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

The balenced equation can be given as:
CnHm...+(n+m/4)O2............>nCO2.....+(m/2)H2O
small amount burns and give .3.38gCO2 and 0.690gH2O
so number of mole of Co2=3.38/44mole
and that of H2Om produced=0.690/18mole
But from the balenced we can see that ratio of moles of CO2 and that of H2O would be n/(m/2)=2*(n/m)=(3.38/44)/(0.690/18)=2
so n/m=1
So the empirical formula can ber given as CH
Now we know that at STP, 22.4L gas is equivalent to 1mol of gas
Hence 10L means 10/22.4 mol=11.6g
So molecular weight=11.6/(10/22.4)=26g/mol
Let molecular formulais(CxHx)=(CH)x
x=Molecular weight/Empirical weight=26/13=2
Hence molecular formula=C2H2

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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 gThe 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 = 7.69Moles of hydrogen in the compound = 7.68/1 = 7.68Simplest 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