A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide ,

Amount of carbon in 3.38 g of CO₂ = 12/44 × 3.38 g = 0.9218 gAmount 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 = 1:1 (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.