Question

When 108.4 mg of a compound is completely combusted, 305.3 mg of CO2 and 125.0 of H2O are produced.

(a) Determine the empirical formula for the compound.

(b) If the molar mass of the compound is 468.8 g, determine the molecular formula.

Answer 1

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

            = 12*305.3*100/(44*108.4)   = 76.8%

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

            = 2*125*100/(18*108.4)   = 12.8%

O%     = 100-(C% + H%)

           = 100-(76.8 + 12.8)

           = 10.4%

Element     %             A.Wt            Relative number              simple ratio

carbon      76.8            12                  76.8/12   = 6.4            6.4/0.65   10

hydrogen   12.8              1                  12.8/1    = 12.8           12.8/0.65   = 20

oxygen      10.4             16                 10.4/16     = 0.65         0.65/0.65   = 1

The empirical formula of compound is C10H20O

b. molar mass   = 468.8g

E.F.Wt   = 10*12 + 20*1 + 16   = 156

n = molar mass/E.F.Wt

      = 468.8/156   = 3

molecular formula = (C10H20O)3

                             = C20H40O3 >>>>>answer