A metal forms two oxides. The higher oxide contains 80% metal. `0.72 g` of the lower oxide gave `0.8 g` of higher oxide when oxidised. Calculate the weight of oxygen the combines with the fixed weight of metal in the two oxides, and show that the data supports the law of multiple proportines

Explanation

First method:
Since, `0.8 g` of higher oxide is obtained from `0.72 g` of lower oxide, therefore, the weight of lower oxide that would produce `100 g` higher oxide on oxidation
`= (0.72 xx 100)/(0.80) = 90 g`
Thus, `90 g` of lower oxide contains as much metal as `100 g` of higher oxide, i.e., `80 g` (given).
Hence, `80 g` of metal combines with `10 g` of oxygen in the lower oxide and `20 g` of oxygen in the higher oxide. The weights of oxygen that combine with the same weight of the metal in the two oxides are in the ratio of `10 : 20` or `1 : 2`, The ratio, being simple, proves the law of multiple proportions.
Second method : Lower oixde `+ O_(2) rarr` Higher oxide
Lower oxide : `0.72 g`
Weight of oxide `= 0.72 g`
Weight of `M` (fixed weight) `= 0.64 g`
Weight of `O_(2) = 0.72 - 0.64 = 0.08 g` of `O_(2)`
Higher oxide : `0.2 g`
`100 g` of oxide contains `implies 80 g of M`
`0.8 g` of oxide contains `implies (80)/(100) xx 0.8 implies 0.64 g of M`
weight of `O_(2) = 0.8 - 0.64 = 0.16 g of O_(2)`
Ratio of oxygen lower and higher oxide
`= 0.08 : 0.16 implies 1 : 2`