$A$ 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 $10.0 \, L$ (measured at $STP$) of this welding gas is found to weigh $11.6 \, g$. Calculate:
$(i)$ empirical formula,
$(ii)$ molar mass of the gas,and
$(iii)$ molecular formula.

(N/A) $(i)$ $1 \, \text{mole} \, (44 \, g)$ of $CO_{2}$ contains $12 \, g$ of carbon.
$\therefore 3.38 \, g$ of $CO_{2}$ contains carbon $= \frac{12 \, g}{44 \, g} \times 3.38 \, g = 0.9217 \, g$.
$18 \, g$ of water contains $2 \, g$ of hydrogen.
$\therefore 0.690 \, g$ of water contains hydrogen $= \frac{2 \, g}{18 \, g} \times 0.690 \, g = 0.0767 \, g$.
Total mass of the sample $= 0.9217 \, g + 0.0767 \, g = 0.9984 \, g$.
Moles of carbon $= \frac{0.9217 \, g}{12.01 \, g/mol} \approx 0.0767 \, \text{mol}$.
Moles of hydrogen $= \frac{0.0767 \, g}{1.008 \, g/mol} \approx 0.0761 \, \text{mol}$.
Ratio of $C:H \approx 1:1$. Hence,the empirical formula is $CH$.
$(ii)$ Given,$10.0 \, L$ of gas at $STP$ weighs $11.6 \, g$.
$1 \, \text{mole}$ of gas at $STP$ occupies $22.4 \, L$.
Molar mass $= \frac{11.6 \, g}{10.0 \, L} \times 22.4 \, L/mol = 25.984 \, g/mol \approx 26 \, g/mol$.
$(iii)$ Empirical formula mass of $CH = 12 + 1 = 13 \, g/mol$.
$n = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{26}{13} = 2$.
Molecular formula $= (CH)_{2} = C_{2}H_{2}$.