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A student analyses an organic compound that contains only carbon, hydrogen and oxygen…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A student analyses an organic compound that contains only carbon, hydrogen and oxygen. She burns a $0.264\ \text{g}$ sample completely in a stream of dry oxygen, in the Liebig combustion train shown in the figure. The copper(II) oxide in the tube makes sure that all the carbon leaves as carbon dioxide. Moist air is kept out, so the U-tubes gain mass only from the products of burning.

Liebig's combustion train, drawn from left to right. Oxygen gas enters a long horizontal combustion tube. Inside the tube, near the inlet, a small boat holds the sample, and further along the tube is packed with copper(II) oxide. Flames under the tube heat it. The gases leave the tube and pass first through U-tube A, packed with anhydrous calcium chloride, and then through U-tube B, which holds potassium hydroxide solution, before leaving on the right. Under U-tube A: before 31.406 g, after 31.622 g. Under U-tube B: before 45.118 g, after 45.646 g.

The gases pass first through U-tube A and then through U-tube B. The figure gives the mass of each U-tube before and after the run. In a separate experiment, the vapour density of the compound is found to be $44$ . Atomic masses in $\text{g mol}^{-1}$ : H 1, C 12, O 16.

What is the percentage of carbon in the compound?
What is the percentage of hydrogen?
Oxygen is not measured directly. Find its percentage by difference.
What is the empirical formula mass of the compound?
How many carbon atoms does one molecule of the compound have?
Write the balanced equation for the complete combustion of the compound, using its molecular formula. What mass of oxygen gas did the sample use?

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Given. Sample $m = 0.264\ \text{g}$ , vapour density $44$ . From the figure, U-tube A (anhydrous calcium chloride) gains $31.622\ \text{g} - 31.406\ \text{g} = 0.216\ \text{g}$ : this is the water. U-tube B (potassium hydroxide) gains $45.646\ \text{g} - 45.118\ \text{g} = 0.528\ \text{g}$ : this is the carbon dioxide.

Plan. Find the mass of carbon in the carbon dioxide and of hydrogen in the water. Oxygen is what is left. Then turn the percentages into a mole ratio, and use the molar mass for the molecular formula.

Step 1. Carbon. Mass of C $= \dfrac{12}{44} \times 0.528\ \text{g} = 0.144\ \text{g}$ . So $\%\ \text{C} = \dfrac{0.144\ \text{g}}{0.264\ \text{g}} \times 100 = 54.55\%$ .

Step 2. Hydrogen. Mass of H $= \dfrac{2}{18} \times 0.216\ \text{g} = 0.024\ \text{g}$ . So $\%\ \text{H} = \dfrac{0.024\ \text{g}}{0.264\ \text{g}} \times 100 = 9.09\%$ .

Step 3. Oxygen by difference. $\%\ \text{O} = 100\% - 54.55\% - 9.09\%$ $= 36.36\%$ .

Step 4. Empirical formula. In $100\ \text{g}$ : C $\dfrac{54.55\ \text{g}}{12\ \text{g mol}^{-1}} = 4.546\ \text{mol}$ ; H $\dfrac{9.09\ \text{g}}{1\ \text{g mol}^{-1}} = 9.09\ \text{mol}$ ; O $\dfrac{36.36\ \text{g}}{16\ \text{g mol}^{-1}} = 2.273\ \text{mol}$ . Dividing by $2.273\ \text{mol}$ gives $2 : 4 : 1$ . The empirical formula is $\text{C}_2\text{H}_4\text{O}$ , with mass $2(12) + 4(1) + 16 = 44\ \text{g mol}^{-1}$ .

Step 5. Molecular formula. Molar mass $= 2 \times 44 = 88\ \text{g mol}^{-1}$ . This is $\dfrac{88}{44} = 2$ empirical units, so the formula is $\text{C}_4\text{H}_8\text{O}_2$ , with $4$ carbon atoms.

Step 6. Oxygen used. $\text{C}_4\text{H}_8\text{O}_2 + 5\,\text{O}_2$ $\rightarrow 4\,\text{CO}_2 + 4\,\text{H}_2\text{O}$ . The sample is $\dfrac{0.264\ \text{g}}{88\ \text{g mol}^{-1}} = 0.0030\ \text{mol}$ . It uses $5 \times 0.0030\ \text{mol} = 0.0150\ \text{mol}$ of $\text{O}_2$ , which is $0.0150\ \text{mol} \times 32\ \text{g mol}^{-1} = 0.480\ \text{g}$ .

Answer: (a) $54.55\%$ . (b) $9.09\%$ . (c) $36.36\%$ . (d) $44\ \text{g mol}^{-1}$ . (e) $4$ carbon atoms. (f) $0.480\ \text{g}$ .

Check. Mass is conserved in the burning. Sample plus oxygen: $0.264\ \text{g} + 0.480\ \text{g} = 0.744\ \text{g}$ . Carbon dioxide plus water: $0.528\ \text{g} + 0.216\ \text{g} = 0.744\ \text{g}$ . They agree.

Watch out. Read the order of the U-tubes. If you swap them, you take $0.216\ \text{g}$ as carbon dioxide and get far too little carbon. And the vapour density is half the molar mass: taking $44\ \text{g mol}^{-1}$ as the molar mass gives only $2$ carbon atoms.

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