Percentage Yield & Purity
Section: 3. Stoichiometry | Syllabus: Cambridge AS Level Physics 9702
What is Percentage Yield? Percentage yield is a measure of how efficient a chemical reaction is. It compares the actual amount of product obtained to the maximum theoretical amount that could be produced.
Percentage yield = (Actual yield ÷ Theoretical yield) × 100% The percentage yield is always less than or equal to 100%. Why is Yield Less Than 100%? In real reactions, we rarely get 100% yield due to: Incomplete reactions: Not all reactants convert to products Side reactions: Some reactants form unwanted products Product lost during transfer: Material stuck to glassware, spillages Product lost during purification: Filtering, washing, crystallization Reversible reactions: Some products convert back to reactants Impure reactants: Starting materials not 100% pure Theoretical Yield Theoretical yield is the maximum amount of product that could be formed if: All reactants converted to products (100% conversion) No side reactions occurred No losses during the process We calculate theoretical yield using stoichiometry (mole ratios from balanced equations).
Calculating Percentage Yield - Method Step 1: Calculate the theoretical yield using mole calculations Step 2: Use the formula: % yield = (actual ÷ theoretical) × 100% Example 1: Basic Percentage Yield Question: 10 g of calcium carbonate decomposes to produce 4.8 g of calcium oxide.
Calculate the percentage yield. (Ar: Ca=40, C=12, O=16) CaCO₃ → CaO + CO₂ Step 1: Calculate theoretical yield 1a. Find moles of CaCO₃ Mr of CaCO₃ = 40 + 12 + (3×16) = 100 Moles = 10 ÷ 100 = 0.1 mol 1b.
Use mole ratio (1:1) Moles of CaO = 0.1 mol 1c. Convert to mass Mr of CaO = 40 + 16 = 56 Theoretical yield = 0.1 × 56 = 5.6 g Step 2: Calculate percentage yield Actual yield = 4.8 g % yield = (4.8 ÷ 5.6) × 100% = 85.7% Example 2: Two Reactants Question: 2.4 g of magnesium reacts with excess oxygen to produce 3.6 g of magnesium oxide.
Calculate the percentage yield. (Ar: Mg=24, O=16) 2Mg + O₂ → 2MgO Step 1: Calculate theoretical yield Moles of Mg = 2.4 ÷ 24 = 0.1 mol Mole ratio 2:2 (1:1), so moles of MgO = 0.1 mol Mr of MgO = 24 + 16 = 40 Theoretical yield = 0.1 × 40 = 4.0 g Step 2: Calculate percentage yield % yield = (3.6 ÷ 4.0) × 100% = 90% Finding Actual Yield from Percentage Yield Rearranging the formula: Actual yield = (Percentage yield × Theoretical yield) ÷ 100 Example 3: A reaction has a theoretical yield of 50 g and a percentage yield of 80%.
What is the actual yield? Actual yield = (80 × 50) ÷ 100 = 40 g What is Purity? Purity is the proportion of a desired substance in a sample, expressed as a percentage. Percentage purity = (Mass of pure substance ÷ Total mass of sample) × 100% Example: A 10 g sample of salt contains 9 g of pure NaCl and 1 g of impurities.
Purity = (9 ÷ 10) × 100% = 90% Why is Purity Important? Medicine: Drugs must be pure to be safe and effective Industry: Impurities can interfere with reactions or damage equipment Research: Pure substances give reliable, reproducible results Food: Purity affects quality, taste, and safety Calculations: Impure reactants give lower yields than expected Example 4: Calculating Purity Question: A 5.0 g sample of limestone is heated.
It produces 2.2 g of calcium oxide. If pure limestone would produce 2.8 g of CaO, calculate the purity of the limestone. Method: Actual CaO produced = 2.2 g CaO from pure sample = 2.8 g Purity = (2.2 ÷ 2.8) × 100% = 78.6% Alternatively: Mass of pure CaCO₃ in sample = (2.2 ÷ 2.8) × 5.0 = 3.93 g Purity = (3.93 ÷ 5.0) × 100% = 78.6% Example 5: Using Purity in Calculations Question: What mass of 80% pure magnesium is needed to produce 4.0 g of magnesium oxide?
2Mg + O₂ → 2MgO (Ar: Mg=24, O=16) Step 1: Find mass of pure Mg needed Mr of MgO = 40 Moles of MgO = 4.0 ÷ 40 = 0.1 mol Ratio 2:2, so moles of Mg = 0.1 mol Mass of pure Mg = 0.1 × 24 = 2.4 g Step 2: Account for purity If sample is 80% pure, we need more than 2.4 g Mass of impure sample = (2.4 ÷ 80) × 100 = 3.0 g Answer: 3.0 g of 80% pure magnesium is needed Relationship Between Yield and Purity Low purity of reactants leads to lower actual yield: Example: You use 10 g of "copper carbonate" but it's only 90% pure.
Only 9 g is actually CuCO₃ 1 g is impurities that won't react You'll get less product than if you used 10 g of pure CuCO₃ This lowers the actual yield Example 6: Combined Yield and Purity Question: 5.0 g of 80% pure zinc reacts with excess acid.
The percentage yield is 90%. What mass of hydrogen gas is produced? Zn + 2HCl → ZnCl₂ + H₂ (Ar: Zn=65, H=1) Step 1: Find mass of pure Zn Mass of pure Zn = (80 ÷ 100) × 5.0 = 4.0 g Step 2: Calculate theoretical yield Moles of Zn = 4.0 ÷ 65 = 0.0615 mol Ratio 1:1, so moles of H₂ = 0.0615 mol Mr of H₂ = 2 Theoretical mass = 0.0615 × 2 = 0.123 g Step 3: Apply percentage yield Actual mass = (90 ÷ 100) × 0.123 = 0.111 g Answer: 0.111 g of hydrogen is produced Atom Economy vs Percentage Yield Aspect Atom Economy Percentage Yield What it measures Efficiency of the reaction itself Efficiency of the proce…
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