A welding fuel gas contains carbon and hydrogen only. Burning a small sample

Solution:

(i) 1 mole $(44 \mathrm{~g})$ of $\mathrm{CO}_{2}$ contains $12 \mathrm{~g}$ of carbon.

$\therefore 3.38 \mathrm{~g}$ of $\mathrm{CO}_{2}$ will contain carbon

$=\frac{12 \mathrm{~g}}{44 \mathrm{~g}} \times 3.38 \mathrm{~g}$

$=0.9217 \mathrm{~g}$

$18 \mathrm{~g}$ of water contains $2 \mathrm{~g}$ of hydrogen.

$\therefore 0.690 \mathrm{~g}$ of water will contain hydrogen

$=\frac{2 \mathrm{~g}}{18 \mathrm{~g}} \times 0.690$

$=0.0767 \mathrm{~g}$

Since carbon and hydrogenare the only constituents of the compound, the total mass of the compound is

= 0.9217 g + 0.0767 g

= 0.9984 g

$\therefore$ Percent of $C$ in the compound

$=\frac{0.9217 \mathrm{~g}}{0.9984 \mathrm{~g}} \times 100$

$=92.32 \%$

Percent of $\mathrm{H}$ in the compound $=\frac{0.0767 \mathrm{~g}}{0.9984 \mathrm{~g}} \times 100$

$=7.68 \%$

Moles of carbon in the compound

$=\frac{92.32}{12.00}$

= 7.69

Moles of hydrogen in the compound

$=\frac{7.68}{1}$

= 7.68

$\therefore$ Ratio of carbon to hydrogen in the compound $=7.69: 7.68=1: 1$

Hence, the empirical formula of the gas is CH.

(ii) Given,

Weight of $10.0 \mathrm{~L}$ of the gas (at S.T.P) $=11.6 \mathrm{~g}$

$\therefore$ Weight of $22.4 \mathrm{~L}$ of gas at STP

$=\frac{11.6 \mathrm{~g}}{10.0 \mathrm{~L}} \times 22.4 \mathrm{~L}$

$=25.984 \mathrm{~g}$

$\approx 26 \mathrm{~g}$

Hence, the molar mass of the gas is 26 g.

(iii) Empirical formula mass of $\mathrm{CH}=12+1=13 \mathrm{~g}$

$n=\frac{\text { Molar mass of gas }}{\text { Empirical formula mass of gas }}$

$=\frac{26 \mathrm{~g}}{13 \mathrm{~g}}$

$n=2$

$\therefore$ Molecular formula of gas $=(\mathrm{CH})_{n}$

$=\mathrm{C}_{2} \mathrm{H}_{2}$