Question

I need help with calculating the percent yield of an experiment. 1.0 g of p-aminobenzoic acid (MW=137.14 g/mol) was mixed with 10.0 mL of absolute ethanol (MW=46.07 g/mol) and after solid dissoved 1.0 mL of conc. sulfuric acid was added (MW=98.07 g/mol). The crude product of benzocaine (MW=165.19 g/mol) weighed 2.289 grams.
What is the theoretical yield?
What is the percent yield?

Answer 1

Sol.

Reaction :

p-aminobenzoic acid + ethanol + Concentrated sulphuric acid ------> benzocaine

In this reaction , concentrated sulphuric acid act as a catalyst for the reaction , so , the main reactants are p-aminobenzoic acid and ethanol  

As Mass of p-aminobenzoic acid = 1 g

Molar Mass of p-aminobenzoic acid = 137.14 g/mol  

So , Moles of p-aminobenzoic acid = 1 g / 137.14 g/mol

= 0.00729 mol

Also , Volume of Ethanol = 10 mL  

Density of Ethanol = 0.789 g/mL

Molar Mass of Ethanol = 46.07 g/mol  

So , Moles of Ethanol

= 10 mL × 0.789 g/mL / 46.07 g/mol

= 0.17126 mol

From reaction , 1 mole of p-aminobenzoic acid combines with 1 mole of ethanol

So , 0.00729 moles of p-aminobenzoic acid combines with 0.00729 moles of ethanol . But we have 0.17126 moles of ethanol . So , ethanol is the excess reactant and p-aminobenzoic acid is the limiting reactant  

Now , 1 mole of limiting reactant , p-aminobenzoic acid gives 1 mole of benzocaine  

So , Moles of benzocaine = 0.00729 mol

Also , Molar Mass of benzocaine = 165.19 g/mol

Therefore , Theoretical yield  

= Mass of benzocaine produced

= 0.00729 mol × 165.19 g/mol

= 1.204 g

As Experimental yield = Actual Mass of benzocaine produced = 2.289 g

Therefore , Percentage yield  

= ( Experimental yield / Theoretical yield ) × 100

= ( 2.289 g / 1.204 g ) × 100

=   190.12 %