Chapter 3
Stoichiometry
1
Stoichiometry

The study of quantities of materials
consumed and produced in chemical
reactions.
2
3.1 Chemical equations

The '+' is read as 'reacts with' and the
arrow '' means 'produces'
3
Because atoms are neither created nor
destroyed in a reaction, a chemical equation
must have an equal number of atoms of
each element on each side of the arrow (i.e.
the equation is said to be 'balanced').
4
Balancing Equations

Write 'un-balanced' equation using formulas of
reactants and products

Write 'balanced' equation by determining
coefficients that provide equal numbers of each
type of atom on each side of the equation
(generally, whole number values)

Note! Subscripts should never be changed when
trying to balance a chemical equation. Changing a
subscript changes the actual identity of a product
or reactant. Balancing a chemical equation only
involves changing the relative amounts of each
product or reactant.
5
Examples
6
7
We seem to be o.k. with our number of carbon atoms in both
the reactants and products, but we have only half the
hydrogen in our products as in our reactants. We can fix this
by doubling the relative number of water molecules in the
list of products:
8
Note that while this has balanced our carbon and
hydrogen atoms, we now have 4 oxygen atoms in our
products, and only have 2 in our reactants. We can
balance our oxygen atoms by doubling the number of
oxygen atoms in our reactants:
9

The physical state of each chemical can
be indicated by using the symbols (g), (l),
(s), and (aq) (for gas, liquid, solid, and
aqueous respectively):
Na (s) + H2O (l)  NaOH (aq) + H2(g)
10
Balance the following equations
C2H5OH (aq) + O2 (g)  CO2 (g) + H2O (g)
Fe (s) + O2 (g)  FeO2 (s)
C2H4 (g) + O2 (g)  CO2 (g) + H2O (l)
11
Answer
2Na (s) + 2H2O (l)  2NaOH (aq) + H2(g)
Fe (s) + O2 (g)  FeO2 (s)
C2H4 (g) + 3O2 (g)  2CO2 (g) + 2H2O (l)
12
Which box represents the reaction between NO
and O2 to produce NO2.
13
Homework
Balancing / Writing Rxn wks
14
Classifying reactions movie
15
Chemical Reactivity
Combination/Synthesis Reaction:
2 or more substances react to form one new
product
A+B C
+

16
solid magnesium and oxygen gas react to
produce solid magnesium oxide
2Mg (s) +
Metal
O2(g) 
nonmetal
Diatomic
2MgO (s)
ionic compound
2+ 2-
17
Decomposition Rxn
One substance undergoes a reaction to
produce two or more substances.
 Typically occurs when things are heated.
AX  A + X


+
18
Solid calcium carbonate reacts to produce
solid calcium oxide and carbon dioxide gas
CaCO3 (s)  CaO (s)
2+ (2-)
2+ 2-
+ CO2 (g)
4+ 2(2-)
19
Single displacement

One element replaces a similar element in
a compound
A + BX  AX + B
BX + Y  BY + X
+

+
20
Solid copper is dissolved in aqueous silver
nitrate to produce solid silver and aqueous
copper II nitrate.
Cu(s) + AgNO3 (aq) Ag(s) + Cu(NO3)2 (aq)
Write the sentence for this reaction:
Fe (s) + Cu(NO3)2 (aq)  Fe(NO3)2 (aq)+ Cu
(s)
21
Activity Series

We need to know what metals are most
likely to oxidize others.

Example: We can’t store nickel nitrate in
an iron container because the solution
would eat through the container.
22
Activity Series

A list of metals
arranged in order of
decreasing ease of
oxidation.

Page 139 table
23
Using activity series

Any metal on the list can be
oxidized by the metal below it.
Give: FeCl2 + Mg

Find: will iron oxidize
Magnesium metal?

1.
2.
3.
4.
5.
I finger on Fe
1 finger on Mg
Is the bound chemical below
Yes Fe is below Mg.
Then complete the reaction
24
Give: NaCl + Mg

Find: will sodium oxidize
Magnesium metal?

1.
2.
3.
4.
5.
I finger on Na
1 finger on Mg
Is the bound chemical below
no
Then the reaction is not
possible
25
What if you don’t have an Activity Series
table?





