Question

An organic compound containing C, H, O, and S is subjected to two analytical procedures. In the first procedure a 9.33 mg sample is burned which gives 19.50 mg of C02 and 3.99 mg of H20. In the second procedure, a separate 11.05 mg sample is fused (melted) with Na202 and the resulting sulfate is precipitated as BaS04, which (when washed and dried) weighs 20.4 mg. (Appropriate amounts of Na202 and a compound containing barium ion are added.) The amount of oxygen in the original sample is obtained by difference. Determine the empirical formula of this compound.

Answer 1

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

         = (12/44) *19.5*10^-3 *100/9.33*10^-3

          = 57%

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

           =(2/18)* 3.99*10^-3 *100/9.33

            = 4.75%

S%       = (32/233) * wt of BaSO4*100/wt of substance

           = (32/233) *20.4*10^-3*100/11.05*10^-3

          = 25.35%

O%    = 100-(C% + H%+ S%)

          = 100-(57+4.75 + 25.35)

           = 12.9%

Element         %                 A.Wt             Relative number        simple ratio

C                 57                    12                 57/12 = 4.75           4.75/0.8   = 6

H             4.75                  1                   4.75/1 = 4.75       4.75/0.8   = 6

S                  25.35               32                   25.35/32   = 0.8      0.8/0.8   = 1

O                 12.9                 16                   12.9/16      = 0.8       0.8/0.8   = 1

          Empirical formula = C6H6SO