals of two H atoms overlap. The red shading in Figure 2 is represents electron density; the probability of finding an electron is highest in areas where the shading is thickest.

For double bonds, the VB method proposes a sigma bond and a pi bond, which is illustrated in Figure 3. A pi bond is the result of the overlapping p orbitals on neighboring atoms. In a pi bond, there are two regions of high electron density: one on each side of the internuclear axis. The internuclear axis refers to the line that goes through both nuclei. For

Figure 1. Molecular Potential Energy for a Diatomic Molecule

Chapter 15: Quantum Mechanical Description of Molecules 237

triple bonds, the VB method proposes a sigma bond and two pi bonds. In Figure 3, the pi bond is shown as having high electron densities above and below the internuclear axis; a second pi bond would have high electron densities in front of and behind the internuclear axis.

▼ Example 1

How many sigma and pi bonds are there between C and N in the HCN?

Answer: one sigma and two pi bonds.

Figure 2. Sigma Bond

http://en.wikibooks.org/wiki/File:Ligatio-covalens.jpgCreative Commons Attribution - Share Alike license

Figure 3. Pi Bond

http://en.wikipedia.org/wiki/File:Pi-Bond.svg(Creative Commons Attribution - Share Alike License)

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http://en.wikipedia.org/wiki/File:Pi-Bond.svg


15.2.1 Explaining Bond AnglesIn polyatomic molecules, the concept of hybridization is invoked in order to make the idea of overlapping orbitals consistent with the observed molecular geometry.

Consider the acetylene molecule:

In this molecule, the H-C-C angle is 180°. This is explained by assuming that in the mol-ecule, the C atom no longer has one 2s and three 2p orbitals. Instead, the 2s and one of the 2p orbitals are said to hybridize. It is now said to have two (2sp) hybrid orbitals and two 2p orbitals. In other words,

(2s, 2p, 2p, 2p) in free C atom becomes

(2sp, 2sp, 2p, 2p) in the molecule Remember that orbitals are mathematical functions that give us information about regions where electrons are likely to be found. For the two sp hybrids, the regions of high electron density are directed 180° apart, as illustrated in Figure 4. In Figure 4, the s orbital is represent-ed by the red sphere and the p orbital is represented by the two lobes. The algebraic sign of the wavefunctions are color-coded (+ for red, – for blue). By adding the s and p orbitals together, we get constructive interference (the big red region) on one side of the nucleus and destructive interference (the small blue region) on the other side. The big red region corresponds to a region of high electron density. A similar result is obtained by subtraction, except that the region of high electron density is on the other side, 180° away.

In acetylene, one of the sp hybrid orbitals of C overlaps with the s orbital of H; we say that the H-to-C bond is an s-to-sp overlap. The other sp hybrid overlaps with an orbital from the other C. Since the other C is also a linear center, we expect it to also have a similar hybridization. Therefore, we expect one of the three C-to-C bonding pairs to be due to an sp-to-sp overlap. The s-to-sp and sp-to-sp overlaps are classified as sigma bonds. Since the C-to-C bond is a triple bond; the two other bonding pairs are said to be due to pi bonds, sideways overlap of p orbitals.

Figure 4. Construction of sp Hybrids


In general, we can correlate the steric number to the hybridization of the atom.• If the steric number is 2 (linear center), the hybridization is sp.• If the steric number is 3 (trigonal planar center), the hybridization is sp2.• If the steric number is 4 (tetrahedral center), the hybridization is sp3.• If the steric number is 5 (trigonal planar center), the hybridization is dsp3

(or sp3d).• If the steric number is 6 (octahedral center), the hybridization is d2sp3 (or sp3d2).

▼ Example 2

According to the valence bond method, what hybrid orbitals overlap in the C-to-N bond in the urea? See structure below.

Answer: sp2-to-sp3

• The steric number of C is 3; therefore, the hybrid orbital it is using for each of its three sigma bonds is sp2.

• The steric number of N is 4; therefore, the four groups of electrons “sticking out” of it are in sp3 hybrid orbitals.

15.2.2 Explaining Bond LengthsHybridization can be used to explain the effect of multiple bonding on bond lengths. For a hybrid orbital with a greater “p character,” the region of high electron density stretches out farther from the nucleus; sp3 hybrids have higher p character than sp2 hybrids; more p orbitals are used to construct the sp3 hybrid orbital. This means that the greater the p character of the orbitals used for bonding, the longer the bond.

▼ Example 3

Explain why the C-to-C bond in H3C-CH3 is longer than that in H2C=CH2. Explain why the C-to-H bond is shorter than the C-to-C bond in both molecules.


Figures from http://en.wikipedia.org/wiki/File:Ethane-staggered-CRC-MW-dimensions-2D.png and http://en.wikipedia.org/wiki/File:Ethylene-CRC-MW-dimensions-2D.png (both public domain)Answer:• Both C atoms in H3C-CH3 are sp3 hybridized; the C-to-C bond

in this molecule is due to an sp3-to-sp3 overlap.• Both C atoms in H2C=CH2 are sp2 hybridized. The C-to-C

sigma bond in this molecule is an sp2-to-sp2 overlap.• The C-to-C bond is longer for the sp3-to-sp3 overlap compared

to an sp2-to-sp2 overlap. An sp3 orbital has greater p character than an sp2 orbital; the region of high electron density stretches out farther from the nucleus.

• The C-to-H bond is shorter than the C-to-C bond in H3C-CH3 because the H atom uses a 1s orbital to bond with C. An electron in a 1s orbital is closer to the nucleus than an electron in a 2s or 2p (or 2sp) orbital. Thus, the s-to-sp3 overlap in the C-to-H bond is expected to be shorter than the sp3-to-sp3 overlap in the C-to-C bond. A similar explanation applies to the bonds in H2C=CH2.

15.3 LCAO-MO MethodIn the valence bond method, each molecular orbital is said to be localized; each molecular orbital describes a distribution of electron density that is significant only between a pair of atoms. In the VB method, the calculation is restricted so that each molecular orbital is constructed only from atomic orbitals of two neighboring atoms.

In the LCAO-MO method, the molecular orbitals can be delocalized. A molecular orbital in the LCAO-MO method can describe an electron distribution that is spread out over the entire molecule. Each molecular orbital is expressed in terms of the atomic orbitals of all the atoms in the molecule. Whether or not a particular atomic orbital

http://en.wikipedia.org/wiki/File:Ethane-staggered-CRC-MW-dimensions-2D.png

http://en.wikipedia.org/wiki/File:Ethane-staggered-CRC-MW-dimensions-2D.png

http://en.wikipedia.org/wiki/


“contributes” significantly to a molecular orbital depends on the symmetry properties of the molecule. Figure 5 illustrates six of the molecular orbitals of benzene (C6H6); the orbit-als illustrated in the Figure are constructed from the six 2pz orbitals (one from each of the C atoms in the molecule).

A molecular orbital diagram is used to summarize how molecular orbitals are related to the atomic orbitals. Such a diagram for F2 is shown in Figure 6. The horizontal lines in middle of the diagram represent the energies of (an electron described by) the molecular orbitals; they are labeled with the names of the orbitals (1sσ, 1sσ*, etc). The horizontal lines on the sides represent the energies of atomic orbitals of the two atoms used to construct the molecular orbitals. The green dashed lines indicate which atomic orbitals have significant contributions to the construction of the pertinent molecular orbitals.

If the energy of a molecular orbital is less than that of the atomic orbitals from which it is constructed, it is said to be a bonding orbital; otherwise it is called an antibonding orbital. An asterisk (*) in the name of the orbital is used to indicate that

Figure 5. Molecular Orbitals of Benzene

http://www.chemtube3d.com/BenzeneMOs.htmlLicense: Creative Commons Attribution - Noncommercial Share Alike 2.0

http://www.chemtube3d.com/BenzeneMOs.html


Figure 6. Molecular Orbitals of F2

from http://www.chemtube3d.com/orbitalsfluorine.htmLicense: Creative Commons Attribution - Noncommercial Share Alike 2.0

the orbital is antibonding. Antibonding orbitals are associated with low electron densities between the nuclei. We assign electrons to molecular orbitals just like we do for atomic orbitals. We follow the Aufbau principle, Pauli’s Exclusion principle, and Hund’s rule to determine the ground state of the molecule. If more electrons are assigned to bonding orbitals, we expect the atoms to form a covalent bond. The bond order is calculated by taking the difference between the number of electrons in bond-ing orbitals and antibonding orbitals, then dividing the result by 2. A larger bond order implies a stronger bond.

http://www.chemtube3d.com/orbitalsfluorine.htm


▼ Example 4

Draw arrows to represent electrons on the molecular orbital diagram for F2. Determine the bond order.Answer: We need to draw 18 arrows since each F atom has 9 electrons. From the diagram, we can see that we have 10 electrons in bonding orbitals and 8 in antibonding orbitals. The bond order is (10-8)/2, or 1.


TEST YOURSELF


1. What type of orbital hybridization explains the linear electronic geometry around an atom with a steric number of 2?

A. sp B. sp2 C. sp3 D. sp3d E. sp3d2

2. What is the hybridization of the valence orbitals of the carbon atom in the structure below?

A. sp B. sp2 C. sp3 D. sp3d

3. In the structure below, the sigma bond between which two atoms is due to an sp2-to-sp2 overlap?

A. C and O B. C and N C. N and H

4. In the structure below, which of these bonds is the longest?

A. the C—C bond B. the C——C bond C. the C—H bond

5. Using Figure 6, explain why O2 is paramagnetic. Determine the bond order.


245

16CHAPTER

Intermolecular Forces

Properties of substances depend on how molecules interact. In this chapter, we examine the types of intermolecular forces and how they are manifested in properties of substances.

16.1 Explaining Properties of SubstancesIn general, properties like boiling point, melting point, viscosity, and surface tension depend on the strength of the attractive forces among the molecules. The stronger the attractive forces, the higher the boiling point, viscosity and surface tension of a liquid, and the higher the melting point of a solid.

Molecules gain more freedom of movement as a solid melts or as a liquid vaporizes. The amount of energy that they need to overcome the forces of attraction among them increases the stronger the forces are. Thus, they need to be heated to a higher temperature before they melt or boil. A higher temperature is indicative of higher molecular kinetic energies. Viscosity refers to the resistance of gases and liquids to flow. The stronger the attractive forces among the molecules of the gas or liquid, the more resistant they would be to flow. Surface tension refers to the energy needed to increase the surface area of a liquid. Molecules in the bulk are completely surrounded by other molecules and, therefore, experience stronger attractions compared to molecules on the surface. This is illustrated in Figure 1, where a molecule in the bulk is highlighted in red and a molecule on the surface is highlighted in yellow.

Figure 1. Representation of Liquid Droplet Molecules


To increase the surface area of a liquid requires movement of molecules from the bulk, where they experience stronger attractions, to the surface. This requires energy. The stronger the attractive forces, the more energy would be required, the higher the surface tension.

▼ Example 1

What can be said about attractive forces among water molecules compared to attractive forces among O2 molecules based on the fact that water is a liquid at room temperature while O2 is a gas?Answer:The fact that O2 is a gas at room temperature means its boiling point is below room temperature. In fact, we can look it up and find it to be −183°C. This means O2 is a liquid below −183°C and a gas above −183°C.The fact that water is a liquid at room temperature means that its boiling point is above room temperature. In fact, we know that water boils at 100°C. It is a liquid below 100°C and is a gas above 100°C.Since H2O boils at a higher temperature than O2, we conclude that attractive forces among H2O molecules must be stronger than attractive forces among O2 molecules.

How two different substances interact can be explained by comparing:• the strength of attraction among molecules of the same kind,• and the strength of attraction among molecules of one substance with molecules of

the other substance.

▼ Example 2

How do we explain why ethyl alcohol is soluble in water, but veg-etable oil is not?Answer:When a substance is dissolved in another liquid, the particles that make up the substance (which is called the solute) disperse throughout the other liquid (which is called the solvent) and each one is surrounded by solvent molecules (“solvated”). The extent to which this happens depends on three factors:1. strength of solute-solute interaction. The stronger the attrac-

tion among the particles that make up the solute, the less likely the solute particles will disperse.

Chapter 16: Intermolecular Forces 247

2. strength of solvent-solvent interaction. The stronger the attraction among the particles that make up the solvent, the less likely that the solute particles can disperse among the solvent molecules.

3. strength of solute-solvent interaction. The stronger the attraction between a solute particle and a solvent particle, the more likely the solute will be solvated.

To explain the solubility of ethyl alcohol and vegetable oil in water, we compare 1 and 3 since our solvent is the same (water) in both cases. We can surmise that solute-solute interactions are stron-ger among vegetable oil molecules or that solute-solvent interac-tions are stronger between water molecules and ethyl alcohol molecules. These are, in fact, the conclusions we will reach if we examine the structures of ethyl alcohol, vegetable oil, and water.

16.2 Types of Intermolecular ForcesThe attractive forces between two molecules can be classified into two major types: van der Waals forces and hydrogen bonding interactions. Van der Waals forces can be further classified into: dipole-dipole interaction, dipole - induced dipole interaction, and London dispersion forces

16.2.1 Dipole-Dipole InteractionDipole-dipole interaction refers to the attractive force that is due to molecular polarity. It exists only between two polar molecules, as illustrated in Figure 2. The bond between H and Cl in HCl is polar. There is a higher electron density in the Cl end of the molecule; this end, we say, is partially negative. The H end of the molecule is partially positive and it is attracted to the partially negative end of a neighboring HCl molecule. The partial charges are indicated by the lower case delta (δ), followed by a + or − sign.

Figure 2. Dipole-dipole Interaction


How can we tell if a molecular is polar? A molecule is polar if:• it has at least one polar bond,• and if it has more than one polar bond, the polarities of the bonds do not cancel out

due to symmetry.

How can we tell if a bond is polar? If the electronegativities of two atoms are different, then any sharing of electrons between the two atoms would be unequal. Shared electrons would spend more time closer to the more electronegative atom. We say that the bond is polarized towards the more electronegative atom. We could also say that the bond is polar covalent. We can refer to a table of electronegativities (as defined by Linus Pauling); see Table 1. On the Pauling scale, atoms are assigned electronegativities ranging from 0.7 for the least electronegative atom (Francium) to 4.0 for the most electronegative atom (Fluo-rine); on this scale, hydrogen has an electronegativity of 2.2. Note that electronegativity increases towards F. Metals have electronegativities less than 1.9 and nonmetals have electronegativities larger than 2.1.

With the electronegativity difference between F and the least electronegative nonmet-als being around 1.8 or 1.9, we can pretty much consider any pair of atoms that differ in electronegativity by 2.0 or higher as being in an ionic bond. Pauling defined the percent ionic character of a bond using the equation:

%ionic character = 1 – exp(–x2/4),where x is the electronegativity difference between the two atoms. Plugging in a value of 2.0 for x gives us:

%ionic character = 1 – exp(–2.02/4) = 0.63, or 63%If we try to calculate the electronegativity difference between pairs of atoms, we

will find that some pairs of atoms that we would normally consider as forming ionic bonds (metal + nonmetal) actually have an electronegativity difference less than 2.0. The bonds between these atoms should be considered as polar covalent bonds, not ionic.

Table 1. Electronegativities

H: 2.2

Li: 1.0 Be: 1.6 B: 2.0 C: 2.5 N: 3.0 O: 3.5 F: 4.0Na: 0.9 Mg: 1.3 Al: 1.6 Si: 1.9 P: 2.2 S: 2.6 Cl: 3.0K: 0.8 Ca: 1.0 Ga: 1.8 Ge: 2.0 As: 2.2 Se: 2.6 Br: 2.8*Data from A.L. Allred, J. Inorg. Nucl. Chem., 17, 215 (1961).


The “metal + nonmetal = ionic bond” rule is just a rule of thumb. There are numerous exceptions. Indeed, an entire field of study (coordination chemistry) deals with covalently bonded transition metal atoms.

If the electronegativity difference between two atoms is larger than zero but less than 0.5, the polarization is negligible; the bond between them is classified as polar covalent, but essentially nonpolar.

▼ Example 3

Classify the bond between C and H, Li and Br, Na and Cl, Na and Na.Answer:• For C and H: electronegativities of C and H are 2.5 and

2.2; difference = 0.3. The bond is polar covalent, but essen-tially nonpolar.

• For Li and Br: electronegativities of Li and Br are 1.0 and 2.8; difference = 1.8. The bond is polar covalent.

• For Na and Cl: electronegativities of Na and Cl are 0.9 and 3.0; difference = 2.1. The bond is ionic.

• For Na and Na: the electronegativity difference is obviously zero; bond is pure covalent. [Note: metallic bonding refers to bonding among atoms in a metallic solid; but the bond between just two atoms of a metallic element is pure covalent].

▼ Example 4

Explain why CO2 is nonpolar.Answer:In CO2, the C-to-O bonds are both polar, but because the two C=O bonds are 180° apart, the polarities cancel out as shown below.

This is true for any molecule where all bonds are identical and have a perfectly symmetrical orientation (2 around a linear center, 3 around trigonal planar, 4 around a tetrahedral center, 5 around a trigonal bipyramidal center, or 6 around an octahedral center). The molecule is nonpolar regardless of whether the bonds are polar or nonpolar.


▼ Example 5

Explain why a water molecule is polar.Answer:In water molecules, the polarities (shown in green in the figure below) do not cancel out. There is a net polarity towards the oxy-gen atom as indicated by the yellow arrow in the figure.

▼ Example 6

For which pair of molecules does dipole-dipole interaction exist: CO2 and CO2, H2O and H2O?Answer: H2O and H2O. Water molecules are polar molecules. CO2 molecules, on the other hand, are nonpolar molecules.

16.2.2 London Dispersion ForcesLondon dispersion refers to the force of attraction that exists between any pair of mol-ecules. It is due to temporary molecular polarizations, which occur because electrons are always moving. The larger the molecule, the more frequently the polarizations occur. Therefore, we expect attractions to be stronger among larger molecules.

▼ Example 7

For which of the following pairs of molecules are attractions due London dispersion forces strongest: CH4 and CH4, C2H6 and C2H6.Answer: C2H6 and C2H6We expect London dispersion forces to be stronger between larger molecules.


London dispersion is the predominant van der Waals interaction among most molecules. It is tempting to think that London dispersion would be weaker than dipole-dipole interaction since it is due to temporary polarizations, while dipole-dipole interactions are due to permanent polarization. However, calculations indicate that dispersion forces are just as strong among small polar molecules (like HCl) as dipole-dipole interaction. The apparent inconsistency is due to quantum effects.

16.2.3 Dipole - induced Dipole InteractionDipole - induced dipole interaction occurs between a polar molecule and a nonpolar molecule. When a nonpolar molecule comes close to the positive end of polar molecule, its electrons would be attracted toward the polar molecule causing temporary (or induced) polarization. Similarly, when a nonpolar molecule comes close to the negative end of a polar molecule, its electrons would be repelled, again causing temporary (or induced) polarization.

▼ Example 8

For which of the following pairs do we expect dipole - induced dipole interaction?A. H2O and H2OB. CO2 and CO2C. H2O and CO2

Answer: CH2O is a polar molecule. We expect two water molecules to have dipole-dipole as well as London dispersion forces between them. CO2 is nonpolar; we expect the only interaction between two CO2 molecules to be London dispersion. We expect dipole - induced dipole interaction to occur between H2O (which is polar) and CO2 (which is nonpolar).

16.2.4 Hydrogen Bonding InteractionIn a molecule, a hydrogen atom covalently bonded to a highly electronegative atom (F, O, or N) is almost stripped bare of electrons. This makes it strongly attracted to a lone pair of a highly electronegative atom (F, O, or N) in a neighboring molecule. The interaction is called hydrogen bonding.

Figure 3 illustrates hydrogen bonding interaction between two water molecules. The H atoms in water (like the H highlighted in green) are bonded to a highly electronegative atom (oxygen). Because it is almost stripped bare of electrons, it is strongly attracted to the lone pair of the oxygen atom (highlighted in purple) in the neighboring water molecule.


Hydrogen bonding interaction is much stronger than dipole-dipole interaction. For small molecules it is also stronger than London dispersion forces. Hydrogen bond-ing accounts for the unusually high boiling point of water. Water molecules are small; substances made of molecules of comparable size are gaseous at room temperature. For example, O2 is a gas at room temperature, while water is a liquid even though H2O mol-ecules are smaller than O2 molecules. One would expect London dispersion forces to be stronger among O2 molecules. The “anomaly” is explained when one takes into account that water molecules are polar and capable of extensive hydrogen bonding as illustrated in Figure 4.

16.3 Ion-Molecule InteractionsInteraction of ions with molecules can be classified as ion-dipole interaction or ion - induced dipole interaction. Ion - induced dipole interaction involves ions and nonpolar molecules.

Figure 3. Hydrogen Bonding Interaction Between Two Water Molecules

Figure 4. Extensive Hydrogen Bonding Interactions in Water


▼ Example 9

What type of interaction is involved between Na+ ions and H2O molecules when NaCl is dissolved in water?Answer: Ion - dipole interactionNa+ is an ion and H2O is a polar molecule. The oxygen end of the H2O molecules will be attracted to the positively charged Na+ ion.

16.4 Hydrophobicity and HydrophilicityAmong large molecules (4 or more atoms besides hydrogen), London dispersion is the predominant type of interaction. However, if these molecules have N or O atoms, these atoms will typically have a lone pair that would be capable of hydrogen-bonding interac-tion with water molecules. Parts of the molecule near or at the N and O atoms are said to be hydrophilic (water-loving) and the rest of the molecule is said to be hydrophobic. The strong London dispersion forces between large molecules are mainly due to the hydro-phobic parts and are, therefore, referred to as hydrophobic interactions. We see major applications of these ideas in biological sciences (lipid bilayers, hydrogen bonding in DNA molecules, enzyme-substrate interaction, etc.).

▼ Example 10

Consider the structure shown below for the molecule found in the food dye known as FD&C Red 40. Each unlabeled corner in the structure represents a C atom with enough H around it so that each C atom has an octet. Is this dye likely to be soluble or insoluble in water?

from http://en.wikipedia.org/wiki/File:Allura_Red_AC.png(public domain)

http://en.wikipedia.org/wiki/File:Allura_Red_AC.png


Answer: SolubleThis molecule has numerous hydrophilic parts (O and N atoms) capable of hydrogen bonding with water molecules.This is a com-monly used food dye; we find it in drink mixes like Kool-AidTM.

▼ Example 11

Consider the model shown below for a molecule of stearic acid, C18H36O2. Based on this information, is stearic acid likely to be soluble or insoluble in water?

from http://en.wikipedia.org/wiki/File:Octadecanoic_acid_(stearic).pngLicense: Creative Commons Attribution: Share-AlikeAnswer: InsolubleMost of the molecule is hydrophobic. The only hydrophilic part is shown in red (the oxygen atoms). Stearic acid is an example of a fatty acid and is found in animal fats and vegetable oils.

http://en.wikipedia.org/wiki/File:Octadecanoic_acid_


TEST YOURSELF


1. Which of the following has the highest electronegativity? A. Na B. Br C. F

2. Which type of element has the lowest electronegativity? A. a metal B. a nonmetal C. a metalloid

3. Which pair of atoms would form a bond where the polarity is directed towards the first atom?

A. C and C B. H and F C. O and C D. I and Cl

4. Which pair of atoms would form a polar covalent but essentially nonpolar bond? A. N and N B. H and C C. Na and Cl D. Be and I

5. Which of the following is/are true? 1) A nonpolar molecule can have polar bonds. 2) A molecule with no polar bonds is nonpolar. A. 1 only B. 2 only C. both D. neither

6. Which of the following molecules is polar? A. H2O B. BF3 C. CH4 D. PF5

7. For which pair of molecules does dipole-dipole interaction occur? A. H2O and CO2 B. H2O and HF C. CO2 and CO2 D. CH4 and CO2

8. For which pair of molecules is London Dispersion strongest? A. H2O and H2O B. H2O and H2S C. H2S and H2S

9. Which of the following has the lowest boiling point? A. O2 B. F2 C. Cl2



10. Which of the following explains why the melting point of iodine is higher than that of iodine chloride?

A. London dispersion forces are stronger among I2 molecules, which are larger and more polarizable

B. ICl molecules are capable of dipole-dipole interaction, whereas I2 molecules are not C. Both D. Neither

11. How many atoms in the acetic acid molecule (CH3COOH) are capable of hydrogen bonding interaction with water molecules?

A. 1 B. 2 C. 3 D. 4

12. Which type of intermolecular forces is present between any pair of molecules? A. London dispersion forces B. dipole-dipole interaction C. hydrogen bonding interaction

13. Which of the following is least soluble in water?

A. NH3 B. HF C. CH4

257

CHAPTER

The Mole Concept

Because atoms, ions, and molecules are very, very small, it is convenient to count them in groups called moles.

17.1 Counting by MolesA mole (officially abbreviated mol) is a group count, just like a dozen. While a dozen count is exactly equal to twelve, a mole count is exactly equal to Avogadro’s number. Avogadro’s number, NA, is a 24-digit exact, whole number. We do not yet know all 24 digits. The first six digits are known to be 6, 0, 2, 2, 1, and 3; these are followed by 18 more digits. For most purposes, it is sufficient to express Avogadro’s number to three significant digits; it is approximately equal to 6.02×1023, or about 602 billion trillion.

▼ Example 1

Which has more atoms: 1 mol 12C or 1 mol 13C?Answer: neither, the number of atoms in either case is the same.

This is like asking “which has more people: 1 dozen children or 1 dozen adults?” We expect 1 mol of C-13 to weigh more than 1 mol C-12, just as we expect 1 dozen adults to weigh more than 1 dozen children. But as long as the question involves counting, they are the same.

The definition of Avogadro’s number. Avogadro’s number is equal to the number of atoms in exactly 12 grams of carbon-12. Imagine we have a sample that is exactly 12 grams, and each and every atom in the sample is C-12. If we could count all the atoms in that sample, the number we get is defined as Avogadro’s number. We could also say that one mole of carbon-12 is defined to have a mass of exactly 12 grams. The unit for Avogadro’s number is mol−1 (which is read as “per mole”). For clarity, it is often useful to explicitly specify what is being counted, as in atoms mol−1, or ions mol−1, or molecules mol−1.

17


If we are given the count in moles, calculating the actual individual count is straight-forward. Essentially, we multiply the mole count by Avogadro’s number to get the individual count. If n is the mole count and N is the individual count, we multiply n by Avogadro’s number (NA) to get N:

N n NA=

The same basic idea is involved when we convert dozens to an individual count. If we have 2 dozen eggs, we multiply 2 by 12 to get the individual count of 24 eggs.

2 0121

24. dozeggsdoz

eggs⎛

⎝⎜⎜

⎞

⎠⎟⎟ =

▼ Example 2

Consider a sample containing 2.0 moles of CO2 molecules. How many CO2 molecules are there in this sample?

Answer: 1.2×1024 molecules

N n N mol molecules mol molecA= = ×( ) ×( ) = ×−2 0 10 6 02 10 1 2 1023 23 1 24. . . uules

N n N mol molecules mol molecA= = ×( ) ×( ) = ×−2 0 10 6 02 10 1 2 1023 23 1 24. . . uules

We can also treat this question as a unit conversion problem; we can set up the calculation as follows:

2 06 02 10

11 2 10

2324.

..mol

moleculesmol

molecules× ×⎛

⎝⎜

⎞

⎠⎟ = ×

Note that we rounded off Avogadro’s number to only 3 significant digits since the amount to be converted (2.0 mol) is only given to two significant digits.

What if we want to convert individual count to group count? All we have to do is divide the number of eggs by 12 to get the dozen count. Similarly, if we know the individ-ual count (N), we just divide it by the Avogadro’s number (NA) to get the mole count (n).

n NNA

=

Chapter 17: The Mole Concept 259

▼ Example 3

How many moles of Mg atoms are there in a sample containing 3.00 × 1023 atoms of Mg.Answer: 0.498 mol Mg atoms or 3.00x1023 Mg atoms

nNN

atomsatoms mol

molA

= = ××

=−

3 00 106 022 10

0 49823

23 1

..

.

Note that we used 4 significant digits for Avogadro’s number here since the amount that needs to be converted has 3 significant digits. We can also treat this as a conversion problem and set up the calculation as follows:

3 00 101

6 022 100 49823

23..

.× ××

⎛

⎝⎜

⎞

⎠⎟ =atoms

molatoms

mol

17.2 Counting Atoms in MoleculesTo count atoms of an element in a molecule, we make use of the fact that subscripts in a formula are counting numbers; they tell us how many atoms of a given element are pres-ent in one formula unit. Therefore, they also tell us how many moles of atoms of a given element are present in a mole of formula units.

▼ Example 4

How many O atoms are in 2.0 mol CO2?Answer: 4.0 mol O atoms, or 2.4x1024 O atomsThe subscripts in the formula tell us that there are 2 atoms of O in one molecule of CO2. Therefore, there are 2 moles of O in 1 mole of CO2 and we can treat this as a simple conversion prob-lem. We set up the solution as follows:

2 02

14 02

2

. .mol COmol O

mol COmol O×

⎛

⎝⎜

⎞

⎠⎟ =

We multiply the number of molecules, 2.0 moles of CO2, by a con-version factor that relates moles of O to moles of CO2.


Essentially, what we are doing is multiplying the moles of mol-ecules by the subscript of the atom we are interested in. If we need the individual count, we simply multiply the mole count by Avogadro’s number.

▼ Example 5

How many atoms are in 2.0 mol CO2?Answer: 6.0 mol atoms, or 3.6x1024 atomsA molecule of CO2 consists of three atoms --- one C atom and two O atoms. Since a mole is just a group count, we can say that a mole of CO2 molecules consists of three moles of atoms --- one mole of C and two moles of O. In other words: there are 6.0 moles of atoms in 2.0 moles of CO2 molecules.

2 031

6 022

. .mol COmol atomsmol CO

mol atoms×⎛

⎝⎜

⎞

⎠⎟ =

17.3 Counting Ions in Ionic CompoundsTo count ions in a sample of an ionic compound, we make use of the fact that the subscript of the ion in the formula tells us how many ions are in one formula unit of the compound. Therefore, the subscript also tells us how many moles of ions are in one mole of formula units.

▼ Example 6

How many cations are in 3.0 moles of ammonium sulfate?Answer. 6.0 molThe formula of ammonium sulfate is (NH4)2SO4. This means that one formula unit of the compound consists of two cations (the subscript for ammonium in the formula is 2) and one anion (the implied subscript for sulfate in the formula is 1). This also means that one mole of ammonium sulfate is made up of 2 moles of ammonium ions and 1 mole of sulfate ions. Therefore:

3 02

16 04 2 4

4

4 2 44. .mol NH SO

mol NHmol NH SO

mol NH( ) × ( )⎛

⎝⎜⎜

⎞

⎠⎟⎟ =

++

Chapter 17: The Mole Concept 261

TEST YOURSELF


1. Which of these is false? A. A mole of atoms contains exactly 6.02 × 1023 atoms B. The exact number of molecules in one mole of molecules is equal to Avogadro’s

number C. The exact number of ions in one mole of ions is equal to Avogadro’s number D. None of the above

2. Which has more atoms? A. 1 mole of Mg-24 B. 1 mole of Mg-26 C. neither

3. Which has more molecules? A. 1 mole of SO2 B. 1 mole of SO3 C. neither

4. How many atoms of carbon are in a 2.50 mol sample of carbon? A. 1.51 × 1024 B. 30.0 C. 4.15 × 10−22

5. How many moles of O2 molecules are in a sample containing 3.0 × 1022 O2 molecules? A. 2.0 mol B. 1.06 × 10−21 mol C. 0.050 mol

6. How many atoms of oxygen are in 1.50 mol CO2? A. 1.50 mol B. 0.300 mol C. 3.00 mol D. 0.750 mol

7. Which has more atoms? A. 3.0 mol SO2 B. 2.0 mol SO3 C. neither

8. Which has more O atoms? A. 3.0 mol SO2 B. 2.0 mol SO3 C. neither

9. How many cations are in 2.00 mol Na3PO4? A. 0.667 mol B. 6.00 mol C. 0.500 mol D. 1.00 mol

10. How many moles of nitrate ions are in a mixture of 2.00 mol calcium nitrate and 1.50 mol aluminum nitrate?


263

18CHAPTER

Molar Mass

Since atoms, molecules and ions are too small and too numerous for us to actually count one by one, we count them in groups called moles. To do this, we need to know the mass of a mole of atoms (or ions or molecules); if we do, then all we would have to do is weigh a sample and do a simple calculation to determine the number of moles.

18.1 Determining Molar MassThe term molar mass refers to the mass per mole. The unit for molar mass is g/mol or g mol−1. By definition:

• one atom of carbon-12 has a mass of exactly 12 u.• one mole of carbon-12 atoms has a mass of exactly 12 grams.

A consequence of these two definitions is that:if something has a mass of X atomic mass units, then a mole of that something has a mass of X grams.

This also means that an Avogadro number of atomic mass units is equivalent to one gram; to three significant digits:

6.02 × 1023 u = 1.00 gor, dividing both sides by 6.02 × 1023, we can show that:

1.00 u = 1.66 × 10−24 g

▼ Example 1

The mass of one proton is 1.00 u. What is the mass of one mole of protons?Answer: 6.02 × 1023 u, or 1.00 g


▼ Example 2

Calculate the mass of one C-12 atom in g.Answer: 1.99 × 10−23 gSince the mass of one C-12 atom is exactly 12 u, then, to three significant digits, the mass in grams is (using factor label):

121 66 10

1 001 99 10

2423u

gu

g( ) × ×⎛

⎝⎜⎜

⎞

⎠⎟⎟ = ×

−−.

..

or (using algebraic substitution):

12 12 1 66 10 1 99 1024 23u g g= ×( ) = ×− −. .

▼ Example 3

What is the molar mass of C-12?Answer: exactly 12 g/mol or 12 g mol–1

By definition, one mole of C-12 has a mass of exactly 12 g.

▼ Example 4

If we look up the mass of a C-13 atom, we find that it is 13.0033548 u. What is the molar mass of C-13?Answer: 13.0033548 g/mol, or 13.0033548 g mol−1

For most purposes, it is sufficient to round off molar masses to four significant figures; in the case of C-13, the molar mass is 13.00 g/mol. In general, the numerical value of the molar mass of a specific isotope should be close to a whole number (the mass number).

▼ Example 5

Which isotope has a molar mass of 34.97 g/mol: Cl-35 or Cl-37?Answer: Cl-35 To the nearest whole number, 34.97 g/mol is 35 g/mol. Indeed, we can look it up and find it to be so.

Chapter 18: Molar Mass 265

Typical periodic tables will not have molar masses listed for specific isotopes. The numerical value of the average mass of the atoms, the atomic weight, is typically given in a periodic table. This is also the numerical value of the molar mass of a naturally-occurring sample of the element.

▼ Example 6

The atomic weight for Cl is given in a typical periodic table as 35.45. What does this mean?Answer: This means that the average mass of Cl atoms in a naturally-occurring sample is 35.45 u, and the molar mass of a naturally-occurring sample of Cl atoms is 35.45 g/mol.

▼ Example 7

What is the molar mass of C?Answer: unless otherwise specified, we have to assume that we’re dealing with a naturally-occurring sample. We can look up this value from a periodic table and find that it is 12.01 g/mol.

▼ Example 8

What is the molar mass of oxide ions (O2−)?Answer: 16.00 g/molThis is essentially the same as the molar mass of O atoms, which we can look up and find to be 16.00 g/mol. An oxide ion has two more electrons compared to a neutral oxygen atom; we would expect its molar mass to be larger. But electrons hardly con-tribute to the total mass of an atom, so we can generally ignore charges when calculating molar masses.

▼ Example 9

What is the molar mass of H2O?Answer: 18.02 g/molA mole of H2O molecules contains two moles of H and one mole of O atoms. Referring to a periodic table, we find that:• The mass of 2 moles of H is 2 × 1.008 g.


• The mass of 1 mole of O is 1 × 16.00 g.Adding up the masses gives us a total of 18.02 g for one mole of H2O. Therefore, the molar mass of H2O is 18.02 g/mol.

NOTE: most modern Chemistry textbooks have standardized on the term “Molar Mass” but other terms are still being used. These other terms include: (average) atomic mass, atomic weight, gram atomic weight, gram atomic mass, formula weight, gram formula weight, molecular mass, molecular weight, gram molecular weight, and gram molecular mass. There are subtle differences in the meanings of these terms, but numerically, they are all the same.

18.2 Counting by WeighingKnowing the molar mass of a substance allows us to count the atoms, molecules or ions in any sample of that substance by simply weighing the sample. If we know that the mass of one mole of something is Mm, then the mass (m) of n moles is:

m = n Mm

Dividing both sides of this equation by n, we can show that the number of moles (n) is just equal to the mass divided by the molar mass:

n mMm

=

▼ Example 10

What is the mass of 2.0 mol H2O?Answer: 36 gTo convert moles to mass, we need to know the molar mass. In an earlier example, we found the molar mass of H2O to be 18.02 g mol−1. Therefore, the mass of 2.0 mol H2O is:

m n M mol g mol gm= = ( ) =−( . ) .2 0 18 02 361

It is often useful to be more descriptive when specifying units. In this case, we can write our units as “g H2O” and “mol H2O.”

2.0 mol H Og H O

mol H Og H O2

2

22

18 021

36×⎛

⎝⎜⎜

⎞

⎠⎟⎟

=.


Note that mol−1 means mol in the denominator. We can think of molar mass as a conversion factor to change moles to grams. If we want to change grams to moles instead, we can just flip this conversion factor, as in the next example.

▼ Example 11

How many molecules are in a 36.04 g sample of water?Answer: 2.000 molEarlier, we found the molar mass of water to be 18.02 g/mol. So:

36 04

18 022 000

1

.

..

g

g molmol

−=

or:

36 041

18 022 0002

2

22.

..g H O

mol H Og H O

mol H O×⎛

⎝⎜⎜

⎞

⎠⎟⎟

=

We can multiply this answer by Avogadro’s number if we want an individual count. An answer in moles is a perfectly acceptable way of answering a “how many” question. A mole is a group count, just like a dozen. Saying you have 2 dozen eggs is the same thing as saying you have 24 eggs.

