Your revision route
What you’ll learn
- Explain conservation of mass and apparent mass changes in open systems.
- Calculate relative formula mass and percentage by mass, and calculate solute mass from a known concentration in g/dm³; Higher Tier: calculate concentration from solute mass and solution volume.
- Represent repeated measurements in a frequency table or dot plot, then estimate uncertainty using the mean, range and a half-range ± value.
- Higher Tier: convert between mass and moles and use balanced-equation mole ratios.
- Higher Tier: identify a limiting reactant and derive equation coefficients from masses.
- Separate Chemistry: calculate percentage yield and atom economy.
- Separate Chemistry Higher Tier: compare pathways, molar concentrations and gas volumes at rtp.
Build the big picture
Key ideas
A reaction rearranges atoms; it does not mint them
Balanced equations are less like decoration and more like customs records for atoms crossing a reaction arrow.
- The total number of each type of atom is the same before and after a chemical reaction.
- Use large coefficients to balance an equation; changing a subscript changes the substance itself.
- In a closed system, total reactant mass equals total product mass.
The punchline: Fix the count with coefficients, never by rewriting a formula.
Mr adds every atom shown in the formula
A bracket is a multiplier with excellent camouflage: miss it and every later calculation inherits the error.
- Obtain Mr by totalling every Ar contribution indicated by the chemical formula.
- Multiply atoms inside brackets by the subscript outside; Ca(OH)₂ contains two oxygen and two hydrogen atoms.
- In a balanced equation, total coefficient × Mr is equal on both sides: 2H₂ + O₂ → 2H₂O gives 2 × 2 + 32 = 2 × 18 = 36.
- Percentage by mass = total Ar of the chosen element in the formula ÷ Mr × 100.
The punchline: Expand the formula atom by atom before adding Ar values.
An open container has an invisible door
When a balance reading changes, ask whether a gas entered or escaped before accusing conservation of mass.
- A metal heated in air gains mass because oxygen from the surroundings joins it.
- A metal carbonate can lose measured mass because carbon dioxide escapes during decomposition.
- Use the balanced equation and particle model to identify which gaseous substance crossed the system boundary.
The punchline: Conservation applies to the complete system, including gases you did not weigh.
Higher Tier: the reliable route from mass to mass
A balanced equation compares particles, not grams. Moles are the bridge between laboratory mass and equation ratio.
- BalanceWrite correct formulae and balance the equation without altering subscripts.
- ConvertFind moles using amount = mass ÷ molar mass M in g/mol. M has the same numerical value as Mr.
- Use ratioApply the whole-number coefficients to find moles of the wanted substance.
- Return to massMultiply wanted moles by molar mass M in g/mol, numerically equal to Mr, and state grams.
Measurements arrive with a spread, not a halo
Repeated values do not become identical through optimism; their spread is evidence about uncertainty.
- Represent repeats with a frequency table or dot plot: put measurement on the horizontal axis and show one count or mark for every result.
- For 10.2, 10.4, 10.4 and 10.6 cm³, record 10.2: 1, 10.4: 2 and 10.6: 1; the clustered centre and full spread are now visible.
- Mean = 10.4 cm³ and range = 0.4 cm³; half-range uncertainty is ±0.2 cm³, so report 10.4 ± 0.2 cm³.
- Suitable repeats reveal random variation; a smaller range means more consistent results. Higher resolution reduces resolution uncertainty, but does not itself guarantee repeatability.
The punchline: Represent every repeat first; then report the mean, range and uncertainty and link improvements to their source.
Higher Tier: the mole counts chemical crowds
A mole lets a balance count particles indirectly, much as a dozen counts eggs without naming each egg.
- One mole contains 6.02 × 10²³ stated particles: atoms, molecules, ions or other specified entities.
- Mr is dimensionless. Molar mass M has units g/mol and the same numerical value, so n = m/M; GCSE questions often write n = m/Mr.
- State the particle type: one mole of CO₂ molecules and one mole of carbon atoms contain equal numbers of stated particles.
The punchline: Convert grams to moles before using the equation ratio.
