[shalikadm]

Question:
There's 19.9% (mass percentage) of Fe in (NH₄)₂SO₄.Fe₂(SO₄)₃.xH₂O
Find x
What I have tried..
I have got this answer in some other way..But I want to try it also in this way..But I have got stuck here..Please tell me where I've gone wrong..
Overview of what I have done..
1.Imagining there's 1000g of compound with water of crystallisation
2.Getting mass of Fe in the compound of 1000g.
3.Calculating moles of Fe in the compound of 1000g.
4.Using the mole ration to get moles of (NH₄)₂SO₄.Fe₂(SO₄)₃ in the compound of 1000g
5.Calculating the mass of water crystals in the compound of 1000g.
6.Calculating mole of water crystals in the compound of 1000g.
7.Using the mole ratio of (NH₄)₂SO₄.Fe₂(SO₄) : xH₂O to find x
Calculation where I've stuck
Let's imagine there's 1000g of (NH₄)₂SO₄.Fe₂(SO₄)₃.xH₂O.
∴mass of Fe in the compound of 1000g=199g
∴moles of Fe in the compound of 1000g=199g/18gmol^-1=11.05 mol
1mol (NH₄)₂SO₄.Fe₂(SO₄)₃ ≡ 2mol Fe
∴moles of (NH₄)₂SO₄.Fe₂(SO₄)₃ in the compound of 1000g=11.05mol/2=5.527mol
M_{(NH4)2SO4.Fe2(SO4)3}=532gmol^-1
∴mass of (NH₄)₂SO₄.Fe₂(SO₄)₃ in the compound of 1000g=5.527mol*532gmol^-1=2940.364g
Ooopse..how can that be ? 2940.364g in 1000g ?
Please make this clear to me..
PS.I want to try this in this way

[Borek]

moles of Fe in the compound of 1000g=199g/18gmol^-1=11.05 mol

Check your numbers.

[Hunter2]

et's imagine there's 1000g of (NH4)2SO4.Fe2(SO4)3.xH2O.
∴mass of Fe in the compound of 1000g=199g
∴moles of Fe in the compound of 1000g=199g/18gmol-1=11.05 mol

Dont understand this part. Where the 199 g Fe comes from and why you devide by the molar weight of water?

[shalikadm]

Check your numbers.

why you devide by the molar weight of water?

Oh...nice thing !...
I've divided it by M of water...maybe I was sleepy...
Yeah we have to divide it by M of iron...thanks for the help..!
 [/quote]

Where the 199 g Fe comes from

I imagined that there's 1000g of the compound..so the percent of 19.9 % gives the mass of iron.

[AWK]

Simply use definition of percetage

19.9=2*M(Fe)*100/(M(ammonium iron alum)+18*x) - solve for x
where x is the number of water molecules, 2*M(Fe) is a mass of iron in stoichiometric unit of alum; denominator contains the mass of hydrate divided into two parts: the mass of anhydrous alum and the mass of water in the form of 18*x).

Using 1000 g instead of 100 is possible, but this is a bit longer way.
And do not calculations errors!