18.3 Counting by Measuring VolumeSubstances that exist as liquids under ordinary conditions are not always easy to weigh. This is especially true for liquids that are volatile, toxic, or corrosive. It would be more convenient to measure their volume in the laboratory. If we know the density of a pure liquid, then we can do volume-to-mass conversion so that we figure out the moles of substance in our sample. The density is defined as the ratio of the mass (m) to volume (V).

d mV

=


▼ Example 12

The density of carbon tetrachloride, CCl4, is 1.59 g/mL. How many moles CCl4 are present in a 200.0 mL sample of this liquid?Answer: 2.07 molFirst, we rearrange the defining equation for density to solve for the mass:

m V d mL g mL g= = ( ) =−( . ) .200 0 1 59 3181

Then, we convert grams to moles. For this, we need the molar mass of CCl4, which is 153.8 g/mol:

nm

M

g

g mol2.07 mol

m

= = =−

318

153 8 1.

or:

318 g CCImol CCI

g CCI2.07 mol CCI4

4

44

1153 8

×⎛

⎝⎜⎜

⎞

⎠⎟⎟

=.

Oftentimes, we know how many moles of a liquid substance we need. To figure out what volume to dispense, we first convert moles to mass, then figure out the volume from the mass.

▼ Example 13

An organic synthesis experiment calls for 0.200 mol of ethanol, C2H5OH. The density of ethanol is 0.785 g/mL. What volume of ethanol do we need?Answer: 11.7 mLFirst, we figure out the mass of ethanol that we need. For this, we need the molar mass, which is 46.07 g/mol.

m n M mol g mol gm= = ( ) =−( . ) . .0 200 46 07 9 2141

Next, we calculate the volume by rearranging the defining equa-tion for density; solving for V, we get:

Vmd

g

g mLmL= = =

−

9 214

0 78511 7

1

.

..

There should only be three significant digits in our calculated mass, but we kept an extra digit since we needed to carry it over to another calculation.


18.4 Amount of Element in a CompoundThe percent composition of a compound refers to a listing of the amounts of all the elements present in the compound, with each amount expressed as percentage of the total mass. The percentage of an element is simply equal to the mass of the element present in a pure sample of the compound, divided by the mass of the sample, multiplied by 100%. Since compounds have definite composition, we should get the same value regardless of the mass of sample, provided the sample is pure. It is convenient to just use the mass of one mole to do these calculations.

▼ Example 14

Determine the percent composition of NO2.Answer: 30.45% N, 69.55% OOne mole of NO2 contains:• one mole of N, or 14.01 g• two moles of O, or 2 × 16.00 g = 32.00 g

for a total mass of 46.01 g. Therefore:

%_

_%

.

.% . %N in NO

mass Nmass NO

g

g22

10014 01

46 01100 30 45= × = × =

%_

_%

.

.% . %O in NO

mass Omass NO

g

g22

10032 01

46 01100 69 55= × = × =

Note that percentages should add up to 100%, so we could have solved for %O by subtracting the %NO2 from 100%.

To calculate the amount of an element in a sample of a compound, we simply multiply the mass of the compound by the percentage of the element.

▼ Example 15

How much oxygen can be obtained from 50.0 g of NO2?Answer: 34.8 gIn the previous example, we saw that the percentage of O in NO2 is 69.55%. Therefore, the amount of oxygen present in 50.0 g of NO2 is 69.55% of 50.0 g.

50.0 g × 69.55% = 50.0 g × 0.6955 = 34.8 g


TEST YOURSELF


1. To two significant figures, the mass of an electron is 9.1 × 10−31 kg. Convert this to atomic mass units.

2. Which of the following is a plausible mass for a diatomic molecule? A. 5.32 × 10−23 g B. 2.33 × 10−31 kg C. 160 g D. 1.66 × 10−21 g

3. Which of the following has a molar mass of 60.06 g/mol? A. CaCO3 B. HC2H3O2 C. none of these

4. What is the molar mass of NH3? A. 8.00 g/mol B. 10.00 g/mol C. 15.02 g/mol D. 17.03 g/mol

5. What is the molar mass of Ca(NO3)2? A. 70.09 g/mol B. 82 g/mol C. 164.10 g/mol

6. What is the molar mass of Na2SO4.10H2O?

A. 160.7 g/mol B. 322.9 g/mol C. 72.05 g/mol

7. What is the mass of one mole of CO2? A. 28.01 g B. 30.01 g C. 44.01 g

8. What is the mass of 2.00 mol NH3? A. 20.0 g B. 34.1 g C. 30.0 g

9. If an experiment calls for 0.500 mol calcium carbonate (CaCO3), how many grams of pure calcium carbonate do we need?

10. If an experiment calls for 0.200 mol acetic acid (HC2H3O2), how many grams of acetic acid do we need?

11. How many moles of CH4 are in 4.00 g of CH4? A. 0.249 mol B. 64.0 mol C. 0.307 mol



12. How many moles of acetic acid (HC2H3O2) are present in a 5.00 g sample of pure acetic acid?

13. Which sample has more O atoms? A. 4.00 g O2 B. 4.00 g O3 C. neither (equal)

14. Which sample has more anions? A. 4.00 g calcium carbonate B. 4.00 g calcium nitrate C. neither (equal)

15. Acetone is a liquid with a density of 0.7925 g mL−1 and its molar mass is 58.08 g mol−1. What volume of acetone should be dispensed if an experiment calls for 0.500 mol of acetone?

16. Concentrated HCl is an aqueous solution that is 36–38% HCl by mass. Suppose a concentrated HCl solution, with a density of 1.18 g mL−1, is 38% HCl by mass. What volume of this solution contains 0.500 mol HCl?

17. What is the percentage (by mass) of O in CO2? A. 27.29% B. 66.67% C. 72.71%

18. How many moles of O atoms are present in 50.0 g of NO2?

19. Calculate the mass of O atoms in 2.00 mol NO2.

20. Calculate the moles of C in a mixture of 10.0 g calcium carbonate and 15.0 g aluminum carbonate.

273

19CHAPTER

Elemental Analysis

Elemental Analysis is the determination of the elemental composition of a compound. The goal of elemental analysis is the determination of the empirical formula of a compound.

19.1 Empirical FormulaThe empirical formula of a compound is the simplest way of representing the relative abundance of the atoms of the elements that make up a compound. The greatest common factor of the subscripts in an empirical formula is 1. If the greatest common factor of the subscripts of a formula is not 1, then just divide all the subscripts by that factor in order to get the empirical formula.

▼ Example 1

What is the empirical formula of glucose (C6H12O6)?Answer: The subscripts are 6, 12, and 6. The greatest common factor is 6. Dividing by 6, we get:• C: 6/6 = 1• H: 12/6 = 2, and• O: 6/6 = 1

The subscripts for the empirical formula are 1, 2, and 1. Therefore, the empirical formula is CH2O.

How can we tell if the greatest common factor is 1? If the smallest number is 1, or if we divide all the numbers by the smallest number and we get at least one non-integer, then the greatest common factor is 1.


▼ Example 2

Is C3H6O2 an empirical formula?Answer: Yes, the greatest common factor for the subscripts (3,6, and 2) is 1. If we were to divide all three numbers by the smallest (which is 2):• 3/2 = 1.5• 6/2 = 3• 2/2 = 1

we find that one of the resulting numbers (1.5) is not an integer.

19.2 Molecular FormulaThe subscripts of the (correct) molecular formula of a compound are multiples of the subscripts of the empirical formula.

▼ Example 3

If the empirical formula of a compound is CH2, what are the possible molecular formulas?Answer: CH2, C2H4, C3H6, C4H8, etc.The subscripts of the empirical formula are 1 and 2 for C and H, respectively. Remember that unwritten subscripts are implied to be 1. To get the multiples:• we multiply both numbers by 1; this gives us 1 and 2.• we multiply both numbers by 2; this gives us 2 and 4.• we multiply both numbers by 3; this gives us 3 and 6.• etc.

The empirical mass is the molar mass associated with the empirical formula of a compound. The actual molar mass of a compound is a multiple of the empirical mass. If we divide the molar mass by the empirical mass we should get a whole number.

▼ Example 4

What are the possible molar masses of a compound with an empirical formula of CH2.Answer: multiples of 14.02 g mol−1

• The molar mass if the molecular formula is CH2 is (12.01 + 2 × 1.008) g mol−1, or 14.02 g mol−1.

Chapter 19: Elemental Analysis 275

• If the molecular formula is CH2, then the molar mass is 14.02 g mol−1.

• If the molecular formula is C2H4, then the molar mass is2 × 14.02 g mol−1, or 28.04 g mol−1 .

• If the molecular formula is C3H6, then the molar mass is3 × 14.02 g mol−1, or 42.06 g mol−1.

• etc…

▼ Example 5

Which of the following is least likely to be obtained as the molar mass of a compound with an empirical formula of CH2? A. 28.1 g/mol, B. 42.1 g/mol, C. 78.0 g/molAnswer: CThe empirical mass is (12.01 + 2 × 1.008) g/mol, or 14.02 g/mol. Let’s examine the choices given for the molar mass. Dividing by the empirical mass, we get• 28.2/14.02 = 2.01 (this is close enough to a whole number).• 42.1/14.02 = 3.00.• 78.0/14.02 = 5.56 (too far from a whole number).

19.3 Determining Empirical FormulaThe general strategy for determining the empirical formula of a compound is to:

• break the compound down into its component elements.• determine the number of atoms of each element obtained (the counts can be

individual counts or group counts in moles).• determine the subscripts of the empirical formula by reducing the count-to-count

ratio to the smallest possible set of whole numbers.

▼ Example 6

A sample of a compound of carbon and hydrogen is found to con-tain 2.500 × 1022 C atoms and 7.490 × 1022 H atoms. What is the empirical formula?Answer: CH3The C-to-H count-to-count ratio is 2.500 × 1022-to-7.490 × 1022. We need to reduce this to the smallest whole number ratio. A good strategy is to divide by the smaller number.


2.500 × 1022/2.500 × 1022 = 1.0007.490 × 1022/2.500 × 1022 = 2.996

When doing this type of calculation, we have to make sure that we follow the rules for rounding involving significant figures. We should not expect to see exact whole numbers, especially from experimental data. Experimental data are subject to random errors. Since 2.996 is very close to 3, the C-to-H count ratio is 1:3. The empirical formula is CH3.

▼ Example 7

A sample of a compound of nitrogen and oxygen contains 1.90 mol N and 3.80 mol O. What is the empirical formula of the compound?Answer: NO2The N-to-O mole-to-mole ratio is 1.90-to-3.80. Reduce this to the smallest whole number ratio. A good strategy is to divide by the smaller number.1.90/1.90 = 1.003.80/1.90 = 2.00

Therefore, the empirical formula is NO2.

In reducing count-to-count or mole-to-mole ratios to whole numbers, we round off only those numbers that are very close to a whole number.

▼ Example 8

A sample of a compound of C and H contains 5.00 mol C and 6.65 mol H. What is the empirical formula of the compound?Answer: C3H4Reduce the mole-to-mole ratio (5.00-to-6.65) to the smallest possible whole number ratio. A good starting strategy is to divide by the smaller number.5.00/5.00 = 1.006.65/5.00 = 1.33

Do not round off 1.33 to 1. The next thing to do is multiply by 2 to see if we can get all the numbers close to a whole number.1.00 × 2 = 2.001.33 × 2 = 2.66


The number 2.66 is still far from a whole number. Try multiplying by 3 instead.1.00 × 3 = 3.001.33 × 3 = 3.99

Now, 3.99 can be rounded off to 4. The empirical formula is C3H4.

Typical laboratory data give us information about masses, not moles. We just need to convert the information to moles to determine the empirical formula.

▼ Example 9

A sample of a compound of C and H contains 1.200 g C and 0.4000 g H. What is the empirical formula of the compound?Answer:First convert given information to moles:• 1.200 g C → 0.09992 mol C• 0.4000 g H → 0.3968 mol H

Then, reduce the mole-to-mole ratio to a whole number ratio. Start by dividing with the smaller number:• C: 0.09992/0.09992 = 1.000• H: 0.3968/0.09992 = 3.971

The number 3.971 is close enough to a whole number, so we can round it off (to 4). The empirical formula is CH4.

We do not need to know the actual amount of compound to determine empirical formula. We just need to know the relative amounts of the elements present. Relative amounts are usually given as percentages by mass. If we have the percentages, we can assume any amount of compound and base our calculations on the amount of each element in that sample. A convenient amount to assume is 100 g.

▼ Example 10

A compound of K, Cl, and O is found to be 31.9% K and 28.9% Cl. Determine the empirical formula.Answer: KClO3Since we are given percentages, we assume we have a 100 g sample of the compound. We calculate the amounts of K, Cl, and O in this sample.• K: 31.9% of 100 g is 31.9 g


• Cl: 28.9% of 100g is 28.9 g• O: Remaining percentage = 100% − (31.9% + 28.9%)

= 39.2% 39.2% of 100 g is 39.2 g

Next, we change mass to moles:• K: 31.9 g → 0.816 mol• Cl: 28.9 g → 0.815 mol• O: 39.2 g → 2.45 mol

Then reduce mole-to-mole ratio to small whole numbers by dividing with the smallest number (0.815):• K: 0.816/0.815 = 1.00• Cl: 0.815/0.815 = 1.00• O: 2.45/0.815 = 3.01 (close enough to a whole number;

round off to 3)Thus, the K:Cl:O ratio is 1:1:3.

19.4 Combustion AnalysisCombustion analysis is a common procedure used to do elemental analysis of organic compounds. In combustion analysis, all the C atoms in the compound end up as CO2 and all the H atoms in the compound end up as H2O. In the laboratory, a sample of the compound is burned; in the process the CO2 and H2O produced are collected.

▼ Example 11

A 2.003 g sample of compound containing C, H, and O is burned, and 4.401 g of CO2 and 2.703 g of H2O were collected. What is the empirical formula of the compound?Answer: C2H6ORemember that the goal is to get the mole-to-mole ratio of the elements. So, the first step is to convert the given information to moles:• 4.401 g CO2 → 0.1000 mol CO2• 2.703 g H2O → 0.1500 mol H2O

From these, we can calculate the moles of C and H that came from the compound:• C: 0.1000 mol CO2 has 0.1000 mol C, which came from the

compound• H: 0.1500 mol H2O has 0.3000 mol H, which came from

the compound


Now, we just need to figure out moles of O. First, we subtract the mass of C and H from the total mass to get the mass of O, so that we can calculate moles of O:• 0.1000 mol C → 1.201 g C• 0.3000 mol H → 0.3024 g H• Therefore:

mass of O = total mass of sample − (mass of C + mass of H) = 2.003 g − (1.201 g + 0.3024 g) = 0.800 g

We can now calculate moles of O in the compound: 0.800 g O → 0.0500 mol O.At this point, we have the moles of all the elements in the compound:• 0.1000 mol of C• 0.3000 mol H, and• 0.0500 mol of O

Divide by the smallest number to try to get whole numbers:• C: 0.1000/0.0500 = 2.00• H: 0.3000/0.0500 = 6.00• O: 0.0500/0.0500 = 1.00

The obvious mole-to-mole ratio is 2:6:1. Therefore, the empirical formula of the compound is C2H6O.


TEST YOURSELF


1. What is the sum of the subscripts in the empirical formula of C6H12? A. 1 B. 2 C. 3 D. 18

2. A compound with an empirical formula of NO2 has a molar mass between 43 and 47 g/mol. What is the molecular formula of the compound?

A. NO2 B. N2O C. N2O3 D. N2O4

3. A sample of a compound contains 3.581 × 1022 C atoms and 1.436 × 1023 H atoms. What is the empirical formula of the compound?

A. CH B. CH4 C. C2H5 D. C3H4

4. A sample of a compound contains 0.5860 moles of P and 1.466 moles of O. What is the empirical formula of the compound?

A. PO B. PO4 C. P2O5 D. P3O4

5. A sample of a compound contains 19.27 g Fe and 7.360 g O. What is the empirical formula of the compound?

A. FeO B. FeO2 C. Fe2O3 D. Fe3O4

6. Which of the following is an empirical formula of a compound that is 14.37% hydrogen by mass?

A. CH2 B. CH3 C. C2H4 D. C3H6

7. What is the formula for the compound composed of 71.62% gold, 13.10% carbon, and 15.28% nitrogen?

A. AuCN B. Au(CN)2 C. Au(CN)3 D. Au(CN)4

8. Combustion analysis of a sample of hydrocarbon produced 0.100 mol CO2 and 0.200 mol H2O. What is the empirical formula of the compound?

9. Combustion analysis of a 3.003 g sample of a compound containing C, H, and O produced 0.100 mol CO2 and 0.100 mol H2O. What is the empirical formula of the compound?


281

20CHAPTER

Stoichiometry

Stoichiometry deals with the relationships between amounts of reactants and products involved in a reaction. Its major application is in the field of analytical chemistry, which is concerned with the determination of the amounts of substances in real-world material samples.

20.1 Relating Amounts of Reactants and Products

If we know at least two things:• the chemical equation for a reaction• and the change in moles of one reactant or product involved in the reaction

then we can calculate the amounts of all other reactants and products that are also involved. This is because the amounts involved are directly proportional and the mole-to-mole ratio is given by the coefficients in the balanced equation. Mathematically, we say that:

nn

cc

1

2

1

2

=

where n1 and n2 are the changes in moles of substances “1” and “2” and c1 and c2 are their coefficients. We can rearrange this equation to a more ready-to-use form:

n n cc1 2

1

2

= ×

In other words, if we know the change in moles of substance “2” that were involved, we assign it to n2, then multiply it by the ratio of coefficients, (c1/c2), to get n1, the change in moles of substance “1”. The ratio of coefficients, (c1/c2), essentially serves as a conversion factor.


▼ Example 1

Consider the reaction 2 H2(g) + O2(g) → 2 H2O(l)How many moles of H2O are produced if 0.20 mol O2 is consumed during the reaction? How many moles of H2 are consumed?Answer: 0.40 mol H2O (produced), 0.40 mol H2 (consumed)The coefficients of H2O and O2 in the balanced equation are 2 and 1. Therefore:

0.20 mol O2molH O1mol O

0.40 mol H O (produced)22

22× =

⎛

⎝⎜⎜

⎞

⎠⎟⎟

The coefficients of H2 and O2 are 2 and 1. Therefore:

0 2021

0 4022

22. . ( )mol O

mol Hmol O

mol H consumed×⎛

⎝⎜⎜

⎞

⎠⎟⎟

=

It is very important to keep in mind that n1 and n2 are changes in amounts (moles). They are not necessarily the amounts that you have at the beginning of the reaction.

▼ Example 2

Consider the combustion of Mg:

2 Mg(s) + O2(g) → 2 MgO(s)

If 0.500 mol Mg is ignited and 0.100 mol of Mg remains after the reaction, how much MgO was produced?Answer: 0.400 mol MgOThe amount of Mg consumed is the difference between the start-ing amount and the remaining amount: (0.500 mol Mg) − (0.100 mol Mg) = 0.400 mol Mg (consumed).The coefficients of Mg and MgO are both 2. Therefore:

0 40022

0 400. .mol Mgmol MgOmol Mg

mol MgO×⎛

⎝⎜⎜

⎞

⎠⎟⎟ =

In a typical laboratory setting, we measure amounts in grams, not moles. In these cases, we first convert known amounts in grams to moles. If we are interested in solving for amounts in grams, we just convert any mole amount we calculate to grams.

Chapter 20: Stoichiometry 283

▼ Example 3

Consider the combustion of Mg: 2 Mg(s) + O2(g) → 2 MgO(s)If 0.486 g Mg is burned, how much MgO is obtained?Answer: 0.0200 mol MgO, or 0.806 g MgOWe first convert the given amount in grams to moles:

0.486 g Mg → 0.0200 mol Mg

From this we can calculate moles of MgO obtained:

0 020022

0 0200. .mol Mgmol MgOmol Mg

mol MgO×⎛

⎝⎜⎜

⎞

⎠⎟⎟ =

If we are interested in the amount of MgO in grams:0.0200 mol MgO → 0.806 g MgO

20.2 Using Moles of ReactionAn alternative approach to solving stoichiometry problems is to use extent of reaction. One mole of reaction is defined as the extent of reaction where the changes in the moles of reactants of products correspond to their coefficients in the balanced equation. In other words, if x moles of reaction has occurred, the change in moles of any reactant or product is equal to its coefficient times x:

n1 = c1xn2 = c2xn3 = c3x

etc.

▼ Example 4

Consider the reaction 3H2(g) + N2(g) → 2 NH3(g)How many moles of the substances are involved if the reaction occurs to the extent of 0.50 mol?Answer: 1.5 mol H2 consumed; 0.50 mol N2 consumed, 1.0 mol NH3 produced• Coefficient of H2 is 3. Therefore,

moles of H2 consumed = 3x = 3(0.50 mol) = 1.5 mol• Coefficient of N2 is 1. Therefore,

moles of N2 consumed = x = 0.50 mol• Coefficient of NH3 is 2. Therefore,

moles of NH3 produced = 2x = 2(0.50 mol) = 1.0 mol


If we want to be more descriptive with our units, here is how we should have done the calculation of moles of H2 in the preceding example:

31

0 50 1 522

mol Hmol rxn

mol rxn mol H⎛

⎝⎜⎜

⎞

⎠⎟⎟ ( ) =. .

In other words, the unit for a coefficient is moles of whatever per mole of reaction. Note: “rxn” is a commonly used abbreviation for reaction.

Typically, we have information to determine the change in moles of one of the reac-tants or products. Dividing this by the appropriate coefficient gives us the value of x. We can then easily solve for change in moles of other reactants or products as shown above.

▼ Example 5

Consider the reaction 3H2(g) + N2(g) → 2 NH3(g)Use moles of reaction to calculate moles of H2 and N2 consumed if 0.60 mol NH3 are produced?Answer: 0.90 mol H2 consumed, 0.30 mol N2 consumedWhen using moles of reaction, it is convenient to set up a table of changes:

H2 N2 NH3

Changes 3x x 2x

Based on the given information:• Change in NH3 = 2x = 0.60 mol; therefore: x = 0.30 mol• Change in H2 = 3x = 3(0.30 mol) = 0.90 mol• Change in N2 = x = 0.30 mol

Moles of reaction are particularly useful where the information available is limited.

▼ Example 6

Consider the reaction 3 H2(g) + N2(g) → 2 NH3(g)Suppose 1.00 mol H2 and 1.50 mol N2 were to allowed to react and, sometime later, a total of 2.30 mol of gas molecules were found in the reaction mixture. How much H2, N2, and NH3 are in the final mixture?Answer: 0.70 mol H2, 1.40 mol N2, 0.20 mol NH3For questions like this, it is convenient to set up an “ICE” table. Tabulate the Initial moles, Change in moles, and Ending moles.


H2 N2 NH3

Initial moles 1.00 mol 1.50 mol 0 mol

Change in moles 3x mol x mol 2x mol

Ending moles (1.00 − 3x) mol (1.50 − x) mol (0 + 2x) mol

Based on the given information, total ending moles = 2.30 mol. Therefore:

(1.00 − 3x) + (1.50 − x) + (0 + 2x) = 2.30

Solving for x, we get: x = 0.10. Therefore:• Ending moles of H2 = (1.00 − 3x) mol = 0.70 mol• Ending moles of N2 = (1.50 − x) mol = 1.40 mol• Ending moles of NH3 = (0 + 2x) mol = 0.20 mol

20.3 Limiting Reactant ProblemsConsider a chemical reaction involving two or more reactants. As soon as one reactant runs out, the reaction would no longer be able to generate additional products. That reac-tant, the one that would run out first, puts a limit on how much product we can generate. That reactant is, therefore, called the limiting reactant or limiting reagent. The amount of product obtained if the limiting reactant were to completely run out is called the reaction’s theoretical yield.

▼ Example 7

A 2.0 g sample of reactant X is completely used up in a reaction with excess Y to yield 18.0 g of product Z. A 4.0 g sample of reactant “Y is completely used up in a reaction with excess X to yield 4.5 g of product Z. If 2.0 g X and 4.0 g Y are mixed, what is the limiting reactant? What is the theoretical yield?Answer: Y is the limiting reactant, 4.5 g Z is the theoretical yieldWhat this problem is essentially saying is that:• 2.0 g X can yield up to 18.0 g of Z.• 4.0 g Y can yield up to 4.5 of Z.

Therefore, a mixture of 2.0 g X and 4.0 g can only yield up to 4.5 g of Z. By the time this much Z has formed, we will have run out of Y. We cannot make any more. Y is the limiting reactant and 4.5 g of Z is the theoretical yield.


▼ Example 8

Consider the reaction: 2 H2(g) + O2(g) → 2 H2O(l)Suppose we have a mixture of 3.0 mol H2(g) and 2.0 mol O2(g). Verify that H2 is the limiting reactant. Calculate how much O2 would remain unreacted when H2 runs out. Calculate the theoretical yield.Answer:If all of the H2 were consumed, we can calculate how much O2 is also consumed by multiplying the amount of H2 (in moles) by the ratio of coefficients (of O2 and H2) in the balanced equation:

3 012

1 522

22. .mol H

mol Omol H

mol O consumed×⎛

⎝⎜⎜

⎞

⎠⎟⎟

= ( )

The amount of O2 consumed (1.5 mol) is less than what we have to begin with (2.0 mol). So we still have unreacted O2 while H2 is completely consumed.

2.0 mol O2 − 1.5 mol O2 = 0.5 mol O2, remaining

We say that O2 is our excess reactant (or excess reagent).We can calculate the theoretical yield by calculating the amount of product (H2O) generated when the limiting reactant runs out. We multiply amount of H2 (in moles) by the ratio of coefficients (of H2O and H2).

3 022

3 022

22. .mol H

mol H Omol H

mol H O×⎛

⎝⎜⎜

⎞

⎠⎟⎟

=

The theoretical yield is 3.0 mol H2O, or 54 g H2O.

20.3.1. Strategy for Determining Excess ReactantThe preceding example illustrates a strategy that we can use to determine excess reactant:

• Assume one of the reactants is the limiting reactant and is completely consumed.• Calculate how many moles of the other reactant would also be consumed.• Subtract amount of excess reagent consumed from the starting amount of excess

reagent to get the remaining amount. If the answer is negative, we made the wrong assumption; start over and assume the other reactant is limiting.

In the preceding example, had we assumed that O2 is the limiting reactant, we would have gotten a negative remaining amount for H2, which would be physically unreasonable.


20.3.2 Strategy for Determining Theoretical YieldThe quickest way to determine the theoretical yield is to do the following:

• For each reactant, calculate how many moles of product would be generated if the reactant were completely consumed.

• The smaller amount of product generated is the theoretical yield. The reactant that would generate the smaller amount of product is the limiting reactant; by the time we generate this much product, one of the reactants would have been completely consumed and it would not be possible to generate any additional product.

In the preceding example, had we assumed that O2 is the limiting reactant, we would have gotten more than 3.0 mol H2O, which would be impossible. We have shown that by the time the amount of H2O produced reaches 3.0 mol, no more H2 will be left; the reaction cannot proceed further.

20.3.3 Using an ICE TableAn ICE table is a convenient way to organize the information to solve stoichiometry prob-lems. It is especially useful for limiting reactant problems. In this table, we list (below the formulas of reactants and products) the initial (I) moles of reactants and products present, the change (C) in moles, and the ending (E) moles.

▼ Example 9

Consider the reaction 3H2(g) + N2(g) → 2 NH3(g)For a mixture of 0.600 mol H2 and 1.000 mol N2, determine the theoretical yield and amount of unreacted excess reactant if the reaction were to go to completion.Answer: Theoretical yield of 0.400 mol NH3, 0.800 mol unreacted N2

H2 N2 NH3

Initial moles 0.600 mol 1.000 mol 0 mol

Change in moles

3x mol x mol 2x mol

Ending moles

(0.600 − 3x) mol (1.000 − x) mol (0 + 2x) mol

The reaction is said to go to completion if the limiting reactant is completely consumed. If the reaction were to go to completion, then the ending amount for at least one of the reactants (the limiting reactant) should be zero, and the other should be zero or positive.


Assuming H2 is the limiting reactant:Ending moles of H2 = (0.600 − 3x) mol = 0Solving for x, we get: x = 0.200 molEnding moles of N2 = (1.000 − x) mol = (1.000 − 0.200) mol = 0.800 molEnding moles of NH3 = (0 + 2x) mol = 2(0.200) mol = 0.400 mol

Assuming N2 is the limiting reactant:Ending moles of N2 = (1.000 − x) mol = 0Solving for x, we get: x = 1.000 molEnding moles of H2 = (0.600 − 3x) mol = [0.600 − 3(1.000)] mol = −2.400 mol (NEGATIVE!)Ending moles of NH3 = (0 + 2x) mol = 2(1.000) mol = 2.000 mol

Obviously N2 cannot be the limiting reactant. If it were consumed completely, we would end up with a negative amount of H2, which is physically unreasonable. Therefore, we conclude that H2 is the limiting reactant, the amount of unreacted N2 if the reactionwent to completion is 0.800 mol, and the theoretical yield is 0.400 mol NH3.

20.3.4 Percent YieldLimiting reactant is defined as the reactant that would (not will) run out first. Strictly speaking, in reality, no reactant is completely consumed. Reactions are reversible; products can react to re-generate the reactants. When a reaction appears to have stopped, what has actually happened is that the forward and reverse reactions are occurring at the same rate; we have reached what is called a state of (dynamic) equilibrium.

There are cases where consumption of the limiting reactant is essentially complete; in these cases we say the reaction has gone to completion. A commonly used strategy to make reactions go to completion is to use a very large excess of one of the reactants (usually, the less expensive one); the basis for this is a principle known as Le Chatelier’s principle.


In practical situations, the amount of product obtained by carrying out a reaction (the actual yield) is less than the theoretical yield. Reasons other than reversibility include loss of product during the recovery process as well as side reactions that lead to the formation of other products. The percent yield of a product is defined as:

Percent Yield Actual YieldTheoretical Yield

= ×100 %

▼ Example 10

Consider a hypothetical reaction: 2A + B → CSuppose 0.300 mol A and 0.200 mol B were allowed to react, but only 0.120 mol C was produced. Calculate the percent yield.Answer: 80.0%The actual yield is given as 0.120 mol C. To calculate percent yield, we need to know the theoretical yield.If A were completely consumed, the amount of C expected would be:

0 30012

0 150. .mol Amol Cmol A

mol C×⎛

⎝⎜⎜

⎞

⎠⎟⎟ =

If B were completely consumed, the amount of C expected would be:

0.200 mol B1mol C1mol B

0.200 mol C× =⎛

⎝⎜

⎞

⎠⎟

Therefore, the limiting reactant is A and the theoretical yield is 0.150 mol C. By the time this much C is formed, the reaction has to stop. It is not possible to completely use up B and get 0.200 mol C. To determine percent yield, we simply divide the actual yield by the theoretical yield, then multiply by 100%:

Percent YieldActual Yield

Theoretical Yieldmol= × =100

0 1200

%...

% . %150

100 80 0mol

× =

Percent YieldActual Yield

Theoretical Yieldmol= × =100

0 1200

%...

% . %150

100 80 0mol

× =


TEST YOURSELF


1. Consider the reaction: N2 + 3 H2 → 2 NH3

How much NH3 is produced if 0.200 mol H2 is consumed? A. 0.133 mol B. 0.300 mol C. 2.00 mol

2. Consider the reaction of calcium carbonate with acetic acid: CaCO3(s) + 2 HC2H3O2(aq) → Ca(C2H3O2)2(aq) + H2O(l) + CO2(g) A solution containing acetic acid is mixed with CaCO3. How much acetic acid is con-

sumed in a reaction that produces 0.400 mol CO2? A. 0.200 mol B. 0.400 mol C. 0.800 mol


Suppose a reaction mixture initially contained 0.800 mol N2, and now contains only 0.500 mol N2 as a result of the reaction, how much H2 was consumed?

A. 0.100 mol B. 0.900 mol C. 0.300 mol

4. Consider the reaction of calcium carbonate with acetic acid: CaCO3(s) + 2 HC2H3O2(aq) → Ca(C2H3O2)2(aq) + H2O(l) + CO2(g) A solution containing acetic acid is mixed with 0.600 mol CaCO3. Suppose the reac-

tion produces 0.400 mol CO2. How much CaCO3 is left unreacted? A. 0.200 mol B. 0.400 mol C. 0.800 mol


How much hydrogen is needed to produce 25.0 g of NH3? A. 2.22 g B. 4.44 g C. 1.97 g


Suppose 0.500 mol of N2 is used up in this reaction. How much NH3 is produced? A. 4.26 g B. 8.50 g C. 17.0 g

7. Consider the reaction: 3 H2(g) + N2(g) → 2 NH3(g) For a mixture of 2.00 mol N2 and 3.00 mol H2, how much of the excess reactant would

remain if the limiting reactant is consumed?



8. Consider the reaction: 3 H2(g) + N2(g) → 2 NH3(g) What is the theoretical yield of NH3 from a mixture of 2.00 mol N2 and 3.00 mol H2?

9. Consider the reaction: 2 H2(g) + O2(g) → 2 H2O(l) What is the limiting reactant in a mixture containing 1.00 g H2 and 1.00 g O2? What

is the theoretical yield for H2O in grams? Hint: convert amounts to moles first.

10. Consider a 5.32 g sample of CaCO3 (99.87% pure) in a flask and a 100.0 mL sample of vinegar (5% acidity) in a graduated cylinder. The combined mass of both reagents and containers is 255.98 g. The vinegar is poured into the flask with the calcium carbonate and the following reaction occurs:

CaCO3(s) + 2 HC2H3O2(aq) → Ca(C2H3O2)2(aq) + H2O(l) + CO2(g) After swirling the reaction mixture for about twenty minutes, the combined mass of

the reaction mixture and containers is found to be 254.46 g. What is the percent yield of carbon dioxide in this experiment?

11. Consider a reaction between substances A and B. Suppose 0.200 mol A and 0.100 mol B are mixed and, a few minutes later, the amounts remaining are 0.160 mol A and 0.090 mol B. Which substance is the limiting reactant? How much of the excess reactant would be left unreacted if the reaction went to completion?

12. Consider a reaction between substances X and Y, to yield product Z. Suppose 0.400 g of X and 3.600 g of Y are mixed and 0.100 g of X and 1.200 g of Y are found unreacted when the reaction stops. What is the limiting reactant? Calculate the theoretical yield. Calculate the percent yield, assuming no side reactions and no product lost during the experiment.

13. Consider the reaction illustrated in the figure below. What is the limiting reactant?


14. A series of experiments (I, II, III, IV, and IV) reacted different amounts of solu-tion A(aq) with 1.00 mL of solution B(aq). The reaction goes to completion and a gaseous product, C(g) is obtained in each case. Suppose the experimental results are as shown below:

What is the limiting reactant in experiment I? II? III? IV? V?

293

21CHAPTER

Reactions in Aqueous SolutionsA very large number of important reactions occur in aqueous solution. In this chapter, we examine what happens to ions and molecules in aqueous solutions.

21.1 The Solution ProcessTo understand aqueous reactions, we need to first understand what happens to com-pounds when they are mixed with water. When a compound dissolves in water, the ions or molecules that make up the compound are evenly dispersed in the water; each ion or molecule is surrounded by water molecules (“solvated” or “hydrated”). These ions or molecules may also react with water molecules.

The dispersal of the ions allows aqueous solutions of ionic compounds to conduct electricity (note: pure water is a poor electrical conductor). Thus, ionic compounds are called strong electrolytes.

If a compound dissolves in water but the resulting solution does not conduct electric-ity, then we conclude that the compound is made up of molecules that do not yield ions; we call the compound a nonelectrolyte.

There are molecular compounds that react with water molecules and yield ions when dissolved in water. Those that yield hydronium ions (H3O

+) are called acids; those that yield hydroxide (OH−) ions are called bases. The reaction of these compounds with water is called ionization.

Acid molecules dissociate (“break up”) into H+ and anions; the resulting H+ ions are transferred to H2O molecules to form hydronium ions (H3O

+). If we represent an acid by “HX”, then we can represent the ionization of HX in water by:

HX(aq) + H2O(l) → H3O+(aq) + X−(aq)

We could say that the acid (HX) “donates” a proton (H+) to a water molecule, leading to the formation of hydronium ion (H3O

+) and an anion (X−). We could also say that a water


molecule causes an HX molecule to dissociate by “pulling” a proton away from the HX molecule. It is customary to use H+(aq) to represent hydronium ions, to abbreviate the ionization reaction to:

HX(aq) → H+(aq) + X−(aq)and to refer to the reaction as a dissociation reaction.

▼ Example 1

Write the chemical equation that for the reaction of HF molecules with water molecules.

Answer: HF(aq) + H2O(l) → H3O+(aq) + X−(aq)

Base molecules, such as NH3, cause H2O molecules to dissociate by “pulling” protons away from water molecules. If we represent a base molecule by “B”, then the ionization is represented by:

B(aq) + H2O(l) → BH+(aq) + OH−(aq)We could say that the base molecule (B) “accepts” a proton (H+) from a water molecule.

▼ Example 2

Write the chemical equation that for the reaction of NH3 mol-ecules with water molecules.

Answer: NH3(aq) + H2O(l) → NH4+(aq) + OH−(aq)

Most molecular acids and bases are only partially ionized; they are, therefore, called weak electrolytes. By partial ionization, we mean that only a small fraction of the molecules are ionized at any given time. There are, in fact, only six known molecular compounds that completely ionize in water; these are the strong acids, (HCl, HBr, HI, H2SO4, HNO3, HClO4); strong acids are strong electrolytes.

Ionic compounds whose anions are oxide or hydroxide yield hydroxide ions when dissolved in water and are, therefore, considered as strong bases. Any oxide ion that makes it into water immediately gains a proton (H+) from a water molecule, forming two hydrox-ide ions.

O2−(aq) + H2O(l) → 2 OH−(aq)

Chapter 21: Reactions in Aqueous Solutions 295

21.2 Chemical Equations for DissolutionTo represent the dissolution of a substance in water, we write:

• the formula of the compound on the reactant side, with the (s), (l), or (g) label to indicate whether the physical state of the substance is solid, liquid, or gas, and

• the same formula on the product side, with the (aq) label.

