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5.1.6Additional lessons

Percentage Yield Calculations — IB Chemistry HL Revision Notes & Topical Questions

5 min read 5 parts Lesson 94 of 172

📚 IB Chemistry HL Hub / Topical Revision
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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.
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Section 1 of 5Yield Terms

Theoretical yield
  • Theoretical yield The maximum mass or amount of product expected from the balanced equation, based on the limiting reactant.
Actual yield
  • Actual yield The mass or amount of product really obtained at the end of the experiment.
Percentage yield
  • Percentage yield (actual yield ÷ theoretical yield) × 100, expressed as a percentage.
Range
  • Range The value is between 0 and 100 percent in practice;
    • a value above 100 percent signals impurities or errors.

Section 2 of 5Calculation Steps

1
Write a balanced equation and identify the limiting reactant if more than one reactant amount is given.
2
Calculate the moles of limiting reactant from its mass or concentration and volume.
3
Use the mole ratio to find the theoretical moles of product, then convert to mass if needed.
4
Take the actual yield from the question, using the same unit as the theoretical yield.
5
Divide the actual by the theoretical yield and multiply by 100, quoting suitable significant figures.

Section 3 of 5Why Yields Fall Short

Why Yields Fall Short
Incomplete reaction
  • Incomplete reaction Reversible reactions reach equilibrium with some reactants left.
Side reactions
  • Side reactions Other products form from the same reactants, reducing the desired product.
Losses
  • Losses Product is lost during transfers, filtering, washing and purification.
Impure samples
  • Impure samples Impure reactants lower the effective amount of the substance that reacts.

Section 4 of 5Worked Example

Question

  • Heating 10.0 g of calcium carbonate gives 4.90 g of calcium oxide: CaCO₃(s) → CaO(s) + CO₂(g).
  • Calculate the percentage yield (M: CaCO₃ = 100.09, CaO = 56.08 g mol⁻¹).

  1. n(CaCO₃) = 10.0 ÷ 100.09 = 0.0999 mol.
  2. Theoretical n(CaO) = 0.0999 mol, so mass = 0.0999 × 56.08 = 5.60 g.
  3. Percentage yield = (4.90 ÷ 5.60) × 100 = 87.5 percent (using the unrounded 5.603 g).

Answer Percentage yield ≈ 87.5 percent

Section 5 of 5Exam Checklist & Tips

5.1.6 – Key Points for the Exam

  • Theoretical yield The maximum product predicted by stoichiometry.
  • Actual yield The product mass actually isolated.
  • Percentage yield Actual yield divided by theoretical yield, multiplied by 100.
  • Side reaction An unwanted reaction that consumes reactants without forming the desired product.

Common Mistakes

  • Dividing theoretical by actual yield, giving a value above 100 percent.
  • Using the excess reactant to calculate the theoretical yield.
  • Mixing mass and moles when comparing actual and theoretical yields.
  • Stating a percentage yield above 100 percent without suggesting an error or impurity.

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