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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 C2H2Since, 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=25.9813≃2 ∴ Molecular formula =2× empirical formula ⇒2×(CH)=C2H2

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