[RespectfulStudent]

formula C2H2O4+2NaOH--> Na2C2O4 +2H2O

How many moles of NaOH were used?

nNaOH=cV
nNaOH=(1.000M)(20.00mL/1000)
=0.02 mol NaOH were used

(below calculations are for another question, the purity)
molar mass of NaOH =40 g/mol
n=m/M
m=nM
m base= (0.02 mol)(40g/mol [the molar mass of NaOH])
=0.8 grams of base

1 mol acid/2mol base = x mol acid /0.02 mol base
^theoretical value              ^actual/experimental value
x mol acid = 0.02 mol NaOH (1mol acid/2 mol base)
x mol acid = 0.01 mol
 0.02 mol of NaOH were used and 0.01 mol of acid were used

How many moles of the oxalic acid were present in the sample?

n=m/M
if the sample isn't pure, then this following procedure may not be correct:
n acid = 1.80g/90.00g/mol
=0.02 mol acid present in the sample

What is the mass of the oxalic acid in the sample?

--> the mass of oxalic acid in the sample is 1.8 grams (? Is this not equal to the mass of the sample)

What is the original sample purity?

0.8 grams of base/1.8g of acid )*100
=44.4% purity

Is this correct?

[mjc123]

if the sample isn't pure, then this following procedure may not be correct:

Since you are asked to calculate the purity, you must assume it isn't pure, so this is not correct.
You have just calculated the amount of oxalic acid from the titration. Assuming you used the whole sample (the wording of the question implies this), this is the number of moles of oxalic acid in the sample. So what is the mass of oxalic acid in the sample?

0.8 grams of base/1.8g of acid )*100
=44.4% purity

What are you doing here? What is the definition of the purity of the sample?

[RespectfulStudent]

The amount of the selected atom in the sample expressed  as a percentage is its percent purity. The amount of the selected atom in the sample is the purity. The selected atom is H2A.

I learned how to solve this at school today with a table that I filled in. I created an image showing the steps that I took:
Problem: Titration of 1.80g of the diprotic acid (with a molar mass of 200.00g/mol) required 24.3mL of 0.110 M NaOH. Find % purity of the acid sample.


https://imgur.com/a/doZHIEo
are these calculations correct?

-also, I learned that no water has to be added to the acid. I place the acid into the erlenmeyer flask, then a few drops of phenolphthalein, before adding the base.

[Borek]

The amount of the selected atom in the sample expressed  as a percentage is its percent purity. The amount of the selected atom in the sample is the purity. The selected atom is H2A.

Not an atom, substance.

If it is about amount of a chosen substance in the sample, why do you calculate purity of H₂A by dividing amount of base by the sample mass? Base is not your chosen substance.

I learned how to solve this at school today with a table that I filled in. I created an image showing the steps that I took:
Problem: Titration of 1.80g of the diprotic acid (with a molar mass of 200.00g/mol) required 24.3mL of 0.110 M NaOH. Find % purity of the acid sample.

I am afraid you have not learned anything, you are making the same mistakes again and again, and introducing new ones.

You were told several times that to find the actual amount of the acid in the sample you have to not use mass of the sample, but result of the titration.

-also, I learned that no water has to be added to the acid. I place the acid into the erlenmeyer flask, then a few drops of phenolphthalein, before adding the base.

Which probably means you are not expected to dilute it, you still need to use water to dissolve the acid prior to the titration.

[Borek]

formula C2H2O4+2NaOH--> Na2C2O4 +2H2O

How many moles of NaOH were used?

nNaOH=cV
nNaOH=(1.000M)(20.00mL/1000)
=0.02 mol NaOH were used

OK so far.

(below calculations are for another question, the purity)

No, it is all the time the same question.

molar mass of NaOH =40 g/mol
n=m/M
m=nM
m base= (0.02 mol)(40g/mol [the molar mass of NaOH])
=0.8 grams of base

Correct, but irrelevant, you don't need mass of NaOH for anything.

1 mol acid/2mol base = x mol acid /0.02 mol base
^theoretical value              ^actual/experimental value
x mol acid = 0.02 mol NaOH (1mol acid/2 mol base)
x mol acid = 0.01 mol
 0.02 mol of NaOH were used and 0.01 mol of acid were used

Yes, that's from the stoichiometry. Titration used 0.02 moles of NaOH to neutralize 0.01 moles of H₂A.

How many moles of the oxalic acid were present in the sample?

n=m/M
if the sample isn't pure, then this following procedure may not be correct:
n acid = 1.80g/90.00g/mol
=0.02 mol acid present in the sample

And here is where you are again going wrong. 1.8 g is not mass of the acid, it is mass of the sample. BUT: you have calculated actual number of moles of the acid moments ago, from the titration result. You already know there were 0.01 moles of the acid present.

Repeat the rest of the calculations using this DETERMINED amount of acid in the sample.