Ch03 lecture

© 2016 Pearson Education, Inc.Instructor’s Resources Materials (Download Only) for Principles of Chemistry: A Molecular Approach, 3/eNivaldo J. Tro

Catherine E. MacGowanArmstrong State University

Lecture Presentation

Chapter 3

Molecules, Compounds, and

Chemical Equations


Hydrogen + Oxygen = Water

• Elements can combine to form compounds.– Dalton’s atomic theory

• Law of definite proportions• Law of multiple proportions

• The properties of the compound are totally different from the constituent elements.


Chemical Bonds• Compounds are made of atoms held together by

chemical bonds.

• Bonds are attraction forces between atoms.– The bonding attraction comes from attractions between

protons and electrons.

• Types of chemical bonds– Ionic

• Bond formed between a metal and a nonmetal element – Covalent

• Bond formed between two nonmetal elements– Metallic

• Bond formed between two metal elements


Ionic Bonds

• Occur between a metal and a nonmetal element– Transfer of electron(s) from one element to another

• A metal atom can transfer an electron to a nonmetal.

• Ionic bonding results in the attraction between the two charges (cation plus anion).

• The resulting cation and anion are attracted to each other by electrostatic forces.– These forces are governed by Coulomb’s law. – Strength of bond can be calculated using Coulomb’s law.

• Compounds formed having ionic bonds are referred to as ionic compounds.


Formation of an Ionic Compound


What Is Covalent Bonding?• It occurs between nonmetal elements.

• The bond arises from the mutual attraction of two nuclei for the same electrons. – Electron sharing results.

• Bond is a balance between – attractive forces.

• (protons of atom A to electrons of atom B) – repulsive forces.

• (electrons of atom A to electrons of atom B) and (protons of atom A to protons of atom B)

Covalently bonded compounds are referred to as molecular compounds.


Chemical Formulas• A chemical formula represents the composition of the

substance.– It indicates the type and number of elements that make up

the molecule (compound).

• Example– C2H6O is the chemical formula for ethanol.

• The compound of ethanol is composed of two moles of carbon atoms, six moles of hydrogen atoms, and one mole of oxygen atoms.

• The molecular mass for ethanol is determined from its molecular formula.– Ethanol’s molecular mass is 46.0 grams/mole.– One mole of ethanol has a mass of 46.0 grams.


Types of Chemical Formulas• Molecular lists the exact number and types of atoms of

which the molecule is composed.• Example

– The hydrogen peroxide molecular formula is H2O2.

• Empirical lists the relative type and number of atoms in the compound.• It is the simplest ratio of atom to atom.• Example

– For the molecule hydrogen peroxide, H2O2, the empirical formula is (HO)n=2, where n means two units of (HO).

• For every H atom there is one O atom present.

• Structural represents how the atoms in the molecule are connected to one another.


Empirical Formulas • Empirical formulas

– do not give information on the compound’s shape or how atoms are attached to each other.

• Empirical formulas may or may not indicate how many of each type of atom is present in the compound.– Will if the molecular formula is the empirical formula

• Example – Water is H2O.

– Will not if the molecular formula is a multiple of the empirical formula • Example

– Oxalic acid is C2H2O4.

• The empirical formula for oxalic acid is CHO2.

• The molecular formula of oxalic acid is C2H2O4, which is two units of CHO2 or (CHO2)n=2.


Types of Chemical Formulas• Representing the molecule oxalic acid

• Molecular C2H2O4

• Empirical CHO2 or (CHO2)n=2

• Structural


Problem:Determine the empirical formula for the following:

1. An ionic compound that has two aluminum ions for every three oxide ions

2. Hexane, C6H14

3. Arabinose, C5H10O5


Answer:Determine the empirical formula for the following:

Al2O3

(CH2O)n where n = 2

1. An ionic compound that has two aluminum ions for every three oxide ions

2. Hexane, C6H14

3. Arabinose, C5H10O5

(C3H7)n where n = 2


Molecular and Empirical Problems


Molecular Models• Ball-and-stick models use balls to represent the atoms

and sticks to represent the attachments between them.– Gray = Carbon– White = Hydrogen

• Space-filling models use interconnected spheres to show the electron clouds of atoms connecting together.

• Models depict the three-dimensional structure for the compound methane. Methane’s chemical formula is CH4.


