What is the percentage of ASA in one tablet?

[pbonnie]

Homework Statement

Calculate the number of moles and the mass of ASA in one tablet. Determine the percentage of the original mass of the tablet that was actually ASA.

(this question was based off an experiment previously in the lesson)

Homework Equations

c = n/v
n = m/M

The Attempt at a Solution

Given:
$HC_9H_7O_4 + NaOH --> C_9H-7O-4 + H_2O$
I'm using the subscript b to represent the base, and subscript a to represent the acid.
$c_b = 0.100 mol/L$
$V_b = 18.0 mL = 0.018 L$
$n_b = c_bV_b = 0.100mol/L x 0.018L = 1.8x10^{-3} mol$

The mol ratio is 1:1, so the number of moles of the acid is also $1.8x10^{-3}$
$n =m/M$
$M_a = 180.17 g/mol$
$m = (1.8x10^{-3})(180.17 g/mol)$
$= 0.324 g$
$% purity = 0.324g/0.500g x 100% = 64.8%$
(The original tablet was 0.500 g)

I was hoping someone could let me know if I'm doing this correctly.
Thank you!

 

[pbonnie]

* % purity = 0.324g/0.500g = 64.8%

 

[Borek]

Looks OK to me.