$A$ welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives $3.38 \ g$ of carbon dioxide,$0.690 \ g$ of water and no other products. $A$ volume of $10.0 \ L$ of this welding gas (measured at $STP$) is found to weigh $11.6 \ g$. Calculate: $(i)$ Empirical formula,$(ii)$ Molar mass of the gas,$(iii)$ Molecular formula.

(N/A) Step $1$: Calculate the mass of $C$ and $H$ in the sample.
Mass of $C = \frac{12}{44} \times 3.38 \ g = 0.9218 \ g$
Mass of $H = \frac{2}{18} \times 0.690 \ g = 0.0767 \ g$
Total mass of sample $= 0.9218 + 0.0767 = 0.9985 \ g$
Step $2$: Calculate empirical formula.

Element	Mass %	Atomic Mass	Moles	Ratio
$C$	$92.32$	$12$	$7.69$	$1$
$H$	$7.68$	$1$	$7.68$	$1$

Empirical formula is $CH$.
Step $3$: Calculate molar mass.
$10.0 \ L$ at $STP$ weighs $11.6 \ g$.
$22.4 \ L$ at $STP$ weighs $\frac{11.6 \times 22.4}{10.0} = 25.984 \ g \approx 26 \ g/mol$.
Step $4$: Calculate molecular formula.
$n = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{26}{13} = 2$.
Molecular formula $= 2 \times CH = C_2H_2$.