Classification of Elements and Compounds• Compounds composed of a metal element and a

nonmetal element are ionic compounds.– The cation and/or anion may be simple or polyatomic.– Examples

• Simple: chloride ion (Cl–), copper(II) ion (Cu2+), silicon(IV) ion (Si4+), or nitride ion (N3–)

• Polyatomic: nitrate ion (NO3–) or ammonium ion (NH4

+)

– Their formulas are sometimes referred to as ionic formulas because the bonding in the compound is ionic in nature.

• Compounds composed of nonmetal elements are referred to as molecular compounds.– Atoms are attached to each other in a molecular compound

through covalent bonding.


Classification of Elements and Compounds


Ionic versus Molecular Compounds

The compound propane contains individual C3H8 molecules.

The sodium chloride molecule, NaCl, is composed of an array of

Na+ ions and Cl– ions.


Predicting Formulas for Ionic Compounds• Ionic compounds are composed of

– metal + nonmetal elements.– metal + nonmetal polyatomic ions.

• Chemical formulas of ionic compounds are “neutral” or have a net charge of zero.

– Question: What is the molecular formula for Mn3+ and O2– (oxide ion)?

• Mn3+ has an oxidation state of 3+ and the oxide ion has an oxidation state of 2–.

• Multiply the absolute value of the respective oxidation states to derive the molecular formula – Mn2O3.

Mn: + 3 × 2 = + 6O: – 2 × 3 = – 6

0


A List of Common Charges (Oxidation States)

Element or Group Common Charge (oxidation state)• Alkali metals 1+ only

• Alkaline earth metals +2 only

• Halogens

Fluorine (F) –1 only

Other halogens –1

• 5A column nonmetals –3

• 6A column nonmetals –2

• Aluminum (Al) +3 only

• Hydrogen (H)

When combined with a nonmetal +1

When combined with a metal –1


List of Common Polyatomic Ions

Name FormulaAcetate C2H3O2

–

Carbonate CO32–

Hydrogen carbonate(aka bicarbonate) HCO3

–

Hydroxide OH–

Nitrate NO3–

Nitrite NO2–

Chromate CrO42–

Dichromate Cr2O72–

Ammonium NH4+

Name FormulaHypochlorite ClO–

Chlorite ClO2–

Chlorate ClO3–

Perchlorate ClO4–

Sulfate SO42–

Sulfite SO32–

Hydrogen sulfate(aka bisulfate) HSO4

–

Hydrogen sulfite(aka bisulfite) HSO3

–


Naming Ionic Compounds• Alkali and alkaline earth metals

– Metal name first, followed by nonmetal

– Nonmetal ending is exchanged with “IDE.”

• Example– MgCl2 magnesium

chloride

– KNO3 potassium nitrate

• Alkali (+1) and alkaline earth (+2) metals oxidation states are known.

That is why their charge is not indicated in the formula name!

Ionic compoundMetal and nonmetal

Name of metal’s cation

Base name of nonmetal’s anion + ide


Problem:Name the following ionic compounds:

1.NaI

2.Mg3N2

3.Al2S3


Answers:Name the following ionic compounds:

1.NaI

2.Mg3N2

3.Al2S3

sodium iodide

magnesium nitride

aluminum sulfide


Naming Ionic Compounds: Metal + Nonmetal for Transition and Main Group (“p” block) Metals• Metal name first, followed by nonmetal

• Metal’s oxidation state is indicated by Roman numerals.

• Nonmetal ending is exchanged with “IDE.”

– Examples• MnBr4 manganese(IV) bromide• Fe2O3 iron(III) oxide• SnF2 tin(II) fluoride

Name of metal’s cation

Base name of nonmetal’s anion + ide

Charge on metal’s cation in Roman numerals within parentheses


Problem:Name the following ionic compounds:

1.SnF2

2.Fe(NO3)3

3.Cu3(PO4)2


Answers:Name the following ionic compounds:

tin(II) fluoride

iron(III) nitrate

copper(II) phosphate

1.SnF2

2.Fe(NO3)3

3.Cu3(PO4)2


Hydrated Compounds versus Anhydrous Compounds

HydrateCuSO4 • 5H2O

AnhydrousCuSO4


Hydrated Compounds• Hydrates are ionic compounds containing a

specific number of waters for each formula unit.– Water of hydration often “driven off” by

heating

• In formula, attached waters follow. – CoCl2 • 6H2O

• In name, attached waters are indicated by prefix + hydrate after the name of the ionic compound.– CoCl2 • 6H2O = cobalt(II) chloride

hexahydrate– CaSO4 • ½H2O = calcium sulfate

hemihydrate

Prefix Number of Waters

hemi ½

mono 1

di 2

tri 3

tetra 4

penta 5

hexa 6

hepta 7

octa 8


Problem:Hydrated Compounds

1.What is the formula of magnesium sulfate heptahydrate?

