Question

a metal forms two oxides. The higher oxide contains 80% metal. 0.72g of the lower oxide gave 0.8g of the higher oxide when oxidised.show that the data illustrate the law of multiple proportions.

Answer 1

80%  of higher oxide 0.80 gms :   0.64 gm  of metal in the higher oxide.
Thus before oxidation, there is 0.64 gm of metal in the lower oxide.

So there is 0.72 gms - 0.64 gms = 0.08 gms of oxygen in lower oxide.
   and  0.80 - 0.64 = 0.16 gms of oxygen in lower oxide.

   The proportions of metal : oxygen in both oxides are:
                 0.64 : 0.08  =  8 : 1
                 0.64 : 0.16  =  8 : 2

So in the higher oxide there is double the amount of oxygen as compared to the lower oxide.  The content of oxygen that reacts with metal are in the integer ratio :  1 : 2.

Answer 2

The mass of metal in both oxides are same so first we have to Calculate mass of metal by using percentage of metal and mass of metal oxide .  For remaining process see the attachment