Two oxides of metal contain 27.6% and 30% of oxygen respectively. If the formula of first oxide is M3O4 what is the formula of second oxide.pls explain

Formula of first oxide = M₃O₄ 

let mass of the metal be == x

percentage of metal in M₃O₄ = (3x/ 3x+64) *100

but % age = (100-27.6) = 72.4 %

so, (3x/ 3x+64)*100 = 72.4

or x = 56.

in 2nd oxide,

oxygen = 30%....so metal = 70%

so, ratio :--

M : O

70/56 : 30/16

1.25 : 1.875

2 : 3

so, 2nd oxide = M₂O₃