Respuesta :
Answer:
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.
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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.
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