Question

A sample of the green crystal compound with a mass of 0.150 g is weighed into an Erlenmeyer flask. 36.14 mL of 0.0100 M KMn04 solution is required to titrate to the equivalence point. How many grams of C2O42- are in the sample of green crystal compound and what is the % oxalate in the compound?

Answer 1

Reaction between oxalic acid and pot. permanganate is as shown below.

2 MnO ₄^- + 5 C₂O₄^2- + 16 H ⁺ 2 Mn ²⁺ + 10 CO ₂ + 8 H₂O

From reaction, Stoichiometric ratio = 2 mol MnO ₄^- / 5 mol C₂O₄^2- = 2/5

We have correlation, No. of moles of MnO ₄^- = No. of moles of C₂O₄^2- Stoichiometric ratio

We have, Molarity = No. of moles of solute / Volume of solution in L

Therefore, No. of moles of MnO ₄^- = Molarity Volume of solution in L = 0.0100 M 0.03614 L

= 3.614 10 ^-04 mol

Hence, 3.614 10 ^-04 mol =   No. of moles of C₂O₄^2- Stoichiometric ratio

No. of moles of C₂O₄^2- = 3.614 10 ^-04 mol /  Stoichiometric ratio

= 3.614 10 ^-04 mol / (2/5)

= 9.035 10 ^-04 mol

We have, No. of moles = Mass / Molar Mass

Mass of C₂O₄^2- = No. of moles Molar Mass

Molar mass of C₂O₄^2- = ( 2 12.01) + ( 4 16.00 ) = 88.02 g/ mol

Therefore, Mass of C₂O₄^2- = 9.035 10 ^-04 mol   88.02 g/ mol = 0.07953 g

ANSWER : grams of oxalate in the sample of green crystal compound = 0.07953 g

% C₂O₄^2- = Mass of C₂O₄^2- / mass of sample 100

= 0.07953 g / 0.150 g 100

= 53.02

ANSWER : % C₂O₄^2- in the sample = 53.02