Question

A reaction was performed in which 3.6 g 3.6 g of benzoic acid was reacted with excess methanol to make 1.3 g 1.3 g of methyl benzoate. Calculate the theoretical yield and percent yield for this reaction.

Answer 1

Answer: The percent yield of the reaction is 32.34 %

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)

• For benzoic acid:

Given mass of benzoic acid = 3.6 g

Molar mass of benzoic acid = 122.12 g/mol

Putting values in equation 1, we get:

`\text{Moles of benzoic acid}=(3.6g)/(122.12g/mol)=0.0295mol`

The chemical equation for the reaction of benzoic acid and methanol is:

`\text{Benzoic acid + methanol}\rightarrow \text{methyl benzoate}`

By Stoichiometry of the reaction

1 mole of benzoic acid produces 1 mole of methyl benzoate

So, 0.0295 moles of benzoic acid will produce =
`(1)/(1)* 0.0295=0.108` moles 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 = 0.0295 moles

Putting values in equation 1, we get:

`0.0295mol=\frac{\text{Mass of methyl benzoate}}{136.15g/mol}\\\\\text{Mass of methyl benzoate}=(0.0295mol* 136.15g/mol)=4.02g`

• 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 = 1.3 g

Theoretical yield of methyl benzoate = 4.02 g

Putting values in above equation, we get:

`\%\text{ yield of methyl benzoate}=(1.3g)/(4.02g)* 100\\\\\% \text{yield of methyl benzoate}=32.34\%`

Hence, the percent yield of the reaction is 32.34 %