CH. 9
Stoichiometry
A. Proportional Relationships
2 1/4 c. flour
1 tsp. baking soda
1 tsp. salt
1 c. butter
3/4 c. sugar
3/4 c. brown sugar
1 tsp vanilla extract
2 eggs
2 c. chocolate chips
Makes 5 dozen cookies.
I
have 5 eggs. How many cookies
can I make?
Ratio of eggs to cookies
5 eggs 5 doz.
2 eggs
= 12.5 dozen cookies
Proportional Relationships
 Stoichiometry
• mass relationships between
substances in a chemical reaction
• based on the mole ratio
 Mole
Ratio
• indicated by coefficients in a
balanced equation
2 Mg + O2  2 MgO
Stoichiometry Steps
1. Write a balanced equation.
2. Identify known & unknown.
3. Line up conversion factors.
• Mole ratio - moles
moles
moles

moles
• Molar mass -moles  grams
• Molar volume - moles  liters gas
Core step in all stoichiometry problems!!
4. Check answer.
Conversions
MASS
IN
GRAMS
Molar Mass
(g/mol)
MOLES
6.02 
1023
particles/mol
NUMBER
OF
PARTICLES
Stoichiometry Problems: Moles Only
 How
many moles of KClO3 must
decompose in order to produce 9
moles of oxygen gas?
2KClO3  2KCl + 3O2
? mol
2 mol KClO3
3 mol O2
9 mol
=
x mol KClO3
9 mol O2
9 mol O2 2 mol KClO3
3 mol O2
= 6 mol KClO3
Stoichiometry Problems: Mass
 How
many grams of KClO3 must
decompose in order to produce 9
grams of oxygen gas?
2KClO3  2KCl + 3O2
?g
9g
a) Molar mass
c) Molar mass
b) Mole ratio
Moles KClO3
Moles O2
Divide problem in steps
a) Calculate O2 moles
2KClO3  2KCl + 3O2
?g
9g
9.00 g 1mol
32.0g
= 0.281
mol O2
b) Calculate KClO3 moles
2KClO3  2KCl + 3O2
? mol
0.281 mol
2 mol KClO3
3 mol O2
0.281 mol O2
=
x mol KClO3
0.281 mol O2
2 mol KClO3
3 mol O2
= 0.187 mol KClO3
c) Calculate KClO3 mass
0.187mol
122.55 g
1 mol
= 22.9 g KClO3
2KClO3  2KCl + 3O2
22.9 g
9.00 g
Stoichiometry Problems: density
 How
many grams of KClO3 must
decompose in order to produce 6.3 mL
of oxygen gas? (density of O2 1.429
g/mL)
2KClO3  2KCl + 3O2
?g
6.3 mL
a) Density
Mass O2
d) Molar mass
b) Molar mass
c) Mole ratio
Moles KClO3
Divide problem in steps
Moles O2
a) Calculate O2 mass
2KClO3  2KCl + 3O2
?g
9g
6.3 mL 1.426 g
1 mL
9 g O2
b) Calculate O2 moles
2KClO3  2KCl + 3O2
?g
9g
9.00 g 1mol
32.0g
= 0.281
mol O2
c) Calculate KClO3 moles
2KClO3  2KCl + 3O2
? mol
0.281 mol
0.281 mol O2
2 mol KClO3
3 mol O2
= 0.187 mol KClO3
d) Calculate KClO3 mass
0.187mol
122.55 g
1 mol
= 22.9 g KClO3
2KClO3  2KCl + 3O2
22.9 g
6.3 mL
Gas Stoichiometry
 Liters of a Gas:
• STP - use 22.4 L/mol
• Non-STP - use ideal gas law
 Moles
1 mol of a gas=22.4 L
at STP
Standard Temperature
&
0°C and 1 atm
Pressure
Molar Volume at STP
LITERS
OF GAS
AT STP
Molar Volume
(22.4 L/mol)
MASS
IN
GRAMS
Molar Mass
(g/mol)
MOLES
6.02 
1023
particles/mol
NUMBER
OF
PARTICLES
Stoichiometry Problems: Volume
many grams of KClO3 are req’d to
produce 9.00 L of O2 at STP?
 How
2KClO3  2KCl + 3O2
?g
9L
a) Molar volume
c) Molar mass
b) Mole ratio
Moles KClO3
Moles O2
Divide problem in steps
a): Calculate O2 moles
2KClO3  2KCl + 3O2
?g
9.00 L
9.00 L 1mol
22.4L
= 0.4 mol O2
b) Calculate KClO3 moles
2KClO3  2KCl + 3O2
? mol
0.4 mol
2 mol KClO3
3 mol O2
=
x mol KClO3
0.4 mol O2
0.4 mol O2 2 mol KClO3
3 mol O2
= 0.27 mol KClO3
c) Calculate KClO3 mass
0.27 mol
122.55 g
1 mol
= 33 g KClO3
2KClO3  2KCl + 3O2
33 g
9.00 L
Stoichiometry Problems:
only gas reactions
C2H2 (g) + 5 O2 (g)  4 CO2 (g) +
2 H2O (g)
 If 79 L of carbon dioxide are
produced in this reaction, how
many liters of ethene (C2H2) were
used?
2
2 L C2H2
4 L CO2
=
x L C2H2
78 L CO2
Stoichiometry Problems: particles
many grams of KClO3 are req’d to
produce 9.044 x 1023 molecules of O2?
 How
2KClO3  2KCl + 3O2
9.044 x 1023
?g
molecules
c) Molar mass
b) Mole ratio
Moles KClO3
a) Avogadro’s
number
Moles O2
Divide problem in steps
Step 1: Calculate O2 moles
2KClO3  2KCl + 3O2
23
9.044
x
10
?g
molecules
9.044 x 1023molecules
1mol
= 1.500 mol O
6.022 x 1023molecules
Step 2: Calculate KClO3 moles
2KClO3  2KCl + 3O2
? mol
1.500 mol
2 mol KClO3
3 mol O2
1.500 mol O2
=
x mol KClO3
1.500 mol O2
2 mol KClO3
3 mol O2
= 1.00 mol KClO3
Step 3: Calculate KClO3 mass
1.00 mol
122.55 g
1 mol
= 122.55 g KClO3
2KClO3  2KCl + 3O2
122.33 g
9.044 x 1023
molecules