C12: Concentration and Yield
Concentration of solutions in g/dm³ and mol/dm³, percentage yield and why actual yield is less than theoretical yield, and atom economy as a measure of reaction efficiency.
Concentration of Solutions
Concentration is a measure of how much solute is dissolved in a given volume of solution. The more solute dissolved, the more concentrated the solution.
A dilute solution has a low concentration (small amount of solute per volume of solvent). A concentrated solution has a high concentration (large amount of solute per volume of solvent).
Concentration (g/dm³) = mass of solute (g) / volume of solution (dm³)
Volume must be in dm³. To convert from cm³ to dm³, divide by 1000. For example, 250 cm³ = 0.25 dm³.
Worked Example: Concentration in g/dm³
5 g of sodium chloride is dissolved in 250 cm³ of solution. Calculate the concentration in g/dm³.
Volume in dm³ = 250 / 1000 = 0.25 dm³
Concentration = 5 / 0.25 = 20 g/dm³
Worked Example: Mass from Concentration
What mass of solute is needed to make 500 cm³ of a 40 g/dm³ solution?
Volume = 500 / 1000 = 0.5 dm³
mass = concentration × volume = 40 × 0.5 = 20 g
Concentration in mol/dm³
Concentration can also be expressed in moles per cubic decimetre (mol/dm³). This is sometimes called molarity and is often written as M.
Concentration (mol/dm³) = moles of solute / volume of solution (dm³)
You can convert between g/dm³ and mol/dm³ using the Mr of the solute:
Concentration (g/dm³) = concentration (mol/dm³) × Mr
Worked Example: Concentration in mol/dm³
2 g of NaOH is dissolved in 200 cm³ of solution. Calculate the concentration in mol/dm³. (Mr of NaOH = 40)
moles of NaOH = 2 / 40 = 0.05 mol
Volume = 200 / 1000 = 0.2 dm³
Concentration = 0.05 / 0.2 = 0.25 mol/dm³
Worked Example: Converting Units
A solution of HCl has a concentration of 0.5 mol/dm³. What is this in g/dm³? (Mr of HCl = 36.5)
Concentration in g/dm³ = 0.5 × 36.5 = 18.25 g/dm³
Always check the units the question asks for. If it wants g/dm³, make sure your final answer is in g/dm³ and not mol/dm³. Watch out for volume given in cm³ — always convert to dm³ first.
Percentage Yield
Theoretical yield is the maximum possible mass of product that could be produced from a given mass of reactant, assuming complete conversion and no losses.
Actual yield is the mass of product actually obtained from a reaction. It is always less than the theoretical yield.
Percentage yield = (actual yield / theoretical yield) × 100
Reasons why actual yield is less than theoretical yield:
- The reaction may be reversible — not all reactants convert to products
- Some product may be lost during transfer between vessels
- Side reactions may produce unwanted by-products
- The reactants may not be pure
- The reaction may not go to completion
Worked Example: Calculating Percentage Yield
The theoretical yield of a reaction is 50 g but only 38 g of product is obtained. Calculate the percentage yield.
Percentage yield = (38 / 50) × 100 = 76%
Worked Example: Calculating Theoretical Yield First
10 g of magnesium is burned in oxygen. The actual yield of MgO is 14.2 g. Calculate the percentage yield.
2Mg + O2 → 2MgO
moles of Mg = 10 / 24 = 0.417 mol
Mole ratio 1:1, so moles of MgO = 0.417 mol
Theoretical yield = 0.417 × 40 = 16.67 g
Percentage yield = (14.2 / 16.67) × 100 = 85.2%
Atom Economy
Atom economy is a measure of how efficiently atoms are used in a reaction. It shows the proportion of reactant atoms that end up in the desired product.
Atom economy = (Mr of desired product / sum of Mr of all products) × 100
Reactions with high atom economy are more sustainable and produce less waste. A reaction with only one product has 100% atom economy (addition reactions).
Worked Example: Atom Economy
Calculate the atom economy for producing hydrogen from methane:
CH4 + H2O → CO + 3H2
Mr of H2 (desired) = 2 (3 mol produced, so total = 3 × 2 = 6)
Sum of Mr of all products = 28 (CO) + 6 (3H2) = 34
Atom economy = (6 / 34) × 100 = 17.6%
Worked Example: High Atom Economy
Calculate the atom economy for producing ammonia:
N2 + 3H2 → 2NH3
Ammonia is the only product, so atom economy = 100%
This is an addition reaction — all reactant atoms become product atoms.
