Question

on analysis it was found that the black oxide of copper,red oxide of copper litharge,the red oxide of lead and peroxide of lead contain 79.90%.88.8%,92.88%.90.6%and 86.6% respectively of the metal .establish the law of multiple proportions with the help of the data

Answer 1

First calculate the  mass of metals combined with fixed mass of oxygen.

For Black oxide of Cu

79.9 parts metal combines with 20.1 parts of O.

So 

1 part Oxygen will combine = 79.9/20.1 =3.9 part of metal..

similarly

For red oxide..

1part O will combine = 88.8/11.2 =7.9 parts metal..

 

There is a fixed mass of O in both the oxides.

Ratio of masses of Cu in both case = 3.9:7.9 

The simplest ratio...

1:2

Hence it establish the low of multiple proportion.

 

Similarly you can calculate for oxides of Pb

Answer 2

Ans - 100gm of Black oxide of copper contains 79.9 gm of copper

and 100gm-79.9gm= 20.1gm of Oxygen

1gm of Oxygen combineswith 79.9/20.1 =3.9 gm of copper

100gm of Red oxide of copper

contains 88.8gm of copper

and 100gm-88.8gm= 11.2gm of Oxygen

1gm of Oxygen combines with 88.8/11.2 = 7.9 gm of of copper

Thus the raito of copper weights which combines with fixed wieght of oxygen (1gm) is 3.9:7.9. or 1:2

Explanation:

You need to first go for weights of copper and oxygen, then fix the oxygen 1 gm and then you will get 2 weights of copper in both data. after doing this make ratio of two weights 3.9:7.9 =1:2

Because 3.9= 4 and 7.9= 8

Thus 4:8= 2:4 = 1:2