Atom economy — AQA GCSE Chemistry
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Atom economy explained
Atom economy covers 7 specification points, including The atom economy (atom utilisation) is a… and It is important for sustainable development….
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The atom economy (atom utilisation) is a measure of the amount of starting materials that end up as useful products.
Atom economy, also called atom utilisation, measures how much of the mass of the reactants becomes the desired product rather than waste or by-products. It is calculated from the balanced equation, not from an experiment, so it is a theoretical measure of a reaction's efficiency. A reaction with high atom economy converts most reactant atoms into the useful product; addition reactions typically have 100% atom economy because all reactant atoms appear in the single product, while substitution and elimination reactions often produce by-products and therefore lower atom economy. For example, in an addition reaction such as ethene reacting with hydrogen to form ethane, all atoms become ethane, so the atom economy is 100%. Atom economy differs from percentage yield: yield depends on how much product is actually collected, whereas atom economy depends only on the reaction pathway chosen.
It is important for sustainable development and for economic reasons to use reactions with high atom economy.
Choosing reactions with high atom economy supports sustainable development because fewer reactant atoms are wasted as by-products, so less raw material is consumed and less waste must be treated or disposed of. There are also economic benefits: raw materials cost money, and separating, purifying and disposing of unwanted by-products adds further cost and energy use. A process that converts most of its starting materials into the desired product uses resources more efficiently, produces less waste and is often cheaper to run. For example, an addition reaction that gives a single product avoids the cost of separating a mixture and the cost of disposing of a co-product. However, atom economy is only one factor; a company must also consider yield, rate, equilibrium position, energy demands, safety and the usefulness of any by-products before selecting a pathway.
The percentage atom economy of a reaction is calculated using the balanced equation for the reaction as follows: Relative f ormula mass o f desired product f rom equation × 100 Sum o f relative f ormula masses o f all reactants f rom equation Students should be able to: calculate the atom economy of a reaction to form a desired product from the balanced equation (HT only) explain why a particular reaction pathway is chosen to produce a specified product given appropriate data such as atom economy (if not calculated), yield, rate, equilibrium position and usefulness of by-products.
Percentage atom economy is calculated from the balanced equation using: percentage atom economy = (relative formula mass of desired product from equation ÷ sum of relative formula masses of all reactants from equation) × 100. Multiply each species by its balancing number before adding. For example, for N₂ + 3H₂ → 2NH₃, the desired product mass is 2 × 17 = 34 and the reactant total is 28 + 6 = 34, giving 100%. Higher tier students calculate this value; all students may explain pathway choice using given data. A pathway is chosen by weighing atom economy against yield, rate, equilibrium position and whether by-products are useful, so the highest atom economy is not automatically selected.
The concentration of a solution can be measured in mol/dm .
Note: This content officially belongs to section 4.3.4 Using concentrations of solutions in mol/dm³ and is strictly Higher Tier only. Concentration indicates how much solute is dissolved in a given volume. Measuring it in mol/dm³ is useful as chemical reactions depend on particle numbers. For example, dissolving 0.5 mol of sodium hydroxide and making the total volume up to 2 dm³ gives a concentration of 0.5 ÷ 2 = 0.25 mol/dm³. The volume used must be the total solution volume, not just the water added. A 1 mol/dm³ solution contains 1 mol of solute per 1 dm³ of solution. Since 1 dm³ = 1000 cm³, volumes in cm³ must be divided by 1000 before calculating. Always read units carefully.
The amount in moles of solute or the mass in grams of solute in a given volume of solution can be calculated from its concentration in mol/dm .
Note: This content officially belongs to section 4.3.4 Using concentrations of solutions in mol/dm³ and is strictly Higher Tier only. If you know a solution's concentration and volume, you can find the amount of solute in moles using amount = concentration × volume in dm³. From moles, you can find mass using mass = moles × molar mass. For example, 250 cm³ of 0.200 mol/dm³ NaOH contains 0.200 × 0.250 = 0.0500 mol. NaOH has molar mass 40.0 g/mol, so the mass is 0.0500 × 40.0 = 2.00 g. Always convert volume to dm³ first. The molar mass must be calculated from the formula using relative atomic masses. This skill links to the mole concept and is essential for titration calculations.
If the volumes of two solutions that react completely are known and the concentration of one solution is known, the concentration of the other solution can be calculated.
Note: This content officially belongs to section 4.3.4 Using concentrations of solutions in mol/dm³ and is strictly Higher Tier only. Titration calculations use concentrations in mol/dm³ and balanced equations to link reacting solutions. First calculate moles of the known solution using amount = concentration × volume in dm³. Then use the mole ratio to find moles of the unknown solution. Finally divide its moles by its volume in dm³ to find its concentration. For example, 25.0 cm³ of 0.100 mol/dm³ HCl neutralises 20.0 cm³ of NaOH. Moles of HCl = 0.100 × 0.0250 = 0.00250 mol. The equation HCl + NaOH → NaCl + H₂O shows a 1:1 ratio, so moles of NaOH = 0.00250 mol. Concentration of NaOH = 0.00250 ÷ 0.0200 = 0.125 mol/dm³.