2.What is the name of NiCl2 • 6H2O?


Answer: Hydrated Compounds

1.What is the formula of magnesium sulfate heptahydrate?

MgSO4 • 7H2O

2. What is the name of NiCl2 • 6H2O?

Nickel(II) chloride hexahydrate


Naming Molecular Compounds• Name the most “metallic” of the nonmetals first.

• Change the ending of the nonmetal to “IDE.”

• Indicate the number of atoms of the nonmetal by Latin prefixes.– Mono, di, tri, tetra, penta, hexa, hepta, octa, nona, deca– Examples

• P2O5 diphosphorus pentoxide• CO carbon monoxide

PrefixName of 1st

nonmetal elementBase name of 2nd nonmetal + idePrefix


Problem: Name the following molecular compounds:

1.CO2

2.SBr6

3.I2F7


Answer: Name the following molecular compounds:

1.CO2

2.SBr6

3.I2F7

carbon dioxide

sulfur hexabromide

diiodine heptafluoride


Problem: Write formulas for the following compounds:

1.Sulfur trioxide

2.Dinitrogen tetrahydride

3.Diarsenic trisulfide


Answer: Write formulas for the following compounds:

1.Sulfur trioxide

2.Dinitrogen tetrahydride

3.Diarsenic trisulfide

SO3

N2H4

As2S3


Molecular Compounds Known by Their Common Names

Water H2O

Ammonia NH3


Acids• Acids release hydrogen ions (H+) when dissolved in water.

– Example• HCl(aq) H+(aq) + Cl–(aq)

• Characteristics– Sour taste– Low pH values (<7)

• Types of acids– Binary acids

• have only two elements:– H+ and nonmetal anion.

– Oxyacids • H+ cation and polyatomic anion that contain oxygen atoms

(e.g., CO32–)


How to Name an Acid• The ide and ate endings of the anions change to ic.

NO3– + H+ HNO3

Nitrate ion Nitric acid

Cl– + H+ HCl Chloride ion Hydrochloric acid

• The ite endings of anions change to ous.

NO2– + H+ HNO2

Nitrite ion Nitrous acid


Problem:Name the following acids:

1.H2S(aq)

2.HClO4(aq)

3.HNO2(aq)


Answer:Name the following acids:

1.H2S(aq)

2.HClO4(aq)

3.HNO2(aq)

hydrosulfuric acid

perchloric acid

nitrous acid


Problem:Write the formula for the following acids:

1.Chlorous acid

2.Phosphoric acid

3.Hydrobromic acid


Answer:Write the formula for the following acids:

1.Chlorous acid

2.Phosphoric acid

3.Hydrobromic acid

H+ and ClO2–

HClO2

H+ and PO43– H3PO4

H+ and Br – HBr


Molecular Weight versus Molar Mass Atomic Weight versus Atomic Mass• They are all describing the same thing: What is the mass/weight

for one mole of something?

• Molecular weight or molecular mass • It is the sum of the atomic weights of all atoms in the molecule.

– Unit: grams/mole

– One mole of H2O has a molecular mass of 18.0 g/mol.

• Atomic weight or atomic mass• It is the atomic mass of one mole of the element.

– Unit: grams/mole– One mole of Cu atoms has an atomic mass of 63.5 g/mol.


Problem:What is the molar mass of ethanol, C2H6O?


Answer:What is the molar mass of ethanol, C2H6O?

1 mole contains

2 mol C (12.01 g C/1 mol) = 24.02 g C

6 mol H (1.01 g H/1 mol) = 6.06 g H

1 mol O (16.00 g O/1 mol) = 16.00 g O

Total = molar mass = 46.08 g/mol


Molecular Formulas, Molecular Mass, and Stoichiometry• Chemical formulas tell you the number and type of

atoms in a molecule.– The molecular mass of a compound is computed from

its chemical formula.

