Percentage yield (Triple Only)

Lajoy Tucker

Teacher

Lajoy Tucker

Introduction & Definitions

  • Percentage Yield measures how much product is obtained in a chemical reaction compared to the theoretical maximum.

Definition: Percentage Yield = (Actual Yield/Theoretical Yield​)×100

  • Actual Yield: The amount of product actually obtained from the reaction.

  • Theoretical Yield: The maximum possible mass of product, based on stoichiometric calculations.

Basic Principles

  • A 100% yield means the reaction produced all the product it theoretically could.

  • Common reasons for a percentage yield below 100% include:

    • Incomplete reactions.

    • Reversible reactions (the products turn back to reactants).

    • Side reactions forming unwanted products.

    • Loss during purification (e.g. filtration, evaporation).

    • Product remains in the reaction vessel.

Percentage Yield Calculations

Example 1:

Titanium can be extracted from titanium chloride by the following react

45g is the maximum mass of Ti that was calculated to be produced from reacting a known mass of

and Mg. Only 20g of Ti is produced from titanium chloride and excess magnesium.

Calculate the percentage yield

Answer:

Percentage Yield = (Actual Yield/Theoretical Yield​)×100

= (20/45) x 100

= 44.4%

In some questions, reacting masses calculations must be used to first calculate the theoretical (maximum) yield).

Example 2 (challenging):

Question:

In the reaction:

Mg + 2 HCl → MgCl₂ + H₂

A student reacts 2.40 g of Mg with excess HCl. The mass of hydrogen gas collected was 1.80 g. What is the percentage yield? (Mr H₂ = 2.0, Mg = 24.0)

Answer:

Step 1: Moles of Mg used; 2.40 g ÷ 24.0 = 0.10 mol

Step 2: Theoretical moles of H₂ (1:1 ratio with Mg); 0.10 mol

Step 3: Theoretical mass of H₂; 0.10 mol × 2.0 g/mol = 0.20 g

Step 4: Use percentage yield formula; (0.18 / 0.20) × 100 = 90%



Practice Questions

Question 1

75g of aluminium was collected on reducing aluminium oxide which was an 80% yield.

Calculate the maximum mass of aluminium that could have been produced. Give the answer to 2 s.f.

Answer:

Percentage Yield = (A/T​)×100

80% = (75/T) x 100

0.80 = 75/T

T = 75/0.80 = 93.75 g = 94g

Question 2

Fe₂O₃ + 3CO → 2Fe + 3CO₂

6.40 g of Fe₂O₃ reacts with excess CO, and 3.80 g of Fe is formed. What is the percentage yield? (Mr Fe₂O₃ = 160, Fe = 56)

Answer:

nFe₂O₃ = 6.40 ÷ 160 = 0.040 mol

Ratio Fe2O3:Fe = 1:2

Theoretical moles of Fe = 0.040 × 2 = 0.080 mol

Mass = 0.080 × 56 = 4.48 g

% Yield = (3.80 / 4.48) × 100 = 84.8%

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