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1. 1
WONDERFUL
WORLD OF CHEMISTRY!
Chemistry 1311
Section D / F
Welcome
to the
Professor Wendy Pell

2. 2
Chemistry 1311D
• PROFESSOR:
Dr. Wendy Pell
MCD 409
562-5800 x6041
wpell@uottawa.ca
• OFFICE HOURS:
Tues. 3 – 4:30 pm
Wed. 4:30– 6 pm
Or by appointment

3. CHEMISTRY 1311 D/F
DGD
530 : 7 pm
FSS 1030
Friday
Thursday
Wednesday
Tuesday
Monday
DGD
530:7 pm
MRN Aud
DGD
230 : 4 pm
BSC 140

4. Textbook
4
You are not required to buy any text,
however, I recommend that you get a
copy of one of the following
recommended texts. Used books are
good.
Chemistry (Canadian Edition) The
Molecular Nature of Matter and Change
by Silberberg, Lavieri, Venkateswaran
General Chemistry (10th edition) Principles
and Modern Applications
by Petrucci, Herring, Madura, Bissonnette

5. 5
LAB (3rd floor MRN)
Lab Instructor: Dr. Rashmi Venkateswaran (Dr. Rashmi’s
email: vrashmi@uottawa.ca)
Access to all lab material will be found on Blackboard – this
includes lab policies and schedules, the lab manual, prelab
exercises, videos etc.
Labs begin: Our undergraduate labs are being renovated and
lab start date is as yet undetermined. We will let you when labs
start (in class and posted to the Blackboard Lab web site) as soon
as possible.
Lab tutorials begin: Keep checking the Blackboard website for
details

6. 6
LAB (3rd floor MRN)
Your timetable will identify which lab section you are registered in.
The first week is a real lab and you are required to have completed the
prelab assignment. If you do not attend 80% of the labs, for whatever
reason, you will not receive credit for the course.
BEFORE THE FIRST LAB
You will need: lab coat
eye protection (glasses, goggles)
proper shoes
You can purchase lab coat/goggles, beginning Sept. 8, in
MRN 08 using your student card.
And have: Accessed the lab website
Completed the prelab exercises

7. 7
CHM 1311D/F
COURSE MARKS:
75% of the course mark will be from the lecture portion of the course
(see below) and 25% of the course mark will be from your lab work (see
lab manual).
Laboratory 25 %
Homework: 10 %
In class (LectureTools) 5 %
Test 1: 15 % Thurs., Oct. 15
Test 2: 15 % Thurs., Nov. 19
Final Exam: 30 %

8. 8
EVALUATION EXERCISES
You must print neatly on all tests and exams your STUDENT
NUMBER, and FULL NAME (FIRST AND LAST). If these are
not LEGIBLE and completed in full, you will be credited 0 for that
test or exam.
Tests:
There will be two in-class tests (October 15; November 19) each worth
15% of your final mark. The projected content of each test is listed in the
course syllabus BUT is subject to modification depending on how we
progress through the syllabus. Tests are not cumulative. A formula sheet
and periodic table will be provided for all tests and for the final exam.
Final Exam:
Final exam dates and locations will be posted later in the term. Your final
exam will be cumulative and will be worth 30% of your lecture mark. A
formula sheet and periodic table will be provided.

9. 9
EVALUATION EXERCISES
In class participation (5 %): Free marks!!!!
We will be using Lecture Tools ..
You have been emailed a link to register to the Lecture
Tools class.
You must use your uOttawa email address
You must use your First and Last names exactly as they
appear on your student card!
During our lectures, you can either sign into Lecture Tools
with a WIFI device or text your answer. If you plan to text
your answers then you must enter you telephone number
during the registration process – I will never see this and
promise not to call you during dinner or early Saturday
morning!

10. 10
EVALUATION EXERCISES
Problem based homework assignments (Sapling Learning)
(10 % of Mark):
Weekly problem based homework to reinforce on material covered during lectures.
You must register with Sapling Learning and sign up for my course:
University of Ottawa – CHM1311 – Fall15 - PELL
Instructor: Wendy Pell
Your “keycode” is your one letter section (either D or F)
There is a $32 fee associated with this homework tool
The objective of the homework is to help you understand and apply course material.
It is not supposed to be an exercise in frustration or torture and, provided you are
prepared, should take approximately one hour to complete.
Prepared means you have reviewed the course notes and completed suggested text
problems.
If the homework is taking you significantly longer than 1 hour to complete – get
some help!!!!

11. 11
Alternate Grading Schemes (lecture)
Sapling homework
And Participation
No Assignments
Participation 5 5
Weekly on line
homework
10 -
Midterm 1
(Oct. 15)
15 17
Midterm 2
(Nov. 19)
15 17
Final Exam 30 36

12. 12
Test Re-Grade Policy
- submitted in writing within 1 week of date that test is
returned in class
- if you find an error
• staple a note to the front of your test indicating:
1. problem number
2. detailed explanation of why you think you
should receive more points
• bring test to me before or after class or to my
office (MCD 409)
- I will not discuss grades/re-grades prior to or
immediately following lecture or DGD
- If after test is returned you have further comments,
you must make an appointment to see me

13. 13
CALCULATORS:
You will need a non programmable calculator for the tests and for the
final exam (TI-30X, Ti-30XA, TI-30XIIS, TI-30SLR) You may NOT use
cell phones, smart phones, ipads, tablets, or other electronic devices in
place of a calculator during tests or finals.
ABSENCE:
If you are ill on the day of a test, please bring me a medical certificate
to receive exemption from the test. If you miss the final exam due to
illness, please bring a medical certificate to the faculty of science to have
your final exam deferred to the study week in February.
You will not be exempt from an online homework due to illness unless
you have proof that you were ill for the full 7 days during which the
homework was available online.
If you miss a lab due to illness, please obtain a medical certificate,
inform your demonstrator and Dr. Rashmi.

