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 10.0 L (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 (ii) molecular
formula.
Answers
Answer:
ANSWER
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)=C2H
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