A chemical equation written this way is a called molecular equation (or complete formula equation). However, when dealing with strong electrolytes, it is best to write an ionic equation to reflect the fact that the ions are not paired up or “stuck” together in solution. In an ionic equation, formulas of strong electrolytes in aqueous solution are written as separate ions.

▼ Example 3

Write the chemical equation for the dissolution of barium chloride (a solid) in water.Answer:The molecular equation is BaCl2(s) → BaCl2(aq)The ionic equation is BaCl2(s) → Ba2+(aq) + 2 Cl−(aq)The term molecular equation is really a poor choice of words in this case, since no molecules are actually involved. However, we need to be familiar with it since it is the commonly used terminology.

▼ Example 4

Write the chemical equation for the dissolution of glucose (C6H12O6, a solid) in water.Answer: C6H12O6(s) → C6H12O6(aq)There is no ionic equation associated with this process since glucose is not a strong electrolyte. How do we know? It is not an ionic compound, and it is not one of the six molecular compounds known to completely ionize in water.

▼ Example 5

Write the chemical equation for the dissolution of potassium hydroxide (a solid) in water.


Answer:The molecular equation is KOH(s) → KOH(aq)The ionic equation is KOH(s) → K+(aq) + OH−(aq)KOH is an example of a base. Bases are substances that yield OH− when dissolved in water. Metal hydroxides like KOH are clas-sified as strong bases because, like any ionic compound, they are completely dissociated into ions when dissolved in water.

▼ Example 6

Calcium carbonate is insoluble in water. Write the ionic equation for the dissolution process.Answer:Molecular equation: CaCO3(s) ⇆ CaCO3(aq)Ionic equation: CaCO3(s) ⇆ Ca2+(aq) + CO3

2−(aq)If a solid compound is insoluble* in water, it is customary to draw a double arrow to represent its “dissolution” in water. This indi-cates that if the compound is mixed with water, most of the ions will be in the solid, not in solution. However, CaCO3, like any ionic compound, is completely ionized in water. Therefore, it is best to write CaCO3(aq) as separate ions.*Note: the term “insoluble” does not mean zero solubility; it actu-ally means very low solubility.

▼ Example 7

Write the chemical equation for the dissolution of hydrogen chloride (HCl, a gas) in water.Answer:The molecular equation for the dissolution is: HCl(g) → HCl(aq)The ionic equation is: HCl(g) → H+(aq) + Cl−(aq)HCl is a strong electrolyte. It is one of the six known strong acids. The proton (H+) is actually transferred to a water molecule, so we can also represent the overall process by HCl(g) + H2O(l) → H3O

+(aq) + Cl−(aq)


▼ Example 8

Write the chemical equation for the dissolution of acetic acid (HC2H3O2, a liquid) in water.

Answer: HC2H3O2(l) → HC2H3O2(aq)HC2H3O2 is a weak electrolyte; it is not one of the six strong acids. Majority of the acetic acid molecules are un-ionized in water; only a small fraction are ionized at any given time.

▼ Example 9

Write the chemical equation for the dissolution of ammonia (NH3, a gas) in water. NH3 is a weak base.

Answer: NH3(g) → NH3(aq)Most of the NH3 molecules remain un-ionized in water.

21.3 Solubility RulesImagine a sample (of a compound) comparable in size to a grain of rice. If you can get this sample to dissolve in about a teaspoon (5 mL) of water, the compound is said to be soluble in water; otherwise, the compound is said to be insoluble in water. A soluble compound mixed with water gives a clear (transparent) solution; an insoluble compound mixed with water gives a cloudy mixture. Note that insoluble does not mean zero solubility; it actually means very, very, very low solubility; a better term for insoluble is slightly soluble.

We can predict the solubility of common ionic compounds in water by applying a few simple rules. You are likely to see different versions of these rules in different textbooks. They are essentially the same, except that some textbooks include more ions than others. It depends on what the authors consider as “common.” In general, the rules tell us:

• which compounds are always soluble (rule 1),• which compounds are generally soluble and what the exceptions are (rule 2), and• that most everything else is insoluble and what the exceptions are (rule 3).

These rules are empirical; they are based on experimental observations. We apply them in the order listed; once we find a rule the applies, we ignore the rest.

• Rule 1. Compounds of the following are always soluble: sodium, potassium, ammo-nium, nitrate, nitrite, perchlorate, acetate.


• Rule 2. Compounds of the following are generally soluble:• chloride, bromide, iodide — except with Ag+, Pb2+, Hg2

2+, and Cu+.• sulfate — except with lead(II), barium, strontium, calcium, and mercury(I).• fluoride — except with lead(II) and group IIA.

• Rule 3. The rest are generally insoluble. Exceptions: sulfides of group IIA are soluble, hydroxides of calcium, strontium and barium are moderately soluble.

▼ Example 10

Classify the following as soluble or insoluble in water:NaC2H3O2, K2CO3, CaCO3, MgCl2, ZnSO4,PbSO4, AgBr, MgS, CuI, AgF, CuS, Ag3PO4

Answer:NaC2H3O2 is soluble (rule 1), K2CO3 is soluble (rule 1), CaCO3 is insoluble (rule 3), MgCl2 is soluble (rule 2), ZnSO4 is soluble (rule 2), PbSO4 is insoluble (rule 2), AgBr is insoluble (rule 2), MgS is soluble (rule 3), CuI is insoluble (rule 2), AgF is soluble (rule 2), CuS is insoluble (rule 3), Ag3PO4 is insoluble (rule 3).

21.4 Precipitation ReactionsWhen aqueous solutions containing ions are mixed, a precipitation reaction may occur. Precipitation occurs when it is possible to form an insoluble compound. We would expect the mixture to suddenly turn cloudy as a result of the formation of insoluble compound, which eventually settles to the bottom of the container.

▼ Example 11

Write the chemical equation for the precipitation reaction that occurs when solutions of KI(aq) and Pb(NO3)2(aq) are mixed.

Answer: Pb(NO3)2(aq) + 2 KI(aq) → PbI2 (s) + 2 KNO3(aq)If we mix KI(aq) and Pb(NO3)2(aq), we would end up with a solution that contains the following ions:K+, I−, Pb2+, and NO3

−

Using the solubility rules, we predict that PbI2 is insoluble in water (rule 2). Therefore, we write PbI2 with (s) in the chemical equation; we expect it to precipitate. We expect the other product (KNO3) of this double replacement reaction to remain in solution; KNO3 is soluble in water (rule1). Thus, we put an (aq) label next to KNO3 it when we write the chemical equation.


21.5 Ionic EquationsWhen strong electrolytes (ionic compounds and strong acids) are dissolved in water, we know that the ions are individually hydrated. Therefore, when writing chemical equations for aqueous reactions, it is better to write the ions of strong electrolytes (in solution) separately. When we do this, we may find that some ions are not actually involved in the reaction; these are called spectator ions. In these cases, we can write a simpler, more accurate representation of the reaction by writing a net ionic equation. The net ionic equation shows only the ions and molecules that are actually involved in the reaction; spectator ions are excluded.

▼ Example 12

Write the net ionic equation for the reaction: Pb(NO3)2(aq) + 2 KI(aq) → PbI2 (s) + 2 KNO3(aq)

Answer: Pb2+(aq) + 2 I−(aq) → PbI2 (s)A more accurate way of representing the reaction is to write formulas for strong electrolytes in aqueous solution as separate ions:Pb2+(aq) + 2 NO3

− (aq) + 2 K+ (aq) + 2 I−(aq)

→ PbI2 (s) + 2 K+ (aq) + 2 NO3− (aq)

This is called a full or complete ionic equation. The ions are not written separately for PbI2 since most of them will be in the solid (precipitate), not in the solution.If we look carefully at the full ionic equation, we can see that noth-ing really happened to the K+ and NO3

− ions. So, an even better representation for the reaction is:Pb2+(aq) + 2 I−(aq) → PbI2 (s)

This called the net ionic equation. We could say that since K+ and NO3

− appear on both sides of the full ionic equation, they cancel out; they are not involved and are called spectator ions.

When writing ionic equations, the formula of a solid reactant should not be written as separate ions even if it is a strong electrolyte.

▼ Example 13

Write the ionic equation for the “dissolution” of sodium oxide (a solid) in water. The molecular equation is: Na2O(s) + H2O(l) → 2 NaOH(aq)


Answer:The ionic equation is: Na2O(s) + H2O(l) → 2 Na+(aq) + 2 OH−(aq)Na2O is the oxide of a metal (Na). Oxides of metallic elements react with water to yield metal hydroxides; they are also known as base anhydrides.

21.6 Reactions of AcidsAn acid is a molecular compound that yields H+ in aqueous solutions. Therefore, reac-tions common to acids in aqueous solution involve interactions of H+ with other ions or molecules present in the solution.

21.6.1 Reactions with Metal Oxides and Metal HydroxidesThe reaction of an acid with a metal oxide or metal hydroxide leads to the formation of salt and water. The H+ ions from the acid combine with O2− or OH− ions to form H2O. The salt consists of cations from the metal oxide (or metal hydroxide) and anions from the acid.

▼ Example 14

Write the net ionic equation for the reaction of NaOH(aq) and HCl(aq).

Answer: OH−(aq) + H+(aq) → H2O(l)The molecular equation is:NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

The products are H2O and a salt (NaCl). Based on empirical solubility rules, we expect the NaCl product to be in aqueous solution; NaCl is soluble in water. To write the full ionic equation, we write the formu-las of strong electrolytes, substances that are completely dissociated into ions, as separate ions. HCl is a strong electrolyte; it is one of the six known strong acids. Ionic compounds in aqueous solutions, such as NaOH and NaCl, are completely dissociated and should be written as separate ions. Therefore, the full ionic equation is:Na+(aq) + OH−(aq) + H+(aq) + Cl−(aq) → Na+(aq) + Cl−(aq) + H2O(l)

The net ionic equation is obtained by removing the spectator ions (Na+ and Cl−) which appear on both sides of the full ionic equation. The net ionic equation is:OH−(aq) + H+(aq) → H2O(l)


In general, the equation:OH−(aq) + H+(aq) → H2O(l)

is the net ionic equation whenever aqueous solutions of a metal hydroxide and a strong acid are mixed. Exceptions would be cases where an insoluble product is formed.

▼ Example 15

Write the net ionic equation for the reaction of Ba(OH)2(aq) with H2SO4(aq).Answer:The molecular equation is:Ba(OH)2(aq) + H2SO4(aq) → BaSO4(s) + 2 H2O(l)

The products are H2O and a salt, BaSO4. The salt is made up of Ba2+ from Ba(OH)2 and SO4

2− from H2SO4. Since H2SO4 is a strong acid, it is completely dissociated in aqueous solution and should be written as separate ions. An aqueous solution of Ba(OH)2 is specified in the question and we expect ionic com-pounds like Ba(OH)2 to be completely dissociated in aqueous solution. Therefore, Ba(OH)2 should be written as separate ions. Based on empirical solubility rules, we expect BaSO4 to be insoluble in water; we expect it to be a precipitate and we do not write it as separate ions. Therefore, the full ionic equation is:Ba2+(aq) + 2 OH−(aq) + 2 H+(aq) + SO4

2− (aq)

→ BaSO4(s) + 2 H2O(l)This is also the net ionic equation since no ion appears by itself on both sides of the equation; there are no spectator ions.

▼ Example 16

Write the net ionic equation for the reaction of Ca(OH)2(aq) with HC2H3O2(aq).

Answer: OH−(aq) + HC2H3O2(aq) → C2H3O2−(aq) + H2O(l)

The molecular equation is:Ca(OH)2(aq) + 2 HC2H3O2(aq) → Ca(C2H3O2)2(aq) + 2 H2O(l)

The products are H2O and a salt, Ca(C2H3O2)2. Based on em-pirical solubility rules, we expect calcium acetate to be soluble in water; since it is a strong electrolyte, we need to write it as separate ions. Calcium hydroxide is also a strong electrolyte and should be written as separate ions. Acetic acid, HC2H3O2, is not


one of the six strong acids; it is a weak electrolyte and should not be written as separate ions. Therefore, the full ionic equation is:Ca2+(aq) + 2 OH−(aq) + 2 HC2H3O2(aq)

→ Ca2+(aq) + 2 C2H3O2−(aq) + 2 H2O(l)

To get the net ionic equation, we omit the spectator ion, Ca2+(aq):2 OH−(aq) + 2 HC2H3O2(aq) → 2 C2H3O2

−(aq) + 2 H2O(l)We can simplify the net ionic equation by dividing all the coeffi-cients by their greatest common factor (which is 2).

▼ Example 17

Write the net ionic equation for the reaction of Ca(OH)2(s) and HCl(aq).

Answer: Ca(OH)2 (s) + 2 H+(aq) → Ca2+(aq) + 2 H2O(l)The molecular equation is:Ca(OH)2 (s) + 2 HCl (aq) → CaCl2(aq) + 2 H2O(l)

The products are H2O and a salt (CaCl2). The salt’s cation is Ca2+, from Ca(OH)2, and its anion is Cl−, from HCl. Since HCl is a strong acid, it is completely dissociated in aqueous solution and should be written as separate ions. Based on empirical solubility rules, we expect the salt to be in aqueous solution; we expect it to be com-pletely dissociated and write it as separate ions. Therefore, the full ionic equation isCa(OH)2 (s) + 2 H+(aq) + 2 Cl−(aq) → Ca2+(aq) + 2 Cl−(aq) + H2O(l)

The net ionic equation is obtained by omitting the spectator ions (Cl−), which appear on both sides of the full ionic equation.

21.6.2 Reactions with Ammonia and AminesAmmonia and amines are compounds made up of nitrogen containing molecules and are called bases. The N atom in these molecules have an unshared pair of electrons that attracts the H+ ion, as illustrated in Figure 1 for NH3. This results in a covalent bond being formed between N and H. Unlike acids, which release H+ ions in water, base molecules attract H+ ions instead. Since H+ (a hydrogen atom without its electron) is just a proton, we say that a base molecule has a tendency to be protonated. Thus, the N-containing molecule is converted into a polyatomic cation; in the case of NH3, the cation formed is ammonium (NH4

+).


The reaction of an acid with these compounds can be classified as a combination reaction; one product is formed, a salt; in the case of ammonia, the salt formed is an ammonium salt.

▼ Example 18

Write the chemical equation for the reaction of HCl(aq) and NH3(aq).

Answer: The H+ from HCl combines with NH3 to form ammonium ions (NH4

+). The product of the reaction is a salt called ammonium chloride. The chloride ion (Cl−) is the anion formed when an HCl molecule dissociates in water. So, the molecular equation is:HCl(aq) + NH3(aq) → NH4Cl(aq)

To write the full ionic equation, we write the formulas of the strong electrolytes as separate ions. The strong electrolytes in this chemical equation are HCl, which is one of the six known strong acids, and NH4Cl. Both of these are completely dissociated in in aqueous solution. Therefore, the full ionic equation is:H+(aq) + Cl−(aq) + NH3(aq) → NH4

+ (aq) + Cl−(aq)The net ionic equation is obtained by taking out the spectator ion (Cl−), which appears on both sides of the full ionic equation.H+(aq) + NH3(aq) → NH4

+ (aq)

21.6.3 Reactions with SaltsSalts are ionic compounds for which the anion is neither oxide nor hydroxide. The common feature of the reaction of acids with salts is the interaction of H+ ions with the anion of the salt, to yield another acid as a product. The remaining ions combine to form another salt. We need to examine the net ionic equation to reasonably predict if there will be a reaction.

Figure 1. Protonation of Ammonia


▼ Example 19

Predict what happens when vinegar, HC2H3O2(aq), is mixed with baking soda (sodium bicarbonate), NaHCO3(s).Answer:The H+ from the acid combines with the anion from the salt to yield another acid, H2CO3 (carbonic acid).HC2H3O2(aq) + NaHCO3(s) → H2CO3(aq) + NaC2H3O2(aq)

However, H2CO3 is known to readily decompose to H2O(l) and CO2(aq). CO2 is a gas under ordinary conditions and it has a low solu-bility in water, about 0.15 g per 100 g of water. Therefore, we would typically expect the bulk of the CO2 produced to be in the gas phase. In fact, when if we carry out this reaction, we will likely observe the CO2 as bubbles (“fizz”). Therefore, the overall reaction is:HC2H3O2(aq) + NaHCO3(s) → H2O(l) + CO2(g) + NaC2H3O2(aq)

To write the full ionic equation, we write NaC2H3O2(aq) as sepa-rate ions; it is the only strong electrolyte in aqueous solution (for this reaction). HC2H3O2 is a weak acid; NaHCO3 would completely dissociate in aqueous solution but, in baking soda, the ions are held together by ionic bonds in the solid. Thus, the full ionic equation is:HC2H3O2(aq) + NaHCO3(s)

→ H2O(l) + CO2(g) + Na+(aq) + C2H3O2 − (aq)

If we examine the full ionic equation, we can see that there are no spectator ions. Therefore, it is also the net ionic equation.

▼ Example 20

Predict what happens when HNO3(aq) is mixed with K2S(aq).Answer:The H+ from the acid combines with the anion from the salt to yield another acid (H2S); the remaining ions (K+ and NO3

− ) combine to form the salt. It is very likely that, if we carry out this reaction in the laboratory, we will observe the odor of rotten eggs due to H2S(g). At room temperature, the solubility of H2S in water is only 0.35 g per 100 g of water. Typically, this concentration is signifi-cantly exceeded. Thus, we specify H2S(g) rather than H2S(aq). The molecular equation is:2 HNO3(aq) + K2S(aq) → H2S(g) + 2 KNO3(aq)

Based on the empirical solubility rules, we predict KNO3 to remain in aqueous solution. KNO3(aq), HNO3(aq) and K2S(aq) are strong


electrolytes in aqueous solution and should be written as separate ions in the full ionic equation.2 H+(aq) + 2 NO3

−(aq) + 2 K+(aq) + S2−(aq) → H2S(g) + 2 K+(aq) + 2 NO3

− (aq)We can see that K+ and NO3

− are spectator ions; they appear on both sides of the equation. The net ionic equation is simply the combination of H+ from the acid and the anion from the salt (S2−) to form a weak acid (H2S):2 H+(aq) + S2−(aq) → H2S(g)

Note: This is not the only reaction that can occur. This is what we predict would happen based on the expected behavior of acids in general. HNO3 is known to react with sulfides in what is called a redox reaction; redox reactions are covered in another lesson.

▼ Example 21

Predict what happens when HCl(aq) is mixed with AgNO3(aq).Answer:We can imagine that the H+ from the acid combines with the anion from the salt (NO3

− ) to yield another acid (HNO3); the remaining ions (Ag+ and Cl−) combine to form another salt (AgCl). Based on the empirical solubility rules, we predict AgCl to be insoluble; we expect it to precipitate. All the other substances in the molecular equation are strong electrolytes in aqueous solution. Therefore, to write the ionic equation, they should be written as separate ions:H+(aq) + Cl−(aq) + Ag+(aq) + NO3

− (aq) → H+(aq) + NO3 − (aq) + AgCl(s)

We can see that H+ and NO3 − are spectator ions; theyappear on both sides of the equation. The net ionic equation,

Cl−(aq) + Ag+(aq) → AgCl(s) has nothing to do with the H+ from the acid.

▼ Example 22

Predict what happens when sulfuric acid, H2SO4(aq), is mixed with MgSO3(aq).Answer:The H+ from the acid combines with the anion from the salt to yield another acid, H2SO3 (sulfurous acid). However, H2SO3 is


known to readily decompose to H2O(l) and SO2(aq). The overall molecular equation is:H2SO4(aq) + MgSO3(aq) → H2O(l) + SO2(aq) + MgSO4(aq)

and the full ionic equation is:2 H+(aq) + SO4

2− (aq) + Mg2+(aq) + SO3 2− (aq)

→ H2O(l) + SO2(aq) + Mg2+(aq) + SO4 2− (aq)

since sulfuric acid, MgSO3(aq), and MgSO4(aq) are all strong electrolytes in aqueous solution. The net ionic equation is obtained by canceling out the spectator ions, Mg2+(aq) and SO4

2− (aq), which appear on both sides of the equation. The net ionic equation is:2 H+(aq) + SO3

2− (aq) → H2O(l) + SO2(aq)At room temperature, if the amount of SO2 produced exceeds 10 g per 100 g of water, we may observe bubbles and a sharp odor due to SO2(g).

▼ Example 23

Predict what would happen when hydrochloric acid, HCl(aq), is mixed with barium sulfate BaSO4(s).Answer:We can imagine that the H+ from the acid combines with the anion from the salt (SO4

2−) to yield another acid, H2SO4 (sulfuric acid) and a salt that is soluble in water (BaCl2). To write the full ionic equation, we write BaCl2(aq) and H2SO4(aq) as separate ions; they are both strong electrolytes in aqueous solution. The full ionic equation is:2 H+(aq) + 2 Cl−(aq) + BaSO4(s) →

2 H+(aq) + SO4 2− aq) + Ba2+(aq) + 2 Cl−(aq)

If we get rid of the spectator ions, H+(aq) + Cl−(aq), we are left with a net ionic equation of:BaSO4(s) ⇆ SO4

2− aq) + Ba2+(aq)which just describes the “dissolution” of BaSO4 in water. This is just what would happen if we added water to solid barium sulfate, even without the acid. We replaced the single arrow with double-headed arrow since we know from our empiri-cal solubility rules that BaSO4 is insoluble in water. This means that the reaction, as written, will occur to a very, very small extent. For practical purposes, we could say that there really is no reaction when we mix HCl(aq) and BaSO4(s). It is probably


just as well since this is what happens when a person about to have a gastric x-ray is given a “barium cocktail.” If the barium sulfate were to react with HCl in the stomach and release a significant amount of toxic Ba2+ ions in solution, the patient could die.


TEST YOURSELF


1. Which figure below best represents an aqueous solution of CaCl2?

2. Which figures below best represent aqueous solutions of HF and HCl?

A. A for both B. B for both C. HF-B, HCl-A D. HF-A, HCl-B

3. Which of the following best represents the dissolution of CaCl2(s) in water? A. CaCl2(s) → Ca2+(aq) + 2 Cl−(aq) B. CaCl2(s) → CaCl2(aq) C. CaCl2(s) → Ca2+(aq) + Cl2

2−(aq)



4. Which of the following best represents the dissolution of methanol, CH3OH(l), in water?

A. CH3OH(l) → CH3OH(aq) B. CH3OH(l) → CH3

+(aq) + OH−(aq) C. CH3OH(l) → CH3O

−(aq) + H+(aq) D. CH3OH(l) + H2O(l) → CH3OH2

+(aq) + OH−(aq)

5. Which of the following best represents the ionization of nitric acid in water? A. HNO3(aq) → H(aq) + NO3(aq) B. HNO3(aq) → H+(aq) + NO3

−(aq) C. HNO3(aq) → H+(aq) + N3−(aq) + 3 O2−(aq) D. HNO3(aq) → H+(aq) + N3−(aq) + O3

−(aq)

6. Which of the following best represents the dissolution of gaseous ammonia (NH3) in water?

A. NH3(g) → NH3(aq) B. NH3(g) + H2O(l) → NH4

+(aq) + OH−(aq) C. NH3(g) + H2O(l) → NH4OH(aq) D. NH3(aq) + H2O(l) → NH4

+(aq) + OH−(aq)

7. Which of the following is a weak base? A. NaOH B. Mg(OH)2 C. K2O D. NH3

8. Which of the following is insoluble in water? A. Na2CO3 B. MgSO4 C. PbCl2 D. CaS

9. Which of the following is insoluble in water? A. CuCl2 B. Hg2Cl2 C. HgBr2 D. Ca(NO3)2

10. Which of the following is soluble in water? A. MgCO3 B. MgF2 C. Mg(OH)2 D. MgSO4

11. Precipitation is expected from which mixture? A. Na2CO3(aq) + KNO3(aq) B. AgNO3(aq) + CaCl2(aq) C. Neither D. Both


12. A student is given two test tubes and is told that one contains NaCl(aq), while the other test tube contains Ba(NO3)2(aq). Which precipitating reagent, if added to both test tubes would allow the student to determine their contents?

A. AgNO3(aq) B. K2SO4(aq) C. neither D. both

13. What is the net ionic equation for the precipitation reaction that occurs when K2SO4(aq) and Pb(NO3)2 are mixed?

A. K+(aq) + NO3 −(aq) → KNO3(s) B. Pb+(aq) + SO4

−(aq) → PbSO4(s) C. Pb2+(aq) + SO4

2−(aq) → PbSO4(s)

14. What is the net ionic equation for the precipitation reaction that occurs when Na2CO3(aq) and AgNO3 are mixed?

A. Na+(aq) + NO3 −(aq) → NaNO3(s) B. Ag+(aq) + CO3

2−(aq) → Ag2CO3(s) C. Ag2+(aq) + CO3

2−(aq) → AgCO3(s) D. 2 Ag+(aq) + CO32−(aq) → Ag2CO3(s)

15. For which of the following mixtures is the net ionic equation for neutralization given by: H+(aq) + OH−(aq) → H2O(l)?

A. HBr(aq) + KOH(aq) B. HCl(aq) + NaOH(s) C. HC2H3O2(aq) + Ba(OH)2(aq) D. HI(aq) + NH3(aq)

16. What is the net ionic equation for the reaction between HF(aq) and Ca(OH)2(aq)? A. 2 HF(aq) + Ca(OH)2(aq) → 2 H2O(l) + CaF2(s) B. 2 H+(aq) + 2 F−(aq) + Ca2+(aq) + 2 OH−(aq) → 2 H2O(l) + 2F−(aq) + Ca2+(aq) C. 2 HF(aq) + Ca2+(aq) + 2 OH–(aq) → 2 H2O(l) + 2F−(aq) + Ca2+(aq) D. 2 HF(aq) + Ca2+(aq) + 2 OH–(aq) → 2 H2O(l) + CaF2(s)

17. What is the net ionic equation for the reaction between Mg(OH)2 in milk of magnesia with HCl (in stomach acid)? The complete formula equation is: Mg(OH)2 + 2 HCl → MgCl2 + 2 H2O

A. Mg2+ + 2 OH− + 2 H+ + 2 Cl– → Mg2+ + 2 Cl– + 2 H2O B. Mg(OH)2 + 2 H+ + 2 Cl– → Mg2+ + 2 Cl– + 2 H2O C. H+ + OH− → H2O D. Mg(OH)2 + 2 H+ → Mg2+ + 2 H2O


18. What is the net ionic equation for the reaction between HNO3(aq) and NH3(aq)? A. HNO3(aq) + NH3(aq) → NH4NO3(aq) B. H+(aq) + NO3

−(aq) + NH3(aq) → NH4 +(aq) + NO3

−(aq) C. H+(aq) + NH3(aq) → NH4

+(aq)

19. Which of the following appears in the net ionic equation for the reaction of HNO2(aq) and Na2CO3(aq)?

A. HNO2(aq) B. H+(aq) C. Na+(aq)

20. Which of the following does not appear in the net ionic equation for the reaction of Ca3(PO4)2(s) and HCl(aq)?

A. Ca2+(aq) B. PO4 3−(aq) C. H+(aq) D. H3PO4(aq)

21. Which of the following does not appear in the net ionic equation for the reaction BaCO3(s) and H2SO4(aq)?

A. Ba2+ B. CO2(g) C. H2O(l) D. BaSO4(s)

313

22CHAPTER

Molarity and Solution StoichiometrySince a large number of important reactions occur in aqueous solution, we need to learn how to count atoms, molecules, and ions present in aqueous solutions.

22.1 MolarityThe predominant component of a solution is called the solvent; all the other components are called solutes. When the solvent is water, we say that the solution is an aqueous solution. Unless otherwise specified, it is customary to assume that a solution is aqueous.

▼ Example 1

Imagine a solution prepared stirring a pinch of sugar into a glass of water. Identify the solute and solvent.Answer: the solvent is water; the solute is sugar.

The concentration of a solute in a solution refers to the ratio of the amount of solute to either the amount of solvent or the amount of solution. Think of a concentration as a measure of how crowded the solute particles are; the higher the concentration, the more crowded.

Since there are many ways of expressing amounts, there are many ways of expressing concentration. The most commonly used way in Chemistry is molarity (or molar concentration). Aqueous solutions in the laboratory are typically labeled with the molar concentration of the solutes. Molarity is defined the moles of solute that would be present in a liter of solution. If M is the molarity, n is the moles of solute, and V is the volume of solution in liters, then:

M = nV

Since one millimole is 1/1000 of a mole and one milliliter is 1/1000 of a liter, molarity can also be defined as the millimoles of solute per milliliter of solution.


The unit for molarity is moles per liter, which can be written as mol/L, mol L−1, or M. It can also be written as mmol/mL or mmol mL−1. If the solution contains only one solute, it is customary to refer to molar concentration of that solute as the molarity of the solution. If there is more than one solute, the molar concentration of a solute can also be referred to as the molarity of the solution with respect to that solute.

▼ Example 2

Describe the solution in the bottle illustrated below.

Answer:NaCl is a solute in this solution. If we have a one-liter sample of this solution, that sample would contain 0.1 moles of NaCl. If we have a one-milliliter sample of this solution, that sample would contain 0.1 millimoles of NaCl. The label is read as “0.1 molar sodium chloride.” We can say that the solution is “0.1 molar in NaCl” or “0.1 molar with respect to NaCl.”

22.1.1 Calculating Amount of SolutesWe are frequently interested in determining the amount of solute present in a given sample of solution. This is easily done by solving for n in the defining equation for molarity:

M = nV

Multiplying both sides by V, we get:

n = VM

Chapter 22: Molarity and Solution Stoichiometry 315

▼ Example 3

How much Na2CO3 is in a 2.00 L sample of 0.200M Na2CO3?Answer: 0.400 mol or 42.4 gBased on the definition of molarity, we conclude that one liter of this solution would contain 0.200 moles of Na2CO3. Common sense tells us that two liters should contain twice as much Na2CO3, and that is what we get when we do the calculation:

n VM 2.00 L 0.200 mol L 0.400 mol1= = ( ) ( ) =−

Two liters, times 0.200 moles for each liter, equals 0.400 moles. In cases where we have to deal with more than one solute, we can avoid ambiguity by being more descriptive with our units. For example, we use “mol Na2CO3” instead of just “mol.” We can also put L in the denominator instead of writing L−1.

2 000 200

0 4002 32 3.

.. L

mol Na CO1 L

mol Na CO×⎛⎝⎜

⎞⎠⎟ =

Note that the ratio of (moles of solute) and (Liters of solution) implied by the molarity is essentially a conversion factor, to change liters of solution to moles of solute.If we are interested in the amount in grams, all we need to do is convert from moles to grams. We would need the molar mass of Na2CO3, which is 105.99 g/mol.

0.400 mol Na CO105.99g Na CO1 mol Na CO

42.4g 22

32 3

3

×⎛

⎝⎜⎜

⎞

⎠⎟⎟ = NNa CO2 3

▼ Example 4

How would you prepare a 2.00 L solution of 0.200M Na2CO3?Answer: Based on the preceding example, we know that this solution should contain 42.4 grams of sodium carbonate. So, we go to the laboratory, weigh out 42.4 grams of sodium carbonate, then add enough water to make a solution with a total volume of 2.00 liters.


▼ Example 5

How many moles of sucrose (C12H22O11) does a 25.0 mL sample of 0.500 M C12H22O11 contain?Answer:Since we’re given the volume in mL, we can first get the amount of solute in millimoles.

n V M 25.0 mL 0.500 mmol mL 12.5 mmol1= = ( ) ( ) =−

Then, we change millimoles to moles:12.5 mmol = 12.5 (10−3 mol) = 0.0125 mol

Alternatively, we can first convert mL to L,25.0 mL = 25.0 (10−3 L) = 0.0250 L,

then multiply the volume in liters by the molarity:

n = V M = 0.0250 0.500 mol 0.0125 molL L1( )( ) =−

22.1.2 Counting Ions in SolutionWhen dealing with aqueous solutions of electrolytes, we are often interested in the amount of specific ions. In these cases, we just multiply moles of compound with the subscript of the ion in one formula unit.

▼ Example 6

How many moles of chloride ions are there in 2.00L of 0.300 M BaCl2?Answer: First, we find the moles of BaCl2:

2.00 L0.300 mol BaCl

1 L0.600 mol BaCl2

2× ⎛⎝⎜

⎞⎠⎟ =

The subscript of chloride in BaCl2 is 2; this means that there are 2 moles of chloride ions for every mole of BaCl2. Therefore, the number of moles of chloride ions is two times 0.600 moles, or 1.20 moles:

0.600 mol BaCl2 mol Cl

1 mol BaCl1.20 mol Cl

22

×⎛

⎝⎜⎜

⎞

⎠⎟⎟

=−

−


22.1.3 Calculating MolarityTo calculate the molarity of a compound in solution, all we have to do is figure out how many moles of the compound is present (in a given volume of solution), then divide that by the volume of solution in liters.

▼ Example 7

Suppose a 2.00 L solution contains 0.400 mol NaOH. What is the molar concentration of NaOH in this solution?Answer:

M =nV

=0.400 mol

2.00 L= 0.200 mol L−1

If we are given the amount of solute in grams, we just need to convert it to moles first.

22.1.4 Molar Concentration of IonsWhen dealing with strong electrolytes, we are usually interested in the molar concentration of specific ions. If we know the molarity of the compound, all we have to do is multiply it by the appropriate subscript to get the molarity of the ion we’re interested in.

▼ Example 8

What is the molar concentration of sulfate ions in 0.500M Na2SO4?Answer: The coefficient of sulfate in Na2SO4 is 1. Therefore,

0.500 mol Na SO

L1 mol SO

1 mol Na SO mol SO2 4 4

2

2 4

42

× =− −

0 500. .LL

The molar concentration of a specific ion or molecule is denoted by putting the formula of the ion or molecule inside square brackets. In the preceding example, we say that the molar concentration of sulfate ions is 0.500 mol L−1 by writing:

[SO42−] = 0.500

or:[SO4

2−] = 0.500 mol L−1

The unit (mol L−1) is implied but some teachers prefer that the unit be explicitly indicated.


If more than one strong electrolyte is present in solution, then the molar concentration of an ion is calculated by adding up contributions from all sources.

▼ Example 9

What is the molar concentration of chloride ions in a solution that is 0.10M with respect to NaCl and 0.20M with respect to BaCl2?Answer: 0.50 mol L−1

The contribution from NaCl is:

0.10 mol NaClL

1 mol Cl1 mol NaCl

0.10mol Cl

L× = ⋅

− −

and the contribution from BaCl2 is:

0.20 mol BaClL

2 mol Cl1 mol BaCl

0.40mol Cl

L2

2× = ⋅

− −

Therefore, the molar concentration of chloride is:[Cl−] = [Cl−] from NaCl + [Cl−] from BaCl2 = 0.10 M + 0.40 M = 0.50 M

22.1.5 DilutionDilution refers to the addition of more solvent to a solution. What happens when we do this? Recall that molarity is defined as:

M = nV

where n is the moles of solute and V is the volume of solution. Adding more solvent increases the volume, but the amount of solute remains the same. Therefore, we will now have a larger denominator (V), but the same n. This means that the molarity decreases. In other words, diluting a solution lowers the concentration.

Dilution is a commonly used procedure in the laboratory to prepare a solution if a solution of higher concentration happens to be in stock. If we use M1 and V1 to represent the molarity and volume of a solution (the “stock solution”), and M2 and V2 as the new molarity and volume after dilution, then:

V1 M1 = V2 M2


This is called the dilution formula. This formula simply says that the amount of solute does not change when we do a dilution. Remember that the amount of solute (in moles) is simply the product of V and M. All we are doing when doing a dilution is adding more solvent.

▼ Example 10

What is the molarity of the resulting solution if 1.00L of 2.00M NaOH solution is diluted to 4.00L?Answer: 0.500 M We divide both sides of the dilution formula by the final volume V2, in order to solve for the final molarity (M2):

MV M

V

1.00 L 2.00 M

4.00 L0.500 M2

1 1

2

= =( ) ( )

=

We can see that increasing the total volume 4-fold reduces the molarity to 1/4 of the original concentration; 0.500 is 1/4 of 2.00. The relationship between M and V in the dilution formula is an example of what is called inverse proportionality. Two quantities are said to be inversely proportional if their product is a constant. In the case of the dilution formula, the product of V and M is the amount of solute, which does not change when we do dilutions. If two quantities are inversely proportional, then when one increases, the other decreases by the same factor; for example, if one doubles (multiplied by 2), the other is halved (divided by 2).

22.2 Solution StoichiometryThe amounts of reactants and products involved in a reaction are directly proportional and the mole-to-mole ratio is given by the coefficients in the balanced equation. If we represent changes in moles of substances “1” and “2” as n1 and n2 , and represent their coefficients as c1 and c2, then:

nn

cc

1

2

1

2

=

which we can rearrange into a more ready-to-use form:

n n cc2 1

1

= × 2


If we know how much (in moles) of one substance was consumed or produced in a reac-tion, we assign it to n1, then we multiply it by an appropriate ratio of coefficients to get the amount involved for any of the other substances involved in the ratio. When dealing with reactions in aqueous solution, it is important to remember that the number of moles of solute is just equal to the VM, where V is the volume of solution in liters and M is the molar concentration of the solute.

▼ Example 11

Consider the neutralization of sulfuric acid by sodium hydroxide.H2SO4(aq) + 2 NaOH(aq) → 2 H2O(l) + Na2SO4

How many moles of H2SO4 will be neutralized by 25.0 mL of 0.200M NaOH?Answer: 2.50 × 10−3 molWe are given molarity and volume of the NaOH solution; therefore, we can use this to calculate moles of NaOH.

