What you'll learn
Percentage yield and percentage purity calculations appear regularly on CXC CSEC Chemistry Paper 2, particularly in Section II stoichiometry questions worth 8-12 marks. These calculations connect theoretical predictions from balanced equations to real-world laboratory results, testing your understanding of mole relationships and practical chemistry limitations.
Key terms and definitions
Theoretical yield — the maximum mass of product that could be formed from given reactants, calculated using balanced equations and stoichiometry, assuming complete reaction with no losses.
Actual yield — the mass of product actually obtained from a chemical reaction when carried out in practice, always less than or equal to the theoretical yield.
Percentage yield — the ratio of actual yield to theoretical yield expressed as a percentage: (actual yield ÷ theoretical yield) × 100%.
Percentage purity — the proportion of a desired substance in an impure sample, calculated as (mass of pure substance ÷ total mass of sample) × 100%.
Limiting reagent — the reactant that is completely consumed first in a chemical reaction, determining the maximum amount of product that can form.
Impurities — unwanted substances present in a sample that reduce its purity; can be unreacted starting materials, side products, or contaminants.
Stoichiometric calculations — mathematical methods using mole ratios from balanced equations to predict quantities of reactants needed or products formed.
Core concepts
Understanding theoretical yield
Theoretical yield represents the ideal outcome of a chemical reaction. Calculating it requires:
- Write and balance the chemical equation
- Convert the given mass of reactant to moles using molar mass
- Use the mole ratio from the balanced equation to find moles of product
- Convert moles of product back to mass
For example, when limestone (calcium carbonate) from the Cockpit Country in Jamaica is heated:
CaCO₃(s) → CaO(s) + CO₂(g)
The equation shows 1 mole of CaCO₃ produces 1 mole of CaO. If starting with 50 g of pure CaCO₃, the theoretical yield of calcium oxide can be calculated using this 1:1 ratio.
The theoretical yield assumes:
- All reactants convert completely to products
- No side reactions occur
- No product is lost during collection or purification
- Perfect laboratory technique
These conditions rarely exist in real experiments, which explains why actual yields are lower.
Calculating percentage yield
The percentage yield formula compares laboratory reality to theoretical prediction:
Percentage yield = (Actual yield ÷ Theoretical yield) × 100%
Both yields must use the same units (typically grams). Values above 100% indicate experimental error, usually from:
- Incomplete drying of product (excess water retained)
- Impurities in the collected product
- Measurement errors in mass determination
Percentage yields in school laboratories typically range from 60-85%, while industrial processes achieve 85-95% through optimized conditions.
Factors affecting percentage yield
Several factors cause actual yields to fall below theoretical values:
Incomplete reactions — Many reactions reach equilibrium before all reactants convert to products. Reversible reactions never achieve 100% conversion.
Side reactions — Alternative reaction pathways consume reactants without forming the desired product. When producing ethanol by fermentation in Trinidad rum distilleries, some glucose converts to unwanted organic acids.
Practical losses — Product remains on filter papers, in reaction vessels, or is lost during transfers between containers. Each transfer reduces yield by 1-3%.
Separation difficulties — Extracting product from reaction mixtures is imperfect. Some product remains dissolved in wash liquids or mixed with unreacted materials.
Competing reactions — Multiple products may form simultaneously, dividing the available reactants.
Understanding percentage purity
Pure substances contain only one chemical compound. Most chemicals purchased or produced contain impurities. Percentage purity quantifies how much of a sample is the desired substance:
Percentage purity = (Mass of pure substance ÷ Total sample mass) × 100%
Determining purity requires knowing either:
- The mass of pure substance directly (from analytical techniques)
- The mass of impurities (then pure substance = total − impurities)
- Results from a reaction where only the pure substance reacts
For example, bauxite ore mined in Jamaica contains 30-60% aluminum oxide (the desired compound) mixed with iron oxides, silica, and other minerals. Processing increases the percentage purity before aluminum extraction.
Purity determination through titration
A common CXC CSEC exam question involves determining purity by reacting the impure sample with a standard solution:
- An impure sample reacts completely with a measured volume of standard solution
- Calculate moles of standard solution used
- Use mole ratios to find moles of pure substance that reacted
- Convert to mass of pure substance
- Calculate percentage purity
This method assumes impurities are inert (don't react with the standard solution).
Combined yield and purity calculations
Examination questions often combine these concepts:
- Calculate theoretical yield from an impure reactant
- Find percentage yield when both purity and actual yield are given
- Determine the mass of impure product needed to obtain a target mass of pure substance
When starting with impure reactants, first calculate the mass of pure reactant:
Mass of pure reactant = (Percentage purity ÷ 100) × Total sample mass
Then proceed with normal stoichiometric calculations using only the pure reactant mass.
Worked examples
Example 1: Calculating percentage yield
Question: When 12.0 g of magnesium reacts with excess hydrochloric acid, 10.8 g of magnesium chloride is collected. Calculate the percentage yield.
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
[Relative atomic masses: Mg = 24, Cl = 35.5]
Solution:
Step 1: Calculate theoretical yield of MgCl₂
Moles of Mg = 12.0 g ÷ 24 g/mol = 0.50 mol
From equation: 1 mol Mg produces 1 mol MgCl₂
Therefore moles of MgCl₂ = 0.50 mol
Molar mass of MgCl₂ = 24 + (2 × 35.5) = 95 g/mol
Theoretical yield = 0.50 mol × 95 g/mol = 47.5 g
Step 2: Calculate percentage yield
Percentage yield = (10.8 g ÷ 47.5 g) × 100% = 22.7%
Example 2: Determining percentage purity
Question: A sample of impure sodium carbonate has a total mass of 15.0 g. When reacted with excess sulfuric acid, 2.64 g of carbon dioxide is produced. Calculate the percentage purity of the sodium carbonate sample.
