Question

You have an unknown substance that contains three elements: carbon, hydrogen, nitrogen. Your burn 7.114 grams of the unknown formula in excess oxygen gas to get 10.03 g of CO2 and 10.26 g of H20. What is the empirical formula of the unknown? what is the molecular formula if the molar mass of the unknown is 248.04 g/mol?

Answer 1

C%     = 12*wt of CO2*100/44*wt of substance

           = 12*10.03*100/(44*7.114)   = 38.45%

H%     = 2*wt of H2O*100/18*wt of substance

           = 2*10.26*100/(18*7.114)    = 16.025%

N%     = 100-(C% + H%)

            = 100-(38.45 + 16.025)

             = 45.525%

Element      %              A. Wt             Relative number                 simple ratio

C               38.45          12                    38.45/12   = 3.2               3.2/3.2 = 1

H               16.025          1                     16.025/1 = 16.025         16.025/3.2   = 5

N              45.525           14                   45.525/14   = 3.25           3.25/3.2   = 1

The empirical formula = CH5N

empirical formula mass = 12 + 5 + 14   = 31g/mole

molecular formula = (empirical formula)n

              n     = molecular mass/empirical formula mass

                     = 248.04/31    = 8

The molecular formula   = (CH5N)8

                                      = C8H40N8