Higher Tier: coefficients are mole ratios, not mass ratios
Two moles are equal chemical amounts, but their masses differ whenever their molar masses M differ.
- Use the balanced coefficients to relate moles of reactants and products.
- For a mass-to-mass calculation: divide by known molar mass M, apply the coefficient ratio, then multiply by wanted M; each M in g/mol is numerically equal to Mr.
- Given masses of all substances, convert each to moles and simplify to find balancing numbers.
The punchline: Mass → moles → ratio → moles → mass.
Make the model move
Interactive checkpoint
Touch the science. Change a state, build a route or test a relationship.
Cross the mole bridge
Higher Tier: how many moles are in a sample?
Change the mass and molar mass M. Its numerical value equals Mr.
n = m ÷ M
Amount1 mol
Higher Tier: Mr is dimensionless. Molar mass M has units g/mol and the same numerical value, so n = m ÷ M; GCSE questions often write n = m ÷ Mr.
Find the missing gas
Why can the measured mass change?
Switch between open and closed systems and follow any gaseous reactant or product.
Oxygen from outside joins the metal. The product contains the original metal plus oxygen that was not initially on the balance.
1 of 3 states explored
Atoms are conserved. A balance reading changes only when matter crosses the boundary of the measured system.
Higher Tier: the smallest batch count calls time
Moles per coefficient reveal the limiting reactant; equal batch counts reveal a perfectly matched mixture instead.
- Convert each reactant to moles, then calculate n ÷ coefficient to compare equation-ratio batches.
- When the values differ, the smaller value identifies the limiting reactant; it is used up and fixes the maximum product.
- Equal values mean an exact stoichiometric mixture: both reactants are exhausted together and neither is in excess.
The punchline: Compare n ÷ coefficient; equal values mean a tie, not an excess.
Concentration measures how much solute occupies a volume
Mass concentration tells you how many grams of solute are present in each cubic decimetre of solution.
- Concentration in g/dm³ = mass of solute in g ÷ volume of solution in dm³.
- Rearrange to mass of solute = concentration × volume; g/dm³ × dm³ gives mass in g.
- Convert cm³ to dm³ by dividing by 1000 before using g/dm³.
- Higher Tier: increasing solute mass at fixed volume raises concentration; increasing volume at fixed mass lowers it.
The punchline: Match the volume unit to the concentration unit before substituting.
Separate Chemistry: yield measures distance from the upper bound
In percentage-yield work, the equation predicts a maximum product; real experiments can mislay product through transfer, incomplete reaction and competing chemistry.
- Percentage yield = actual product ÷ maximum theoretical product × 100.
- Yield may be below 100% because a reversible reaction is incomplete, product is lost, or side reactions occur.
- Separate Chemistry HT: within percentage-yield work, use the shared Higher-Tier reacting-mass method to calculate the maximum theoretical product mass.
The punchline: Put the amount actually collected over the maximum theoretically possible.
Separate Chemistry: two percentages answer different questions
Yield describes what the experiment collected. Atom economy describes where the equation sends the reactant atoms.
- Percentage yieldActual product ÷ maximum theoretical product × 100; losses and incomplete reaction can lower it.
- Atom economy(Coefficient × Mr) of desired product ÷ total (coefficient × Mr) of reactants × 100; unwanted products lower it.
- Pathway choiceSeparate Chemistry HT: also consider rate, equilibrium, energy and useful by-products when data are supplied.
Know which calculation belongs to which route
The public guide shows the whole map, so route labels prevent a separate-only calculation becoming false revision advice.
- All routesConservation, Mr, gas-related mass changes, uncertainty and concentration in g/dm³.
- Combined and separate HTUse mole ratios to obtain a reactant or product mass from one supplied mass; also derive coefficients, identify limiting reactants and explain concentration relationships.
- Separate ChemistryPercentage yield and atom economy at both tiers.
- Separate HTWithin percentage-yield work, calculate the maximum theoretical product mass; also pathway choice, mol/dm³ and gas-volume calculations.