• The chemical formula and its corresponding molar mass can be used to convert between amounts of constituent elements and molecules.– Examples

• Percent composition• Moles


Mass to Mole Conversions• To go from mass to mole

• Mass (g) × (1 mol/atomic mass of element) = mole of

element

• Example

– 24.00 grams C × (1 mole carbon/12.011 grams)

= 2.00 moles of carbon atoms

Mass of element (g) Mole of element


Mole to Mass Conversions

• To go from mole to mass

• Mole of element × (atomic mass (element)/1 mole) =

mass (g)

• Example

– 8.00 mole He × (4.00 grams/1 mole He) = 32.0 grams He

atoms

Mass of element (g)Mole of element


Mass to Number of Atoms Conversions• To go from mass to number of atoms you must go through

the mole.

• Mass (g) × (1 mol/atomic mass) × (6.02 × 1023 atoms/1 mol) = # of atoms

• Example

– 45.99 g Na × (1 mol Na/22.99 g) ×

(6.02 × 1023 Na atoms/1 mol) = 1.204 × 1024 Na atoms

Mole of element

Mass of element (g)

Number of atoms of the element


Atoms to Mass Conversions• To go from atoms to mass you must go through the mole.

• (# of atoms) × (1 mol/6.02 × 1023 atoms) ×

(atomic mass/1 mol) = mass (g)

• Example

– 8 atoms Am × (1 mol/6.02 × 1023 Am atoms) ×

(243 g/1 mol Am) = 3.23 × 10–21 grams of Am

Mole of element

Mass of element (g)

Number of atoms of the element


Problem: How many Mg atoms are in a 0.200 gram sample?


Answer: How many Mg atoms are in a 0.200 gram sample?

1. Convert mass to moles.Mg has an atomic mass of 24.3050 g/mol.

0.200 g Mg × (1 mol Mg/24.31 g) = 8.23 × 10–3 mol Mg

2. How many atoms in this piece of Mg?

8.23 × 10–3 mol Mg × (6.02 × 1023 atoms Mg/1 mol)= 4.95 × 1021 atoms Mg


Molecules to Mass Conversions• To go from atoms to mass you must go through the mole.

• (# of molecules) × (1 mole/6.02 × 1023 molecules)

= mole of molecules

• Mole of molecules × (1 mol mass/1 mole) = mass (g)

• Example

– 8 molecules CH4 × (1 mole/6.02 × 1023 CH4 molecules) = 1.33 × 10–23

– 1.33 × 10–23 molecules of CH4 × (16.0 g/1 mole CH4)

= 2.13 × 10–22 molecules of CH4

Mole of molecules

Mass of compound (g)

Number of molecules


Problem:What is the mass of 4.78 × 1024 NO2 molecules?


Answer:What is the mass of 4.78 × 1024 NO2 molecules?

1. Convert molecules to moles.

4.78 × 1024 NO2 molecules × (1 mol NO2 molecules) (6.02 × 1023 molecules)

= 7.94 mol NO2

2. Mole to mass of NO2

7.94 mol NO2 × (46.0 g/1 mol NO2) = 365 g NO2


Problem: Determine the mass of table salt (NaCl) containing 2.4 g of Na.


Answer: Determine the mass of table salt (NaCl) containing 2.4 g of Na.1. Find the % Na in NaCl.

Na: Atomic mass is 22.99 g/mol.

A mole of salt has molecular mass of 58.44 grams.

22.99 g Na × (1 mol/58.44 g/mol NaCl) × 100 = 39.34%

This means that, in a 100.0 gram sample of NaCl, you have 39.34 grams Na.

2. Set up a ratio to determine mass of NaCl if you have 2.40 grams Na.

(2.4 g Na/x g NaCl) = (39.94 g Na/100 g NaCl)

x = 6.10 grams NaCl


Percent Composition • A pure compound always consists of the same elements

combined in the same proportions by weight.

• Therefore, we can express molecular composition as a– percent by weight or percent composition.

• Example– Ethanol’s molecular formula is C2H6O.– Its molecular mass is 46.0 g/mol.– Percent composition

52.31% C13.15% H34.72% O


Problem: Find the mass percent of Cl in C2Cl4F2.


Answer: Find the mass percent of Cl in C2Cl4F2.1. Determine the molecular mass of C2Cl4F2.

