Problem 2

A mixture of a gaseous hydrocarbon and oxygen is placed in a 1 dm

^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 = 8.314

J mol

^{-1}

^{-1}

)

Step 2 of 4: Water stoichiometry

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n(\ce{H2O}) = \dfrac{0.162}{18} = 0.009\ \text{mol} \Rightarrow n(\ce{C_xH_y}) = \dfrac{0.018}{y}\ \text{mol}

Analysis

From $\ce{C_xH_y + (x + y/4)O2 -> x CO2 + (y/2)H2O}$ , 1 mole of hydrocarbon yields $y/2$ moles of water, so $n(\ce{C_xH_y}) = 0.009 / (y/2) = 0.018/y$ mol.