Question

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 $11.6 \,g$. The molecular formula of gas is

• A. $C_2H_2$
• B. $CH_4$
• C. $C_2H_4$
• D. $C_2H_6$

Answer 1

Answer: (A)

Number of moles of $CO_2 = \frac{3.38}{44} = 0.0768$
No. of moles of $C = 0.0768 $; No. of moles of $H_2O$
$ = \frac{0.690}{18}$
$ = 0.0383$
$\therefore $ No. of moles of $H = 2 \times 0.0383 = 0.0766$
(i) The ratio of moles of $C$ to $H$ is $0.0768 : 0.0766$ or $1 : 1$
Therefore, empirical formula $= CH$
(ii) $10.0\, L$ of fuel gas at STP weighs
$ = \frac{11.6 \times 22.4}{10} = 25.98\,g$
$\therefore $ Molar mass of gas $= 25.98\, g = 26 \,g \,mol^{-1}$
(iii) $n = \frac{\text{Molar mass}}{\text{Empirical formula mass}}$
$ = \frac{26}{13} = 1$
$\therefore $ Molecular formula = (empirical formula)$_n$
$ = (CH)_2 = C_2H_2$