A tablet sample of mass $0.50 g$ containing aspirin (acetylsalicylic acid, $C X_{9} H X_{8} O X_{4}$ , molar mass $180$ ) is boiled with $30.0 mL$ of $0.200 M$ $NaOH$ so that each aspirin molecule consumes 2 $OH X^{-}$ . The excess base needs $10.0 mL$ of $0.200 M$ $HCl$ . Find the percentage of aspirin in the sample to the nearest integer.

Numerical answer type, answer in %, tolerance 0.6.

Answer: 72 %

NaOH taken

= 30 \times 0.200 = 6.00

mmol; excess

= 10 \times 0.200 = 2.00

mmol; consumed

= 4.00

mmol. Aspirin

= 4.00/2 = 2.000

mmol

= 360.0

mg. Percentage

= 360.0/500 \times 100 = 72.0%