Chemistry Labs

Problem 2

A mixture of a gaseous hydrocarbon and oxygen is placed in a 1 dm3^3 vessel at 406.5 K and 101 325 Pa. The oxygen present is twice the stoichiometric amount required for combustion. After combustion, the pressure in the vessel at the same temperature increases by 5 %. Determine the molecular formula of the hydrocarbon if the mass of water formed was 0.162 g. (R=8.314R = 8.314 J mol−1^{-1} K−1^{-1})
Step 1 of 4: Initial and final gas amounts
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Intuition

406.5 K is 133.35 °C, well above 100 °C, so all water remains gaseous throughout.

n1=p1VRT=101325×10−38.314×406.5=0.0300 mol;n2=1.05×n1=0.0315 moln_1 = \dfrac{p_1 V}{R T} = \dfrac{101325 \times 10^{-3}}{8.314 \times 406.5} = 0.0300\ \text{mol}; \quad n_2 = 1.05 \times n_1 = 0.0315\ \text{mol}
Analysis

Ideal gas law gives 0.0300 mol before reaction. Because temperature and volume are constant, a 5 % pressure rise means the total mole number increased by 5 % to 0.0315 mol.