Chemistry practical 6:
Aspirin analysis
Aim
The aim of this experiment was to research the 2-ethanoylhydroxybenzoic acid CH3COOC6H4COOH (acetyl-salicylic acid, the main constituent of aspirin), to further learn molarity calculations and to become familiar with excess-acid calculations.
Materials&Apparatus
-weighing apparatus
-three 500 mg tablets of aspirin
-1.1 M sodium hydroxide (NaOH)
-0.1 M hydrochloric acid
-a burette and a stand
-a 20 ml pipette
-several beakers
-indicator(phenolphtalein)
-200 ml conical flask
-distilled water
-bunsen burner and a wire stand
-a small funnel
-eye protection
Method
Part 1 The sodium hydroxide needed to be standardized first, so exactly 20 ml of the approximately 1.1 M sodium hydroxide was pipetted into a 200 ml standard flask and filled to the mark with distilled water, making the mixture be roughly 0.11 molar NaOH.
Part 220 ml of this solution was titrated against 0.10 M hydrochloric acid using phenolphtalein to find the accurate molarity.
Part 3 The three aspirin tablets were weighed, and the tablets hydrolysed as follows: 20 ml of the roughly 1.1 molar sodium hydroxide was pipetted on to the tablets in a conical flask, followed with the same amount of distilled water. The mixture was then gently simmered for ten minutes with the bunsen burner to hydrolyse the acetyl-salicylic acid. After the mixture had cooled, it was poured (with washings) to a 200 ml flask and filled up to the mark with distilled water.
Part 4 Finally, 20 ml of the hydrolysed solution was pipetted into a beaker and titrated against 0.10 M hydrochloric acid using phenolphtalein as an indicator.
Data
1. Weight of aspirin tablets: 1.67 g
2. The average amount of hydrochloric acid needed against the non-hydrolysed
solution (Step 2): 21.5 ml (21.3, 21.8, 21.4)
3. The average amount of hydrochloric acid needed against the hydrolysed solution:
6.6 ml (Step 4) (6.5, 6.7, 6.8, 6.4)
Calculations
formula for molarity calculations: c1v1 = c2v2
In Part 2:
reaction formula: NaOH + HCl NaCl + H2O
From the titration we know that 21.5 ml of hydrochloric acid was needed to neutralize 20 ml of the solution. Therefore the molarity of the NaOH in the 20 ml of solution is
0.0215 l * 0.1 M * l-1 / 0.020 l = 0.1075 M.20 ml of the solution is 1/10th of the whole 200 ml of solution, so the molarity of the whole solution is 0.1075 M *10 =1.075 M.
In Part 4:
reaction formula: NaOH + HCl NaCl + H2O
In the final titration we needed an average of 0.0066 l hydrochloric acid to neutralize the 20 ml sample of the solution. The amount of NaOH left is then 0.0066 l of HCl * 0.1 M/l 0.00066 M. If 20 ml of the solution contains 0.00066 moles of NaOH, 200 ml will contain 0.00066 * 10 = 0.0066 moles of NaOH.
In Part 3:
reaction formula: CH3COOC6H4COOH + 2 NaOH
CH3COONa + HOC6H4COONa + H2O
Originally, we added 0.020 l * 1.075 M/l = 0.0215 moles of NaOH. In Part 4 we had 0.0066 moles of NaOH left, which means that 0.0215 moles - 0.0066moles in excess = 0.0149 moles of NaOH are used in the hydrolysation of the acetyl-salicylic acid. We can see from the formula that we need two molecules of NaOH per molecule of acetyl-salicylic acid, so there were 0.0149 M / 2 = 0.00745 moles of acetyl-salicylic acid. From the formula we can count that the mass of one molecule of acetyl-salicylic acid is 9*12.011 u (C) + 4*15.9994 u (O) + 8*1.00794 u (H) = 180.16012 u. We have 0.00745 moles = 4.48649... * 1021 molecules, which weigh 180.16012 u *
1.66 * 10 -27 kg/u each and together add up to a total of 0.00134219... kg = 1.34219... grams. This means that the tablets contain 1.34219... / 1.67 = 80.37...% ~ 80% of acetyl-salicylic acid in the tablets.
Observations
The reason why excess-acid calculations are used is that in some cases, such as this, the substances are such that it is hard to be certain of uniform concentration and reactions. The hydrolysis of the acetyl-salicylic acid to the products sodium ethanoate CH3COONa and sodium-hydroxybenzoate HOC6H4COONa does not happen readily. It may take a long time until all acid has reacted, so heating and excess sodium hydroxide are used. Boiling must be avoided, however, as evaporation is not wanted since we need the amounts of substances be as close as possible to the original amounts. Also it might occur that some undesirable reactions would take place, such as the COO--groups reacting with either NaOH or water.
Error Analysis
The low percentage can be explained by the fact that the tablets used were old, and hvae probably absorbed moisture within the course of time. New tablets, mass 500 mg each, would have at least 89.48% acid, and probably more, since it is likely that the acid has been reacting. If the added weight (0.17 grams) is solely due to atmospheric moisture, according to the reaction CH3COOC6H4COOH + 2 OH-CH3COO- + HOC6H4COO- + H2O each molecule of acid broken uses up two hydroxide ions and produces one molecule of water. The excess mass was 0.17 grams and the molecular mass of OH- is 17.00734. Per one mole of broken-down acetyl-salicylic acid we have
absorbed two hydroxide ions and produced one mole of water, one of ethanoate ions and one of 2-hydroxybenzoate ions. As their total mass is counted within the total mass of the tablets and away from the amount of acetylsalicylic acid, we have to take into account their total mass, 9 * 12.011 + 8 * 15.9994 + 12 * 1.00794 = 216.19068 g/mol. When we count that with the extra mass, 0.17 g, we find out that 0.17 g /216.19068 g / mol ~ 0.00078634 moles. That amount weighs 0.1416676... grams, which brings the total amount of acetylsalicylic acid up to 1.48386052 grams, or 98.9%.