Answer:
This is a ratio of two small integers, 1.500 = 3:2. Let's assume that in the molecule of ferrous oxide there is one oxygen atom and one iron atom, and the ratio of the mass of the oxygen atom to the mass of the iron atom is 0.2865.
Step-by-step explanation:
Molecular weight of Iron (II) oxide= FeO Molecular weight of Iron (III) oxide= Fe2O3
In FeO , every one iron atom contains one oxygen atom.
In Fe2O3, every one iron atom contains 1.5 oxygen atoms.
So FeO(II) will contain more Iron than FeO(III).
Calculation of percentage of iron is not necessary. I think this argument
alone is enough to prove this.
However on calculating the % of Fe, in FeO and Fe2O3 we get,
% of Iron in FeO= 55.85/(55.85+16) =55.85/71.85= 77.73%
% of Iron in Fe2O3= (2 x55.85)/(111.7+48)
=111.7/159.48 = 69.94%