Mass Relationships in
Chemical Reactions
Chapter 3
Empirical Formula of a Hydrate
Hydrated copper (II) sulfate has the formula
CuSO4· x H2O.
To find x, 1.023 g of the blue solid is heated in
a crucible until its mass no longer decreases.
The mass of the anhydrous, white CuSO4 is
0.654 g.
How many moles of water (x) are there per
mole of CuSO4?
Mass Changes in Chemical Reactions
a.k.a. “Stoichiometry”
Methanol burns in air according to the equation
2CH3OH + 3O2
2CO2 + 4H2O
If 209 g of methanol are used up in the combustion,
what mass of water is produced?
grams CH3OH
moles CH3OH
molar mass
CH3OH
209 g CH3OH x
moles H2O
grams H2O
molar mass
coefficients
H2O
chemical equation
4 mol H2O
18.0 g H2O
1 mol CH3OH
=
x
x
32.0 g CH3OH
2 mol CH3OH
1 mol H2O
235 g H2O
Given the following balanced equation,
2Mg + O2 → 2MgO
(a) how many grams of oxygen are required
to completely react with 0.145 g of
magnesium?
(b) How many grams of magnesium oxide
are formed?
Limiting Reagent
• Reactant that is present in the smaller or smallest
required stoichiometric ratio
2 H2 (g) + O2 (g) → 2 H2O (v)
Fig 3.15
2 H2 (g) + O2 (g) → 2 H2O (v)
Fig 3.15
• i.e., in this case O2 is in excess
and H2 is the limiting reagent (LR)
• The amount of product depends on LR alone!!
How do we determine the limiting reagent?
Calculate the theoretical yield twice:
Assume first reagent is LR and calculate
Assume second reagent is LR and calculate
The smaller result gives LR and yield
Sample exercise 3.19 p 104
How many grams of water will be formed
from 150 g H2 and 1500 g O2 in a fuel cell?
2 H2 (g) + O2 (g) → 2 H2O (v)
Ans: 1400 g H2O
This is the theoretical yield
• Theoretical yield - the maximum amount of
product that can be made
– In other words it’s the amount of product
possible as calculated through the
stoichiometry problem.
• This is different from the actual yield, which is
the amount one actually produces and
measures.
Actual Yield
Percent Yield =
Theoretical Yield
x 100%
Sample exercise 3.19 p 104
How many grams of water will be formed
from 150 g H2 and 1500 g O2 in a fuel cell?
2 H2 (g) + O2 (g) → 2 H2O (v)
Ans: 1400 g H2O
This is the theoretical yield
Assume, say, 1250 g H2O are formed.
1250 g H2 O
(100%)  89%
Then percent yield =
1400 g H2 O