25.0 10 L0.200 mol NaOH

L5.00 10 mol NaOH3 3− −( ) × = ×

Note that we first replaced mL by 10−3 L in order to have the answer in moles. From this, we can solve for the moles of H2SO4:

5.00 10 mol NaOH1 mol H SO2 mol NaOH

2.50 10 3 2 4 3×( ) × ⎛⎝⎜

⎞⎠⎟ = ×− − mmol H SO2 4

Titration is a typical laboratory procedure for determining unknown amounts of substances or ions in a sample. The two reactants in a titration are called the standard (the “known”) and the analyte.(the “unknown”). Titration involves the gradual addition of one reactant (the titrant) to another. The titrant is usually the standard; it is a solution dispensed with a buret as illustrated in Figure 1. The other reactant, usually the analyte, is placed in a flask.

The addition of titrant is stopped when a drastic change is observed in the reaction mixture; the point when this happens is called the endpoint. Titrations are designed so that the endpoint very closely approximates the equivalence point. At the equivalence point, the mole-to-mole ratio of the standard and analyte is equal to the ratio of their coefficients in the balanced equation.

Traditionally, a small amount of another reagent is added to the flask before the start of a titration in order to be able to determine the endpoint; this reagent is called an indicator.


For example, in the titration of an acid by a base, one or two drops of phenolphthalein so-lution is added to the acid in the flask. The endpoint of the titration occurs when the phe-nolphthalein changes from being colorless to having a very light pink color. With modern instrumentation, titration endpoints can be determined by monitoring some property of the reaction mixture, such as pH, electrical conductivity and (light) absorbance.

Sometimes, the concentration of a solution meant to be used a standard must first be determined. In this case, we do what is called a standardization—a titration is carried out where the solution (titrant) is the unknown and the sample in the flask is the standard. Once this is done, the solution can then be used as a standard for other (unknown) samples in the flask. For example, a solution of NaOH can be used as a standard for determining unknown amounts of acids. But before it can be used, we need to determine its concentration by using it to titrate a precisely known amount of the acid called “KHP” (potassium hydrogen phthalate, or potassium biphthalate, KHC8H4O4). In this case, we say that the NaOH solution is being standardized against KHP, KHP is the primary standard, and NaOH is a secondary standard.

Figure 1. Experimental Setup for Titration


TEST YOURSELF


1. How many moles of NaCl are in a 400.0 mL sample of 0.18M NaCl? A. 0.072 mol B. 72 mol C. 0.45 mol D. 2.2 mol

2. How many grams of NaCl are in a 400.0mL sample of 0.18M NaCl? A. 2.2 g B. 4.2 g C. 7.2 g

3. An aqueous solution in the lab is labeled “0.100 M HC2H3O2”. How many moles of acetic acid are in a 25.0 mL sample of this solution?

4. An aqueous solution in the lab is labeled “6.00 M HC2H3O2”. How many grams of acetic acid are in a 2.00 L sample of this solution?

5. How many moles of sodium ions are in 50.0 mL of 0.400 M Na2CO3? A. 0.0200 mol B. 0.0400 mol C. 0.25 mol D. 6.67×10−3 mol

6. What is the molarity of a solution if a 2.0 L sample contains 3.0 mol NaOH? A. 1.5 M B. 0.67 M C. 6.0 M

7. What is the molar concentration of NaCl in a solution prepared by dissolving 11.6 g NaCl in enough water to make a 500.0 mL solution?

A. 0.0232 M B. 0.397 M C. 5.80 M

8. What is the molar concentration of chloride ions in 0.30M BaCl2? A. 0.30M B. 0.15M C. 0.60M D. 0.20M

9. What is the molar concentration of sodium ions in a solution that is 0.20M in NaCl and 0.30M in Na2SO4?

A. 0.50M B. 0.25M C. 0.70M D. 0.80M

10. A solution is prepared by mixing 10.0 mL of 0.100M NaNO3, 20.0 mL of 0.300M Ca(NO3)2, and enough water to bring the total volume up to 50.0 mL. What is the molar concentration of nitrate ions in the resulting solution?

A. 0.200M B. 0.260M C. 0.130 M D. 0.700 M



11. Consider a 0.500 M NaOH(aq). If enough water is added to a 10.0 mL sample of this solution in order to bring up the total volume to 100.0mL, what is the molarity of the resulting solution?

A. 0.0500M B. 1.00M C. 2.00M D. 5.00M

12. How much water should be added to a 100.0mL sample of 6.000M NaOH in order to dilute the solution to 3.000M?

A. 100.0 mL B. 200.0 mL C. enough to bring the total volume to 200.0 mL D. none of the above

13. Consider the neutralization of Mg(OH)2(s) by HCl(aq).Mg(OH)2(s) + 2 HCl(aq) → MgCl2(aq) + 2 H2O(l)

What volume of 0.200M HCl(aq) is required to neutralize 0.200 mol Mg(OH)2? A. 500.0 mL B. 1.00 L C. 2.00 L D. 4.00 L

14. What mass of silver chloride will be formed if 50.0 mL of 0.750 molar silver nitrate is mixed with 70.0 mL of 0.500 molar sodium chloride?

A. 2.04 g B. 5.02 g C. 5.37 g D. 6.37 g

15. Consider the neutralization of sulfuric acid by sodium hydroxide.H2SO4(aq) + 2 NaOH(aq) → 2 H2O(l) + Na2SO4

What if a 25.00 mL sample of sulfuric acid solution is titrated with 0.500M NaOH. If 25.00 mL of the base was required to reach the endpoint,what is the molarity of the sulfuric acid solution?

A. 0.125M B. 0.250M C. 0.500M D. 1.00M

325

23CHAPTER

Ideal Gas Behavior

The behavior of gases is best summarized by the equation:PV = nRT

This equation is called the ideal gas equation. In this equation P, V, and T are the pressure, volume, and temperature of the gas, n is the amount of gas in moles, and R is a constant. In this chapter, we will examine these quantities in depth.

23.1 PressurePressure is defined as force per unit area. If a force, f, is applied to a surface with an area of A, then the pressure, P, is given by:

P fA

=

The SI unit for pressure is the Pascal (Pa), or Newton per square meter (N m−2). In terms of the base units,

1 N = 1 kg m s−2

1 Pa = 1 N m−2 = 1 (kg m s−2)(m−2) or 1 (kg m−1 s−2)

▼ Example 1

Calculate the pressure at the bottom of a column of mercury (Hg) liquid that is 76.0 cm high. The density of liquid Hg is 13.6 g/mL or 13.6×103 kg m−3.Answer: 1.01×105 PaWe figure out the pressure that the Hg column exerts on what-ever is at the bottom of the column as follows:• The earth’s gravitational pull exerts a downward force on any-

thing; this is called the weight. The weight of the Hg column is equal to its mass (m) times the acceleration due to gravity (g):f = mg, where g = 9.81 m s−2

• According to Newton’s second law, if the Hg column is not moving, then the net force on it is zero. This means that


whatever is at the bottom of the column must be exerting an upward force on the column that counteracts the pull of gravity; this is called the “normal force” and is due to net repulsions between the atoms of Hg and those on the sur-face once the atoms get too close. Thus, the normal force is also equal to mg, but directed upward.

• According to Newton’s third law, the Hg column exerts an equal and opposite force on whatever is at the bottom. There-fore, we conclude that the Hg column exerts a downward force of mg on whatever is at the bottom.

• The pressure on the surface at the bottom of the Hg column is, therefore, mg/A, where A is the cross sectional area of the Hg column. If we assume that the column has a uniform cross sectional area, then the volume (V) of the column is just equal to the cross sectional area times the height (h); V = Ah. Therefore, the cross sectional area is equal to V/h. The pressure due to a column of liquid of height h and cross sectional area A is:

PfA

mgA

mgVh

mghV

= = =⎛⎝⎜

⎞⎠⎟

=

• Density is defined as the ratio of mass (m) to volume (V); d = m/V. Substituting d for (m/V) in the preceding equation, we get:

P = dghFor a column of Hg that is 76.0 cm (or 0.760 m) high,

P dgh 13.6 10 kg m 9.81 ms 0.760 m

1.01 10 Pa

= ×( ) ( ) ( )= ×

= − −3 3 2

5

Although we assumed a uniform cross sectional area in the preced-ing example, it turns out that for any column of liquid, the pressure only depends on the height and density. The formula we derived, P = dgh, is valid whether or not the cross sectional area is uniform.

In Chemistry, we typically deal with pressure associated with gases. Pressures of gases are commonly expressed in terms of equivalent heights of Hg column (cm Hg or mm Hg), Torr, atmosphere, bar, mbar (millibar), and psi (pounds per square inch). These units are related as follows:

1 atm = 76 cm Hg = 760 mm Hg = 760 Torr= 1.01325×105 Pa = 1.01325 bar = 1013.25 mbar = 14.7 psi

Chapter 23: Ideal Gas Behavior 327

It is obvious from the preceding that:1 mm Hg = 1 Torr1 bar = 105 Pa = 1000 mbar

▼ Example 2

Express the pressure due to a 1.00 m column of Hg in cm Hg, mm Hg, Torr, atm, Pa, and bar.Answer:We can use heights of Hg in cm or mm as units of pressure. Since 1.00 m is equivalent to 100 cm or 1000 mm,

P m Hg100 cm Hg

1 m Hg10 cm Hg2= × = ×1 00 1 00. .

P m Hg1000 m Hg

1 m Hg10 mm Hg= × = ×1 00 1 00 3. .

In other units:

P mm Hg1 Torr

1 mm Hg10 Torr= × × = ×1 00 10 1 003 3. .

P mm Hg1 atm

760 mm Hg atm= × × =1 00 10 1 323. .

P mm Hg1.01325 10 Pa

760 mm Hg10 Pa 1.33

5

= × × × = × =1 00 10 1 333 5. . bbar

▼ Example 3

What is the pressure underwater at a depth of 50.0 m? Assume the water surface is at sea level; take the density of water as 1.00 g/mL, or 1.00 × 103 kg m−3.Answer. 5.84 atmThe pressure due to water is:

P dgh kg m 9.81 m s 50.0 m Pa= = ×( ) ( ) ( ) = ×− −1 00 10 4 905 103 3 2 5. .

In atmospheres, the pressure due to water is:

P1 atm

Pa4.84 atm= × ×

×=4 905 10

1 01325 105

5..

There is an additional pressure of 1.00 atm due to the air above the water surface. So the total pressure is 5.84 atm.


An alternative way of determining the pressure due to water is to use the fact that Hg is 13.6 times denser than water. Therefore, the pressure due to 1 mm Hg is equivalent to the pressure of 13.6 mm H2O.

P 50.0 m H 01000 mm H O

1 m H 0

1 mm Hg

13.6 mm H O1 atm

76= × × ×2

2

2 2 00 mm Hg4.84 atm=

P 50.0 m H 01000 mm H O

1 m H 0

1 mm Hg

13.6 mm H O1 atm

76= × × ×2

2

2 2 00 mm Hg4.84 atm=

Essentially, we divide the height of water column by 13.6 to get the equivalent height of mercury column.

23.2 Gas Pressure MeasurementThe pressure of a gas is due to the force exerted by gas particles on a surface (such as the wall of the gas container); the force is the result of collisions of gas particles with the surface.

Nowadays, gas pressure is conveniently measured using electronic sensors. Traditional methods of measuring the pressure of a gas use columns of mercury. Atmospheric pressure is measured using a barometer, which is illustrated in Figure 1.

Imagine a tube with no gas in it (a vacuum), sealed at the top, is inverted and opened into a pool of liquid mercury. The atmospheric pressure pushing down on the pool will cause the liquid to rise up the tube. This is similar to what happens if you suck the air out of a straw that is dipped in a glass of water. The atmospheric pressure on the water causes it to rise up the straw. At sea level, the typical atmospheric pressure will cause the Hg to rise to a height (h) of about 76 cm above the surface of the pool, provided the space above the Hg column is a vacuum. It is for this reason that a pressure of 1 atmosphere (abbreviated

Figure 1. Barometer


as 1 atm) is defined to be equivalent to the pressure due to a mercury column that is exactly 76 cm high. The actual atmospheric pressure varies depending on where you are and the pre-vailing weather conditions. In other words, atmospheric pressure is hardly ever exactly 1 atm; to avoid confusion, it is preferable to refer to atmospheric pressure as barometric pressure.

For samples of gases, the pressure is measured using a manometer, a U-shaped tube filled with mercury. In a close-end manometer, Figure 2, the gas is connected to one arm and the space above the Hg in the other arm is a vacuum. The pressure of the gas is manifested as a difference in the height of the Hg in the arms. The mercury level in the arm connected to the gas would be lower. If it is lower by h cm, then the pressure of the gas is h cm Hg.

Figure 2. Close-end Manometer

▼ Example 4

A gas is connected to a close-end manometer. If the difference in mercury levels is 20.0 cm, what is the pressure of the gas in Torr?

Answer: 2.00 × 102 TorrThe pressure of the gas is 20.0 cm Hg, which is equivalent to 200 mm Hg, or 2.00 × 102 mm Hg. But 1 mm Hg is equivalent to 1 Torr. So, the pressure of the gas is 2.00 × 102 Torr.

Another common setup for measuring gas pressure involves an open-end manometer, as illustrated in Figures 3 and 4. One arm of the manometer is connected to the gas and the other arm is open to the atmosphere. If the pressure of the gas is equal to the prevail-ing barometric pressure, then the mercury levels will be the same in both arms. If the pressure of the gas is higher than the prevailing barometric pressure, the Hg level will be


lower in the arm connected to the gas as illustrated in Figure 3. If the difference in height is h, then the pressure of the gas is: Pgas = Pbarometric + h

Figure 3. Open-end Manometer; Pgas > Pbarometric

Figure 4. Open-end Manometer; Pgas < Pbarometric

If the pressure is lower than the prevailing barometric pressure, the Hg level will be higher in the arm connected to the gas as illustrated in Figure 4. If the difference in height is h, then the pressure of the gas is: Pgas = Pbarometric − h

▼ Example 5

What is the pressure of the gas in the figure illustrated below if the right arm of the manometer is open to the atmosphere and the prevailing barometric pressure is 1.00 atm.


Answer: 21.0 cm HgThe pressure of the gas is lower than the barometric pressure:P = Pbarometric − h.

The barometric pressure is given as 1.00 atm, which is equiva-lent to 76.0 cm Hg. Therefore:

P = 76.0 cm Hg − 55.0 cm Hg = 21.0 cm Hg

Blood pressure measurements are, in fact, measurements of air pressure in the cuff that is wrapped around a person’s arm. Air is pumped into the cuff and the air pressure in the cuff is monitored as the air is released. The device used to monitor the pressure is called a sphygmomanometer.

23.3 Gas Laws Relating P, V, and TThe ideal gas equation, PV = nRT, summarizes several laws that describe the behavior of gases at low pressures. In this section, we discuss laws based on studies of how the pressure (P), volume (V), and temperature (T) are related for a fixed amount of gas (i.e. constant n).

23.3.1 Boyle’s LawRobert Boyle discovered the relationship between P and V for a fixed amount of gas at constant T. He found that P and V are inversely proportional. When two quantities are inversely proportional, it means that if one increases, the other decreases by the same factor. Mathematically, we say that two quantities are inversely proportional if their product is a constant:

PV = constantWe can see that this is consistent with the ideal gas equation; for a fixed amount of gas (constant n) at constant T, the right-hand side of the ideal gas equation is constant since R is also a constant. If we were to plot P vs. V at constant T, we would get a curve similar to that shown in Figure 5. Any plot at constant T is called an isotherm. The isotherm shown in Figure 5 is for one mole of an ideal gas at 1000K. At any point on the isotherm, the


product PV has the same value; that value is nRT. For the points labeled “1” , “2”, and “3” we can say that:

P1V1 = P2 V2 = P3 V3

It is easy to from the graph that the volume doubles from point 1 to point 2 while the pressure is halved. Similarly, from point 2 to point 3, volume is doubled while pressure is halved.

▼ Example 6

A 25.0 L sample of gas, initially at 1.00 atm expands to 50.0 L at constant temperature. Calculate the final pressure.Answer. 0.500 atmThe problem explicitly states that T is constant. There is nothing in the problem that indicates a possible change in amount of gas; so, n is also constant. Therefore, Boyle’s Law is applicable for this problem; if we refer to the initial pressure and volume as P1 and V1, and to the final pressure and volume as P2 and V2, then:

P1 V1 = P2 V2

Solving for the final pressure, P2, we get:

PP VV

1.00 atm 25.0 L

50.0 L0.500 atm1 1

22 = =

( ) ( )=

We can see that doubling the volume causes the pressure to be cut in half.

If we rearrange the ideal gas equation so that only P is on the left-hand-side, we get:

P nRTV

= ⎛⎝⎜

⎞⎠⎟

1

Figure 5. PV Isotherm


If nRT is constant, then a plot of P vs. 1/V is a straight line. The standard form for the equation of a straight line is:

y = mx + bwhere m is the slope and b is the y-intercept. If we let y = P and x = (1/V), we can see that the slope, m, is equal to nRT, and the y-intercept, b, is equal to zero.

▼ Example 7

Which of the four graphs below illustrates Boyle’s Law? Assume the bottom left corners of the graphs shown below correspond to (0,0).

Answer. AIf we plot P vs. 1/V, we are letting y = P and x = 1/V and we get a straight line with a slope of nRT, and a y-intercept of zero.

P nRTV

y m x b

= ⎛⎝⎜

⎞⎠⎟

= +

1

Since n, R, and T are positive numbers, the slope is positive. The graph that fits this description is A. Graph B has a negative slope. Graph C has a zero slope. Graph D has a positive slope, but its y-intercept is not zero. Graphs B and C also have nonzero y-intercepts.

23.3.2 Charles’ LawJacques Charles discovered the relationship between V and T for a fixed amount of gas at constant P. Joseph Gay-Lussac first published the discovery; thus, the law is sometimes known as Charles’ and Gay-Lussac’s law. The modern statement of Charles’ law states that V and T are directly proportional at constant n and P. When two quantities are directly proportional, it means that when one changes, the other also changes by the same factor.


Mathematically, two quantities are directly proportional when their ratio is a constant. Therefore:

VT

constant=

We can see that this is consistent with the ideal gas equation. If we rearrange the ideal gas equation to solve for V/T:

VT

nRP

=

we can see that the right-hand-side is constant if n and P are constants. If we were to plot V vs. T, we expect to get a straight line.

It is very important to note that T in the equations above refer to the absolute temperature in Kelvin. The temperature in Kelvin is obtained by adding 273.15 to the Celsius temperature,

▼ Example 8

A 10.0 L sample of gas initially at 300.0K is heated to 600.0K at constant pressure. What is the final volume of the gas?Answer. 20.0 LThe problem explicitly states that P is constant. There is nothing in the problem that indicates a possible change in amount of gas; so, n is also constant. Therefore, Charles’ Law is applicable for this prob-lem; if we refer to the initial volume and temperature as V1 and T1, and refer to the final volume and temperature as V2 and T2, then:

VT

VT

1

1

2

2

=

Solving for V2, we get:

VVT

T1.00 atm300.0 K

600.0 K 2.00 L21

12= =

⎛

⎝⎜

⎞

⎠⎟ ( ) =

We can see that doubling of the temperature leads to a doubling of the volume.

23.3.3 Amontons’ LawGuillaume Amontons discovered the relationship between P and T at constant n and V. He found that P and T are directly proportional at constant n and V. Amontons’ Law is erroneously called Gay-Lussac’s law in some textbooks. Mathematically, this means that:

PT

constant=


We can see that this is consistent with the ideal gas equation if we rearrange the ideal gas equation to solve for P/T. We get:

PT

nRV

=

and see that the right-hand-side is, indeed, constant when n and V are constant.

▼ Example 9

A fixed-volume container contains a gas at 300.0K, 1.00 atm. If the temperature is increased to 600.0K, calculate the final pressure.Answer. 2.00 LThe problem says that V is constant (‘fixed volume”) and there is nothing to indicate that there is a change in the amount of gas (so n is also constant). If we refer to the initial pressure and temperature as P1 and T1, and to the final temperature as P2 and T2, Amontons’ law says that:

PT

PT

1

1

2

2

=

Solving for P2, we get:

PPT

T1.00 atm300.0 K

600.0 K 2.00 L21

12= =

⎛

⎝⎜

⎞

⎠⎟ ( ) =

23.3.4 Combined Gas LawThe relationship between P, V, and T for a fixed amount of gas can be combined into just one equation.

PVT

constant=

Once again, we can see that this is consistent with the ideal gas equation, which we can rearrange to:

PVT

nR=

to show that PV/T is indeed equal to a constant for a fixed amount of gas (constant n). The combined gas law is useful for determining how one of the three quantities change if changes in the other two are known; we simply use the equation:

P VT

P VT

1 1

1

2 2

2

=


TEST YOURSELF


1. What is the pressure due to a vertical column of mercury that is 760.0 mm long? A. 760.0 Torr B. 760.0 Pa C. 760.0 atm D. 760.0 bar

2. What is the pressure due to a vertical column of water that is 760 mm long? (use the following densities: water 1.00 g/mL, Hg is 13.6 g/mL)

A. 7.6 × 102 Torr B. 1.0 × 105 Torr C. 7.4 × 10−2 atm

3. A gas bulb is connected to an open-end mercury manometer and the liquid level in the arm connected to the gas bulb is found to be 15.0 cm lower than in the arm that is open to the atmosphere. The prevailing barometric pressure is 755 mm Hg. What is the pressure of the gas in the bulb?

A. 905 Torr B. 770 Torr C. 740 Torr D. 605 Torr

4. A gas bulb is connected to an open-end mercury manometer and the liquid level in the arm connected to the gas bulb is found to be 20.0 mm higher than in the arm that is open to the atmosphere. The prevailing barometric pressure is 758 mm Hg. What is the pressure of the gas in the bulb?


5. What is the final pressure of a gas, originally at (30.0 L, 250.0 Torr) if it is compressed to 15.0 L at constant temperature?


6. If the pressure of a gas is tripled at constant temperature, the volume... A. is tripled B. is reduced to 1/3 of the original volume C. increases 9-fold D. decreases by 3 L

7. If the volume of a gas increases from 8.0L to 2.0L at constant temperature, the pressure...

A. is quadrupled B. is reduced to (1/4) of the original pressure C. increases by 6.0 atm



8. Which of the two PV isotherms shown below represents data at a higher temperature (for the same sample of gas)?

9. What is the final volume of a gas, originally at (25.0 L, 25.00°C) if it is heated to 50.00°C at constant pressure?

A. 12.5 L B. 27.1 L C. 50.0 L

10. What is the final volume of a gas, originally at (30.0 L, 250.0 Torr, 25°C) if its pressure and temperature are changed to (500.0 Torr, 115°C)?

A. 19.5 L B. 46.0 L C. 69.0 L D. 78.1 L

339

24CHAPTER

Counting Gas Particles

When dealing with any pure substance, in any physical state (solid, liquid, or gas), we can count the particles that make up a sample by simply dividing the mass of the sample by the molar mass. Unfortunately, weighing gases is a cumbersome process.

It is easier to count particles in a gas sample by measuring three properties of the sample: pressure (P), volume (V), and temperature (T). If we know these three, we can calculate the number of moles of particles in the sample.

24.1 Gas ParticlesA gas sample is a collection of particles, which can either be molecules or free atoms. Strictly speaking, the term molecule refers to a group of two or more atoms that are sharing electrons (or covalently bonded). By free atom, we mean a single neutral atom that is not bonded to any other atom. When discussing gases, people tend to use molecule as a generic term for both molecule and free atom; the more precise term to use is particle.

▼ Example 1

Which gas sample has more gas particles: 1 mol CH4 or 1 mol He?Answer: Same. Each CH4 molecule counts as one particle; each He atom counts as one particle.

▼ Example 2

Which gas sample has more atoms: 1 mol CH4 or 1 mol He?Answer: CH4

One mole of CH4 has 5 moles of atoms, one mole of C and 4 moles of H. One mole of He atom has one mole of atoms.


24.2 The Ideal Gas EquationIf we know the pressure (P), volume (V), and temperature (T) of a gas sample, we can calculate the number of moles of gas particles in the sample by rearranging the ideal gas equation:

PV = nRT

to solve for n:

n PVRT

=

The value and unit for R depends on the units used for P and V. The most commonly used units for P and V in Chemistry are atmosphere and liter. Therefore, it is convenient to memorize the value of R as:

R = 0.08206 L atm mol-1 K-1

If we have P and V values expressed in other units, we just convert those to atmosphere and liter.

▼ Example 3

How many moles of particles are in a 22.4 L sample of air at STP. STP stands for standard temperature and pressure, which corre-sponds to the 1 atm and 273.15K (which is the freezing point of water at 1 atm).Answer: 1.00 molTo solve for n, we divide both sides of the ideal gas equation by (RT), and we get:

nP VR T

1 atm 22.4 L

0.08206 L atm mol K K1.00 = =

( ) ( )( ) ( )

=− −1 1 273 15.

mmol

For determining significant figures in the final answer, treat the STP values (1 atm and 273.15K) as exact.

The ideal gas equation is, strictly speaking, valid only at very low pressures. At typical conditions, room temperature and a pressure close to 1 atmosphere, calculations based on the ideal gas equation could be in error by up to about 1%. Thus, calculations of n at ordinary conditions are probably reliable to only about 3–4 significant figures.

Chapter 24: Counting Gas Particles 341

24.3 Avogadro’s LawThe molar volume (Vm) of a gas is defined as the volume that one mole would occupy; mathematically, it is just the ratio of V to n:

V Vnm =

If the ideal gas equation is applicable, then the molar volume is given by:

V Vn

R TPm = =

We can see that at a given temperature and pressure, the molar volume is constant; all the terms on the right-hand side of the preceding equation (R, T, and P) are constants.

▼ Example 4

What is the molar volume of any gas at STP?

Answer: 22.4 L mol-1

VVn

R TP

0.08206 L atm mol K K

atm22.4 L mol

m = =

=( ) ( )

=− −

−1 1 273 15

1

.11

Whenever the ratio of two things is constant, we say that they are directly propor-tional. Therefore, we can say that at a given T and P, the volume of a gas sample is directly proportional to the number of moles of gas particles in the gas sample. This is known as Avogadro’s Law. When we say that V and n are directly proportional, it means that:

• if n = 0, then V = 0, and• if n changes, so does V (by the same factor).

If we have twice as many gas particles, the gas would occupy twice as much volume (at the same T, P). If we have 3 times as many gas particles, the gas would occupy 3 times as much volume (at the same T, P). Another way of stating Avogadro’s law is to say that: at the same T and P, equal volumes of gas contain equal number of particles.


▼ Example 5

Which of the following gas samples contains the most number of particles at the same temperature and pressure?Sample A. 2.00 L CH4Sample B. 1.00 L H2Sample C. 2.00 L HeSample D. 4.00 L He

Answer: Sample D Since volume is directly proportional to n (at the same T,P), equal volumes mean equal number of gas particles. Samples A and C have the same number of particles. Sample D has twice as many; its volume is twice that of sample A or C. Sample B has half as many gas particles; its volume is only half that of sample A or C.

▼ Example 6

Which of the following gas samples contains the most number of atoms at the same temperature and pressure?Sample A. 2.00 L CH4Sample B. 1.00 L H2Sample C. 2.00 L HeSample D. 4.00 L He

Answer: Sample ASince we do not have enough information to calculate number of particles, let “x” be the number of particles in sample B (which has the fewest).• For sample A, there are 2x particles since its volume is twice

that of sample B. Since each particle is a CH4 molecule, each particle is made up of 5 atoms (one C and four H). Therefore, sample A has:

2 10x particles5 atoms

1 particlex atoms=

• For sample B, the number of particles is x; this times 2 atoms per particle gives us 2x atoms.

• For sample C, there are 2x particles since its volume is twice that of sample B. Since each particle is a free He atom, we also have 2x atoms.

• For sample D, there are 4x particles since its volume is 4 times that of sample B. Each particle is a free He atom; therefore, we also have 4x atoms.


24.4 Gas MixturesThe ideal gas equation is a very powerful equation in that if it is applicable, it does not depend on the nature of the gas particles. We can use it for pure gaseous substances as well as gaseous mixtures.

24.4.1 Partial PressurePartial pressure is defined as the pressure that a component of a gas mixture would exert if it were alone in the container. Therefore, if “A” is a component of a gas mixture, its partial pressure, PA, is simply:

Pn RT

VAA=

assuming the ideal gas equation is applicable. In a fixed volume container at constant temperature, we can see that the partial pressure and number of moles of any component are directly proportional. The more moles present, the higher the partial pressure.

▼ Example 7

What are the partial pressures of O2 and N2 in a 25.0 L sample containing 1.00 mol O2 and 4.00 mol N2 at 300.0 K? What is the total pressure?Answer: 0.985 atm, 3.94 atm, and 4.92 atmThe partial pressure of O2 is:

P nRTV

1.00 mol0.08206 L atm mol K 300.0

O O2 2= ⎛

⎝⎜⎞⎠⎟

= ( ) ( )− −1 1 KK

25.0 L0.985 atm

( )=

Similarly, we can calculate the partial pressure of N2 and find it to be 3.94 atm.

P n RTV

4.00 mol0.08206 L atm mol K 300.

N N2 2

1 1

= ⎛⎝⎜

⎞⎠⎟

= ( ) ( )− − 00 K

25.0 L3.94 atm

( )=


and the total pressure is:

Ptotal totalnRTV

5.00 mol0.08206 L atm mol K

= ⎛⎝⎜

⎞⎠⎟

= ( ) ( )− −1 1 3300.0 K

25.0 L4.92 atm

( )=

24.4.2 Dalton’s Law of Partial PressuresDalton’s Law of Partial Pressures states that the pressure of a gas mixture is equal to the sum of the partial pressures of its components. We can see that this is consistent with the ideal gas equation. In general, for a mixture of A, B, C, etc..., the total pressure is:

P n RTV

n n n .... RTVtotal A B C= ⎛

⎝⎜⎞⎠⎟

= + + +( )⎛⎝⎜

⎞⎠⎟

or:

P n RTV

n RTV

n RTV

P P PA B C A B C= ⎛⎝⎜

⎞⎠⎟

+ ⎛⎝⎜

⎞⎠⎟

+ ⎛⎝⎜

⎞⎠⎟

+ = + + +..... ....

▼ Example 8

In any enclosed vessel containing liquid water, the partial pres-sure of water vapor is called the vapor pressure of water; it is a constant value (at a given temperature) that we can look up. At 298K, it is 23.8 Torr. A sample of H2 is collected over water at 298K. If the pressure of the sample is 750.0 Torr, what is the partial pressure of H2?Answer: 726.2 TorrDalton’s law says that the total pressure is equal to the sum of the partial pressures of the components. The components in this case are hydrogen and water vapor. Therefore:

P P + Ptotal hydrogen water_vapor=

which we can rearrange to solve for the partial pressure of hydrogen:


P P P

Torr 23.8 Torr 726.2 hydrogen total water_vapor = −

= − =750 0. TTorr

24.4.3 Partial Pressure and Mole FractionMole fraction is defined as the ratio of the number of moles of a component to the total number of moles in a mixture. If a mixture contains n1 moles of component “1” and total moles of ntotal, then the mole fraction of component “1” is:

χ11= n

ntotal

Using the ideal gas equation, we can show that the mole fraction of a component of any gas mixture is equal to the ratio of its partial pressure to the total pressure. For component “1”:

χ11

1

= =

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

=nn

P VRT

P VRT

PPtotal

total

1

total

▼ Example 9

About 21% of molecules in a typical air sample are O2 molecules. We say that the mole percent of O2 is 21%, or that the mole fraction of O2 is 0.21. What is the partial pressure of O2 in an air sample if the total pressure of the sample is 1.20 atm?Answer. 0.25 atmWe are not given the volume and temperature of this mixture, so we cannot calculate the partial pressure of O2 using the defining formula for partial pressure. However, we can determine the partial pressure from the mole fraction and total pressure. If χ1 is the mole fraction of O2 and P1 is its partial pressure, then:

11=

PPtotal

χ

which we can rearrange to solve for P1: P1 = χ1 Ptotal = (0.21)(1.20 atm) = 0.25 atm


24.5 Stoichiometry Involving GasesThe amounts of reactants and products involved in a reaction are directly proportional and the mole-to-mole ratio is given by the coefficients in the balanced equation. If we represent changes in moles of substances “1” and “2” as n1 and n2, and represent their coef-ficients as c1 and c2, then:

nn

cc

1

2

1

2

=

which we can rearrange into a more ready-to-use form:

n n cc2 1

2

1

= ×

If we know how much (in moles) of one substance was consumed or produced in a reac-tion, we assign it to n1, then we multiply it by an appropriate ratio of coefficients to get the amount involved for any of the other substances involved in the ratio. When dealing with reactions involving gases, it is important to remember that the number of moles of gas particles can be calculated from the ideal gas equation.

▼ Example 10

Suppose 22.4 L of CO2 were collected at STP from the decomposition of NaHCO3: 2 NaHCO3 → Na2O(s) + H2O(g) + 2 CO2(g)How much Na2O(s) was also produced?Answer: First, calculate moles of CO2 produced. STP means T = 273.15K, P = 1.00 atm

nP VRT

1 atm 22.4 L

0.08206 L atm mol K 273.15 K

= =

( ) ( )( ) ( )

=− −1 1

11.00 mol

Therefore, we can calculate moles of Na2O, by multiplying moles of CO2 with the ratio of the coefficients of Na2O and CO2:

1.00 mol CO1 mol Na O2 mol CO

0.500 mol Na O22

22× =


It is often convenient to use pressure measurements to monitor the progress of a reaction that involves gases. When the reaction mixture contains more than one gaseous substance, it is useful to know partial pressures.

▼ Example 11

A strip of Mg is reacted with HCl(aq):Mg(s) + 2 HCl(aq) → H2(g) + MgCl2(aq)

and the H2 gas was collected over water as illustrated below. As such, the gas is a mixture of H2 and water vapor, H2O(g). Suppose the volume of gas collected at 298K is 2.50 L, the water level inside the tube is 10.0 cm above the water level outside, and the prevail-ing barometric pressure is 755 Torr, how much HCl was consumed in the reaction? The vapor pressure of water at 298K is 23.8 Torr.

Answer: 0.195 mol HClThe pressure of the gas is less than the barometric pressure; the difference is equal to the pressure due the water column (which is given as 10.0 cm high).

P 755 Torr 10.0 cm H O1 cm Hg

13.6 cm H O

0 mm Hg

1 cm Hg= − × × ×2

2

1 11 Torr1 mm Hg

Torr

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= 747 6.

This should only have 3 significant digits, but we keep an extra digit to minimize round-off error in subsequent calculations. This is equal to the sum of the partial pressures of H2(g) and H2O(g). We can therefore solve for the partial pressure of H2.

P P P 747.6 Torr 23.8 Torr 723.8 Torrhydrogen water_vapor= − = − =


In atmospheres, the partial pressure if H2 is:

P Torratm

Torratmhydrogen = × =723 8

1760

0 9524. .

Finally, we can calculate moles of H2, using PV = nRT:

nP V

RT0.9524 atm 2.50 L

0.08206 L atm m

hydrogenhydrogen=

=( ) ( )

ool K 298 K0.09737 mol

− −( ) ( )=

1 1

To calculate moles of HCl consumed, we use the coefficients of H2 and HCl in the balanced equation.

0.0974 mol H2 mol HCl1 mol H

0.195 mol HCl 2

2 × =

24.6 Determination of Molar Mass: Dumas Method

Molar mass is defined as the mass per mole of a substance. Therefore, if we know the mass (m) of any sample (solid, liquid, or gas) and and the moles of particles (n) present in that sample, then we can calculate the molar mass (Mm).

M mnm =

▼ Example 12

A sample of unknown gas, with a mass of 0.8802 g contains 0.0200 mol of particles. What is the molar mass of the unknown gas?Answer:

Mmn

gmol

g molm = = = −0 88020 0200

44 0 1..

.


If we have a gaseous sample that can be assumed to be behaving ideally, then we can easily determine the number of moles of gas particles in our sample from measurements of pressure (P), volume (V), and temperature (T). We simply rearrange the ideal gas equa-tion, PV = nRT, to solve for n:

n PVRT

=

Although cumbersome, it is possible to weigh gases. The procedure for determining the molar mass of a gas from measurement of its mass, P, V, and T is known as the Dumas method.

▼ Example 13

A sample of unknown gas with a mass of 0.140 g occupies a volume of 153.1 mL at 373K and 1.00 atm. Which of these is a possible identity of the compound?A. CH4, B. CO, C. Cl2Answer: BFirst, we calculate the number of moles:

nPVRT

atm LL atm mol K

mol= = =− −

( . )( . ).

.1 00 0 1533

0 082060 005001 1

Therefore, the molar mass is:

Mmn

gmol

g molm = = = −0 1400 00500

28 0 1..

.

The molar masses of the compounds given are 16.0 g/mol for CH4, 28.0 g/mol for CO, and 70.9 g/mol for Cl2.

If we combine the two equations:

M mnm =

and:

n PVRT

=

by substituting (PV/RT) for n in the first equation, we get:

M mn

mPVRT

mRTPVm = =

⎛⎝⎜

⎞⎠⎟

=


Since density, d, is the ratio of mass to volume (m/V), we can simplify the preceding equation to:

M dRTPm =

Therefore, if we know the density of a gas at a given T and P, we can calculate its molar mass. Thus, the Dumas method is also known as the vapor density method.

If we define molar volume (Vm) as the volume per mole, then:

V Vn

RTPm = =

Therefore, we can also say that:

M dVm m=

or that the molar mass is equal to the density times the molar volume. We can rearrange this equation to solve for d, and we find that:

d MV

m

m

=

or that the density is just the ratio of (the mass of one mole) to (the volume of one mole). We can see that this is consistent with the definition of density as the ratio of a sample’s mass to its volume.

▼ Example 14

The density of an gas at STP is 3.17 g/L. What is the molar mass of the gas?

Answer: 71.0 g mol-1

The molar volume at STP is 22.4 L mol-1. Therefore, the molar mass is:

MdRTP

dV g L L mol g molm m= = = =− − −( . )( . ) .3 17 22 4 71 01 1 1

Note that we could have plugged in the values for R, T, and P into the formula and we will still get the same answer; RT/P at STP is 22.4 L mol-1.

If our gas sample is a mixture, the molar mass that we end up calculating from the density is called the average molar mass. If we have a 2-component mixture and we know


the molar masses of the components of our mixture, then we can figure out the composi-tion of our mixture.

▼ Example 15

Ignore trace components and assume that air is only made up of N2 and O2. What are the mole fractions of N2 and O2 in air given that its density at STP is 1.29 g/L?

Answer. 28.9 g mol-1

From the density at STP, we can calculate the average molar mass.

MdRTP

dV g L L mol g molm m= = = =− − −( . )( . ) .1 29 22 4 28 91 1 1

To figure out mole fractions, imagine we have a sample that con-tains 1 mole of air molecules. If the mole fraction of O2 is x, then:This sample would contain x mol O2 and (1-x) mol N2.The mass of x mol O2 is (x mol) (32.00 g mol-1).The mass of (1-x) mol N2 is [ (1-x) mol ] (28.02 g mol-1).The total mass of the sample is: [ (x)(32.00) + (1-x)(28.02) ] g

Since we already know that the mass of one mole of air molecules is 28.9 g, we can solve for x:

(x)(32.00) + (1-x)(28.02) = 28.9

Solving for x gives us the mole fraction of O2: x = 0.221The mole fraction of N2 is (1-x), or 0.779

As we can see from the preceding example, the average molar mass of a mixture of O2 and N2 is:

Mm = (x)(32.00 g mol-1) + (1-x)(28.02 g mol-1)if the mole fraction of O2 is x and the mole fraction of N2 is (1-x). In other words, if

we refer to O2 as component “1” and to N2 as component “2”, then the average molar mass is just:

M M Mm m m= +χ χ1 1 2 2

We can generalize this, for any mixture of gases:M M M Mm m m m= + +χ χ χ1 1 2 2 3 3 ...

This formula actually works for any homogeneous mixture, not just gases.


TEST YOURSELF


1. Which gas sample has more gas particles? A. 1.0 mol CH4 B. 2.5 mol O2 C. same

2. What is the mass of a 248.0 mL sample of oxygen gas at 100.00°C and 0.500 atm? A. 0.130 g B. 0.484 g C. 2.03 g

3. A mixture of 1.00 g H2 and an unknown amount of He occupies 22.4 L at STP. What is the mass of helium in the sample?

A. 0.496 g B. 0.503 g C. 2.02 g D. 4.00 g Hint: calculate total moles (n), subtract moles of H2 from n to get moles of He;

convert moles of He to grams.

4. A flask is open to the atmosphere and contains 300.0 mL of air at room temperature (298K); the prevailing barometric pressure is 755 Torr. The flask is heated to 596K. How many moles of air molecules are in the flask initially? How many moles of air molecules remain after the flask is heated? Hint: P and V are constant.

5. Suppose a fixed-volume container initially contains 2.00 moles of gas at 0.500 atm. The gas decomposes and, hours later, the pressure (at the same temperature) is found to be 0.800 atm. How many moles of gas are present in the final mixture?

6. Which of the following gas samples has the smallest volume at 300K and 125 Torr? A. 1.00 mol H2 B. 2.00 mol He C. 2.00 mol CH4

7. Which of the following gas samples has the smallest volume at 300K and 125 Torr? A. 1.00 g H2 B. 2.00 g He C. 2.00 mol CH4

8. Hydrogen gas is collected over water at 25°C. The vapor pressure of water at this temperature is 23.8 Torr. If the total pressure of the gas is 755.0 Torr, what is the partial pressure of hydrogen?

A. 731.2 Torr B. 778.8 Torr C. 23.8 Torr D. None of the above

9. A 5.0 mg sample of gas “X” is found to have a pressure of 0.100 atm at 298K in a given container. A 10.0 mg sample of gas “Y,” placed in the same container is found to have



a pressure of 0.150 atm at 298K. What would be the total pressure of a mixture of 5.0 mg “X” and 10.0 mg “Y” in the same container at 298K? (assume no reaction)

A. 0.050 atm B. 0.105 atm C. 0.250 atm D. 2.0 atm

10. A mixture containing 0.200 mol H2 and 0.300 mol N2 has a total pressure of 720.0 Torr. What is the partial pressure of H2 in the mixture?


11. What is the partial pressure of O2 in a mixture consisting of 16.0 g O2 and 4.0 g He if the total pressure is 850.0 Torr?

12. A piece of Mg ribbon is reacted with hydrochloric acid. If 224 mL of hydrogen gas is produced at 25.00°C, 1.00 atm, how much HCl was consumed? The reaction is:

Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)

13. Consider the combustion of urea: 2 CO(NH2)2 (s) + 3 O2(g) → 2 CO2(g) + 2 N2(g) + 4 H2O(l)

Suppose 3.00 g of urea were burned in excess oxygen in a fixed-volume 1.00-L sealed container that initially contained 2.00 atm of O2 at 298K. Estimate the final pressure in the container at 298K.

14. Consider the hypothetical reaction: 2 A(g) → B(g) + 3 C(g)Suppose a fixed-volume container initially contains 2.00 mol of pure A(g) at a pressure of 0.500 atm. Hours later, the pressure, at the same temperature, is found to be 0.800 atm. How many moles of B(g) are present in the final mixture?

15. An evacuated 63.8 mL glass bulb is filled with a gas at 295K and 0.983 atm. The mass of this gas is 0.104 g. Which of these is the possible density of this gas?

A. N2 B. Ne C. Ar D. O2

16. What is the density of Freon-11 (CFCl3) at 120oC and 1140 Torr? A. 1.59 x 104 g/L B. 20.9 g/L C. 6.39 g/L D. 22.4 g/L

17. Calculate the density of NH3(g) at 25oC, 2.00 atm?

18. The density of a mixture of He and Ne at STP is found to be 8.00 g L-1. Calculate the average molar mass and the mole fractions of He and Ne in the mixture?

355

25CHAPTER

Kinetic Molecular Theory

The behavior of gases can be explained using Kinetic Molecular Theory, which applies the basic principles of physics to relate the motion of gas particles to what we observe as temperature and pressure.

25.1 Assumptions of Kinetic Molecular TheoryWe can use basic physics concepts to explain ideal gas behavior if we treat gas particles as “point masses” and make the following assumptions:

• Gas particles are in continuous random motion.• No energy transfer occurs when gas particles collide with walls or with each other

(collisions are elastic); although this is not valid for individual collisions, the total energy transfer for all the collisions does average out to zero.

• The gas particles exert no force on each other.

By “point mass,” we mean that the particles themselves have no volume and a colli-sion between them occurs when their locations coincide. Real gas particles do have finite volumes and do exert a force on each other; the extent to which these are negligible deter-mines the extent to which the gas obeys the ideal gas equation.

25.2 Molecular Interpretation of PressureGas pressure is due to collisions of gas particles with a surface. When a particle collides with a surface, its velocity changes and its speed and direction of motion change. A change of velocity is called acceleration and is an indication that a force has acted on the particle. Newton’s second law says that the force is equal to the mass of the particle times its acceleration. Since the surface exerts a force on the particle during the collision, we can also say that the particle exerts a force on the surface because Newton’s third law says that for every action, there is an equal and opposite reaction. If we add up the forces due to all the collisions of gas particles with the surface, and divide the result by the area of the surface, we get the pressure.


▼ Example 1

Consider one particle, an N2 molecule with a mass of 4.65 × 10−26 kg, moving at a speed of 300.0 m s−1 colliding with a wall (with an area of 10.0 cm2) and reversing direction during a 1.0 s time period. Assume the speed remains the same. Calculate the pressure on the wall due to this one particle.Answer: The average acceleration during this time period, per molecule is:

avt

v vt

m s m ss

final initial= =−

= − − = − ×− −Δ

Δ Δ( . )

..

300 0 3001 0

6 0 101 1

2 mm s −2

avt

v vt

m s m ss

final initial= =−

= − − = − ×− −Δ

Δ Δ( . )

..

300 0 3001 0

6 0 101 1

2 mm s −2

Note that we plugged in a final velocity of −300.0 m s−1 and an initial velocity of +300.0 m s−1. We are assuming that the original direction of motion is the positive direction; so the final velocity is negative since the particle has reversed direction. The force expe-rienced by the molecule is:

f ma kg m s N= = × − × = − ×− − −( . )( . ) .4 65 10 6 0 10 2 8 1026 2 2 23

The acceleration and force are both negative. This means that they are directed in the negative direction, opposite the original direction of motion of the molecule. By Newton’s third law, the force that the molecule exerted on the wall is equal and opposite: +2.8 × 10−23 N. The pressure on the wall is just the force on the wall divided by the area of the wall:

PN

mPa= × = ×

−

−−2 8 10

10 0 102 8 10

23

2 220.

. ( ).

Note that we replaced cm2 by (10−2 m)2 so that all the units in the preceding calculation are base SI units.

25.3 Molecular Interpretation of Temperature

The translational kinetic energy of a particle is defined as:

K mvtrans = 12

2

where m is the mass of the particle and v is its speed. Molecules, which are made up of more than one atom, have additional kinetic energy due to vibration and rotation.

Chapter 25: Kinetic Molecular Theory 357

A single atom does not rotate or vibrate; its total kinetic energy is equal to its translational kinetic energy.

▼ Example 2

Which has more kinetic energy?A. a molecule with mass of 32 u, moving at 300 m s−1

B. an atom with a mass of 32 u, moving at 300 m s−1

Answer: ABoth atom and molecule have the same translational kinetic energy because they have the same mass and the same speed. However, the molecule has additional kinetic energy because it can rotate and vibrate.

In the preceding section, we calculated the pressure due to just one particle colliding with a wall. In a typical gas sample, we have a very large number of particles moving in three-dimensional space at different velocities. Using the assumptions of Kinetic Molecular Theory, it can be shown that the pressure, P, is:

PK

V

trans=

∑23

where ΣKtrans is the sum of the translational kinetic energies of the gas particles. This equation can be rearranged to:

PV Ktrans= ∑23

Comparing the right-hand-side of this equation with the right-hand-side of the ideal gas equation:

PV nRT=

we can see that:23

K nRTtrans =∑or that the total translational kinetic energies of all the particles is:

K nRTtrans =∑ 32

and the translational kinetic energy per mole of particles, Ktrans,m, is:

KKn

nRT

nRTtrans m

trans, = = =∑

32 3

2


Thus, we can see that temperature is a measure of the average translational kinetic energy of the gas particles: the higher the average translational kinetic energy, the higher the temperature.

▼ Example 3

Which sample of gas has more translational kinetic energy per mole at 298K?

A. 2.00 mol Ne B. 1.00 mol H2 C. sameAnswer: CThe translational kinetic energy per mole is (3/2)RT. Both samples have the same T. It does not matter how many moles of particles either sample has.

▼ Example 4

Which sample of gas has a higher total translational kinetic energy at 298K?

A. 2.00 mol Ne B. 1.00 mol H2 C. sameAnswer: ASince both samples are at the same temperature, the average translational kinetic energy is the same, (3/2)RT (per mole). But the total translational kinetic energy is (3/2)nRT. Since sample A has more moles of particles (larger n), it has a higher total translational energy. Note that one H2 molecule is counted just one particle; it has additional kinetic energy due to rotation and vibration.

When dealing with energy, the preferred unit is Joule (J). In terms of base SI units:1 J = 1 kg m2 s−2

The product of the units L and atm is an energy unit:1 L atm = (10−3 m3) (1.01325 × 105 Pa) = 101.325 m3 Pa = 101.325 J

From this, we can show that R, which is 0.08206 L atm mol−1 K−1, is also equal to 8.314 J mol−1 K−1.


▼ Example 5

Calculate the translational kinetic energy per mole of gas particles at room temperature (298K).Answer:

K RT J mol K K J moltrans m, ( . )( ) .= = = ×− − −32

32

8 314 298 3 72 101 1 3 1

We can calculate the average translational kinetic energy (per particle), <Ktrans> by divid-ing the translational kinetic energy per mole by the number of particles in one mole. The number of particles in one mole is Avogadro’s number (NA = 6.022 × 1023 mol−1). Therefore:

< >KRT

NRTN

k TtransA A

B= = =

32 3

232

where kB, the ratio of R to NA, is called Boltzmann’s constant:

k RN

J mol Kmol

J KBA

= =×

= ×− −

−− −8 314

6 022 101 381 10

1 1

23 123 1.

..

▼ Example 6

Calculate the average kinetic energy of He atoms at 300.0K.

Answer: 1.036 × 10–20 JThe average translational kinetic energy is:

< >K k T J K K Jtrans B= = × = ×− − −32

32

1 381 10 500 0 1 036 1023 1 20( . )( . ) .

Since an atom does not have rotational or vibrational energy, the average translational kinetic energy is also the average kinetic energy.

▼ Example 7

Which gas sample has higher average kinetic energy?A. 2.0 mol He at 298K B. 1.0 mol O2 at 298K C. same

Answer: B. When comparing average values, the number of particles do not matter. Therefore, we ignore the given number of moles. At the


same temperature, the average translational kinetic energy of gas particles is the same. However, O2 molecules have additional kinetic energy due to rotation and vibration. He atoms have no rotational or vibrational energy.

In the preceding discussion, we see that the temperature of a gas is related to the average translational kinetic energy of the gas particles: the higher the temperature, the higher the average translational kinetic energy. Obviously, we can also say that gas particles move faster when a gas is heated. In fact, we can determine the average speed of gas particles if we know the temperature. Since, for any particle:

K mvtrans = 12

2

and, for a collection of gas particles:

< >K k Ttrans B= 32

,

then for a sample where all the gas particles have the same mass:

32

12

2k T m vB = < >

where <v2> is the average of the square of the speed, which we can solve for by rearranging the preceding equation to:

< >vk Tm

B2 3=

The mass of each individual (m) particle is equal to the mass of one mole (the molar mass, Mm) divided by Avogadro’s number (NA), and the Boltzmann’s constant (kB) is equal to R divided by NA. Therefore:

< >v k Tm

RN

T

MN

RTM

BA

m

A

m

2 33

3= =

⎛

⎝⎜

⎞

⎠⎟

=

If we take the square root of the average of the square of the speed, we get what is called the root-mean-square speed, vrms, which is sort of like an average speed:

v v RTMrms

m

= =< >2 3


The derivation for the true average speed, <v>, is somewhat more complicated (for an introductory course), but it can be shown to be slightly less than the root-mean-square speed.

< >vRTM

RTMm m

= =8 2 5465π

.

What the expressions for vrms and <v> tell us is that the average speed of the gas particles is directly proportional to the square root of the temperature, and inversely proportional to the square root of the particle mass. The higher the temperature, the faster the particles are moving. But at the same temperature, particles with more mass move more slowly (on average).

▼ Example 8

Which of these gas samples has particles with the slowest average speed?

A. 2.0 mol He at 298K B. 1.0 mol He at 500KC. 2.0 mol O2 at 298K

Answer: C. O2 at 298K A higher temperature means faster average speed; therefore the average speed is higher for He atoms at 500K; the number of particles does not matter since we are comparing averages. Therefore, choice B is not the correct answer. The question is asking for the sample with the slowest average speed.Choices A and C have the same temperature, but the O2 mol-ecules have larger masses than He atoms. A larger mass, at the same temperature, means slower average speed. Therefore, the particles in O2 at 298K have slower average speed than those in He at 298K.

▼ Example 9

Calculate the average speed of O2 molecules at 298K.Answer:

< >vRTM

RTM

J mol K K

m m

= = =×

− −8 2 5465 2 5465 8 314 29832 00 10

1 1

π. . ( . )( )

. −− −−=3 11444

kg molm s

< >vRTM

RTM

J mol K K

m m

= = =×

− −8 2 5465 2 5465 8 314 29832 00 10

1 1

π. . ( . )( )

. −− −−=3 11444

kg molm s

For unit consistency, we use SI units for all values in this formula; for example, the molar mass is entered as kg mol−1


instead than g mol−1. Here’s how the units work out to meters per second (m s−1). By definition:

1 J = 1 kg m−2 s−2

Therefore, the unit inside the square root term is:

J mol K K

kg mol

Jkg

kg m s

kgm s

− −

−

−−= = =

1 1

1

2 22 2

The square root of m2 s−2 is m s−1.

25.4 Molecular Interpretation of Empirical Gas Laws

Boyle’s law states that, at constant temperature, the pressure of a gas is inversely proportional to its volume. In other words, as volume decreases, pressure increases, and vice versa. How do we explain this? Since pressure is due to collisions of gas particles with the walls, we expect more frequent collisions (higher P) when the walls are closer (smaller V); particles will have less distance to travel before hitting a wall.

Amontons’ law states that at constant volume, the pressure of a gas is directly pro-portional to its temperature. In other words, the higher the temperature, the higher the pressure. An increase in temperature means faster average speed, which means more frequent collisions with the walls and greater force exerted on the walls. Thus, we expect an increase in pressure with an increase in temperature at constant volume.

Charles’ law states that, at constant pressure, the volume of a gas is directly propor-tional to its temperature. In other words, the higher the temperature, the higher the vol-ume. An increase in temperature means faster average speed, which means greater force exerted on the walls and more frequent collisions with the walls, which would lead to a higher pressure; see explanation for Amontons’ law. If we want to maintain the pressure at the higher temperature, we need to lower the frequency of collisions to compensate for the greater force exerted on the walls; we move the walls farther apart (i.e., increase volume).

Avogadro’s law states that at constant T and P, the volume of a gas is directly propor-tional to the number of gas particles. What would happen if we had more particles in the same container at the same temperature? There would be more particles colliding with the walls; we expect a higher pressure. If we want pressure to also remain the same, we need to move the walls farther apart (larger volume). Thus, having more particles at constant T and P means a larger volume.


▼ Example 10

If the most probable speed of particles in a gas sample is 400 m s−1, then the average speed is…

A. faster than 400 m s−1

B. slower than 400 m s−1

Answer: ABased on the formulas, we expect the average speed to be larger than the most probable speed:

vRT

Mmpm

= 2

25.5 Distribution of Molecular SpeedsParticles in a gas sample do not all move at the same speed. The distribution of molecular speeds, first derived by Maxwell in 1860, is:

f v mk T

v eB

mvk TB( ) ( )

/

=2

43 2

2 2

2

��

⎛

⎝⎜⎜

⎞

⎠⎟⎟

−

For O2 at 500K, a plot of f(v) vs. v is shown in Figure 1. The speed distribution function, f(v), for O2 peaks at around 510 m s−1. This is the most probable speed, which can be shown to be:

vRT

Mmpm

=2

Figure 1. Speed Distribution in O2 at 500K


▼ Example 11

Consider the molecular speed distributions for the same gas sample at two different temperatures, T1 and T2, as shown in the figure below. Which distribution corresponds to a higher temperature?

Answer: T1 corresponds to a higher temperature. The T1 curve is shifted to the right compared to the curve at T2. The

< >vRTM

RTMm m

= =8 2 5465π

.

The speed distribution function is said to be normalized; this means that the total area under the curve is equal to 1. We can use the function to determine the fraction of molecules moving at speeds within a given interval by simply taking the area under the curve for that interval. For example, if we want to determine the fraction of O2 molecules at 500K that have speeds between 500 and 1000 m s−1, we would determine the area that is shaded in Figure 2.

Figure 2. Fraction of O2 Molecules with Speeds 500–1000 m s−1


most probable speed is higher at T1; therefore the average speed is also higher at T1. More molecules have higher speeds at T1.

▼ Example 12

Consider the speed distribution for the two gases at the same temperature as shown in the figure below. Which curve, A or B, represents the gas made up of lighter particles?

Answer: BThe average speed is directly proportional to the square root of temperature and inversely proportional to the square root of the molar mass. At the same temperature, the samples have the same average translational kinetic energy, but the sample with smaller particle masses will have a faster average speed.

25.6 Diffusion and EffusionDiffusion refers to the movement of particles from a region of high concentration (where they are more crowded) to a region of low concentration (less crowded). Graham’s Law of Diffu-sion states that the rate of diffusion of a gas at a given temperature is inversely proportional to the square root of the molar mass. It is easy to explain this law using Kinetic Molecular Theory. We simply refer to the formula for the average speed of gas particles: at a given tem-perature, the average speed is inversely proportional to the square root of the molar mass; we expect lighter molecules to move (hence, diffuse) faster than heavier molecules.

▼ Example 13

Cotton balls dipped in concentrated aqueous solutions of NH3 and HCl are simultaneously placed in opposite ends of a glass tube. The NH3 and HCl gas from the cotton balls diffuse inside the tube


and react to form NH4Cl(s) where they meet. The NH4Cl(s) will be expected to form

A. closer to the NH3 end of the tubeB. closer to the HCl end of the tubeC. exactly halfway

Answer: B. NH3 molecules (molar mass = 17.04 g mol−1) are lighter than HCl molecules (molar mass = 36.45 g mol−1). Therefore, NH3 will diffuse faster and will have covered more distance than HCl by the time they meet; the gases will meet at a point closer to the HCl.

The rate of diffusion of a gas is typically much slower than the speeds that we calculate for gas particles. The reason for this is that the movement each gas particle is hindered by collisions with other gas particles. The average distance that a gas particle, assuming it is a sphere of diameter d, travels between collisions is called the mean free path (λ); for a gaseous sample where all the particles have the same mass and size, it can be shown to be:

��

= =⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

12

12

12

12 2 2dRTPN d

RTnRTV

N d nV

NAA A

We can see that the mean free path is shorter the larger the molecule (larger d means smaller λ) and the higher the concentration of particles, n/V. A larger n/V means a higher concentration of particles and a smaller λ. A good analogy for remembering this is to imagine people walking about aimlessly in a room; a larger person (larger d) is more likely to bump into another and people are more likely to bump into one another in a more crowded room (larger n/V).

▼ Example 14

In which sample do gas particles have a larger mean free path?A. 1 mol He at 300K, 1 atm B. 1 mol O2 at 600K, 2 atm

Answer: HeSince:

��

=1

2 2dRTPNA

we can say that, at the same T/P , a smaller molecular diameter (d) would mean a larger mean free path. • T/P in both cases is the same: (300 K/1 atm) = (600 K/2 atm). • He atoms are smaller than O2 molecules.


Therefore, on average, at the same temperature and pressure, a He atom can travel farther before colliding with another He atom. Another way of looking at it is that, since PV = nRT,

PT

nV

R= ⎛⎝⎜

⎞⎠⎟

P/T is proportional to concentration (n/V). The same P/T means the same concentration. If the concentration is the same, then a difference in diffusion rate is due to a difference in molecular size.

25.6.1 Separation of Gas MixturesOne application of Graham’s law is in the separation of a mixture of gases by effusion. Effusion is diffusion through a pinhole into a vacuum chamber. Effusion can also be used to determine molar masses. In both of these applications, we take advantage of the fact that, at a given temperature, lighter molecules move faster than heavier molecules.

▼ Example 15

Which component of a mixture of He and O2 would effuse faster?Answer: HeOn average, the lighter gas particles (He atoms) move faster than heavier gas particles (O2 molecules) at the same temperature and pressure.

25.6.2 Determination of Molar MassEffusion can be used to determine molar mass. It can be shown that the rate of effusion (r) through a pinhole of area A is:

r nV

v A PRT

RTM

Am

= =14

14

8

< >p

If we use this formula, we can calculate the moles of gas particles that would pass

through the pinhole per unit time. We can see that at a given T, P, and A, the rate of effusion is inversely proportional to the square root molar mass (Mm). This is known as Graham’s Law of Effusion.

If we compare the effusion rate of two gases (one with known Mm, the other with unknown Mm) in the same container, at the same temperature, through the same pinhole, we can determine the molar mass of the unknown. If we refer to the rates and molar masses of these two gases as r1 , r2, Mm1, and Mm2, then:


r M r Mm m1 1 2 2=or:

r M r Mm m12

1 22

2=

▼ Example 16

CO2 effuses at a rate of 2.00 mmol min−1. At the same condi-tions, an unknown gas effuses at a rate of 3.32 mmol min−1. What is a possible identity of the unknown gas?

A. CH4 B. O2 C. Cl2 D. CCl4Answer: CIf we take CO2 as gas “1”, we can solve for the molar mass of the unknown gas (Mm2) using:

r M r Mm m12

1 22

2=which we can rearrange to

M Mrr

Mrr

g molmmol

m m m2 112

22 1

1

2

2

144 012 00

=⎛

⎝⎜

⎞

⎠⎟ =

⎛

⎝⎜

⎞

⎠⎟

= −( . ). minn. min

.−

−−⎛

⎝⎜

⎞

⎠⎟ =

1

1

2

1

3 3216 0

mmolg mol

M Mrr

Mrr

g molmmol

m m m2 112

22 1

1

2

2

144 012 00

=⎛

⎝⎜

⎞

⎠⎟ =

⎛

⎝⎜

⎞

⎠⎟

= −( . ). minn. min

.−

−−⎛

⎝⎜

⎞

⎠⎟ =

1

1

2

1

3 3216 0

mmolg mol

Among the choices given, CH4 has the closest molar mass (16.04 g mol−1).

▼ Example 17

How much faster would He effuse compared to CH4 under the same conditions?Answer: two times fasterThe molar masses are 16.04 g mol−1 for CH4 and 4.00 g mol−1 for He. The mass of CH4 is about four times that of He. The square root of four is two. Therefore, we expect the rate of effusion of the smaller particles (He) to be two times faster. If we refer to He as gas “1” and CH4 as gas “2”, then:

rr

MM

g molg mol

m

m

1

2

2

1

1

1

16 044 00

2 00= = =−

−

..

.


TEST YOURSELF


1. Which has more kinetic energy? A. a CH4 molecule moving at 150.0 m s−1

B. a He atom moving at 150.0 m s−1

C. same

2. Which gas sample has a higher average translational kinetic energy? A. 1.0 mol CH4 at (298K, 1.0 atm) B. 2.0 mol O2 at (298K, 0.5 atm) C. same

3. Which gas sample has a higher translational kinetic energy? A. 1.0 g CH4 at (298K, 1.0 atm) B. 2.0 g O2 at (298K, 0.5 atm) C. same

4. Which gas sample has a higher average particle speed? A. 1.0 mol CH4 at (298K, 1.0 atm) B. 2.0 mol O2 at (298K, 0.5 atm) C. same

5. The average speed, most probable speed, and root mean square speed of He at 298.15K are. . .

A. 1113, 1256, 1363 m s−1, respectively B. 1256, 1113, 1363 m s−1, respectively C. 1363, 1256, 1113 m s−1, respectively

6. In a mixture of H2 and CH4, which of these is true? A. a CH4 molecule has a lower speed than an H2 molecule. B. a CH4 molecule has a lower kinetic energy than an H2 molecule. C. both A and B are true D. none of these

7. At constant temperature, the pressure of gas increases when its volume decreases. Kinetic Molecular Theory explains the increase in pressure as being due to:

A. more force exerted by gas particles on the wall of the container B. more frequent collisions of gas particles with the wall of the container C. both A and B D. none of these



8. At constant volume, the pressure of gas increases when the temperature increases. Kinetic Molecular Theory explains the increase in pressure as being due to:

A. more force exerted by gas particles on the wall of the container B. more frequent collisions of gas particles with the wall of the container C. both A and B D. none of these

9. For which sample do gas particles have a larger mean free path at STP? A. 1 mol H2 B. 2 mol H2 C. same

10. Consider a gas containing 0.7% 235UF6 and 99.3% 238UF6. If the gas is contained in chamber A, as illustrated below, and is allowed to escape through a pinhole into vacuum chamber B, the percentage of 235UF6 in the gas that initially emerges in chamber B will be…

A. higher than 0.7% B. lower than 0.7% C. still 0.7%

11. Which gas sample will effuse faster than a sample containing 5.00 g CH4 at STP? A. 1.00 g CH4 at STP B. 5.00 g H2 at STP C. 5.00 g CO2 at STP

12. Calculate the rate of effusion of CO2 if the rate of effusion of H2 under the same conditions is 2.50 mmol s−1.

371

CHAPTER

Real Gases, Liquids, and Solids

In this chapter, we take a more in-depth look at the physical states of matter.

26.1 The Compression FactorThe deviation of a gas from ideal behavior can be determined by comparing its actual vol-ume (V) to what its volume would be if it were behaving ideally (Videal). This ratio is called the compression factor (Z):

Z VV

VnRT

P

PVnRTideal

= =⎛⎝⎜

⎞⎠⎟

=

▼ Example 1

What is the value of the compression factor for a gas that is behaving ideally?Answer: Z=1If a gas is behaving ideally, then we expect Z to be equal to 1 since PV=nRT for an ideal gas:

ZPVnRT

= = 1

▼ Example 2

One mole of a gas sample at STP has a volume of 22.0 L. Calculate the compression factor.

26


Answer: 0.982Since the volume of one mole of ideal gas at STP is 22.4 L:

ZV

VLLideal

= = =22 022 4

0 982..

.

In general, we can calculate Z by plugging in the values of P, V, n, R, and T.

26.2 Atomic Interpretation of Compression Factor

Ideal gas behavior is expected if we assume that gas particles are non-interacting point masses. This assumption is a very good approximation when gas particles are very, very far apart. We explain deviations from ideal behavior by considering that gas particles do interact when they are closer together.

The compression factor tells us how the volume of a gas compares with the ideal vol-ume. If V is the volume of a gas and Videal is its volume if it were ideal.

• If Z = 1, V = Videal• If Z >1, then V > Videal• If Z < 1, then V < Videal

To explain why a gas would deviate from ideal behavior, it is useful to see at what conditions these deviations typically occur. Figure 1 shows a typical plot of Z vs. pressure (red curve) for a gas at room temperature. The curve is called an isotherm because the temperature is the same for all the points on the curve. For comparison, the ideal Z value (1) is shown as a dotted blue line. Note that:

• Z approaches 1 at very low pressures. At very low pressures, the concentration of gas particles is very low and the average distance between particles is very large compared to the size of the particles. Interactions are minimal, so the particles can be expected to behave ideally.

• Z > 1 at very high pressures. At very high pressures, the average distance between the particles becomes comparable to their size, which leads to strong repulsions among the nuclei. The repulsions cause the particles to spread out more (compared to particles that do not interact). Thus, the volume of the gas is larger than ideal.

Chapter 26: Real Gases, Liquids, and Solids 373

• Z < 1 at intermediate pressures. The average distance between the particles is such that the predominant interactions are attractive. The attractions cause the particles to occupy a smaller volume compared to particles that do not interact.

▼ Example 3

The pressure dependence of the compression factor (Z) of CO2 and N2 at 350K is shown below. Which is which?

Figure 1. Typical ZP Isotherm at Room Temperature


Answer: The red curve is the isotherm for N2; the green curve is the isotherm for CO2.Attractive forces are responsible for the compression factor dipping below the ideal value of 1 at intermediate pressures. We expect Z to dip more for CO2 because intermolecular forces of attraction are stronger among CO2 molecules compared to N2 molecules. Why? Both N2 and CO2 are nonpolar molecules. We expect the intermolecular forces in these gases to be mainly London dispersion. London dispersion forces are stronger among larger, more polarizable molecules. CO2 molecules are larger than N2 molecules.

26.3 Liquefaction, Deposition, and Vapor Pressure

Increasing the pressure of a gas eventually causes it to turn into a liquid as long as the tem-perature is below a certain threshold known as the critical temperature (Tc). The typical behavior is illustrated in Figure 2 using PV isotherms. A PV isotherm is a plot that shows the relationship between pressure and molar volume at a constant temperature. The blue curve in the Figure is an example of a PV isotherm for an ideal gas; this curve summarizes how the pressure and molar volume are related a constant temperature. The red curve resembles the typical relationship between pressure and molar volume for a real gas at the same temperature. How do we interpret this Figure?

• The two curves appear to be “merging” at the bottom right corner (large volume, low pressure). This is where we expect the gas to behave ideally. Z approaches 1 at very low pressures. If the real gas curve (the red curve) is to the left (or below) the blue curve, Z < 1. If it is to the right (or above) the blue curve, Z > 1.

• Imagine a gas sample at very low pressure. This corresponds to a point at the bottom right part of the red curve. What would happen if we were to compress the gas while maintaining its temperature? If we compress the gas while main-taining the temperature, we are essentially “tracing the red curve towards the left where volume is smaller.” We can see that the pressure will rise, but not as much as it would if it were ideal (“the red curve is below the blue curve”). Eventually, we reach a point where additional compression does not cause a change in pressure; this is the flat part of the red curve (just below the “real”


label). The reason for this is that the gas now starts to change into a liquid (liquefaction) or a solid (deposition). Whether it changes into a solid or a liquid depends on the temperature; deposition is favored at lower temperatures. The pressure where liquefaction or deposition occurs is called the vapor pressure of the substance at this temperature. If we were to continue compressing the gas, enough gas will be converted to solid or liquid so that the pressure of the gas is maintained.

• Once all the gas is completely converted into liquid (or solid), any additional attempt to compress it will require a drastic increase in pressure. The steep rise in the red curve on the left side tells us that a very large increase in pressure will hardly cause a change in volume. This is because the atoms in a liquid or solid are so close to each other that repulsions between neighboring nuclei will strongly resist additional squeezing.

▼ Example 4

Consider the PV isotherm shown below. At this temperature, what is the vapor pressure of the substance? Which labeled point on the diagram represents a condition where the substance would

Figure 2. Typical PV Isotherm at T < Tc


be a liquid (or solid)? Which labeled point represents a condition where the substance would be gaseous and deviates least from ideal behavior?

Answer: The vapor pressure is the pressure that corresponds to points K and L. This is the pressure where liquid and gas coexist. Point J (higher pressure and smaller volume compared to points K and L) represents a state or condition where the substance exists as a liquid or solid. Points M and N represent gaseous states (lower pressure and larger volume compared to points K and L); point N represents a state which would deviate less from ideal behavior (lower pressure compared to point M).

26.4 The Critical PointFigure 3 shows the typical dependence of PV isotherms on temperature at the critical temperature (Tc) and at temperatures below it. The green curve represents behavior at the critical temperature. The blue curve represents behavior at a higher temperature than the red curve. How do we interpret this figure?

• The pressure where liquefaction occurs, the vapor pressure of the substance, is higher at a higher temperature. The flat part of the blue curve corresponds to a higher pressure than the flat part of the red curve.

• The shrinkage during liquefaction is less drastic at higher temperature. The flat part of the blue curve is shorter compared to flat part of the red curve. The difference


between the volumes at the red and blue dots is smaller for the blue curve (compared to the red curve).

• The trend of less shrinkage (due to liquefaction) as temperature increases continues until we reach the critical temperature (the green curve). If we were to overlay additional isotherms on the Figure, the blue and red dots would eventually merge at the critical temperature; this is indicated as a purple dot in the Figure. We say that the critical temperature is the temperature above which a gas can no longer be liquefied. The point indicated by the purple dot is called the critical point. Mathematicians call this an inflection point. If the substance is at any condition where the pressure and tempera-ture is higher than that at the critical point, we say that it is a supercritical fluid.

▼ Example 5

Consider the plot below of a ZP isotherm for a substance. Based on the dependence of the compression factor on pressure, we can say that temperature of the substance is: A. above the critical temperature B. below the critical temperature

Figure 3. Temperature Dependence of PV Isotherms @ T ≤ Tc


Answer: BRemember that Z is the ratio of the volume to the ideal volume. The vertical drop from point K to point L suggests significant shrinkage without a change in pressure, which is observed during the conversion of the gas to a liquid or solid. Therefore, this iso-therm must be for a temperature below the critical temperature.

26.5 The van der Waals EquationWe can introduce “correction terms” into the ideal gas equation (PV = nRT) in order to summarize real gas behavior. One way of doing this, as proposed by Johannes van der Waals, is to write:

P nRTV nb

a nV

=−

− ⎛⎝⎜

⎞⎠⎟

2

The terms a and b in the equation are called van der Waals parameters and account for attractive and repulsive interactions, respectively. Note that if a and b are both zero, the van der Waals equation reduces to the ideal gas equation:

P nRTVideal =

To understand the correction terms, let’s take an alternative look at the meaning of com-pression factor. We can see from the derivation below that we can also think of Z as the ratio of the gas pressure to what it would be if it were ideal (at the same T, V):

Z PVnRT

PnRT

V

PPideal

= =⎛⎝⎜

⎞⎠⎟

=


Therefore, Z < 1 means the pressure is less than ideal and Z > 1 means the pressure is higher than ideal (at the same T and V).

For now, let us ignore attractive interactions by removing the second term from the van der Waals equation. We get:

P nRTV nb

=−

which is essentially the ideal gas equation except for nb being subtracted from the volume in the denominator. How do we justify this correction? Repulsions essentially lessen the amount of space available to the gas particles; thus we subtract something from V; the parameter b is called the excluded volume (per mole). We expect larger b values for larger gas particles. Because of the unavailable space, gas particles would have less distance to travel before colliding with the walls of the container; this leads to more frequent colli-sions with the walls of the container. This leads to a higher pressure (Z > 1) for the gas (compared to the ideal pressure at the same T and V).

How do we justify the second correction term? Imagine a gas particle on its way to colliding with the wall. Attractive forces from other gas particles will slow it down. It will, therefore, impart a smaller pressure on the wall. The overall effect on pressure depends on the concentration of particles (n/V). Within a given volume, the effect depends on the number of particles affected and the number of other particles affecting each particle. Thus, we subtract a correction term that depends on the square of (n/V).

P nRTV nb

a nV

=−

− ⎛⎝⎜

⎞⎠⎟

2

We expect larger a values for gas particles that have stronger attractions.

▼ Example 6

The van der Waals parameters for H2S are 4.484 atm L2 mol−1 and 4.34 L mol −1.Which one is a? Which one is b?

Answer: a = 4.484 atm L2 mol −1 and b = 4.34 L mol −1.Why? The parameter b is called the excluded volume. Since nb is subtracted from volume, the unit for nb must be a volume unit like L. The unit for n is mol. Therefore, the unit for b must be a volume unit per mole, or L mol −1.The a(n/V)2 correction term should have a pressure unit such as atm. Since the unit for (n/V)2 is mol2 L− 2, the parameter a should have (a pressure unit) times (units that cancel out when multiplied by mol2 L−2).


