Question

3. A 10.1-mL volume of a 0.00940 M KMnO4 solution was used to reach the stoichiometric point in the titration of 0.1040 g of an unknown sample containing oxalate ion.

a. Calculate the moles and mass of C2042- in the sample.

b. Determine the moles and mass of 02042- that would be present in a 100 g sample.

Answer 1

Volume of KMnO₄ solution used = 10.1 mL = 0.0101 L

Molarity of KMnO₄ solution = 0.00940 M

Thus, moles of KMnO₄ used = Molarity*Volume in liters = 0.00940*0.0101 = 9.494*10^-5

Thus, moles of MnO₄^- ions produced = 9.494*10^-5

Basic reaction between oxalate ion and permanganate ion is given by :

2 MnO₄^- + 5 H₂C₂O₄ + 6 H⁺ =>10 CO₂ + 2 Mn²⁺ + 8 H₂O

Thus we see in the balanced equation :

2 permanganate ions react with 5 oxalate ions

So, 9.494*10^-5 moles of MnO₄^- will react with : 5/2*9.494*10^-5 = 2.3735*10^-5 moles oxalate ions

Thus, moles of oxalate ion in sample = 2.3735*10^-5

MW of oxalate ion = 88

Thus , mass of oxalate ion in sample = 88*2.3735*10^-5 = 208.868*10^-5 = 2.08*10^-3 g

(b)

Now, 0.104 g sample contains 2.3735*10^-5 moles oxalate ions

So, 100 g sample will contain = (2.3735*10^-5)/0.104*100 = 0.0228 moles oxalate ions

Thus , mass of oxalate ions in 100 g sample = 0.0228*88 = 2.0064 g oxalate ions