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A gas burette holds 80.0 mL of a mixture of methane, ethane and helium…

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A gas burette holds $80.0\ \text{mL}$ of a mixture of methane, ethane and helium. After $200.0\ \text{mL}$ of oxygen (more than enough) is added, the mixture is sparked. The vessel is cooled to the starting temperature and pressure, when all the steam has become liquid. The gas volume is then $90.0\ \text{mL}$ smaller than before the spark. Shaking with potassium hydroxide solution removes a further $57.0\ \text{mL}$ . All volumes are measured at the same temperature and pressure. Helium does not react, and its molar mass is $4.0\ \text{g mol}^{-1}$ . Atomic masses in $\text{g mol}^{-1}$ : H 1.0, C 12.0.

What volume of methane was in the mixture?
What volume of ethane was in the mixture?
What is the mole fraction of helium in the original mixture?
What volume of oxygen was used up?
What is the average molar mass of the original mixture?

The worked solution

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Contraction per volume. $\text{CH}_4 + 2\,\text{O}_2 \rightarrow \text{CO}_2 + 2\,\text{H}_2\text{O}(l)$ : $3$ volumes of gas become $1$ , so the volume falls by $2$ for each volume of methane. $\text{C}_2\text{H}_6 + 3\tfrac{1}{2}\,\text{O}_2 \rightarrow 2\,\text{CO}_2 + 3\,\text{H}_2\text{O}(l)$ : $4\tfrac{1}{2}$ volumes become $2$ , so it falls by $2\tfrac{1}{2}$ for each volume of ethane. Helium changes nothing.

Equations. $x + y + z = 80.0$ , then $2x + 2.5y = 90.0$ , and the $\text{CO}_2$ absorbed gives $x + 2y = 57.0$ . From the last, $x = 57.0 - 2y$ . Substituting, $2(57.0 - 2y) + 2.5y = 90.0$ , so $57.0 \times 2 - 1.5y = 90.0$ and $y = 16.0$ .

(a), (b) $y = 16.0\ \text{mL}$ of ethane and $x = 57.0 - 2(16.0) = 25.0\ \text{mL}$ of methane.

(c) Helium is $80.0 - 25.0 - 16.0 = 39.0\ \text{mL}$ , so its mole fraction is $39.0 / 80.0 = 0.488$ .

(d) Oxygen used $= 2(25.0) + 3.5(16.0) = 106.0\ \text{mL}$ , which is less than the $200.0\ \text{mL}$ added, so the oxygen was in excess as stated.

(e) Mole fractions: methane $0.3125$ , ethane $0.2000$ , helium $0.4875$ . $\bar{M} = 0.3125(16.0) + 0.2000(30.0) + 0.4875(4.0) = 12.95\ \text{g mol}^{-1}$ .

Check. The gas left after the potassium hydroxide should be the unused oxygen plus helium: $(200.0 - 106.0) + 39.0 = 133.0\ \text{mL}$ , which equals $80.0 + 200.0 - 90.0 - 57.0$ . A matrix solution of the three equations gives the same volumes.

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