Percentage of oxygen=27.6
Precentage of metal=100-27.6=72.4
Formula of the metal oxide=M3O4
Let the atomic mass of the metal= x
Percentage by weight of the metal in the oide M3O4=3x *100/3x+64=72.4
3x=0.724*3x+0.724*64
0.828x=46.336
x=56
The atomic mass of the metal=56
Element Percentage Atomic Mass Atomic ratio Simplest ratio whole no.ratio
M 70 56 70/56=1.25 1.25/1.25=1 2
O 30 16 30/16 1.88/1.25=1.5 3
M2O3... Or...
In the case of the 1st oxide, Mass of oxygen = 27.6
Therefore,
Mass of metal = 100 - 27.6 = 72.4
In the case of the 2nd oxide, Mass of oxygen = 30
Therefore,
Mass of metal = 100 - 30 = 70
Also,
Formula of the 1st oxide is M3O4
Therefore,
Number of atoms of metal in the 2nd oxide = (3/72.4) x 70
= 2.9
Number of atoms of oxygen in the 2nd oxide = (4/27.6) x 30
= 4.35
Thus,
Ratio of metal : oxygen in the 2nd oxide = 2.9 : 4.35 or ≈ 2 : 3
Hence,
The Formula of the 2nd oxide is M2O3. Hope I helped u.. Plz mark as the BRAINLIEST answer if helpful..