Question

Consider the Fischer ester synthesis of methyl benzoate from benzoic acid and methanol in the presence of sulfuric acid as a catalyst. A reaction was performed in which 3.6 g of benzoic acid was reacted with excess methanol to make 1.4 g of methyl benzoate. Calculate the theoretical yield and percent yield for this reaction.

Answer 1

Answer: The percent yield of the reaction is 32.04 %.

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

Given mass of benzoic acid = 3.6 g

Molar mass of benzoic acid = 112.12 g/mol

Putting values in equation 1, we get:

`\text{Moles of benzoic acid}=(3.6g)/(112.12g/mol)=0.0321mol`

The chemical equation for the formation of methyl benzoate from benzoic acid follows:

`\text{Benzoic acid + Methanol}\rightarrow \text{Methyl benozate}`

By Stoichiometry of the reaction:

1 mole of benzoic acid produces 1 mole of methyl benzoate

So, 0.0321 moles of benzoic acid will produce =
`(1)/(1)* 0.0321=0.0321mol` of methyl benzoate

Now, calculating the mass of methyl benzoate from equation 1, we get:

Molar mass of methyl benzoate = 136.15 g/mol

Moles of methyl benzoate = 0.0321 moles

Putting values in equation 1, we get:

`0.0321mol=\frac{\text{Mass of methyl benzoate}}{136.15g/mol}\\\\\text{Mass of methyl benzoate}=(0.0321mol* 136.15g/mol)=4.37g`

To calculate the percentage yield of methyl benzoate, we use the equation:

`\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100`

Experimental yield of methyl benzoate = 1.4 g

Theoretical yield of methyl benzoate = 4.37 g

Putting values in above equation, we get:

`\%\text{ yield of methyl benzoate}=(1.4g)/(4.37g)* 100\\\\\% \text{yield of methyl benzoate}=32.04\%`

Hence, the percent yield of the reaction is 32.04 %.