14. 14
EMAIL
I will happily answer email inquiries subject to the following
restrictions:
1. the subject of the email message is clearly indicated
in the message header, this should include the course code
(CHM 1311D) and a brief description of body of the email.
eg. Subject: CHM 1311D, request for an appt
2. you identify yourself – include your NAME
3. your email is a professional correspondence – treat it
as such (no gunnas or wannas or yos).
4. I will try to answer all email messages with 24 h, do
not expect an answer to email messages after 5pm or on
weekends – so budget your time wisely
You are encouraged to use the website message board!!! You
can post questions and have either myself or your classmates
answer them.

15. 15
ATTENDANCE
LECTURES and LABS
The Faculty of Science has an 80% attendance
requirement. If you miss a lecture, it is your
responsibility to acquire any notes and assignments
for that day from another member of the class.
Lectures will be posted to the course website in pdf
format. Posted lectures do not include worked
problems.
The 80 % attendance requirement is applicable to
laboratory exercises; any absence from the lab must
be justified with appropriate medical documentation.
If you do not complete 80% of the labs, for
whatever reason, you will not receive credit for the
course.

16. 16
ATTENDANCE
Lecture DGDs are optional, but highly recommended. In
Lecture DGDs, you will be guided through practice problems
relevant to the current lecture material and similar to what
you might see on midterms. Material covered in DGDs will
not be posted to the course website but any material
covered in the DGD will be fair game for midterms and
exams.
DGDs provide an opportunity for you to ask questions!!
There are three DGDs, you may attend any or all of them
Monday 530 – 7 pm FSS 1030
Wednesday 230 – 4 pm BSC 140
Thursday 530 – 7 pm MRN AUD

17. 17
What you need to do to succeed!
1. Don’t get behind!!!
2. Complete all graded assignments, midterms, labs
3. Do suggested problems from the text – reading your notes
and the textbook isnt enough; like anything else, if you want command
of the material you need to practice! Exam questions will be of similar
difficulty to the higher numbered problems
4. Attend lectures, participate and collect your 5 %
5. Participation + Homework = 1 grade point!!!
• That’s the difference between an E(fail) and a D(pass) or a C+
and a B!!!!
6. Get help sooner
• DGDs
• Lab tutorials
• Office hours
• Chemistry help center (hours to be posted)

18. 18
Goals in Chemistry
We want to understand the big picture in terms of electrons,
atoms, molecules that make up matter and the changes they
undergo
macroscopic property: rust
microscopic explanation:
redox reaction involving iron ions
and atoms, oxygen molecules and
water molecules.
macroscopic property:
fire!
microscopic explanation:
Hydrogen and oxygen molecules
undergo a combustion reaction,
releasing a large amount of
energy (and water).

19. 19
and what happened to Mr. Gummy
Bear???!!!
Good bye Mr. Gummy Bear

20. www.theodoregray.com/PeriodicTable
Why else do we study chemistry???
Why do you store
Li under inert gas?
Why is it even more
important that you
store Cs under inert
gas?

21. 21
Chapter 1: Keys to Study of Chemistry
In particular
1.4 Chemical Problem Solving
1.5 Measurement in Scientific Study
and 1.6 Significant Figures
You are responsible for all of Chapter 2: The Components
of Matter
and all of Chapter 3: Stoichiometry of Formulae and
Equations, but we will briefly look at a few key
topics as a review
You should review:

22. 22
What is Matter???
Antoine Lavoisier
1743-1794
Law of Conservation of Mass:
mass before
reaction
mass after
reaction
=
Matter is neither created
nor destroyed.

23. 23
What is Matter???
Law of Definite
(or Constant)
Composition If elements A and B react to form
two compounds, the different
masses of B that combine with a
fixed mass of A can be expressed as
a ratio of small whole numbers.
Example: Carbon Oxides I & II
Carbon Oxide II : 72.7% oxygen and 27.3% carbon
Carbon Oxide I : 57.1% oxygen and 42.9% carbon
The Law of Multiple
Proportions

24. 24
Dalton’s Atomic Theory:
1) All matter is made of small, indivisible particles called
atoms
2) Atoms of the same element are identical in mass and
other properties, atoms of different elements are
different.
3) Atoms of different elements combine in simple, whole
number ratios (e.g. 1:1 as in AB; or 1:2 as in AB2)
4) Chemical reactions result in separation, combination, or
rearrangement of atoms, but do not result in creation
or destruction of atoms.
What is Matter???
John Dalton (1766 – 1844)

25. 25
Do Atoms Really Exist?
Scanning Tunneling Microscopy (STM) allows us to see
individual atoms!
Bright spots are
sulfur atoms in
MoS2
For more info on STM, see:
http://www.iap.tuwien.ac.at/www/surface/STM_Gallery/stm_schematic.html
And
http://www.nobelprize.org/educational/physics/microscopes/scanning/

