A 1.038 g sample of unknown containing C, H, and O yielded 2.48 g of CO2 and 0.510 g of H2O during combustion analysis. Determine the empirical formula of the compound. A) C3H3O
B) C6H6O
C) CH3O
D) C6H6O2
E) C2H6O2

A. C3H3O

Explanation:

We have the following data:

m = mass of organic compound = 1.038g

Then mass of CO2 is 2.48g

Also mass of H2O is 0.510g

To calculate percentage of carbon = 12 x mass of CO2 x 100/44 x mass of organic compound

This gives us 65.1 percent

Formula to calculate percentage of H=

2 x mass of water x 100/18 x mass of organic compound.

= 2 x 0.510 x 100/18x 10038

This gives 5.4%

Percentage of Oxygen =100-(percentageof carbon +percentageof H)

= 100-(65.1+5.4)

= 100-70.5

= 29.5

We have to calculate emperical formula:

For C = 65.1

65.1/12

= 5.4

= 5.4/1.8

=3

For H=5.4

5.4/1

= 5.4

= 5.4/1.8

=3

For O = 29.5

= 29.5/16

= 1.8

= 1.8/1.8

=1

So the emperical formula of organic compound is given as

C3H3O1

= C3H3O

(The atomic weights of C, H, O are 12,1,16)

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