EMPIRICAL AND MOLECULAR FORMULAS
1. An organic compound was found on analysis to contain 47.37% carbon and 10.59% hydrogen.
The balance was presumed to be oxygen. What is the empirical formula of the compound?
Ans.: C3H8O2
C
47.37 g
×
1 mol
3.944 mol
= 3.944 mol
12.01 g
H
10.59 g
×
1 mol
10.506 mol
= 10.506 mol
1.008 g
O
42.04 g
×
1 mol
= 1.501 mol
× 2 = 3.002 mol
≈ 3 mol
= 3.998 mol
× 2 = 7.996 mol
≈ 8 mol
= 1.000 mol
× 2 = 2.000 mol
≈ 2 mol
2.6275
2.6275
2.6275 mol
= 2.6275 mol
16.00 g
2.6275
2. A hydrocarbon containing 92.3% C and 7.74% H was found (by measuring its gas density) to
have a molar mass of approximately 79 g/mol. What is the molecular formula?
Ans.: C6H6
C
92.3 g
×
1 mol
7.69 mol
= 7.69 mol
12.01 g
H
7.74 g
×
1 mol
= 1.00 mol
≈ 1 mol
= 1.00 mol
≈ 1 mol
7.68
7.68 mol
= 7.68 mol
1.008 g
7.68
Molar Mass from Emp. Formula = 12.01 g/mol × 1 mol + 1.008 g/mol × 1 mol = 13.02 g/mol
n = (79 g/mol) / (13.02 g/mol) ≈ 6
Molec. Formula: (CH)n
3. A 1.500-g sample of a compound containing only C, H, and O was burned completely. The only
combustion products were 1.738 g CO2 and 0.711 g H2O. What is the empirical formula of the
compound?
Ans.: C2H4O3
1 mol CO2
1.738 g CO2
×
1 mol C
×
44.01 g CO2
1 mol CO2
1 mol H2O
0.711 g H2O
×
12.01 g C
×
2 mol H
×
18.01 g CO2
= 0.4743 g C
1 mol C
1.008 g H
×
1 mol H2O
= 0.0796 g H
1 mol H
mass of O = 1.500 g – 0.4743 g – 0.0796 g = 0.946 g
C
0.4743 g
×
1 mol
= 0.03949 mol
12.01 g
H
0.0796 g
×
1 mol
0.946 g
×
1 mol
16.00 g
= 1.000 mol
× 2 = 2.00 mol
≈2 mol
= 2.000 mol
× 2 = 4.00 mol
≈ 4 mol
= 1.50 mol
× 2 = 3.00 mol
≈ 3 mol
0.03949
= 0.07897 mol
1.008 g
O
0.03949 mol
0.07897 mol
0.03949
= 0.0591 mol
0.0591 mo
0.03949
4. One of the earliest methods for determining the molar mass of proteins was based on chemical
analysis. A hemoglobin preparation was found to contain 0.335% iron. (a) If the hemoglobin
molecule contains 1 atom of iron, what is its molar mass? (b) If it contains 4 atoms of iron, what
is its molar mass?
Ans.: (a) 1.67104 g/mol;
(b) 6.67104 g/mol
(a)
g Fe
=
g protein
0.335 g
=
55.85 g
100 g
Xg
(100 g protein)
(1 mole of protein)
Molar mass of the protein = 1.67104 g/mol
X = 16,672
(b)
g Fe
g protein
Z = 4X = 66,686
=
0.335 g
=
4×55.85 g
100 g
Zg
(100 g protein)
(1 mole of protein)
Molar mass of the protein = 6.67104 g/mol