By
David B. Oestreicher

What is Stoichiometry?
Stoichiometry is the study of
relationships between the amounts of reactants used and products formed by a chemical reaction; based on the law of conservation of mass.

1. You can write mole ratios from balanced chemical equations
2. You can calculate the number of moles and the mass of a reactant or product when given the number of moles or the mass of another reactant or product
3. You can identify the limiting reactant in a chemical reaction
4. You can determine the percent yield of a chemical reaction

Cooking Examples
• How many eggs will we need to make 3
Western Omelets?
• How many peppers will we need to make 4
Western Omelets?
• If we made 40 Western Omelets, how many pieces of cheese did we use?
2 eggs + 1 slice of cheese + 1 pepper  1 W. Om

Stoichiometry and balanced equations
4Fe (s) + 3O
2
------ 2Fe
2
O
3
(s)
4 Moles of Fe = 223.4 g of Fe
3 Moles of O
2
2 Moles Fe
2
= 96.00 g of O
2
O
3
= 319.4 g
223.4 g + 96.0 g = 319.4 g
Law of the Conservation of Mass or
One side of the equation equals the other

2Al + 3Br
2
 2AlBr
3
How do all the molecules relate to one another?
2 moles Al
3 moles Br
2
2 moles Al
2 moles AlBr
3
3 moles Br
2
2 moles Al
3 moles Br
2
2 moles AlBr
3
2 moles AlBr
3
2 moles Al
2 moles AlBr
3
3 moles Br
2

How many moles of carbon dioxide are produced when 10.0 moles of propane are burned in excess oxygen in a gas grill?
1. First, what do we know? What do we need to find?
a.
We have 10.0 moles of propane b.
We want to find out how much carbon dioxide will be produced.
2. Write a balanced equation for the problem:
C
3
H
8
(g) + 5O
2
- 3CO
2
+ 4H
2
O(g)

Ratios of reactants to each other and the products
C
3
H
8
(g) + 5O
2
- 3CO
2
+ 4H
2
O(g)
The ratio of CO
2 to C
3
H
8 is 3:1 or
3 moles CO
2 =
3
1 mole C
3
H
8
1
So 10.0 moles C
3
H
8 x
3 moles CO
2 =
30.0 moles of CO
2
1 mole C
3
H
8

Determine the mass of sodium chloride or table salt
(NaCl) produced when 1.25 moles of chlorine gas (Cl
2
) reacts vigorously with sodium.
1. What do we have? What do we want to find out?
a.
We have 1.25 moles of Cl
2 b.
We want to find out the mass of NaCl produced
2. Write a balanced equation for the reaction
2Na(s) + Cl
2
(g)--- 2NaCl(s)
2. What is the ratio of NaCl to Cl
2
?
2 moles NaCl or 2:1
1 mole Cl
2

Solve using ratios
1.25 moles Cl
2 x
2 moles NaCl = 2.5 moles NaCl
1 mole Cl
2
Is that the answer? NO! We were asked for the mass of NaCl so we need to convert from moles to grams
2.5 moles NaCl x 58.44 g NaCl = 146 g NaCl
1 mole NaCl

The reaction between solid white phosphorus and oxygen produces solid tetraphosphorous decoxide (P
4
O
10
). This compound is often called diphosphorous pentoxide because its empirical formula is P
2
O
5
.
a. Determine the mass of tetraphosphorous decoxide formed if 25.0 g of phosphorous
(P
4
) and 50.0 g of oxygen are combined.
b. How much of the excess reactant remains after the reaction stops?

How do we solve this?
1. What do we know?
mass of phosphorous = 25.0 g P4 mass of oxygen = 50.0 g O
2
2. Write a balanced equation for the reaction
P
4
(s) + 5O
2
(g) -- P
4
O
10
(s)

Determine the number of moles of reactants
P
4
(s) + 5O
2
(g) -- P
4
O
10
(s)
25.0 g P
4 x 1 mol P
4
123.9
g P
4
= 0.202 mol P
4
50.0 g O
2 x 1 mol O
2
= 1.56 mol O
2
32.00 g O
2

P
4
(s) + 5O
2
(g) -- P
4
O
10
(s)
Calculate the actual ratio of available moles of
O
2 and available moles of P
4
.
1.56 moles O
2
0.202 moles P
4
= 7.72 moles O
2
1 mol P
4

Determine the mole ratio of the two reactants from the balanced equation
P
4
(s) + 5O
2
(g) -- P
4
O
10
(s)
5 mol O
2
1 mol P
4
Because 7.72 mole of O
2 are available, but only 5 moles are needed to react with 1 mole of P
4
, the
O
2 is in excess and P
4 is the limiting reactant.

Use the limiting reactant’s moles to determine how much product will be produced.
0.202 mol P
4 x 1 mol P
4
O
10
1 mol P
4
To find the mass
= 0.202 mol P
4
O
10
0.202
mol P
4
O
10 x
283.9 g P
4
O
10
1 mol P
4
O
10
= 57.3 g P
4
O
10

b. How much of the excess reactant remains after the reaction goes to completion?
0.202 mol P
4 x 5 mol O
2
= 1.01 mol O
2
(needed)
1 mol P
4
1.01 mol O
2 x 32.00 g O
2
= 32.3 g O
2 needed
1 mol O
2
50.0 g O
2 given – 32.3 g O
2 needed = 17.7 g O
2 in excess

Percent yield = actual yield (from an experiment)______________ x 100 theoretical yield (from stoichiometric calculations)

Calculating Percent Yield
When potassium chromate (K
2
CrO
4
) is added to a solution containing 0.500 g of silver nitrate
(AgNO
3 formed.
), solid silver chromate (Ag
2
CrO
4
) is a. Determine the theoretical yield of the silver chromate precipitate.
b. If 0.455 g of silver chromate is obtained, calculate the percent yield.

To Solve
1. What do we know?
a. mass of silver nitrate = 0.500 g AgNO
3 b. percent yield = ? % yield of Ag
2
CrO
4
2. Write a balanced equation for the reaction
2AgNO
3
+ K
2
CrO
4
-- Ag
2
CrO
4
+ 2KNO
3

2AgNO
3
+ K
2
CrO
4
-- Ag
2
CrO
4
+ 2KNO
3
Convert grams to moles of AgNO
3
0.500 g AgNO
3 x 1 mol AgNO
3
169.9 g AgNO
3
= 2.94 x 10 -3 mol AgNO
3
Use the appropriate mole ratio to convert mol
AgNO
3 to mol Ag
2
CrO
4
2.94 x 10 -3 mol AgNO
3 x
1 mol Ag
2
CrO
4
2 mol AgNO
3
=
1.47 x 10 -3 mol Ag
2
CrO
4

Calculate the mass of Ag
2
CrO
4
(theoretical yield) by multiplying mol Ag
2
CrO
4 by the molar mass
From last slide
2.94 x 10 -3 mol AgNO
3 x
1 mol Ag
2
CrO
4
2 mol AgNO
3
=
1.47 x 10 -3 mol Ag
1.47 x 10 -3 mol Ag
2
CrO
4 x
331.7 g Ag
2
CrO
4 =
0.488 g Ag
2
CrO
4
1 mol Ag
2
CrO
4
2
CrO
4
Divide the actual yield by the theoretical yield and multiply times 100
0.455 g Ag
2
CrO
4 x 100 = 93.2 % Ag
2
CrO
4
0.488 g Ag
2
CrO
4