Lecture1 CH315 2013 Winter Answers

Welcome to CH 315

Analytical Instrumentation

Course Description

� This course allows students to apply their skills and knowledge of analytical chemistry and instrumentation to the theory and practice of chemical sampling and analysis used by the MOE (Ministry of the Environment).

� The problems presented in the course will emphasis the chemical analysis of environmental samples utilizing MOE procedures and equipment available at the college.

Course Learning Outcomes

� Upon successful completion students will be able to:

� Identify and follow appropriate procedures for sample collection, preparation and analysis.

� Solve mathematical problems related to concentration and dilution.

� Describe, explain and solve problems related to redoxtitration.

� Operate AAS, GC, HPLC, Ion Selective Electrodes and Spectrophotometers.

� Collect, record, analyze, interpret and present data obtained from experiments with the above instruments according to MOE procedures.

Dr. Andrew BaerOffice Hours

Monday: 12:30 pm – 2:00 pm (HPSTC 428)

Wednesday: 10:30 am – 12:00 pm (HPSTC 428)

Friday: 12:30 pm – 2:00 pm (HPSTC 428)or by appointment

Email: [email protected]

Lecture:Thursday: 10:30 am – 12:20 pm (Sec 061-062) (HPSTC 422)

Textbooks:

Harris, Daniel C., Exploring Chemical Analysis, 5th Ed., W. H. Freeman and Company, 2012.

Harvey, David, Analytical Chemistry 2.0

Available Online at: http://www.asdlib.org/onlineArticles/ecourseware/Analytical%20Chemistry%202.0/Text_Files_files/AnalChem2.0.pdf.

ReviewApril 15th

Final Exam (Exam Week)April 22nd

Ion Selective ElectrodesApril 8th

Week Topic Notes

Jan. 7th Concentration, Dilution and Statistics

Jan. 14th Spectrophotometry/AAS

Jan. 21st Quality Assurance

Jan. 28th Sampling and Sample preparation

Feb. 4th Introduction to Chromatography

Feb. 11th Midterm 1 Tentative Date

Feb. 18th Introduction to Chromatography

Feb. 25th GC

March 4th Reading Week

March 11th HPLC

March 18th Redox and Redox titrations

March 25th Midterm 2 Tentative Date

April 1st Redox and Redox titrations

Course Overview

Evaluation Assignments/Quizzes = 5.0 %

Midterm test 1 = 12.5 %

Midterm test 2 = 12.5 % Final Exam = 25.0 %

Laboratory = 45.0 %

Course MaterialsCourse lectures will be available on the course

eCentennial site. I would like everyone to print out a copy

of the notes and bring them to the lecture. You will really struggle if you don’t have the notes. You will need to

scribble and add things to the notes during class.

Course Evaluation

Labs:Friday: 2:30 pm – 5:20 pm (Sec 061) (HPSTC 440)

Friday: 5:30 pm – 8:20 pm (Sec 062) (HPSTC 440)

Laboratory Requirements:

Lab Manual: CH315 Analytical Instrumentation

(available for download from eCentennial)

Lab Coat, Safety Glasses, Long Pants, Closed Shoes

A notebook and pen

Locker (contact Enrolment Services for rental)

Laboratory

Final Lab due.Makeup LabApril 19th

Exam Week (No Labs)April 22nd

Labs due as appropriate.End Lab Rotation (Lab 3 – 8)April 12th

Labs due as appropriate.Lab Rotation (Lab 3 – 8)April 5th

No labs dueGood FridayMarch 29th

Labs due as appropriate.Lab Rotation (Lab 3 – 8)March 22nd

Labs due as appropriate.Lab Rotation (Lab 3 – 8)March 15th

No labs dueReading weekMarch 8th

Labs due as appropriate.Lab Rotation (Lab 3 – 8)March 1st

Labs due as appropriate.Lab Rotation (Lab 3 – 8)Feb. 22nd

Labs due as appropriate.Lab Rotation (Lab 3 – 8)Feb. 15th

Lab 2 dueBegin Lab Rotation (Lab 3 – 8)Feb. 8th

Lab 2 (all students)Feb. 1st

Lab 1 dueLab 2 (all students)Jan. 25th

Lab 1 (all students)Jan. 18th

Introduction / SafetyJan. 11th

NotesTopicDate

Laboratory Overview

Ways of Expressing

Concentrations of Solutions

Concentration expressions

• Molarity

• Normality

• Parts per million

• Parts per billion

• Parts per trillion

• The concentration of a solution is the amount of

solute present in a given quantity of solvent or solution.

• High to Mid-range Concentrations:

� Molarity (mol/L) or Normality (mol/L)

• Very Low Concentrations:

� ppm or mg/L

� ppb or µg/L

� ppt or ng/L

Concentration Expressions

Molarity

• You should be able to calculate molarity:

solution L

solute mol molarity M ==

Calculating Molarity

� What is the molarity of a solution made by dissolving 2.355g of sulfuric acid (H2SO4) in water and diluting to a

final volume of 50.0 mL?

• What mass of KCl would you need to prepare 250.0 mL of a 0.340 M solution of KCl (MW = 74.55 g/mol) in water:

Concentration Units (Molarity)

• Calculate the molarity of a solution containing 34.2 g of NaCl(s) (MW = 58.5 g/mol) in 250.0 mL of solution.

Concentration Units (Molarity)

• The Normality of a solution is defined as the number

of equivalents of solute per liter of solution.

• What is an equivalent?

solution L

equivalent of # normality N ==

Concentration Units (Normality)

• An equivalent of a substance is the amount of that substance in moles which supplies or consumes one

mol of reactive species.

• The equivalent weight is the mass of a substance which supplies or consumes one mol of reactive

species.

capacity reaction

weightmolecular weightequivalent =

weightequivalent

mass equivalent of =#


• In acid-base chemistry:

• For an acid:

� Reaction capacity = # of H+ produced by the acid.

� The equivalent weight of acid is equal to its molecular weight divided by the number of hydrogen atoms produced by the acid.

• For a base:

� Reaction capacity = # of H+ that the base would react with.

� The equivalent weight of a base is equal to its molecular weight divided by the number of H+ the base would react with.


• Determine the reaction capacity of the following:

– HCl

– H2SO4

– H3PO4

– CH3OOH

– NaOH

– Ba(OH)2

– Na2CO3


• What is the Molarity and Normality of a solution made by dissolving 17.5 grams of NaOH in 200 ml of water?



• What is the Molarity and Normality of a solution made by dissolving 49.0 grams of Na2CO3 in 1 L of water?

Dilutions

• The ability to prepare a solution of a specific concentration from a stock is very important in analytical chemistry:

2211 VC VC ×=×

Calculating dilutions

� How would you prepare 500.0 mL of 0.2500 M NaOHsolution starting from a concentration of 1.000 M?

Balanced equation:

aA + bB cC + dD

1. Find the number of moles of the first reactant (A).

2. Use the coefficients of the balanced equation to find the mole ratio of the first reactant (A) to the second reactant (B).

3. Calculate the number of moles of the second reactant (B).

4. Use this to find the amount of the second reactant.

Solution Stoichiometry

• Molarity is useful in performing stoichiometry calculations on substances in solutions.


Stomach acid, a dilute solution of HCl in water, can be neutralized by reaction with NaHCO3, according to the equation:

HCl(aq) + NaHCO3(aq) NaCl(aq) + H2O(l) + CO2(g)

How many milliliters of 0.125 M NaHCO3 solution are needed to neutralize 18.0 mL of 0.100 M HCl?


2. Check the coefficients of the balanced equation:

1 mol HCl reacts with 1 mol NaHCO3

1. Find the number of moles of HCl in 18.0 mL of a 0.100 M solution:

mol10 x 1.80L 1

mol 0.100

mL 1000

L 1mL 18.0 HCl of Moles 3-=××=

L 01440NaHCO mol 0.125

solution L 1

HCl mol 1

NaHCO mol 1HCl mol 10 x 1.80

3

33-.=××

3. Calculate the volume of the 0.125 M NaHCO3 solution needed:

4. 14.4 mL of 0.125 M NaHCO3 solution is needed to

neutralize 18.0 mL of the 0.100 M HCl solution.

Calculating Dilutions

• If 15.0 mL of a 12.0 M HCl solution were diluted to a volume of 100.0 mL, what would the concentration of the new solution be?

• How would you prepare 60.0 mL of 0.20 M HNO3 from a stock solution of 4.00 M HNO3?

Calculating Dilutions

• In chemistry sometimes substances are at very low concentrations.

• They are often expressed as parts per million (ppm), parts per billion (ppb) or parts per trillion (ppt).

610×=sample of mass

substance of massppm


substance of massppb


substance of massppt

Concentration Units (ppm, ppb, ppt)

• 1 ppm = 1 piece in 1,000,000 pieces

• In an aqueous solution 1 ppm = 1 mg/L (assuming 1

kg water = 1 L water).

• It is the equivalent of 30 seconds out of a year.

How much is a part per million (ppm)?


substance of massppm

How much is a part per billion (ppb)?

