Question

An unknown compound has a molecular mass of 180.15 g/mole and an elemental analysis showing that it is 60.00% C and 4.48% H. Determine the molecular formula.

Answer 1

O% = 100-(C% + H%)

        = 100-(60+4.48)   = 35.52%

Element            %               A.Wt               Relative number          simplest ratio    simple ratio    

C                     60                 12                    60/12 = 5                    5/2.22 = 2.25         2.25*4 = 9

H                    4.48                1                    4.48/1 = 4.48                4.48/2.25 =2           2*4     = 8

O                   35.52              16                   35.52/16 = 2.22            2.22/2.22 = 1        1*4       = 4

      Empirical formula = C9H8O4

    molecular formula = (empirical formula)n

                      n       = molar mass/empirical formula mass

                              = 180.15/180 = 1

molecular formula = (C9H8O4)1 = C9H8O4