▼ Example 7

Explain the difference between the van der Waals parameters for H2O and H2S.For H2O: a = 5.464 atm L2 mol −1, b = 3.05 L mol −1.For H2S: a = 4.484 atm L2 mol −1 and b = 4.34 L mol −1.Answer:H2S molecules are larger than H2O molecules. We can predict this based on the location of O and S in the periodic table. O and S belong to group VIA; size increases going down a column. Therefore, we expect a larger b value for H2S.Based on size alone, we would expect H2S to also have a larger a value since larger molecules are more polarizable and have stronger London dispersion forces among them. However, H2O molecules are capable of hydrogen bonding interaction which, in this case, is strong enough that its a value is actually higher than that of H2S.

▼ Example 8

Shown below (in red) are the PV isotherms for a gas at its critical point (TC) and at two higher temperatures; T2 > T1 > Tc. Corre-sponding ideal gas isotherms are shown (in blue) for comparison. Using the van der Waals equation, explain the difference in the curves at T1 and T2.


Answer:When the red and blue curves are almost matched, it means the gas is behaving ideally. This happens at low pressure (bottom right corner) at both T1 and T2.When the real (red) curve is below the ideal (blue) curve, it means that P < Pideal, or Z < 1. This means that the deviation is primarily due to attractive interactions. This happens at intermediate pres-sures for T1 but not for T2.If the red curve is above the blue curve, it means that P > Pideal or Z > 1. This happens at very high pressures for T1 and at all pressures for T2. We can explain this using the van der Waals equation:

PnRT

V nba

nV

=−

− ⎛⎝⎜

⎞⎠⎟2

Depending on the value of a, at sufficiently high T, the a(n/V)2 term becomes negligible and we expect P > Pideal and Z > 1 at all pressures, which is the case at T2. In other words, at high temperatures, deviation from ideal behavior is primarily due to repulsions.

26.6 Phase DiagramsA phase diagram summarizes experimental information about the stable phases of a sample of matter at different conditions. The common features of phase diagrams for pure substances are illustrated in Figure 4. Here’s how we interpret the diagram.

• Any point on the diagram corresponds to a specific condition (pressure and temperature).

• The four main regions, labeled solid, liquid, gas, and SCF (supercritical fluid), represent conditions where these are the most stable phases observed.

• The blue curve, the boundary between the liquid and gas regions, represents condi-tions where liquid and gas coexist. Points on the blue curve represent conditions where the tendencies for transformation from liquid to gas and gas to liquid are equal. This curve tells us how the vapor pressure of the liquid varies with tempera-ture. Alternatively, we can say that this curve tells us how the boiling point of the liquid varies with pressure. The temperature where the vapor pressure is 1 atm, if applicable, is called the normal boiling point of the liquid, the temperature at which we expect the liquid to boil under ordinary conditions.


Figure 4. Phase Diagram of Pure Substance

• The red curve, the boundary between the solid and gas regions, represents condi-tions where solid and gas coexist. This curve tells us how the vapor pressure of the solid varies with temperature. Alternatively, we can say that this curve tells us how the sublimation point of the solid varies with pressure. The temperature where the vapor pressure is 1 atm, if applicable, is called the normal sublimation point of the solid.

• The green line, the boundary between the solid and liquid regions, represents conditions where solid and liquid coexist. This curve tells us how the melting point of the solid (which is the same as the freezing point of the liquid) depends on pressure. The temperature that corresponds to 1 atm pressure is called the normal melting point of the solid, or normal freezing point of the liquid. It is important to note that the solid-liquid boundary is very steep; the melting point does not change significantly with increasing pressure. Typically, as shown in the diagram, this curve has a positive slope (“leans to the right”); the melting point increases slightly as pressure increases. One notable exception is water; the solid-liquid boundary for water has a negative slope (“leans to the left”).

• The point labeled CP represents the critical point; the pressure and temperature at this point are called the critical pressure and temperature. As discussed earlier, no


liquefaction is observed above the critical temperature. When the pressure of the gas above the critical temperature is increased beyond the critical pressure, we have what is called a supercritical fluid; this is the region labeled SCF.

• The label TP stands for “triple point.” This point represents a condition where three phases coexist (solid, liquid, and gas). For water, this occurs at 273.16K (or 0.01°C) and 4.58 Torr.

▼ Example 9

Why does food cook faster in a pressure cooker?Answer:Heating gives more water molecules the ability to leave the liquid phase and go into the gas phase. Since the container is sealed and has a fixed volume, this allows us to achieve a high pressure. The boiling point of water increases with increasing pressure. This allows us to achieve higher temperatures for the liquid water inside the pressure cooker.

▼ Example 10

Explain why ice melts under an ice skater’s blades.Answer:The slope of the solid-liquid boundary in the phase diagram for water is negative. This means that the melting point of ice decreases with increasing pressure. The ice skater’s weight is concentrated on a very small area, the area under the sharp edge of the blades. This subjects the ice underneath to a very high pressure; pressure is force per unit area. The temperature of the ice (which does not change) suddenly becomes higher than its melting point, which is now lower due to the increase in pressure.

▼ Example 11

If left alone, why does water completely dry out in an open container but not in a sealed container.Answer:In a sealed container, movement of water molecules into the gas phase will eventually cause the partial pressure of water mole-cules in the gas phase to reach the vapor pressure at the prevail-ing temperature. At this point, the rate of molecules transferring


from the liquid to the gas phase equals the rate of molecules transferring from the gas to the liquid phase. We say that we eventually reach a state of equilibrium; the equilibrium is said to be a dynamic equilibrium (as opposed to one that is static) since movement of molecules between the liquid and gas continues even if no apparent changes are observed.In an open container, nothing stops the water molecules in the gas phase from leaving the container. Some water molecules will be leaving the container at all times. Thus, it is not possible to achieve equilibrium between the liquid and gaseous phase.

▼ Example 12

Consider the phase diagram below. Under ordinary conditions, what is the physical state of the substance? What is the normal boiling point?

Answer: solid, T1“Ordinary conditions” would correspond to a pressure close to 1 atm and room temperature (298K). This point is in the solid region of the phase diagram. When heated at 1 atm, this sub-stance is expected to melt and eventually boil at temperature labeled T1 on the diagram. T1, the temperature where liquid and gas coexist at 1 atm, is the normal boiling point.

The most stable arrangement of particles in a solid depends on temperature and pres-sure. Each unique arrangement is considered as a different phase. Thus, depending on the range of conditions included, the solid region in a phase diagram may be subdivided into


two or more sub-regions, as illustrated in Figure 5 for water. Note that, in the Figure, the pressure and temperature axes are switched and a logarithmic scale is used for pressure. The diagram in this Figure is from the website of a research group (based in University College London) that studies “the structure and behaviour of the constituent materials of icy moons, in order to better understand their thermal and geological evolution.”

▼ Example 13

The phase diagram for HgI2 is illustrated below. Identify the physi-cal states of the stable phases in regions A, B, C, and D. Solid HgI2 is known to be red at room temperature and turns yellow when heated to 127°C. Which region of the phase diagram repre-sents conditions where solid HgI2 is red?

Figure 5. Phase Diagram for Waterfrom http://www.es.ucl.ac.uk/research/pig/pig/research.htm

http://www.es.ucl.ac.uk/research/pig/pig/research.htm


Answer:A - solidB - solidC - liquidD - gas, HgI2 is a red solid in region A.

We figure out that that C and D represent liquid and gaseous phases since the boundary between them terminates at the criti-cal point (CP). The other phases (A and B) must be solid phases. Since region B is to the right of region A, it represents condi-tions at higher temperature. The question says that HgI2 is red at lower temperature (room temperature) and yellow at higher temperature (127°C).

26.7 Effects of HeatingA phase diagram is useful for determining how much heat is involved when a substance undergoes changes under ordinary conditions. Heating a pure substance under ordinary conditions means heating it at a constant pressure of 1 atm. We can break down our calculations into a series of steps, based on the changes that we can deduce from the phase diagram.

• For each step that involves only a temperature change (no phase change), we use the formula

q = CΔT where C is the heat capacity of the substance, which we need to look up, and ΔT is

the temperature change.• For each step where a phase transition occurs, we need to look up ΔtrsH, the

enthalpy of phase transition or heat of phase transition.

When we look up heat capacities and enthalpies of phase transitions, we will find that values are usually given on a “per gram” or “per mole” basis; we must multiply these values by the amount of substance (in grams or moles, whichever is appropriate). Heat capacity per mole is called molar heat capacity; heat capacity per gram is called specific heat. [NOTE: If we do look up these quantities, we will find that they depend on temperature; we can do approximate calculations by using the average values for the temperature range we are interested in. More precise calculations involve calculus and are beyond the scope of a typical introductory course.]


▼ Example 14

Consider a substance whose phase diagram is illustrated below. What information do we need to calculate the heat needed to change it from state A to state C?

Answer:We can see from the diagram that we start with a solid at point A and end up with a gas at point C. From the phase diagram, we deduce that this involves 3 steps if done at constant pressure.• Step 1. Heat the solid from TA to TB. We need to look up the

heat capacity of the substance when it is a solid (Csolid) and calculate the heat needed to raise its temperature from TA to TB. The heat required for this is:

q1 = CsolidΔT, where ΔT = TB – TA

• Step 2. Once our substance’s temperature reaches TB, the phase diagram tells us that it will sublime. The heat required for this phase transition (ΔtrsH) is also known as the heat of sublimation (ΔsubH):

q2 = ΔsubH The temperature of the substance will remain constant while it

is undergoing a phase transition.• Step 3. Heat the substance, now a gas, from TB to TC. We need

to look up the heat capacity of the substance when it is a gas (Cgas), and the heat required is:

q3 = CgasΔT, where ΔT = TC – TB

The total heat required is q1 + q2 + q3.


Note that when the triple point (where solid, liquid and gas coexist) of a substance is above one atmosphere, as in this example, we expect it to sublime rather than melt under ordinary conditions.

A heating curve summarizes all the heat transfers that occur as a substance undergoes heating at constant pressure. If we plot of the amount of heat absorbed by a sample vs. temperature, we will see vertical sections at temperatures where phase transitions occur; temperature remains constant during a phase transition. If we plot temperature vs. amount of heat, we will see horizontal sections where phase transitions occur.

▼ Example 15

Sketch the heating curve for the sublimation described in the preceding example. Explain.Answer:The heating curve is shown in below (in blue).

We expect the amount of heat absorbed to increase as tempera-ture increases from TA to TB. The amount of heat required for this is shown on the graph as x:

x = q1 = CsolidΔTWe expect the amount of heat to increase some more while the temperature remains at TB while the substance is subliming.


We show this on the plot as vertical line. The amount of heat required for this is:

y – x = q2 = ΔsubHFinally, we expect the amount of heat to increase again as the temperature of the gas increases to TC. The amount of heat required for this is:

z – y = q3 = CgasΔT

If we were to look up ΔtrsH for freezing, deposition, and condensation, we would not likely find it. This is because these are just the reverse of melting (fusion), sublimation, and vaporization, respectively. The values are the same, except for the opposite algebraic sign. Melting, sublimation, and vaporization are endothermic processes; the ΔtrsH for these are positive. Freezing, deposition, and condensation are exothermic processes; the ΔtrsH for these are negative.

▼ Example 16

The heat of vaporization of water at 100°C is 40.7 kJ mol −1. Calculate q for the condensation of 180.2 g of steam to liquid water at 100°C.Answer: − 407 kJSince the heat of vaporization is 40.7 kJ mol −1, the heat of con-densation is the negative of this value (− 40.7 kJ mol −1). Since the value is given on a “per mole” basis, we multiply it by the number of moles of water. The number of moles of water is equal to the mass divided by the molar mass (which is 18.02 g/mol):

ng

g molmolwater = =−

180 218 02

10 01

..

.

Therefore:q = (− 40.7 kJ mol −1)(10.0 mol) = − 407 kJ

▼ Example 17

A 5.00 g ice cube at 0.00°C is mixed with 100.0 g of water at 25.00°C in a calorimeter. The final temperature is found to be 20.00°C. Calculate the heat of fusion of water.Answer: 6.03 kJ/molA calorimeter is a thermally-insulated container; it does not allow heat to flow in or out. Any heat lost by something inside the


container must be gained by something else inside the calorimeter. If we let:• q1 = heat of fusion, the heat required to melt 5.00 g of ice

into 5.00 g of liquid water at 0.00°C• q2 = heat needed to raise the temperature of 5.00 g of

melted ice from 0.00°C to 20.00°C.q2 = mcΔT = (5.00 g)(4.184 J °C −1 g −1)(20.00°C − 0.00°C)

= 418.4 J• q3 = heat flow for cooling 100.0 g of water from 25.00°C to

20.00°Cq3 = mcΔT

= (100.0 g)(4.184 J °C −1 g −1)(20.00°C − 25.00°C) = −2092 J

Then:q1 + q2 + q3 = 0

orq1 = − q2 − q3 = −(418.4 J) − (−2092 J) = +1674 J

The heat of fusion of 5.00 g of ice is 1674 J. [Note: digits in boldface are the last significant digits. We keep additional digits in the intermediate results to avoid buildup of round off errors.]On a “per gram” basis, the heat of fusion is:

q1 = 1674 J/5.00 g = 334.8 J/g, or 335 J/gOn a “per mole” basis, the heat of fusion is:

q1 = (334.8 J/g)(18.02 g/mol) = 6.03 × 103 J/mol, or 6.03 kJ mol−1

The literature value is 6.01 kJ/mol.

26.8 Structure and Properties of SolidsThe properties of a solid depend on what type of particles (atoms, ions, or molecules) it has and how these particles are arranged (the structure). If we find a repeating pattern in the way the particles are arranged in a solid, we say that the particles are in a crystalline lattice, or that we have a crystalline structure. Otherwise, we say that the solid is amorphous.

26.8.1 Amorphous SolidsThere is no order in the arrangement of particles in an amorphous solid. Materials made up of long molecules (polymers) tend to form amorphous materials. Proteins (in muscles) and cellulose (in plants) are examples of natural polymers. Everyday plastics and fabrics


are examples of synthetic (man-made) polymers. Rapid cooling of a liquid tends to pro-duce an amorphous solid because the particles are unable to have enough time to settle into an orderly arrangement; such materials are called glass. Cooling a liquid could lead to the formation of an amorphous solid if impurities interfere with the particles that would normally form a crystal upon cooling. For example, a molten mixture of soda ash (sodium carbonate) and sand (silicon dioxide) produces the glass that we encounter everyday in things like windows, mirrors, and wine bottles.

26.8.2 Crystalline SolidsThe regular arrangement of particles in a crystalline solid leads to the minimization of total potential energy of interactions of the particles. In other words, it is the most stable arrangement. This is why an overwhelming majority of naturally-occurring solids are crystalline. Examples of crystalline solids include minerals (salts), metals, carbon (diamond). This also explains why crystals tend to have smooth, flat surfaces when cut.

Properties of crystals depend on the type of particles in the lattice. Crystals can be classified as ionic, molecular, or atomic. Atomic crystals can be classified as nonbonding, metallic, or covalent network.

Ionic crystals tend to have very high melting points due to strong attractions among ions. They are also brittle. This is due to the strong repulsions that result when ions of like charge are momentarily brought closer together as ions are slightly displaced from their locations when the crystal is, say, hit by a hammer.

Molecular crystals tend to have low melting or sublimation points. The attractive forces among the molecules are relatively weak (mainly van der Waals forces).

Nonbonding atomic crystals are formed when noble gases are frozen to very low tem-peratures. These atoms are held together by very weak London dispersion forces.

Metallic crystals are made up of atoms of metallic elements. If more than one element is present, the solid is a solution and is called an alloy. Because electrons are loosely held in metal atoms, bonding among metal atoms in a solid (metallic bonding) can be described as one that involves extensive delocalization of electrons. A strong metallic bond is the reason metals have high melting points and boiling points; most metals are solids at room temperature. Because metal atoms can readily slip and roll over each other without breaking the metallic bond, metals are malleable (easily hammered into thin sheets) and ductile (easily pulled out into a thin wire). Irregularities in the structure, introduced by hitting the metal at low temperature, can cause a metal to become hard and brittle; the irregularities hinder the ability of the atoms to roll over each other. Irregularities can also be introduced by mixing metals. Pure gold is too soft for jewelry, but can be made harder if alloyed with other metals. Heating a metal “shakes” the atoms, causing them to “relax” into a more regular structure and makes the metal soft.

A covalent network crystal can be thought of as one giant molecule; the atoms are held together by very strong covalent bonds. Thus, a network covalent crystal like diamond


(where each C atom is an sp3-hybridized tetrahedral center, bonded to four other C atoms) have very high melting points and are among the hardest materials known. Graphene, a single layer of covalently bonded carbon atoms (where each C atom is an sp2-hybridized tri-gonal planar center, bonded to three other C atoms) is stronger than steel. Graphene, which has numerous interesting and potentially useful properties, has been the subject of much research in recent years. The 2010 Nobel Prize in physics was awarded to scientists who did pioneering work on graphene. Graphite, the “lead” in pencils, consists of layers of graphene.

▼ Example 18

Classify the following crystals: I2(s), brass, Ar(s), CaCl2(s), graphite.Answer: molecular, metallic, atomic (nonbonding), ionic, covalent network

A crystalline structure is said to have a long-range order. The overall structure can be thought of in terms of a repeating pattern, called a unit cell. The simplest type of unit cell, which is not very common, is called the simple cubic cell; Figure 6 shows a simple cubic cell (left) and an array of these cells (right); an actual crystal would have a very, very large array of these cells. The rods in the Figure are intended to help you visualize the cube; they are not to be interpreted as something physically tangible. In a simple cubic unit cell, a particle is located at each of the 8 corners of the cube. However, we can see that each corner is actually shared by 8 neighboring cells; only 1/8 of each corner particle belongs to

Figure 6. Simple Cubic Cellsfrom http://www.molsci.ucla.edu/pub/explorations.html#Crystalline Solids



Figure 8. Face-Centered Cubic Cellsfrom http://www.molsci.ucla.edu/pub/explorations.html#Crystalline Solids

Figure 7. Body-Centered Cubic Cellsfrom http://www.molsci.ucla.edu/pub/explorations.html#Crystalline Solids

any one cell. Thus, we say that each simple cubic cell contains only one particle. Figures 7 and 8 show two more-commonly observed cubic cells:

• the body-centered cubic (bcc) cells; each bcc cell has a particle in the center of the cube in addition to the particles at the eight corners

• the face-centered cubic (fcc) cells; each fcc cell has a particle in the center of each of the six faces of the cube in addition to the particles at the eight corners




▼ Example 19

How many particles are there inside a bcc unit cell? How many are in an fcc unit cell?Answer: 2, 4In a body-centered cubic cell, only 1/8 of the 8 corner particles can be considered as being inside the cell; 1/8 of 8 is 1. This plus the particle inside the cell gives us a total of two.In a face-centered cubic cell, only 1/8 of the 8 corner particles can be considered as being inside the cell; 1/8 of 8 is 1. Each of the particles in the six faces is shared with a neighboring cell. Therefore, only half of the six can be considered as being inside the cell; 1/2 of 6 is 3. Therefore, the total number of particles in an fcc cell is 1 + 3, or 4.

The type and size of unit cell can be experimentally determined using x-ray diffrac-tion. We can use this information to define the size of particles in a crystal; we define the size of the particles as the size of spheres that are as tightly packed as possible. In the case of a simple cubic cell, the tightest packing of equal sized spheres is achieved by having each one touch its six nearest neighbors. Each particle can be thought of as touching one neighbor on the left, one on the right, one above, one below, one in front, and one behind; we say that each particle has a coordination number of 6. Figure 9 illustrates the tightest packing of four spheres in a simple cubic cell as seen from the front of one of its faces. We can see that the length of the cube is just equal to twice the radius of each sphere.

Figure 9. Tightest Packing in Simple Cubic Cell


▼ Example 20

Polonium, Po, is the only element known to have a simple cubic crystal structure. The length of one side of the unit cell is 335.9 pm. Calculate the radius and volume of a Po atom.

Answer: 168.0 pm, 1.984 x 10−29 m3

For a simple cubic cell with equal sized spheres at each corner, the length of the side (a) is equal to twice the radius of the sphere. So, considering each Po atom as a sphere, the radius of each atom is:

ra pm

pm pm= = = →2

335 92

167 95 168 0.

. .

The formula for the volume of a sphere is:

V r m m= = × = ×− −43

43

167 95 10 1 984 103 12 3 29 3� � ( . ) .

Packing efficiency is defined as the ratio of the volume of the spheres inside a unit cell to the volume of the unit cell itself.

pV

Vparticles

cell

=

Since a simple cubic cell contains one particle, the packing efficiency is:

pr

ror= = =

432 6

0 523 52 33

3

� �( )

. . %

▼ Example 21

Polonium, Po, is the only element known to have a simple cubic crystal structure. What is the packing efficiency of atoms in solid polonium?Answer: 52.3%

Density is the ratio of the mass of a sample to its volume. The observed density of a crystalline solid should be equal to the particles in a unit cell to the volume of the unit cell.


For a simple cubic unit cell with a side of length a, the density is equal to the mass of one particle divided by the volume of the unit cell (a3).

dm

aone particle= 3

▼ Example 22

Polonium, Po, is the only element known to have a simple cubic crystal structure. The length of one side of unit cell (a) is 335.9 pm. There is no naturally occurring isotope of Po, but the most stable radioisotope is Po-209, with a mass of 208.98 u. What would be the density of a solid sample of Po?Answer:

dm

Vkg

mone particle

cell

= = ××

=−

−

208 98 1 6605 10335 9 10

27

12 3

. ( . )( . )

99 156 10 9 1563 3. .× =−kg m g mL/

dm

Vkg

mone particle

cell

= = ××

=−

−

208 98 1 6605 10335 9 10

27

12 3

. ( . )( . )

99 156 10 9 1563 3. .× =−kg m g mL/

For a bcc cell, in the tightest packing of equal-sized spheres, the corner particles do not touch each other. In a bcc cell, the middle particle touches the 8 corner particles; its coordination number is 8. [This is actually true of every particle in the unit cell since we could think of each corner particle as touching the middle particle of the eight cells that have it as a common corner.] This means that the length of the diagonal for the cube is four times the radius of each sphere (4r), as illustrated in Figure 10. Using some trigonometry, we can show the length of one side of the cube is:

a rbcc = 4

3

Since a bcc cell has 2 particles in it, the packing efficiency is:

pVV

r

a

r

rbcc

particle

cell bcc

= = =2

2 43

2 43

43

3

3

3⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎠⎟

3

38

0 680 68 0= = . . %or��

For an fcc unit cell, in the tightest packing of equal-sized sphere, the corner particles do not touch each other. Each particle has a coordination number of 12. Consider the fcc


unit cell shown in Figure 11. Imagine looking at it from the top. The middle particle, in the face shown, touches the 4 corner particles plus the four particles in the layer below it. It also touches four particles in another layer above it. It is easy to see that the length of the diagonal, for the face of the cube, is four times the radius of the spheres (4r). Using some trigonometry, we can show the length of one side of the cube is:

a rfcc = 8

Figure 10. Diagonal of Cube

Figure 11. Top View of fcc Unit Cellfrom http://www.molsci.ucla.edu/pub/explorations.html#Crystalline Solids



The packing efficiency, since an fcc cell has 4 particles in it, is:

pVV

r

a

r

rfcc

particle

cell fcc

= = = =4

4 43

4 43

8

23

3

3

3

� �⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

( )��

60 7405 74 05= . . %or

▼ Example 23

The unit cell for Cu metal is face-centered cubic. The density of Cu is 8.92 g/mL. Calculate the length of one side of a unit cell of Cu. Calculate the radius of a Cu atom.Answer: 127.8 pm

Here’s a strategy for solving this problem:• Assume any amount of Cu (in grams). Try one mole: 63.55 g• Calculate moles of Cu in the amount assumed. n = 1.000 mol• Calculate the actual number of Cu atoms by multiplying moles

by Avogadro’s Number. N = 6.022 × 1023

• Calculate the number of unit cells given that each unit fcc cell has 4 atoms.

Ncells = N/4 = 6.022 × 1023 / 4 = 1.506 × 1023

• Calculate the volume of the amount of Cu assumed; use the density. V = (63.55 g) / (8.92 g/cm3) = 7.124 cm3

• Divide the volume by the number of unit cells to get the vol-ume of each unit cell.

Vcell = V/Ncells = (7.124 cm3) / (1.506x1023) = 4.730 × 10 −23 cm3

• Take the cube root of the calculated volume of each unit cell to get the length of the side of the cube, afcc.

afcc = (Vcell)1/3 = (4.730 × 10 −23 cm3)1/3 = 3.616 × 10 −8 cm

• Solve for r knowing that

a rfcc = 8

ra cm

cm

m pm

fcc= = × = ×

= × =

−−

−

8

3 616 10

81 278 10

1 278 10 127 8

88

10

( . ).

. .

ra cm

cm

m pm

fcc= = × = ×

= × =

−−

−

8

3 616 10

81 278 10

1 278 10 127 8

88

10

( . ).

. .


TEST YOURSELF


1. The compression factor for a 2.50 L gas sample is 1.20. If the gas were ideal, at the same temperature and pressure, the volume would be:

A. higher than 2.50 L B. smaller than 2.50 L C. the same D. different, but information is insufficient

2. The compression factor for a gas sample is 0.90; its pressure is 20.0 atm. If the gas were ideal, at the same temperature and volume, the pressure would be:

A. higher than 20.0 atm B. lower than 20.0 atm C. the same D. different, but information is insufficient

3. A 2.00 mol sample of gas occupies a volume of 11.0 L at 298K and 5.00 atm. These values suggest that:

A. the gas is behaving ideally B. the gas is not behaving ideally and the deviation is due to predominance of

attractive forces C. the gas is not behaving ideally and the deviation is due to the predominance of

repulsive forces

4. A gas occupies a volume of 22.0 L/mol at STP. Which of the following explains the deviation from ideal behavior?

A. predominance of attractive forces among the molecules B. predominance of repulsive forces among the molecules C. absence of attractive forces among the molecules D. absence of repulsive forces among the molecules

5. At the same temperature and pressure, which of the following would you expect to have the highest compression factor?

A. H2 B. N2 C. H2O



6. Consider the plot below of a ZP isotherm for a substance. Which of the labeled points on this diagram corresponds to a gas? What is the vapor pressure of this substance at this temperature?

7. Which gas would you expect to have a larger van der Waals a value? Explain. A. CH4 B. C2H6

8. Which gas would you expect to have a larger van der Waals b value? Explain. A. F2 B. Cl2

9. Consider the ZP isotherms shown (in red) below for a gas at temperatures T1 and T2 above the critical temperature. Which temperature is higher? Explain.


10. Consider the phase diagram below. The vapor pressure of the substance at 298K is: A. less than 1 atm B. 1 atm C. higher than 1 atm

11. Consider the phase diagram shown below. Match the Descriptions to the Processes listed below.

Descriptions: A. solid melting at constant pressure B. gas liquefying as the system is pressurized at constant temperature C. liquid changing to gas at constant pressure D. solid subliming at constant pressure E. solid subliming as the pressure is lowered at constant temperatureProcesses:____ 1. starts at point A and ends at point C?____ 2. starts at point D and ends at point E?


____ 3. starts at point E and ends at point F?____ 4. starts at point D and ends at point B?

12. Consider the phase diagram shown below.

Which point on the diagram describes conditions where the following phases are in equilibrium?____1. two solid phases only____2. solid, liquid, and gas____3. liquid and gas only____4. two solid phases and gas____5. solid and liquid

13. Explain why the water level in the tube will rise when the temperature of liquid in-creases in the set up shown below.


14. Imagine a solid sample of the substance in an initial state represented by point A in the phase diagram below. The blue line and red lines are not part of the phase diagram; they are intended to indicate the pressure and temperature at various points in the dia-gram. Describe how you would calculate the heat needed to vaporize it to point B at constant pressure. Sketch the heating curve for the entire process.

15. The molar heat capacities of ice, liquid water, and steam are 38.1, 75.3, and 37.5 J K-1 mol-1, respectively. The heat of fusion of water is 6.01 kJ mol-1 at its normal melting point (0.00°C). The heat of vaporization is 40.7 kJ mol-1 at its normal boiling point (100.00°C). Calculate q for the conversion of 25.0 g steam at 110.0°C to ice at -5.00°C.

16. What type of solid is paper? A. crystalline B. amorphous

17. Which type of crystalline solid most likely has a low melting point? A. molecular B. metallic C. ionic D. covalent network

18. Explain why solid C has a higher melting point than solid Ar even though an Ar atom has a larger mass compared to a C atom.

19. Which type of unit cell has 4000 particles in 1000 unit cells? A. simple cubic B. body-centered cubic C. face-centered cubic


20. What type of unit cell has the highest packing efficiency? A. simple cubic B. body-centered cubic C. face-centered cubic

21. The unit cells for Fe is body-centered cubic. The length of the side of the unit cell is 286.65 pm. Calculate the radius of each Fe atom.

22. The volume per mole of atoms in a body-centered cubic crystal is 37.9 cm3/mol. Calculate the radius of each atom.

405

CHAPTER

Redox Reactions

27.1 Reduction and OxidationThe term redox is short for reduction-oxidation. Redox reactions are reactions that involve electron transfer. We can think of a redox reaction as consisting of two half reactions.

• Reduction = gain of electron(s)• Oxidation = loss of electron(s)

In a redox reaction, the reactant that is reduced is called the oxidizing agent while the reactant that is oxidized is the reducing agent.

▼ Example 1

Consider the reaction between copper metal, Cu(s), with an aqueous solution of silver nitrate, AgNO3(aq):

Cu(s) + 2 AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq)

Explain why Cu atoms are said to be oxidized and that Ag+ ions are said to be reduced in this reaction. For this reaction, what is the oxidizing agent? What is the reducing agent?Answer: In this reaction, a neutral Cu atom in Cu(s) ends up as a copper(II) ion (Cu2+) in Cu(NO3)2. How does a neutral atom become positively charged? It is by losing electron(s). Loss of electrons is oxidation. On the other hand, a silver(I) ion (Ag+) in AgNO3 ends up as a neutral Ag atom in Ag(s). A positive ion becomes a neutral atom by gaining electron(s). Gain of electrons is reduction. Since Cu atoms in copper metal are oxidized, we say that the copper metal, Cu(s), is the reducing agent. Since Ag+ ions in AgNO3(aq) are reduced, we say that the silver nitrate solution is the oxidizing agent.

Since oxidation involves a loss of electron, we can think of electrons as “products” of an oxidation half reaction. Similarly, we can think of electrons as “reactants” of a reduction half

27


reaction. Therefore, if we were to write a chemical equation for just a half reaction, we would specify the number of electrons involved on the right-hand-side if the half reaction is oxidation; electrons are written on the left-hand-side if the half reaction is reduction. We represent electrons by writing e−.

▼ Example 2

Write half reactions for: Cu(s) + 2 AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq)

Answer: The oxidation half reaction is:

Cu(s) → Cu2+(aq) + 2e−

and the reduction half reaction is:Ag+(aq) + e− → Ag(s).

Since each Cu atom loses two electrons, as explained in the previous example, we indicate 2e− on the right-hand-side of the oxidation half reaction. Similarly, the conversion of Ag+ to Ag involves a gain of only one electron. Therefore, we write e− on the left-hand-side of the reduction half reaction.

27.2 Recognizing Redox Reactions27.2.1 Reactions Involving Elemental SubstancesSome reactions can be easily recognized as redox reactions. These are reactions that involve an elemental substance as either a reactant or a product of the reaction. Note, however, that not every redox reaction involves an elemental substance.

▼ Example 3

Which of the following chemical reactions is not a redox reaction?A. 2 Mg(s) + O2(g) → 2 MgO(s)B. 2 HgO(s) → 2 Hg(l) + O2(g)C. Zn(s) + HCl(aq) → ZnCl2(aq) + H2(g)D. Ba(OH)2(aq) + H2SO4(aq) → BaSO4 (s) + 2 H2O(l)E. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)

Answer: DThe reactions in choices A, B, C, and D can be easily identified as redox reactions because they involve at least one elemental substance as either reactant or product. Choice A is a synthesis reaction that involves two elemental substances as reactants,

Chapter 27: Redox Reactions 407

Mg and O2. Choice B is a decomposition reaction that involves two elemental substances as products, Hg and O2. Choice C is a displacement reaction that involves two elemental substances, Zn (as a reactant) and H2 (as a product). Choice E is a combustion reaction; it involves one elemental substance as a reactant, O2.At this point in the chapter, we have not discussed how to deter-mine whether or not the reaction in choice D is a redox reaction. However, by process of elimination, we conclude that the reaction in choice D is not a redox reaction.

▼ Example 4

Which reaction is a redox reaction?A. Ca2+(aq) + 2 F−(aq) → CaF2(s)B. Ca(s) + F2(g) → CaF2(s)C. bothD. none of the above

Answer: BThe reaction in choice B is a synthesis reaction involving two elemental substances.Choice A is a net ionic equation for a precipitation reaction. Ca2+ and F− do not represent an elemental substance; they come from ionic compounds. We know that calcium ions are always found in nature to have a charge of +2. Its charge in CaF2 is +2; there-fore, the Ca2+ ion shown on the reactant side has not gained or lost any electron; it is still Ca2+ in CaF2. Similarly, the F in CaF2 represents fluoride ions (F−); the F− ions shown on the reactant side did not lose or gain any electrons.

27.2.2 Using Oxidation NumbersConsider the combustion of carbon:

C(s) + O2(g) → CO2(g)

This reaction does not involve any ions, but is considered a redox reaction. Why? Even though C and O atoms are sharing electrons in CO2, the sharing is not equal. Shared elec-trons are, on average, actually spending more time closer to the O atom. We say that the covalent bond between C and O is polar, or that the electron density is polarized towards the O atom. We can say there is, in fact, a transfer of electrons from C to O, albeit a partial one. Therefore, in this reaction, we say that C atoms lost electrons (oxidized), and that the O atoms gained electrons.


Keeping track of electrons lost and gained is useful for identifying and balancing redox reactions, but dealing with partial charges would be quite challenging and cumber-some. Therefore, as a matter of convenience, the concept of oxidation number (or oxida-tion state) was invented. An atom is said to be oxidized if its oxidation number increases, and reduced if its oxidation number decreases.

The oxidation number of a monatomic species is its charge. A monatomic species is either a neutral atom (with oxidation number defined as zero), or a monatomic ion. For atoms involved in covalent bonding (that is, in a molecule or polyatomic ion), charge is not defined; the entire species has a charge, but the atoms that make it up do not have a charge. In these cases, the oxidation number is defined as fictitious charge assigned to each atom assuming that shared electrons “belong” to the more electroneg-ative atom. Based on this, we can assign oxidation numbers, in most cases, by following these rules:

Rule 1. The oxidation number of atoms in an elemental substance is zero.

▼ Example 5

What is the oxidation number of F atoms in F2?

Answer: In an F2 molecule, neither of the two F atoms is expected to attract shared electrons more strongly than the other; we expect equal sharing of electrons and neither atom can be thought of as having gained or lost an electron. The oxidation number of the fluorine atoms in the F2 molecule is zero.

Rule 2. In a compound,• the oxidation number of a monatomic ion is equal to its charge• the oxidation number of the most electronegative atom in a molecule or polyatomic

ion is the charge it would have if it were to gain enough electrons to be isoelectronic with the closest noble gas; in other words:• −1 if it belongs to group VIIA (F, Cl, Br, I) or if it is H• −2 if it belongs to group VIA (O, S, Se)• −3 if it belongs to group VA (N, P)• −4 if it belongs to group IVA (C)

• the sum of oxidation numbers of all atoms in a molecule is zero• the sum of oxidation numbers of all atoms in a polyatomic ion is equal to the

charge of the ion


▼ Example 6

What is the oxidation number of chloride ion?Answer: −1Chloride ion, Cl−, which is found in ionic compounds like NaCl and CaCl2, is a monatomic ion with a charge of −1. Therefore, the oxidation number is −1, the charge of monatomic ion.

▼ Example 7

What are the oxidation numbers of Na and O in Na2O?Answer: +1 and −2In Na2O, sodium exists as monatomic Na+ ions, while oxygen exists as oxide ions, O2−. Therefore, the oxidation numbers of Na and O in Na2O are +1 and −2.

▼ Example 8

What are the oxidation numbers of B and H in BH3?Answer: +3 and −1H is more electronegative than B. Therefore, we assign it an oxidation number of −1; it would acquire a −1 charge if it becomes isoelectronic with the nearest noble gas (He). If x is the oxidation number of B, then:

x + 3(−1) = 0 since the oxidation numbers of atoms in a molecule add up to zero. Solving for x, we get x = +3.The oxidation of B in BH3 is +3.

▼ Example 9

What are the oxidation numbers of O and F in OF2?Answer: +2 and −1.The oxidation number of F in any compound is −1 since it is the most electronegative atom. If the oxidation number of O is x, then:

x + 2(−1) = 0since the sum of all oxidation numbers must be equal to zero. Solving for x, we get x = +2.


▼ Example 10

What are the oxidation numbers of K, N, and O in KNO3?Answer: +1, +5, −2KNO3 is made up of potassium (K+) and nitrate (NO3

−) ions. Therefore, the oxidation number of K is +1.O is more electronegative than N; therefore, in nitrate ion, its oxidation number is −2. If the oxidation number of N is x, then:

x + 3(−2) = −1since the oxidation numbers of N and the three O atoms should add up to the charge of nitrate, which is −1. Solving for x, we get x = +5.

▼ Example 11

In a chemical reaction, the Cl in HCl molecules end up as Cl2 molecules. Is Cl oxidized or reduced?Answer: oxidizedThe oxidation number of Cl in an HCl molecule is −1 (since Cl is more electronegative than H). In Cl2 (an elemental substance), the oxidation number of Cl is 0. Therefore, Cl is oxidized; its oxidation number increases from −1 to 0.

Rule 2 should allow us to assign oxidation numbers to atoms in most compounds. However, there are exceptions. Sometimes, the best way to determine oxidation numbers is to draw the structure of the molecule or polyatomic ion. Oxidation numbers are deter-mined in a manner similar to formal charges. The difference is that in determining formal charges, atoms sharing electrons are assigned equal ownership of the shared electrons regardless of electronegativities. For oxidation numbers, ownership of shared electrons is assigned to the more electronegative atom.

