C10: Concentration and Yield
Concentration, percentage yield and atom economy
Concentration And Yield
📋 Key Definitions
Concentration: The amount of solute dissolved in a given volume of solution. Measured in g/dm³ or mol/dm³.
Percentage yield: The percentage of the theoretical maximum amount of product actually obtained from a reaction.
Atom economy: The percentage of atoms in the reactants that end up in the desired product. A measure of how efficiently atoms are used.
Key Formulae:
Concentration (g/dm³) = mass of solute (g) ÷ volume of solution (dm³)
Concentration (mol/dm³) = moles of solute ÷ volume of solution (dm³)
Percentage yield = (actual yield ÷ theoretical yield) × 100
Atom economy = (Mr of desired product ÷ sum of Mr of all products) × 100
📝 Concentration
The concentration of a solution tells you how much solute is dissolved in a certain volume of solvent. A more concentrated solution has more solute per unit volume.
| Concentration Unit | Meaning | Formula |
|---|---|---|
| g/dm³ | Grams of solute per dm³ of solution | mass (g) ÷ volume (dm³) |
| mol/dm³ | Moles of solute per dm³ of solution | moles ÷ volume (dm³) |
Note: 1 dm³ = 1000 cm³. To convert cm³ to dm³, divide by 1000.
Worked Example 1: Calculating concentration in g/dm³
8 g of sodium hydroxide is dissolved in 250 cm³ of solution. Calculate the concentration in g/dm³.
Volume = 250 ÷ 1000 = 0.25 dm³
Concentration = 8 ÷ 0.25 = 32 g/dm³
Worked Example 2: Calculating concentration in mol/dm³
0.2 mol of HCl is dissolved in 500 cm³ of solution. Calculate the concentration in mol/dm³.
Volume = 500 ÷ 1000 = 0.5 dm³
Concentration = 0.2 ÷ 0.5 = 0.4 mol/dm³
Worked Example 3: Calculating mass from concentration
A solution of NaCl has a concentration of 20 g/dm³. What mass of NaCl is in 150 cm³ of this solution?
Volume = 150 ÷ 1000 = 0.15 dm³
Mass = concentration × volume = 20 × 0.15 = 3 g
📊 Percentage Yield
The theoretical yield is the maximum amount of product possible from the given amounts of reactants (calculated from the balanced equation). The actual yield is the amount of product actually obtained. The actual yield is always less than the theoretical yield.
Why actual yield is always less than theoretical:
- Incomplete reactions: Not all reactants are converted to products
- Product lost: Some product is lost during separation and purification (e.g. left on filter paper)
- Reversible reactions: Some products react back to form reactants
- Side reactions: Unexpected reactions produce unwanted by-products
Worked Example 4: Calculating percentage yield
The theoretical yield of a reaction is 50 g but only 40 g of product is obtained. Calculate the percentage yield.
Percentage yield = (40 ÷ 50) × 100 = 80%
Worked Example 5: Percentage yield from reacting masses
10 g of calcium carbonate is heated. The theoretical yield of calcium oxide is 5.6 g, but only 4.2 g is collected. Calculate the percentage yield.
Percentage yield = (4.2 ÷ 5.6) × 100 = 75%
♻️ Atom Economy
Atom economy measures how efficiently the reactant atoms end up in the desired product. A high atom economy means less waste, which is better for the environment and more sustainable.
| Feature | Percentage Yield | Atom Economy |
|---|---|---|
| Measures | How much product you actually make vs theory | How efficiently atoms are used |
| Formula | (actual ÷ theoretical) × 100 | (desired product Mr ÷ total products Mr) × 100 |
| Can be 100%? | Rarely — always less in practice | Yes, for addition reactions |
| Related to | Practical efficiency | Sustainability and waste |
Worked Example 6: Calculating atom economy
In the reaction CaCO₃ → CaO + CO₂, the desired product is CaO. Calculate the atom economy.
Mr(CaO) = 56, Mr(CO₂) = 44
Atom economy = (56 ÷ (56 + 44)) × 100 = (56 ÷ 100) × 100 = 56%
Worked Example 7: Atom economy comparison
Compare the atom economy of two routes to produce hydrogen:
Route 1: CH₄ + H₂O → CO + 3H₂ (desired product: H₂)
Atom economy = (3 × 2) ÷ (28 + 6) × 100 = 6 ÷ 34 × 100 = 17.6%
Route 2: Zn + 2HCl → ZnCl₂ + H₂ (desired product: H₂)
Atom economy = 2 ÷ (136 + 2) × 100 = 2 ÷ 138 × 100 = 1.4%
Route 1 has the higher atom economy and is more sustainable.
