# Titration, finding percentage of compound in unknown substance

A .750 g sample of an uknown solid is dissolved in 100 mL of water and acidified with 25 mL of 3 M H2SO4 then titrated with a .0200 M KMn04 solution. If the unknown solid requires 12.5 mL of the KMnO4 solution to reach the endpoint, what is the % sodium oxalate in the unknown solid? Assume the only compound in the unknown solid that reacts with the KMnO4 solution is sodium oxalate. Report your answer as a percentage.

Can someone please help?? i can't figure this one out! I'm not even sure where to begin. If someone could explain how to start I think I could figure out the rest.

You need to start with a balanced equation.  To check myself, I googled it and found this:

http://www.bgsu.edu/departments/chem/faculty/endres/ch128/oxalate.htm

2 MnO4- + 16 H+ + 5 C2O42- -----10e-----> 2 Mn+2 + 8 H2O + 10 CO2 (g)

Next you need to find the moles of permanganate:

moles of permanganate* =      .0125L  x  .0200 M

Next find the moles of oxalate by using the coefficients in the balanced equation to set up a proportion:

5 moles oxalate / 2 moles permanganate =  x moles oxalate / moles of permanganate*

Solve for x moles of oxalate

Now find the mass of the sodium oxalate =  moles of oxalate       x      molar mass of Na2C2O4

% = part / whole x 100 so in this case

% sodium oxalate = mass of sodium oxalate*  / .750 g  x 100

Hope that helps!