Molecular mass2 × (12.01 g/1 mol C) = 24.02 g4 × (35.45 g/1 mol Cl) = 141.80 g2 × (19.00 g/1 mol F) = 38.00 g

203.8 g/mol

2. Determine the mass Cl contributes to the compound’s mass.

4 × (35.45 g/1 mol Cl) = 141.80 g

3. Calculate Cl mass percent in the compound.

(141.80 g Cl/203.8 g/mol of C2Cl4F2) × 100 = 69.58%


Strategy for Determining an Empirical Formula and Its Molecular Formula by Percent Composition

Step 1: Determine the empirical formula from percent composition.

– To determine the empirical formula• Percent to mass• Mass to moles• Divide by the smallest.

– And if not whole (number), multiply until whole

– This strategy gives you the empirical formula of the compound.


Strategy for Determining an Empirical Formula and Its Molecular Formula by Percent Composition (continued)Step 2: Determine the molecular formula from the empirical formula

and molecular mass of the compound.

– To determine the molecular formula from percent composition you also need the molecular mass (g/mol) of the substance.• Empirical formula from percent composition• Determine the unit mass of the empirical formula. • Divide the mass of the empirical formula by the molecular

mass of the compound.• The whole number (round to the nearest whole number) is

the number of empirical units that make up the molecule.

– This gives you the molecular formula of the substance.


Problem: A compound of B (81.10%) and H has a molecular mass of 53.3 g/mol. Determine its molecular formula.


Answer: A compound of B (81.10%) and H has a molecular mass of 53.3 g/mol. Determine its molecular formula.Step 1: Determine the compound’s empirical formula.

1. Because it contains only B and H atoms, it must contain 18.90% H.100% – 81.10% = 18.90%

2. In a 100.0 gram sample there would be 81.10 g of B and 18.90 g of H according to the above percentages.

3. Calculate the number of moles of each constituent.81.10 g B × (1 mol B/10.81 g) = 7.502 mol B18.90 g H × (1 mol H/1.008 g) = 18.75 mol H


Answer (continued): A compound of B (81.10%) and H has a molecular mass of 53.3 g/mol. Determine its molecular formula.4. Now find the ratio of moles of elements in the compound.

Take the ratio of moles of B and H.

Always divide by the smaller number. (18.75 mol H/7.502 mol B) = 2.5 atoms H to 1 atom B(7.502 mol B/7.502 mol B) = 1 atom B to 1 atom B

5. But we need a whole-number ratio. (2.5 H atoms) × 2 = 5 H atoms(1 B atom) × 2 = 2 B atoms

Empirical formula = B2H5


Answer (continued): A compound of B (81.10%) and H has a molecular mass of 53.3 g/mol. Determine its molecular formula.Step 2: Determine the molecular formula from the empirical formula and

the compound’s molecular mass.

1. The empirical formula is B2H5.

– The mass of the empirical formula is 26.66 g/unit. 2. To determine the molecular formula you need to find the ratio of the masses

of the empirical formula to the molecular formula.

(53.0 g/mol/26.66 g per unit B2H5) = 2 units of B2H5

Molecular formula = B4H10


Combustion Analysis

• A common technique for analyzing compounds is to burn a known mass of compound and weigh the amounts of product made.– Generally used for organic compounds containing C, H, O

• By knowing the mass of the product and composition of constituent elements in the product, the original amount of constituent elements can be determined.

– All the original C forms CO2, the original H forms H2O, and the original mass of O is found by subtraction.

• Once the masses of all the constituent elements in the original compound have been determined, the empirical formula can be found.


Determining the Empirical and Molecular Formulas of a Compound from a Combustion Reaction

Problem: What is the empirical formula for an unknown compound CxHy given the following information:

A 0.115 g sample of a hydrocarbon, CxHy, is burned in oxygen and produces 0.379 g of CO2 and 0.1035 g of H2O.


Determining the Empirical and Molecular Formulas of a Compound from a Combustion ReactionProblem Strategy: What is the empirical formula for an unknown compound CxHy given the following information:A 0.115 g sample of a hydrocarbon, CxHy, is burned in oxygen and produces 0.379 g of CO2 and 0.1035 g of H2O.

1. Write a chemical equation describing the reaction.2. Assume

a) all of the carbon in CO2 comes from the combustion of the CxHy compound.b) all of the hydrogen in water (H2O) comes from the combustion of the CxHy

compound.

3. Steps 1 and 2 give you the mass of carbon and hydrogen in the unknown, so from this information you can compute the CxHy formula.


