Percentage Purity & Percent Yield | Stoichiometry
By Abhishek Sengar | CHEMCA – JEE & NEET Chemistry
Recommended Prerequisite
To master purity and yield calculations, you must be comfortable with the basic Stoichiometric methods (like the Basic Ratio Method). Revise the core foundations here: Some Basic Concepts of Chemistry for Class 11 and JEE/NEET.
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Lecture 7: Some Basic Concepts of Chemistry
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Hello students! Welcome to Lecture 7. Up until now, our chemical equations have assumed perfect scenarios: 100% pure reactants creating 100% of the expected products. In the real world (and in JEE/NEET exams!), samples contain impurities, and reactions rarely go to perfect completion. Today, we conquer Percentage Purity and Percentage Yield.
1. Percentage Purity
When you are given a sample in a lab, it often contains inert impurities (dirt, other chemicals) that do not participate in the reaction. Only the pure substance reacts.
Percentage Purity =
(Mass of Pure Substance / Total Mass of Sample) × 100
*You can also use Moles instead of Mass: (Moles of Pure Substance / Total Moles of Sample) × 100
Problem Type A: Finding Product from an Impure Sample
Question: What amount of CO2 is formed upon the thermal decomposition of a 60% pure sample of CaCO3 weighing 2 kg?
- Total mass of sample = 2 kg = 2000 g.
- Mass of pure CaCO3 = 60% of 2000g = 1200 g.
- Moles of pure CaCO3 = 1200 / 100 (molar mass) = 12 moles.
- Equation: CaCO3(s) → CaO(s) + CO2(g).
- Since 1 mole CaCO3 gives 1 mole CO2, 12 moles of CaCO3 will produce 12 moles of CO2.
Problem Type B: Finding Purity from Product Formed
Question: What is the purity of a 5 kg sample of CaCO3 if it produces 660 g of CO2?
- Moles of CO2 formed = 660g / 44g/mol = 15 moles.
- By stoichiometry (1:1), this means exactly 15 moles of pure CaCO3 reacted.
- Initial total moles of the sample (assuming it was 100% pure) = 5000g / 100g/mol = 50 moles.
- Percentage Purity = (Pure Moles / Total Moles) × 100 = (15 / 50) × 100 = 30%.
2. Percentage Yield
In a chemical plant, you calculate that mixing X and Y should give you 100 kg of product. But in reality, you only get 80 kg. Why? Side reactions, loss during transfer, or incomplete reactions. The 100 kg is the Theoretical Yield, and the 80 kg is the Actual Yield.
Percentage Yield =
(Actual Yield / Theoretical Yield) × 100
- Theoretical Yield: The expected amount calculated via Stoichiometry.
- Actual Yield: The real observed amount formed in the experiment.
Solving Single-Step Yield Problems
Question: For the reaction 2A → 3B, if we start with 100 moles of A and the reaction has an 80% yield, what is the actual amount of B formed?
- Find Theoretical Yield: 2 moles A → 3 moles B. Therefore, 100 moles A → 150 moles B.
- Find Actual Yield: 80% of Theoretical Yield.
- Actual Yield = (80 / 100) × 150 = 120 moles of B.
3. Multi-Step Sequential Reactions
Advanced problems will string multiple reactions together. The actual yield of Step 1 becomes the starting reactant for Step 2. Let's solve a complex chain reaction!
Sequential Problem: Find the final amount of D formed.
- Step 1: 2A → B (60% Yield)
- Step 2: B → 3C (80% Yield)
- Step 3: 3C → 4D (90% Yield)
- Starting Material: 100 Moles of A.
Solution Breakdown:
Step 1: A to B
2 moles A gives 1 mole B.
Theoretical: 100 moles A → 50 moles B.
Actual: 60% of 50 = 30 moles B.
Step 2: B to C
1 mole B gives 3 moles C.
Theoretical: 30 moles B → 90 moles C.
Actual: 80% of 90 = 72 moles C.
Step 3: C to D
3 moles C gives 4 moles D. (Ratio is 4/3).
Theoretical: 72 moles C → (4/3) × 72 = 96 moles D.
Actual: 90% of 96 = 86.4 moles D.
Test Your Understanding! 🎯
Take this 10-question MCQ quiz to verify your grasp of Lecture 7. Explanations and study recommendations will be revealed upon submission.
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