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 of gas.

• A. $C_2{H}_4$
• B. $C_4{H}_8$
• C. $C_2{H}_2$
• D. None of the above

Answer 1

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

Since, 3.38 g of carbon dioxide are obtained, the mass of C present in the sample of welding fuel gas is
$\frac{12\times 3.38}{44}=0.92g$
Since, 0.690 g of water are obtained, the mass of hydrogen present in the welding fuel gas sample is
$\frac{2\times 0.690}{18}=0.077g$



The percentage of C is $\frac{0.92\times 100}{0.0770+.92}=92.30$%


The percentage of H is $\frac{.077\times 100}{0.0770+.92}=7.70$%


(i) The number of moles of carbon =$\frac{92.30}{12}=7.7$.

The number of moles of hydrogen =$\frac{7.70}{1}=7.7$.


The mole ratio $C:H=7.7:7.7=1:1$.

Hence, the empirical formula of the welding fuel gas is $CH$.


The empirical formula of welding gas is $CH$ and 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
$\Rightarrow 2\times \left(CH\right)=C_2{H}_2$