Making Ethyl Ethanoate
Ethyl Ethanoate is a colourless liquid which has a characteristic of a sweet smell (similar to pear drops) and is used in glues, nail polish removers, decaffeinating tea and coffee. It is manufactured on a large scale and used as a solvent. The chemical manufacturing company that you work for produces ethyl ethanoate and the director of product development has asked you to investigate how the company can produce it more effectively.
Preparing an organic compound – preparation of Ethyl Ethanoate.
In order to determine a more effective way to produce ethyl ethanoate I produced ethyl ethanoate using a method combining two techniques, duct development has asked you to investigate how the company can produce it morrefluxing and distilling.
The equation for this method is;
Carboxylic acid + alcohol.
Ethanoic acid + ethanol Ethyl ethanoate + water.
Firstly, I mixed 50cm3 of ethanol with 50cm3 of glacial ethanoic acid in a 250cm3 round bottom flask then began slowly adding 10cm3 of concentrated sulphuric acid with cooling. I ensured the mixture of the 3 chemicals was homogeneous (well mixed) before fitting the flask with ta reflux condenser and allowing it to boil gently for 10 minutes. At this point the reflux condenser was in a vertical orientation so I rearranged the condenser for distillation and distilled around two thirds of the mixture off.
I transferred the distillate to a separating funnel and added 25cm3 of 30% sodium carbonate solution. I placed the stopped in the funnel, inverted it and shook it while ensuring I regularly opened the tap. I allowed the two layers formed to separate and carefully discarded the lower layer. I prepared a solution of 25g of calcium chloride in 25cm3 of water then added this to the crude ethyl ethanoate in the funnel and shook vigorously. I allowed the mixture to settle and ran off the lower aqueous layer.
I ran the ethyl ethanoate into a small conical flask and added a few lumps of granular anhydrous calcium chloride and shook.
I next decanted the liquid into a dry, clean 100cm3 round bottom flask, added anti bumping granules and arranged the equipment for distillation again. The flask was in a cold water bath that was gradually heated so the fraction that is created at 35-40⁰C. I then collected the fraction that boiled between 74 and 79⁰C.
Percentage Yield Using this method provided me with a fairly low percentage yield of CH3COOC2H5 produced. I calculated the percentage yield using the formula: Figure 1 Percentage yield = actual number of moles/expected number of moles * 100.
In order to calculate the overall percentage yield and firstly had to calculate the actual number of moles of ethyl ethanoate produced using the formula:
Actual number of moles = m/Mr
(where m = mass of product and Mr = mass of one mole).
Substituting the values 3.77674,as the mass of ethyl ethanoate (density*volume produced = mass), and 88, as the mass of one mole, into the equation gave me the number of moles of ethyl ethanoate produced being 0.0428 mol.
To calculate the number of expected moles I looked at the equation of the reaction CH3COOH + C2H5OH CH3COOC2H5 + H20 which shows one mole of C2H5OH is produced for every one mole of CH3COOC2H5. So to calculate the number of moles of ethanol used I used the same formula as previously used:
Actual number of moles = m/Mr.
The number of moles of C2H5OH used was 0.428 (the same as the number of moles of ethyl ethanoate being produced) and the expected number of moles of CH3COOC2H5 produced is 0.853.
Using the figures gained from these equations leads me to calculating the percentage yield of CH3COOC2H5 produced as 5.02% which overall is a fairly low percentage. A higher percentage yield would make the experiment more economical meaning it is more effective overall.
As this reaction is reversible the percentage yield will be low as equilibrium will be reached meaning the reaction will never be fully complete.