What are the structures of trimesic acid and isopropanol, and what is the structure, name
and CAS number of the ester that results from the complete Fisher esterification of trimesic
acid in an excess of isopropanol? If ten molar equivalents of isopropanol are required for
each reactant carboxylic acid equivalent to achieve an 80% reaction yield, what volume of
isopropanol is required to prepare 10.0 g of the ester of trimesic acid described above?
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The structure of trimesic acid is:
O=C(O)C1=CC=C(C=C1)C(=O)O
The structure of isopropanol is:
CC(C)O
The structure of the ester formed from the complete Fisher esterification of trimesic acid in an excess of isopropanol is:
O=C(OCC(C)O)C1=CC=C(C=C1)C(=O)O
The name of the ester is isopropyl trimesate.
The CAS number of the ester is 2943-75-1.
To calculate the volume of isopropanol required to prepare 10.0 g of the ester, we need to use the molar ratio and the molecular weights of the reactants.
Molecular weight of trimesic acid = 270.22 g/mol
Molecular weight of isopropanol = 60.10 g/mol
Molecular weight of isopropyl trimesate = 348.40 g/mol
First, we need to calculate the number of moles of isopropyl trimesate:
moles of isopropyl trimesate = mass / molecular weight
moles of isopropyl trimesate = 10.0 g / 348.40 g/mol
moles of isopropyl trimesate = 0.0287 mol
Since we know that 10 molar equivalents of isopropanol are required for each reactant carboxylic acid equivalent, we need to determine the number of moles of isopropanol required.
moles of isopropanol = 10 x 0.0287 mol
moles of isopropanol = 0.287 mol
Finally, we can calculate the volume of isopropanol using the molar volume:
volume of isopropanol = moles of isopropanol x molar volume
volume of isopropanol = 0.287 mol x 60.10 g/mol
volume of isopropanol = 17.27 g
Therefore, approximately 17.27 ml of isopropanol is required to prepare 10.0 g of the ester of trimesic acid.