Question

Molecule: Molindone

0.5000g of an organic compound containing C, H, O, and N was combusted. The resulting analysis revealed that 1.274 grams of CO₂ and 0.3902 grams of H₂O were produced. A subsequent elemental analysis revealed that the same sample contained 10.14% nitrogen by mass. If the compund has a molar mass of 276.37 g/mol, what are the empirical and molecular formulas of the compound?

Answer 1

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

           = (12/44) *1.274*100/0.5

           = 69.5%

H%      = (2/18) Wt of H2O *100/Wt of organic ompounf

            = (2/18) *0.3902*100/0.5

           = 2*0.3902*100/18*0.5

            = 8.67%

N%        = 10.14%

O%         = 100-(C%+ H%+ N%)

               = 100-(69.5+8.67+10.14)

               = 11.69%

Element             %              A.Wt                  relative number             simple ration

C                    69.5             12                         69.5/12   = 5.79        5.79/0.724    = 8

H                     8.67             1                           8.67/1    = 8.67        8.67/0.724    = 12

O                   11.69           16                         11.69/16   = 0.73        0.73/0.724      = 1

N                   10.14          14                          10.14/14   = 0.724        0.724/0.724   = 1

Empirical formula   = C8H12NO

empirical formula weight   = 138

molecular formula   = (empirical formul)n

                      n        = M.Wt/E.F.Wt

                                 = 276.37/138   = 2

molecular formula   = (C8H12NO)2

                                 = C16H24N2O2