A student prepares benzoic acid by hydrolyzing 6.47 g of methyl benzoate (FW 136.15 g/mol) using excess sodium hydroxide. If the student recovers 3.71 g of benz

Question

Ig the student recovers 3.71 g of benzoic acid (FW 122.12 g/mol), what was the percent yield of the reaction?

Answer 1

Hello Aspen

 

The balanced chemical reaction is:

 

C₆H₅−COOCH₃ + NaOH → C₆H₅−COOH + CH₃OH

 

Starting with 6.47 g of methyl benzoate, we will calculate the theoretical yield of benzoic acid using stoichiometry.  I will abbreviate MB for methyl benzoate and BA for benzoic acid.  The mole ratio of MB to BA is 1:1.

 

6.47 g MB x (1 mol MB / 136.15 g MB) x (1 mol BA / 1 mol MB) x (122.12 g BA / 1 mol BA) = 5.80 g BA

 

%-Yield = actual yield / theoretical yield x 100%

 

Actual yield of BA = 3.71 g

 

%-Yield = (3.71 / 5.80) x 100% = 64.0%