Question

A reaction was performed in which 3.4 g of benzoic acid was reacted with excess methanol to make 1.2 g of methyl benzoate. Calculate the theoretical yield and percent yield for this reaction.

Answer 1

Answer : The theoretical yield and percent yield for this reaction is, 3.78 grams and 31.7 % respectively.

Explanation : Given,

Mass of
`C_6H_5COOH` = 3.4 g

Molar mass of
`C_6H_5COOH` = 122.12 g/mol

First we have to calculate the moles of
`C_6H_5COOH`

`\text{Moles of }C_6H_5COOH=\frac{\text{Given mass }C_6H_5COOH}{\text{Molar mass }C_6H_5COOH}`

`\text{Moles of }C_6H_5COOH=(3.4g)/(122.12g/mol)=0.0278mol`

Now we have to calculate the moles of
`C_6H_5COOCH_3`

The balanced chemical equation is:

`C_6H_5COOH+CH_3OH\rightarrow C_6H_5COOCH_3`

From the reaction, we conclude that

As, 1 mole of
`C_6H_5COOH` react to give 1 mole of
`C_6H_5COOCH_3`

So, 0.0278 mole of
`C_6H_5COOH` react to give 0.0278 mole of
`C_6H_5COOCH_3`

Now we have to calculate the mass of
`C_6H_5COOCH_3`

`\text{ Mass of }C_6H_5COOCH_3=\text{ Moles of }C_6H_5COOCH_3* \text{ Molar mass of }C_6H_5COOCH_3`

Molar mass of = 136.14 g/mole

`\text{ Mass of }C_6H_5COOCH_3=(0.0278moles)* (136.14g/mole)=3.78g`

The theoretical yield of
`C_6H_5COOCH_3` produced is, 3.78 grams.

Now we have to calculate the percent yield of the reaction.

Theoretical yield of the reaction = 3.78 g

Experimental yield of the reaction = 1.2 g

The formula used for the percent yield will be :

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

Now put all the given values in this formula, we get:

`\text{Percent yield}=(1.2g)/(3.78g)* 100=31.7\%`

The percent yield of the reaction is, 31.7 %