A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g of carbon dioxide. 0.69 g of water and no other products. A volume of 10 L (measured at STP) of this gas is found to weigh 11.6g. Calculate a) empirical formula b) molar mass of the gas c) molecular formula
Answers
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
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%.
(i) The number of moles of carbon =
92.2/12 = 7.7
The number of moles of hydrogen =
7.7/1 = 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
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
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