Combustion Analysis

A 1.50 g sample of hydrocarbon undergoes complete combustion to produce 4.40 g of CO₂ and 2.70 g of H₂O. What is the empirical formula of this compound?

First, it is requred that you determine the grams of carbon and hydrogen in the CO₂ and the H₂O.

This can be done by multiplying the mass of the carbon dioxide by the molar mass of carbon in carbon dioxide.

This can also be done for hydrogen in water.

The steps are shown in the next card

C = 4.40 x (12.0/44.0)

   = 1.20g

H = 2.70 x (2.02/18.0)

   = 0.302g

simples no?

Carbon = 1.20/12

             = 0.1

Hydrogen = 0.302/1

                = 0.302

Divide through by the smallest number and round to integers.

This gives a final answer of CH3.

Note: this process does not account for the presence of oxygen as it wasn't mentioned in the question.

Also, this is an EMPERICAL formula.

If you know that oxygen is definetely present, then you can work out the mass of oxygen by a simple subtraction. If you knwo the mass of the sample burnt, say 0.670g and you have the rest of the information. (i.e. the weights of CO₂ and the H₂O.) 

You work out the masses of the carbon and hydrogen and then subtract this from the total weight. 

The remaining weight must be oxygen.

Then you can work out the moles, and divide through. 

This can be done if you can deduce or are given the molecular mass.

With this you can follow all the previous steps and then divide the molecular mass by the emperical  mass

For example, If you work out a molecualar mass of 162 and your emperical formula is C₃H₂O, then you can work out your emperical mass = 54

Then divide your molar mass by the the emperical mass. 162/54 = 3

So you multiply all the co efficients of you atoms by that number

C₃H₂O --> C9H6O3

Mass of carbon and hydrogen

--> Mass of Oxygen and everything else (could be nitrogen in there too D:)

-->Moles

-->divide through

--> if possible work out molecular formula