Students should be able to explain how the concentration of a solution in mol/dm is related to the mass of the solute and the volume of the solution.
Concentration in mol/dm³ links solute mass to solution volume. Although the metadata lists this under atom economy, it directly addresses using concentrations of solutions in mol/dm³. To explain the relationship, first convert solute mass into moles using: moles = mass ÷ relative formula mass. Next, divide the moles by the volume in dm³ to find concentration. For a fixed solute mass, increasing the solution volume decreases concentration (dilution). Conversely, for a fixed volume, increasing solute mass increases concentration. For example, 4.0 g of NaOH (molar mass 40.0 g/mol) is 0.100 mol. Dissolved to make 500 cm³ (0.500 dm³) of solution, the concentration is 0.100 ÷ 0.500 = 0.200 mol/dm³.
Your focus
- Define atom economy and explain how it differs from percentage yield.
- Identify the desired product and all reactants from a balanced equation.
- Compare reaction pathways in terms of atom economy and by-products formed.
Show all 21 objectives
- Explain how high atom economy contributes to sustainable development.
- Describe the economic advantages of reducing by-product formation.
- Evaluate a reaction pathway using atom economy alongside other practical and economic factors.
- Calculate percentage atom economy from a balanced equation, including coefficients.
- Interpret atom economy values alongside yield, rate and equilibrium data.
- Justify the selection of a reaction pathway using several pieces of given evidence.
- Define concentration in mol/dm³ and identify the solute and solution volumes involved.
- Convert between cm³ and dm³ accurately in concentration calculations.
- Calculate concentration from a given amount in moles and volume of solution.
- Calculate the amount in moles of solute from concentration and volume.
- Convert between moles and mass using molar mass.
- Carry out multi-step concentration calculations accurately with correct units.
- Use a balanced equation to determine the mole ratio between two reacting solutions.
- Calculate the concentration of an unknown solution from titration volume and concentration data.
- Convert volumes correctly and present a clear multi-step calculation.
- Explain the relationship between the mass of a solute, the volume of the solution, and its concentration in mol/dm³.
- Convert the mass of a solute to moles and calculate the resulting concentration.
- Predict and explain how changing the mass of the solute or the volume of the solution affects the concentration.
Atom economy exam tips
Marking Points
- Define atom economy as the proportion of reactant mass that becomes the desired product, distinguishing it from percentage yield.
- Identify the desired product and all reactants from a balanced equation before any calculation.
- Explain that atom economy is a theoretical value determined by the equation and is independent of experimental losses.
- Compare two reaction pathways and state which has higher atom economy from the balanced equations, linking this to fewer unwanted by-products.
- Recognise that addition reactions generally have higher atom economy than substitution or elimination reactions because no atoms are wasted.
- Explain that high atom economy reduces waste because a greater proportion of reactant atoms become the desired product.
- Link high atom economy to sustainable development by describing reduced consumption of raw materials and reduced waste disposal.
- Link high atom economy to economic benefits such as lower raw material costs and lower separation, purification and disposal costs.
- Recognise that atom economy is one of several factors, including yield, rate, equilibrium position and by-product usefulness, that influence pathway choice.
- Apply the argument to a specific reaction by identifying the by-product that would be avoided or the cost that would be reduced.
- Write the balanced equation and identify the desired product before calculating.
- Calculate the relative formula mass of the desired product and multiply by its balancing number.
- Sum the relative formula masses of all reactants, each multiplied by its balancing number, for the denominator.
- Substitute into percentage atom economy = (desired product mass ÷ total reactant mass) × 100 and evaluate the result.
- Explain pathway choice by comparing atom economy with yield, rate, equilibrium position and the usefulness of by-products, using the data provided.
- State that concentration in mol/dm³ is the amount of solute in moles dissolved per cubic decimetre of solution.
- Use the relationship concentration = amount in moles ÷ volume in dm³, and rearrange it correctly when needed.
- Convert volumes between cm³ and dm³ by dividing or multiplying by 1000 before substituting.
- Recognise that the volume used is the total volume of the solution, not the volume of solvent added.
- Interpret a stated concentration such as 0.100 mol/dm³ as 0.100 mol of solute in every 1 dm³ of solution.
- Calculate moles of solute using amount = concentration × volume in dm³.
- Convert a volume in cm³ to dm³ by dividing by 1000 before multiplying by concentration.
- Calculate molar mass from the chemical formula using relative atomic masses.
- Calculate mass of solute using mass = moles × molar mass, with correct units.
- Show a clear sequence: volume conversion, moles, then mass, so each step can be followed.
- Calculate moles of the solution with known concentration using amount = concentration × volume in dm³.
- Use the balanced equation to identify the mole ratio between the two reactants.
- Apply the mole ratio to find moles of the second solution.
