Question

The percent composition of oxygen in a molecule that contains 2 oxygen atoms is 27.6%. This molecule contains only carbon, hydrogen and oxygen. Combustion of 225 mg of the compound yields 512 mg of carbon dioxide and 209 mg of water. What is the molecular formula of this compound?

Answer 1

    

    

C%    = 12*weight of CO2*100/44*weight of organic compound

          = 12*512*100/44*225   = 62%

H%    = 2*weight of H2O * 100/18*weight of organic compound

         = 2*209*100/18*225   = 10.32%

O%   = 100-(C%+ H%)

         = 100-(62+10.32)   = 27.68%

Element             %           At.wt              Relative number              simple ratio

C                        62           12                  62/12   = 5.16                  5.16/1.73   = 3

H                       10.32        1                  10.32/1 = 10.32            10.32/1.73    = 6

O                        27.68       16                27.68/16 = 1.73              1.73/1.73     = 1

Empirical formula = C3H6O

molecule contains two oxygen atoms.

molecular formula     = C6H12O2 >>>>>>answer