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 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.
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 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.
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 $=(\mathrm{CH})_{n}$
$=\mathrm{C}_{2} \mathrm{H}_{2}$
(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 $=(\mathrm{CH})_{n}$
$=\mathrm{C}_{2} \mathrm{H}_{2}$