Question

What is the empirical chemical formula of a compound that is 69.9% of iron and 30.0% oxygen?

Answer 1

`Fe_2O_3`

Explanation:

We assume there are `100*g` of compound, and thus there are:

`(i)` `(69.9*g)/(55.85*g*mol^-1)` `=` `1.25*mol*Fe`, and

`(ii)` `(30.0*g)/(15.999*g*mol^-1)` `=` `1.88*mol*O`.

If we divide thru by the lowest molar quantity we get a formula of `FeO_(1.5)`. Because, by definition the empirical formula is the simplest WHOLE number ratio, we double this result to give `Fe_2O_3`.