Two oxides of metal contain 50% and 40% of metal M respectively,if the formula of the first oxide is MO_{​​​​​2},the formula of the second oxide is!?

M :- mass of metal

M+(16 x 2) = M+32 :- Mass of ${MO}_2$
( 1 Metal + 2 oxygen )

{ mass of 1 oxygen =16 g }

${MO}_2$ contains 50% M,

Mass of metal / mass of ${MO}_2$ = 0.5

M/(M+32) = 0.5

M = 0.5M + 16

0.5M = 16

So, M = 32


2nd oxide contain 40%

{ 'n' indicate no: of oxygen }

32/(32 + 16n) = 0.4

32 = 0.4(32 + 16n) = 12.8 + 6.4n

n = (32-12.8)/6.4 = 19.2/6.4 = 3

no: of oxygen in oxide =3

So the formula for the 40% M is ${MO}_3$