Welcome to Chem 321
A key course to your future success
ANALYTICAL CHEMISTRY
COURSE OUTLINE
Instructor: Dr. X. Nancy Xu
Office: Alfriend Chemistry Building Room 201
Office Hours: MW 7:10-8:00 PM (OCNPS 200)
Right after the class
Teaching Assistant: Dr. Tao Huang and Jill Lowman
Email: [email protected]
MW 5:45 – 7:00 PM
OCNPS Room 200
Class Time & Place
MATH 102 or equivalent
CHEM 115, 116.
Pre-requisites
Textbook
Quantitative Chemical Analysis, 7th edition, D. Harris, Freeman, 2007.
Reference Book
Fundamentals of Analytical Chemistry, 7th Ed.
Skoog/West/Holler, Saunders Publishing, 1996
One copy of the book is placed in the reserved desk of the library.
About Textbook
PrefaceChapter 1-16 (Chem 321) Summary
Questions and Problems
Answers to the problems: in the reserved desk in the library
Glossary G-1… on the back of the book
Chapters heavily associated with Chem 322
Course Website• Class notes, assigned problems and announcements
are posted at http://www.odu.edu/sci/xu/chem321/chem321.htmhttp://www.odu.edu/sci/xu/chem321/chem321.htm
• Visit this site as frequently as possible, especially before and after every class
• You will need Acrobat Reader to download large documents
Honor Code
• Students registered for CHEM 321 are
expected to obey the ODU Honor Code!!!
AttendanceAttendance at class meeting is compulsory.
Regular and punctual class attendance is required of all students.
If you are absent, you will be responsible for everything covered in class including any handouts (e.g., problem sets, answers, sample tests, etc).
Absence from ExamsMissed exams may be made up only with a medical excuse (written by physician or health center), death or hospitalization in the family.
Without these reasons, one will earn a “zero”grade on that test.
No exceptions will be made. Please contact me prior to the exam if you must be absent.
HomeworkAssigned homework is mandatory.
Working on these assignments in a timely manner is the best way to learn the material and get good grades.
I may check your assignments and sample your homework on the due date even though it may not be graded.
The brief answers to assigned problems are available in the solution book, which is placed in the reserved desk in the library .
Exams
• 4 unit tests and a comprehensive final examination
Grading• Average cumulated grade = unit tests
(70%) and final exam (30%)
100-93 = A 92-90 = A-
89-87 = B+ 86-83 = B 82-80 = B-
79-77 = C+ 76-73 = C 72-70 = C-
69-67 = D+ 66-63 = D 62-60 = D-
<60 = F
Tentative Timeline
Date Chapter Topic Homework
01/08 1 Measurement 1-12, 17, 22, 28, 31, 32, 34 01/10 3 Experimental Error 3-1, 2, 5, 10, 12, 13, 15, 16, 22 01/15 Martin Luther King Holiday- No classes 01/17, 22 4 Statistics 4-A, E, F, 11, 14, 17, 20 01/24 1st Unit Test (Chapter 1, 3, 4) 01/29, 31 6 Chemical Equilibrium 6-4, 15, 16, 19, 22, 25, 30, 37, 47, 48, 5302/5, 7 7 Let the Titrations Begin 7-B, F, 7, 22, 23, 28, 36 02/12, 14, 19
8 Activities and Systematic Treatment of Equilibrium
8-A, B, C, E, F, G, 2, 4, 11, 12, 18, 20, 24, 26
02/21 9, 11 Monoprotic Acid-Base Equilibrium 9-B, D, E, 2, 5, 6, 22, 23, 27, 28, 33 02/26 6-9 Recitation and help session 02/28 2nd Unit Test (Chapter 6-9) 03/05-10 Spring Break- No classes 03/12, 14 10, 11 Polyprotic acid-Base Equilibrium 10-A, B, D, 4, 11, 17, 18, 23 03/19, 23 11 Acid-Base Titration 11-A, B, F, 3, 6, 14, 19, 23 03/26,28 12 EDTA Titration 12-C, D, 1, 2, 3, 7, 8, 24, 34 04/02 3rd Unit Test (Chapter 10-12) 04/04, 09 14 Fundamentals of Electrochemistry 14-B, C, D, 2, 10, 25 04/11,16 15 Electrodes and Potentiometry 15-4, 8, 9, 21, 26 04/18 16 Redox Titration 16-A, 1, 2, 6, 13 04/23 4th Unit Test (Chapter 14-16) 04/30 3:45-6:45 PM Comprehensive Final Examination (Chapter 1-16, except 2, 5 and 13)
Note to the class
• Welcome to Chem 321, a key course to your future success in analytical data analysis and experimental design of any chemical related problems in every scientific disciplinary including chemistry, biology, biotechnology, forensic science, food science, material science, medicine, environmental science, etc.
