Stoichiometry
Mass Calculations
•
Yesterday: Mole to Mole Ratios – relating reactants and
products in a chemical equation
•
If we have a balanced chemical equation, we can calculate the
number of moles of a substance using the known ratio of
reactants and products when given the number of moles of one
of the reactants or products.
•
Previous classes:
• Relationship between mass (g), molar mass (grams/mol)
and mols of a substance  # of mols = mass / MM
• Molar Mass is the atomic mass of an element or formula
weight of a compound (in grams/mol)
• 1 mole = 6.022 X 1023 particles/atoms/molecules
Today: Mass to Mass Calculations
• If we know the reaction (from the balanced
chemical equation) and the amount of one of
the substances (mass, # of moles, or particles)
in the reaction, we can calculate the amount of
the other substances in the reaction.
Mass to Mass Calculations:
The general strategy:
Step 1: Write a balanced chemical equation
Step 2: If you are given the mass or number of particles of a
substance, convert it to the number of moles.
Step 3: Calculate the number of moles of the required substance
based on the number of moles of the given substance, using the
appropriate mole ratio
Step 4: Convert the number of moles of the required
substances to mass or number of particles.
1.) Write out the balanced chemical
equation
2.) Convert the given
mass of a substance
to moles of the
substance  Use the
molar mass of the
given substance
Mass (g)
of given
substance
3.) Calculate the
number of mols of
the required
substance  use mol
to mol ratio from the
balanced equation
4.) Convert the
number of moles of
the required
substances to a mass
or number of particles
 Use the molar
mass of the required
substance
Moles
Mass (g) of
the required
substance
Mass to Mass Calculations for Products and Reactants
Example 1:
Iron can be produced from iron ore, Fe2O3 by reacting the ore with carbon
monoxide (CO). Carbon dioxide is also produced. What mass of iron can
be formed from 425g of iron ore?
Step 1: Write out the balanced chemical equation
Fe2O3 + 3CO  2Fe + 3CO2
m= 425g
m= ?
• Step 2: Fill in chart with information you know
Balanced
Equation
Mole
Ratio
Mass (m)
Fe2O3 3CO
1
425g
Molar
159.7
Mass (M) g/mol
Moles (n)
3
2Fe
3CO2
2
3
?g
Step 3: Convert given mass into moles (n=m/M)
Moles of Fe2O3 = 425g
159.7g/mol
= 2.66 mol of Fe2O3
• Fill in chart with information
Balanced
Equation
Mole
Ratio
Mass (m)
Fe2O3 3CO
1
425g
Molar
159.7
Mass (M) g/mol
Moles (n) 2.66mol
3
2Fe
3CO2
2
3
?g
Mass to Mass Calculations for Products and Reactants
Step 4: Calculate the number of mols of the required substance (mol to mole ratio)
Fe2O3 + 3CO  2Fe + 3CO2
x mols of Fe
= 2 mol of Fe
2.66 mols of Fe2O3
1mol of Fe2O3
= 5.32 mols of Fe
or
2.66 mols of Fe2O3
X
2 mol of Fe
1 mols Fe2O3
= 5.32 mols of Fe
• Fill in chart with information
Balanced
Equation
Mole
Ratio
Mass (m)
Fe2O3 3CO
1
425g
Molar
159.7
Mass (M) g/mol
Moles (n) 2.66mol
3
2Fe
3CO2
2
3
?g
55.85
g/mol
5.32 mol
Step 5: Convert moles of required substance to
the mass
mass of Fe = 5.32 mols of Fe X 55.85g/mol of Fe
= 297g of Fe
Mass to Mass Calculations for Reactants
Example 2:
What mass of hydrazine (N2H4) is required to react completely with 1000g
of dinitrogen tetraoxide (N2O4)?
Step 1: Write out the balanced chemical equation
2N2H4
m= ?
+
N2O4  3N2
m= 1000g
+ 4H2O
• Step 2: Fill in chart with information you know
Balanced
Equation
2N2H4
N2O4
3N2
4H2O
Mole
Ratio
Mass (m)
2
1
3
4
?g
1000g
Molar
Mass (M)
Moles (n)
92 g/mol
Step 3: Convert given mass into moles (n=m/M)
moles of N2O4 = 1000g
92g/mol
= 10.87 mol of N2O4
• Fill in chart with information you know
Balanced
Equation
2N2H4
N2O4
3N2
4H2O
Mole
Ratio
Mass (m)
2
1
3
4
?g
1000g
Molar
Mass (M)
Moles (n)
92 g/mol
10.87mol
Mass to Mass Calculations for Reactants
Step 4: Calculate the number of mols of the required substance (mol to mole ratio)
2N2H4
+
N2O4  3N2
x mols of N2H4
10.87 mols of N2O4
+ 4H2O
= 2 mols of N2H4
1 mol of N2O4
= 21.74 mols of N2H4
OR
10.87 mols of N2O4
X
2 mol of N2H4
1 mols N2O4
= 21.74 mols of N2H4
• Fill in chart with information you know
Balanced
Equation
2N2H4
N2O4
3N2
4H2O
Mole
Ratio
Mass (m)
2
1
3
4
?g
1000g
Molar
32 g/mol 92 g/mol
Mass (M)
Moles (n) 21.74 mol 10.87mol
Step 5: Convert moles of required substance to the mass
mass of N2H4 = 21.74 mols of N2H4 X 32g/mol
= 695.7 of N2H4
Mass to Mass Calculations for particles/molecules
Example 3:
Ammonia gas react with oxygen gas to produce water and nitrogen
monoxide. How many molecules of oxygen are required to completely
react with 23.0g of ammonia?
Step 1: Write out the balanced chemical equation
4NH3
m= 23.0g
+
5O2

6H2O
m= ?
Step 2: Convert given mass into moles
Mols of NH3 =
23.0g of NH3
17g/mol of NH3
= 1.35 mols of NH3
+
4NO
Mass to Mass Calculations for particles/molecules
Step 3: Calculate the number of mols of the required substance (mol to mole ratio)
4NH3
+
5O2
x mols of O2 =
1.35 mols of NH3

6H2O
+
4NO
5 mols of O2
4 mols of NH3
= 1.68 mols of O2
Step 4: Convert moles of required substance to molecules
# of molecules of O2 = 1.68 mols of O2 X 6.022 X 1023 molecules of O2
1 mol
= 1.02 X 1024 molecules
Practice Questions
p. 244 #11,12, 14
p. 246 #15, 16, 18
p. 248 # 19, 20
p. 249 #1, 4, 6, 7