Hey guys, i've been searching the forums and i've seen this question maybe once or twice but i've failed to comprehend the explanation given.
An impure sample of Iron(II) sulfate, weighing 1.545g, was treated to produce a precipitate of Fe203. If the mass of the dried precipitate was 0.315g, calculate the percentage of iron in the sample.
I was wondering if anyone could give me an easier explanation =/ as i always end up with 19.3% rather then the true answer, 14.3%
Appreciate any help that i can get
Cheers, Rob.