[FerndaleTiger]

I believe I have a simple problem. Oddly, I cannot figure it out.

A bottle contains 30.8 grams of pure iron(III) sulfate. How many sulfate ions are present in this sample?

So I started with what was given 30.8 g Fe(SO4)3 then multiplied this by 1 mol of iron(III)sulfate over 344.03 g of iron(III)sulfate then by 3 moles SO4 over 1 mol iron(III)sulfate and got .2686 mol of sulfate. Finally I multiplied the .2686 by avagadro's number 6.02 x 10^23 and got 1.62 x 10^23 ions.

The problem is, that isn't an answer that was given and I don't see what I did wrong.

[Demotivator]

The formula is Fe2(sO4)3

[AWK]

and molecular mass is 400.

[FerndaleTiger]

really? I fixed my equation and used the same molecular mass and got it right.