two oxides of a metal , one contain 27.6% and 30% oxygen respictively . if the formula of the first oxide is M3O4 , find that of second

Given that, formula of first oxide = M₃O₄
Let mass of the metal = x
% of metal in M₃O₄ = (3x/ 3x+64) *100
but as given % 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, the ratio is
M : O
70/56 : 30/16
1.25 : 1.875
2 : 3
so, 2nd oxide is M₂O₃

Percentage of oxygen=27.6Precentage of metal=100-27.6=72.4Formula of the metal oxide=M3O4Let the atomic mass of the metal= xPercentage by weight of the metal in the oide M3O4=3x *100/3x+64=72.43x=0.724*3x+0.724*640.828x=46.336x=56The atomic mass of the metal=56 Element Precentage Atomic Mass Atomic ratio Simplest ratio whole no.ratioM 70 56 70/56=1.25 1.25/1.25=1 2O 30 16 30/16 1.88/1.25=1.5 3 M2O3

Formula 1 = M3O4O2% = 27.6Metal = 72.4Metal/Oxygen = 3x/64= 72.4/27.6x=72.4*64/27.6*3=56Formula 2 = 70%/30%= 56/16=> 112:168= 2:3=> M2O3

Hello student,
the solution is as follows
Given that, formula of first oxide = M3O4
Let mass of the metal = x % of metal in M3O4
= (3x/ 3x+64) *100
given % age of metal = (100-27.6) = 72.4 %
so, (3x/ 3x+64)*100 = 72.4
x = 56
in 2nd oxide, oxygen = 30%
so metal = 70%
so, the ratio is M : O = 70/56 = 2 : 3
so, 2nd oxide is M2O3
Thanks
I hope the solution will solve all your doubts.
All the for exams.