Question

Assuming the atomic weight of a metal M to be 56, find the emperical formula of its oxide containing 70.0% of M.

Answer 1

Empirical formula is:-

M: 70/56=1.25
O:30/16=1.875
Ratio is 10:15=2:3
It’s empirical formula is M2O3

Answer 2

Atomic weight of M = 56

Percentage of M = 70.0 %

Since, the oxide contains 70.0 % of M, the amount of oxygen will be 30.0%

The atomic weight of Oxygen = 16

1) Moles of M

= % of M / Atomic mass of M

= 70.0 / 56

= 1.25

2) Moles of O

= % of O / Atomic mass of O

= 30.0 / 16

= 1.875

Hence, the ratio of the number of moles of the metal with oxygen would be,

• For M = 1.25 / 1.25 = 1

• For O = 1.875 / 1.25 = 1.5

i.e., 2 : 3

Hence, the empirical formula of the compound would be M₂O₃.

Hope it helps!!✔✔