Two oxides of a metal contain `50%` and `40%` metal `M` respectively. If the formula of the first oxide is `MO_(2)`, the formula of the second oxide will be

Options

A. `MO_(2)`
B. `MO_(3)`
C. `M_(2)O`
D. `M_(2)O_(5)`

Answer

B

Explanation

`{:(,"Oxide I","Oxide II"),("Metal",M" " 50%,40%),("Oxygen",O" "50%,60%):}`
As first oxide is `MO_(2)`
Let atomic mass of M = x
`:. % O = (32)/(x+32)xx100`
or `(50)/(100) = (32)/(x+32)`
or `0.5 = (32)/(x+32)` or `0.5x+16 = 32`
`0.5 x =16`
`x = 32`
At. mass of metal M, x = 32
Let formula of second oxide `M_(2)O_(n)`
`%M = (2x)/(2x+16n)xx100 = (64)/(64+16n)xx100`
`(40)/(100) = (64)/(64+16n) or (100)/(40) = (64+16n)/(64)`
2.5 = 1 +0.25n
2.5n = 2.5-1=1.5
`n = (1.5)/(0.25) = 6`
Therefore, formula of second oxide
`= M_(2)O_(6) or MO_(3)`