Determination of Molecular Formula of Hydrocarbons using Combustion Data

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:

C_xH_y (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 C_xH_y (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

C_xH_y (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.

PS: Let me know how you find it. Drop me a comment. I would love to hear from you.