Answer to Question #87729 in General Chemistry for Sophie

Question

Find the percentage of aspirin in tablet calculate by titration using the formula mass of aspirin calculated by titration divided my mass of weighed aspirin tablet multiplied by 100
The mass of the tablet is 0.132g

Answer 1

To calculate the percentage of aspirin in a tablet one needs to perform some steps:

1.To dissolve one tablet (0.132 g) in water and add a known amount of standard sodium hydroxide solution in excess.

2.To perform back-titration of soldium hydroxide to a phenolphthalein end-point with definite volume of hydrochloric acid solution.

As there is no some data in the task (volume of hydrochloric acid solution, it's molarity, volume of sodium hydroxide solution and it's molarity) we can take some data to show example calculations.

For example, let's take 0.1 M sodium hydroxide, 50.00 mL, the volume of 0.10375 M hydrochloric acid solution used for back-titration to be 34.30 mL.

1.

$$CH_3COOC_6H_4COOH + 2NaOH \rightarrow CH_3COONa + HOC_6H_4COONa + H_2O$$

2. Number of moles of NaOH used is

$$n_{total}(NaOH) = 0.050 L \times 0.1 \frac{mol}{L} = 0.005 mol$$

3. Number of moles of HCl used for back-titration

$$n(HCl) = 0.0343 L \times 0.10375 \frac{mol}{L} = 0.00356 mol$$

4. According to equation of back-titration

$$NaOH + HCl = NaCl + H_2O$$

$n_{excess}(NaOH) = n(HCl) = 0.00356 mol$ as mole ratio $n(NaOH):n(HCl) = 1:1$

5.

$$n_{total}(NaOH) = n_{reacted}(NaOH) + n_{excess}(NaOH)$$

$$\therefore n_{reacted}(NaOH) = n_{total}(NaOH) - n_{excess}(NaOH) = 0.00500 mol - 0.00356 mol = 0.00144 mol$$

6. According to equation

$$CH_3COOC_6H_4COOH + 2NaOH \rightarrow CH_3COONa + HOC_6H_4COONa + H_2O$$

mole ratio $n(CH_3COOC_6H_4COOH):n(NaOH) = 1:2$ , then

$$n(CH_3COOC_6H_4COOH) = \frac{n(NaOH)}{2} = \frac{0.00144 mol}{2} =0.000721 mol$$

7.

$$m(CH_3COOC_6H_4COOH) = 0.000721 mol \times 180.16 \frac{g}{mol} = 0.130 g$$

8. Percent by mass:

$$\frac{0.130 g}{0.132g} \times 100\% = 98.48\%$$