Question

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?

Answer 1

M - Mass of metal

1. Molecular mass of MO₂ = M+(16 x 2) = M+32 g {Mass of 1 oxygen = 16 g}
(1 Metal + 2 oxygen )
MO₂ contains 50% M,

% of M in MO₂ =
Mass of metal / Molecular mass of MO₂ = 50/100

M g/(M+32) g = 0.5

M = 0.5M + 16

0.5M = 16

So, M = 32 g

2. Molecular mass of MO_n = M+(16 x n) = (M+16n) g {'n' indicate No. of oxygen}

Second oxide MO_n contain 40% M,

% of M in MO_n =
Mass of metal / Molecular mass of MO_n = 40/100

32 g/(32 + 16n) g = 0.4

32 = 0.4(32 + 16n)

32 = 12.8 + 6.4n

n = (32-12.8)/6.4

n = 19.2/6.4

n = 3
No. of oxygen in oxide = 3

So, the formula for the 40% M is MO₃