• Whether you aim to be a brilliant scientist, a medical doctoral, a technician or a good cook, you will find this course essential and helpful for your future career.
What are we going to do in the class ??
• We will introduce basic analytical concepts and theories and focus on data analysis and interpretation.
Six tips for success in this class• actively attending every class and take good
notes • improving your math (good math is absolutely
essential)
• completing all homework and study textbook, class notes and slides
• asking questions
• coming to Dr. Xu’s office hours
• visiting course website frequently.
Good-luck to you!!
---Dr. Nancy Xu
Chapter 1: Introduction & Measurements
What is Analytical Chemistry?
“Analytical Chemistry is what analytical chemists do.”
--- C. N. Reilley
Roles of Analytical ChemistryPlay a vital roles in all sciences, just to name of a few: chemistry, biology, biotechnology, forensic science, food science, material science, medicine, environmental science, etc.
Examples of achievement of Analytical chemistry:
Breakthrough of the century: Sequence of Human Genome, Analytical chemists-capillary electrophoresis.
Classification of AnalysisQualitative Analysis
determination of chemical identity of the species in the sample.
Quantitative Analysisdetermination of the amount of species or analytes, in numerical terms. Hence, Math is heavily involved.
In order to perform quantitative analysis, typically one needs to complete qualitative analysis.
One needs to know what it is and then select the means to determine the amount
Classification of Quantitative Methods of Analysis
Gravimetric Method: mass is measured.
Volumetric Method: volume is measured or used to determine amount of sample via concentration.
Instrumental Method: use an instrumental technique to assay the amount of sample:
Such as: Electroanalytical based upon electron-transferSpectroscopy including mass spectrometry
(***hot) (proteomic)
Steps in a Typical Quantitative Analysis
Units
SI Units
1 attoliter = 1x10-18 L
1 L = 1x1018 attoliter
1 mL = 1x10-3 L1 L = 1x103 mL
1 μL = 1x10-6 L1 L = 1x106 μL
Weight & Mass
Weight is the force of gravitational attraction between that matter and the earth.Location dependenceMass is an invariant measurement of the amount of matter.Location independenceW = mg
I. Review of Stoichiometry - please refer to your freshman chemistry text to review this concept.
A. Empirical vs molecular or structural formulas:1. Empirical formulas give information only about the simplest ratio between the different elements composing the molecule.
Example: HO, H2CO
2. Molecular formulas give information about the numbers of atoms of each element found in the molecule.
Example: H2O2, H4C2O2, C2H5OH, C2H4O2, C3H6O3, C6H12O6
3. Structural formulas give information about the structure of the molecule as well as the numbers of atoms of each element
Example: HOOH, (CH3)3COH, CH3CH2OCH2CH3,
B. The Mole (mol):
A unit which defines the number of units of a chemical species (molecules, atoms, ions, etc.) and from which we can calculate the weight of the species if we have a knowledge of the chemical formula of that species.
Avogadro Number:
6.022 x 1023 particles is one mole.
Example I-1
Calculate molecular weight (molar mass) of CH2O(Formaldehyde) and C6H12O6 (glucose)1 mole of C = 12.0 g1 mole of H = 1.0 g1 mole of O = 16.0MW of CH2O =12.0 + 2x1.0 + 16.0 =30.0 g/mole
MW of C6H12O6=6x12.0+12x1.0+6x16.0=180.0 g/mole
Example I-2
How many moles and millimoles are contained in 2.00 g of pure benzoic acid (C6H5COOH) (MW = 122.1 g/mol)
Moles of benzoic acid = mol0164.01.122
00.2MW
Weight ==
mmoles of benzoic acid = moles x1000 =16.4 mmol
Solution
Solute: a minor species in a solution
Solvent: a major species in a solution
(e.g., water is a solvent for aqueous solution)
Chemical ConcentrationsMolar Concentrations:
Molarity (M) = solute/solution (mol/L)Molality (m) = solute/solvent (mol/kg)
Percent Composition
Weight percent
Parts per million (ppm)Parts per billion (ppb)
Volume percent
MolarityMolarity
The number of moles of species (x) dissolved in 1 L of solution
Molar ConcentrationMolar Concentration
mol/L = M(L)Solution of Volume
(mole) Solute of Moles ofNumber Cx =
(g/mol) solute ofMW
(g) Solute ofWeight (mole) Solute of Moles ofNumber =
Analytical Analytical MolarityMolarity::The total number moles of a solute (regardless its chemical state) in one liter of solution
Example-I-3:117.0 g of NaCl dissolved in 1.00 L of water has an analytical concentration of _____ mol /L MW of NaCl = 58.5 g/mol
ConcentrationConcentration
mole00.25.580.117
MWW NaCl of Mole
NaCl
NaCl ===
M00.2 mole/L00.2L00.1
moles00.2(L) V
MoleC NaClNaCl ====
Example I-4:294.0 g of H2SO4 dissolved in 1.00 L of water has an analytical concentration of ____ mol/L MW of H2SO4 = 98.0 g/mol
moles00.3g/mol0.98
g0.294MWW
SOH of Mole42
42
SOH
SOH42 ===
M00.3 mole/L00.3L00.1
moles00.3(L) V
MoleC lSOH
SOH42
42====
Molality
(kg)solvent of Weight (mole) Solute of moles ofNumber Cweight =
Density:Density:
Expresses the mass of a substance per unit volume.
In SI Units: density unit --- g /mL or kg/LExample: Density of water is approximately 1.00 g/mL at 40C.
Weight = Volume x density
DensityDensity
Practice at Home
Assume density of solution in Examples I-3 and I-4 = 1 g/ml3,
what is the molality of the solution in Examples I-3 and I-4, respectively?
Percent Concentration
1.
2.
3.
%100xsolution of mass
solute of mass (w/w)percent Weigh =
%100xsolution of volume
solute of volume (v/v)percent Volume =
%100x (mL)solution of volume
(g) solute of mass
(w/v)percent umeweight/vol
=
Example I-5:What is the w/w % of an aqueous ammonia (NH3) solution at
14.3 M, with density = 0.900 g/mL (900 g/L)?
Solute : NH3
1) MW of NH3 = 17.0 (g/mol)
2) Mole of NH3 at 14.3 M in 1.00 L=14.3 (mol/L) x 1.00 L=14.3 mol
3) Weight of NH3 at14.3 M in 1.00 L = Mole of NH3 x MW of NH3= 14.3 mol x 17.0 (g/mol) = 243 g
4) Weight of 1.00 L solution = volume x density = 1.00 (L) x 900(g/L) = 900 g
27.0%x100%900
243x100%
solution of mass
solute of mass (w/w)percent Weigh ===
Example I-6:What is the molar concentration of an aqueous ammonia (NH3)
solution with density = 0.900 g/mL (900 g/L) and 27.0% (w/w)?
Solute : NH3MW of NH3 = 17.0 (g/mol)
Weight of NH3 = Mole of NH3 x MW of NH3 =CNH3VNH3 x MW of NH3
Weight of 1.00 L solution = volume x density = 1.00 (L) x 900 (g/L) = 900 g
27.0%x100%900
Wx100%
solution of mass
solute of mass (w/w)percent Weigh 3NH
===
W NH3 = CNH3VNH3 x MW of NH3 = 243 g
CNH3 = 14.3 mol/L = 14.3 M
Example I-7: What is the v/v% of ethanol in a solution prepared by mixing 5.00 mLof ethanol with enough water to give 1.00 L of solution?