Down Group 1 (I) the "Alkali Metals" the activity
increases Cs > Rb > K > Na > Li
Down Group 2 (II) the activity increases e.g. Ca >
Mg
On the same period, the Group 1 metal is more
reactive than the group 2 metal
the group 2 metal is more reactive than the Group
3 metal,
All three are more reactive than the "Transition
Metals". e.g. Na > Mg > Al (on Period 3) and K > Ca
> Ga > Fe/Cu/Zn etc. (on Period 4)
26
Double Replacement Rxn/
Metathesis

The ions of two compounds exchange places
in an aqueous solution to form two new
compounds.
AX + BY  AY + BX
One of the compounds formed is usually a
precipitate, an insoluble gas that bubbles out
of solution, or a molecular compound,
usually water.

27
Double Replacement Rxn/
Metathesis
AX + BY  AY + BX
+

+
28
Write the sentence for these double replacement
reactions
KOH (aq) + H2SO4 (aq)  K2SO4 (aq) + H2O (l)
FeS (aq) + HCl (aq)  FeCl2 (aq) + H2S (aq)
29
Combustion Reaction
A substance combines with oxygen, releasing a
large amount of energy in the form of light and
heat.
C3H8 (g)+ 5O2 (g)  3CO2 (g) + H2O (g)
Usually CO2 (carbon dioxide) / CO (carbon monoxide)
and water are produced.
30

Reactive elements combine with
oxygen
P4(s) + 5O2(g)  P4O10 (s)
(This is also a synthesis reaction)
The burning of natural gas, wood,
gasoline
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)

31
Homework

Classifying types of Rxns worksheet
32
3.3 Formula Weights

Although we can’t “count atoms” in a
molecule directly, we can count them
indirectly if we know their masses.
33
Formula Weights/ Molecular
Weight

Sum of atomic masses of each atom in a
molecule.
F.W of H2SO4 = 2(H) + S + 4(O)
2(1) + 32 + 4(16) = 98amu
98 g/mol
34
Percent Composition
We can describe
composition in two
ways
1. number of atoms
(amu)
2. % (by mass) of its
elements.
35
Percent Composition
We can find % mass of an atom in a
compound from formula mass, by
comparing each element present in 1 mole
of compound to the total mass of 1 mole of
compound
36
Percent Composition Equation
% element = # of atoms element (atomic weight of the element )
Formula Weight
* 100
37
Example

Calculate the percentage of nitrogen in
Ca(NO3)2
38
Think:
% N = # N atoms (m.w N)
m. w Ca(NO3)2
X 100
39
Answer
% N = 2(14.02 N amu)
164.12 Ca(NO3)2amu
X 100
= 17%
40
Question

Calculate the percent composition of each
element in C12H22O11
41
Homework
Molar mass wks
Percent composition wks
42
3.4 The Mole!!!!!

The unit for dealing
with, atoms,
molecules, ions

Abbreviation = mol
(oh the time you will
save!)
43
History



Avogadro
Italian
1776-1856
44
Molar Mass



A dozen eggs = 12
A dozen elephants =
12
But 12 eggs has a
different weight than
12 elephants
45

Thus 1 mole of carbon is 6.02 x 1023
molecules but weighs 12 grams

1 mole of sodium is 6.02 x 1023 molecules
but weighs 23 g
46

If you had Avogadro's
number of unpopped
popcorn kernels, and
spread them across
the United States of
America, the country
would be covered in
popcorn to a depth of
over 9 miles.
47
If we were able to count
atoms at the rate of
10 million per second,
it would take about 2
billion years to count
the atoms in one
mole.
48

An Avogadro's
number of standard
soft drink cans would
cover the surface of
the earth to a depth of
over 200 miles.
49

Molar mass of any substance is equal to its
mass in atomic mass units (amu)
50
Grams
Moles
Molecules
Atoms
Application
How many moles of water are in 5.380g of
water?
1.
Molar mass of water = 2(1) + 1(16) = 18g/mol
5.380g H2O x
1 mol =
18g H2O
0.2989 moles H2O
52
Using Moles in calculations
How many oxygen atoms are present in
4.20 grams NaHCO3?
53
4.20 g NaHCO3 x (1mole NaHCO3)
atoms
84 g NaHCO3
NaHCO3
x (6.02e 23molec)
1 mol
3 Oxygen
1 molec
= 9.03 x 10 22 atoms of Oxygen in 4.20 grams NaHCO3
54
Using Moles in calculations

How many nitrogen atoms are in 0.25 mol
of Ca(NO3)2
55
Answer
0.25mol x
6.02 x 1023 molec Ca(NO3)2 x _2N atoms___
1 mol
1 molec Ca(NO3)2
= 3.0 x 10 23 moles
56
Homework
Molar mass worksheet
 G-mol-molec atoms wks

57
3.5 Determining empirical formula
from mass percent

Recall: Empirical formula: simplest whole
# ratio of atoms in a compound.

