Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Percent of sodium oxalate help

  1. Sep 18, 2008 #1
    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

    ---------------------------------------------------------
    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?
     
  2. jcsd
  3. Sep 18, 2008 #2

    Borek

    User Avatar

    Staff: Mentor

    What are you trying to do here?
     
  4. Sep 18, 2008 #3
    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.
     
  5. Sep 18, 2008 #4

    Borek

    User Avatar

    Staff: Mentor

    You have to use these things to calculate number of moles of permanganate - but not dividing them...
     
  6. Sep 19, 2008 #5
    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.
     
  7. Sep 19, 2008 #6

    Borek

    User Avatar

    Staff: Mentor

    Seems OK. I got 0.5269g, but that's most likely because of the difference in molar masses used.
     
  8. Sep 19, 2008 #7
    Thanks so much!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook