Question

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?

Answer 1

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 `Fe_((0.76*g)/(55.8*g*mol^-1))O_((0.222*g)/(16.00*g*mol^-1))-=Fe_0.01362O_0.013875`

We divide thru by the lowest molar quantity to get a trial empirical formula of....`Fe_((0.01362)/(0.01362))O_((0.013875)/(0.01362))-=FeO`... i.e. we got ferrous oxide...