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





Q.34:- 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:-
Amount of carbon in 3.38 g of CO= 12/44 × 3.38 g = 0.9218 g
Amount of hydrogen in 0.690 g H2O = 2/18 × 0.690 g = 0.0767 g
The compound contains only C and H, therefore total mass of the compound = 0.9218 + 0.0767 = 0.9985 g
% of C in the compound = (0.9218/0.9985)×100 = 92.32
% of H in the compound = (0.0767/0.9985)×100 = 7.68
(i) Calculation of empirical formula,
Moles of carbon in the compound = 92.32/12 = 7.69
Moles of hydrogen in the compound = 7.68/1 = 7.68
Simplest molar ratio = 7.69 : 7.68 = 1(approx)
∴ Empirical formula CH
(ii) 10.0 L of the gas at STP weigh = 11.6 g
∴ 22.4 L of the gas at STP = 11.6/10.0 × 22.4 = 25.984 = 26 (approx)
∴ Molar mass of gass = 26 g mol-1
(iii) Mass of empirical formula CH = 12+1 = 13
∴ n = Molecular Mass/Empirical Formula = 26/13 = 2
∴ Molecular Formula = C2H2