26. 26
What are Atoms???
Dalton –Atoms are small and indivisible.
Thomson – Atoms contain negatively charged electrons
Rutherford – Atoms contain a positively charged nucleus

27. 27
Scale of Atoms
Useful units:
 1 amu (atomic mass unit) = 1.66054 x 10-24
g
 1 pm (picometer) = 1 x 10-12 m
 1 Å (Angstrom) = 1 x 10-10 m = 100 pm = 1
x 10-8 cm
The heaviest atom has a mass of only 4.8 x 10-22 g
and a diameter of only 5 x 10-10 m.
Biggest (naturally occurring) atom is ca. 240 amu and 2 Å
Typical C-C bond length 154 pm (1.54 Å)

28. 28
Standard Notation for Atoms
E
A
Z
mass # = # protons + # neutrons
atomic # = # protons
No two elements have the same Z!
If Z = 7, what is the element?
Nitrogen!
Most chemical properties result from
the # and arrangement of electrons.
How is # of electrons determined?

29. 29
Isotopes
atoms having the same # of protons
but different # of neutrons
H
1
1
H
1
2
H
3
1
In nature, elements occur as a mixture of isotopes
How many electrons does each isotope have?
Hydrogen Deuterium Tritium

30. 30
To measure masses of atoms, a new unit is defined: Unified
Atomic Mass Units (u).
Definition: mass of 12C = 12 u
The mass of every other element in amu is determined by
measuring its mass relative to the mass of 12C.
What is the mass of one atom of Carbon??
(Where 12 u is an exact number. It has an infinite
number of zeros past the decimal)
e.g. mass 16O/12C = 1.33291
mass 16O = 1.33291 x mass 12C = 1.33291 x 12 amu
= 15.9949 amu

31. 31
Why is the mass number for carbon
12.011 (not 12.00) in the periodic table??
Carbon exists in nature as carbon-12 and carbon-13.
isotope mass %abundance in nature
12C 12.000 u 98.89%
13C 13.003 u 1.11%
(0.9889 x 12.000) + (.0111 x 13.003) = 12.011
MASS NUMBER = average of the isotopic masses, weighted
according to the naturally occurring abundances of each
isotope.
However, no single carbon atom has a mass of 12.011 u!
All carbon atoms have masses of 12.000 or 13.003 u!

32. 32
There are five naturally occurring isotopes of Zn.
64Zn (? u, ? %), 66Zn (65.926 u, 27.90 %); 67Zn
(66.927 u,4.10 %); 68Zn (67.925 u, 18.75 %); 70Zn
(69.925 u, 0.62 %).
a) What is the % natural abundance of 64Zn?
b) What is the atomic mass of the isotope having the
highest natural abundance?

33. 33
Periodic Table
Elements are arranged in the periodic table according to atomic
number. Elements with similar properties are grouped together.
Group or Family
Period

34. The Periodic Table
Alkali Metals
Alkaline Earths
Transition Metals
Halogens
Noble Gases
Lanthanides and Actinides
Main Group
Main Group

35. 35
Molecules
Only the noble gases exist in nature as single atoms
The atoms of all other elements combine to form MOLECULES
e.g. H2 a DIATOMIC molecule
H2O a POLYATOMIC molecule
In a neutral atom or molecule # protons = # electrons

36. 36
Ions
CATIONS have
fewer electrons
than protons so are
positively charged.
e.g. Na+
ANIONS have
more electrons than
protons so are
negatively charged.
e.g. OH-
cartoon from http://nearingzero.net/
Atoms and molecules can gain and lose electrons.
When # protons ≠ # electrons, the species has a net
charge and is called an ION.

37. Chemical Formulas
The identity and ratio of elements in a molecule are expressed
by a CHEMICAL FORMULA
Molecular Formula: Shows the exact # and ratio of each
element
H2O water
O2 oxygen
Structural Formula: Shows how atoms are bonded together
using a line to represent a bond
H-O-H water
Empirical Formula: Shows the simplest whole # ratio of
elements
Hydrogen peroxide
molecular formula: H2O2
Empirical formula: HO

38. Connectivty makes a difference!!! Consitutional Isomers
Property Ethanol Dimethyl Ether
M (g/mol)
Boiling Point
Density at 200C
Structural
formula
46.07
78.50C
0.789 g/mL
(liquid)
46.07
-250C
0.00195 g/mL
(gas)
Butane 2-Methylpropane
C4H10 C2H6O
58.12 58.12
Space-filling
model
-0.50C
0.579 g/mL
(gas)
-11.060C
0.549 g/mL
(gas)
C C C C
H H
H
H H
H
H
H
H
H
C C C
H H
H
H
C
H
H
H
H H
H
C C
H H
H
H H
OH C O
H
H
H
C
H
H
H
©2013 McGraw-Hill Ryerson

39. The formation of
an ionic
compound.
Ionic Compounds
Formula Mass

40. 40
Naming Compounds
Review this on your own, section 2.8
You should know and understand nomenclature for:
Inorganic coumpounds
including: binary and polyatomic compounds, acids,
hypo and per concepts, ous and ic suffixes
Organic compounds

41. 41
Avogadro’s Number (The mole)
In real life, we deal with macroscopic samples containing
large numbers of atoms. We need a unit to define a very
large numbers of atoms.
12 1 dozen
144 1 gross
6.02214 x 1023 1 mole
12 atoms = 1 dozen atoms
144 atoms = 1 gross of atoms
6.02214 x 1023 atoms = 1 mol of atoms
number unit
Definition: 1 mole is defined as the number of 12C atoms
found in exactly 12 g of 12C (6.02214 x 1023 ).

