April 7, 2014
• Today: Stoichiometry and % Yield
Percent Yield
• Remember, stoichiometry is used to tell you how
much product you can form from X amount of
product without doing the reaction
• Percent yield tells you how much product you
actually got in the lab compared to how much you
could have got
Theoretical yield
• The maximum amount of product that can be
formed from a given amount of reactant.
– This is a value you calculate on paper
• In other words: (Write your own definition of
Theoretical yield here)
Actual yield
• The measured amount of a product obtained
from a reaction
– This is a measurement of a product formed in an
actual chemical reaction
• In other words: (Write your own definition of
Actual yield here)
Percent Yield Formula
Actual yield x 100 =
Theoretical yield
Example 1
Mg(s) + 2H2O(g)  Mg(OH)2(s) + H2(g)
A. If 16.2 g Mg are heated with excess H2O how many grams of hydrogen gas
could theoretically be formed?
Known: mass Mg
Unknown: mass H2
Plan: g Mg  mol Mg  mol H2  g H2
Relationships: molar mass Mg = 24.31 g/mol, molar mass H2 = 2.02 g/mol,
mole ratio = 1 mol Mg : 1 mol H2
16.2 g Mg | 1 mol Mg | 1 mol H2 | 2.02 g H2 = 16.2 x 2.02 = 1.35 g
24.31 g
1 mol Mg 1 mol H2
24.31
Example 1

B. If only 0.905 g H2 are actually formed, what is the percent
yield of this reaction?
Actual yield
x 100 = % yield
Theoretical yield
0.905 g H2
1.35 g H2
x 100 = 67.0%
(This means that you produced or collected only 67% of the
product that it was possible to form with the amount of reactant
you started with.)
Example 2
CO(g) + 2H2(g)  CH3OH(l)
If 11.0 g H2 reacts with CO to produce 68.4 g CH3OH, what is the
percentage yield of CH3OH?
We need to determine the theoretical yield.
Known: mass H2
Unknown: mass CH3OH
Plan: g H2  mol Mg  mol H2  g H2
Relationships:
molar mass H2 = 2.02 g/mol
molar mass CH3OH = 32.04 g/mol
mole ratio = 2 mol H2 : 1 mol CH3OH
Example 2
CO(g) + 2H2(g)  CH3OH(l)
If 11.0 g H2 reacts with CO to produce 68.4 g CH3OH, what is the
percentage yield of CH3OH?
Known: mass H2
Unknown: mass CH3OH
Plan: g H2  mol Mg  mol H2  g H2
Relationships: molar mass H2 = 2.02 g/mol, molar mass CH3OH = 32.04
g/mol, mole ratio = 2 mol H2 : 1 mol CH3OH
11.0 g H2 | 1 mol H2 | 1 mol CH3OH | 32.04 g CH3OH = 11.0 x 32.04 = 87.2
2.02 H2 2 mol H2
1 mol CH3OH
2.02 x 2
g
Example 2
CO(g) + 2H2(g)  CH3OH(l)
If 11.0 g H2 reacts with CO to produce 68.4 g
CH3OH, what is the percentage yield of CH3OH?
Actual yield
x 100 = % yield
Theoretical yield
68.4 g CH3OH
87.2 g CH3OH
x 100 = 78.4%