Question

Combustion analysis of a 0.2608 g sample of a compound containing carbon, oxygen, and hydrogen only produced 0.5501 g of carbon dioxide, and 0.2703 g of water.

Determine the empirical formula of this compound. If the molar mass for this compound is 104.1 g/mole, find the molecular formula. Write a complete balanced equation for the combustion reaction of this compound.

Answer 1

c%       = 12* weight of CO2 *100/44*weigh of organic compound

         = 12*0.5501*100/44*0.2608   = 57.53%

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

       = 2*0.2703*100/18*0.2608 = 11.52%

O%   = 100-(C%+ H%)

         = 100-(57.53+11.52) = 30.95%

element               %                               A.Wt                 Relative number            simple ratio       simplest ratio

C                       57.53                             12                    57.53/12 = 4.79           4.79/1.934   =2.5      2.5*2 = 5

H                      11.52                                 1                  11.52/1     = 11.52            11.52/1.934 = 6       6*2   = 12

O                      30.95                            16                  30.95/16     = 1.934            1.934/1.934 = 1       1*2 = 2

       Empirical formula = C5H12O2

     molecular formula = ( empirical formula)n                          { E.F.Wt Of C5H12O2 = 12*5 + 1*12 + 16*2 = 104}

                       n          = E.F.Wt/M.wt

                                  = 104/104 = 1

           molecular formula = ( C5H12O2)1 = C5H12O2

C5H12O2 + 7O2 ----------> 5CO2 + 6H2O