Chapter 9: Stoichiometry
9.1 Mole to Mole
Objective: To perform mole to
mole conversion problems.
Stoichiometry

Stoichiometry is the branch of chemistry
that deals with quantities of substances in
chemical reactions.
Mole to Mole Problems: The Steps
1. Write chemical equation
2. Balance chemical equation using coefficients
3. Use the following setup to perform
calculation:
___ m olB(eqn.)
___ m olA
 ___ m olB
___ m olA(eqn.)
A is the known quantity
Mole
Ratio
B is the unknown quantity
4. Don’t forget units on your final answer!!
9.2 Mole to Mass and Mass to Mole

Objective: To perform mole to mass and
mass to mole conversion problems.
Mole to Mass
Moles
Given
Mole Ratio
Unknown
Moles (Eqn.)
Molar Mass
Unknown
Given
Moles
1 mole
Unknown
(Eqn.)
Grams
Unknown
Mass to Mole
Grams
Given
Unknown
1 mol Given Moles (Eqn.)
Molar Mass
Given
Given
Moles
Moles
Unknown
(Eqn.)
Mole Ratio
9.3 Mass to Mass

Objective: To perform mass to mass
conversion problems.
Stoichiometry Roadmap
Grams
Given
Unknown
1 mol Given Moles (Eqn.)
Molar Mass
Unknown
Molar Mass
Given
1 mole
Unknown
Given
Moles
(Eqn.)
Grams Unknown
Mole Ratio
9.4 Limiting Reactant
Objectives:
(1) To calculate the theoretical yield of a
chemical reactions.
(2) To determine the limiting reactant and
excess reactant in a chemical reaction.

Limiting Reactant


Any reactant that is used up first in a
chemical reaction.
It determines the amount of product that
can be formed in the reaction.
Excess Reactant

The reactant that is not completely used
up in a reaction.
Limiting Reactant Problems

Use the mass to mass
conversions
Grams
Given
1 mol Given
Molar Mass
Given
Unknown
Moles
Molar Mass of
Unknown
Given Moles
1 mol of Unknown
Example

Copper reacts with sulfur to form copper(I)
sulfide according to the following balanced
equation:
2Cu + S  Cu2S
What is the limiting reactant when 80.0 grams
of Cu reacts with 25.0 grams of S?
Example
2Cu + S  Cu2S
The general equation is:
Grams
Given
1 mol Given
Molar Mass
Given
Unknown
Moles
Molar Mass of
Unknown
Given Moles 1 mol of Unknown
Start with Copper:
80.0 g
Cu
1 mol Cu
63.55 g Cu
1 mol
159.17 g Cu2S
Cu2S
= 100.19 g Cu2S
2 mol Cu 1 mol of Cu2S
Now use Sulfur:
25.0 g
S
1 mol S
1 mol
Cu2S
159.17 g Cu2S
32.07 g S
1 mol S
1 mol of Cu2S
= 124.08 g Cu2S
Example

The limiting reactant is copper.

The excess reactant is sulfur.

The amount of Cu2S that is produced is
100.19 g Cu2S.
9.5 Percent Yield

Objective: To calculate percent yield.
Introduction to Percent Yield…

If you get 15 out of 20 questions correct on a
test, what percentage did you receive on the
test? How did you figure this out?
15/20 x 100 = 75%
If Sammy Sosa gets 25 hits in the month of
May and has 113 bats, what is his batting
average for the month of May? Explain how
you arrived at your answer.
25/113 = 0.221
As a percentage, this is written, 25/113 x 100 =
22.1%

Percent Yield
·Percent yield is the ratio of the actual yield to the
theoretical yield for a chemical reaction
expressed as a percentage.
· It is a measure of efficiency of a reaction.
Percent Yield =
actual yield
theoretical yield
x 100%
Actual Yield

The amount of product that forms when a
reaction is carried out in the laboratory.
Theoretical Yield


The amount of product that could form
during a reaction calculated from a
balanced chemical equation.
It represents the maximum amount of
product that could be formed from a given
amount of reactant.
Example #1
The equation for the complete combustion of ethene
(C2H2) is
2C2H2 + 5O2  4CO2 + 2H2O
1.
2.
3.
If 0.10 g of C2H2 is reacted with 201.60 g of O2,
identify the limiting reactant.
What is the theoretical yield of H2O?
If the actual yield of H2O is 0.05 g, calculate the
percent yield.
Example #1
2C2H2 + 5O2  4CO2 + 2H2O
Start with C2H2:
0.10 g
C2H2
1 mol C2H2
2 mol H2O
26.02 g C2H2 2 mol C2H2
Limiting Reactant = C2H2
18.00 g H2O
1 mol H2O
= 0.07 g
H2O
Theoretical Yield
Now start with O2:
201.60 g
O2
1 mol O2
18.00 g H2O
2 mol H2O
32.00 g O2
1 mol H2O
5 mol O2
= 45.36 g
H2O
Example #1
2C2H2 + 5O2  4CO2 + 2H2O
Actual Yield = 0.05 g H2O
Theoretical Yield = 0.07 g H2O
Percent Yield = actual yield
x 100%
theoretical yield
Percent Yield = 0.05 g H2O x 100% = 71.43 %
0.07 g H2O
Example #2



Determine the percent yield for the
reaction between 3.74 g of Na and excess
O2 if 5.34 g of Na2O2 is recovered?
First, write the chemical equation:
Na + O2  Na2O2
Second, balance the chemical equation:
2Na + O2  Na2O2
Example #2
2Na + O2  Na2O2

Solve the mass-mass problem, starting with Na:
3.74 g
Na
1 mol Na
22.99 g Na
1 mol
Na2O2
2 mol Na
Actual Yield = 5.34 g Na2O2
Theoretical Yield = 6.34 g Na2O2
Percent Yield = actual yield
theoretical yield
x 100%
Percent Yield = 5.34 g x 100% = 84.23%
6.34 g
77.98 g Na2O2
1 mol Na2O2
=
6.34 g
Na2O2