Separate Chemistry: atom economy follows every atom
A high-yield reaction can still manufacture a mountain of unwanted product; atom economy exposes that hidden traffic.
- Atom economy = (coefficient × Mr) of desired product ÷ total (coefficient × Mr) of reactants × 100.
- High atom economy can reduce waste, resource use and separation costs.
- Separate Chemistry HT: use supplied yield, rate, equilibrium and by-product data when comparing pathways.
The punchline: Yield asks how much you collected; atom economy asks where the atoms went.
Separate Chemistry HT: solutions can be counted in moles
Molar concentration replaces grams with chemical amount, making reacting-solution ratios visible.
- Concentration in mol/dm³ = amount in mol ÷ volume in dm³.
- Amount = concentration × volume; convert cm³ to dm³ before calculating.
- Solute mass = concentration × volume × molar mass M; M is in g/mol and numerically equal to Mr.
- When two solutions meet in stoichiometric amounts, transfer from known moles through the equation ratio to obtain the unknown concentration.
The punchline: Convert volume, find moles, then apply the balanced ratio.
Separate Chemistry HT: one mole of gas occupies 24 dm³ at rtp
At the same temperature and pressure, equal mole amounts of different gases occupy equal volumes.
- At room temperature and pressure, 20 °C and 1 atmosphere, one mole of gas occupies 24 dm³.
- Gas volume at rtp = amount in moles × 24 dm³, or × 24,000 cm³.
- Use equation coefficients to relate gaseous reactant and product volumes under the same conditions.
The punchline: Choose 24 dm³ or 24,000 cm³ and keep that unit throughout.
Words worth knowing
Key definitions
- relative formula mass (Mr)
- A dimensionless total obtained from the Ar contribution of every atom represented in a formula.
- Higher Tier: molar mass (M)
- Mass per mole of a substance, with units g/mol; its numerical value equals the substance's dimensionless Mr.
- Higher Tier: mole
- The amount of substance containing 6.02 × 10²³ specified particles.
- Higher Tier: Avogadro constant
- 6.02 × 10²³ particles per mole.
- Higher Tier: limiting reactant
- When reactants are not in exact stoichiometric proportion, the one with the smaller n ÷ coefficient value; it is used up first and caps product.
- Higher Tier: excess reactant
- A reactant present in more than the balanced equation requires, so some remains.
- Separate Chemistry: yield
- The amount of product obtained from a reaction.
- Separate Chemistry: atom economy
- The percentage of total reactant mass represented by the desired product in the balanced equation.
- concentration
- The mass of solute per unit volume of solution; Separate Chemistry Higher Tier also uses amount in moles per dm³.
- Separate Chemistry Higher Tier: room temperature and pressure (rtp)
- The specified conditions 20 °C and 1 atmosphere, where one mole of gas occupies 24 dm³.
Calculate with confidence
Equations
Higher Tier: amount from mass
n = m ÷ M
Amount in moles equals mass in grams divided by molar mass in g/mol.
| Symbol | Meaning | Unit |
|---|---|---|
| n | amount | mol |
| m | mass | g |
| M | molar mass, numerically equal to Mr | g/mol |
Exam tip: Higher Tier: GCSE questions often write m/Mr; use Mr's numerical value as M in g/mol and preserve the unrounded mole value.
Mass of solute from concentration
m = c × V
Solute mass equals mass concentration multiplied by solution volume.
| Symbol | Meaning | Unit |
|---|---|---|
| m | solute mass | g |
| c | mass concentration | g/dm³ |
| V | solution volume | dm³ |
Exam tip: Convert cm³ to dm³ first; g/dm³ multiplied by dm³ gives g.
Separate Chemistry: percentage yield
% yield = actual yield ÷ theoretical yield × 100
The proportion of the maximum predicted product that is actually obtained.
| Symbol | Meaning | Unit |
|---|---|---|
| actual | product obtained | same unit as theoretical |
| theoretical | maximum predicted product | same unit as actual |
Exam tip: Separate Chemistry: actual goes on top; the result should normally not exceed 100%.
