Empirical formula:
1
Hydroquinone, used as a photographic developer is 65.4% C,
5.5% H, and 29.1% O, by mass. What is the empirical
formula?
If we work with 100 grams, we would have 65.4 g C, 5.5 g H
and 29.1 g O.
Then: Change g to moles and convert to whole numbers.
C:
65.4 g
m ol
12.0 g
= 5.42 mol C
 1.82  2.97
 1.82  2.99
5.5 g
m ol
1.01 g
= 5.45 mol H
O: 29.1 g
m ol
16.0g
= 1.82 mol O
H:
C3H3O1
 1.82  1
Divide by smallest to get whole number
2
A compound was sent out for analysis and was found to contain
6.7% H, 39.9% C, and 53.4% O. It was also found to have a
molecular mass of 60.0 g/mol.
Determine both the empircal and the molecular formulas.
If we work with 100
grams, we could call
the %’s grams!
3
A compound was sent out for analysis and was found to contain
6.7% H, 39.9% C, and 53.4% O. It was also found to have a
molecular mass of 60.0 g/mol.
Determine both the empirical and the molecular formulas.
H:
6.7 g m ol
= 6.64
1.01g
C:
39.9 g mol
12.01g
O: 53.4g
= 3.32
mol = 3.33
16.00g
 3.32 = 2.00
 3.32 = 1.00
 3.32 = 1.00
empirical formula:
CH2O
empirical formula wt.:
30.03
molecular formula:
60.0  30.03  2.0
so: C2H4O2
We need
whole
numbers
4
Stoichiometry: “working with ratios”
5
When N2O5 is heated, it decomposes:
2N2O5(g)  4NO2(g) + O2(g)
a. How many moles of NO2 can be produced from 4.3 moles of N2O5?
2N2O5(g)  4NO2(g) + O2(g)
4.3 mol
? mol
4.3 mol N2O5
4mol NO 2
2mol N 2O 5
= 8.6
moles NO2
b. How many moles of O2 can be produced from 4.3 moles of N2O5?
2N2O5(g)  4NO2(g) + O2(g)
4.3 mol
4.3 mol N2O5
1mol O 2
2mol N 2O 5
? mol
= 2.15
mole O2
6
When N2O5 is heated, it decomposes:
2N2O5(g)  4NO2(g) + O2(g)
210 g
a. How many moles of N2O5 were used if 210g of NO2 were produced?
210 g NO2
mol NO 2
46.0g NO 2
2mol N 2O 5
= 2.283 moles N2O5
4mol NO 2
b. How many grams of N2O5 are needed to produce 75.0 grams of O2?
75.0 g O2
mol O 2 2mol N 2O 5
32.0 g O 2 1mol O 2
108g N 2O 5
= 506.25 grams N2O5
m ol N 2O 5
7
Limiting/excess/theoretical yield Reactant Problems:
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
Determine the limiting reactant.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
First copy down the
the BALANCED
equation!
8
Limiting/excess/theoretical yield Reactant Problems:
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
Determine the limiting reactant.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
Now place
numerical the
information below
the compounds.
9
Limiting/excess/theoretical yield Reactant Problems:
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
Determine the limiting reactant.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
0.15 mol
? moles
0.10 mol
Hide
Two starting
amounts?
Where do we
start?
one
10
Limiting/excess/theoretical yield Reactant Problems:
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
Determine the limiting reactant.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
0.15 mol
? moles
Hide
0.10
mol
Based on:
0.15 mol KO2
KO2
3molO 2
4molKO 2
= 0.1125 mol O2
11
Limiting/excess /theoretical yield Reactant Problems:
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
Determine the limiting reactant.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
0.15
mol
? moles
Hide
0.10 mol
Based on:
0.15 mol KO2
KO2
Based on:
H2 O
3molO 2
4molKO 2
0.10 mol H2O 3molO 2
2molH 2O
= 0.1125 mol O2
= 0.150 mol O2
12
Limiting/excess /theoretical yield Reactant Problems:
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
Determine the limiting reactant.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
0.15 mol
? moles
0.10 mol
Based on:
0.15 mol KO2
KO2
Based on:
H2 O
3molO 2
4molKO 2
0.10 mol H2O 3molO 2
2molH 2O
= 0.1125 mol O2
it was limited!
= 0.150 mol O2
H2O = XS reactant!
What is the theoretical
yield? Hint: which reactant
was the limiting reactant?
13
Theoretical yield vs. Actual yield
Suppose the theoretical yield for an
experiment was calculated to be
19.5 grams, and the experiment was
performed, but only 12.3 grams of
product were recovered. Determine
the % yield.
Theoretical yield = 19.5 g based on limiting reactant
Actual yield = 12.3 g experimentally recovered
actualyie ld
% yie ld
x 100
the ore tica
l yie ld
% yield 
12.3
x 100  63.1% yield
19.5
14
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O,
how many grams of O2 can be produced?
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
?g
120.0 g
47.0 one
g
Hide
Based on: 120.0 g KO2
KO2
mol 3mol O 2 32.0g O 2
71.1g 4mol KO 2 mol O 2
= 40.51 g O2
15
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O,
how many grams of O2 can be produced?
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
?g
120.0
47.0 g
Hideg
Based on: 120.0 g KO2
KO2
mol 3mol O 2 32.0g O 2
71.1g 4mol KO 2 mol O 2
mol H 2O 3 molO 2 32.0g O 2
Based on: 47.0 g H2O
18.02 g H 2O 2 mol H 2O mol O 2
H2 O
= 40.51 g O2
= 125.3 g O2
Question if only 35.2 g of O2 were recovered, what was the percent yield?
35.2
x 100  86.9% yield
40.51
16
If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O,
how many grams of O2 can be produced?
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
?g
120.0 g
47.0 g
Based on: 120.0 g KO2
KO2
mol 3mol O 2 32.0g O 2
71.1g 4mol KO 2 mol O 2
= 40.51 g O2
mol H 2O 3 molO 2 32.0g O 2
Based on: 47.0 g H2O
18.02 g H 2O 2 mol H 2O mol O 2
H2 O
= 125.3 g O2
Determine how many grams of Water were left over.
The Difference between the above amounts is directly RELATED to the XS H2O.
125.3 - 40.51 = 84.79 g of O2 that we have XS water to form.
2 mol H 2O 18.02 g H 2O
32.0 g O 2 3 mol O 2 1 mol H 2O
84.79 g O2 mol O 2
= 31.83 g XS H2O
17