Question 89ca6

Jun 18, 2015

Percent weight of aspirin in the drug: 42.9%.

Explanation:

So, you know that your drug has a total mass of 0.230 g and that it contains acetylsalicycli acid, $H {C}_{9} {H}_{7} {O}_{4}$.

In order to determine exactly how much acetylsalicycli acid your drug contains, you titrate it with sodium hydroxide, $K O H$.

The neutralization reaction that takes place between the two compounds will help you determine how many moles of acetylsalicyclic acid your drug contained.

The balanced chemical equation for this reaction looks like this

$H {C}_{9} {H}_{7} {O}_{4 \left(a q\right)} + K O {H}_{\left(a q\right)} \to K {C}_{9} {H}_{7} {O}_{4 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)}$

Notice that you have a $1 : 1$ mole ratio between acetylsalicyclic acid and potassium hydroxide. This means that, in order for a complete neutralization to take place, you need to mix equal numbers of moles of each compound.

Use the molarity and volume of the potassium hydroxide solution to determine how many moles of base were needed.

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{K O H} = \text{0.0819 M" * 6.689 * 10^(-3)"L" = "0.0005478 moles}$ $K O H$

This means that the drug contained 0.0005478 moles of acetylsalicyclic acid.

0.5478*10^(_3)cancel("moles"KOH) * ("1 mole "HC_9H_7O_4)/(1cancel("mole"KOH)) = 0.5478 * 10^(-3)"moles" $H {C}_{9} {H}_{7} {O}_{4}$

To determine the mass of acetylsalicyclic acid, use its molar mass

0.0005478cancel("moles") * "180.16 g"/(1cancel("mole")) = "0.09869 g"

The percent composition by weight of aspirin in the drug will be

(0.09869cancel("g"))/(0.230cancel("g")) * 100 = color(green)("42.9%")#

The answer is rounded to three sig figs, the number of sig figs you gave for the mass of the drug.