# Combustion Analysis of Gaseous Hydrocarbons

In this video we want to discuss how to use combustion analysis to determine the molecular formula of an unknown gaseous hydrocarbon C_{x}H_{y}.

The balanced equation of the combustion of hydrocarbon in excess oxygen to give carbon dioxide and water is as follows:

Notice that all the carbon in the hydrocarbon is converted to carbon dioxide CO_{2} which is acidic in nature.

This means we can use a base to absorb acidic CO_{2} to measure the amount of CO_{2} produced and determine the amount of carbon in the hydrocarbon.

Also, all the hydrogen is converted to water H_{2}O which is a liquid at room temperature or standard temperature.

We can simply cool the gaseous product mixture to collect the liquid water, or pass the gaseous mixture through an anhydrous salt to absorb water to determine the amount of water produced and the amount of hydrogen in the hydrocarbon.

Therefore combustion analysis is a simple method to determine the amount of carbon and hydrogen present in the hydrocarbon, and we can deduce the molecular formula of that hydrocarbon.

Let's take a look at an example question.

To help visualise the question we can use the following diagram to determine the volumes of gases involved:

1. Volume of hydrocarbon C_{x}H_{y}

The volume of C_{x}H_{y} is given in the question which is 20 cm^{3}

2. Volume of CO_{2} produced

The 130 cm^{3} volume of residual gases consists of volume of CO_{2} produced and volume of unreacted O_{2} in excess.

Water is a liquid at room temperature and pressure hence its volume is not considered as part of the 130 cm^{3} of residual gases.

After shaking with alkaline sodium hydroxide, carbon dioxide is absorbed and final volume of 90 cm^{3} is the volume of O_{2} in excess.

This means that volume of CO_{2} will just be the difference which is 40 cm^{3}.

3. Volume of O_{2} reacted

Total volume of O_{2} is 150 cm^{3} and volume of O_{2} in excess is 90 cm^{3}. Therefore the volume of O_{2} reacted will be the difference which is 60 cm^{3}.

We can now fill up the following table which lists down the mole ratio and volume ratio of the gases.

The mole ratio is always fixed since it is based on the balanced equation.

The volume ratio is the one that varies and we have already determined the volumes of C_{x}H_{y}, O_{2} reacted and CO_{2} produced from the information given in the question.

Again since water is a liquid at rtp, we do not need to compare the mole and volume ratio of water hence its mole and volume are not required.

We can now compare mole ratio and volume ratio of these gases to solve for x and y.

1. Solve for x

We compare mole ratio and volume ratio of CO_{2} and C_{x}H_{y} to determine that x=2.

2. Solve for y

We compare mole ratio and volume ratio of O_{2} and C_{x}H_{y} to solve for y. We need to substitute x=2 to determine that y=4.

Finally we can determine the molecular formula for this hydrocarbon is C_{2}H_{4}.

For the detailed step-by-step discussion on how to determine the molecular formula of an unknown hydrocarbon from combustion analysis, check out this video!

Topic: Mole Concept, Physical Chemistry, A Level Chemistry, Singapore

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