42. 42
Avogadro’s number
The mole is an amount of substance that
contains the same number of elementary
entities as there are carbon-12 atoms in
exactly 12 g of carbon-12.
NA = 6.02214199 x 1023 mol-1

43. 43
Molar Mass: mass in grams of 1 mole (6.02214 x 1023
atoms or molecules) of a substance
Molar Mass, M
molar mass of 12C = 12.0000 g
1 mole of O = 15.9994 g
1 mole of H = 1.0079 g
1 mole of C = 12.011 g
In nature, elements occur as a mixture of their isotopes.
The molar mass of an element is the mass of 1 mole of the
naturally occurring mixture of isotopes.
Mass Number for the
Element
(as shown on the periodic table)
Molar mass of an element
(including all naturally occurring
isotopes)
=

44. 44
Conversion Between Mass, Moles, and Atoms
How many moles of Al are in a 10.0 g sample?
How many atoms of Al are in this sample?

45. 45
Molecular Mass: the average mass (in amu) of one molecule.
Molecular Mass
molecular
mass =
sum of atomic mass of each
atom in the molecule (in u)
Determine the molecular mass of trimethyl amine, (CH3)3N :
Determine the molar mass of (CH3)3N :

46. 46
Molecular and Molar Mass
For H2O:
Molecular Mass: mass of one molecule 18.015 u
Molar Mass: mass of one mole of molecules 18.015 g mole-1
For ionic compounds we refer to a formula mass since
ionic compounds do not consist of molecules.
Formula Mass
For NaCl:
Formula Mass: mass of 1 formula unit 58.44 u
Molar Mass: mass of 1 mole of formula units 58.44 g mole-1

47. The mole
If a mole of moles were digging a mole
of holes, what would you see?
(images: /www.mole-catcher.co.uk)

48. 48
A sample of MgCl2 is analyzed and found to contain 26
µg of Mg. How many magnesium atoms and chlorine
atoms are present in the sample?
Strategy:
µ g → g → moles Mg → moles Cl
→ # atoms Mg and Cl
What mass of chlorine is present in the sample of
MgCl2?
Strategy:
mg → g → moles MgCl2 → moles Cl
(Same steps as previous problem!)
What percent of the mass of the MgCl2
sample is from the magnesium atoms?

49. Mass % of element X =
atoms of X in formula x atomic mass of X (amu)
molecular (or formula) mass of compound (amu)
x 100
Mass % of element X =
moles of X in formula x molar mass of X (g/mol)
mass (g) of 1 mol of compound
x 100
Mass Percent from the Chemical Formula

50. 50
What is the percent composition by mass of H2O2?
Percent Composition by Mass
****Pretend you have a sample containing 1 mole H2O2****
mass % of H =
mass of H
mass of H2O2
x 100%
mass % of O =
mass of O
mass of H2O2
x 100%

51. 51
Overall: Determining Molecular Formula
from % Composition By Mass
Step 1: % composition by mass can be determined
experimentally
Step 2: % composition by mass can be converted to an
empirical formula based on molar masses of
elements.
Step 3: The molecular formula can be determined from
the empirical formula using an experimentally
determined molar mass of the unknown.

52. 52
Empirical formula
1. Choose an arbitrary sample size (100g).
2. Convert masses to amounts in moles.
3. Write a formula.
4. Convert formula to small whole numbers.
5. Multiply all subscripts by a small whole
number to make the subscripts integral.
5 Step Approach:

53. 53
Experimental Determination of Empirical
Formula from % Composition by Mass
When an unknown compound is decomposed into its constituent
elements, it is found to contain 71.65% Cl, 24.27% C, and
4.07% H by mass. What is the empirical formula for the
unknown?
However, the % composition is in mass units.
To determine the empirical formula, we must convert from
mass to moles.
Recall: The empirical formula shows the simplest whole
# ratio of elements. (ratio of # of atoms, not mass of
atoms!)
***If a quantity is not given, assume you start with 100 g****

54. 54
Determining the Empirical and Molecular Formulas
of a Compound from Its Mass Percent Composition.
Dibutyl succinate is an insect repellent used against
household ants and roaches. Its composition is
62.58% C, 9.63% H and 27.79% O. Its experimentally
determined molecular mass is 230 u. What are the
empirical and molecular formulas of dibutyl succinate?
Dibutyl succinate is an insect repellent used against
household ants and roaches. Its composition is
62.58% C, 9.63% H and 27.79% O. Its experimentally
determined molecular mass is 230 u. What are the
empirical and molecular formulas of dibutyl succinate?
Example

55. 55
Chemical Reactions and Chemical Equations
A chemical reaction is a process by which a substance is
changed into one or more new substances

56. 56
A chemical reaction is depicted using a chemical equation:
P4 (s) + 6 Cl2 (g) → 4 PCl3 (l)
reactants products
arrow represents chemical reaction
The numbers in front are called stoichiometric coefficients
Letters (s), (l), (g), and (aq) represent physical states
(solid, liquid, gas, and dissolved in water).
Chemical Reactions and Chemical Equations

