Question

Burning a 1.50-g sample of carbon-hydrogen-oxygen compound in oxygen yeilds 2.997g CO2 and 1.227g H2O. A separate experiment shows that the molecular mass of the coumpound is 88u. Determine mass percent compostion. Determine empirical Formula, and demtermine the molecular formula of the compound.

Answer 1

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

                = 12*2.997*100/(44*1.50)   = 54.5%

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

                = 2*1.227*100/(18*1.50)   = 9.1%

O%         = 100-(C% + H%)

               = 100-(54.5+9.1)   = 36.4%

Element                 %                                 A.wt              Relative number              simple ratio

C                           54.5                              12                 54.5/12   = 4.54              4.54/2.275   = 2

H                         9.1                                  1                   9.1/1      = 9.1                  9.1/2.275   = 4

O                          36.4                              16                    36.4/16   = 2.275          2.275/2.275    = 1

          The empirical formula   = C2H4O

empirical formula weight   = 2*12 + 4*1 + 16   = 44g/mole

molar mass of compound     = 88g/mole

molecular formula   = (empirical formula)n

      n                    =   molar mass of compound /empirical formula weight

                                = 88/44 = 2

molecular formula   = (C2H4O)2

                                   = C4H8O2