Theoretical and percentage yield: Class 11 practice
Theoretical yield comes from stoichiometry. Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Mole concept
- Balanced equations
The idea behind theoretical and percentage yield
Theoretical yield comes from stoichiometry. Percentage yield = actual ÷ theoretical × 100; it is below 100% because of side reactions and losses.
Step-by-step method
- Convert the reactant mass to moles.
- Use the mole ratio for product moles → mass (theoretical).
- Percentage yield = actual ÷ theoretical × 100.
Worked example 1 (Basic)
CaCO₃ → CaO + CO₂: find the theoretical mass of CaO from 800 g of reactant.
- Moles = 800/100 = 8
- CaO: 8 mol × 56 = 448 g
- Answer: 448 g
Worked example 2 (Standard)
2H₂ + O₂ → 2H₂O (from H₂): 16 g of reactant gives 123.84 g of H₂O. Find the percentage yield.
- Theoretical = 144 g
- 123.84 ÷ 144 × 100
- Answer: 86%
Worked example 3 (Challenge)
N₂ + 3H₂ → 2NH₃ (from N₂): 252 g of reactant gives 195.84 g of NH₃. Find the percentage yield.
- Theoretical = 306 g
- 195.84 ÷ 306 × 100
- Answer: 64%
Common mistakes and misconceptions
- Using the reactant mass as the theoretical yield
- Inverting the fraction
Practice questions
- (Basic) 2Mg + O₂ → 2MgO (from Mg): find the theoretical mass of MgO from 96 g of reactant.
- (Basic) 2Mg + O₂ → 2MgO (from Mg): find the theoretical mass of MgO from 144 g of reactant.
- (Standard) 2Mg + O₂ → 2MgO (from Mg): 216 g of reactant gives 327.6 g of MgO. Find the percentage yield.
- (Standard) CaCO₃ → CaO + CO₂: 300 g of reactant gives 149.52 g of CaO. Find the percentage yield.
- (Challenge) 2Mg + O₂ → 2MgO (from Mg): 48 g of reactant gives 68 g of MgO. Find the percentage yield.
- (Challenge) 2Mg + O₂ → 2MgO (from Mg): 48 g of reactant gives 59.2 g of MgO. Find the percentage yield.
Answer key with full working
- Q1: Moles = 96/24 = 4 → MgO: 4 mol × 40 = 160 g → Answer: 160 g
- Q2: Moles = 144/24 = 6 → MgO: 6 mol × 40 = 240 g → Answer: 240 g
- Q3: Theoretical = 360 g → 327.6 ÷ 360 × 100 → Answer: 91%
- Q4: Theoretical = 168 g → 149.52 ÷ 168 × 100 → Answer: 89%
- Q5: Theoretical = 80 g → 68 ÷ 80 × 100 → Answer: 85%
- Q6: Theoretical = 80 g → 59.2 ÷ 80 × 100 → Answer: 74%
Syllabus note
This skill is listed under "Some Basic Concepts of Chemistry" for Class 11 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
- Some Basic Concepts of Chemistry chapter notes (Class 11)
- Molarity of solutions
- Mass–mass stoichiometry
- Limiting reagent
- Empirical and molecular formula
- Formula from combustion data
- Quantum numbers and orbitals
- Hydrogen spectrum and Rydberg formula
- Hybridisation and molecular shape (VSEPR)
- Hess's law and enthalpy of reaction
- All Class 11 practice
Frequently asked questions
Which chapter is theoretical and percentage yield in?
"Some Basic Concepts of Chemistry", Class 11 Chemistry (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.