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
The structure of trimesic acid is:
HOOC-C6H3(COOH)3
The structure of isopropanol is:
CH3-CH(OH)-CH3
The ester that results from the complete Fisher esterification of trimesic acid in an excess of isopropanol is called isopropyl trimesate.
The structure of isopropyl trimesate is:
CH3-CH(OOC-C6H3((CH3)2)3)-CH3
The CAS number for isopropyl trimesate is 5343-92-0.
To determine the volume of isopropanol required to prepare 10.0 g of the ester, we need to determine the molar mass of isopropyl trimesate. Isopropyl trimesate has a molar mass of 412.48 g/mol.
Using the equation: moles = mass / molar mass, we can calculate the moles of isopropyl trimesate:
moles = 10.0 g / 412.48 g/mol = 0.0242 mol
Since it is stated that ten molar equivalents of isopropanol are required for each reactant carboxylic acid equivalent, we need ten times the moles of isopropyl trimesate in isopropanol. Therefore:
moles of isopropanol = 0.0242 mol * 10 = 0.242 mol
Now, we need to convert the moles of isopropanol to volume using the molar volume of isopropanol. The molar volume of isopropanol is 58.09 mL/mol.
volume of isopropanol = 0.242 mol * 58.09 mL/mol = 14.07 mL
Therefore, approximately 14.07 mL of isopropanol is required to prepare 10.0 g of the ester of trimesic acid described.