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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 10.0 litre (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate molecular formula of gas.

A
C2H4
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B
C4H8
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C
C2H2
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D
None of the above
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Solution

The correct option is C C2H2
Since, 3.38 g of carbon dioxide are obtained, the mass of C present in the sample of welding fuel gas is
12×3.3844=0.92 g
Since, 0.690 g of water are obtained, the mass of hydrogen present in the welding fuel gas sample is
2×0.69018=0.077 g



The percentage of C is 0.92×1000.0770+.92=92.30%


The percentage of H is .077×1000.0770+.92=7.70%


(i) The number of moles of carbon =92.3012=7.7.

The number of moles of hydrogen =7.701=7.7.


The mole ratio C:H=7.7:7.7=1:1.

Hence, the empirical formula of the welding fuel gas is CH.


The empirical formula of welding gas is CH and empirical formula mass of welding gas =13.
Molecular mass = (empirical formula mass)×n
n=25.98132
Molecular formula =2× empirical formula
2×(CH)=C2H2

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