[steph_r]

Hi all,
I have a question that im unsure of..

An impure sample of Iron(II) Sulfate, weighing 1.545g was treated to produce a precipitate of Fe₂O₃. If the mass of the dried precipitate was 0.315g, calculate the percentage of iron in the sample.

Ive worked out the mol of Fe₂O₃
n(Fe₂O₃) = 0.315/159.7 = 0.00197 mol.

Im not sure how to write an equation either. Do I need it anway? How would i work out the mol ratio?

Cheers
Thank u in advance

[JGK]

1. The equation is not necessary.

2 Conversion of weight of Fe₂O₃ to moles is not necessary.

3. This a simple mathematical problem.

4. Based on atomic and molecular weights, what is the percentage of Fe in Fe₂O₃?

5. Using the percentage from 4, what is the weight iron in the mass of Fe₂O₃ produced by the reaction?

6. Taking the answer from 5, express it as a percentage of the original sample mass, Job done

[steph_r]

Oh i see! I was trying to make a simple question harder..

Thank you very much for your help