In: Chemistry
sample of an Iron Oxalato complex salt weighting 0.145 grams requires 31.13 mL of 0.015 M KMnO4 to turn the solution a very light pink color at the quivalence point.
Calculate the number of moles of KMnO4 added.
Calculate the number of moles of C2O42− in the sample of the oxalate salt.
Calculate the percent weight of C2O42− in the original Iron Oxalato Complex salt sample
Calculate the number of grams of C2O42− in the sample.Calculate the number of moles of C2O42− in the sample of the oxalate salt.Calculate the number of grams of C2O42− in the sample.
a) Millimols of KMnO4 added = (31.13 mL)*(0.015 M) = 0.46695 mmol.
Mol(s) of KMnO4 added = (0.46695 mmol)*(1 mol)/(1000 mmol) = 4.6695*10-4 mol,
b) The balanced chemical equation for the reaction between oxalate and MnO4- is given as below.
MnO4- (aq) + 8 H+ (aq) + 5 e- --------> Mn2+ (aq) + 4 H2O (l)
C2O42- (aq) ---------> 2 CO2 (g) + 2 e-
Multiply the first equation by 2 and the second by 5 to give the balanced equation as
2 MnO4- (aq) + 5 C2O42- (aq) + 16 H+ (aq) -------> 2 Mn2+ (aq) + 10 CO2 (g) + 8 H2O (l)
As per the stoichiometric equation,
2 mols MnO4- = 5 mols C2O42-.
Mols C2O42- in the sample = (4.6695*10-4 mol)*(5 mols C2O42-)/(2 mols MnO4-)
= 1.167375*10-3 mol.
c) The atomic masses are
C: 12.011 u
O: 15.999 u
The MW of C2O42- = (2*12.011 + 4*15.999) g/mol
= 88.018 g/mol.
Mass of oxalate in the sample = (1.167375*10-3 mol)*(88.018 g/mol)
= 0.10275 g ≈ 0.103 g (ans, correct to 3 sig. figs).
Mass percent oxalate in the sample = (0.103 g)/(0.145 g)*100
= 71.03% ≈ 71.0% (ans, correct to 3 sig. figs).
d) Number of grams of C2O42- in the sample = 71.0% (ans).