57. 57
Chemical Reactions and Chemical Equations
This equation means:
P4 (s) + 6 Cl2 (g) → 4 PCl3 (l)
1 mole of P4 molecules in the solid state
and
6 moles of chlorine in the gas state
react to form
4 moles of phosphorus trichloride in the liquid state
P4 (s) + 6 Cl2 (g) → 4 PCl3 (l)

58. 58
Balancing Chemical Equations
Law of conservation of mass: mass is neither created
nor destroyed. The same number and kind of atoms must
be present before and after a chemical reaction.
Before reaction
1 mole of C
2 moles of O
4 moles of H
CH4 + O2 CO2 + H2O
After reaction
1 mole of C
3 moles of O
2 moles of H
Overall: 1 mole O created and 2 moles H destroyed!
VIOLATES law of conservation of mass!
EQUATION IS NOT BALANCED

59. 59
Balancing Chemical Equations
2 2
Before reaction
1 mole of C
4 moles of O
4 moles of H
After reaction
1 mole of C
4 moles of O
4 moles of H
BALANCED!
Practice balancing chemical equations!
(For now, you do not need to be able to predict products of
reactions! Just balance the equations.)
CH4 + O2 CO2 + H2O

60. 60
Chemical Reaction - Balancing
Equations
Nitrogen monoxide + oxygen → nitrogen dioxide
Step 1: Write the reaction using chemical
symbols.
NO + O2 → NO2
Step 2: Balance the chemical equation.
2 1 2

61. 61
Balancing Equations
• Never introduce extraneous atoms to balance.
NO + O2 → NO2 + O
• Never change a formula for the purpose of
balancing an equation.
NO + O2 → NO3

62. 62
Balancing Equation Strategy
• Balance elements that occur in only
one compound on each side first.
• Balance free elements last.
• Balance unchanged polyatomics as
groups.
• Fractional coefficients are
acceptable and can be cleared at the
end by multiplication.

63. 63
Example
Writing and Balancing an Equation: The Combustion
of a Carbon-Hydrogen-Oxygen Compound.
Liquid triethylene glycol, C6H14O4, is used a a
solvent and plasticizer for vinyl and polyurethane
plastics. Write a balanced chemical equation for
its complete combustion.

64. Balancing Redox equations
PAY ATTENTION – This is Wendy’s favourite
Inspection doesn`t always work!!!
Chapter 19 Section 1
Mass balance works, but this is WRONG:
8H+ (aq) + C2O4
2-(aq) + MnO4
-(aq) → 2CO2(g) + Mn2+(aq) +4H2O(l)
64

65. 65
Recognizing Redox Reactions
How do we know if electron transfer occurs in a reaction??
2 Ca (s) + O2 (g) → 2CaO (s)
Step 1: Determine the oxidation number (or
oxidation state) of each element in the
reactants and products.
Step 2: If the oxidation state of a given element is
different in the products than it is in the
reactants, then electrons were transferred.

66. 66
Oxidation Numbers
2 Ca (s) + O2 (g) → 2CaO (s)
0 0
+2 -2
Oxidation numbers are a mathematical tool for dividing
up the electrons in a molecule.
Element Initial Final Change______
Ca 0 +2 Ca lost 2 electrons
O 0 -2 O gained 2 electrons
Gaining electrons cause oxidation number to decrease
Losing electrons causes oxidation number to increase

67. 67
Rules for Assigning Oxidation Numbers
1) The oxidation number, Nox, for a free elements (not in a
compound) is zero.
e.g. Na Nox = 0
Mg Nox = 0
Al Nox = 0
2) The oxidation number of a monatomic ion (ion composed of
only 1 atom) is equal to the ion charge:
e.g. Na+ Nox = +1
Mg2+ Nox = +2
O2- Nox = -2
Cl- Nox = -1

68. 68
3) The oxidation number of each atom in a compound
containing only 1 type of element is zero.
e.g. H2 each H atom has Nox = 0
O2 each O atom has Nox = 0
P4 each P atom has Nox = 0
4) The oxidation number of a group 1a element in a molecule
is always +1
e.g. NaCl Nox (Na) = +1
KClO3 Nox (K) = +1
5) The oxidation number of a group 2A element in a molecule
is always +2
e.g. MgCl2 Nox (Mg) = +2
Ba(OH)2 Nox (Ba) = +2

69. 69
7) For a polyatomic ion, the sum of oxidation numbers of each
atom in the compound is equal to the overall ion charge:
e.g. CO3
2- [Nox (C) + (3 x Nox (O))]= -2
HSO4
- [Nox (H) + Nox (S) + (4 x Nox (O))]= -1
8) The oxidation number of oxygen in most compounds is -2.
Exception: In H2O2 and in O2
2-, each oxygen has Nox = -1
e.g MgO Nox (O) = -2
H2O Nox (O) = -2
CO2 Nox for each O = -2
H2O2 Nox for each O = -1
6) In a neutral, polyatomic compound, the sum of the oxidation
numbers of each atom in the compound is zero:
e.g. NaCl [Nox (Na) + Nox (Cl)] = 0
SO2 [Nox (S) + (2 x Nox (O)] = 0
CO [Nox (C) + Nox (O)] = 0

70. 70
9) The oxidation number for a halogen in a molecule is
usually -1. There are exceptions.
e.g. NaCl Nox (Cl)= -1
MgF2 Nox (F)= -1
KClO3 Nox (Cl) = +5
10)The oxidation number of hydrogen in most compounds is +1.
Exception: When H is bound directly to a group 1A or group
2A metal, it has Nox = -1.
e.g. H2O Nox for each H = +1
CH4 Nox for each H = +1
HNO2 Nox (H) = +1
NaH Nox (H) = -1
MgH2 Nox for each H = -1

71. 71

72. 72
Examples Assigning Oxidation Numbers (Nox)
Assign oxidation states to all atoms in CO2
Assign oxidation states to all atoms in NO3
-
Assign oxidation states to all atoms in SF6

73. 73
The only SURE WAY to identify a
redox reaction is to compare
oxidation numbers of each element
in reactants and products!