Recall: We can find % mass from formula
mass, by comparing each element
present in 1 mole of compound to the
total mass of 1 mole of compound
58
Example
Example: Vitamin C is composed of 40.92%
C, 4.58% H, and 54.50% O by mass. What
is the empirical formula?
59
How to attack the problem
1. Convert mass % into grams (assume
100g ie: 40.92% = 40.92 g)
2. Convert grams to moles using molar
mass.
3. Divide moles of each element by the
smallest number of moles present. You
may round to nearest whole # . (This
establishes a ratio of comparison)
60
Answer
40.92 g C 1mol C
12 g C
= 3.4 moles C /3.4 = 1 C
54.40 g O 1mol O
16g O
= 3.4 moles O / 3.4 = 1 O
4.58 g H
= 4.58 moles H/ 3.4 = 1.3 = 1 H

1mol H
1gH
CHO
61
Given:
Mass % elements
Find:
Calculate
mole
ratio
Assume
100g of
sample
Grams of
each element
Empirical
formula
Use
atomic
weights
Moles of each
element
62
Question
Compound X is composed of 55.3% K,
14.6% P , and 30.1% O.
What is the empirical formula of compound
X?
63
55.3 g K 1mol K
39 g K
= 1.4 moles K /.47 = 3 K
14.6 g P
= 0.47 moles P/ .47 = 1 P
1mol P
31 g P
30.1 g O 1mol O
16g O

= 1.9 moles O / .47 = 4 O
K3PO4
64
Determine Molecular formula from
Empirical Formula
Recall:
Molecular formula: the exact formula of a molecules, giving
types of atoms and the number of each type.
1. Using mass % and molar mass, determine mass of each
element in 1 mole of compound (same)
2. Determine number of moles of each element in 1 mole of
compound. (same)
3. The integers from the previous step represent the subscripts
in the molecular formula ( you just don’t divide by smallest
65
mole ratio)
Let’s look back at our work
40.92 g C 1mol C
12 g C
= 3.4 moles C /3.4 = 1 C
54.40 g O 1mol O
16g O
= 3.4 moles O / 3.4 = 1 O
4.48 g H
= 4.48 moles H/ 3.4 = 1 H
1mol H
1gH
C3H4O3 = molecular formula
66
Shortcut
n = Molecular Formula Weight
Empirical Formula Weight
(where n = # of atoms)
67
Try it …..
The molecular weight of butyric acid is
88 amu. If the empirical formula is C2H4O.
What is the molecular formula?
68
1. The empirical formula was given, us it to find the empirical
formula weight.
C2H4O = 12 + 12+ 1+1+1+1+16 = 44 amu
2. The molecular formula weight was given (88amu) plug it into the
short cut formula.
n = 88 amu = 2
44
3. Apply the new number of atoms.
Molecular formula = (empirical) n
(C2H4O)2 = Molecular Formula = C4H8O2
69
Homework

Chan pg 108 43, 44, 45, 49, 50, 52, 53

BLPg 106-107: 37a, 39, 42, 44, 45, 46
70
3.6 Qualitative information from
balanced equations.

Stoichiometry: mixing exactly enough chemical so that all is
used
Mass-Mass problems a new highway!!!!!!
g given  mol given  mol required  g required
Think: (grams to moles to moles to grams)
71
Time out for mole ratios
2H2 (g) + O2 (g)  2H2O (l)
Coefficients tell us that 2 molecules of H2 react with
each molecule of O2 to form 2 molecules of H2O.
(recall: 6.02 x 1023 molecule = 1 mol)
These coefficients can be used to convert between
quantities of reactants and products.
72
Example of mole ratios