Atom economy uses the Mr of PRODUCTS, not reactants. Percentage yield uses the mass of product compared to the theoretical mass. Don't confuse the two!
Percentage Yield vs Atom Economy
| Feature | Percentage Yield | Atom Economy |
|---|---|---|
| What it measures | How much product is actually obtained vs the theoretical maximum | How efficiently atoms are converted into the desired product |
| Formula | (actual yield / theoretical yield) × 100 | (Mr desired product / sum of Mr all products) × 100 |
| Affected by | Experimental conditions, losses, side reactions, reversible reactions | The balanced equation — which products form |
| Can it vary? | Yes — same reaction can give different yields under different conditions | No — fixed for a given reaction pathway |
| Ideal value | 100% (never achieved in practice) | 100% (addition reactions achieve this) |
| Improving it | Optimise conditions, reduce losses, use excess of cheaper reactant | Choose a different reaction pathway with fewer by-products |
Both high percentage yield and high atom economy are desirable for sustainable industrial processes. A reaction can have a high yield but low atom economy (lots of waste) or vice versa.
Exam questions often ask you to evaluate a reaction using BOTH percentage yield and atom economy. A good industrial process needs both to be as high as possible.
Practice Questions
4 g of sodium hydroxide is dissolved in 100 cm³ of solution. Calculate the concentration in g/dm³.
Volume = 100 / 1000 = 0.1 dm³
Concentration = 4 / 0.1 = 40 g/dm³
A solution of H2SO4 has a concentration of 0.2 mol/dm³. Calculate the concentration in g/dm³. (Mr of H2SO4 = 98)
Concentration in g/dm³ = 0.2 × 98 = 19.6 g/dm³
12 g of calcium carbonate is heated. The theoretical yield of CaO is 6.72 g but only 5.6 g is obtained. Calculate the percentage yield.
Percentage yield = (5.6 / 6.72) × 100 = 83.3%
Calculate the atom economy for the production of ethene from crude oil cracking, where ethene (Mr = 28) and propene (Mr = 42) are both produced as products. Ethene is the desired product.
Atom economy = (28 / (28 + 42)) × 100 = (28 / 70) × 100 = 40%
Explain why a reaction with 100% atom economy may still have a percentage yield less than 100%.
Atom economy depends only on the equation — it measures whether by-products form. Percentage yield depends on practical factors such as reversible reactions, product losses during transfer, and incomplete reaction. Even if all atoms theoretically go to the desired product, losses and reversible reactions mean actual yield is always less than theoretical.
Required Practical
Preparing a Pure Dry Sample of a Salt (Titration Method)
This required practical involves using titration to prepare a pure, dry sample of a soluble salt from the reaction between an acid and an alkali. The titration method is used because both reactants are soluble, so you cannot simply add excess of one.
Method:
- Use a pipette to measure 25.0 cm³ of the alkali (e.g. sodium hydroxide) into a clean conical flask.
- Add a few drops of phenolphthalein indicator to the flask. The solution turns pink.
- Fill a burette with the acid (e.g. hydrochloric acid) of known concentration and record the initial reading.
- Slowly add the acid from the burette, swirling the flask constantly, until the indicator just turns colourless. Record the final burette reading.
- Repeat the titration until concordant results are obtained. Calculate the mean titre.
- Now repeat the reaction using the exact same volumes of acid and alkali but without the indicator (to avoid contaminating the salt with indicator dye).
- Evaporate some water from the neutral solution by gentle heating.
- Leave the solution to cool and crystallise. Filter off the crystals, wash with cold distilled water, and dry between filter paper.
The titration method ensures exactly the right amounts of acid and alkali are mixed so that the solution is neutral. This is essential because any excess acid or alkali would remain dissolved and contaminate the salt crystals.
Common errors: Rinsing the burette with water dilutes the acid, giving a falsely high titre. Forgetting to remove the indicator in the final crystallisation step produces contaminated crystals. Not recording burette readings to 0.05 cm³ reduces accuracy.
Maths Skills
Concentration Calculations in g/dm³ and mol/dm³
Concentration in g/dm³ = mass of solute (g) / volume of solution (dm³). To convert volume from cm³ to dm³, divide by 1000. For example, 250 cm³ = 0.25 dm³.