74. 74
Oxidation and Reduction
The element that loses electrons is said to be oxidized
The element that gain electrons is said to be reduced
LEO the lion says GER
Loss of Electrons is
Oxidation
Gain of Electrons is
Reduction

75. 75
Half Reactions
2 Ca (s) + O2 (g) → 2CaO (s)
Step 1: The oxidation half-reaction shows electrons
being lost by an element (electrons are a
product)
2 Ca → 2 Ca2+ + 4 e-
Each Ca loses 2 electrons
Step 2: The reduction half-reaction shows electrons
being gained by an element (electrons are a
reactant)
O2 + 4 e- → 2 O2-
Each O gains 2 electrons
To visualize electron flow, we can write any redox reaction
as if it occurs in two different steps called half-reactions.

76. 76
Half Reactions
The sum of the half reactions gives the overall reaction:
Oxidation: 2 Ca → 2 Ca2+ + 4 e-
Reduction: O2 + 4 e- → 2 O2-
Overall: 2 Ca + O2 + 4 e- → 2 Ca2+ + 2 O2- + 4 e-
Cancel electrons that appear on both sides:
2 Ca + O2 → 2 Ca2+ + 2 O2-
Combine Ca2+ with O2- to make CaO
2 Ca + O2 → 2 CaO

77. 77
Conservation of Electrons
# of electrons consumed in the reduction
=
# of electrons produced by oxidation
Oxidation: 2 Ca → 2 Ca2+ + 4 e-
Reduction: O2 + 4 e- → 2 O2-
Overall: 2 Ca + O2 + 4 e- → 2 Ca2+ + 2 O2- + 4 e-
# electrons in reactant = # electrons in products
NO ELECTRONS ARE CREATED OR DESTROYED

78. 78
Balancing Redox Reactions by the Ion-
Electron Method
Step 2: Divide the reaction into the contributing half-reactions.
Step 3: Balance each half reaction (We will see how!)
Step 4: Equalize the # of electrons in the two half reactions. (#
electrons produced by oxidation must equal # electrons
consumed by reduction.)
Step 5: Add the balanced half reactions together to obtain the
overall reaction.
Step 1: Determine oxidation states of each atom in the reaction.

79. 79
Balance the following redox reaction in acidic solution:
H+ (aq) + Cr2O7
2- (aq) + C2H5OH (l) → Cr3+ (aq) + CO2 (g) + H2O (l)

80. 80
Step 3: Balance each half reaction
Rules for Balancing Half-Reactions in ACIDIC solution
Treat each half-reaction separately
1. Balance all elements except hydrogen and oxygen
2. Balance oxygen using H2O
3. Balance hydrogen using H+
4. Balance charge using electrons
charge on right
side of arrow
=
charge on left
side of arrow

81. 81
Rules for Balancing Half-Reactions in BASIC solution
Treat each half-reaction separately
1. Balance all elements except hydrogen and oxygen
2. Balance oxygen using H2O
3. Balance hydrogen using H+
In basic solution, H+ is not available! Thus it cannot appear in our
final balanced equation.
Basic Solutions
4. Balance charge using electrons
• For every H+ that is added, add an equal number
of OH- ions to both sides of the equation.
• Wherever H+ and OH- appear on the same side of
the equation, combine them to make water. (No
H+ should appear anywhere in the equation!)

82. 82
Balance the following chemical equation in basic solution:
MnO4
- (aq) + SO3
2- (aq) → MnO4
2- (aq) + SO4
2- (aq)

83. 83
Balance the following redox reaction in acidic solution:
Cu + NO3
-  Cu2+ + NO
Balancing Redox Reactions in Acidic Solution
Another Example
Balance: Same example in base
Cu + NO3
-  Cu2+ + NO in basic solution

84. 84
3. Balance each half-reaction following this sequence in order:
a) Balance elements (except H, O)
b) Balance O using H2O
c) Balance H using H+
For basic solutions
i) For every H+ ion, add on OH- to both sides of the equation
ii) When H+ and OH- appear on the same side of the equation,
combine them to make H2O.
d) Balance charges of half-reactions using electrons
2. Identify the two half reactions (oxidation and reduction).
1. Determine oxidation numbers of each element:
Summary of Steps for Balancing Redox Reactions:
4. Equalize the electron count in oxidation and reduction half-reactions
5. Add together the oxidation and reduction half-reactions
6. Check that all atoms and overall charges are balanced

85. 85
I know your high school teacher / your dad / your dog showed you a
different way, and yes you can use it BUT
i) if you are wrong, and it isn’t clear where you went wrong,
you will get ZERO because I can’t tell if you understand the
significance of changing oxidation states and have no idea if you know
what you are doing
ii) the other method involves simple math – so it is easy to
make silly mistakes
iii) sometime equations are really hard to balance, and my
method always!!! works.
My method works flawlessly on:
Fe(CN)6
4- (aq) + Ce4+ (aq) 
Ce(OH)3(s) + Fe(OH)3 (s) + CO3
2-(aq) + NO3
-(aq)

86. 86
STOICHIOMETRY
The study of the quantitative aspects of chemical reactions
The quantity of product formed from a reaction
depends upon the amount of reactant available.

