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

73.10% the percent purity of aspirin.

Step-by-step 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 :

`(1)/(1)* 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:

`\%=(0.132 g)/(0.18056 g)* 100=73.10\%`

73.10% the percent purity of aspirin.