Calculate the number of moles of H2O
produced from 1.57 mol of O2?
2H2 (g) + O2 (g)  2H2O (l)
73
2H2 (g) + O2 (g)  2H2O (l)
1.57 mol O2 ( 2 mol H2O) =
1 mol O2
= 3.14 mole H2O
74
Silicon carbide is made by heating silicon dioxide
to high temperatures.
SiO2 (s) + 3C (s)  SiC(s) + 2CO (g)
How many grams of CO are formed by complete
rxn of 5.00 g SiO2?
HINT: always make sure your equation is
balanced first or mole ratios will be wrong.
75
Given:
Find:
Grams substance A
Grams of
substance B
Use
molar
mass of
A
Use
molar
mass of
B
Moles of
substance A
Use coefficients
of A and B from
balanced
equation
Moles of
substance B
Every line of dimensional analysis should have a unit AND a
chemical formula!!!!!
76
Given: 5.00 g SiO2
Find : CO g
SiO2 (s) + 3C (s)  SiC(s) + 2CO (g)
think: grams to moles to moles to grams
5.00 g SiO2 1mol SiO2 2 mol CO
28 g CO = 4.67 g CO
60 g SiO2 1 mol SiO2 1 mol CO
Mole ratio
77
How many moles of sulfuric acid would be
needed to produce 4.80 moles of
molecular iodine (I2) according to the
following balanced equation.
10HI + 2KMnO4 + 3H2SO4  5I2 + 2MnSO4 + K2SO4 + 8H2O
78
4.80 mol I2
3 mol H2SO4
5 mol I2
=
2.88 mol H2SO4
79
Homework

Chan pg: pg 109 46, 66,67,68,69

BLPg: 107 #’s 51, 53, 55, 57, 59, 60
Note: you must be able to pass a pop quiz with
80% accuracy to be able to move on to the next
section. If not additional problems and help will
be mandatory.
80
3.7 Limiting Reactant
The number of products that can form is
limited by the amount of reactant present.
The limiting reactant is the one that gives
the least amount of product.
Reactants  Products
81

Lets say you want to
make a sandwich
using 2 slices of
bread (Bd), 1 slice of
ham (Hm).
2Bd + 1Hm  Bd2Hm
82
If you have 10 slices of bread and 7 slices of ham,
how many sandwiches will you be able to make
according to the previous reaction?
2Bd + 1Hm  Bd2Hm
What is your limiting reagent?
83
Extra Tasty Ham
Slabs
84
What is the limiting reactant, Blue
or Red?
85
Attacking limiting reactant problems
1.
2.
3.
Using the grams of each reactant and
their mole ratios calculate how many
grams of product will be formed.
Compare the amount of product formed
by each reactant.
The reactant that give the lesser amount
of product is the limiting reactant.
86
Try One….
When a mixture of silver and sulfur is heated,
silver sulfide is formed:
16 Ag (s) + S8 (s)  8 Ag2S (s)
What mass of Ag2S is produced from a
mixture of 2.0 g of Ag and 2.0 g of S8?
87
2 g Ag 1 mol Ag 8 mol Ag2S 248 g Ag2S
108 g Ag 16 mol Ag
1 mol Ag2S
= 2.3g Ag2S can be formed from 2 g Ag
2 g S8 1 mol S8 8 mol Ag2S 248 g Ag2S
256 g S8 1 mol S8
1 mol Ag2S
= 15.5 g Ag2S can be formed from 2.0 g S8
88
= 15.5 g Ag2S can be formed from 2.0 g S8
Most amount of product can be formed (aka:
ham slabs!!!)
= 2.3g Ag2S can be formed from 2 g Ag
Least amount of product can be formed Thus Ag
is the limiting reactant!!!! (aka: bread)
89
Try one again…. (man these are
fun!!!)
How many grams of urea can be produced
from 10.0 g of NH3 and 10.0 g of CO2?
2NH3 + CO2  CH4N2O + H2O
(Urea)
90
Answer
17.6 g of Urea from 10g NH3
13.6 g of urea from 10 g CO2 LIMITNG
REACANT
91
Theoretical Yield

The amount of product calculated based on the
limiting reactant.
% yield = collected yield
Predicted yield
X 100
Actual/collected = what you really get
Theoretical/Predicted = What you might get or
predict you will get.
92
Try one!
In a chemical reaction the theoretical yield is
145 g. If the percent yield was 92.0%,
what was the actual yield?
93
Answer
92.0 = actual x 100
145
= 133 g
94
Calculating Theoretical Yield
When you are doing a limiting reactant equation,
(g – mol-mol-g) you are calculating the theoretical
yield. It is not until you actually run the
experiment in the lab that you will get your actual
yield.
Some times not all the reactants react, or they may
react in a way different than you desired.
Chemistry is not perfect ( unlike you guys)
95
Homework 3.7
Chang: pg 107-108 # 63, 67, 69, 70, 77, 78
BL: Page
96