[kandyfloss]

Here is the question :-
2.0 g of an oxide of iron contains approximately 0.60 g of oxygen and 1.4 g iron.What is the empirical formula of the oxide?
I know the question is easy but i dunno why i'm totally stuck on this one.
Here is what i did

I found the amount of 0₂ and amount of Fe
0.60/32 for 0₂ and 1.4/55.95 for Fe which gives me 0.018 mol and 0.25 mol respectively.
Now since there are two oxygen atoms in 0₂ i multiplied 0.018 x 2 = 0.036 mol
Now what?

[Borek]

So far so good, just 0.25 or 0.025 mole?

Now, assuming it was 0.025 - you know that iron oxide contains 0.025 mole iron per 0.036 mole of oxygen. What is their ratio? What is the closest pair of whole numbers with the same ratio?

[kandyfloss]

The ratio comes to be around 1:1.4

[DrCMS]

OK its FeO_1.44 but we normally write these things with whole numbers. 

So now you multiply both sides until they are both whole numbers (or very close).

[kandyfloss]

Oh ok so it seems if i multiply the ratio by 2 i get 2:3.8 ~3 > 2:3
Fe₂O₃
But why can't i use 1:1.4 as 1:~1 which gives FeO

[sjb]

1:1.4 -> 1:1 is about a 40% loss, so is perhaps not really the *best* that you can do, whereas 1:1.4 -> 2:3 is a much closer match.

It's good that you recognise FeO as a known compound, but especially with the transition metals there are a wide variety of oxidation states that they achieve.

[kandyfloss]

Ok that seems to have solved my problem.
Thanks a lot to you all  Actually i was preparing for my chemistry test tomorrow.