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:

1. empirical formula,
2. molar mass of the gas, and
3. molecular formula

Answer 1

(i) 1 mole (44 g) of CO2 contains 12 g of carbon.
∴ 3.38 g of CO2 will contain carbon = $\frac {12 \ g }{ 44\ g} × 3.38\ g$
= 0.9217 g
18 g of water contains 2 g of hydrogen.
∴ 0.690 g of water will contain hydrogen = $\frac {2 \ g }{18\ g} × 0.690$
= 0.0767 g
Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is:
= 0.9217 g + 0.0767 g = 0.9984 g
∴ Percent of C in the compound = $\frac {0.9217\ g }{ 0.9984\ g }× 100$
= 92.32%
Percent of H in the compound = $\frac {0.0767\ g }{ 0.9984\ g} × 100$
= 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
∴ Ratio of carbon to hydrogen in the compound = 7.69: 7.68 = 1: 1

Hence, the empirical formula of the gas is CH.

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(ii) Given, Weight of 10.0 L of the gas (at S.T.P) = 11.6 g
∴ Weight of 22.4 L of gas at STP = $\frac {11.6 \ g}{10.0\ L} × 22.4 \ L$
= 25.984 g
≃ 26 g

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

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(iii) Empirical formula mass of CH = 12 + 1 = 13 g
n = $\frac {\text {Molar\ mass \ of\ gas }}{\text {Empirical \ formula \ mass\ of\ gas}}$

= $\frac {26\ g}{13 \ g}$
n = 2
∴ Molecular formula of gas = (CH)n = C2H