If 1.87 g‎ of acetic acid (ch3co2h)‎ reacts with 2.31 g‎ of isopentylalcohol (c5h12o)‎ to give‎ 3.20 g of isopentylacetate (c7h14o2)‎, what is the percent yield of the reaction?

The reaction of acetic acid with isopentyl alcohol is

CH3COOH + C5H12O  ---> C7H14O2  + H2O

Thus one mole of acetic acid reacts with one mole of isopentyl alcohol to give one mole isopentyl acetate

Molar mass of acetic acid = 60.05 g / mole

Moles of acetic acid used = Mass used  / Molar mass = 1.87/ 60.05 = 0.0311

Molar mass of isopentyl alcohol = 88.15 g / mole

Moles of isopentyl alcohol = Mass / Molar mass = 2.31 / 88.15 = 0.0262

Molar mass of isopentyl acetate = 130.19 g / mole

Moles of isopentyl acetate = 3.20 / 130.19 = 0.0246

As per equation 0.0262  should of isopentyl alcohol should react with 0.0262 moles of acetic acid to give 0.0262 isopentyl acetate

Thus theoretical yield = 0.0262 moles

Actual yield = 0.0246

% yield = Actual yield X 100 / theoretical yield = 0.0246 X 100 / 0.0262 = 93.89%