Wt% Question for NaC2O4

Question

Hello I am struggling to calculate the weight percentage for NaC2O4 inn a precipitate.

the molarity of KMnO4 is 0.0290M

volume of Kmno4 = 25.90mL and molarity of KMnO4 is 0.02153

the mass of unknown sample = 1.018g

I know that its mass solute/mass of solvent *100

the wt% is = 18.35% but I'm having a hard time trying to figure out the steps to calculate it to get this answer.

Answer 1

There are a few problems with the current question.

First, Na₂C₂O₄ (sodium oxalate) is not a precipitate. It is soluble.

Next, you give 2 different values for molarity of KMnO₄ (0.0290 M and 0.02153 M)

Next, what is the original sample composed of?

I'm guessing this is a redox reaction between Na₂C₂O₄ and KMnO₄ and you are trying to find the mass% of Na₂C₂O₄ in solution. If that is indeed the question, here is how to go about it.

First, write the correctly balanced redox equation:

2MnO₄^- + 5C₂O₄^2- + 16H⁺ ==> 2Mn²⁺ + 10CO₂ + 8H₂O

Next, find moles of MnO₄^- used (we'll do it for each of the concentrations given):

25.90 ml x 1 L / 1000 ml x 0.0290 mol/L = 7.511x10^-4 moles MnO₄^-

25.90 ml x 1 L / 1000 ml x 0.02153 mol/L) = 5.576x10^-4 moles MnO₄^-

Next, find moles of C₂O₄^2- present in the sample (we'll do it twice for the 2 different MnO₄^- values)

7.511x10^-4 moles MnO₄^- x 5 mols 5C₂O₄^2- / 2 mols MnO₄^- = 1.878x10^-3 moles 5C₂O₄^2-

5.576x10^-4 moles MnO₄^- x 5 mols 5C₂O₄^2- / 2 mols MnO₄^- = 1.394x10^-3 moles 5C₂O₄^2-

Convert moles to grams using molar mass of Na₂C₂O₄^2- (134 g/mole)

1.878x10^-3 moles 5Na₂C₂O₄^2- x 134 g / mol = 0.2517 Na₂C₂O₄^2-

1.394x10^-3 moles 5Na₂C₂O₄^2- x 134 g / mol = 0.1868 g Na₂C₂O₄^2-

Finally, calculate mass % Na₂C₂O₄:

0.2517 g / 1.018 g (x100%) = 24.72%

0.1868 g / 1.018 g (x100%) = 18.35%

SO APPARENTLY THE CORRECT MOLARITY OF KMnO₄ WAS 0.02153 M. NOT SURE WHERE YOU GOT 0.0290 M