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