Percentage composition Questions - Shailendra Kumar Chemistry

PHYSICAL CHEMISTRY
By: Shailendra Kumar
1. An organic substance gave on analysis the following percentage composition, C = 60%,
H = 6.4%, N = 10.1%, O = 23.5%. The empirical formula of the compound is :
(a) C 6
H 8
NO 2
(b) C 6
H 5
NO 2
(c) C 7
H 9
NO 2
(d) C 7
H 10
NO 2
2. An organic compound has the following percentage composition, C = 40% H = 6.7%,
and O = 53.3%. The empirical formula of the compound is :
(a) CH 2
O (b) CHO
(c) CHO 3
(d) C 2
H 2
O 2
3. 0.30 g of an organic substance on combustion gave 0.66 g of carbon dioxide and 0.36
g of water. Its V.D. is 30. The molecular formula is :
(a) C 3
H 8
O (b) C 2
H 4
O 2
(c) C 8
H 3
O 2
(d) C 4
H 4
O
4. An organic compound on analysis gave the following results :
C = 65%, H = 3.5%, N = 9.6%, O = 21.9%
0.365 g if the substance displaced in a Victor Meyer’s apparatus 56 mL of dry air at
S.T.P. The molecular formula of the substance is :
(a) C 8
H 5
NO 2
(b) C 4
H 5
NO 2
2
(c) C 6
H 5
N 2
O 2
(d) C 4
H 6
N 4
O 4
5. A compound contains the following percentage of elements :- C = 12.76, H = 2.13 and
Br = 85.11. Its vapour density is 95. Its molecular formula is :
(a) CH 3
Br (b) C 2
H 4
Br 2
(c) C 3
H 6
Br 2
(d) C 4
H 4
Br 4
6. The empirical formula of a compound is CH 2
. 0.083 mole of the compound contains 2.0
g of carbon, the molecular formula of the compound is :
(a) C 4
H 8
(b) C 3
H 6
(c) C 2
H 4
(d) C 2
H 2
7. 0.1 mole of a compound of S and Cl on oxidation with conc. HNO 3
and subsequent
treatment gave 0.1 mole of PbCl 2
and 0.2 mole of BaSO 4
as precipitates. The molecular
formula of the compound is :
(a) SCl (b) S 2
Cl 2
(c) SCl 4
(d) SCl 3
8. Two elements A (at.wt. = 12) and B (at. wt. = 16) combine to yield a compound. The
percentage by weight of A in the compound is 27.3 %. The formula of the compound will
be :
(a) A 2
B 2
(b) AB
(c) A 2
B (d) AB 2
9. On analysis of an organic compound the following results were obtained :
C = 31.9%, H = 6.8%, N = 18.5%
The empirical formula of the compound is :
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PHYSICAL CHEMISTRY
1. Opp. Khuda Baksh Library, Ashok Rajpath, Patna - 4
2. House no. 5A/65, Opp. Mahual Kothi, Alpana Market, Patna
By: Shailendra Kumar
(a) C 2
H 5
NO (b) CH 3
NO 2
(c) C 2
H 4
N (d) C 2
H 5
NO 2
10. 0.0833 mole of the carbohydrate (empirical formula CH 2
O) contains 1.00 g of hydrogen.
The molecular formula of the carbohydrate is :
(a) C 6
H 12
O 6
(b) C 5
H 10
O 5
(c) C 2
H 5
N (d) C 2
H 5
NO 2
11. 0.220 g of an organic compound gave upon analysis 0.6179 g of CO 2
and 0.124 g of
H 2
O. The empirical formula of the compound is :
(a) C 6
H 6
O 2
(b) C 5
H 6
O 2
(c) C 6
H 6
O (d) C 6
H 4
O 3
12. A gaseous compound of carbon and nitrogen containing 53.8 percent by weight of
nitrogen was found to have a vapour density of 25.8. The molecular formula of the
compound is :
(a) CN (b) C 2
N 2
(c) CN 2
(d) C 2
N 3
13. The percentage of oxygen in heavy water is :
(a) 80 (b) 60
(c) 88.9 (d) 50
14. 0.2154 g of an organic compound gave on combustion 0.4308 g of CO 2
and 0.1766 g of
H 2
O. The vapour density of the compound is 44. The molecular formula of the compound
is :
(a) C 2
H 2
O (b) C 4
H 8
O 2
(c) C 3
H 4
O 3
(d) C 2
H 2
O 2
15. A quantity of 2.40 g of the oxide of a metal X (atomic mass = 55.9 a.m.u) was heated
in carbon monoxide. The mass of the metal formed is 1.68 g. The simplest formula of
the oxide is :
(a) XO (b) X 2
O
(c) X 2
O 3
(d) XO 2
16. 64 g of an organic compound consisting of carbon, hydrogen and oxygen contains in
the ratio 3 : 1 : 4 by weight, respectively. The empirical formula of the compound is (a)
CH 2
O (b) C 2
H 4
O
(c) C 2
H 6
O 2
(d) CH 4
O
17. An organic compound containing C, H and O is found to have 32% C, 4% H, the
remaining being oxygen. The molecular formula of the compound if it contains six
atoms of oxygen per molecule is :
(a) C 2
H 2
O 6
(b) C 4
H 6
O 6
(c) C 6
H 6
O 6
(d) C 3
H 6
O 6
18. A compound contains 28% nitrogen and 72% metal by mass. 3 atoms of the metal
