Question

A welding fuel gas contains carbon and
hydrogen only. Burning a small sample of it in
oxygen gives 3.38 g carbon dioxide, 0.690 g of
water and no other products. A volume of 10.0L
(measured at STP) of this welding gas is found
to weigh 11.6 g. Calculate (i) empirical formula,
(ii) molar mass of the gas, and (iii) molecular
formula.​

Answer 1

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

.