Unit 8:
Stoichiometry:
Part 1
Stoichiometry
• Means “element measuring”
• The study of quantitative
relationships between the
amounts of reactants used and
the products formed by a
chemical reaction
• Based on the law of conservation
of mass
Mole Ratio
•In a balanced equation, the
ratio between the numbers of
moles of any 2 substances
•Indicated by the coefficients
2 Mg + O2  2 MgO
Mg : O2
1 mol O2
2 mol Mg
or
1 mol O2
2 mol Mg
2
mol
MgO
2
mol
Mg
Mg : MgO
or
2 mol MgO 2 mol Mg
MgO : O2
2 mol MgO 1 mol O2
or
1 mol O2
2 mol MgO
Stoichiometry Steps
1. Write a balanced equation.
2. Identify the given &
unknown.
3. Draw a roadmap
4. Set up railroad tracks
5. Calculate
Mole  Mole
Mole to Mole
Ratio
mol Given
___ mol
Unknown
___ mol
Given
Mole Ratio:
Use
coefficients
from
equation
Ex: How many moles of KClO3 must decompose in
order to produce 9 moles of oxygen gas?
2KClO3  2KCl + 3O2
? mol
mol O2 
9 mol O2
9 mol
mol of KClO3
2 mol KClO3
3 mol O2
= 6 mol
KClO3
Mole  Mass
Molar Mass
from
Periodic
table
Mole to Mole
Ratio
mol Given
___ mol
Unknown
Molar mass (g)
of Unknown
___ mol
Given
1 mol
Unknown
Mole Ratio:
Use
coefficients
from
equation
Ex: How many grams of H2O would be required
to produce 5 moles of O2?
2H2O

2H2 +
O2
5 mol
? grams
mol O2  mol of H2O  grams of H2O
5 mol
O2
2 mol
H 2O
18.02 g
H2O
1 mol
1 mol
O2
H2O
= 180.2 g
H 2O
Mass  Mole
Mole to Mole
Ratio
g Given
1 mol Given
___ mol
Unknown
Molar mass
(g) of Given
___ mol
Given
Molar Mass
from
Periodic
table
Mole Ratio:
Use
coefficients
from
equation
Ex: How many moles of NO would be formed
with 824 g of NH3?
4NH3 + 5O2  4NO + 6H2O
824 g
? mol
grams NH3  mol NH3  mol NO
824 g
NH3
1 mol
NH3
4 mol
NO
17.04 g
NH3
4 mol
NH3
= 48.4 mol
NO
Mass  Mass
Molar Mass
from
Periodic
table
Mole to Mole
Ratio
g Given
1 mol Given
___ mol
Unknown
Molar mass (g)
of Unknown
Molar mass
(g) of Given
___ mol
Given
1 mol
Unknown
Molar Mass
from
Periodic
table
Mole Ratio:
Use
coefficients
from
equation
Ex: How many grams of silver will be formed
from 12.0 g copper?
Cu + 2AgNO3  2Ag + Cu(NO3)2
12.0 g
?g
g Cu  mol Cu  mol Ag  g Ag
12.0
g Cu
1 mol
Cu
63.55
g Cu
2 mol
Ag
1 mol
Cu
107.87
g Ag
= 40.7 g
1 mol
Ag
Ag
Unit 8:
Stoichiometry:
Part 2
How many sandwiches can
you make?
• Available Ingredients
– 4 slices of bread
– 1 jar of peanut butter
– 1/2 jar of jelly
• What limits the amount?
– bread
• What is left over?
– peanut butter and jelly
Limiting vs. Excess
• Limiting Reactant • Excess Reactant
–The reactant that
–The reactant
used up first in a
that is left over
reaction
after the
–Determines the
reaction stops
amount of
product
Percent Yield
•The ratio of actual yield
(from an experiment) to
theoretical yield (from
stoichiometric caculations)
expressed as a percent
Percent Yield
• Actual Yield:
• Theoretical Yield:
The amount of
The maximum
product actually amount of
produced when
product that can
a reaction is
be produced from
carried out in an a given amount of
experiment
reactant
Percent Yield Equation
measured in lab
actual yield
% yield 
 100
theoretical yield
calculated on paper
When 45.8 g of K2CO3 react with
excess HCl, 46.3 g of KCl are
formed. Calculate the theoretical
and % yields of KCl.
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
?g
actual: 46.3 g
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
?g
actual: 46.3 g
Theoretical Yield:
45.8 g
K2CO3
1 mol
K2CO3
138.21 g
K2CO3
2 mol
KCl
74.55
g KCl
1 mol
K2CO3
= 49.4
1 mol
g KCl
KCl
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
49.4 g
actual: 46.3 g
Theoretical Yield = 49.4 g KCl
% Yield =
46.3 g
49.4 g
 100 = 93.7%