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 $C_6H_5COOH$

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

$\text{Moles of }C_6H_5COOH=\frac{3.4g}{122.12g/mol}=0.0278mol$

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 $C_6H_5COOCH_3$

$\text{ Mass of }C_6H_5COOCH_3=\text{ Moles of }C_6H_5COOCH_3\times \text{ Molar mass of }C_6H_5COOCH_3$

Molar mass of  = 136.14 g/mole

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

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 :

$\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

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

$\text{Percent yield}=\frac{1.2g}{3.78g}\times 100=31.7\%$

The percent yield of the reaction is, 31.7 %