If a student reacts 4.40 mL of acetic acid with 3.35 mL of isopentyl alcohol and obtains 3.25 mL of isopentyl acetate as the product, what was the percent yield

Question

Isopentyl acetate can be prepared by an esterification reaction using acetic acid and isopentyl alcohol. The balanced reaction is shown below. (Acetic acid has MW of 60.0 g/mol & density of 1.05 g/ML, Isopentyl alcohol has MW of 88.2 g/mol & density of 0.81 g/ML, Isopentyl acetate has MW of 130.2 g/mol & density of 0.867 g/ML

Answer 1

C5H12O + CH3COOH ==> C7H14O2 +H2O

Note a 1:1 mole ratio of isopentyl alcohol to acetic acid to product.

First, find the limiting reactant:

moles of IPA present = 3.35 ml x 0.81 g/ml x 1 mole/88.2 g = 0.03077 moles

moles acetic acid present = 4.40 ml x 1.05 g/ml x 1 mole/60.0 g = 0.077 moles 

Limiting reactant is IPA

 

Moles of isopentyl acetate that can theoretically form = 0.03077 moles

Mass of theoretical yield = 0.03077 moles x 130.2 g/mole = 4.006 g 

Actual yield = 3.25 ml x 0.867 g/ml = 2.818 g

Percent yield = actual yield/theoretical yield (x100%) = 2.818 g/4.006 g (x100%) = 70.3% (to 3 significant figurers)