Question

A student found that the titration had taken 10.00 ml of 0.1002 m naoh to titration 0.132 g of aspirin, a monoprotic acid. calculate the percent purity of aspirin (c9h8o4, molar mass = 180.2 g/mol) sample.

Answer 1

The balanced equation for the reaction between NaOH and aspirin is as follows;
NaOH + C₉H₈O₄ --> C₉H₇O₄Na + H₂O
stoichiometry of NaOH to C₉H₈O₄ is 1:1
The number of NaOH moles reacted - 0.1002 M / 1000 mL/L x 10.00 mL 
Number of NaOH  moles - 0.001002 mol
Therefore number of moles of aspirin - 0.001002 mol
Mass of aspirin reacted - 0.001002 mol x 180.2 g/mol = 0.18 g
However the mass of the aspirin sample is 0.132 g but 0.18 g of aspirin has reacted, therefore this question is not correct.

Answer 2

Answer:

73.10% the percent purity of aspirin.

Explanation:

$NaOH+HAsp\rightarrow NaAsp+H_2O$

Moles of NaOH:

$0.1002 M=\frac{\text{moles of NaOH}}{0.010 L}$

Moles of NaOH = 0.001002 moles

According to reaction. 1 NaOH reacts with 1 mole of aspirin .

Then 0.001002 moles NaOH will reacts with :

$\frac{1}{1}\times 0.001002 moles=0.001002 moles$

Mass of 0.001002 moles of aspirin:

= 0.001002 moles × 180.2 g/mol = 0.18056 g

Percentage of purity of aspirin:

$\%=\frac{0.132 g}{0.18056 g}\times 100=73.10\%$

73.10% the percent purity of aspirin.