Question

A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives $3.38g$ carbon dioxide, $0.690g$ of water and no other products. A volume of $10.0L$ (measured at STP) of this welding gas is found to weigh $11.6g$.

Calculate (i) empirical formula, (ii) molar mass of the gas and (iii) molecular formula.

Answer 1

Since, 3.38 g of carbon dioxide are obtained, the mass of C present in the sample of welding fuel gas is
$\frac{12}{44}\times 3.38=0.92$ g.
Since, 0.690 g of water are obtained, the mass of hydrogen present in the welding fuel gas sample is
$\frac{2}{18}\times 0.690=0.077$g.
The percentage of C is $\frac{0.92}{0.92+0.077}\times 100=92.3$ %.
The percentage of H is $\frac{0.077}{0.92+0.077}\times 100=7.7$%.

(i) The number of moles of carbon $=\frac{92.2}{12}=7.7$.
The number of moles of hydrogen $=\frac{7.7}{1}=7.7$.
The mole ratio $C:H=\frac{7.7}{7.7}=1:1$.
Hence, the empirical formula of the welding fuel gas is CH.

(ii) The weight of 10.0 L of gas at N.T.P is 11.6 g.
22.4 L pf gas at N.T. P will weigh $\frac{11.6}{10.0}\times 22.4=26g/mol$
This is the molar mass.

(iii) Empirical formula mass is $12+1=13$.
Molecular mass is 26.
The ratio, n of the molecular mass to empirical formula mass is $\frac{26}{13}=2$.
The molecular formula is $2\left(CH\right)=C_2{H}_2$.