combine with 2 atoms of nitrogen. The atomic mass of the metal is :
(a) 23 (b) 24
(c) 39 (d) 40
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PHYSICAL CHEMISTRY
By: Shailendra Kumar
19. A substance on analysis gave the following percentage composition : Na = 27.38%, H =
1.19%, C = 14.28%, O = 57.14%. The empirical formula of the compound is :
(a) NaHC 3
(b) Na 2
CO 3
(c) Na 2
CO 3
.10 H 2
O (d) Na 2
CO 3
.7 H 2
O
20. Three of the following formulae might be either empirical or molecular formula but one
of the four must be a molecular formula. The one is :
(a) N 2
O (b) N 2
O 4
(c) NH 3
(d) Mg 3
N 2
21. A gas is found to have the formula (CO)n. Its vapour density is 70. The value of n will be
:
(a) 1 (b) 2 (e) 5
(c) 3 (d) 4
22. A gaseous compound contains carbon and hydrogen in atomic ratio 1 : 2. 0.32 g of
oxygen and 0.42 g of the conpound occupies the same volume under similar conditions
of temperature and pressure. The molecular formula of the compound is :
(a) C 2
H 4
(b) C 3
H 6
(c) C 4
H 8
(d) C 5
H 10
23. The compound C 7
H 7
NO 2
:
(a) contains 17 atoms per mole
(b) contains equal percentage of C and H by weight
(c) contains twice the percent by mass of oxygen as of nitrogen
(d) contains twice the percent by mass of nitrogen as of hydrogen
24. A compound consisting of only carbon and chlorine shows on analysis 10.15% carbon
and 89.85% chlorine . If each molecule weighs 3.93 x 10 –22 g, the molecular formula of
the compound would be :
(a) CCl 4
(b) C 2
Cl 6
(c) C 2
Cl 4
(d) C 2
Cl 2
25. 10 g of hydrofluric acid occupies 5.60 litres at S.T.P. The empirical formula of the gas
is HF, the molecular formula in gaseous state will be :
(a) HF (b) H 2
F 2
(c) H 3
F 3
(d) H 4
F 4
26. The elements A (at.wt. = 75) and B (at. wt. = 25) combine to yield a compound. The % by
weight of A in the compound was found to be 75.0. The formula of the compound is :
(a) AB (b) AB 2
(c) A 2
B 2
(d) A 2
B
27. The formula which represents the simple ratio of atoms is :
(a) molecular formula
(b) structureal formula
(c) empirical formula
(d) rational formula
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2. House no. 5A/65, Opp. Mahual Kothi, Alpana Market, Patna
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PHYSICAL CHEMISTRY
By: Shailendra Kumar
28. A 24.00 g sample of an oxide of iron contains 6.63 g of oxygen. The simplest formula of
the compound (Fe = 56) would be :
(a) Fe 3
O 4
(b) FeO
(c) Fe 2
O 3
(d) Fe 4
O 6
29. Atomic composition of entire universe is approximately 93% hydrogen and 7% helium.
The percentage by weight of hydrogen is :
(a) 72.8 (b) 76.8 (c) 93 (d) 95.8
30. A sample of hydrocarbon on combustion gave 7.92 g of CO 2
and 3.24 g of water. The
weight of the sample that was burnt was :
(a) 2.52 g (b) 2.0 g
(c) 1.52 g (d) 1.0 g
31. The relative number of atoms in a hydrocarbon is C = 0.75 and H =3. The empirical
formula of the hydrocarbon is :
(a) C 3
H 8
(b) C 2
H 6
(c) CH 4
(d) none
32. The empirical formula of an organic compound containing carbon and hydrogen is
CH 2
. The mass of 1.0 L of this compound is exactly equal to that of 1.0 L of N 2
under
identical conditions. The molecular formula of the hydrocarbon is :
(a) C 2
H 4
(b) C 3
H 6
(c) C 6
H 12
(d) C 4
H 8
33. An organic compound has empirical formula CH 2
O and its molecular weight is 90. Its
molecular formula is :
(a) C 2
H 4
O (b) C 3
H 4
O 3
(c) C 3
H 6
O 3
(d) C 2
H 4
O 2
34. Molecular weight of a compound is 180. 0.0833 mole of the compound contains 5 g of
carbon. The molecular formula of the compound is :
(a) C 5
H 10
O 5
(b) C 9
H 8
O 4
(c) C 5
H 8
O 7
(d) C 10
H 12
O 3
35. The empirical formula of the compound containing 50% of element A (atomic weight
10) and 50% of element B (at.wt.20) is :
(a) AB (b) A 2
B
(c) AB 2
(d) A 2
B 2
36. An organic compound contians x% C, y% H, and z% O. If x =6y and (x-y) = 3 (z-y), the
empirical formula of the compound is :
(a) CH 2
O (b) CHO
(c) C 3
H 8
O 3
(d) C 3
H 6
O
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