Three students are asked to prepare samples of benzoic acid, C6H5COOH, from the alkaline hydrolysis of methyl benzoate, C6H5COOCH3. Sodium benzoate, C6H5COONa, is the salt formed, which is then acidified with hydrochloric acid to form C6H5COOH. C6H5COOCH3(l) + NaOH(aq) → C6H5COONa(aq) + CH3OH(l) C6H5COONa(aq) + HCl(aq) → C6H5COOH(s) + NaCl(aq) The students use the following method. step 1 Put 1.242 g of liquid methyl benzoate into a 50cm³ round-bottomed flask. step 2 Prepare 100.0cm³ of 1.00 moldm⁻³ aqueous sodium hydroxide, NaOH(aq). step 3 Add 10cm³, an excess, of the prepared NaOH(aq) to the round-bottomed flask. step 4 Add some anti-bumping granules to the round-bottomed flask. step 5 Fit a condenser to make reflux apparatus. Reflux the reaction mixture for 20 minutes. step 6 Allow the reaction mixture to cool and carefully pour the liquid into a beaker. step 7 Acidify the liquid with dilute hydrochloric acid. step 8 Filter the mixture formed in step 7. step 9 Purify the benzoic acid by recrystallisation from hot water. Filter, dry and record the mass of pure benzoic acid obtained.
- (a)(i)Calculate the volume of methyl benzoate used in step 1. Give your answer to the nearest 0.05 cm³. [density: methyl benzoate, 1.08gcm⁻³] volume of methyl benzoate =[1]
- (a)(ii)Identify a suitable piece of apparatus to measure the volume of methyl benzoate required in step 1.[1]
- (a)(iii)Calculate the mass of NaOH(s) that is needed to prepare the solution in step 2. mass of NaOH(s) =[1]
- (b)Student 1 added the mass of NaOH(s) calculated in (a)(iii) into a beaker. Describe the steps the student should take to make 100.0cm³ of 1.00moldm⁻³ NaOH(aq). Give the name and capacity of any apparatus that should be used. Write your answer using a series of numbered steps.[3]
- (c)Student 2 prepared 0.100 moldm⁻³ NaOH(aq) instead of 1.00 moldm⁻³ NaOH(aq) in step 2. State how this would affect the final mass of benzoic acid formed. Explain, using calculations, how you came to this conclusion. [Mᵣ: methyl benzoate, 136; benzoic acid, 122] effect on mass explanation[2]
- (d)(i)Explain why it is necessary to reflux the mixture in step 5.[1]
- (d)(ii)Explain why a naked flame is not used in step 5.[1]
- (d)(iii)Explain the purpose of transferring the liquid in step 6.[1]
- (e)(i)Explain what the students should do to confirm that the mixture has been acidified in step 7.[1]
- (e)(ii)Describe what you would expect to observe as the sodium benzoate mixture is acidified in step 7.[1]
- (f)Suggest why it is necessary to cool the mixture before filtering in step 9.[1]
- (g)Pure benzoic acid has a melting point of 122 °C. The product made by student 1 has a melting point of 119°C. This student suggests the melting point of the product was lower than expected because it contained some water. Explain what the student should do to ensure that the product no longer contains water.[1]
- (h)(i)Calculate the maximum mass of benzoic acid that can be formed from 1.242 g of methyl benzoate. [Mᵣ: methyl benzoate, 136; benzoic acid, 122] maximum mass of benzoic acid =[1]
- (h)(ii)Student 3 produces 0.825 g of benzoic acid from 1.242 g of methyl benzoate. Calculate the percentage yield of benzoic acid produced by student 3. percentage yield =[1]
Mark scheme
Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.
- [1] 1.242 g / 1.08 = 1.15 cm³
- [1] burette
- [1] 1 × 100 / 1000 × 40 = 4.0(0) g
- [3] M1 dissolve the solid / sodium hydroxide (in the beaker) (using a small volume of) (distilled) water
- [3] M2 transfer / add to a (100 cm³ volumetric) flask AND rinse (with distilled water)
- [3] M3 top the 100 cm³ volumetric flask up to mark (with distilled water)
- [2] M1 mol C6H5COOCH3 = 1.242 / 136 = 0.00913 AND mol of NaOH = 0.100 × (10 / 1000) = 0.001(00)
- [2] M2 mass of benzoic acid decreases AND NaOH no longer in excess / NaOH (now) the limiting (reagent) / not all C6H5COOCH3 reacts / statement that mol NaOH < mol ester
- [1] allows heating to increase the rate of reaction without the loss of volatile substances
- [1] (substances) are flammable
- [1] to remove the anti-bumping granules (from the reaction mixture)
- [1] add (a little) (named) (acid-base) indicator AND the reaction mixture shows the colour for a (strong) acid in that indicator
- [1] precipitate forms
- [1] benzoic acid is less soluble in cold water
- [1] dry the precipitate and reweigh until constant mass
- [1] 1.242 x 122 / 136 = 1.11
- [1] 74.0(%)
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