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2.7 Calculation of % Yield from theoretical and experimental yield

2.7 Calculation of Percentage Yield

🎯 Learning outcomes

After the completion of the chapter, the students should be able to:

  • Calculate the theoretical yield and percentage yield from the given actual yield (experimental yield)

Introduction to Yields

When performing a chemical synthesis, we use a balanced chemical equation to compute exactly how much product should theoretically form from our starting materials.

Theoretical Yield (Stoichiometric Yield): The maximum amount of a particular product that could be formed from the given amount of reactants, calculated strictly based on the stoichiometry of the balanced chemical reaction. This calculation must always be based on the limiting reagent (the reactant that is completely consumed first).
Experimental Yield (Actual Yield): The actual, physical amount of a product obtained when the reaction is carried out in a laboratory setting. Due to various known and unknown factors—such as side reactions, incomplete conversions, or loss of material during filtration and purification—the experimental yield is almost always lower than the theoretical yield.

Percent Yield

To measure the efficiency of a synthetic procedure, chemists calculate the per cent yield. It is the ratio of the experimental yield to the theoretical yield, expressed as a percentage.

Percent yield = (Experimental yield / Theoretical yield) × 100%

Percent yield is very important in the calculation of products in chemical synthesis.

Worked Examples (Yield Calculations)

Example 1
1.274 g of CuSO₄ reacts with an excess of zinc metal. If 0.392 g of Cu metal is obtained experimentally, calculate the percentage yield.
Equation: CuSO₄(aq) + Zn(s) → Cu(s) + ZnSO₄(aq)
(Atomic Masses: Cu = 63.5, S = 32, O = 16, Zn = 65.4)
1. Find the molar mass of CuSO₄:
Molar Mass = 63.5 + 32 + (16 × 4) = 159.5 g/mol
2. Calculate the Theoretical Yield of Cu:
From the balanced equation, 1 mole of CuSO₄ (159.5 g) produces 1 mole of Cu (63.5 g).
Theoretical Yield of Cu = (63.5 g / 159.5 g) × 1.274 g ≈ 0.507 g
3. Calculate Percent Yield:
Given Experimental Yield = 0.392 g
Percent Yield = (0.392 g / 0.507 g) × 100% ≈ 77.3%
Example 2
6.55 g of pure zinc is reacted with an excess of dilute hydrochloric acid. If the percentage yield of the reaction is 75%, calculate the actual mass of hydrogen gas produced.
Equation: Zn + 2HCl → ZnCl₂ + H₂
1. Calculate the Theoretical Yield of H₂:
1 mole of Zn (65.4 g) produces 1 mole of H₂ (2.016 g ≈ 2 g).
Theoretical Yield of H₂ = (2 g / 65.4 g) × 6.55 g = 0.20 g
2. Calculate the Experimental Yield using Percent Yield:
Percent Yield = (Experimental Yield / Theoretical Yield) × 100%
75% = (Experimental Yield / 0.20 g) × 100%
Experimental Yield = (75 × 0.20) / 100 = 0.15 g
Practice Questions
  1. 1.274 g of CuSO4 reacts with an excess of zinc metal, 0.392 g of Cu metal was obtained according to the equation
    CuSO4(aq.) + Zn → Cu + ZnSO4(aq.)
    Calculate the percentage yield.
  2. 6.55 g of pure zinc is reacted with an excess of dilute hydrochloric acid. If the percentage yield of the reaction is 75%, calculate the amount of hydrogen gas produced. (ans = 0.15 g)
  3. 4.6 g of pure sodium was reacted with an excess of chlorine to produce sodium chloride. If the amount of sodium chloride obtained was 10 g, what is the percentage yield of this reaction? (ans = 85.47%)
🧪 Test Your Understanding

Try this quick quiz to check how well you’ve grasped percentage yield calculations.

📋 Percentage Yield Quiz

Welcome to your 2.% yield, molecular and empirical formula

Download PDF 2.7 % yield from experimental & theoretical yield -Notes

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