Separate Chemistry: percentage atom economy
% atom economy = (coefficient × Mr of desired product) ÷ sum of (coefficient × Mr for all reactants) × 100
The equation-based percentage of reactant mass ending in the desired product, including every balanced coefficient.
Exam tip: Separate Chemistry: multiply the desired-product Mr and every reactant Mr by their balanced coefficients.
Separate Chemistry Higher Tier: molar concentration
c = n ÷ V
Amount concentration equals moles of solute divided by solution volume.
| Symbol | Meaning | Unit |
|---|---|---|
| c | amount concentration | mol/dm³ |
| n | amount | mol |
| V | solution volume | dm³ |
Exam tip: Separate Chemistry HT: convert cm³ to dm³ first.
Separate Chemistry Higher Tier: gas volume at rtp
V = n × 24
At rtp, gas volume in dm³ equals amount in moles multiplied by 24 dm³/mol.
| Symbol | Meaning | Unit |
|---|---|---|
| V | gas volume | dm³ |
| n | amount | mol |
Exam tip: Separate Chemistry HT: multiply by 24,000 instead when the answer is required in cm³.
Separate Chemistry Higher Tier: mass from molar concentration
m = c × V × M
Solute mass equals molar concentration multiplied by volume in dm³ and molar mass in g/mol.
| Symbol | Meaning | Unit |
|---|---|---|
| m | solute mass | g |
| c | molar concentration | mol/dm³ |
| V | solution volume | dm³ |
| M | molar mass, numerically equal to Mr | g/mol |
Exam tip: Convert cm³ to dm³, calculate n = cV, then use m = nM; M in g/mol has the same numerical value as Mr.
Follow it step by step
Processes to remember
How to balance an equation
- Write correct formulae for every substance.
- Count each element on both sides.
- Add whole-number coefficients until every count matches.
- Reduce to the smallest whole-number ratio and check again.
Exam tip: Do not change subscripts: that invents different substances.
Higher Tier: how to solve a reacting-mass question
- Balance the equation and calculate each required Mr; use the same numerical value for molar mass M in g/mol.
- Convert the given mass to moles using n = m/M.
- Apply the coefficient ratio to obtain moles of the wanted substance.
- Convert wanted moles using m = nM and state grams.
Exam tip: The coefficient ratio acts on moles, not directly on masses.
Separate Chemistry HT: how to compare pathways
- Identify the desired product and the data supplied.
- Compare atom economy, yield and reaction rate without treating them as synonyms.
- Use equilibrium position, energy demand and usefulness of by-products where given.
- Make a supported choice and state the trade-off.
Exam tip: There may be no universally best route; justify the choice from the supplied criteria.
Separate Chemistry: worked atom-economy example
- 2Na + 2H₂O → 2NaOH + H₂; the desired product is NaOH. Ar(Na) = 23, Ar(H) = 1 and Ar(O) = 16.
- Equation mass of desired NaOH = 2 × Mr(NaOH) = 2 × 40 = 80.
- Total reactant equation mass = (2 × 23) + (2 × 18) = 82.
- Atom economy = 80 ÷ 82 × 100 = 97.6%.
Exam tip: Answer: 97.6%. Include every balanced coefficient; the hydrogen product is not desired.
Separate Chemistry HT: worked reacting-solutions example
- H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O. 20.0 cm³ of 0.150 mol/dm³ H₂SO₄ reacts exactly with 30.0 cm³ KOH.
- Convert 20.0 cm³ to 0.0200 dm³; n(H₂SO₄) = cV = 0.150 × 0.0200 = 0.00300 mol.
- The 1:2 ratio gives n(KOH) = 2 × 0.00300 = 0.00600 mol.
- Convert 30.0 cm³ to 0.0300 dm³; c(KOH) = n/V = 0.00600 ÷ 0.0300 = 0.200 mol/dm³.
Exam tip: Answer: 0.200 mol/dm³. Convert both volumes, then apply the coefficient ratio before dividing by the unknown volume.
