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Moles/molecular formula question

1 (e) (iii) The amount of hydrocarbon burned was 0.050 mol.

Use this information and your answers to parts (e)(i) and (e)(ii) to calculate themolecular formula of the hydrocarbon.

If you could not answer parts (e) (i) or (e) (ii) use the values of 0.20 moles carbondioxide and 0.50 moles hydrogen. These are not the answers to parts (e) (i)and (e) (ii).

I can see the answer and working on the markscheme but I don't understand how it works so please could someone explain it to me?
Original post by nettogrof
1 (e) (iii) The amount of hydrocarbon burned was 0.050 mol.

Use this information and your answers to parts (e)(i) and (e)(ii) to calculate themolecular formula of the hydrocarbon.

If you could not answer parts (e) (i) or (e) (ii) use the values of 0.20 moles carbondioxide and 0.50 moles hydrogen. These are not the answers to parts (e) (i)and (e) (ii).

I can see the answer and working on the markscheme but I don't understand how it works so please could someone explain it to me?


When a hydrocarbon burns (complete combustion) all of its carbon turns to carbon dioxide and all of its hydrogen turns to water.

CxHy + nO2 --> xCO2 + 0.5yH2O

So, how ever many times the volume of carbon dioxide is of the hydrocarbon, that is the number of carbon atoms.

In this case mol hydrocarbon = 0.05
mol of carbon dioxide = 0.2

hence the number of carbon atoms in the hydrocarbon = 0.2/0.05 = 4

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