Question

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

Answer 1

Answer:

M2O3.

Explanation:

From the question we know that the atomic weight of metal M is given as    56 gram. Again we get that the metal oxide contains 70% of the metal, so the remaining (100-70)=30% will be the oxide present in it.

So, in 1 atom of metal if we take n atoms of oxygen then the mass proportion of the metal and the oxygen will be :-

56:n*16 = 70:30.

16n=56*30/70.

n=24/16 = 1.5. So, the whole number ratio will be 2:3.

Therefore, we get that there will be two atoms of metal against the 3 atoms of Oxygen. So, the empirical formulae will be 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!!✔✔