A certain compound, used as a welding fuel, contains carbon and hydrogen only. Burning
completely a small sample of the compound in oxygen gives 3.38 g of CO2, 0.692 g of water
and no other products. Calculate the empirical formula of the compound.
Answers
Answer:
Since, 3.38 g of carbon dioxide are obtained, the mass of C present in the sample of welding fuel gas is
44
12
×3.38=0.92 g.
Since, 0.690 g of water are obtained, the mass of hydrogen present in the welding fuel gas sample is
18
2
×0.690=0.077g.
The percentage of C is
0.92+0.077
0.92
×100=92.3 %.
The percentage of H is
0.92+0.077
0.077
×100=7.7%.
(i) The number of moles of carbon =
12
92.2
=7.7.
The number of moles of hydrogen =
1
7.7
=7.7.
The mole ratio C:H=
7.7
7.7
=1:1.
Hence, the empirical formula of the welding fuel gas is CH.
(ii) The weight of 10.0 L of gas at N.T.P is 11.6 g.
22.4 L pf gas at N.T. P will weigh
10.0
11.6
×22.4=26g/mol
This is the molar mass.
(iii) Empirical formula mass is 12+1=13.
Molecular mass is 26.
The ratio, n of the molecular mass to empirical formula mass is
13
26
=2.
The molecular formula is 2(CH)=C
2
H
2
.
Answer:
Empirical formula of the compound = CH
Now molecular formula calculation
Empirical formula mass = 12 + 1 = 13 amu
Also molecular mass = 26 g
Therefore n = molecular mass/empirical formula mass = 26/13 = 2
Explanation:
Since, 3.38 g of carbon dioxide are obtained, the mass of C present in the sample of welding fuel gas is
4412×3.38=0.92 g.
Since, 0.690 g of water are obtained, the mass of hydrogen present in the welding fuel gas sample is
182×0.690=0.077g.
The percentage of C is 0.92+0.0770.92×100=92.3 %.
The percentage of H is 0.92+0.0770.077×100=7.7%.
(i) The number of moles of carbon =1292.2=7.7.
The number of moles of hydrogen =17.7=7.7.
The mole ratio C:H=7.77.7=1:1.
Hence, the empirical formula of the welding fuel gas is CH.
(ii) The weight of 10.0 L of gas at N.T.P is 11.6 g.
22.4 L pf gas at N.T. P will weigh 10.011.6×22.4=26g/mol
This is the molar mass.
(iii) Empirical formula mass is 12+1=13.
Molecular mass is 26.
The ratio, n of the molecular mass to empirical formula mass is 1326=2.
The molecular formula is 2(CH)=C2H2.