Experiment: The Preparation of Benzoic Acid
A student carried out the preparation of benzoic acid by oxidizing 1cm3
of phenylmethanol (benzyl alcohol, C6H5CH2OH).
The oxidation was carried out in alkaline conditions using sodium
carbonate (Na2CO3) and an excess potassium manganate (VII) solution
in a conical flask.
After heating the solution for 20 minutes, it was cooled by running cool
water over the outside of the flask. It was then acidified with
concentrated hydrochloric acid and a brown precipitate formed.
Sodium sulfite was then added until the precipitate disappeared and
white crystals of benzoic acid became visible. The flask was placed in ice
and the crystals were filtered off by vacuum filtration. The crystals were
dried and weighed. It was found that 0.58g of benzoic acid was
obtained.
The balanced equation for the oxidation reaction is:
3C6H5CH2OH + 4KMnO4
3C6H5COOH + 4MnO2 + H2O + 4KOH
(a) Why was an excess solution of potassium manganate (VII) used? (6)
(b) Identify the brown precipitate formed in the reaction. (6)
(c) Give two reasons why the solution was acidified with concentrated
hydrochloric acid. (8)
(d) How would you test the solution to find out if enough acid had been
added? (3)
(e) What reaction occurred when the sodium sulfite was added? (6)
(f) Why was the flask placed in ice before filtration? (3)
(g) How could the crystals have been dried? (3)
(h) The phenylmethanol was the limiting factor in this reaction. Using
the fact that the student used 1 cm3 of phenylmethanol (density =
1.04g/cm3 ) calculate:
(i) the theoretical yield (12)
(ii) the percentage yield if 0.58g of benzoic acid was obtained (3)
Worked Answers
(a) To ensure there was a sufficient amount of KMnO4 to oxidise all of
the benzyl alcohol to benzoic acid.
(b) A brown precipitate of MnO4 is formed as the Mn7+ ion is reduced
to Mn4+ .
(c) 1. To convert sodium benzoate to benzoic acid.
2. To neutralise any excess sodium carbonate.
(d) Dip a glass rod into the conical flask and touch the end of the glass
rod to a piece of damp blue litmus paper. The solution should be acidic
(blue litmus paper turns red).
(e) A reduction reaction (Mn4+ reduced to Mn2+).
(f) To bring the benzoic acid crystals out of the solution. Benzoic acid
has low solubility in cold water.
(g) Place filter paper with crystals on clock glass and put them in an
oven at a low temperature.
(h)
(i) mass = density x volume
(1.04 g/cm3 ) (1 cm3 ) = 1.04g phenylmethanol.
molarity = mass / molar mass.
molar mass phenylmethanol = 108 g/mol.
1.04g / 108 g/mol = 0.00962 moles phenylmethanol.
Look at the ratio in the balanced equation. There are three moles of
phenylmethanol to every three moles of benzoic acid.
So the theoretical yield of benzoic acid is 0.00962 moles. Let’s get that
in grams.
mass = molarity x molar mass
molar mass benzoic acid = 122 g/mol.
(0.00962) (122) = 1.17g benzoic acid (theoretical yield).
(ii) percentage yield = ( actual yield / theoretical yield ) (100/1)
(0.58g / 1.17g) (100 / 1) = 49.57% yield.