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?

1 Answer
anor277
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 #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...

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