87. 87
General Plan for Stoichiometric Calculations
Mass Reactant
Moles Reactant Moles Product
Mass Product
stoichiometric
factor
MOLES
MOLES will never lead you wrong – NOTHING
ELSE can be trusted!!

88. 88
Carbohydrates are compounds containing C, H and O in which
the H to O ratio is 2:1. A 4.220 g sample of a carbohydrate
is burned in excess oxygen to yield 6.189 g of CO2.
a) (5 points) What is the empirical formula for the
carbohydrate.
b) (2 points) Determine the molecular formula of the
compound if the approximate molar mass of the compound is
185 g mole-1.
c) (2 points) Combustion of the carbohydrate yields only
H2O and CO2. Write the balanced combustion reaction.
d) (2 points) What mass of H2O is produced from the
combustion of the 4.220 g sample?

89. 89
Limiting Reagents
Often there is not enough of one reagent to completely
consume the other reagent. The reactant that is completely
consumed determines the quantities of the products formed
The consumed reagent is the limiting reagent.
The leftover reagent is called excess.
excess Zn:
limiting
reagent is HCl
not enough Zn:
limiting reagent
is Zn
Zn (s) + 2 HCl (aq) → ZnCl2 (aq) + H2 (g)

90. 90
Limiting Reagents
Often there is not enough of one reagent to completely
consume the other reagent
The consumed reagent is the limiting reagent.
The leftover reagent is called excess.
excess Zn:
limiting
reagent is HCl
not enough Zn:
limiting reagent
is Zn
Zn (s) + 2 HCl (aq) → ZnCl2 (aq) + H2 (g)

91. 91
Limiting Reagents
WARNING: You can not determine the limiting reagent
just by looking at the stoichiometric coefficients!!
Zn (s) + 2 HCl (aq) → ZnCl2 (aq) + H2 (g)
In the first flask, hydrochloric acid was the limiting reagent.
In the third flask, zinc was the limiting reagent.

92. 92
Limiting Reagent Problem: What mass of H2 is produced
when 0.50 g Zn is combined with 1.3 g HCl? Which picture
does this correspond to?
Limiting Reagent Problem: What mass of is left
unreacted when 0.50 g Zn is combined with 1.3 g HCl?

93. 93
After the reaction of 92.3 g P4 with 286 g Cl2, what are
the masses of P4, Cl2, and PCl3? The chemical reaction that
occurs is shown below:
P4 (s) + 6 Cl2 (g) → 4 PCl3 (l)
Another Example

94. 94
Theoretical Yield, Actual Yield,
and Percent Yield
Theoretical Yield: Calculated quantity of products
expected from a given quantity of
reactants.
Actual Yield: Quantity of products actually produced.
Percent Yield:
% yield = actual yield
theoretical yield
x 100%

95. 95
If the maximum theoretical yield of CO2 from a given
reaction is 2.7 g, but only 1.8 g CO2 was recovered in the
experiment, what is the percent yield?
% yield = actual yield
theoretical yield
x 100%

96. 96
Limiting Reactant Problem
(i.e. any problem!!!)
Determine the limiting reactant
Use the mass of the limiting reactant to determine moles of
limiting reactant
Use moles of limiting reactant and the appropriate
stoichiometric ratio to determine any of (assuming all of
the limiting reactant is used up):
Moles of other reactants consumed
Moles of product formed
Use the moles of the product to determine the mass of the
product formed if all of the limiting reactant were used up
or
Use the moles of the other reactant (3.1 above) to determine
the mass of the other reactant used, and therefore how
much is left over…..
% yield = actual mass formed/theoretical maximum (calculated
above) x 100

97. 97
Reasons for Less than 100% Yield
Reasons for Greater than 100% Yield

98. 98
Concentration Units
Molarity (c) =
solution
Liters
solute
moles
Molality (b or m) =
solvent
kilograms
solute
moles
Mole fraction (XA) =
C
moles
B
moles
A
moles
A
moles
+
+
Weight percent = 100%
x
solution
grams
solute
grams
Parts by volume =
solution
volume
solute
volume
Parts by mass =
solution
mass
solute
mass

99. 99
Concentration Units
If a solution contains 0.035 moles solute in 2.0 L of
water, what is the molarity (assuming the solute occupies
a negligible volume)?
Molarity =
moles solute
L solution
0.035 moles
2.0 L
= = 1.8 x 10-2 M
Concentration Notation:
Often instead of saying the molarity of A is... , we will
write: [A] =
Where [A] means molarity of A
In this problem, [Solute] = 1.8 x 10-2 M

100. 100
Preparing Solutions of Specific
Concentration
TWO METHODS:
• Weigh out a solid solute and
dissolve in a given quantity of
solvent.
• Dilute a concentrated solution
to give one that is less
concentrated.
Molarity (c) =
solution
Liters
solute
moles

