Question

An unknown compound contains only C, H, and O. Combustion of 6.10 g of this compound produced 14.9 g of CO2 and 6.10 g of H2O. What is the empirical formula of the unknown compound?

Answer 1

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

            = 12*14.9*100/44*6.1   = 66.62%

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

          = 2*6.1*100/18*6.1   = 11.11%

O%      = 100-(C% +H%)

            = 100-(66.62+11.11)   = 22.27%

Element            %            A.Wt          Relative number                simple ratio

C                  66.62            12               66.62/12= 5.55              5.55/1.39   = 4

H                  11.11               1                11.11/1   = 11.11          11.11/1.39   = 8

O                22.27              16               22.27/16   = 1.39          1.39/1.39 = 1

   Empirical formula   = C4H8O