Separate Chemistry HT: worked mass-to-gas example
- CaCO₃ → CaO + CO₂. At rtp, find the CO₂ volume from 5.0 g CaCO₃; M(CaCO₃) = 100 g/mol.
- n(CaCO₃) = 5.0 g ÷ 100 g/mol = 0.050 mol.
- The 1:1 coefficient ratio gives n(CO₂) = 0.050 mol.
- V(CO₂) = n × 24 = 0.050 × 24 = 1.2 dm³ at rtp.
Exam tip: Answer: 1.2 dm³. Convert mass to moles, apply the coefficient ratio, then use 24 dm³/mol.
See the thinking
Worked example
Worked example: limiting reactant to product mass
Higher Tier: 28 g N₂ reacts with 3.0 g H₂. N₂ + 3H₂ → 2NH₃. Using Ar(N) = 14 and Ar(H) = 1, identify the limiting reactant and calculate the maximum ammonia mass.
- Mr(N₂) = 28, Mr(H₂) = 2 and Mr(NH₃) = 17, so their molar masses M are 28, 2 and 17 g/mol respectively.
- n(N₂) = 28 g ÷ 28 g/mol = 1.0 mol; n(H₂) = 3.0 g ÷ 2 g/mol = 1.5 mol.
- Equation-ratio batches: N₂ gives 1.0 ÷ 1 = 1.0; H₂ gives 1.5 ÷ 3 = 0.50, so H₂ is limiting.
- The 3:2 ratio gives n(NH₃) = 1.5 × 2/3 = 1.0 mol.
- m(NH₃) = nM = 1.0 mol × 17 g/mol = 17 g; 0.50 mol N₂, or 14 g, remains in excess.
Answer: H₂ is limiting, so the maximum ammonia mass is 17 g; 14 g N₂ remains in excess.
The smaller n ÷ coefficient value identifies the limiter. Equal values would instead mean an exact stoichiometric mixture: both reactants would finish together and neither would be in excess.
Protect the marks
Common mistakes
Watch out: Changing a subscript to balance an equation.
Do this instead: Keep formulae fixed and add coefficients in front.
Watch out: Saying mass disappears when gas escapes.
Do this instead: The gas carries mass outside the measured open system; total mass remains conserved.
Watch out: Higher Tier: applying equation coefficients directly to grams.
Do this instead: Higher Tier: convert grams to moles before using the coefficient ratio.
Watch out: Higher Tier: choosing the smaller reactant mass as limiting.
Do this instead: Higher Tier: compare n ÷ coefficient. The smaller value is limiting; equal values mean both finish together and neither is in excess.
Watch out: Using cm³ in an equation requiring dm³.
Do this instead: Divide cm³ by 1000 before using g/dm³. Separate Chemistry HT: the same conversion applies to mol/dm³.
Watch out: Separate Chemistry: treating yield and atom economy as the same percentage.
Do this instead: Yield uses actual versus theoretical product; atom economy follows equation mass into the desired product.
Try it before you move on
Quick check
Say your answer first, then open the card to check it.
Calculate Mr of CaCO₃ using Ar values Ca = 40, C = 12 and O = 16.
Answer: 100
Mr = 40 + 12 + (3 × 16) = 100; relative formula mass has no unit.
Balance this equation using the smallest whole-number coefficients: Al + O₂ → Al₂O₃.
Answer: 4Al + 3O₂ → 2Al₂O₃.
Four Al atoms and six O atoms appear on each side; the formula subscripts stay unchanged.
MgO has Ar(Mg) = 24 and Ar(O) = 16. What percentage by mass is oxygen?
Answer: 40%
Mr(MgO) = 24 + 16 = 40. Oxygen contributes 16, so percentage by mass = 16 ÷ 40 × 100 = 40%.
Results are 10.2, 10.4, 10.4 and 10.6 cm³. Represent the distribution, then give the mean, range and half-range uncertainty.
Answer: Frequency table or dot plot: 10.2 cm³: 1 result; 10.4 cm³: 2 results; 10.6 cm³: 1 result. Mean 10.4 cm³; range 0.4 cm³; uncertainty ±0.2 cm³.
