Percentage yield (Triple Only)
Lajoy Tucker
Teacher
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%