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 (iii) molecular formula.

Explanation

Amount of carbon in 3.38 g `CO_(2) = 12/44 xx 3.38 = 0.9218 g`
Amount of hydrogen in 0.690 g `H_(2)O =2/18 xx 0.690 = 0.0767 g`
As the fuel gas contains only carbon and hydrogen, total mass of the gas burnt =` 0.9218 + 0.0767 = 0.9985` g
`therefore` of C in the fuel gas = `(0.9218)/0.9985 xx 100 = 92.318`
% of H in the fuel gas = `0.0767/0.9985 xx 100 = 7.682`
(i) Calculation of empirical formula:

`therefore` Empirical formula = CH
(ii) Calculation of molar mass:
`therefore 10.0 L` of the gas at STP weigh = 11.6 g
`therefore 22.4 L` of the gas at STP will weigh `=11.6/10.0 xx 22.4 = 25.984 g = 26 g`
Molar mass of the gas `=26 g "mol"^(-1)`
(iii) Calculation of molecular formula :
`n=("Molar mass")/("Empirical formula mass") = 26/(12 + 1) = 26/13 = 2`
`therefore` Molecular formula of the gas `=2 xx (CH) = C_(2)H_(2)`