Answer:What is the empirical formula for an unknown compound CxHy given the following information:

A 0.115 g sample of a hydrocarbon, CxHy, is burned in oxygen and produces 0.379 g of CO2 and 0.1035 g of H2O. 1. Equation describing reaction

CxHy + some oxygen 0.379 g CO2 + 0.1035 g H2O

2a. Calculate the amount of C in CO2 (all carbon in CXHy went to the formation of CO2).

0.379 g CO2 × (1 mol CO2/44.0 g) = 8.61 × 10–3 mol CO2

8.61 × 10–3 mol CO2 × (1 mol C/1 mol CO2) = 8.61 × 10–3 mol C

8.61 × 10–3 mol C × (12.0 g C/1 mol) = 0.103 grams C

2b. Calculate the amount of H in H2O (all hydrogen in CXHy went to the formation of H2O).

0.1035 g H2O × (1 mol H2O/18.0 g) = 5.744 × 10–3 mol H2O

5.744 × 10–3 mol H2O × (2 mol H/1 mol H2O) = 1.149 × 10–2 mol H

1.149 × 10–2 mol H × (1.00 g H/1 mol) = 1.149 × 10–2 grams H


Answer (continued):What is the empirical formula for an unknown compound CxHy given the following information:

A 0.115 g sample of a hydrocarbon, CxHy, is burned in oxygen and produces 0.379 g of CO2 and 0.1035 g of H2O. 3. Determine the empirical formula.

0.103 grams C × (1 mol C/12.0 g) = 0.00861 mol C atoms0.01149 grams H × (1 mol H/1.00 g) = 0.01149 mol H atoms

The smaller value is mol of C.0.00861 C/0.00861 C = 1 C atom0.01149 H/0.00861 C = 1.337 H atom

It cannot have a 1/3 (0.33) part of an atom, so multiply by 3.1.337 H atoms × 3 = 4 H atoms

1 C atom × 3 = 3 C atomsThe empirical formula for CxHy is C3H4.


Problem: Determine the molecular formula of a CxHyOz by combustion.

An unknown compound is composed of only C, H, and O atoms. Determine the molecular formula for the compound given the following information:A 13.42 gram sample was ignited.It produced 39.61 g of CO2 and 9.01 g of H2O.

The unknown’s molecular mass is 268.34 g/mol.

CxHy Oz + some oxygen CO2 + H2O


Problem Strategy: Determine the molecular formula of a CxHyOz by combustion.1. Write a chemical equation describing the reaction.2. Assume

a) all of the carbon in CO2 comes from the combustion of the CxHyOz compound.

b) all of the hydrogen in water (H2O) comes from the combustion of the CxHyOz compound.

c) To calculate the oxygen content (grams), use the following:

[mass of compound – C (grams) – H (grams)] = mass of O (grams)

3. Steps 1 and 2 give you the mass of carbon, hydrogen, and oxygen in the unknown so, from this information, you can compute the empirical formula for CxHyOz.

4. Determine the molecular formula.


Answer:1. 13.42 g CxHy Oz + some oxygen 39.61 g CO2 + 9.01 g H2O

2a. Calculate the amount of C in COC2.

39.61 g CO2 × (1 mol CO2/44.0 g) = 0.900 mol CO2

0.900 mol CO2 × (1 mol C/1 mol CO2) = 0.900 mol C

0.900 mol C × (12.0 g C/1 mol) = 10.80 grams C

2b. Calculate the amount of H in H2O.

9.01 g H2O × (1 mol H2O/18.0 g) = 0.501 mol H2O

0.501 mol H2O × (2 mol H/1 mol H2O) = 1.00 mol H

1.00 mol H × (1.00 g H/1 mol) = 1.00 grams H

2c. To calculate the amount of O in the sample, use the following:10.80 g + 1.00 g = 11.80 g of the 13.42 gram CxHyOz compound is accounted for; therefore, the amount (grams) of O in the sample is 13.42 – 11.80 = 1.60 g O.


Answer (continued):3. Determine the empirical formula.

10.80 grams C × (1 mol C/12.0 g) = 0.900 mol C atoms1.00 grams H × (1 mol H/1.00 g) = 1.00 mol H atoms1.60 grams O × (1 mol O/16.0 g) = 0.100 mol O atoms

The smaller value is mol of O.0.100 O/0.100 O = 1 oxygen atom1.00 H/0.100 O = 10 hydrogen atoms0.900 C/0.100 O = 9 carbon atomsAll are whole integers, so the empirical formula for CxHyOz is C9H10O.

