Question

Two oxides of a metal contain 36.4% and 53.4% of oxygen by mass respectively.


If the formula of first oxide is M20, then that of the second is -


a) M203


c) MO 2


d) M 205


b) MO


​

Answer 1

Answer:

The answer is M2O.

Explanation:

The ratio of metal and oxygen in the first oxide, M3O4 = 46.6: 53.4

The ratio of metal and oxygen in the second oxide =63.6: 36.4

Let the molecular mass of metal = M

Now calculate the percentage by weight of the metal in the oxide:

 M x 100 / M + 16 = 46.6

100M = 46.6M + 745.6

100M - 46.6M = 745.6

53.4M = 745.6

M = 745.6/53.4

M =13.96 = 14 g

Now,

Moles of metal in second oxide = 63.6 / 14 = 4.54

Moles of oxygen in second oxide = 36.4 / 16 = 2.275

Ration of moles of metal and oxygen = 4.54: 2.275

= 2 :1

Thus formula of secong oxide is M2O.