Question

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

Answer 1

Given:
Amount of metal in higher oxide = 80%
0.72g of lower oxide give=0.8g of higher oxide
Now,
Metal in higher oxide = 80% of 0.8
= 0.64g
Oxygen in higher oxide = 20% of 0.8
= 0.16g
Now, amount of oxygen in lower oxide = (0.8-0.72)g
= 0.08g
Also, metal in lower oxide = (0.72-0.08)g
= 0.64g
Now, according to law of multiple proportion, ratio of amount of oxygen in higher oxide to the amount of oxygen in lower oxide should be a whole number, i.e.
= 0.16/0.08
= 2
Hence, this is in accordance with law of multiple proportion.