▼ Example 12

What are the oxidation numbers of the atoms in CH3CH2OH? The Lewis structure of CH3CH2OH is shown below.


Answer: first C (−3), second C (−1), O (−2), each H (+1)Consider the C atom on the left. C is more electronegative than H. Therefore, it is assigned ownership of the six electrons shared with the hydrogen atoms. The two C atoms have the same elec-tronegativity, so each is assigned ownership of one of the two electrons they are sharing. Therefore, the C atom on the left appears to own 7 valence electrons: the 6 shared with the H atoms, plus one of the two shared with the other C. These are enclosed by the red box in the figure below:

Since a neutral C atom has 4 valence electrons, the C atom on the left appears to have gained 3 electrons. Its oxidation number is, therefore, −3.The C atom on the right appears to own 5 valence electrons, shown enclosed by the blue box in the figure above: the 4 shared with the H atoms, plus one shared with the other C; it does not own any of the electrons shared with O since O is more electro-negative. Therefore, compared to a neutral C atom (which has 4 valence electrons), the C atom on the right appears to have gained one electron; its oxidation number is −1.The O atom, being more electronegative than C and H, is assigned ownership of all eight valence electrons surrounding it. A neutral O atom has 6 valence electrons. Therefore, the O atom in this structure appears to have gained two electrons. Its oxida-tion number is −2.Since H is less electronegative than C and O, none of the H atoms are assigned ownership of any electron. Since a neutral H atom has one electron, each H atom in this structure appears to have lost one electron. Each H atom in this structure has an oxidation number of +1.We can see that, in general, a positive oxidation number means that the atom appears to have fewer electrons than a neutral atom; an increase in oxidation number means that an atom appears to have lost electrons.


Note that the C atom on the right has a higher oxidation number (−1) compared to the one of the left (−3). The more electrons an atom shares with a more electronegative atom, the higher its oxidation number is. We expect the C atom on the right to have a higher oxidation number (less negative in this case) because it is sharing electrons with O, which is more electronegative.

▼ Example 13

What is the oxidation number of O in Na2O2?Answer: −1Sodium is always found in compounds as a cation with a charge of +1 (Na+). Therefore, its oxidation number in Na2O2 is +1. Two sodium ions in Na2O2 suggest that the total positive charge in one formula unit of the compound is 2(+1), or +2. This suggests that the total anion charge in one formula unit of the compound is −2. Therefore, one formula unit of the compound cannot have two oxide ions; two oxide ions will have a total charge of −4. The anion must be a peroxide ion, O2

2−. The Lewis structure of peroxide is shown below.

Each O atom in the structure can be thought of as owning 7 valence electrons --- six in the three lone pairs, plus one of the two electrons shared in the single bond. Since a neutral O atom has 6 valence electrons, each O atom in peroxide can be thought of having gained one electron; the oxidation number of each O atom in peroxide ion is, therefore, −1.In this particular case, we actually do not need to know the Lewis structure of peroxide to figure out that the oxidation number of each O is −1. Since neither of the two O atoms is expected to attract electrons more strongly than the other, we expect equal sharing of the two additional electrons by the two O atoms. Thus, we can imagine each O atom as having gained one electron.


27.3 Balancing Redox ReactionsChemical equations for redox reactions can be difficult to balance by inspection if more than four substances are involved. Using algebra can make the task easier.

▼ Example 14

Balance the following reaction using algebra:HCl + KMnO4 → Cl2 + MnCl2 + KCl + H2O

Answer:We start by assigning letters to the coefficients:

a HCl + b KMnO4 → c Cl2 + d MnCl2 + e KCl + f H2Othen set up algebraic equations that must be satisfied if the equa-tion is balanced:• Eq. 1, to balance H: a = 2f• Eq. 2, to balance Cl: a = 2c + 2d + e• Eq. 3, to balance K: b = e• Eq. 4, to balance Mn: b = d• Eq. 5, to balance O: 4b = f

We then assign any number to one of the coefficients, and figure out the others by making a series of simple substitutions. For example, we can let a=1. Then:• Substitute a = 1 into Eq. 1: 1 = 2f; therefore: f = 1/2• Substitute f = 1/2 into Eq. 5: 4b = 1/2; therefore: b = 1/8• Substitute b = 1/8 into Eq. 3: 1/8 = e; therefore: e = 1/8• Substitute b = 1/8 into Eq. 4: 1/8 = d; therefore: d = 1/8• Substitute a = 1, d = 1/8, and e = 1/8 into Eq. 2:

1 = 2c + 2(1/8) + 1/8; therefore: c = 5/16Thus,

a = 1, b = 1/8, c = 5/16, d = 1/8, e = 1/8, f = 1/2or, if we want the smallest set of whole number coefficients, we just multiply all the coefficients by 16 (the least common multiple of the denominators) to get:

a = 16, b = 2, c = 5, d = 2, e = 2, and f = 8.Our balanced equation is:

16 HCl + 2 KMnO4 → 5 Cl2 + 2 MnCl2 + 2 KCl + 8 H2O

In general, we can simplify the balancing of redox reactions if we add an additional constraint to our coefficients. This additional constraint is based on the fact that, in a redox reaction, electrons lost must be equal to electrons gained.


▼ Example 15

Balance the following redox reaction by taking into account that electrons lost equals electrons gained.

H2S + HNO3 → H2SO4 + NO2 + H2O

Answer:We start by assigning letters to the coefficients:

a H2S + b HNO3 → c H2SO4 + d NO2 + e H2Othen set up algebraic equations that must be satisfied if the equa-tion is balanced:• Eq. 1, to balance H, Eq. 1: 2a + b = 2c + 2e• Eq. 2, to balance S, Eq. 2: a = c• Eq. 3, to balance N, Eq. 3: b = d• Eq. 4, to balance O, Eq. 4: 3b = 4c + 2d + e

If we examine the equations, we find that setting a value for e will not easily allow us to figure out the others. Setting a value for a, b, c, or d, will allow us to easily figure out one other coefficient, but we will be left with 3 simultaneous equations in 3 unknowns. We can make it easier by adding the constraint that electrons lost must be equal to electrons gained. We do this by examining changes in oxidation numbers.By assigning oxidation numbers, we find that N is reduced and S is oxidized.• The oxidation number of N is +5 in HNO3 and its oxidation

number in NO2 is +4. A decrease in oxidation number by 1 (from +5 to +4) means a gain of 1 electron for each N. Therefore, we can say that: 1 electron is gained for every NO2 produced.

• The oxidation number of S in H2S is −2, and it is +6 in H2SO4. An increase in oxidation number by eight (from −2 to +6) means a loss of 8 electrons for each S. Therefore, we say that 8 electrons are lost for every H2SO4 produced.

Applying the constraint that electrons lost must equal electrons gained, we can say that the coefficients of NO2 and H2SO4 must be in an 8-to-1 ratio. Why?• (8 molecules of NO2) times (1 electron gained per NO2) =

8 electrons gained• (1 molecule of H2SO4) times (8 electrons lost per H2SO4) =

8 electrons lost


Thus, we can simplify our problem by letting:• coefficient of NO2 = d = 8• coefficient of H2SO4 = c = 1

Substituting values of c and d into Eqs. 1–4,• Eq. 1 becomes: 2a + b = 2(1) + 2e, [Eq. 5]• Eq. 2 becomes: a = 1• Eq. 3 becomes: b = 8• Eq. 4 becomes: 3b = 4(1) + 2(8) + e, [Eq. 6]

We can see right away from Eqs. 2 and 3 that a = 1 and b = 8. This leaves just one unknown to determine (e); we can solve for it using Eq. 5 or Eq. 6.• Using Eq. 5: 2(1) + 8 = 2(1) + 2e, we find that e = 4• Using Eq. 6: 3(8) = 4(1) + 2(8) + e, we also find that e = 4

The balanced equation is:H2S + 8 HNO3 → H2SO4 + 8 NO2 + 4 H2O

The strategy described here will work as long as there is only one reduction product, only one oxidation product, and the reduction and oxidation products are not the same. If there is one and the same reduction and oxidation product, try applying the electron-balance constraint on the coefficients of the reactants. Applying the electron-balance constraint to the reactants may not always work; for example, consider the reaction:

HCl + KMnO4 → Cl2 + MnCl2 + KCl + H2OIn this particular case, not all the Cl in HCl ended up in Cl2; some ended up in KCl and MnCl2 and are, thus, not being oxidized (same oxidation number, −1).

For redox reactions in aqueous solutions, it is preferable to write net ionic equations. When writing ionic equations, we write formulas of strong electrolytes in aqueous solu-tion as separate ions. Then we can cancel spectator ions to get the net ionic equation.

▼ Example 16

Write the net ionic equation for the redox reaction:H2S(aq) + 8 HNO3 (aq) → H2SO4 (aq) + 8 NO2 (g) + 4 H2O(l)

Answer: H2S(aq) + 6 H+(aq) + 8 NO3− (aq)

→ SO42−(aq) + 8 NO2 (g) + 4 H2O(l)


The strong electrolytes in aqueous solution in this reaction are HNO3 and H2SO4, which are two of the six known strong acids. Therefore, we rewrite the equation as:

H2S(aq) + 8 H+(aq) + 8 NO3− (aq)

→ 2 H+(aq) + SO42−(aq) + 8 NO2 (g) + 4 H2O(l)

to get the full ionic equation. We can see that 2 H+(aq) is common to both sides, so the net ionic equation is:

H2S(aq) + 6 H+(aq) + 8 NO3− (aq)

→ SO42−(aq) + 8 NO2 (g) + 4 H2O(l)

▼ Example 17

Write the net ionic equation for the redox reaction:16 HCl(aq) + 2 KMnO4(s)

→ 5 Cl2(g) + 2 MnCl2 (aq) + 2 KCl(aq) + 8 H2O(l)Answer:

16 H+(aq) + 10 Cl−(aq) + 2 KMnO4(s) → 5 Cl2(g) + 2 Mn2+(aq) + 2 K+(aq) + 8 H2O(l)

The strong electrolytes in aqueous solution in this reaction are HCl, MnCl2 and KCl. We write these formulas as separate ions. Note that KMnO4 is a strong electrolyte but the chemical equation suggests that solid KMnO4, not an aqueous solution of KMnO4, is being used as a reactant. Therefore, we do not write KMnO4 as separate ions. The full ionic equation is:

16 H+(aq) + 16 Cl−(aq) + 2 KMnO4(s)→ 5 Cl2(g) + 2 Mn2+(aq) + 4Cl−(aq) + 2 K+(aq)

+ 2 Cl−(aq) + 8 H2O(l)and the net ionic equation is obtained by canceling the spectator ions. 6 of the 16 Cl− ions are spectator ions; only 10 are actually involved in the reaction.

▼ Example 18

Write the net ionic equation for the redox reaction:16 HCl(aq) + 2 KMnO4(aq)

→ 5 Cl2(g) + 2 MnCl2(aq) + 2 KCl(aq) + 8 H2O(l)Answer:

16 H+(aq) + 10 Cl−(aq) + 2 MnO4−(aq)

→ 5 Cl2(g) + 2 Mn2+(aq) + 8 H2O(l)


This is similar to the preceding example, except that KMnO4 is in aqueous solution. Since KMnO4 is an ionic compound, it is completely dissociated in aqueous solution; we need to write it as separate ions. The net result is that K+ ions need to be cancelled out as spectator ions.

Writing the net ionic equation is easy if you already have the balanced molecular equation as in the preceding examples. However, if you have to start from an unbal-anced equation, it is easier to directly obtain the net ionic equation using the ion-electron method. In this method, we identify the ions and molecules that contain the atoms that are actually oxidized or reduced and use those to construct half reactions. We balance the half reactions and combine them so that electrons lost equals electrons gained. To balance a half reaction in acidic solution we:

• balance all atoms besides H and O,• balance O by adding H2O to the side that needs it,• balance H by adding H+ to the side that needs it, and• balance charges by adding electrons to the more positive (or less negative) side.

▼ Example 19

Write the balanced half reaction for the conversion of permanga-nate ion (MnO4

−) to Mn2+ in acidic solution. Is this a reduction or oxidation half reaction?

Answer: 5e− + 8 H+ + MnO4− → Mn2+ + 4 H2O; this is a reduction

half reactionBalance all atoms besides H and O:

MnO4− → Mn2+ (Mn is already balanced)

Balance O: (need 4 more O atoms on the right side, so add 4 H20 to the right side):

MnO4− → Mn2+ + 4 H2O

Balance H: (need 8 H atoms on the left side, so add 8 H+ to the left side):

8 H+ + Mn04− → Mn2+ + 4 H2O

Balance charges:• Total charge on the left = 8(+1) + 1(−1) = +7• Total charge on the right = 1(+2) + 4(0) = +2


• Add electrons to the left side, which is the more positive side. Add 5 electrons to the left side so that the total charge on that side will equal the total charge on the right side:

5e− + 8 H+ + MnO4− → Mn2+ + 4 H2O

Now, the total charge on the left with 5 electrons included is also +2:

8(+1) + 1(−1) + 5(−1) = +2Note that the number of electrons needed to balance charges is equal to the change in the oxidation number; the oxidation number of Mn is +7 in MnO4

−, +2 in Mn2+.Since electrons appear on the left side, we say that this process involves a gain of electrons; the half reaction is a reduction.

When balancing half reactions in basic solution, we do one more step, replace H+ by H2O, then add as many OH− ions to the other side. Then, cancel common terms.

▼ Example 20

Write the corresponding net ionic equation if the following half reaction (balanced in acidic solution) were to occur in basic solution:2 H2O(l) + SO2(g) → SO4

2−(aq) + 4 H+(aq) + 2e−

Answer: 4 OH−(aq) + SO2(g) → SO42−(aq) + 2 H2O(l) + 2e−

Starting with:2 H2O(l) + SO2(g) → SO4

2−(aq) + 4 H+(aq) + 2e−

Replace 4 H+(aq) by 4 H2O(l), then add 4 OH−(aq) to the other side:

4 OH−(aq) + 2 H2O(l) + SO2(g) → SO42−(aq) + 4 H2O(l) + 2e−

Simplify by canceling out common terms: subtract 2 H2O from both sides.

4 OH−(aq) + SO2(g) → SO42−(aq) + 2 H2O(l) + 2e−

An alternative procedure for balancing half reactions in basic solutions is to balance H and O by doing the following:

• to the side with extra H, add OH−(aq), then add H2O(l) to the other side• to the side with extra O, add H2O(l), then add 2 OH−(aq) to the other side


▼ Example 21

Balance the following half reaction in basic solution: MnO4

−(aq) → MnO2(s)Answer:• Balance all atoms besides H and O: Mn balanced as is• Balance H: not necessary (no H on either side)• Balance O: add 2 H2O(l) to the left side since it has 2 more O

compared to the right, then add 4 OH−(aq) to the other side2 H2O(l) + MnO4

−(aq) → MnO2(s) + 4 OH−(aq)• Balance charges by adding electrons to the less negative side

(3e− to the left side):3e− + 2 H2O(l) + MnO4

−(aq) → MnO2(s) + 4 OH−(aq)

Once we have balanced half reactions, getting the overall net ionic equation is a simple matter of combining the two half reactions so that electrons lost equals electrons gained.

▼ Example 22

What is the overall net ionic equation for these two half reactions:

Cu(s) → Cu2+(aq) + 2e−

Ag+(aq) + e− → Ag(s)

A. Cu(s) + Ag+(aq) → Cu2+(aq) + Ag(s)B. 2 Cu(s) + Ag+(aq) → 2 Cu2+(aq) + Ag(s)C. Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s)D. all of the above are correct

Answer: CAlthough atoms are balanced in each of the three equations shown, the charges are balanced only in the equation given in choice C. If charges are not balanced, as in choices A and B, it means that electrons gained and electrons lost are not equal. In choice A, for example, the total charge on the left is +1 (due to one Ag+) and the total charge on the right is +2 (due to one Cu2+).To construct the overall equation from half reactions, we must make sure electrons lost equals electrons gained.• The oxidation of Cu involves a loss of 2 electrons.• The reduction of Ag+ involves a gain of only 1 electron.


This means that the reduction must occur twice every time the oxidation occurs, so that electrons lost equals electrons gained. We multiply the reduction half reaction by two:

2 Ag+(aq) + 2e− → 2 Ag(s)before we combine it with the oxidation half reaction.

▼ Example 23

Use the ion-electron method to determine the net ionic equation for the following reaction:HCl(aq) + KMnO4 (aq) → Cl2(g) + MnCl2(g) + KCl(aq) + H2O(l)

Answer:By assigning oxidation numbers, we find that Mn is reduced. Its oxidation number is +7 in KMnO4 and +2 in MnCl2. Since both KMnO4 and MnCl2 are ionic compounds, we expect them to be completely dissociated in aqueous solution. So the (unbalanced) half reaction for reduction will only have the ions that actually contain Mn.

MnO4− → Mn2+

In an earlier example, we saw that in acidic solution, the balanced half reaction is:

5e− + 8 H+ + MnO4− → Mn2+ + 4 H2O

Similarly, we find that Cl is oxidized. Its oxidation number in HCl is −1 and its oxidation number in Cl2 is 0. HCl is a strong acid and it is completely dissociated in aqueous solution. Cl2, on the other hand, is a molecular substance and in the gas phase. Thus, the (unbalanced) half reaction for oxidation is:

Cl− → Cl2To balance this half reaction, we balance Cl and add electrons to balance charges; no H or O atoms are involved in this half reac-tion. Our balanced oxidation half reaction is:

2 Cl− → Cl2 + 2e−

Finally, we combine the two half reactions, keeping in mind that electrons lost must equal electrons gained. Since the reduction half reaction involves a gain of 5 electrons, and the oxidation half reaction involves a loss of 2 electrons, we can balance electrons lost and gained by multiplying the reduction half reaction by 2 and


multiplying the oxidation half reaction by 5. The reduction half reaction, times 2:

2 (5e− + 8 H+ + MnO4− → Mn2+ + 4 H2O)

becomes:10 e− + 16 H+ + 2 MnO4

− → 2 Mn2+ + 8 H2Oand the oxidation half reaction, times 5:

5 (2 Cl− → Cl2 + 2e− )becomes:

10 Cl− → 5 Cl2 + 10 e−

Putting the two half reactions together, we get:16 H+(aq) + 10 Cl−(aq) + 2 MnO4

−(aq) → 5 Cl2(g) + 2 Mn2+(aq) + 8 H2O(l)

Note that the 10e− term, which appears on opposite sides in the half reactions, disappears in the net ionic equation.


TEST YOURSELF


1. In a reaction where magnesium metal reacts to form a compound, magnesium is said to be…

A. oxidized B. reduced C. the oxidizing agent

2. When Cl2 (g) reacts with a metallic element to form an ionic compound, we say that chlorine is:

A. oxidized B. reduced C. the reducing agent

3. When Br2(l) is produced from the reaction involving a bromine-containing binary ionic compound, the half reaction leading to the formation of Br2 is classified as:

A. oxidation B. reduction C. oxidation or reduction depending on what the compound is

4. Which of the following chemical reactions is not a redox reaction? A. C(s) + O2(g) → CO2(g) B. Cl2(g) + 2 KBr(aq) → 2 KCl(aq) + Br2(l) C. 2 KClO3(s) → 2 KCl(s) + 3 O2(g) D. NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

5. Which of the following chemical reactions is a redox reaction? A. NH3(g) + HCl (g) → NH4Cl(s) B. CaCO3(s) → CaO(s) + CO2(g) C. 2 CaO(s) → 2 Ca(s) + O2(g) D. Ag+(aq) + Cl−(aq) → AgCl(s)

6. Which of the following is false? A. The charge of Hg in mercury(I) is +1 B. The oxidation number of Hg in mercury(I) is +1 C. The charge of Hg in mercury(II) is +2 D. The oxidation number of Hg in mercury(II) is +2



7. For which of the following is the oxidation number of F equal to zero? A. CaF2 B. CF4 C. NaF D. F2

8. Which of the following has the largest oxidation number for chlorine?. A. MgCl2 B. CCl4 C. Cl2 D. ClO2

−

9. Which of the following is false? A. The charge of oxygen in carbon monoxide is −2 B. The oxidation number of oxygen in carbon monoxide is 2 C. The charge of oxygen in magnesium oxide is 2 D. The oxidation number of oxygen in magnesium oxide is −2

10. What is reduced in the following reaction? Cu(s) + HNO3(aq) → NO(g) + Cu(NO3)2(aq) + H2O(l)

. A. Cu B. H C. N D. O

11. Which of the three structures below has the C atom with the highest oxidation number?

12. When the following chemical equation is balanced, what is the ratio of the coefficients of NO and Cu(NO3)2?

Cu(s) + HNO3(aq) → NO(g) + Cu(NO3)2(aq) + H2O(l) A. 1:1 B. 1:2 C. 2:1 D. 2:3 E. 3:2

13. When the following chemical equation is balanced, what is the ratio of the coefficients of MnCl2 and Cl2?

KMnO4 (s) + HCl (aq) → MnCl2(aq) + Cl2(g) + KCl(aq) + H2O(l) A. 1:2 B. 2:3 C. 2:4 D. 1:5 E. 2:5


14. Consider the half-reaction (unbalanced) in acidic solution: HClO2(aq) → Cl2(g) When this half reaction is balanced using the smallest set of whole number coeffi-

cients, what is the coefficient of H+(aq) and how many electrons are involved? A. 2, 3 B. 4, 7 C. 6, 6 D. 8, 3

15. Consider the half-reaction (unbalanced) in basic solution: ClO2

−(aq) → Cl2(g) When this half reaction is balanced using the smallest set of whole number coeffi-

cients, what is the coefficient of OH−(aq) and how many electrons are involved? A. 2, 3 B. 4, 7 C. 8, 6 D. 8, 3

16. When the following half reaction: H2SO3(aq) → SO4

2−

is balanced in acidic solution using the smallest of whole numbers, what is the coef-ficient of H2O and how many electrons are involved?

A. 1, 2 B. 1, 3 C. 2, 3 D. 3, 1

17. Which of the following is a reduction half reaction? A. SO2(g) → SO3

2−(aq) B. SO4

2−(aq) → SO32−(aq)

C. S2−(aq) → SO32−(aq)

18. Consider the conversion of ethanol (the intoxicating compound in wine) to acetic acid (the sour tasting compound in vinegar):

C2H6O(aq) → C2H4O2(aq) This half reaction is classified as: A. oxidation B. reduction C. neither

19. Consider the following balanced half reactions: Al(s) → Al3+(aq) + 3e−

Ag+(aq) + e− → Ag(s) For the overall reaction, in the balanced equation with the smallest whole number

coefficients, the coefficient for Ag(s) is… A. 1 B. 2 C. 3 D. 4


20. Consider the following balanced half reactions: 14e− + 16 H+(aq) + 2 ClO4

−(aq) → Cl2(aq) + 8 H2O(aq) 2H2O(l) + SO2(g) → SO4

2−(aq) + 4 H+(aq) + 2e−

For the overall reaction, in the balanced equation with the smallest set of whole number coefficients, the coefficient for H2O(l) is…

A. 2 B. 6 C. 8 D. 14

21. Which of these is not balanced? A. Zn(s) + Ag+(aq) → Zn2+(aq) + Ag(s) B. Zn(s) + 2 Ag+(aq) → Zn2+(aq) + 2 Ag(s) C. 3 Zn(s) + 6 Ag+(aq) → 3 Zn2+(aq) + 6 Ag(s) D. None of the above

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427

CHAPTER

Solutions

Several times during the course of a day, we use or make solutions. We wake up to a coffee solution consisting of solids (sugar and coffee) dissolved in a liquid (water). When we drive our car, it consumes a solution of gasoline (a solution of several liquid hydrocarbons). The soda we drink is a solution of a gas, carbon dioxide, in a liquid, water. The majority of chemical reactions occur in solution. Solutions have particular properties that are useful such as solubility, boiling point, melting point, etc. The most familiar solutions are those in the liquid state, especially ones in which water is the solvent. Aqueous solutions are the most commonly used in chemistry.

A solution is a homogeneous mixture of two or more substances or components. Solutions may exist in any of the three states of matter (gases, liquids, solids). The solute is the substance being dissolved. It is the gas or solid being dissolved in a liquid; if it is of the same state, it is the component of lesser amount. The solvent is the substance doing the dissolving. In the case of a gas or solid being dissolved in a liquid, the solvent is the liquid; if it is of the same state, it is the component of greater amount. For example, in a cup of coffee, coffee and sugar are the solute and water is the solvent.

28.1 SolubilitySolubility is the amount of solute that can dissolve in a given amount of solvent at given conditions. There is a limit as to how much of a given solute will dissolve at a given tem-perature. When we say something is soluble, it does not mean that it has an infinite solubil-ity, nor does it mean when we say something is insoluble, that it has zero solubility. There is a specific amount of a particular solute that will dissolve in a particular solvent and is referred to as the saturation point. A saturated solution is one holding as much solute as is allowed at a stated temperature; one that is in equilibrium with respect to a given dissolved substance. An unsaturated solution is a solution that is not in equilibrium with respect to a given dissolved substance and in which more of the substance can be dissolved. For example, the solubility of NaCl in water is 36 g NaCl/100 mL H2O at 20°C. If we add less than 36 g of NaCl to 100 mL of H2O, the solution would be unsaturated meaning all solute would dissolve. When we reach 36 g of NaCl added, we would have a saturated solution.

28


Any amount of NaCl added above 36 g would be undissolved and an equilibrium would be established at the given temperature. If the temperature is changed, the equilibrium would shift either to more solute dissolving or less until equilibrium is re-established at the new temperature. The solubility of most solids increase as the temperature is increased.

Cooling a saturated solution usually causes a solid to crystallize out of solution since the solubility is less at the lower temperature. Sometimes this doesn’t happen, and the excess solute remains in the solution. When this occurs, we refer to the solution as a supersaturated solution. A supersaturated solution is one that contains more solute than is possible at a given temperature. Supersaturated solutions are very unstable and the slightest disturbance will cause the excess solute to crystallize out.

The extent to which a solute dissolves in a particular solvent depends upon several factors. The most important of these are the nature of solvent and solute particles and the interactions between them, the temperature at which the solution is formed, and the pressure of a gaseous solute.

28.1.1 Solute-Solvent InteractionsFluids that dissolve in each other in all proportions are said to be miscible fluids. Typically when discussing solubility the phrase “like dissolves like” is given. A more appropriate way of expressing this is to state that two substances with intermolecular forces of about the same type and magnitude are likely to be very soluble in one another. For example, ethanol is completely miscible in water. Both substances are polar with intermolecular forces of similar magnitude. An ethanol molecule experiences little or no change in intermolecu-lar forces when it goes into solution in water. Basically, this means polar solvents dissolve polar solutes (or ionic substances since ionic compounds are the extreme in polarity) and nonpolar solvents dissolve nonpolar solutes. The relative force of attraction of the solute for the solvent is a major factor in their solubility. The dipole-dipole interactions of polar solutes with polar solvents can be easily explained as electrostatic attractions (δ−, δ+) between the dipoles of the molecules. When water is the solvent, the attraction of ions to the water molecules is referred to as hydration.

If two fluids do not mix, the fluids are said to be immiscible. For example, oil and water do not mix and are immiscible in each other. There will be two distinct and separable layers formed with the less dense species (oil in this case) on top. For the oil to dissolve in water, the oil would have to break the hydrogen bonding holding the water molecules together; however, there is no comparable attractive force between the oil and water to supply the required energy to overcome the attraction between H2O molecules.

28.1.2 Effects of Temperature on SolubilityThe solubility of solutes is temperature dependent. As temperature increases, the solubility of gases dissolved in liquids decreases. The opposite effect usually occurs for most solids

Chapter 28: Solutions 429

dissolved in liquids. As temperature increases, the solubility of solids dissolved in liquids increases. Basically, an increase in temperature always shifts the position of an equilibrium towards the endothermic process.

Usually dissolving a solid in a liquid is an endothermic process because heat must be absorbed to break down the crystal lattice.

solid + liquid solution ΔH > 0 (endothermic)Any additional heat would shift the equilibrium to the right (favor forward

endothermic reaction) causing more solid to dissolve. For example, more sugar dissolves in hot coffee than cold coffee.

The process of a gas condensing to a liquid is always an exothermic process.gas + liquid solution ΔH < 0 (exothermic)

Any additional heat would shift the equilibrium to the left (favor reverse endothermic reaction) causing less gas to dissolve. For example, a warm beer goes flat faster than a cold beer.

28.1.3 Effects of Pressure on SolubilityPressure also has effects on solubility. The higher the pressure of a gas above the liquid, the more soluble the gas is in the liquid. This relation is referred to as Henry’s Law which states that the solubility of a gas is directly proportional to the partial pressure of the gas in direct contact with the liquid. Henry’s Law is expressed as follows:

S k Pg H g=

where Sg is the solubility of the gas, kH is a proportionality constant (Henry’s Law constant) that depends on the specific solute and solvent at a particular temperature, and Pg is the partial pressure of the gas. For example, the fizz that occurs when a soda can is opened results from reduced pressure of carbon dioxide over the liquid. At lower pressure, the carbon dioxide bubbles out of solution due to lower solubility.

28.2 Solution Concentration ExpressionsThere are several different methods of expressing relative amounts of solute to the amount of solvent or solution. The quantity of solute, solvent, or solution can be expressed in mass, volume, or moles. The common ways to express concentration are molarity, molality, mass percent, and mole fraction.

28.2.1 MolarityThe molarity (M) of a solution is the moles of solute in a liter of solution.

Molarity Mmoles of solute

liters of solution( ) =

→←

→←


A solution can be prepared to a specific molarity by weighing out the mass of the solute and dissolving in enough solvent to obtain the needed volume of solution. Note that molarity is moles of solute per liter of solution (volume of solute and solvent). Typically, solutions are made by diluting with the solvent to the desired volume of solution and not by adding a certain amount of solvent. The volume of the solute must also be accounted for in the total volume of the solution.

28.2.2 Mole FractionThe mole fraction (χΑ) of a component in a solution is defined as the moles of the compo-nent divided by the total moles of solution.

mole fractionmoles of component Atotal moles of solution

χA( ) =

Mole fraction is a unitless quantity with the sum of mole fractions of all the compo-nents of the solution equaling to 1.

28.2.3 Mass PercentIt is often convenient to report solution concentration as a ratio of masses. Mass percent is a ratio of the mass of the solute to the mass of the solution multiplied by 100.

Mass percentageof solutemass of solute

mass of solution= × 100%

For example, a 6.5% by mass solution contains 6.5 g of solute per 100 g of solution. Note: the “%” in the formula is a unit like grams, etc. Also note the unit of mass doesn’t matter as long as we use the same unit for solute and solution.

▼ Example 1

How many grams of water are needed to prepare 555 g of an aqueous solution containing 3.75% by mass of sodium chloride?Answer: 534 2g H OWe know there is 3.75 g of NaCl needed per every 100 g of solu-tion by the mass percent; therefore, we can determine how much NaCl is in 555 g of solution.

mass of NaCl g solutiong NaCl

g solution

=⎛

⎝⎜

⎞

⎠⎟ =( )

..555

3 75100

20 88 g NaCl


We also know that of the 555 g of solution that 20.8 g consists of NaCl and the remainder is the solvent water.

total mass of solution mass of solute NaCl mass of solvent = +, ,,H O2

Rearranging,

mass of H O total mass of solution mass of NaClmass of H O

2

2 5= −= 555 20 8 534 2g g g H O− =.

Note: masses are additive and temperature independent while volumes are not additive and temperature dependent.

28.2.4 MolalityThe molality (m) of a solution is the moles of solute per kilogram of solvent. This expression is useful in situations when concentrations must be compared over a range of different temperatures because it is based on mass which is temperature independent.

molality m moles of solutekilograms of solvent

( ) =

Note: molality is defined with respect to the amount of solvent, not solution.

▼ Example 2

A solution used for intravenous feeding contains 5.20 g of glucose in 90.0 g of water. What is the molality of glucose?Answer: 0 321 6 12 6. m C H OWe will need to convert the mass of glucose to moles and mass of water to kg:

moles of C H O g C H Omol C H O

g C H O 6 12 6 6 12 6

6 12 6

6 12 6

5 201

180 2=

⎛

⎝⎜

⎞( . )

. ⎠⎠⎟ = 0 0289 6 12 6. g C H O

moles of C H O g C H Omol C H O

g C H O 6 12 6 6 12 6

6 12 6

6 12 6

5 201

180 2=

⎛

⎝⎜

⎞( . )

. ⎠⎠⎟ = 0 0289 6 12 6. g C H O

kg of H O g H Okg

gkg H O 2 2 290 0

11000

0 0900=⎛

⎝⎜

⎞

⎠⎟ =( . ) .

Plugging these values into the definition of molality:

mg C H O

kg H Om C H O= =

0 02890 0900

0 3216 12 6

26 12 6

..

.


28.3 Colligative PropertiesThe properties of a solution differ from those of the pure solvent. The colligative proper-ties of solutions are those properties that depend on the number of particles dissolved in solution rather than their nature. These properties include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure. The relationships among colligative properties and solute concentrations are regarded as limiting laws. The more dilute the solution, the closer results are to expected values.

28.3.1 Vapor Pressure Lowering (Nonelectrolytes)The vapor pressure of a liquid is the pressure of the gas above the liquid. Vapor pressure is a colligative property that decreases by the addition of a nonvolatile solute. The vapor pressure of the solution (nonvolatile solute and nonelectrolyte solvent) will be lower than the vapor pressure of the pure solvent. The vapor pressure lowering is independent of the nature of the solute but directly proportional to its concentration. Vapor pressure is a mea-sure of the gas molecules exerting pressure above the liquid, and any interference with the ability of the solvent particles to vaporize would result in a decrease in the amount of gas molecules and hence lower the vapor pressure. By adding a nonvolatile solute to the solu-tion, we decrease the surface area formerly occupied by just solvent (now mixture of solute and solvent particles on surface) and diminish the rate of vaporization of the solvent. The net result is a decrease in vapor pressure for the solution as compared to the pure solvent.

This relationship of vapor pressure and concentration of solute particles is called Raoult’s Law. Raoult’s Law states that the vapor pressure of a solution containing a non-electrolyte nonvolatile solute is proportional to the mole fraction of the solvent:

P Psolution solvento

solvent= ( )( )χ

where Psolution is the vapor pressure of the solution (solute and solvent), csolvent is the mole fraction of the solvent, and Psolvent

o is the pure vapor pressure of the solvent. Basically, the vapor pressure of the solution is a fraction of that of the pure solvent and depends on the percentage of solvent making up the solution (partial vapor pressure).

If a solution contains a volatile solute, the vapor pressure of the solution will be a com-bination of the partial vapor pressures of each volatile component:

P P Psolution solvento

solvent soluteo

solute= +( )( ) ( )( )χ χ

28.3.2 van’t Hoff factorColligative properties of solutions are directly proportional to the concentration of solute particle; therefore, the effect that solutes have on colligative properties depends on the quantity of solute particles present in the solution. When one mole of glucose, C6H12O6,


dissolves in water, one mole of solute molecules is obtained. Glucose is a molecular compound composed of covalent bonding and is a nonelectrolyte that does not ionize in water. For each mole of solute, there is one mol of solute particles:

C6H12O6 (s) → C6H12O6 (aq)

Electrolytes ionize in water and for each mole of solute, there could be several moles of solute particles. For example, CaCl2 is an ionic compound that is an electrolyte that dis-sociates into three solute particles in water (1 mol of Ca2+ and 2 mol of Cl−).

CaCl2 (s) → Ca2+ (aq) + 2Cl− (aq)

Since calcium chloride has three times the number of solute particles as glucose, we would expect CaCl2 to have approximately three times the effect on the colligative proper-ties as compared to glucose. The ratio of moles of solute particles in solution to moles of formula units dissolved is referred to as the van’t Hoff factor (i):

i moles of solute particles in solutionmoles of formula un

= iits dissolved in solution

We expect the moles of solute particles in solution and the van’t Hoff factor to be the same, but this actually only occurs for very dilute solutions. We expect the dissociation of an ionic compound to be complete, but typically it is not complete and at any particular moment, some cations and anions may still be paired giving a slightly lower number of solute particles. For our purposes, we will assume that the ions of an electrolyte behave independently meaning i should be equal to the number of moles of ions per mole of electrolyte and 1 for nonelectrolytes. For example, i should be 2 for NaCl, 5 for Al2(SO4)3, and 1 for C2H6O2.

In the previous section, we discussed the vapor pressure of nonelectrolytes. When we calculate the vapor pressure of a solution containing an ionic solute (electrolyte), we must account for the number of solute particles when we calculate the mole fraction of the solvent.

▼ Example 3

A solution contains 0.155 mol NaCl and 0.756 mol H2O. Calculate the vapor pressure of the solution at 55°C given the vapor pressure of pure water at this temperature is 118.1 mm Hg.


Answer: 83 7. mm Hg

NaCl has 2 particles per every one mol of substance. This must be accounted for in the mole fraction of water before we use Raoult’s Law.

χ

χ+

H OH O

NaCl H O

n

n n

molmol mol

2

2

22

20 756

0 155 0 756

=× +

=×

=

( )

.( . ) .H O2

00 709

118 1

.

( )( )

( .

P P

P mm Hg

solution solvento

solvent

solution

=

=

χ

)) ( . ) .0 709 83 7= mm Hg

χ

χ+

H OH O

NaCl H O

n

n n

molmol mol

2

2

22

20 756

0 155 0 756

=× +

=×

=

( )

.( . ) .H O2

00 709

118 1

.

( )( )

( .

P P

P mm Hg

solution solvento

solvent

solution

=

=

χ

)) ( . ) .0 709 83 7= mm Hg

χ

χ+

H OH O

NaCl H O

n

n n

molmol mol

2

2

22

20 756

0 155 0 756

=× +

=×

=

( )

.( . ) .H O2

00 709

118 1

.

( )( )

( .

