2 metallic oxides contain 27.6% and 30.0 % oxygen respectively. if the formula of the first oxide os M3O4 that of the second will be...?

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