Since, 3.38 g of carbon dioxide are obtained, the mass of C present in the sample of welding fuel gas is
12/44×3.38=0.92 g.
Since, 0.690 g of water are obtained, the mass of hydrogen present in the welding fuel gas sample is
2/18×0.690=0.077g.
The percentage of C is
0.92/0.92+0.077×100=92.3 %.
The percentage of H is
0.077/0.92+0.077 ×100=7.7%.
.
Explanation:
(i) The number of moles of carbon =
92.2/12
=7.7.
The number of moles of hydrogen =
7.7
=7.7.
The mole ratio C:H=
7.7/1
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
11.6/10.0×22.4=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
26/13
=2.
The molecular formula is 2(CH)=C2H2