Put measurement on the horizontal axis and one mark per result. Mean = 41.6 ÷ 4 = 10.4 cm³; range = 10.6 − 10.2 = 0.4 cm³, so report 10.4 ± 0.2 cm³.
Higher Tier: 5.4 g Al reacts with 4.8 g O₂ to form Al₂O₃. M values are 27, 32 and 102 g/mol. Derive the coefficients.
Answer: 4Al + 3O₂ → 2Al₂O₃.
n(Al) = 0.20 mol and n(O₂) = 0.15 mol. Product mass = 10.2 g, so n(Al₂O₃) = 0.10 mol. Divide 0.20:0.15:0.10 by 0.05 to get 4:3:2.
Higher Tier: 0.50 mol H₂ reacts with 0.25 mol O₂ in 2H₂ + O₂ → 2H₂O. Which is limiting, and how much water forms?
Answer: The batch counts tie: both finish together, neither is in excess, and 0.50 mol H₂O forms.
H₂ gives 0.50 ÷ 2 = 0.25 batches; O₂ gives 0.25 ÷ 1 = 0.25. Equal values mean an exact mixture, and the 2:2 ratio gives 0.50 mol H₂O.
Higher Tier: what is the concentration of 5.0 g solute in 250 cm³ of solution?
Answer: 20 g/dm³
250 cm³ = 0.250 dm³, so c = 5.0/0.250 = 20 g/dm³.
A solution has concentration 12 g/dm³ and volume 250 cm³. What solute mass does it contain?
Answer: 3.0 g
250 cm³ = 0.250 dm³, so m = cV = 12 × 0.250 = 3.0 g.
Separate Chemistry: 8 g is collected from a theoretical 10 g. What is the yield?
Answer: 80%
Percentage yield = 8/10 × 100 = 80%.
Separate Chemistry: CaCO₃ → CaO + CO₂. Mr values are 100, 56 and 44. What is the atom economy for desired CaO?
Answer: 56%
The desired-product equation mass is 1 × 56 and total reactant equation mass is 1 × 100, so atom economy = 56 ÷ 100 × 100 = 56%.
Separate Chemistry HT: 25.0 cm³ of 0.100 mol/dm³ HCl neutralises 20.0 cm³ NaOH. Find c(NaOH); the ratio is 1:1.
Answer: 0.125 mol/dm³
n(HCl) = 0.100 × 0.0250 = 0.00250 mol. The 1:1 ratio gives 0.00250 mol NaOH; c = n/V = 0.00250 ÷ 0.0200 = 0.125 mol/dm³.
Separate Chemistry HT: 2H₂O₂ → 2H₂O + O₂. What O₂ volume at rtp forms from 6.8 g H₂O₂, M = 34 g/mol?
Answer: 2.4 dm³
n(H₂O₂) = 6.8 g ÷ 34 g/mol = 0.200 mol. The 2:1 ratio gives 0.100 mol O₂; V = 0.100 × 24 = 2.4 dm³ at rtp.
Good questions, clear answers
Frequently asked questions
Why can measured mass change if mass is conserved?
In an open system, a gas may enter or escape. Include that gas and the total mass of all reactants still equals the total mass of all products.
Higher Tier: why are equation coefficients not mass ratios?
Coefficients compare particles and therefore moles. Different substances have different molar masses M, numerically equal to their Mr values, so equal mole amounts usually have different masses.
Separate Chemistry: what is the difference between yield and atom economy?
Yield compares actual product with the theoretical maximum. Atom economy uses the balanced equation to measure how much reactant mass becomes the desired product.
Which quantitative content is Higher Tier?
For Combined and separate routes, moles, reacting masses, coefficients from masses, limiting reactants and the concentration relationship are Higher Tier.
Which calculations are separate Chemistry Higher Tier only?
For percentage yield, calculating the maximum theoretical product mass is separate Chemistry HT. Pathway choice, mol/dm³ concentration and gas volumes at rtp are too. General product-mass calculations from a given reactant mass are HT for Combined and separate routes.
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