The stochiometric equation for the reaction between KMnO4 and oxalate is
2MnO4- (aq) + 5 C2O42-(aq) + 14 H+ (aq) ---------> 2Mn2+(aq) + 10CO2(g) + 7H2O(l)
This equation indicates two moles of permanganate is equivalent to 5 moles of oxalate.
Given that
The volume of KMnO4 = 28.70 mL
= 0.0287 L
The molarity of KMnO4 = 0.020 M
∴ The no. of moles of KMnO4 = molarity * volume
= 0.020 M * 0.0287 L
= 0.000574 mol
Hence the no. of moles of oxalate = 0.000574 mol * (5/2)
= 0.001435 mol
We know that the molar mass of oxalate = 88 g/ mol
∴ The mass of oxalate = number of moles* molar mass
= 0.001435 mol * 0.001435 g/mol
= 0.12628 g
The percentage of oxalate = mass of oxalate *100 / mass of sample
= 0.12628 g *100/ 0.25 g
= 50.5 %
Actual mass of oxalate in the sample = 0.25g * ( 3*molar mass of oxalate / molar mass of complex)
= 0.25g*(3*88g/mol /491.145g/mol)
= 0.1344 g
The percentage purity of the complex = (calculated mass of oxalate / actual mass of oxalate)*100
= (0.12628 g/0.1344g)*100
= 93.95 %