A 2 L sample of a gaseous hydrocarbon is burnt in excess oxygen - VCE - SSCE Chemistry - Question 7 - 2004 - Paper 1

Using the ideal gas law, we can relate volume (V) to moles (n) since the temperature and pressure are consistent. The volume of 1 mole of gas at these conditions is approximately 22.4 L. Hence, the moles of products can be calculated as follows:

For CO₂: $n_{CO₂} = \frac{8 \text{ L}}{22.4 \text{ L/mol}} = 0.357 \text{ mol}$

For H₂O: $n_{H₂O} = \frac{10 \text{ L}}{22.4 \text{ L/mol}} = 0.446 \text{ mol}$

The general combustion reaction of a hydrocarbon (CₓHᵧ) can be represented as: $CₓHᵧ + O₂ \rightarrow xCO₂ + \frac{y}{2}H₂O$ From the stoichiometry we can conclude that:

• Each mole of carbon produces one mole of CO₂,
• Each mole of hydrogen produces half a mole of H₂O.

From the products:

• To produce 8 L (0.357 mol) of CO₂, we need 0.357 moles of carbon, so: x = 0.357.
• To produce 10 L (0.446 mol) of H₂O, we need 0.446 moles of hydrogen, so: (y = 2 \times 0.446 = 0.892). Thus, we have approximately 1 mole of carbon (C) for every 2 moles of hydrogen (H), leading to the empirical formula of C₂H₁₀.

The hydrocarbon formula that matches C₂H₁₀ is option C.