I think ( I hope) I am getting what you are saying. Could you see if this is what you were talking about?

**Part One**Mass Container plus Fe(NH4)2SO4.6H2O 12.824g

Mass Of container less some Fe(NH4)2SO4.6H2O 3.030g

Therefore Mass of Fe(NH4)2SO4.6H2O 9.794g

Volume of Graduated flask used 250 cm³

Mr of Fe(NH4)2SO4.6H2O 392.16n(Fe2+) = mass / Mr therefore: 9.794/55.8 = 0.1755197133

n(KMnO4)=0.1755197133/5 ( because n(KMnO4)= n(Fe2+) x 5)

therefore: n(KMnO4)=0.0351039426

C(Fe2+) =nx1000/v therefore: 0.1755197133x1000/250 = 0.7020788532

So now: m1 ( conc. of Fe2+) = 0.7020788532

v1 (Volume of Fe2+) = 25.00

v2 ( Volume of MnO4-) = 24.82

m2( Conc. Of KMno4-) =

Using the equation m2=m1v1/5v2

m2= 0.141434069

**Part Two**Mass of compound used 0.200 g

Final Bruette reading 32.82 cm³

Final Volume of KMnO4 solution used 32.82 cm³n( FeC2O4) = mass /mr = 0.200/143.8= 0.001390

3 x n(MnO4-)=5 x n( FeC2O4)

Therefore:

n(MnO4-) = 0.002318

c(MnO4-)= nx1000/32.80 = 0.072438

v( FeC2O4) =n x 1000/0.141434069 = 9.833702

**Thus if a sample of W(g)of the Iron (II) ethanedioate hydrate requires a volume v ( cm³ ) of MnO4- solution of concentration m (mol dm-3), then the number of moles of MnO4- reacting is mv/1000 and this number of moles will react with 5mv/3000 moles of FeC2O4.xH2O.**

Hence W/Mr=5mv/3000

So that Mr=600W/mv

Moreover, since Mr=143.87 + 18.02x

Then x= (Mr-143.87)/18.02Mr = 600 x 0.200 / 9.833702 x 0.072438 = 172.6705083

Therefore X = 1.059825

Sal.