C2O42- H2O2, K2C2O4 H2C2O4
Fe2+ ⎯⎯⎯⎯⎯→ FeC2O4 ·2H2O ⎯⎯⎯⎯⎯→ Fe(OH)3 ⎯⎯⎯⎯⎯→ K3Fe(C2O4)3 ·3H2O
2H2O + K3Fe(C2O4)3
Thus all the iron in the iron(II) ammonium sulphate is converted to ferrioxalate, one third of the ferrioxalate being formed from the iron(III) hydroxide. Both H2O2 and Fe(OH)3 are unstable to heat. Potassium trioxalatoferrate(III) is photosensitive, i.e. it decomposes when exposed to light, reforming iron(II) oxalate.
Procedure:
a. Dissolve 7.5 g of iron(II) ammonium sulphate, FeSO4 · (NH4)2SO4· 6H2O in 25 cm3 of warm water(40oC). Add to this warm solution, with stirring, 32.5 cm3 of an oxalic acid solution made by dissolving 10 g of the acid in 100 cm3 of water.
b. Cautiously heat the mixture to boiling, and then allow the yellow granular ppt to settle.
c. Decant the supernatant liquid and wash the ppt by stirring it with 50 cm3 of hot water (~80oC). Allow the ppt to settle and again decant the clear liquid. Repeat with a further 50 cm3 of hot water.
d. Suspend the washed ppt in a warm solution of 5 g of potassium oxalate in 15 cm3 of water. Add slowly from a burette 12.5 cm3 of 1.8 M H2O2, stirring continuously and keeping the temperature at about 40 oC.
e. Heat the mixture to boiling.
f. Add the boiling mixture, in one portion, 10 cm3 of a solution of oxalic acid containing 10 g of the acid in 100 cm3 of water. If the mixture has not turned to a clear green, add oxalic acid solution, drop by drop, until it becomes clear (up to 2.5 cm3 may be required).
g. The solution should now be green due to the presence of the ferrioxalate ion. Filter the hot solution if necessary and add 15 cm3 of ethanol.
h. Set aside in the dark to crystallize. After crystallization, decant off the supernatant liquid, washed by decantation with a little water, and finally with 10 cm3 ethanol.
i. Spread the crystals on a fresh filter paper and allow them to dry in the dark. Weigh the dry crystal and record the result.
Calculation:
- Calculate the theoretical yield, i.e. the mass of potassium trioxalatoferrate(III) expected from
complete conversion of the mass of iron(II) ammonium sulphate used.
FeSO4(NH4)2SO4·6H2O ⎯⎯⎯→ K3Fe(C2O4)3· 3H2O
391.8 490.8
Theoretical yield = x 490.8 = 9.395 (g)
2. Calculate the percentage yield in this exercise, i.e. the percentage of the theoretical yield that has actually been achieved.
% yield = x 100%
= %
Questions for discussion:
1. What is the name of the compound precipitated in procedure a?
Write an equation for the reaction.
Hydrated Iron (II) oxalate
Fe2+(aq) + C2O42-(aq) + 2H2O(l) ⎯⎯⎯→ FeC2O4 2H2O(s)
2. In procedure d, why should the temperature be kept at about 40 oC?
Both H2O2 and Fe(OH)3 are unstable to heat.
3. Write an equation for the reaction.
Colourless gas (O2) is evolved; brown ppt [Fe(OH)3] is observed.
FeC2O4 + 6K2C2O4 + 3H2O2 ⎯⎯⎯→ 4K3Fe(C2O4)3 + 2Fe(OH)3
4. Why excess oxalic acid should be avoided in procedure f?
Write an equation for the reaction.
The crystals would be contaminated with oxalic acid.
2Fe(OH)3 + 3H2C2O4 + 3K2C2O4 ⎯⎯⎯→ 2K3Fe(C2O4)3 3H2O + 3H2O
5. In procedure g, what is the action of the ethanol?
Why is it effective?
The crystals are soluble in water, but very sparingly soluble in ethanol.
(Decrease the solubility of the product)
6. In procedure h, why is it necessary to exclude light?
Why are the crystals finally washed with ethanol?
Because the crystals are photosensitive.
To facilitate drying.