4. Determine the molecular formula.One mole unit of (C9H10O)n has a molecular mass of 134.0 g/mol.

The molecular mass of the compound was given as 268.34 g/mole.To get the molecular formula, use the following: (268.34/134) = n = 2 unitsThe molecular formula for the compound is C18H20O2.


What Are Chemical Equations?• They are a shorthand way of describing a chemical

reaction.

• They must be massed and their charges balanced.

• They provide information about the reaction’s– formulas of reactants and products.– states of reactants and products.– relative numbers of reactant and product molecules that

are required.

• They can be used to determine weights of reactants used and products that can be made.


Balanced Chemical Equations

• Because the same atoms are present in a reaction at the beginning and at the end, the amount of matter in a system does not change.

• The law of the conservation of matter– Because of the principle of the conservation of

matter, an equation must be balanced.– It must have the same number of atoms of the same

kind on both sides.

• Stoichiometry is based on the law of the conservation of matter.


A Balanced Chemical Equation

• Given this chemical reaction

4 Fe(s) + 3 O2(g) 2 Fe2O3(s)

• This equation means the following:

4 Fe atoms + 3 O2 molecules 2 molecules of Fe2O3

• Or you can say it says the following:

4 moles of Fe + 3 moles of O2 2 moles of Fe2O3


A Balanced Chemical Equation• Depict the kind of reactants and products and

their relative amounts in a reaction.

4 Fe(s) + 3 O2(g) 2 Fe2O3(s)

• The numbers in the front are called stoichiometric coefficients.

• The letters (s), (g), (l), and (aq) are the physical states of compounds.– Solid (s), gas (g), liquid (l), and aqueous (aq)

• Aqueous means “in solution with water.”


Chemical Equilibrium:Reaction Reversibility

• Some chemical reactions do not go to completion but reach a state where the forward rate of the reaction is the same as the reverse rate of the reaction. – The reaction is said to be at equilibrium.

• When writing a chemical reaction that is at equilibrium, a double arrow is used to indicate the reaction is at equilibrium.

Ca2+(aq) + 2 HCO3–(aq) CaCO3(s) + CO2(g) + H2O(l)


Balancing Equations Exercise

1. __C3H8(g) + __O2(g) __CO2(g) + __H2O(g)

2. __B4H10(g) + __O2(g) __B2O3(g) + __H2O(g)

3. __Pb(NO3)2(aq) + __KI(aq) __PbI2(s) + __KNO3(aq)

4. __Na3PO4(aq) + __Ca(NO3)2(aq) __Ca3(PO4)2(s) + __NaNO3(aq)

5. __NCl3(g) __N2(g) + ___Cl2(g)


Balancing Equations: The Answers

1. C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)

2. B4H10(g) + 1 1/2 O2(g) 2 B2O3(g) + 5 H2O(g)

This one: 2 B4H10(g) + 11 O2(g) 4 B2O3(g) + 10 H2O(g)

3. Pb(NO3)2(aq) + 2 KI(aq) PbI2(s) + 2 KNO3(aq)

4. 2 Na3PO4(aq) + 3 Ca(NO3)2(aq) Ca3(PO4)2(s) + 6 NaNO3(aq)

5. 2 NCl3(g) N2(g) + 3 Cl2(g)


A Quick Review of Organic Chemistry

Classifying compounds: organic versus inorganic

• In the eighteenth century, compounds from living things were called organic; compounds from the nonliving environment were called inorganic.

• Organic compounds easily decomposed and could not be made in an eighteenth-century lab.

• Inorganic compounds are very difficult to decompose, but can be synthesized.


Organic Compounds• Organic compounds are made mainly of C and H,

sometimes with O, N, P, S, and trace amounts of other elements.

• The main element that is the focus of organic chemistry is carbon.

• Carbon atoms bond almost exclusively covalently.– Compounds with ionic bonding C are generally

inorganic.

• When C bonds, it forms four covalent bonds.– Four single bonds, two double bonds, one triple plus

one single, etc.


Simple Organic Compounds

• Carbon is unique in that it can form limitless chains of C atoms, both straight and branched, and rings of C atoms.

• Organic compounds with only carbon and hydrogen atoms in their structure are classified as hydrocarbons.–The simplest hydrocarbon is the

molecule methane, CH4.


Hydrocarbons

Ch03 lecture

Economy & Finance

Transcript of Ch03 lecture