A sample of an iron oxide weighing 0.982g is found to contain 0.763 g of iron. The rest of the mass is oxygen. What is the % composition of each element?

Jun 21, 2018

Well, we should actually work out the empirical formula of the oxide here...

Explanation:

We gots iron oxide....%Fe=(0.76*g)/(0.982*g)xx100%=77.4%

%O=(0.982-0.76*g)/(0.982*g)xx100%=22.6%

Clearly, the percentages must sum to 100% in a binary compound. Do they?

On these data we could also assess the empirical formula...by finding the molar quantities of each element...

We get $F {e}_{\frac{0.76 \cdot g}{55.8 \cdot g \cdot m o {l}^{-} 1}} {O}_{\frac{0.222 \cdot g}{16.00 \cdot g \cdot m o {l}^{-} 1}} \equiv F {e}_{0.01362} {O}_{0.013875}$

We divide thru by the lowest molar quantity to get a trial empirical formula of....$F {e}_{\frac{0.01362}{0.01362}} {O}_{\frac{0.013875}{0.01362}} \equiv F e O$... i.e. we got ferrous oxide...