Question

3.00g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 60./gmol, is burned completely in excess oxygen, and the mass of the products carefully measured:

product	mass
carbon dioxide	4.40g
water	1.80g

Use this information to find the molecular formula of X

.

Answer 1

c%      = 12* wt of CO2 *100/44*wt of compound

          = 12*4.4*100/(44*3)   = 40%

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

           = 2*1.8*100/(18*3)   = 6.66%

O%    = 100-(C% + H%)

          = 100- (40 + 6.66)   = 53.34%

Element          %                    A.Wt                    Relative number              simple ratio               

C                     40                  12                        40/12 = 3.33                   3.33/3.33 = 1

H                   6.66                1                          6.66/1   = 6.66                 6.66/.3.33 = 2

O                     53.34             16                         53.34/16 = 3.33             3.33/3.33   = 1

The empirical formula = CH2O

mass of empirical formula = 30g/mole

mole mass of Compound    = 60g/mole

molecular formula   = (empirical formula)n

                  n           =   molecular mass/mass of empirical formula

                                = 60/30 = 2

molecular formula   = (CH2O)2 = C2H4O2