101. 101
USING MOLARITY
What mass of oxalic acid, H2C2O4, is required to make
250.0 mL of a 0.0500 M solution?
Molarity =
solution
Liters
solute
moles
THEREFORE
Moles solute = Molarity of solution • Volume of solutio
MOLES = c • V

102. Aqueous Ionic Reactions
The molecular equation shows all reactants and
products as if they were intact, undissociated
compounds.
This gives the least information about the species in solution.
2AgNO3 (aq) + Na2CrO4 (aq) → Ag2CrO4 (s) + 2NaNO3 (aq)
When solutions of silver nitrate and sodium chromate mix, a
brick-red precipitate of silver chromate forms.
©2013 McGraw-Hill Ryerson Limited

103. The total ionic equation shows all soluble ionic
substances dissociated into ions.
This gives the most accurate information about species in solution.
2Ag+ (aq) + 2NO3
- (aq) → Ag2CrO4 (s)
+ 2Na+ (aq) + CrO4
2- (aq) + 2Na+ (aq) + NO3
- (aq)
Spectator ions are ions that are not involved in the
actual chemical change. Spectator ions appear
unchanged on both sides of the total ionic equation.
2Ag+ (aq) + 2NO3
- (aq) → Ag2CrO4 (s)
+ 2Na+ (aq) + CrO4
2- (aq) + 2Na+ (aq) + 2NO3
- (aq)
©2013 McGraw-Hill Ryerson Limited
The net ionic equation eliminates the spectator ions
and shows only the actual chemical change.
2Ag+ (aq) + CrO4
2- (aq) → Ag2CrO4 (s)

104. 104
PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O
in enough water to make 250 mL of
solution. Calculate the molarity.
A
•Weigh the solid
needed.
•Transfer the solid to a
volumetric flask that
contains about half the
final volume of solvent.
B Dissolve the
solid thoroughly by
swirling.
C Add solvent until the
solution reaches its final
volume.

105. 105
Preparing Solutions by Dilution

106. 106
PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH. What
do you do?
0.50 M NaOH
3.0 M NaOH
Concentrated Dilute
H2O
Add water to dilute
the solution

107. 107
cinitial•Vinitial = cfinal•Vfinal
Preparing Solutions by
Dilution
A shortcut

108. 108
Concentration Units
Why do we need units of molality?
Two good reasons:
1) Sometimes we need to know the number of
solute particles per solvent particle.
2) Molarity is temperature dependent because
solvent volume is temperature dependent. We
need a measurement that is temperature
independent.
Molality (b) = moles solute
kg solvent
Molarity (c) = moles solute
L solution

109. 109
Converting Between
Concentration Units
General Strategy
Original
Concentration
Units
Moles of
Solute
and
Quantity
of Solvent
New
Concentration
Units

110. Interconverting Concentration Terms
• To convert a term based on amount (mol) to one based
on mass, you need the molar mass.
• To convert a term based on mass to one based on
volume, you need the solution density.
• Molality involves quantity of solvent, whereas the other
concentration terms involve quantity of solution.
©2013 McGraw-Hill Ryerson Limited

111. 111
Laboratory ammonia is 14.8 M NH3 (aq) with a
density of 0.8980 g/mL. What is the mole
fraction of NH3 in the solution?
Molarity (c) = moles solute
L solution
Xsolute =
moles solute + moles solvent
moles solute

112. 112
Calculate the molality of a 5.24 M NaHCO3
solution. (Density of solution = 1.19 g/mL)
Molality (b) = moles solute
kg solvent
Molarity (c) = moles solute
L solution

113. Sample Problem 3.27 Solving Limiting-Reactant Problems for
Reactions in Solution
PROBLEM: Mercury and its compounds have many uses, from fillings for
teeth ( as a mixture containing silver, copper and tin)in the
past to the current production of chlorine. Because of their
toxicity, however, soluble mercury compounds, such as
mercury(II) nitrate must be removed from industrial
wastewater. One removal method reacts the wastewater
with a sodium sulfide solution to produce solid mercury(II)
sulfide and sodium nitrate solution. In a laboratory
simulation, 0.050 L of 0.010 mol/L mercury(II) nitrate reacts
with 0.020 L of 0.10 mol/L sodium sulfide.
(a)What mass of mercury(II) sulfide forms?
©2013 McGraw-Hill Ryerson Limited

114. Sample Problem 12.5 Interconverting Concentration Terms
PROBLEM: Hydrogen peroxide is a powerful oxidizing agent used in
concentrated solution in rocket fuels and in dilute solution
as a hair bleach. An aqueous solution H2O2 is 30.0% by
mass and has a density of 1.11 g/mL. Calculate its
(a) Molality (b) Mole fraction of H2O2 (c) Molarity
PLAN: (a) To find the mass of solvent we assume the % is per 100 g
of solution. Take the difference in the mass of the solute
and solution for the mass of peroxide.
(b) Convert g of solute and solvent to moles before finding X.
(c) Use the density to find the volume of the solution.
©2013 McGraw-Hill Ryerson Limited

115. 115
What next?
Determine the mass of CO2(g) is produced when
10 g of a pyrolusite ore which is 35 % (w/w)
MnO2 is dissolved in 150 ml of 2 M HCl (aq) and
250 ml 0.5 M oxalic acid, H2C2O4(aq) given the
following unbalanced chemical equation ?
H2C2O4 + MnO2 + H+  Mn2+ + CO2
What kind of problem is this?