Two oxides of the metal contain 27.6 % and 30.0 % of oxygen respectively, If the formula of the first oxide is $M_{3}O_{4}$, find that of the second

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#1

In the first oxide, % of oxygen = 27.6 Metal = 100 - 27.6 = 72.4 parts by mass.
As the formula of the oxide is $M_{3}O_{4}$, it means
72.4 parts by mass of metal = 3 atoms of metal and
27.6 parts by mass of oxygen = 4 atoms of oxygen In the second oxide, % of oxygen = 30.0 = 100-30 = 70 parts by mass.
But, 72.4 parts by mass of metal = 3 atoms of metal
70 parts by mass of metal=3/72.4 x70 atoms of metal=2.90 atoms of metal
Also,
27.6 part by mass of oxygen=4 atoms of oxygen

30 part by mass of oxygen=4/27.6 x 30 atoms of oxygen.
Hence,
Ratio of M : O in the second oxide=2.90 : 4.35=1 : 1.5=2 : 3
Formula of the other metal oxide is $M_{2}O_{3}$