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.
In the article on “Chemistry” in the Ninth Edition of the Encyclopaedia Britannica (published in
1878) the author (H. A. Armstrong) says that Mendeleev had recently proposed that uranium
be assigned the atomic weight 240 in place of the old value 120 that had been assigned to it
by Berzelius, but that he himself preferred 180. Mendeleev was right. The correct formula of
pitchblende, an important ore of uranium, is U3O8.What formula was written for pitchblende by
(a) Berzelius, (b) Armstrong?(From General Chemistry by L. Pauling)
Ans.: (a) U3O4; (b) UO2
Chemist
Accepted Atomic
Weight of
Uranium
Uranium-toOxygen Mole
Ratio
Accepted
Formula for
Pitchblende
Mendeleev
240
3:8
U3O8
Berzelius
120
6:8
U3O4
Armstrong
180
4:8
UO2
5. 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