- Divide moles of the second solution by its volume in dm³ to calculate its concentration.
- Convert both volumes from cm³ to dm³ before any calculation.
- State that concentration in mol/dm³ represents the number of moles of solute dissolved per cubic decimetre of solution.
- Explain the conversion of solute mass to moles using the formula: moles = mass ÷ relative formula mass.
- Describe how to calculate concentration using the formula: concentration = moles ÷ volume in dm³.
- Explain that for a fixed mass of solute, increasing the volume of the solution decreases the concentration.
- Explain that for a fixed volume of solution, increasing the mass of the solute increases the concentration.
Examiner Tips
- 💡Write the balanced equation first and label the desired product clearly before calculating anything.
- 💡Use the term atom utilisation if it appears in the question, showing you recognise it as a synonym for atom economy.
- 💡When comparing pathways, quote the atom economy values or the by-products formed to justify your choice rather than asserting one is better.
- 💡Use the phrase sustainable development and then explain it in terms of conserving raw materials and reducing waste.
- 💡Give both an environmental and an economic reason when a question asks why high atom economy is important.
- 💡Refer to the specific by-product named or implied in the question rather than discussing waste in general terms.
- 💡Show the relative formula mass of each species separately before combining them, so arithmetic errors are easy to spot and method marks can be awarded.
- 💡Check that the numerator and denominator use the same units and that the final answer is a percentage.
- 💡For pathway-choice questions, structure the answer around each piece of data given, stating whether it favours the pathway and why.
- 💡Write the equation concentration = moles ÷ volume in dm³ at the start of a calculation so the rearrangement is clear.
- 💡Check the unit of every volume and convert to dm³ before substituting values.
- 💡Give the unit mol/dm³ with your final answer, and round only at the end of the calculation.
- 💡Write down the values you know with units, then choose the equation that links them.
- 💡Convert cm³ to dm³ in a separate step so you do not lose track of the factor of 1000.
- 💡Check that your final mass is sensible: a small volume of dilute solution should give a small mass.
- 💡Write the balanced equation and state the mole ratio before doing any arithmetic.
- 💡Label each calculated amount clearly, for example 'moles of HCl' and 'moles of NaOH'.
- 💡Convert all volumes to dm³ at the start and keep units with every number.
- 💡Always check the units given in the question; if volume is in cm³, convert it to dm³ immediately before using it in the concentration formula.
- 💡Use the step-by-step chain: mass to moles to concentration, showing your working clearly for each stage to secure method marks.
Common Mistakes
- Confusing atom economy with percentage yield; correct by stating that atom economy comes from the equation while yield comes from measured amounts.
- Including only some reactants in the denominator; correct by summing the relative formula masses of all reactants shown in the balanced equation.
- Assuming a high atom economy guarantees a high yield; correct by explaining that a reaction can be efficient in theory but slow, incomplete or difficult to separate in practice.
- Claiming that high atom economy alone guarantees a profitable process; correct by stating that yield, rate, energy costs and equilibrium position must also be considered.
- Treating sustainability as only an environmental issue; correct by explaining that reducing waste and raw material use also lowers costs.
- Assuming all by-products are useless; correct by noting that a by-product may be sold or used elsewhere, which can make a lower atom economy pathway acceptable.
- Ignoring balancing numbers when finding relative formula masses; correct by multiplying each relative formula mass by its coefficient in the equation.
- Placing the total reactant mass in the numerator; correct by keeping the desired product mass on top and the reactant total underneath.
- Choosing a pathway on atom economy alone; correct by discussing yield, rate, equilibrium position and by-product usefulness as well.
- Using the volume of water added instead of the total solution volume; correct this by making the solution up to the final volume in a volumetric flask before calculating.
- Forgetting to convert cm³ to dm³; correct this by dividing the volume in cm³ by 1000 before using the concentration equation.
- Writing the unit as mol/dm² or mol dm³; correct this by using mol/dm³, which means moles per cubic decimetre.
- Multiplying concentration by volume in cm³ without converting; correct this by dividing the volume by 1000 first.
- Using relative formula mass as if it were the mass in grams without multiplying by moles; correct this by applying mass = moles × molar mass.
- Adding relative atomic masses incorrectly when finding molar mass; correct this by using the formula and checking each element's subscript.
- Ignoring the mole ratio and assuming a 1:1 reaction; correct this by writing and balancing the equation first.
- Dividing by the volume in cm³ instead of dm³; correct this by converting both volumes to dm³.
- Using the volume of the known solution for the unknown solution; correct this by keeping each solution's volume with its own calculation.
- Forgetting to convert the volume from cm³ to dm³ before calculating; correct this by always dividing the volume in cm³ by 1000.
- Treating concentration as mass per volume without converting mass to moles; correct this by dividing the mass by the relative formula mass first to find moles.
- Believing that adding more solvent increases the concentration; correct this by explaining that adding solvent increases the total volume, which decreases the concentration.