Chapter 9
Chemical Quantities
Information Given by the
Chemical Equation
• Balanced equations show the relationship
between the relative numbers of reacting
molecules and product molecules.
2 CO + O2  2 CO2
2 CO molecules react with 1 O2 molecules to
produce 2 CO2 molecules
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Information Given by the
Chemical Equation
• Since the information given is relative:
2 CO + O2  2 CO2
1. 200 CO molecules react with 100 O2 molecules to produce
200 CO2 molecules
2. 2 billion CO molecules react with 1 billion O2 molecules to
produce 20 billion CO2 molecules
3. 2 moles CO molecules react with 1 mole O2 molecules to
produce 2 moles CO2 molecules
4. 12 moles CO molecules react with 6 moles O2 molecules to
produce 12 moles CO2 molecules
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Information Given by the
Chemical Equation
• The coefficients in the balanced chemical
equation show the molecules and the mole
ratio of the reactants and products.
• Since moles can be converted to masses, we
can determine the mass ratio of the reactants
and products as well.
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Information Given by the
Chemical Equation
2 CO + O2  2 CO2
2 moles CO = 1 mole O2 = 2 moles CO2
Since 1 mole of CO = 28.01 g, 1 mole O2 = 32.00 g,
and 1 mole CO2 = 44.01 g
2(28.01) g CO = 1(32.00) g O2 = 2(44.01) g CO2
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Example #1
Determine the number of moles of carbon monoxide
required to react with 3.2 moles oxygen, and
determine the moles of carbon dioxide produced.
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Example #1 (cont.)
• Write the balanced equation:
2 CO + O2  2 CO2
• Use the coefficients to find the mole
relationship: (Wanted/Given)
2 moles CO = 1 mol O2 = 2 moles CO2
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Example #1 (cont.)
• Use dimensional analysis:
2 moles CO
3.2 moles O2 x
 6.4 moles CO
1 mole O2
2 moles CO 2
3.2 moles O 2 x
 6.4 moles CO 2
1 mole O 2
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Steps to Solve Problems
• 1) Balance Equation!
• 2) Take a ride on the mole road (Dimensional
Analysis)
• 3) Wanted / Given
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Example #2
Determine the number of moles of carbon monoxide
required to react with 3.2 moles oxygen, and
determine the moles of carbon dioxide produced.
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Example #2 (cont.)
• Write the balanced equation:
2 CO + O2  2 CO2
• Use the coefficients to find the mole relationship:
2 moles CO = 1 mol O2 = 2 moles CO2
• Determine the molar mass of each:
1 mol CO = 28.01 g
1 mol O2 = 32.00 g
1 mol CO2 = 44.01 g
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Example #2 (cont.)
• Use the molar mass of the given quantity to
convert it to moles.
• Use the mole relationship to convert the
moles of the given quantity to the moles of
the desired quantity:
1 mol O2 2 mol CO
48.0 g O2 x
x
32.00 g 1 mol O2
1 mol O2 2 mol CO2
48.0 g O2 x
x
32.00 g
1 mol O2
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Example #2 (cont.)
• Use the molar mass of the desired quantity
to convert the moles to mass:
1 mol O2 2 mol CO 28.01 g
48.0 g O2 x
x
x
 84.0 g CO
32.00 g 1 mol O2 1 mol CO
1 mol O2 2 mol CO2
44.01 g
48.0 g O2 x
x
x
 132 g CO2
32.00 g
1 mol O2 1 mol CO2
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Limiting and Excess Reactants
• Limiting reactant: a reactant that is
completely consumed when a reaction is run
to completion
• Excess reactant: a reactant that is not
completely consumed in a reaction
• Theoretical yield: the maximum amount of a
product that can be made by the time the
limiting reactant is completely consumed
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Example #3
Determine the number of moles of carbon
dioxide produced when 3.2 moles oxygen
reacts with 4.0 moles of carbon monoxide.
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Example #3 (cont.)
• Write the balanced equation:
2 CO + O2  2 CO2
• Use the coefficients to find the mole
relationship:
2 moles CO = 1 mol O2 = 2 moles CO2
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Example #3 (cont.)
• Use dimensional analysis to determine the
number of moles of reactant A needed to
react with reactant B:
2 moles CO
3.2 moles O2 x
 6.4 moles CO
1 mole O2
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Example #3 (cont.)
• Compare the calculated number of moles of
reactant A to the number of moles given of
reactant A.
• If the calculated moles is greater, then A is the
limiting reactant; if the calculated moles is less,
then A is the excess reactant.
• The calculated moles of CO (6.4 moles) is greater
than the given 4.0 moles; therefore CO is the
limiting reactant.
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Example #3 (cont.)
• Use the limiting reactant to determine the
moles of product:
2 moles CO2
4.0 moles CO x
 4.0 moles CO2
2 mole CO
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Example #4
Determine the mass of carbon dioxide
produced when 48.0 g of oxygen reacts
with 56.0 g of carbon monoxide.
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Example #4 (cont.)
• Write the balanced equation:
2 CO + O2  2 CO2
• Use the coefficients to find the mole relationship:
2 moles CO = 1 mol O2 = 2 moles CO2
• Determine the molar mass of each:
1 mol CO = 28.01 g
1 mol O2 = 32.00 g
1 mol CO2 = 44.01 g
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Example #4 (cont.)
• Determine the moles of each reactant:
1 mol O2
48.0 g O2 x
 1.50 moles O2
32.00 g
1 mol CO
56.0 g CO x
 2.00 moles CO
28.01 g
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Example #4 (cont.)
• Determine the number of moles of reactant
A needed to react with reactant B:
1 moles O2
2.00 moles CO x
 1.00 moles O2
2 mole CO
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Example #4 (cont.)
• Compare the calculated number of moles of
reactant A to the number of moles given of
reactant A.
• If the calculated moles is greater, then A is the
limiting reactant; if the calculated moles is less,
then A is the excess reactant.
• The calculated moles of O2 (1.00 moles) is less
than the given 1.50 moles; therefore O2 is the
excess reactant.
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Example #4 (cont.)
• Use the limiting reactant to determine the
moles of product, then the mass of product:
2 moles CO2 44.01 g CO2
2.00 moles CO x
x
 88.0 g CO2
2 mole CO
1 mol CO2
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Percent Yield
• Most reactions do not go to completion.
• Actual yield: the amount of product made
in an experiment
• Percent yield: the percentage of the
theoretical yield that is actually made
Actual Yield
Percent Yield =
x 100%
Theoretical Yield
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Example #4a
If in the last problem 72.0 g of carbon dioxide
is actually made, what is the percentage yield?
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Example #4a (cont.)
• Divide the actual yield by the theoretical
yield, then multiply by 100%
– The actual yield of CO2 is 72.0 g
– The theoretical yield of CO2 is 88.0g
72.0 g CO2
x 100%  81.8%
88.0 g CO2
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