Preparation of Ethyl Ethanoate
Introduction
In this report I will be discussing how to prepare and test the purity of ethyl ethanoate. I will
also be discussing my results and evaluating my method to give reasons as to why I have
chosen to do certain tasks in certain ways. I will be working out the yield and present ways
on how to improve it. I will be discussing the boiling point test I carried out and share the
results. I will also be showing the data that I obtained from the IR spectroscopy machine.
What is Ethyl Ethanoate and What is it Used For?
Ethyl ethanoate is a commonly used ester which is a colourless and flammable liquid
substance. Esters are usually formed by reacting a carboxylic acid and alcohol in a
reversible reaction to produce the ester and water. In this case, a reaction between ethanoic
acid and ethanol can be represented in the chemical equation:
CH3COOH(l) + C2H5OH(l) ⇔ CH3COOC2H5(l) + H2O(l)
Ethyl ethanoate is usually used in perfumes, flowers, various food flavourings and
pharmaceutical products. However, its most well known attribute is that of manufacturing nail
polish remover.
Equipment:
● Ethanol
● Ethanoic Acid
● Phosphoric Acid
● Clamp and Stand
● Round Bottomed Flask
● Anhydrous kjejCalcium Chloride
● Liebig Condenser
● Heating Mantle
● Stopper
● Conical Flask
Method:
1) Mix 50cm3 of ethanol and 50cm3 of glacial ethanoic acid thoroughly in a
250cm3 round-bottomed flask. Make sure this is done in a fume cupboard.
2) Add slowly with cooling and shaking 10cm3 of concentrated phosphoric
acid.
Separation of the product
3) Rearrange the position of the condenser for distillation, and distil off about
two-thirds of the mixture.
, 4) Transfer the distillate to a separating funnel and add about 25 cm3 of 30%
sodium carbonate solution. Stopper the funnel, invert it, and shake, opening
the tap from time to time.
5) Allow the two layers to separate and carefully run off and reject the lower
layer, ensuring that the sodium carbonate is removed as completely as
possible.
Purification of the Product
6) Add some saturated calcium chloride solution to the ethyl ethanoate, shake
well, and remove the lower aqueous layer.
7) Pour the ethyl ethanoate into a beaker, add a few lumps of solid anhydrous
calcium chloride and shake well. Decant the ethyl ethanoate into a flask.
8) Then decant the liquid into a clean, dry 100 cm3 round-bottom flask, add some
anti-bumping granules, and arrange for distillation including a 0-100oC
thermometer in the apparatus. The distilling flask should be placed in a cold
water bath, which is gradually heated. The ether that is always formed in this
reaction will distil off at 35-40oC, and may be discarded. Continue to heat, and
collect the fraction that boils between 74o and 79oC.
Results - Percentage Yield:
Percentage yield = actual yield / theoretical yield
(34.26g / 0.856 mol) x 100
Calculation of theoretical yield:
Mol = mass/Mr
Mol = conc x vol
Convert 50cm3 of ethanol > mass (g)
D=m/v
Mass = density x vol
Mass(ethanol) = 0.789 x 50 = 39.45g
Mol = mass / Mr
Mol of ethanol = 39..07 = 0.856 mol
So, we have 0.856 moles of ethanol
Mol = mass / Mr
Mass = mol x Mr
Mass = 0.856 x 88.11
Mass = 75.42g
% yield = (34..42) x 100
Introduction
In this report I will be discussing how to prepare and test the purity of ethyl ethanoate. I will
also be discussing my results and evaluating my method to give reasons as to why I have
chosen to do certain tasks in certain ways. I will be working out the yield and present ways
on how to improve it. I will be discussing the boiling point test I carried out and share the
results. I will also be showing the data that I obtained from the IR spectroscopy machine.
What is Ethyl Ethanoate and What is it Used For?
Ethyl ethanoate is a commonly used ester which is a colourless and flammable liquid
substance. Esters are usually formed by reacting a carboxylic acid and alcohol in a
reversible reaction to produce the ester and water. In this case, a reaction between ethanoic
acid and ethanol can be represented in the chemical equation:
CH3COOH(l) + C2H5OH(l) ⇔ CH3COOC2H5(l) + H2O(l)
Ethyl ethanoate is usually used in perfumes, flowers, various food flavourings and
pharmaceutical products. However, its most well known attribute is that of manufacturing nail
polish remover.
Equipment:
● Ethanol
● Ethanoic Acid
● Phosphoric Acid
● Clamp and Stand
● Round Bottomed Flask
● Anhydrous kjejCalcium Chloride
● Liebig Condenser
● Heating Mantle
● Stopper
● Conical Flask
Method:
1) Mix 50cm3 of ethanol and 50cm3 of glacial ethanoic acid thoroughly in a
250cm3 round-bottomed flask. Make sure this is done in a fume cupboard.
2) Add slowly with cooling and shaking 10cm3 of concentrated phosphoric
acid.
Separation of the product
3) Rearrange the position of the condenser for distillation, and distil off about
two-thirds of the mixture.
, 4) Transfer the distillate to a separating funnel and add about 25 cm3 of 30%
sodium carbonate solution. Stopper the funnel, invert it, and shake, opening
the tap from time to time.
5) Allow the two layers to separate and carefully run off and reject the lower
layer, ensuring that the sodium carbonate is removed as completely as
possible.
Purification of the Product
6) Add some saturated calcium chloride solution to the ethyl ethanoate, shake
well, and remove the lower aqueous layer.
7) Pour the ethyl ethanoate into a beaker, add a few lumps of solid anhydrous
calcium chloride and shake well. Decant the ethyl ethanoate into a flask.
8) Then decant the liquid into a clean, dry 100 cm3 round-bottom flask, add some
anti-bumping granules, and arrange for distillation including a 0-100oC
thermometer in the apparatus. The distilling flask should be placed in a cold
water bath, which is gradually heated. The ether that is always formed in this
reaction will distil off at 35-40oC, and may be discarded. Continue to heat, and
collect the fraction that boils between 74o and 79oC.
Results - Percentage Yield:
Percentage yield = actual yield / theoretical yield
(34.26g / 0.856 mol) x 100
Calculation of theoretical yield:
Mol = mass/Mr
Mol = conc x vol
Convert 50cm3 of ethanol > mass (g)
D=m/v
Mass = density x vol
Mass(ethanol) = 0.789 x 50 = 39.45g
Mol = mass / Mr
Mol of ethanol = 39..07 = 0.856 mol
So, we have 0.856 moles of ethanol
Mol = mass / Mr
Mass = mol x Mr
Mass = 0.856 x 88.11
Mass = 75.42g
% yield = (34..42) x 100
