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 CxHy.
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 CO2 which is acidic in nature.
This means we can use a base to absorb acidic CO2 to measure the amount of CO2 produced and determine the amount of carbon in the hydrocarbon.
Also, all the hydrogen is converted to water H2O 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 CxHy
The volume of CxHy is given in the question which is 20 cm3
2. Volume of CO2 produced
The 130 cm3 volume of residual gases consists of volume of CO2 produced and volume of unreacted O2 in excess.
Water is a liquid at room temperature and pressure hence its volume is not considered as part of the 130 cm3 of residual gases.
After shaking with alkaline sodium hydroxide, carbon dioxide is absorbed and final volume of 90 cm3 is the volume of O2 in excess.
This means that volume of CO2 will just be the difference which is 40 cm3.
3. Volume of O2 reacted
Total volume of O2 is 150 cm3 and volume of O2 in excess is 90 cm3. Therefore the volume of O2 reacted will be the difference which is 60 cm3.
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 CxHy, O2 reacted and CO2 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 CO2 and CxHy to determine that x=2.
2. Solve for y
We compare mole ratio and volume ratio of O2 and CxHy 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 C2H4.
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!
Sometimes questions will ask about combustion analysis of other organic compounds too! Check out this video on determining the molecular formula for alcohol J.
Topic: Mole Concept, Physical Chemistry, A Level Chemistry, Singapore
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