Esters
6 marks · NormalPropanoic acid reacts with ethanol to form an ester. You are given 1.5 cm³ of propanoic acid (density = 1.00 g/cm³) and an excess of ethanol.
Draw the structural formula of the ester formed in this reaction and name the ester.
In a laboratory experiment, you attempt to prepare the ester in part (a). (i) Calculate the maximum mass of the ester that could be formed from 1.5 cm³ of propanoic acid (M_r of propanoic acid = 74, M_r of the ester = 102).
Describe how you could separate and identify the smell of the ester after the reaction, referring to the laboratory method and your answer to part (b).
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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(a) Structural formula of ethyl propanoate; name: ethyl propanoate. · must show Correct structure showing COO functional group between ethyl and propanoate, No extra atoms; correct carbon and hydrogen counts
(b) Maximum mass of ethyl propanoate = 2.07 g. g · must show Calculation of moles of propanoic acid, Assume full conversion (limiting reactant), Correct Mr values used
(c) Add the reaction mixture to sodium carbonate solution to neutralise acid. The ester forms a layer floating on the solution. Waft the odour gently towards your nose to identify the smell. · must show Mention use of sodium carbonate (or water layer), Layer separation, Smell by wafting
Common mistakes: Incorrect joining of structural formulae; Incorrect calculation using wrong Mr; Confusing alcohol and acid names; Missing step in separation method
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Structural formula attempts correct ester linkage (–COO–) between correct groups.
A1 (a) · 1 mark · All carbon and hydrogen atoms correct for ethyl propanoate.
B1 (a) · 1 mark · Correct name: ethyl propanoate.
M2 (b) · 1 mark · Converts 1.5 cm³ of propanoic acid to mass, then to moles using Mr = 74.
A2 (b) · 1 mark · Correctly calculates maximum mass of ester using moles × Mr = 102.
C1 (c) · 1 mark · Describes correct post-reaction steps: add to sodium carbonate, observe floating layer, use of wafting to smell.
Worked solution
a) The ester formed from propanoic acid and ethanol is ethyl propanoate.
Structural formula:
CH3CH2COOCH2CH3
b) Calculate the mass of ester:
1.5 cm³ propanoic acid × 1.00 g/cm³ = 1.5 g
Moles of propanoic acid = 1.5 g / 74 = 0.0203 mol
Assuming complete reaction, moles of ester = 0.0203 mol
Mass of ester = 0.0203 mol × 102 = 2.07 g
c) To separate and identify the ester: After the reaction, add the mixture to sodium carbonate solution (to neutralise the acid), a layer of ester will form on top (since ester is insoluble in water), and carefully waft the odour towards your nose to detect the characteristic smell.