Na₂CO₃(s) + H₂SO₄(aq) → Na₂SO₄(aq) + H₂O(l) + CO₂(g)
[Relative atomic masses: C = 12, O = 16, Na = 23]
Solution:
Step 1: Calculate moles of CO₂ produced
Molar mass of CO₂ = 12 + (2 × 16) = 44 g/mol
Moles of CO₂ = 2.64 g ÷ 44 g/mol = 0.060 mol
Step 2: Calculate mass of pure Na₂CO₃ that reacted
From equation: 1 mol Na₂CO₃ produces 1 mol CO₂
Therefore moles of pure Na₂CO₃ = 0.060 mol
Molar mass of Na₂CO₃ = (2 × 23) + 12 + (3 × 16) = 106 g/mol
Mass of pure Na₂CO₃ = 0.060 mol × 106 g/mol = 6.36 g
Step 3: Calculate percentage purity
Percentage purity = (6.36 g ÷ 15.0 g) × 100% = 42.4%
Example 3: Combined calculation
Question: A bauxite processing plant in Jamaica uses impure aluminum oxide (75% pure) to produce aluminum. If 400 kg of this impure sample is used and 142 kg of aluminum is obtained, calculate the percentage yield.
2Al₂O₃(s) → 4Al(s) + 3O₂(g)
[Relative atomic masses: Al = 27, O = 16]
Solution:
Step 1: Calculate mass of pure Al₂O₃
Mass of pure Al₂O₃ = (75 ÷ 100) × 400 kg = 300 kg
Step 2: Calculate theoretical yield of Al
Molar mass of Al₂O₃ = (2 × 27) + (3 × 16) = 102 g/mol
Moles of Al₂O₃ = 300,000 g ÷ 102 g/mol = 2941 mol
From equation: 2 mol Al₂O₃ produces 4 mol Al
Moles of Al = 2941 mol × (4 ÷ 2) = 5882 mol
Molar mass of Al = 27 g/mol
Theoretical yield = 5882 mol × 27 g/mol = 158,814 g = 158.8 kg
Step 3: Calculate percentage yield
Percentage yield = (142 kg ÷ 158.8 kg) × 100% = 89.4%
Common mistakes and how to avoid them
Mistake: Confusing percentage yield with percentage purity. Students calculate yield when asked for purity and vice versa.
Correction: Percentage yield compares how much product you made versus how much you could have made. Percentage purity describes how pure your starting material or product is. Read the question carefully to identify which is required.
Mistake: Using the mass of impure reactant directly in stoichiometric calculations without accounting for purity.
Correction: Always calculate the mass of pure substance first by multiplying total mass by (percentage purity ÷ 100), then use this pure mass in mole calculations.
Mistake: Mixing units — using actual yield in grams but theoretical yield in kilograms, or comparing moles to grams.
Correction: Ensure both yields have identical units before calculating percentage yield. Convert all values to the same unit (typically grams) before dividing.
Mistake: Dividing theoretical yield by actual yield instead of actual by theoretical.
Correction: Remember the formula structure: what you got (actual) divided by what you expected (theoretical), then multiply by 100. The smaller number goes on top.
Mistake: Forgetting to multiply by 100 in percentage calculations, giving answers like 0.85 instead of 85%.
Correction: Both percentage yield and percentage purity formulas end with "× 100%". Make this the final step in every calculation and check your answer is between 0 and 100.
Mistake: Assuming impurities react in purity calculations. When determining purity through reaction, students sometimes include impurity mass in mole calculations.
Correction: Impurities are typically inert (non-reactive) in these problems. Only the pure substance reacts, so only calculate moles for the compound that participates in the chemical equation.
Exam technique for Percentage Yield and Percentage Purity
Command word recognition: "Calculate the percentage yield" requires you to show the theoretical yield calculation first, then apply the percentage yield formula. "Determine the percentage purity" needs either a direct comparison of pure to total mass or a reaction-based calculation. "Suggest reasons" for low yield requires explanation of practical factors, not calculations.
Show all working: CXC CSEC Chemistry awards marks for method even if the final answer is incorrect. Write out each calculation step: mole calculation, mole ratio application, mass calculation, then percentage formula. A 6-mark question typically awards 1 mark per calculation step.
Unit awareness: Examiners deduct marks for missing or incorrect units. Include grams (g), moles (mol), or kilograms (kg) after every numerical answer. When converting between units, show the conversion explicitly.
Read data carefully: Examination questions provide relative atomic masses, which may differ slightly from data booklet values. Use the values given in the question. Check whether you're given actual yield, theoretical yield, or must calculate both. Identify whether samples are pure or impure before calculating.
Quick revision summary
Percentage yield compares actual product obtained to theoretical maximum: (actual ÷ theoretical) × 100%. Calculate theoretical yield using stoichiometry from balanced equations. Actual yields are lower due to incomplete reactions, side reactions, and practical losses. Percentage purity measures pure substance content: (pure mass ÷ total mass) × 100%. When reactants are impure, calculate pure mass first before stoichiometric calculations. Both percentages should fall between 0-100%; values outside this range indicate errors. Always show complete working, use correct units, and apply mole ratios carefully for full marks on CXC CSEC Chemistry examinations.