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 litre (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate : Molecular formula.

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

Answer 1

The correct option is A $C_2{H}_2$

Molecular mass of welding gas $=\frac{11.6\times 22.4}{10}=25.98$ $C{O}_2$ and $H_{2}O$ are obtained from welding gas combustion in water. $\therefore$ g-atoms of $C$ in gas $=\frac{3.38}{44}=0.077$ g-atoms of $H$ in gas $=\frac{0.690\times 2}{18}=0.077$ $\therefore$ Ratio of $C$ and $H$ atoms in gas is $1:1$ Thus, empirical formula of welding gas is $CH$ Empirical formula mass of welding gas $=13$ Molecular mass $=$ (empirical formula mass)$\times n$ $\therefore n=\frac{25.98}{13}\simeq 2$ $\therefore$ Molecular formula $=2\times$ empirical formula $=2\times \left(CH\right)=C_2{H}_2$