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.

A welding fuel gas contains carbon and hydrogen only.Burning a small s

1.	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.
Answer» Solution :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)`