Question

An unknown compound contains only C, H, and O. Combustion of 3.50 g of this compound produced 7.96 g of CO2 and 3.26 g of H2O.What is the empirical formula of the unknown compound?

Answer 1

C%     = (12/44) * wt of CO2*100/wt of organic compound

          = (12/44) *7.96*100/3.5

             = 12*7.96*100/44*3.5    = 62%

H%     =(2/18) * wt of H2O*100/wt of organic compound

         = (2/18) *3.26*100/3.5

        = 2*3.26*100/18*3.5   = 10.35%

O%   = 100-(C%+ H%)

         = 100-(62+10.35)

          = 27.65%

Element               %                A.Wt                  Relative number              simple ratio

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

H                        10.35              1                          10.35/1 = 10.35           10.35/1.73 = 6

O                       27.65             16                          27.65/16 = 1.73            1.73/1.73   = 1

                Empirical formula = C3H6O