Two oxides of metal were found to contain 31.6% and 48% of oxygen, respectively. If the formula of first is represented by `M_(2) O_(3)`, find that of the other

Explanation

First methor
`{:("First oxide","Second oxide"),(M_(2)O_(3), M_(2) O_(x)),(O = 31.6%, O = 48%),(M = 68.4%, M = 50%):}`
Let the atomic weight of the metal `= Mg`
`(2 M + 48) g of M_(2) O_(3)` contains `= 2 M g` of metal
`100 g of M_(2) O_(3)` contains `= ((2 M xx 100)/(2 M + 48)) = 68.4`
Solve for `M = 52`
Atomic weight of metal = 52
Therefore, in second oxide:
`O = (48)/(16) = 3` `M = (52)/(52) = 1`
Therefore, formula `= MO_(3)`
second method:
First oxide:
`68.4 g of M` contains `= 31.6 g` of oxygen
`52 g of M` contains `= (31.6 xx 52)/(68.4) = 24 g of O_(2)`
Ratio of `O` in `M_(2) O_(3)` and `M_(2) O_(x) = 24:48` (given) `= 1 : 2`
Therefore, the ratio of `O` in second oxide should be twice the oxygen in first oxide and the % of `M` is fixed.
Therefore, first oxide `= M_(2) O_(3)`
Second oxide `= M_(2) O_(6)` or `MO_(3)`
Third method:
a. `68.4 g of M` in first oxide = 2 atom of `M`
`50 g` of oxygen in first oxide contains = 3 atom of oxygen
`48 g` of oxygen contains `= (3 xx 48)/(31.6)`
= 4.5 atom of oxygen
Ratio of `M : O = 1.5 : 4.5 = 1: 3`
Thus, the formula is `MO_(3)`