Question

QUESTION 3 Following the titration procedure used in this experiment, 0.0966 g of the green oxalate containing compound was titrated to the end point using 26.433 mL of 0.0100 M KMnO4 solution. Based on this result, what was the percent oxalate present in the sample? Report your answer to three significant figures

Answer 1

Solution :-

Balanced reaction equation

2MnO4^- + 16H^+ 5C2O4^2- --- > 2Mn^2+ + 8H2O + 10 CO2

Lets calculate moles of KMnO4

Volume of KMnO4 = 26.433 ml* 1 L / 1000 ml = 0.026433 L

Moles = molarity x volume in liter

           = 0.0100 mol per L * 0.026433 L

          = 0.00026433 mol KMnO4

Now using the moles of KMnO4 we can find the moles of oxalate

Mole ratio is 2 mol MnO4^- : 5mol C2O4^2-

0.00026433 mol MnO4^- * 5 mol C2O4^2- / 2 mol MnO4^- = 0.0006608 mol C2O4^2-

Now lets calculate the mass of oxalate ions

Mass = moles x molar mass

Molar mass of oxalate = (C2O4^2-) = 88.018 g/mol

Mass of oxalate = 0.0006608 mol * 88.018 g per mol

                        = 0.0582 g oxalate

Now lets calculate the percent of the oxalate

% oxalate = [mass of oxalate / mass of sample]*100%

                  = [0.0582 g / 0.0966 g]*100%

                  = 60.25 %

Therefore the percent of the oxalate in the sample is 60.25%