P P

P mm Hg

solution solvento

solvent

solution

=

=

χ

)) ( . ) .0 709 83 7= mm Hg

χ

χ+

H OH O

NaCl H O

n

n n

molmol mol

2

2

22

20 756

0 155 0 756

=× +

=×

=

( )

.( . ) .H O2

00 709

118 1

.

( )( )

( .

P P

P mm Hg

solution solvento

solvent

solution

=

=

χ

)) ( . ) .0 709 83 7= mm Hg

28.3.3 Boiling Point Elevation and Freezing Point DepressionFor the same reason the vapor pressure reduces by the addition of a nonvolatile solute, the boiling point of a solution will be elevated. The temperature at which the vapor pressure of a liquid equals 1 atm is called the normal boiling point. Since the addition of a nonvolatile solute will diminish the rate of vaporization of a solvent, the temperature of the solution must be increased to a value greater than the normal boiling point of the pure solvent to achieve a vapor pressure of 1 atm. The opposite effect occurs for freezing. When a solution is cooled, it does not begin to freeze until a temperature is achieved that is below the freezing point of the pure solvent. Boiling point elevation and freezing point depression are colligative properties that are directly proportional to the concentration of solute particles. A good example of taking advantage of both properties is the addition of antifreeze solution to the radiator of a car. Typically, ethylene glycol is used as antifreeze and is a nonvolatile solute. By adding antifreeze to our radiator, we have raised the boiling point and lowered the freezing point of our engine cooling system. It allows us to continue to use our cars on hot summer days and cold winter days that otherwise would not be possible with pure water.

The relevant equations associated with these colligative properties are:

ΔT K m ib b= (boiling point elevation)

ΔT K m if f= (freezing point depression)

where ΔT K m ib b= and ΔT K m if f= are the change in temperature relative to that of the pure solvent in °C, Kb and Kf are proportionality constants called the molal boiling point elevation constant and molal freezing depression point constant that are dependent on the solvent,


m is the molality of the solution in moles per kilogram solvent, and i is the number of moles of solute particles per mole of solute. Once the change in temperature is calculated, the actual boiling point or freezing point of a solution is calculated by adding the change to the boiling point of the pure solvent or subtracting the change from the freezing point of the pure solvent.

Tb solution = +T Tb solvent bΔ

Tf solution = −T Tf solvent fΔ

▼ Example 4

Estimate the freezing point of 0.35 m solution of Cr(NO3)3. Assume the value of i is based on the number of solute particles per mole of solute and the Kf of water is 1.86°C/m.Answer: –2.6°CCr(NO3)3 has 4 particles per every one mol of substance which means i = 4. Plugging into the change in temperature equation we obtain:

ΔT K m if f=

ΔTm

mf = =1 860 35 4 2 6

.( . ) ( ) .

ooCC

Tf solution

f solutiono o oT C C C

= −

= − = −

T Tf solvent fΔ

0 00 2 6 2 6. . .

Tf solution

f solutiono o oT C C C

= −

= − = −

T Tf solvent fΔ

0 00 2 6 2 6. . .

▼ Example 5

A 1.25 g sample of a nonelectrolyte compound was dissolved in 75.0 g of benzene. The solution froze 1.20°C below that of pure benzene. Determine the molar mass of the compound if the Kf of benzene is 5.12°C/m.Answer: 71.2 g/mol

ΔT K m if f=

Re-arranging for the molality of the solution:

mT

K if

f

=Δ


m

m

m=⎛

⎝⎜

⎞

⎠⎟

=1 205 12

10 234

..

( ).

o

o

CC

Knowing the relationship between moles and molar mass we can rewrite molality:

mmoles solutekg solvent

g solute Mkg solvent

m= =

/

Re-arranging for molar mass:

Mg solute

m kg solventm = ( ) ( )

Mg solute

mol solutekg solvent

kg som =

⎛

⎝⎜

⎞

⎠⎟

1 250 234

0 0750

..

( .

llvent

g mol)

.= 71 2 /

28.3.4 Osmotic PressureOsmosis is a process whereby a solvent passes from a dilute solution into a more concen-trated one through a membrane permeable only to the solvent. The equalization of con-centration between the two solutions in contact with one another across the membrane is what drives the process of osmosis. The osmotic pressure, π, is equal to the external pres-sure, P, just sufficient to prevent osmosis. Osmotic pressure is a colligative property and is directly proportional to the molar concentration of solute:

� = MRT

where M is the molarity of the solute, R is the ideal gas constant (0.0821 L . atm/mol . K), and T is absolute temperature (K).


TEST YOURSELF


1. Consider the following: A salt solution is able to dissolve more salt. The solution is: A. a saturated solution B. an unsaturated solution C. a supersaturated solution

2. The solubility of “X” in water is 0.472 g/mL at 17.0°C. A solution where the concen-tration of “X” in water is 0.472 g/mL at 17.0°C is said to be:

A. saturated B. unsaturated C. supersaturated

3. The solubility of calcium hydroxide in water at 100°C is 7.70 × 10−4 g/mL. Suppose 0.1273 g of calcium hydroxide and 77.2 mL of water are mixed. At equilibrium, at 100°C, how much solid calcium hydroxide will remain undissolved? (assume no supersaturation)

4. The solubility of calcium hydroxide in water at 100°C is 7.70 × 10−4 g/mL. Suppose 0.02775 g of calcium hydroxide and 50.9 mL of water are mixed. At equilibrium, at 100°C, how much solid calcium hydroxide will remain undissolved? (assume no supersaturation)

5. Calculate the molarity of an aqueous solution that has a mass of 807 grams and contains 47.4 grams of KIO3. The density of the solution is 1.05 g/mL.

6. A solution is prepared by dissolving 5.10 grams of NaCl in 154 mL of H2O, what is the molarity of the NaCl in the solution? The density of H2O is 0.997 g/mL, and the density of the solution is 1.07 g/mL.

7. What is the mass percent of KIO3 in a solution prepared by dissolving 38.3 grams of KIO3 in 53.4 grams of H2O?

8. A solution is prepared by dissolving 4.96 grams of NaCl in 244 mL of H2O. What is the mass percent, wt%, of the NaCl in the solution? The density of H2O is 0.997 g/mL.



9. What is the mole fraction of CH3OH in a solution prepared by dissolving 20.4 grams of CH3OH in 37.8 grams of water?

10. Calculate the molality of an aqueous solution that has a mass of 888 grams and contains 51.4 grams of K2SO4.

11. A solution is prepared by dissolving 4.46 grams of benzoic acid, C6H5CO2H, in 150 mL of benzene, C6H6, what is the molality of the benzoic acid in the solution? The density of benzene is 0.879 g/mL.

12. A solution at 85°C contains 235 grams of glucose, C6H12O6, dissolved in 344 grams of H2O. The C6H12O6 is nonvolatile; the vapor pressure of pure H2O at 85°C is 433.6 mm Hg. What is the vapor pressure of water in this solution?

13. Calculate the boiling point in °C of a solution that contains 57.7 grams of C6H12O6 dissolved in 108 grams of pure H2O. The boiling point and molal boiling point constant, Kb, of pure H2O are 100.00°C and 0.52°C/m, respectively.

14. Calculate the boiling point in °C of a solution that contains 38.6 grams of Cr(NO3)3 dissolved in 93.0 grams of pure H2O. The boiling point and molal boiling point constant, Kb, of pure H2O are 100.00°C and 0.52°C/m, respectively.

15. Calculate the freezing point in °C of a solution that contains 68.6 grams of C12H22O11 dissolved in 105 grams of pure H2O. The freezing point and molal freezing point constant, Kf, of pure H2O are 0.00°C and 1.86°C/m, respectively.

16. Calculate the freezing point in °C of a solution that contains 37.4 grams of NaCl dissolved in 123 grams of pure H2O. The freezing point and molal freezing point constant, Kf, of pure H2O are 0.00°C and 1.86°C/m, respectively.

17. A student dissolves 1.27 grams of a newly prepared nonelectrolyte compound in 66.0 grams of C6H12. The freezing point of this solution was found to be 2.86°C. The freezing point and molal freezing point constant, Kf, of pure C6H12 are 6.50°C and 20.2°C/m, respectively. Calculate the molar mass of the new compound.

439

Thermochemistry is the study of the energy effects that accompany chemical reactions. In this chapter, we will examine the relationship between energy and chemical reactions.

29.1 Reaction EnthalpyWhen discussing energy, it is useful to be familiar with the following terms: system, surroundings, and state function. The term system refers to any part of the universe we are interested in. The rest of the universe is referred to as the surroundings. In Chemistry, the systems we typically deal with are samples of matter. When these samples undergo a chemical or physical change, the change is typically accompanied by a transfer of energy between the system and surroundings.

When a system is at a fixed temperature, pressure, and composition, we say that it is in a state of equilibrium; left alone, it will not undergo any change. State functions are properties associated with a system in equilibrium; for any given state, a state function is uniquely defined. Examples of state functions include temperature (T), pressure (P), volume (V), and energy (E). If a system is in a state of equilibrium, these properties are constant.

If a system does undergo a change (from one equilibrium state to another), and X is a state function, then ΔX (the change in the value of X) is the same regardless of how the change occurred. It only depends on the final state, which has a uniquely defined value for X, and the initial state, which also has a uniquely defined value for X.

ΔX = Xfinal − Xinitial

Every system has a property called enthalpy, H, which is defined as

H = E + PV

Since E, P and V are state functions, H is also a state function. Note that H has the same unit as E. We cannot measure H itself but, whenever a system undergoes a change

CHAPTER

Thermochemistry29


at constant pressure, the amount of heat that flows between the system and surroundings is equal to the change in the enthalpy (ΔH) of the system. Unless otherwise specified, for the rest of this chapter, we will assume that we are dealing with processes occurring at a constant pressure of 1 atm.

The movement of atoms during a chemical change typically leads to net conversion of potential energy of the atoms to kinetic (or thermal) energy, or vice versa. The result is that the products would be at a different temperature than the original reactants and will no longer be at the same temperature as the surroundings. This leads to heat flow between the reaction mixture and the surroundings until the temperature of the reaction mixture is restored to its original value; this heat flow is called the heat of reaction (qrxn). If the reaction is carried out at constant pressure, then the heat of reaction is equal to the reaction enthalpy (ΔHrxn), which is the enthalpy change associated with the conversion of reactants to products (at the same temperature).

ΔHrxn = Hproducts − Hreactants

ΔHrxn = qrxn (at constant P)

If a reaction leads to products that are hotter than the reactants, we say that the reaction is exothermic; heat will flow out of the reaction mixture to restore the system to the original temperature and qrxn or ΔH is assigned a negative value. Otherwise, the reaction is endothermic; heat will flow into the reaction mixture and qrxn or ΔHrxn is assigned a positive value.

▼ Example 1

Solutions of NaOH(aq) and HCl(aq), both at 25°C, are mixed and the temperature of the resulting reaction mixture is found to be 28°C. Is the reaction endothermic or exothermic? What is the algebraic sign of ΔHrxn?Answer: exothermic, negativeThe temperature increased. Therefore, the reaction is exothermic; heat has to flow out of the system in order to restore the original temperature. Since the reaction is exothermic, ΔHrxn is assigned a negative value.

Chapter 29: Thermochemistry 441

29.2 Interpreting Symbols for Reaction Enthalpy

While the symbol for reaction enthalpy is usually written as ΔHrxn in introductory courses, you need to be aware that the standard symbol is ΔrH especially if you need to look up literature values.

If a degree symbol (°) is written, as in ΔrH°, the value is known as the standard reaction enthalpy; by “standard”, we mean that the pressure is 1 bar. The old standard pressure is 1 atm, which is equivalent 1.01325 bar; most introductory textbooks and exams still cling to this definition of standard pressure. Since 1 bar and 1 atm differ by only about 1.3%, values of ΔrH at 1 bar and 1 atm are not significantly different in most cases.

Unless otherwise specified, we can assume that literature values of reaction enthalpy are for a temperature of 25°C or 298.15K. In cases where the temperature is specified, it is indicated by a subscript, as in ΔrH°298K or ΔrH°500K.

The “rxn” or “r” subscript is omitted when it is obvious that the enthalpy change being referred to is reaction enthalpy.

29.3 Thermochemical EquationsA thermochemical equation is a chemical equation that shows directly to the right of the balanced chemical equation the value of the reaction enthalpy for one mole of reaction. For example, the thermochemical equation for the combustion of hydrogen is:

O2(g) + 2 H2(g) → 2 H2O(g) ΔH = −483.7 kJ

A thermochemical equation must indicate the physical states (s, l, g, aq) of the reactants and products involved. The ΔH for the formation of liquid water from hydrogen and oxygen gases is not the same as for the formation of gaseous water.

If more than one form of the solid is known to exist at a temperature of the reaction, the form is also specified in the thermochemical equation; otherwise, we assume that we are dealing with the most stable form. For example, solid carbon can exist as graphite or diamond at room temperature; graphite is the more stable form. If a reaction involves graphite, it can be represented as C(s) or C(graphite). But if a reaction involves diamond, it should be represented as C(diamond).

Enthalpy is an extensive property. This means that the value of reaction enthalpy is directly proportional to the amount of reactants and products involved in the reaction.

Consider the following:

N2( g) + 3 H2( g) → 2 NH3( g) ∆H = −91.8 kJ


This equation basically gives a recipe of what and how much to mix (1 mole of N2 and 3 moles H2) to produce 2 moles NH3 and release 91.8 kJ of heat. Note that a negative value of ΔH means that the reaction is exothermic; it releases heat. Since H is an extensive prop-erty, if the recipe is cut in half, the amount of heat generated will also be cut in half. If only 1/2 mole of N2(g) were involved in the reaction, only 1/2 of 91.8 kJ of heat would be released.

▼ Example 2

Given the following thermochemical equation:O2(g) + 2 H2(g) → 2 H2O(g) ΔH = −483.7 kJ

Complete the following:2 O2(g) + 4 H2(g) → 4 H2O(g) ΔH = ?

Answer: −967.4 kJIn general, if we multiply all the coefficients of a thermochemical equation, we must also multiply ΔH by the same factor. In the example here, we multiplied all the coefficients by 2. We should also double the reaction enthalpy; we multiply the ΔH by 2:2 O2(g) + 4 H2(g) → 4 H2O(g) ΔH = 2(−483.7) = −967.4 kJ

Since enthalpy is an extensive property, we can set up conversion factors based on the coefficients and ΔH values in a thermochemical equation and perform heat calculations similar to stoichiometric calculations.

▼ Example 3

Consider:O2(g) + 2 H2(g) → 2 H2O(g) ΔH = −483.7 kJ

What conversion factors do we need to calculate the heat of reaction if we know the moles of O2, H2, or H2O that are involved in a reaction?Answer: Since the coefficients of O2, H2, and H2O in the thermochemical equation are 1, 2, and 2, we multiply known changes in moles of these substances by these conversion factors, respectively, in order to calculate the corresponding amount of heat involved.

− − −483 71

483 72

483 722 2 2

. . .kJmol O

kJmol H

kJmol H O


▼ Example 4

Consider the following reaction:C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g), ΔH = −2044 kJ

Calculate the heat associated with the production of 25.0 g of carbon dioxide from this reaction.Answer: −387 kJThis calculation is similar to those done in the stoichiometry chapter except that it involves heat. We will convert the mass of carbon dioxide to moles and then use the relationship between heat and moles of carbon dioxide from the thermochemical equation to calculate the final answer.

( . )

. 25 0

144 01

204432

2

2

g COmol CO

g COkJ

−mmol CO

kJ 2

387

−=

If the ΔH for a reaction is x, then the ΔH for the reverse reaction is −x. This is a consequence of H being a state function. If a reaction is exothermic, its reverse is endo-thermic; the magnitude of ΔH is the same but the algebraic sign is reversed. Similarly, if a reaction is endothermic, its reverse is exothermic.

▼ Example 5

Given the following thermochemical equation:O2(g) + 2 H2(g) → 2 H2O(g) ΔH = −483.7 kJ

Write the thermochemical equation for the reverse reaction.Answer: 2 H2O(g) → O2(g) + 2 H2(g) ΔH = +483.7 kJSince the given reaction is exothermic and has a negative ΔH value. The reverse reaction is endothermic and has a positive ΔH value. The negative of (−483.7) is +483.7.

29.4 CalorimetryA calorimeter is a device used to measure heats of reaction. The apparatus is designed with thermally insulated walls. If a temperature change of ΔT is observed, then the heat flow that would be needed to reverse the temperature change is (Ccal)(−ΔT),


where Ccal is the heat capacity and all of its contents (after the reaction has occurred). Therefore:

qrxn = −Ccal ΔT

▼ Example 6

Solutions of NaOH(aq) and H2SO4(aq), both at 25.00°C, are mixed and the temperature of the resulting reaction mixture is found to be 27.00°C. If the heat capacity of the calorimeter and its contents is 2.75 kJ/°C. Calculate qrxn.Answer: –5.50 kJFor a reaction in a calorimeter:qrxn = –Ccal ΔT

Therefore,qrxn = –(2.75 kJ/°C) (27.00°C – 25.00°C)

= –(2.75 kJ/°C) (2.00°C)= –5.50 kJ

▼ Example 7

What is the value of x in the thermochemical equation:2 NaOH(aq) + H2SO4(aq) → 2 H2O(I) + Na2SO4(aq), ΔH = X

if the heat of reaction is found to be −5.58 kJ when 0.100 mol of NaOH is consumed?Answer: x = −112 kJThe value of ΔH attached to a thermochemical equation is for one mole of reaction. Since the given balanced equation has a coefficient of 2 for NaOH, one mole of this reaction, as written, involves the consumption of 2 mol NaOH. In other words, we need to calculate the value of ΔH for the consumption of 2 mol NaOH.Since heat released and amount of NaOH are directly proportional, the ratio of the two quantities is constant. If x is the amount of heat associated with the consumption of 2 mol NaOH, then

−−

=5 580 100 2

..

kJmol NaOH

xmol NaOH


Solving for x, we find that

x mol NaOH kJmol NaOH

= −

= ( ) .

. 2 5 58

0 100−−112 kJ

Note that the ratio of two directly proportional quantities in one particular instance is essentially a conversion factor that we can use to calculate corresponding amounts in other instances.

▼ Example 8

In a calorimeter, 32.4 grams of calcium chloride was dissolved in water. The temperature rose from 25.0°C to 43.2°C. If the heat capacity of the solution and calorimeter is 1250 J/°C, how much heat is released when 45.0 g of calcium chloride are dissolved?Answer: 31600 J of heat is released, qrxn = −3.16 × 103 JFirst, this problem involves plugging the proper values into the equation to determine the ratio of heat generated in this reaction:qrxn = –Ccal ΔTqrxn = –(1250 J/°C) (43.2°C – 25.0°C)

= –22750 J

This is the amount of heat released when 32.4 g of calcium chloride are dissolved. Since heat released and amount of CaCl2 are directly proportional, we can set up a ratio to give us a conversion factor to calculate the amount of heat associated with a different quantity of calcium chloride.

− = −2275032 4

70212 2

Jg CaCl

Jg CaCl.

Using this conversion factor, we can determine how much heat is associated with 45.0 g CaCl2.

( . )

45 0702

1316002

2

g CaClJ

g CaClJ

−

−=


Water has an unusually high specific heat capacity (cwater = 4.18 J g−1 °C−1). So, if the calorimeter contains a large amount of water, Ccal can be approximated as the heat capacity of the water:

Ccal = (masswater)(cwater)

Dilute aqueous solutions can also be approximated to have the same density (1.00 g/mL) and specific heat as pure water. Aqueous solutions with molarities of 1.0 M or less can be considered as dilute. We can use these values to estimate heats of reaction.

▼ Example 9

Two fiftymL dilute aqueous solutions are mixed in a calorimeter and a temperature drop of 5.0°C was observed. Estimate the heat of reaction.Answer: +2.1 kJThe total volume of the reaction mixture is approximately 100 mL; 50 + 50 = 100. The density of water is 1.00 g/mL. Therefore, the total mass of the reaction mixture is approximately 100 g. Therefore, the heat capacity of the calorimeter and its contents is approximately:Ccal = (100 g)(4.18 J/g°C) = 418 J/°C

Therefore the heat of reaction is approximately:qrxn = −Ccal ΔT = −(418 J/°C)(−5°C) = +2090 J → 2.1 × 103 J or 2.1 kJ

Note that the ΔT is negative (the temperature dropped) and qrxn is positive (the reaction is endothermic). Heat has to flow into the reaction mixture to restore the temperature to the original temperature of the reactants.

29.5 Hess’ LawReaction enthalpies can be determined using calorimetry only for reactions that go to completion, are fast, and have no side reactions. For reactions that do not meet these criteria, the reaction enthalpy can be determined indirectly using Hess’ Law.

Hess’ law states when a reaction can be expressed as the algebraic sum of two or more reactions, then the ΔH for the reaction is the algebraic sum of the ΔH of the reactions added. Hess’ law is a consequence of enthalpy being a state function.


▼ Example 10

Given Equation 1) A + D → E + C ΔH = X kJ Equation 2) 2A + B → 2C ΔH = Y kJ

Determine the enthalpy change for the equation below by using the given equations. 2D → B + 2E ΔH = ?Answer: (2X − Y) kJHere’s a strategy for figuring this out. Typically, we look for a species in the desired equation that occurs in only one of the given equations and manipulate the equation to place the species on the correct side and correct number of moles. If we examine the desired equation, we will notice that D occurs in only equation 1. We ask ourselves two questions to determine how to manipulate equation 1: First, is the species on the correct side of the equation? In this case, D is on the correct side in equation 1 as compared to the desired equation; therefore, we will use the equation just as written. Second, does the given equation have the correct number of moles of the species as in the desired equation? In this case, D needs 2 moles and it has one mole in equation 1; therefore, we will multiply the whole reaction including ΔH by 2: 2A + 2D → 2E + 2C ΔH = 2X kJ

Now we look for a species in equation 2 that we can examine and do the same process; we see that B only occurs in equation 2.We ask ourselves two questions to determine how to manipulate equation 2: First, is the species on the correct side of the equation? In this case, B is not on correct side in equation 2 as compared to the desired equation; therefore, we must reverse the equation and change the sign of ΔH. Second, does the given equation have the correct number of moles of the species as in the desired equation? In this case, B needs 1 mole and it has only one mole in equation 2; therefore, we do not need to multiply the equation by any factor: 2C → 2A + B ΔH = −Y kJ

We then add the two equations to determine if we obtained the desired equation. To obtain the ΔH for the desired equation, we add the manipulated ΔH’s. If we do this, we obtain


2A 2D 2E 2C H 2X kJ2C 2A B H Y kJ

+ → + =→ + = −

∆∆

__________________________________________________________________________________

2D B 2E H (2X→ + =∆ Y) kJ−

Notice you can easily check the manipulation of the given equations. If we did things correctly, the overall equation will be exactly the same as the desired equation.

▼ Example 11

Given 1) N2(g) + 3 H2(g) → 2 NH3(g) ΔH = −92.2 kJ 2) C(s) + 2 H2(g) → CH4(g) ΔH = −74.7 kJ 3) 2 C(s) + H2(g) + N2(g) → 2 HCN(g) ΔH = 270.3 kJ

Determine ΔH for the following reaction:CH4(g) + NH3(g) → HCN(g) + 3 H2(g) ΔH = ?

Answer: 256.0 kJIf we examine the desired equation, we will notice that CH4 occurs in only equation 2. We ask ourselves two questions to determine how to manipulate equation 2: First, is the species on the correct side of the equation? In this case, CH4 is not on the correct side in equation 2 as compared to the desired equation; therefore, we must reverse the equation and change the sign of ΔH. Second, does the given equation have the correct number of moles of the species as in the desired equation? In this case, CH4 needs 1 mole and it has one mole in equation 2; therefore, we do not need to multiply the equation by any factor:CH4(g) → C(s) + 2 H2(g) ΔH = +74.7 kJ

Next, we see that NH3 only occurs in equation 1. We ask ourselves two questions to determine how to manipulate equation 1: First, is the species on the correct side of the equation? In this case, NH3 is not on correct side in equation 1 as compared to the desired equation; therefore, we must reverse the equation and change the sign of ΔH. Second, does the given equation have the correct number of moles of the species as in the desired


equation? In this case, NH3 needs 1 mole and it has 2 moles in equation 1; therefore, we will multiply the whole reaction including ΔH by ½ to fix this:

NH N H H3 2 2( ( () ) ( ) ( (g g g→ ( / ) / ) / )1 2 3 2 1 2 92+ +=∆ .. .2 46 1kJ) kJ= +

Next, we see that HCN only occurs in equation 3. We ask ourselves two questions to determine how to manipulate equation 3: First, is the species on the correct side of the equation? In this case, HCN is on correct side in equation 3 as compared to the desired equation; therefore, we will use the equation just as written. Second, does the given equation have the correct number of moles of the species as in the desired equation? In this case, HCN needs 1 mole and it has 2 moles in equation 3; therefore, we will multiply the whole reaction including ΔH by ½ to fix this:

C H N HCN H2 2( ( ( (s g g g) ) ) )( / ) ( / ) ( /+ + → =1 2 1 2 1 2∆ ))( . )270 3 kJkJ= +135 2.

We then add these equations to determine if we obtained the desired equation. To obtain the ΔH for the desired equation, we add the manipulated ΔH values. If we do this, we obtain

CH C 2 H H 74 7NH N

4 2

3

( ) ( ) ( ) kJ( )g s gg

→→

+ = +∆ .( / )1 2 22 2

2

H HC H

( ) ( ) kJ( ) ( )

g gs g

+ = ++ +

( / )( / ) (

.3 21 2

46 1∆11 2 135 2/ ) .

_______________

N HCN H2( ) ( ) kJg g→ ∆ = +_________________________________________________________

CH NH HCN 3 H4 3 2( ) ( ) ( ) ( )g g g g+ +→

ΔH = 74.7 kJ + 46.1 kJ + 135.2 kJ = 256.0 kJ

29.6 Standard Enthalpies of FormationUsing Hess’ law, we can determine standard enthalpies of formation; tabulations of these values are convenient for calculating reaction enthalpies. A good source of literature values is the NIST Chemistry Webbook (http://webbook.nist.gov).

http://webbook.nist.gov


The standard enthalpy of formation of something, ΔHf° or Δf H°, is equal to the enthalpy change when one mole of that something at standard pressure and 298.15K (25°C) from its constituent elements in their standard state. The something can be a formula unit (molecular or ionic), an atom, or an ion. An element is in its standard state if it is in its most stable form at standard pressure and 298.15K.

▼ Example 12

Which of the following best represents the standard state of bromine?A. Br(g), B. Br2(g), C. Br2(l), D. Br2(s)

Answer: CThe most stable form of elemental bromine at standard pressure at 298.15K is a liquid consisting of diatomic molecules.

▼ Example 13

The standard enthalpy of formation of elements in their standard state is zero. Explain.Answer: To form an element from itself means doing nothing.

▼ Example 14

Write the thermochemical equation for which ΔH is called the “standard enthalpy of formation” of liquid water. The ΔHf° of H2O(l) is −285.8 kJ/mol.Answer: H2(g) + (1/2) O2(g) → H2O(I) ΔH = −285.8 kJThe coefficient of the product, H2O(l), must be 1. Therefore, we have to use a fractional coefficient for O2 in order to balance the chemical equation. Note that, under standard pressure and 298.15K, the most stable elemental forms of hydrogen and oxygen are gaseous and diatomic.


▼ Example 15

The standard enthalpy of formation of Na2O(s), CO2(g), H2O(l), and NaHCO3(s) are w, x, y, and z kJ/mol, respectively. Using Hess’ Law, verify that ΔH for the reaction:2 NaHCO3(s) → Na2O(s) + 2 CO2(g) + H2O(g)

is [(w + 2x + y) − (2z)] kJAnswer:Based on the given standard enthalpies of formation, we can write the following thermochemical equations:

Na(s) + O2(g) → Na2O(s) ΔH = w2 C(s) + 2 O2(g) → 2 CO2(g) ΔH = 2xH2(g) + (1/2) O2(g) → H2O(g) ΔH = y2 NaHCO3(s) → 2 Na(s) + (1/2) H2(g)

+ C(s) + (3/2) O2(g) ΔH = −2z

If we add up all these equations, we find get the desired overall thermochemical equation. Note that the overall ΔH is equal to the sum of the standard enthalpies of the products minus the sum of the standard enthalpies of the reactants.

As illustrated in the preceding example the standard reaction enthalpy is equal to the total formation enthalpies of the products minus that of the reactants:

∆ ∆ ∆H n H n Hf products f reactants° o o= −∑ ∑, ,

where ∑ represents “the sum of ”, and n is the coefficient of each substance in the chemical equation. The nice thing about this is that we do not need to tabulate the ΔH for each and every possible reaction. All we need is a table of standard enthalpies of formation.

▼ Example 16

Use the standard enthalpies of formation to determine ΔH° for the reaction:CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)

Given:ΔHf

° for CH4(g) = −74.8 kJ/molΔHf

° for CO2(g) = −393.5 kJ/molΔHf

° for H2O(l ) = −285.8 kJ/mol


Answer: −890.3 kJThe standard reaction enthalpy is equal to the total formation enthalpy of the products (2 mol CO2(g) and 2 mol H2O(l)) minus that of the reactants (1 mol CH4(g) and 2 mol O2(g)):

∆H mol CO g kJmol CO g

m° = +1 393 5 222

( ) . ( )

(−

ool H O l kJ

mol H O lm2

2

285 8 1( )) .( )

(−

− ool CH g kJ

mol CH gmol O ( )) .

( )( 4

42

74 8 2−

+ (( ))

g kJ

mol O0

2

∆H mol CO g kJmol CO g

m° = +1 393 5 222

( ) . ( )

(−

ool H O l kJ

mol H O lm2

2

285 8 1( )) .( )

(−

− ool CH g kJ

mol CH gmol O ( )) .

( )( 4

42

74 8 2−

+ (( ))

g kJ

mol O0

2

∆H kJ kJ kJ° = − + − +[( . ) ( . )] [( . ) ( )]393 5 571 6 74 8 0− − .= −890 3 kJ

Using the law of summation of heats of formation, we can calculate the heat of this reaction. Note that the heat of formation of oxygen is not given. Recall that pure elements at standard state have ΔHf° values of 0.

For an ionic compound in aqueous solution, the standard enthalpy of formation is the sum of the standard enthalpies of formation of the ions in one mole of formula units; by definition a value of zero is assigned for H+(aq).

▼ Example 17

Calculate the standard reaction enthalpy for the dissolution of CaCl2(s) in water given the following:ΔHf

° for CaCl2(s) = −795.8 kJ/molΔHf

° for Cl−(aq) = −167.2 kJ/molΔHf

° for Ca2+(aq) = −542.8 kJ/molAnswer: −81.4 kJThe chemical equation for the dissolution is:CaCl2(s) → CaCl2(aq)

orCaCl2(s) → Ca2+(aq) + 2 Cl−(aq)


The standard reaction enthalpy is equal to the total formation enthalpy of the products (one mol Ca2+(aq) and 2 mol Cl−(aq)) minus that of the reactant (1 mol CaCl2(s)). Therefore:

∆H mol Ca aq kJmol Ca aq

° = −

+

+( ( )) .( )

1 542 822 ++ −

−

−( ( )) .( )

2 167 2mol Cl aq kJmol Cl aq

− −

( ( )) .

( )1 795 8

22

mol CaCl s kJmol CaCl s

= −81 4. kJ

▼ Example 18

A student is asked to calculate the standard reaction enthalpy for the neutralization of NaOH(aq) by HCl(aq) given only the following:ΔHf

° for OH−(aq) = −230.0 kJ/molΔHf

° for H2O(l) = −285.8 kJ/mol

Does the student have enough information?Answer: Yes, the standard reaction enthalpy is −55.8 kJ.The net ionic equation for the reaction is: H+(aq) + OH−(aq) → H2O(I)The standard reaction enthalpy is equal to the total formation enthalpy of the product (one mol H2O(l)) minus that of the reactants (1 mol H+ (aq) and 1 mol OH−(aq)). Therefore:

∆H mol H O l kJmol H O l

° = −

( ( )) .

( )1 285 8

22

−

++

+−( ( ))

( )( (1 0 1mol H g kJ

mol H gmol OH aqq kJ

mol OH aq

kJ

)) .( )

.

−

= −

−

230 0

55 8

Note that the standard enthalpy of formation of H+(aq) is zero by definition.


TEST YOURSELF


1. Use the thermochemical equation below to determine ΔH when 83.9 grams of BrF3 is formed.

Br2(l) + 3 F2(g) → 2 BrF3(l) ∆H = −511.2 kJ

2. Use the thermochemical equation below to determine ΔH when 103 grams of F2 is consumed.

Br2(l) + 3 F2(g) → 2 BrF3(l) ∆H = −511.2 kJ

3. Use the thermochemical equation below to determine ΔH when 189 grams of H2 is formed.

2 H2O(l) → 2 H2(g) + O2(g) ∆H = 571.6 kJ

4. When 9.39 grams of NaCl dissolves in 115 grams of water, the temperature of the water drops from 31.72°C to 26.90°C. Calculate q (heat, J/gsolute) for the solution pro-cess. Assume that heat capacity of the calorimeter and its contents is essentially that of the water.

5. When 10.0 grams of KBr dissolves in 119 grams of water, the temperature of the water drops from 34.34°C to 29.44°C. Calculate q (heat, J) for the solution process when 0.176 moles of KBr are dissolved. Assume that heat capacity of the calorimeter and its contents is essentially that of the water.

6. Calculate ΔH°rxn for the reaction: CaCl2(aq) + H2O(l) + CO2(g) → CaCO3(s) + 2 HCl(aq) from the following enthalpies of reaction at 25°C and 1 atm: (1) CaH2(s) + 2 HCl(aq) → CaCl2(aq) + 2 H2(g) ∆H° = −271 kJ (2) CaH2(s) + 2 H2O(l) → Ca(OH)2(aq) + 2 H2(g) ∆H° = −242 kJ (3) Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l) ∆H° = −96 kJ



7. Calculate ΔH°rxn for the reaction: N2O5(g) → N2(g) + 5/2 O2(g) from the following enthalpies of reaction at 25°C and 1 atm: (1) 2 H2(g) + O2(g) → 2 H2O(l) ∆H° = −571.6 kJ (2) N2O5(g) + H2O(l) → 2 HNO3(l) ∆H° = −73.7 kJ (3) 1/2 N2(g) + 3/2 O2(g) + 1/2 H2(g) → HNO3(l) ∆H° = −174.1 kJ

8. Calculate ΔH°rxn for the reaction: KClO3(s) + 3 PCl3(g) → 3 POCl3(g) + KCl(s) from the following enthalpies of reaction at 25°C and 1 atm: (1) 2 KCl(s) + 3 O2(g) → 2 KClO3(s) ∆H° = 78 kJ (2) P4(s) + 6 Cl2(g) → 4 PCl3(g) ∆H° = −1148.0 kJ (3) P4(s) + 2 O2(g) + 6 Cl2(g) → 4 POCl3(g) ∆H° = −2168.8 kJ

9. Calculate ΔH°rxn for the reaction: Ni2+(aq) + SO2(g) + 2 H2O(l) → SO4

2−(aq) + 4 H+(aq) + Ni(s) from the following standard enthalpies of formation ∆H°f for Ni2+(aq) = −54.0 kJ/mol ∆H°f for SO2(g) = −296.8 kJ/mol ∆H°f for H2O(l) = −285.8 kJ/mol ∆H°f for SO4

2−(aq) = −909.3 kJ/mol

10. Calculate ΔH°rxn for the reaction: 2A + 2B → 2C + 2D from the following standard heats of formation ∆H°f for A = −27 kJ/mol ∆H°f for B = −32 kJ/mol ∆H°f for C = −180 kJ/mol ∆H°f for D = −113 kJ/mol


11. Calculate ΔH°f for D given the following information: 3A + B → 2C + 3D ∆H°rxn = −504 kJ ∆H°f for A = −37 kJ/mol ∆H°f for B = −44 kJ/mol ∆H°f for C = −139 kJ/mol

12. Which of the following has ∆H°f = 0? A. Cu(s) B. Br2(g) C. CO2(g) D. Cl(g) E. Ag(l)

13. True or False: By definition, the standard enthalpy of formation of Br2(g) at 298K is zero.

ChemistryThe Core Concepts

About the Authors

Glenn V. LoDr. Glenn V. Lo is a Professor of Chemistry and Assistant Dean of the College of Arts and Sciences at Nicholls State University in Thibodaux, Louisiana. He has a B.S. in Chemistry (cum laude) from the University of the Philippines (1982), and a Ph.D. (Physical Chem-istry) from Kansas State University (1989). He did postdoctoral work at the Universite’ Joseph Fourier in France. His graduate work involved laser-induced fluorescence studies and spectral simulations. More recent research projects involved semiclassical dynamics simulations.

Dr. Lo’s main interest is in improving science education. He has published in the Journal of Chemical Education and the Chemical Educator. In 1999, he co-developed an online homework system used by thousands of high school and college chemistry students worldwide. An improved version of the system is currently accessible at http://i-assign.com. Dr. Lo’s recent projects include a professional development program for high school physical science teachers (sponsored by the Louisiana Systemic Initiatives Program) and the development of video assessment library for Chemistry (http://chemqa.blogspot.com). He is the director for the Region I tournament of the Louisiana science olympiad (http://www.nicholls.edu/scienceolympiad), which serves New Orleans and surrounding parishes (counties).

Michael A. JanusaMichael A. Janusa received his B.S. from Louisiana Tech University in 1985 and a Ph.D. from Louisiana State University in 1993. He taught chemistry at Nicholls State Univer-sity for ten years before accepting the position of Professor and Chair of the Chemistry Department at Stephen F. Austin State University in 2003. Dr. Janusa teaches General Chemistry, Inorganic, Analytical, and Environmental Chemistry. Known for his clear, well organized lectures, Dr. Janusa has received several awards for excellence in teaching. His research interest focuses on the environmental science area. He holds a U.S. patent in complexing heavy metals with bagasse (sugar cane by-product).

http://i-assign

http://chemqa.blogspot.com

http://www.nicholls.edu/scienceolympiad

http://www.nicholls.edu/scienceolympiad

Chemistry the Core Concepts Part I 2e - Janusa 2010

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