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.
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
.