Percentage yield of an ester

Worked example

Ethyl ethanoate is made from ethanoic acid and excess ethanol: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O. 22.3 g of ethanoic acid gave 25.0 g of ester. Calculate the percentage yield. M(CH₃COOH) = 60.0 g mol⁻¹, M(CH₃COOC₂H₅) = 88.0 g mol⁻¹.

1. 1. Moles from mass: solve for n(acid)

   $n = \dfrac{m}{M}$

   = 0.3717 mol

2. 2. Moles from mass: solve for Theoretical mass of ester

   $m = nM$

   = 32.71 g

3. 3. Percentage yield: solve for Y

   $Y = \dfrac{\text{actual}}{\text{theoretical}} \times 100\%$

   = 76.44 %

Answer: Percentage yield = 76.4 %

Method

n(acid) = m/M. The reaction is 1 : 1, so the theoretical mass of ester is n × M(ester). Percentage yield = actual mass ÷ theoretical mass × 100. The yield is well below 100% partly because esterification is an equilibrium.

Common mistakes

• Compare like with like: the theoretical yield is the mass of ester the acid could make, not the mass of acid.

Formulas used

• Moles from mass

  $n = \dfrac{m}{M}$

• Percentage yield

  $Y = \dfrac{\text{actual}}{\text{theoretical}}$

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