Calculating Benzoic Acid Moles & Percent Purity

[feelau]

chem question please help

Homework Statement

Ok, so there's .9734 g sample which contained monoprotic benzoic acid C6H5COOH and it's dissolved in 75 mL of water and titrated to the equivalence point with 31.78 mL of .09251 M NaOH solution. a) how many moles of benzoic acid were titrated b) Was this a pure sample of benzoic acid? c) If the sample is no, Calculate its percent purity.

The attempt at a solution
ok so for the first problem part a is pretty easy but then I have no idea how to determine if this sample is pure and how to calculate percent of purity.

 

[chaoseverlasting]

In this case, for both the acid and base, normality equals molarity. Hence you can use n1v1=n2v2.

You know for sure the 31.78ml of .09251M soln were used. You are unsure of the weight of benzoic acid. Hence, N*75=31.78*.09251. From this you can calculate N. This much in any case you've figured out... right?

Now, molarity=normality, therefore you know the number of moles. You have the formula for benzoic acid so you know the weight of benzoic acid. You know the number of moles, so you can find the weight of the sample of benzoic acid titrated. If that weight equals .9743 grams, then the sample is 100% pure, else (weight/.9734)*100 is your percentage.

If you don't know the weight of 1 mol of carbon/oxygen/hydrogen:

C=12gms
0=16gms
H=1gm

 

[feelau]

Thank you so much. I've never seen any problems with this question so I had no clue. Thanks!