Concentration in mol/dm³ = moles of solute / volume of solution (dm³). To convert between g/dm³ and mol/dm³, use: concentration (g/dm³) = concentration (mol/dm³) × Mr. This is essential for titration calculations where concentration may be given or required in either unit.
Percentage Yield
Percentage yield = (actual yield / theoretical yield) × 100. The theoretical yield is calculated from the balanced equation using the three-step method. The actual yield is measured experimentally. Always check that the actual yield is less than the theoretical yield.
Atom Economy
Atom economy = (Mr of desired product / sum of Mr of all products) × 100. Unlike percentage yield, atom economy is fixed for a given reaction pathway and cannot be changed by improving experimental technique. Addition reactions always have 100% atom economy. Remember to multiply by the number of moles of each product shown in the balanced equation.
Rearranging Formulae
Each formula in this topic can be rearranged: concentration = mass/volume means mass = concentration × volume, and volume = mass/concentration. Similarly, % yield = (actual/theoretical) × 100 means actual yield = (% yield/100) × theoretical yield. Always show your rearrangement clearly in calculations.
Common Misconceptions
Yield and Atom Economy
Wrong: A high percentage yield means a high atom economy Correct: Percentage yield and atom economy measure different things. A reaction can have 95% yield but only 40% atom economy if lots of waste by-products form. Yield measures how much product you actually get; atom economy measures how efficiently reactant atoms end up in the desired product
Concentration and Amount
Wrong: Concentration and amount of solute are the same thing Correct: Concentration is the amount of solute per unit volume (concentration = amount/volume). Two solutions can have the same concentration but different amounts if their volumes differ. A larger volume of the same concentration contains more solute
Wrong: Atom economy tells you about profitability Correct: Atom economy measures how efficiently atoms are used — it does not consider costs of raw materials, energy, or product value. A reaction with 100% atom economy could still be expensive to run
Wrong: If you halve the volume of a solution, you halve the concentration Correct: If you remove solvent (e.g. by evaporation), the concentration increases because the same amount of solute is now in a smaller volume. The amount of solute stays the same but the concentration = amount/volume increases
6-Mark Extended Question
Yield vs Atom Economy
Explain the difference between percentage yield and atom economy. Why might a reaction with high yield have low atom economy?
Percentage yield measures the actual mass of product obtained compared to the theoretical maximum. It is affected by experimental factors such as product losses during transfer, side reactions, and reversible reactions not going to completion. Atom economy measures the proportion of reactant atoms that end up in the desired product rather than in by-products. It is determined entirely by the balanced equation. [2 marks]
The two measures are independent. A reaction can have a high percentage yield but a low atom economy. For example, in the production of aspirin from salicylic acid, the yield might be 85%, but the atom economy may be low because the reaction also produces a by-product (ethanoic acid) that is not wanted. [2 marks]
A reaction with high yield but low atom economy means most of the reactant is converted to products (good yield), but a large fraction of those products are unwanted by-products (poor atom economy). This generates a lot of chemical waste, making the process unsustainable. For a sustainable industrial process, both percentage yield and atom economy should be as high as possible. [2 marks]
AO3: Analyse and Evaluate
Evaluating Sustainable Processes
Three different methods are used to produce ethylene oxide (Mr = 44), an important industrial chemical. Method A has a percentage yield of 80% and an atom economy of 25%. Method B has a percentage yield of 60% and an atom economy of 100%. Method C has a percentage yield of 70% and an atom economy of 80%. Evaluate which method is the most sustainable and explain your reasoning.
Method A produces 80% of the theoretical maximum product, but only 25% of the reactant atoms end up in the desired product — 75% becomes waste. This is very wasteful despite the high yield. Method B has 100% atom economy (no waste at all — all atoms become product), but only 60% yield. Method C is a compromise: 70% yield and 80% atom economy. The most sustainable method depends on what matters more: minimising waste (Method B, 100% atom economy) or maximising product output per unit of reactant (Method A, highest yield). In practice, Method B is the most sustainable because 100% atom economy means zero waste, and the lower yield can potentially be improved by optimising reaction conditions.
📝 Exam Questions by Topic
🎬 Video Resources
Ready to ace your GCSE Chemistry exams?
Get the best revision books and guides to boost your grades.