Given the following data, determine the % by mass of oxalate (C 2 O 4 2- ) in a

Question

Given the following data, determine the % by mass of oxalate (C2O42-) in a sample of an iron oxalate complex with the general formula Kz[Fex(C2O4)y].wH2O.

The reaction between oxalate and permanganate is as follows:

6H+ + 5(COOH)2 + 2MnO4- 10CO2 + 2Mn2+ + 8H2O

m(iron oxalate complex) used for titration = 0.100 g

C(KMnO4) standard solution = 0.0200 M

Mw(C2O42-) = 88.02 g/mol

Titre volume = 11.82 mL

Enter your answers in the boxes provided to 3 significant figures. Do not enter units. Very large or small numbers can be entered in scientific notation - for example enter 1.23E-4 to represent 1.23x10-4.

Step 1: Determine the number of moles of KMnO4(and thus MnO4-) in the titre:

n(KMnO4) = n(MnO4-) = C x V(L) = Answer_____________ mol

Step 2: Determine the number of moles of C2O42- reacted with the KMnO4:

n(C2O42-) = n((COOH)2) = n(MnO4-) x 5/2 = __________ mol

Step 3: Determine the mass of C2O42- reacted, and thus the mass percentage of C2O42- in the sample:

m(C2O42-) = n(C2O42-) x Mw(C2O42-) = ___________ g

%m(C2O42-) = [m(C2O42-)/m(iron oxalate complex)] x 100 = ____________ %

Explanation / Answer

The reaction between oxalate and permanganate is as follows:

6H+ + 5(COOH)2 + 2MnO4- 10CO2 + 2Mn2+ + 8H2O

m(iron oxalate complex) used for titration = 0.100 g

C(KMnO4) standard solution = 0.0200 M

Mw(C2O42-) = 88.02 g/mol

Titre volume = 11.82 mL

Step 1: Determine the number of moles of KMnO4(and thus MnO4-) in the titre:

n(KMnO4) = n(MnO4-) = C x V(L) = 0.02 X 11.82 / 1000 mol = 0.000236 moles

Step 2: Determine the number of moles of C2O42- reacted with the KMnO4:

As per equation , 5 moles of oxalate reacts with two moles of KMnO4-

n(C2O42-) = n((COOH)2) = n(MnO4-) x 5/2 = 0.000236 X 5/2 mol = 0.000591 moles

Step 3: Determine the mass of C2O42- reacted, and thus the mass percentage of C2O42- in the sample:

m(C2O42-) = n(C2O42-) x Mw(C2O42-) = 0.000591 moles X 88.02 g/mol = 0.052 grams

%m(C2O42-) = [m(C2O42-)/m(iron oxalate complex)] x 100 = 0.052 X 100 / 0.1 = 52.02 %

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