Question

Methyl benzoate can be synthesized by a Fisher Esterification reaction, and then this ester can be used as a starting material in the grignard synthesis of triphenyl carbinol. Suppose that in the synthesis of methyl benzoate, a student started with 13.45 grams of benzoic acid and 32.5 ml of methanol. After finishing the preparation of the ester, 10.65 grams of methylbenzoate was obtained, and the benzoic acid reacted completely. What is the percent yield of the ester? ( density methanol = 0.791 g/ml, methanol = 32.04 g/mol, methyl benzoate = 136.15 g/mol, benzoic acid = 122.12 g/mol )

Answer 1

Answer : The percent yield of ester is, 71.05 %

Solution : Given,

Mass of benzoic acid = 13.45 g

Volume of methanol = 32.5 mL

Molar mass of benzoic acid = 122.12 g/mole

Molar mass of methanol = 32.04 g/mole

Molar mass of methyl benzoate = 136.15 g/mole

First we have to calculate the moles of benzoic acid.

`\text{ Moles of benzoic acid}=\frac{\text{ Mass of benzoic acid}}{\text{ Molar mass of benzoic acid}}=(13.45g)/(122.12g/mole)=0.1101moles`

Now we have to calculate the moles of methyl benzoate.

The balanced chemical reaction is,

`C_7H_6O_2+CH_3OH\rightarrow C_6H_5CO_2CH_3`

From the balanced reaction we conclude that

As, 1 mole of benzoic acid react to give 1 mole of methyl benzoate.

So, 0.1101 mole of benzoic acid react to give 0.1101 mole of methyl benzoate.

Now we have to calculate the mass of methyl benzoate.

`\text{ Mass of methyl benzoate}=\text{ Moles of methyl benzoate}* \text{ Molar mass of methyl benzoate}`

`\text{ Mass of methyl benzoate}=(0.1101moles)* (136.15g/mole)=14.99g`

Theoretical yield of methyl benzoate = 14.99 g

Experimental yield of methyl benzoate = 10.65 g

Now we have to calculate the percent yield of ester.

`\% \text{ yield of ester}=\frac{\text{ Experimental yield of methyl benzoate}}{\text{ Theretical yield of methyl benzoate}}* 100`

`\% \text{ yield of ester}=(10.65g)/(14.99g)* 100=71.05\%`

Therefore, the percent yield of ester is, 71.05 %