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# 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 _{x}H_{y} (g) + (x + y/4) O_{2} (g) –> xCO_{2} (g) + y/2 H_{2}O (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_{x}H_{y} (g), O_{2} (g) and CO_{2} (g).

Using Volume Ratio;

Hence, if 1cm^{3} of is completely burned in oxygen,

Volume of oyxgen used = (x + y/4) cm^{3}

Volume of carbon dioxide produced = x cm^{3}

Volume of water produced (as liquid) = 0 cm^{3}

Let’s check out an exam-based question to see how we can make use of the above to solve question.

Example:

10cm

^{3}of a gaseous hydrocarbon required 20 cm^{3}of oxygen for complete combustion. 10 cm^{3}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

_{x}H_{y}(g) + (x + y/4) O_{2}(g) –> xCO_{2}(g) + y/2 H_{2}O (l)10cm

^{3}…… 20cm^{3}……………….10cm^{3}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 CH

_{4}.

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. 😀

hey. that was really helpful.

It was helpful but would need to know how to cal.. more complex ones..

thanks could you please put up more complex examples?

thanx, very useful

what if you consider the water evolve as water vapour and it is a considerable amount??

i have been searching 4 a solution for a very long time,thanks that was really helpful!

CUD U PLZ PUT UP SOME MORE DIFFICULT EXPLS LIEK TEH ONE AM STUCK WITH IT GOES LIKE: A HYDROCARBON OF 20.CM^3 AND 100C^3 OF OXYGEN IS EXPLODED. WHEN COOLED THE VOL OF TEH GAS MIX WAS LEFT TO 70.0CM^3 . WHEN PASSING THROUGH aq NAOH THE VOL WAS REDUCED BY 60CM^3. DETERMINE THE EMPIRICAL FORMULA OF THE HYDROCARBON.

ITLL BE HELPFUL IF U CAN SEND AN EX OF THIS KIND!

HOPE IAM NT OFFENDING U IN ANYWAY

THANKU

IT IS VERY EASY MATTER……..BUT WE NEAD MORE EXAMPLE TO PRACTISE..

Thank u very much.. This helped me solve a question I stacked at..

Thank u..:-)

thanks a lot

Hmmmmm! I actually enjoyed the way u simplified it thanks, but pls could u give more examples on some difficult one

Thanks everyone for the encouragement and support.

I’m glad you like my simple way of presentation.

Yup, i always do my best to make abstract complicated A-Level Chemistry concepts to be as simple as possible, so that my students in my JC A Level H2 Chemistry classes can benefit from it.

That’s how i will continue to write on this blog also.

Hope it can help other students that are studying overseas as well as other Singapore JC students who are not able to attend our tuition classes.

Keep sharing, Keep learning!

Sean Chua 🙂

i think this is a very useful webpage and i hereby notice that this encourage most of the students to resolve their problems regarding molecular formulae of hydrocarbon.Thank you!