i just had an Epiphany. Part 1 of my calculations were wrong i think.
It should look like this: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)2(SO4)2.6H2O 392.16
n(Fe2+) = mass / Mr therefore: 9.794/392.16 = 0.024974
C(Fe2+) =nx1000/v ( volume is 250 because concentration of Fe(NH4)2(SO4)2.6H2O is the same as Fe2+) = 0.099896
m1 ( conc. of Fe2+) = 0.099896
v1 (Volume of Fe2+) = 25.00
v2 ( Volume of MnO4-) = 24.82Using the equation m2=m1v1/5v2 (Using the following equation ...m2=m1v1/5v2
Where m1 and m2 is concentration of Fe(2+) and MnO4- and v1 and v2 is the corresponding volumes used.)
m2= 0.0201240934Part Two
Mass of compound used 0.200 g
Final Bruette reading 32.82 cm³
Final Volume of KMnO4 solution used 32.82 cm³Thus 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.
So that Mr=600W/mv
Moreover, since Mr=143.87 + 18.02x
Then x= (Mr-143.87)/18.02
Therefore using Mr=600W/mv = 600 x 0.200/ 32.80 x 0.0201240934Mr= 181.7988275
And so, using x= (Mr-143.87)/18.02 = (181.7988275-143.87)/18.02= 2.104818398 which is 2.1
If you spot any mistakes, could you let me know please?