Suppose that in the synthesis of isoamyl acetate by Fisher esterification, a student began with 6.103 grams of acetic acid and 3.728 grams of isoamyl alcohol. If the mass of isoamyl acetate obtained were 4.993, what was the percent yield? (If your answer is not quite the same as one of the answers below, then choose the closest value.)

The % yield is 90.68 %

Explanation:

Step 1: Data given

Mass of acetic acid = 6.103 grams

Mass of isoamyl alcohol = 3.728 grams

Mass isoamyl acetate obtained = 4.993 grams

Molar mass acetic acid = 60.05 g/mol

Molar mass of isoamyl alcohol = 88.15 g/mol

Molar mass of isoamyl acetate = 130.19 g/mol

Step 2: The balanced equation

C5H12O + CH3COOH → C7H14O2 + H2O Isoamyl alcohol + acetic acid - --> Isoamyl acetate + water

Step 3: Calculate moles of acetic acid

Moles acetic acid = Mass acetic acid / Molar mass acetic acid

Moles acetic acid = 6.103 grams / 60.05 g/mol

Moles acetic acid = 0.1016 moles

Step 4: Calculate moles isoamyl alcohol

Moles isoamyl alcohol = 3.728 grams / 88.15 g/mol

Moles isoamy lalcohol = 0.04229 moles

Step 5: Calculate limiting reactant

The mole ratio is 1:1 so isoamyl alcohol has the smallest number of moles. So isoamyl alcohol is the limiting reactant. It will be completely be consumed (0.04229 moles).

Acetic acid is in excess, there will be consumed 0.04229 moles.

There will remain 0.1016 - 0.04229 = 0.05931‬ moles

Step 6: Calculate moles of isoamyl acetate

For 1 mole acetic acid, we need 1 mole of isoamyl alcohol, to produce 1 mole isoamyl acetate and 1 mole of H2O

For 0.04229 moles isoamyl alcohol consumed, there will be produced 0.04229 moles of isoamyl acetate.

Step 7: Calculate mass of isoamyl acetate

Mass isoamyl acetate = moles isoamyl acetate * molar mas isoamyl acetate

Mass isoamyl acetate = 0.04229 moles * 130.19 g/mol

Mass isoamyl acetate = 5.506 grams = theoretical yield

Step 8: Calcuate % yield

% yield = actual yield / theoretical yield

% yield = (4.993 / 5.506) * 100%

% yield = 90.68 %

The % yield is 90.68 %