# Determine formula of hydrocarbon from combustion data

Source: Oct/Nov 2015 Paper 33 IGCSE Chemistry exam (discounted during marking)

Can someone explain how to find the answer for the last part of the question?

Note: I got the answer pentane with 5 carbon atoms. Is it the correct answer?

Hydrocarbons burn in excess oxygen to form carbon dioxide and water. $\pu{20cm^3}$ of a gaseous hydrocarbon burned in an excess of oxygen ($\pu{200cm^3}$). After cooling, the volume of the residual gas at r.t.p. was $\pu{150{cm}^3}$, $\pu{50{cm}^3}$ of which was x=oxygen.

(i) Determine the volume of the oxygen used. [1]
(ii) Determine the volume of the carbon dioxide formed. [1]
(iii) If the hydrocarbon was an alkane, deduce a formula of the hydrocarbon.

• Without trying to solve the problem, it seems a bit odd to me. There is a very limited ability of water to be gaseous at room temperature. To assume all the water formed is gaseous seems wrong. You could assume liquid water, and that the volume of liquid is negligible to the final volume of 150 $\mathrm{cm}^3$. So it seems you're left flipping a coin to get an answer. – MaxW Mar 2 '16 at 5:39
• I found the no. of moles of CO2 and assumed is the amount of Carbon in the original 20cm3 of the hydrocarbon. Then I found number of moles of Carbon in one mole of the Hydrocarbon. The result is 5 Carbon atoms so it must be Pentane. I just want to know if this is correct answer because there is no answer in marking scheme – Charlie Mar 2 '16 at 6:06
• That doesn't really make sense. The initial volume of the alkane is 20 and oxygen is 150 (200-50). So the molar ratio of gaseous alkane to oxygen is 2 to 15 isn't it? (have to assume ideal gas with 22.4 liters per mole which cancels when you take the ratio.) There isn't an alkane that burns in that ratio. – MaxW Mar 2 '16 at 6:13
• n-Pentane and isopentane are also both liquids at room temp. Butane is last n-alkane that would be a gas at room temp. Butane burns in ration of 2 to 13 which is close. – MaxW Mar 2 '16 at 6:23
• The question mentions that the system was cooled after the reaction, so initially the reactants werent at rtp right? – Charlie Mar 2 '16 at 6:26

In order to "solve" the problem the following assumptions are necessary:

1. The initial 20 $\mathrm{cm}^3$ volume of the gaseous hydrocarbon and 200 $\mathrm{cm}^3$ volume of oxygen are at rtp also.
2. The burning of the alkane (from part iii) also creates water, but at rtp the volume of water is negligible since it condenses into the liquid phase.
3. All the gases follow the ideal gas law so that volumes are directly proportional to moles.

(i) Determine the volume of oxygen used.

The starting volume of oxygen was 200 $\mathrm{cm}^3$, but 50 $\mathrm{cm}^3$ of the final 150 $\mathrm{cm}^3$ was oxygen,
so $200 - 50 = 150$ $\mathrm{cm}^3$ of oxygen was used.

(ii) Determine the volume of the carbon dioxide formed.

With assumption (2), the final volume was 150 $\mathrm{cm}^3$, but 50 $\mathrm{cm}^3$ was oxygen
so $150 - 50 = 100$ $\mathrm{cm}^3$ of carbon dioxide are formed.

(iii) The hydrocarbon was an alkane. Determine the formula of the hydrocarbon.

Since each carbon atom in an alkane will burn to produce one carbon dioxide molecule,
then 100/20 = 5 carbon atoms must be in the alkane. So the alkane might be $\ce{C5H12}$. (Thus the hydrocarbon could be isopentane, neopentane, or n-pentane.) $$\ce{C5H12 + 8O2 -> 5CO2 + 6H2O}$$

Problems with $\ce{C5H12}$

(A) Given that the volumes were supposed to follow the ideal gas law, 20 $\mathrm{cm}^3$ of hydrocarbon reacts with 150 $\mathrm{cm}^3$ of oxygen. That gives a molar ration of 2 to 15. But the molar ratio of $\ce{C5H12}$ burning is 1 molecule of hydrocarbon to 8 molecules of oxygen, or 2 to 16.

(B) The initial total volume of reactants was 150+20 = 170 $\mathrm{cm}^3$. The final volume of product was 100 $\mathrm{cm}^3$. So the ration of reactants to products, neglecting water, is 17 to 10. From the chemical equation the ration is 9 to 5, or 18 to 10.

So the answer must be a cycloalkane $\ce{C5H10}$ which would react with right ratio (2 to 15) and have the right ratio of products to reactants (17 to 10).

$$\ce{2C5H10 + 15O2 -> 10CO2 + 10H2O}$$

CHECK:

Assuming 22.4 liters for a mole of the hydrocarbon, then each mole of hydrocarbon would produce 10 moles of water.

$$10 \times \dfrac{0.020}{22.4} = 0.0089 \text{ moles of water}$$

Since each mole has a mass of 18 grams and water has a density of $1 ~\mathrm{g}/\mathrm{cm}^3$, then there is: $$0.0089\times 18 = 0.16 \mathrm{cm}^3$$ of water which is indeed negligible.

Also at rtp water will have a vapor pressure of about 18 torr out of 760 torr so the partial pressure of water is pretty small also in the gaseous phase.

PS: Thanks to DGS for pointing out that the alkane must be a cycloalkane.

• 1,1-dimethylcyclopropane. There's your 2:15 ratio, right? But I don't know where it boils, and it kind of seems like a goofy answer. – SendersReagent Mar 3 '16 at 20:06
• @DGS yes! - A great insight!! A cycloalkane would have right burn ratio, and right product ratio. $$\ce{2C5H10 + 15O2 -> 10CO2 + 6H2O}$$. – MaxW Mar 3 '16 at 20:18
• That and trans-1,2-dimethylcyclopropane were the only ones I found even close to being gaseous at room temperature. I'm not sure 1,1-dimethylcyclopropane has a known bp, but I would suspect it to be below the 1,2- isomers. Anyway, trans-1,2-dimethylcyclopropane has a known bp of 28.2 °C and, as you said of isopentane, that room would have to be fairly warm. – SendersReagent Mar 3 '16 at 20:22
• For the problem we didn't need to identify the particular alkane, I was just using that as a sanity check. I knew that n-pentane was a liquid a reasonable room temp. Cyclopentane would fit the formula but it boils at an unpleasant 49 °C (121 °F). – MaxW Mar 3 '16 at 20:42