Question

On analysis it was found that the black oxide of copper and red oxide of copper contains 80% and 89% of copper respectively. Show that this data is in agreement with the law of multiple proportions.

Answer 1

in black oxide the % of Cu = 80% , so % of O = 20% 
hence if % of Cu = 1 , the % of O =  20/80 =  0.25 

in red oxide % of Cu = 89% , so % of O = 11% 
hence if % of Cu = 1 , the % of O = 11/ 89 = 0.1235

ratio of O:O in the two oxide are = .25/.25 : 0.1235/0.25 = 1: 0.5  = 2: 1 
which is a ssimple whole no. ratio hence the law of multiple proportion .

Answer 2

Answer:

This data is in agreement with the Law of multiple proportions.

Explanation:

in black oxide,

the percentage of $Cu$ $= 80$%$% of Cu = 80%$ , so percentage of $O = 20$%

hence if percentage of $Cu = 1$ , the percentage of $O = 20/80 = 0.25$ $%$%

in red oxide,

the percentage of $Cu = 89$% , so percentage of $O = 11$%

hence if percentage of $Cu = 1$ , the percentage of $O = 11/ 89 = 0.1235$ %

ratio of $O:O$ in the two oxide are :

$\frac{0.25}{0.25} : \frac{0.1235}{0.25} \\= 1: 0.5 \\ = 2: 1$

which is a simple whole number ratio.

Hence the law of multiple proportion is satisfied in this condition.