chemistry-Dr.Bob

ok for a lab we had to titrate a solution with KMnO4 and for each trial i got 22.83 mL,23.92, and 22.48

I need to find the mass percent of the oxalate ion. How would I do this?

The equation is MnO4-+C2O4 2- -> Mn 2+ + CO2

In the lab we used about .1 g of the oxalate salt in each titration.

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asked by Trisha
1. forget to tell u that the M of KMnO4 is .011393M

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posted by Trisha
2. About 0.1 g of the oxalate salt won't do it if you're looking for any kind of quantitative answer. And with buret readings to the second place, you must be looking for a decent answer.
First balance the equation.
Have you determined the molarity of the KMnO4? or is that the purpose of the about 0.1 oxalate ion? I will assume you have the molarity and go from there. If not, then what follows is not correct.
Find mols KMnO4. That will be M x L = ??
(You don't list the molarity but I'm sure you have it there somewhere.)
Then use the balanced equation to convert mols KMnO4 to mols oxalate.
Convert mols oxalate ion to grams oxalate ion.
Then %oxalate = [(grams oxalate)/mass of the initial sample)]*100

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posted by DrBob222

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