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

1 Answer
Apr 24, 2016

Fe_2O_3Fe2O3

Explanation:

We assume there are 100*g100g of compound, and thus there are:

(i)(i) (69.9*g)/(55.85*g*mol^-1)69.9g55.85gmol1 == 1.25*mol*Fe1.25molFe, and

(ii)(ii) (30.0*g)/(15.999*g*mol^-1)30.0g15.999gmol1 == 1.88*mol*O1.88molO.

If we divide thru by the lowest molar quantity we get a formula of FeO_(1.5)FeO1.5. Because, by definition the empirical formula is the simplest WHOLE number ratio, we double this result to give Fe_2O_3Fe2O3.