Homework Help: Percentage of ASA in tablet

1. Jun 21, 2013

pbonnie

1. The problem statement, all variables and given/known data
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)

2. Relevant equations
c = n/v
n = m/M

3. 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!

2. Jun 21, 2013

pbonnie

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

3. Jun 22, 2013

Staff: Mentor

Looks OK to me.