❓ Practice Questions
Q1: Foundation 10 g of sugar is dissolved in 200 cm³ of solution. Calculate the concentration in g/dm³.
Q2: Foundation A reaction has a theoretical yield of 80 g. The actual yield is 60 g. Calculate the percentage yield.
Q3: Higher In the reaction C₂H₄ + Br₂ → C₂H₄Br₂, the desired product is C₂H₄Br₂. Calculate the atom economy. (Mr: C₂H₄Br₂ = 188)
Q4: Higher Explain two reasons why the actual yield of a reaction is always less than the theoretical yield.
Q5: Foundation A solution of HCl has a concentration of 0.5 mol/dm³. How many moles of HCl are in 100 cm³ of this solution?
✅ Answers
- Volume = 200 ÷ 1000 = 0.2 dm³. Concentration = 10 ÷ 0.2 = 50 g/dm³
- Percentage yield = (60 ÷ 80) × 100 = 75%
- This is an addition reaction — there is only one product. Atom economy = (188 ÷ 188) × 100 = 100%
- Any two from: incomplete reactions (not all reactants convert to products); product lost during separation/purification; reversible reactions (some products reform reactants); side reactions producing unwanted by-products.
- Volume = 100 ÷ 1000 = 0.1 dm³. Moles = 0.5 × 0.1 = 0.05 mol
🎯 Exam Tips
- Always convert cm³ to dm³ before calculating concentration (divide by 1000)
- Atom economy uses the Mr of products only, not reactants
- Percentage yield and atom economy are different — a reaction can have a high yield but low atom economy
- Addition reactions always have 100% atom economy because all reactant atoms end up in the product
- In atom economy questions, identify which product is the "desired" one carefully — the rest are waste
- When asked to explain why yield is less, give specific reasons not just "some is lost"
🔢 Maths Skills
Mathematical Skills
Concentration: Concentration (g/dm³) = mass (g) ÷ volume (dm³). Convert cm³ to dm³ by dividing by 1000. For mol/dm³: concentration = moles ÷ volume (dm³).
% yield: % yield = (actual yield ÷ theoretical yield) × 100.
Atom economy: Atom economy = (Mr of desired product ÷ sum of Mr of all products) × 100.
⚠️ Common Misconceptions
Watch Out!
High yield means high atom economy. Wrong: high yield = high atom economy Correct: they measure different things — yield measures practical efficiency (how much you actually make), atom economy measures how efficiently atoms are used (how much waste is produced)
Concentration and amount are the same thing. Wrong: concentration = amount Correct: concentration = amount per unit volume — a small volume of concentrated solution may contain the same amount of solute as a large volume of dilute solution
✍️ 6-Mark Question
Extended Answer
6 marks: Explain the difference between percentage yield and atom economy.
Percentage yield measures how much product is actually obtained compared to the theoretical maximum. It is calculated as (actual yield ÷ theoretical yield) × 100. A yield less than 100% occurs because of incomplete reactions, product lost during purification, reversible reactions or side reactions. Atom economy measures how efficiently the reactant atoms end up in the desired product. It is calculated as (Mr of desired product ÷ sum of Mr of all products) × 100. A reaction can have a high percentage yield but low atom economy (e.g. making H₂ from Zn + 2HCl has high yield but low atom economy because most atoms end up in waste ZnCl₂). Both are important: yield for practical efficiency and atom economy for sustainability and waste reduction.
Mark scheme: 1 mark for defining % yield; 1 mark for formula; 1 mark for reasons yield is less than 100%; 1 mark for defining atom economy; 1 mark for formula; 1 mark for explaining they measure different things with example.
📊 AO3: Analyse & Evaluate
Analysis and Evaluation
Two processes produce the same product:
| Process | % Yield | Atom Economy | Energy Use | Waste |
|---|---|---|---|---|
| A | 95% | 40% | High | Lots |
| B | 70% | 100% | Low | None |
Question: Which process is more sustainable? Justify your answer using all the data.
Answer: Process B is more sustainable. Although it has a lower yield, it has 100% atom economy (all atoms end up in the desired product, so no waste), lower energy use and no waste production. Process A wastes 60% of atoms and produces lots of waste, requiring disposal. Sustainability prioritises minimising waste and resource use, making B the better choice despite its lower yield.
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