Percent of sodium oxalate help

[sami23]

I have been suck on a problem for days:

We dissolve 3.778g of a sample that contains some sodium oxalate, Na_2C_2O_4, in water and acidify the solution with excess sulfuic acid. The sample requires 18.74 mL of 0.08395 M KMnO_4, potassium permanganate, for complete reaction according to the reaction below. What is the percent Na_2C_2O_4 in the sample? Assume that no other component reacts with the potassium permanganate.

8H_2SO_4 + 2KMnO_4 + 5Na_2C_2O_4 --> 8H_2O + 2MnSO_4 + 10CO_2 + 5Na_2SO_4 + K_2SO_4

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Now, I know the objective is to find %sodium oxalate by dividing (grams of sodium oxalate over 3.778 g of solution) times 100.

I know in one mole of sodium oxalate there's 133.996 g of sodium oxalate.

In order to find g of sodium oxalate:
(0.08395 M permanganate / 0.01874 L) * (5 mol sodium oxalate / 2 mol permanganate) * (133.996 g sodium oxalate / 1 mol sodium oxalate) but this gives me sodium oxalate/L

How do I find just the grams so I can plug that into my % formula?

 

[Borek]

(0.08395 M permanganate / 0.01874 L)

What are you trying to do here?

 

[sami23]

Since the sample requires 18.74 mL of 0.08395 potassium permanganate, I thought that's how it was related. I've omitted that part from my equation now.

 

[Borek]

You have to use these things to calculate number of moles of permanganate - but not dividing them...

 

[sami23]

So I found the moles of permanganate as 1.5732 x 10^-3 by multiplying 0.08395M and 0.0187 L. Then I got my sodium oxilate grams as 0.52701 g. I plugged that to my % formula and got 13.95%. I hope I did it correct. Seems reasonable to me.

 

[Borek]

Seems OK. I got 0.5269g, but that's most likely because of the difference in molar masses used.

 

[sami23]

Thanks so much!