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 4 of 4: Solve for x and y

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\dfrac{0.018}{y}\left(1 + \frac{y}{4}\right) = 0.0075 \Rightarrow \frac{0.018}{y} + 0.0045 = 0.0075 \Rightarrow y = 6; \quad x = 3 \;(\ce{C3H6})

Analysis

Solving $0.018/y = 0.0030$ gives $y = 6$ . Then $n = 0.018/6 = 0.003$ mol, and $2x + 6/4 = 0.0225/0.003 = 7.5 \Rightarrow 2x = 6 \Rightarrow x = 3$ . The hydrocarbon is $\ce{C3H6}$ (propene or cyclopropane).