Earlier, we have discussed on the strategy of Determining the Molecular Formula of Compounds using Composition by Mass.
Today, we will discuss on the Determination of Molecular Formula of Hydrocarbons using Combustion Data. This is a new concept for those that making their transition from GCE O-Levels to GCE A-Levels and thus will be one of the key questions to be asked in GCE A-Level H1 and H2 Chemistry Examinations.
Let’s look at the method involved.
Hydrocarbons burns in excess oxygen based on the following equation:
CxHy (g) + (x + y/4) O2 (g) –> xCO2 (g) + y/2 H2O (l)
At room temperature conditions or s.t.pm the water product is actually a liquid. Thus, the volume is negligible compared to volumes of CxHy (g), O2 (g) and CO2 (g).
Using Volume Ratio;
Hence, if 1cm3 of is completely burned in oxygen,
Volume of oyxgen used = (x + y/4) cm3
Volume of carbon dioxide produced = x cm3
Volume of water produced (as liquid) = 0 cm3
Let’s check out an exam-based question to see how we can make use of the above to solve question.
Example:
10cm3 of a gaseous hydrocarbon required 20 cm3 of oxygen for complete combustion. 10 cm3 of carbon dioxide was produced. Calculate the molecular formula of the hydrocarbon.
and
Suggested Solution:
Using the general equation and applying volume ratio, we have
CxHy (g) + (x + y/4) O2 (g) –> xCO2 (g) + y/2 H2O (l)
10cm3…… 20cm3 ……………….10cm3
1mol 2 mol 1 mol
Based on the above comparision, we have
x = 1 and (x + y/4) = 2
Solving it gives y = 4
Hence, the molecular formula of the hydrocarbon is CH4.
Hope the above explanation is useful to you. Do note that this is a very important concept when you are in JC1 and will still be tested when you are in JC2. I realised that many of my students in my GCE A-Level H2 Chemistry Tuition Classes are not well trained in their Junior Colleges when it comes to this basic concept.
PS: Let me know how you find it. Drop me a comment. I would love to hear from you. 😀