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 $11.6g$. The molecular formula of gas is

• A. $C_2{H}_2$
• B. $C{H}_4$
• C. $C_2{H}_4$
• D. $C_2{H}_6$

Answer 1

Answer: (A)

Number of moles of $C{O}_2=\frac{3.38}{44}=0.0768$
No. of moles of $C=0.0768$; No. of moles of $H_{2}O$
$=\frac{0.690}{18}$
$=0.0383$
$\therefore$ 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) $10.0L$ of fuel gas at STP weighs
$=\frac{11.6\times 22.4}{10}=25.98g$
$\therefore$ Molar mass of gas $=25.98g=26gmo{l}^{-1}$
(iii) $n=\frac{\text{Molar mass}}{\text{Empirical formula mass}}$
$=\frac{26}{13}=1$
$\therefore$ Molecular formula = (empirical formula)${}_n$
$=(CH{)}_2=C_2{H}_2$