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Lelu [443]
3 years ago
5

Aspirin (C9H8O4) is an acid which can be titrated with a base to determine purity. If an aspirin tablet weighing 0.615 g is titr

ated with standardized 0.1121 M KOH, the endpoint is reached after 20.52 mL of KOH have been added. What is the percentage of aspirin in the tablet
Chemistry
1 answer:
Iteru [2.4K]3 years ago
6 0

Answer:

67.4 % of C₉H₈O₄

Explanation:

To make titrations problems we know, that in the endpoint:

mmoles of acid = mmoles of base

mmoles = M . volume so:

mmoles of acid = 20.52 mL . 0.1121 M

mmoles of acid = mg of acid / PM (mg /mmoles)

Let's determine the PM of aspirin:

12.017 g/m . 9 + 1.00078 g/m . 8 + 15.9994 g/m . 4 = 180.1568 mg/mmol

mass (mg) = (20.52 mL . 0.1121 M) . 180.1568 mg/mmol

mass (mg) = 414.4 mg

We convert the mass to g → 414.4 mg . 1g / 1000mg = 0.4144 g

We determine the % → (0.4144 g / 0.615 g) . 100 = 67.4 %

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Explanation:

Hello,

In this case, for the given chemical reaction, we first identify the limiting reactant by noticing that due to the 1:1 mole ratio for magnesium to iodine the reacting moles must the same, nevertheless, there are only 2.68 moles of magnesium versus 3.56 moles of iodine, for that reason, magnesium is the limiting reactant, so the theoretical turns out:

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