5.1.6Additional lessons
Percentage Yield Calculations — IB Chemistry SL Revision Notes & Topical Questions
5 min read 6 parts Lesson 64 of 98
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Section 1 of 6Yield
Percentage yield measures how efficiently a reaction converts reactants into the desired product.
Theoretical yield
- Theoretical yield The maximum mass of product predicted by the balanced equation from the limiting reactant, assuming complete reaction.
Actual yield
- Actual yield The mass of product really obtained, which is measured after purification.
Percentage yield
- Percentage yield Percentage yield = actual yield ÷ theoretical yield × 100.
Range
- Range The percentage yield cannot exceed 100% unless the product is impure or wet.
Section 2 of 6Calculation Steps
1
Write the balanced equation and find the limiting reactant.
2
Calculate the moles of the desired product from the moles of the limiting reactant.
3
Convert to a theoretical mass using the molar mass of the product.
4
Divide the actual mass obtained by the theoretical mass and multiply by 100.
5
State the answer to the appropriate number of significant figures.
Section 3 of 6Why Yields Are Below 100%
Why Yields Are Below 100%
Reversible reactions
- Reversible reactions If the reaction reaches equilibrium, some reactants remain and the reaction does not go to completion.
Side reactions
- Side reactions Other products may form from the same reactants, reducing the amount of the desired product.
Losses
- Losses Product is lost during transfer, filtering, washing and recrystallisation.
Impure reactants
- Impure reactants Starting materials that are not pure contain less of the reacting substance than expected.
Section 4 of 6Improving Yield
Improving Yield
Reaction conditions
- Reaction conditions Altering temperature, pressure or concentration can favour the desired reaction.
Better technique
- Better technique Careful transfers and washing minimise losses of product.
Excess of cheap reactant
- Excess of cheap reactant Using an excess of an inexpensive reactant can drive the reaction closer to completion.
Industrial value
- Industrial value Even a small improvement in percentage yield saves large amounts of raw material on an industrial scale.
Section 5 of 6Worked Example
Question
- 6.00 g of ethanol (Mr 46.08) reacts with excess ethanoic acid to give ethyl ethanoate (Mr 88.12): C₂H₅OH + CH₃COOH → CH₃COOC₂H₅ + H₂O.
- 8.50 g of ester is obtained.
- Calculate the percentage yield.
- n(ethanol) = 6.00 ÷ 46.08 = 0.1302 mol.
- The ratio ethanolester is 1 : 1, so the theoretical amount of ester is 0.1302 mol.
- Theoretical mass = 0.1302 × 88.12 = 11.47 g.
- Percentage yield = 8.50 ÷ 11.47 × 100 = 74.1%.
Answer Percentage yield = 74.1%
Section 6 of 6Exam Checklist & Tips
5.1.6 – Key Points for the Exam
- Theoretical yield The maximum amount of product calculated from the limiting reactant.
- Actual yield The amount of product obtained experimentally.
- Percentage yield Actual yield as a percentage of theoretical yield.
- Reversible reaction A reaction in which products can react to re-form the reactants.
Common Mistakes
- Calculating the theoretical yield from the excess reactant instead of the limiting reactant.
- Dividing the theoretical yield by the actual yield, which inverts the ratio.
- Believing that a yield above 100% means the experiment was perfect, rather than that product is impure.
- Mixing up moles and grams when comparing yields.
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