Calculate Percent of Oxalic Acid H2C2O4
 

I need help in a college chemistry class and am really struggling.

The problem is as follows:

Calculate the percent of oxalic acid (H2C2O4) in a solid sample, given that a 0.7984 g of that sample required 37.98 mL of 0.2283 M NaOH solution for neturalization.

I think I have the first part which is the percent composition for H2C2O4 which I calculated at 2.24% for H, 26.68 % for C, 71.08% for O.

Not sure if this is correct or even how to proceed from here.

I can't even find anything similar to even give me a starting point.

Any help would be much appreciated. - Thanks

Another approach
 

First u must know that oxalic acid is a diprotic acid which means it gives 2 hydrogen ions, therefore, 1 mole of oxalic acid will react with 2 moles of oxalic acid.

My method is:

Calculate the mole of NaOH used: (37.98ml)(0.2283M) = 8.6708mmol

mmol of NaOH = 2 X mmol of oxalic acid

Therefore,
mmol of oxalic acid = 0.5 X 8.6708 =4.3354 mmol =0.0043354 mole

Mass of oxalic acid present = (0.0043354)(2X1.00794+2X12.011+4X15.9994)
=0.3910g#

Percent of Oxalic acid in sample = (0.3910/0.7984) X 100% =48.97%##

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Sorry, the 1st part is 1 mole of oxalic acid wil react with 2 moles of NaOH.