• 1 ppb = 1 piece in 1,000,000,000 pieces

• In an aqueous solution 1 ppb = 1 µg/L (assuming 1 kg water = 1 L water).

• It is the equivalent of 3 seconds out of a century.


substance of massppb

How much is a part per trillion (ppt)?

• 1 ppt = 1 piece in 1,000,000,000,000 pieces.

• In an aqueous solution 1 ppt = 1 ng/L (assuming 1 kg

water = 1 L water).

• It is the equivalent of 3 seconds out of one hundred

thousand years.


substance of massppt

• An 11.0 kg sample of soil was extracted and produced 2.732 g of oil. How many ppm of oil were in the soil?

Mass of total sample = 11.0 kg = 1.10 x 104 g

Mass oil = 2.732 g

ppm = (2.732 g / 1.10 x 104 g) x 106

= 248.4 ppm = 248. ppm

Using ppm in a calculation

• Calculate the ppm concentration of 2.7 x 10-3 mg of gold (Au) in 450 mL of aqueous solution:

• The density of a dilute water sample is 1.00 g/mL.

Mass of gold in g = 2.7 x 10-3 mg / 1000 = 2.7 x 10-6 g

Mass of water = 450 mL (1.00 g/mL) = 450. g

ppm = (2.7 x 10-6 g / 450. g) x 106

= 0.006 ppm


• Calculate the concentration in ppm of K present in a solution prepared by dissolving 2.0 g of pure KNO3 and

diluting to a volume of 1L

• MW(KNO3) = 101.11 g/mol, MW (K) = 39.1 g/mol


• Calculate the concentration of Na+ in ppm, present in a solution prepared by dissolving 1.002 g of pure NaBr and

diluting to a volume of one liter.

• Calculate the molarity of Na+ as well.

• NaBr = 23.0 g/mol + 79.9 g/mol = 102.9 g/mol


• Calculate the mass of KCl required to make 1 L of a 1000 ppm K+ stock solution.

• K = 39.1g/mol

• KCl = 74.6g/mol


• Calculate the mass of CaCl2 needed to prepare 1000mL of a solution that is 100 ppb in “Ca”.

Using ppb in a calculation

Statistics

Evaluation of Analytical Data

Mean ( ):

• Also known as the average.

• Obtained by dividing the sum of replicate measurements by the number of measurements in the set.

• Given mathematically as:

where xi represents the individual values of x making up the

set of N replicate measurements.

x

N

x

x

N

i

i∑== 1


• Formula for sample and population means:

• For an infinite set of data the mean is called the

population mean (µ) and is calculated in the same manner as the sample mean.

• Since an infinite set of data cannot exist, the population mean can never actually be measured.

• However as the number of measurements in a set of data increases x approaches µ.

N

x

x

N

i

i∑== 1

N

x

N

i

i∑== 1µ


Standard deviation:

• Standard deviation is a measure of how closely data is clustered around the mean.

• A small standard deviation means the data is clustered close to the mean. A large standard deviation means the data can deviate significantly from the mean.

• The standard deviation is a measure of the precision of the data.


• The population standard deviation (σ) is the standard deviation of an infinite set of data. It can never be measured but is approached as the number of measurements increases.

( )

N

x

N

i

i∑=

−

= 1

2µ

σ


• The sample standard deviation (s) should be used in place of the population standard deviation (σ) when analyzing small sample sets:

• In the sample standard deviation N (population size) is replaced in the denominator of the standard deviation equation with (N-1), which is known as the number of degrees of freedom.

• As a rule of thumb, the sample set is considered small when N < 30.

( )

1

1

2

−

−

=∑

=

N

xx

s

N

i

i

Calculating Standard Deviations

� 5 samples of copper ore were analyzed to determine their copper content yielding values of 16.54%, 16.64%, 16.61%, 16.67% and 16.70%.

Calculating Standard Deviations

� 5 samples of aspirin were analyzed to determine their aspirin content yielding values of 75.54%, 75.34%, 75.61%, 74.67% and 74.70%.


Confidence Intervals:

• It is impossible to determine the population mean (µ) or population standard deviation (σ) from a small data set.

• A confidence interval (CI) is a range around a measured mean in which there is a specific

probability of finding the population mean.

Confidence Intervals

s = measured standard deviation

n = number of observations

t = Value of Student’s t

n

tsxµ ±=

Confidence IntervalsValues of Student’s t:

Calculating Confidence Intervals

� 5 samples of aspirin were analyzed to determine their aspirin content yielding values of 75.54%, 75.34%, 75.61%, 74.67% and 74.70%.

� Find the 50% and 90% confidence intervals for the aspirin content:

Lecture1 CH315 2013 Winter Answers

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