Solute: ethanol
1) Volume of solute (ethanol) = 5.00 mL = 5.00 x10-3 L2) Volume of solution = 1.00 L
3)
%500.0%100L00.1
L1000.5
%100xsolution of volume
solute of volume (v/v)percent Volume
3
==
=
−
xx
pph10x solution of mass
solute of mass (w/w) C 2pph =
ppt10x solution of mass
solute of mass (w/w) C 3ppt =
parts per hundred (pph) & parts per thousand (ppt)
ppm10x solution of mass
solute of mass (w/w) C 6ppm =
ppb10x solution of mass
solute of mass (w/w) C 9ppb =
parts per million (ppm) & parts per billion (ppb)
ppt10x solution of mass
solute of mass (w/w) C 12ppt =
parts per trillion (ppt)
Stoichiometric Calculations
II. Stoichiometric Calculations:
stoichiometric calculations are based on the combining ratios of reactants which result in specific products .
They are expressed in terms of moles.
when you are given the mass of a reactant or product, you should first convert the mass to moles to determine the amount of reactant that will be consumed or product that will be produced for a given reaction.
If the final answer is to be given in a mass unit, then the moles must be converted to grams.
Example I-8(a)What mass of AgNO3 (MW= 169.9 g/mol) is needed toConvert 2.33 g of Na2CO3 (MW=106.0 g/mol) to Ag2CO3 ?
2 mol1 mol
0.10633.2
9.169W
3AgNO
9.169W
0.10633.2
21
3AgNO
= 0.10633.22
9.169W
1 3AgNO xx =
g47.79.1690.106
33.22W3AgNO == xx
Na2CO3 (aq) + 2AgNO3 (aq) Ag2CO3(s) ↓+ 2 NaNO3 (aq)
Example I-9
1 mol1 mol
0.10633.2
7.275W
32COAg
7.275W
0.10633.2
11
32COAg
= 0.10633.21
7.275W
1 32COAg xx =
g06.67.2750.106
33.21W32COAg == xx
Na2CO3 (aq) + 2AgNO3 (aq) Ag2CO3(s) ↓+ 2 NaNO3 (aq)
(b) What mass of Ag2CO3 (MW= 275.7 g/mol) will be
formed?
Preparation of Solution
Solute
Solvent
Total Solution
Example I-10:To prepare a solution with 0.500 M of Cl- from BaCl2 • 2H2O. How much of the BaCl2 • 2H2O must be used to prepare 1.00 liter of solution? Assume BaCl2 completely dissociates:BaCl2 -----> Ba+2 + 2 Cl-1 mole 2 molex mole 0.500 (mol/L) x 1.00 (L)
X mole = Mole of BaCl2 (mole) = (0.500/2) = 0.250 (mole)
MW of BaCl2 • 2H2O = 244.2 g/mol
W of BaCl2 • 2H2O = Mole of BaCl2 (mole) x MW of BaCl2 • 2H2O = 0.250 mol x 244.2 g/mol= 61.1 g
0.500x1.00mole
VCmole
molemole
21 2
--
2
-
2 BaCl
ClCl
BaCl
Cl
BaCl ===
Dilution Dilution the number of moles are the same in dilute and concentrated solutions!
moles = CconcentratedVconcentrated = CdiluteVdilute
Units: V: either in L and mLC: M (mol/L) or mM (mmol/L)
Be sure to match units for both dilute and concentrated solutions
Volumetric CalculationsVolumetric Calculations
Example I-11:
An aliquot of a HCL solution (1.00 M) has been diluted with three equal aliquots of H2O to prepare 1:4 aq HCl. In other words, the HCl has been diluted fourfold. The dilute concentration of the HCl is:
M)(25.041x00.1
V xV(M)C
(M)Cdilute HCl,
edconcentrat HCl,edconcentrat HCl,dilute HCl,
==
=
moles = CconcentratedVconcentrated = CdiluteVdilute
Example I-12:
How many L of a 50 mol /L solution are How many L of a 50 mol /L solution are required to make 200 required to make 200 mL mL of a 1 mol Lof a 1 mol L--11
solution?solution?
50 (mol /L) * Vconc. = 1 (mol/L) * 0.200 L
Vconc. (L) = (1 M * 0.200 L) / 50 M
= 0.004 (L)
moles = CconcentratedVconcentrated = CdiluteVdilute
Summary
Course Overview:
Classification of Analysis :
Units:
Molecular Weight:
Mole:
Concentrations:
Stoichiometric Calculations:
Preparation of Solution:
Homework
Chapter-1: 1-12, 17, 22, 28, 31, 32, 34
Practice with all examples that we discussed in the classand related examples in textbook!!
Before working on Homework,
Any Questions???
The End!