determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.?

Question

please explain in detail

Answer 1

Please give some more info.

Need to know what you mean by dioxygen. Most questions of this type will just quote the level of oxygen.

Or are you saying that the Iron and Oxygen are reacted in those quantities

Addition: I have just run a couple of calcs and I think that the dioxygen should be oxygen.

For the empirical formula you need to know the molar relationship not the weights.

Moles of Iron = mass/MW
= 69.9/55.8
= 1.25

Moles O = 30.1/16
= 1.88

Therefore the empirical relationship is
Iron: Oxygen
1.25:1.88
1:1.5
2:3
So the empirical formula would be Fe2O3

Answer 2

suppose the formula is Fe(x)O
the value of x:
x=(69.9/56)/(69.9/56 + 30.1/32)

Answer 3

Fe2O3