This is the topic where chemistry turns into calculation. It all rests on conservation of mass: because no atoms are lost or made during a reaction, the masses on both sides of an equation must add up. From there you can predict how much product a reaction will make, judge how efficient an industrial process is, or find the formula of an unknown compound. It’s also where careless working costs the most marks, so being methodical with your units and steps matters more here than almost anywhere else.
This is one of the most tier-dependent topics in the course, so check what your tier needs:
- Everyone — conservation of mass, balanced equations, relative formula mass, mass changes involving a gas, uncertainty, and concentration in g/dm³.
- Higher Tier H — the mole, reacting-mass calculations, balancing from masses, and limiting reactants.
- Triple (Chemistry only) T — percentage yield, atom economy, concentration in mol/dm³, titration calculations and gas volumes (these last two are also Higher Tier).
1Conservation of Mass
In a chemical reaction, the atoms in the reactants are rearranged to make the products. Bonds break and new bonds form, but not a single atom is lost or made — they are simply swapped around into new combinations.
The law of conservation of mass — no atoms are lost or made during a chemical reaction, so the total mass of the products is equal to the total mass of the reactants.
Because the same atoms are present before and after, a reaction can be written as a balanced symbol equation: one with the same number of atoms of each element on both sides.
🧪 Exam-style questions
In an experiment, 48 g of magnesium reacts completely with 32 g of oxygen. What mass of magnesium oxide is produced?
Show answer
- No atoms are lost or made, so the mass of product equals the total mass of the reactants: 48 + 32 = 80 g. 1 mark
2Balancing Equations
A balanced symbol equation has the same number of atoms of each element on both sides. Because no atoms are lost or made, we balance an equation by adjusting the big numbers in front of each substance — never by changing the substances themselves.
Reading the numbers in a formula
There are two kinds of number in equations, and they do different jobs:
- A big number in front of a formula (a multiplier, written in normal script) multiplies the whole formula. So 2H₂O means two water molecules — 4 hydrogen atoms and 2 oxygen atoms in total.
- A small subscript number only multiplies the atom (or bracket) immediately before it. So in H₂O the subscript 2 applies only to the hydrogen.
| Formula | Atoms in one formula unit |
|---|---|
| CO₂ | 1 carbon, 2 oxygen |
| Ca(OH)₂ | 1 calcium, 2 oxygen, 2 hydrogen (the 2 multiplies everything in the bracket) |
| Ca(NO₃)₂ | 1 calcium, 2 nitrogen, 6 oxygen (2 × NO₃ = 2 N and 6 O) |
Remember that seven non-metals are written as diatomic molecules whenever they appear on their own: H₂, N₂, O₂, F₂, Cl₂, Br₂ and I₂.
Work across the equation one element at a time, and adjust only the big numbers in front — never change a subscript, because that would change the substance itself.
For example, hydrogen burning in oxygen balances to:
2H₂ + O₂ → 2H₂O
There are now 4 hydrogen atoms and 2 oxygen atoms on each side — balanced.
Balance the equation for methane (CH₄) burning completely in oxygen:
CH₄ + O₂ → CO₂ + H₂O
Work across one element at a time, changing only the big numbers in front:
- Carbon — 1 on each side. Already balanced.
- Hydrogen — 4 on the left (CH₄) but only 2 on the right (H₂O). Put a 2 in front of H₂O, giving 4 H on the right.
- Oxygen — the right now has 2 (in CO₂) + 2 (in 2H₂O) = 4 O, but the left has only 2 (O₂). Put a 2 in front of O₂, giving 4 O on the left.
- Re-check — C 1 = 1, H 4 = 4, O 4 = 4. ✓ Balanced.
CH₄ + 2O₂ → CO₂ + 2H₂O
🧱 Interactive — build the balanced equation
Balance the complete combustion of ethane
Change the big number in front of each molecule. That many molecules appear in the particle picture, and the atom tally updates. Keep adjusting until carbon, hydrogen and oxygen each have the same number of atoms on both sides. (Never change a small subscript — that would change the substance.)
| Element | Reactant side | Product side | |
|---|---|---|---|
| C | 2 | 1 | |
| H | 6 | 2 | |
| O | 2 | 3 |
Change the numbers so each element matches on both sides.
🧪 Exam-style questions
Balance the equation for aluminium burning in oxygen.Type a balancing number in each box (leave it as 1 if no number is needed), then press Check. Any correct set of numbers is accepted.
Show answer
4Al + 3O₂ → 2Al₂O₃
- Balancing aluminium and oxygen gives the numbers 4 : 3 : 2. 1 mark
- Check: Al 4 = 4 · O 6 = 6 ✓ (any equal multiple, e.g. 8 : 6 : 4, also balances)
3Mass Changes Involving a Gas
Sometimes a reaction looks as though it breaks the law of conservation of mass — the mass on the balance goes up or down. This almost always happens because one of the reactants or products is a gas in an open (non-enclosed) container, so its mass is not being measured.
The two classic cases
Mass appears to increase — when a metal reacts with oxygen, oxygen from the air is added to the solid, so the metal oxide weighs more than the metal did:
2Mg + O₂ → 2MgO
Mass appears to decrease — when a metal carbonate is heated (thermal decomposition), carbon dioxide gas escapes into the air, leaving only the solid metal oxide behind:
CaCO₃ → CaO + CO₂
In a closed system (a sealed container), no gas can enter or leave, so the total mass always stays exactly the same — even when a gas is made.
Mass is never actually lost or created — atoms are not destroyed. You must explain the change in terms of the particle model: the gas particles are still there, they have just left (or joined) the substance on the balance. Saying “the mass disappeared” or “atoms were lost” earns no marks.
A carbonate (marble chips) reacts with dilute acid in an open test tube.
Green copper carbonate is heated in an open crucible until it fully decomposes:
CuCO₃ → CuO + CO₂
The mass of the solid in the crucible decreases. Explain why, in terms of the particle model.
Model answer: One of the products, carbon dioxide, is a gas. In an open crucible the CO₂ molecules escape into the air, so they are no longer sitting on the balance. The mass that is measured therefore falls. No atoms are destroyed — if the gas were trapped in a sealed container, the total mass would stay exactly the same.
🧪 Exam-style questions
A piece of magnesium is heated strongly in an open crucible, where it reacts with oxygen from the air.
The mass of the contents of the crucible increases. Explain this increase in terms of the particle model.
Show answer
- Oxygen molecules from the air combine with (bond to) the magnesium to form magnesium oxide. 1 mark
- The mass of this added oxygen is now part of the solid, so the mass measured on the balance increases. 1 mark
- No atoms are created — the extra mass comes from oxygen that was already present in the air.
Which statement explains why the mass appears to decrease when a metal carbonate is thermally decomposed in an open container?Tick (✓) one box.
4Relative Formula Mass (Mr)
Relative atomic mass (Ar) — the average mass of the atoms of an element, compared with carbon-12. It is the larger of the two numbers given for each element on the periodic table.
Relative formula mass (Mr) — the sum of the relative atomic masses of all the atoms shown in the formula of a substance.
Mr = sum of (Ar × number of atoms) for every element in the formula
For example, water (H₂O): (2 × 1) + 16 = 18. Calcium hydroxide, Ca(OH)₂: 40 + (2 × 16) + (2 × 1) = 74.
Because mass is conserved, the sum of the Mr of the reactants equals the sum of the Mr of the products in a balanced equation (remembering to multiply each by its big number).
Percentage by mass of an element (supporting maths — not a separately listed 4.3 point, but can come up in calculations)
You can use Mr to work out what fraction of a compound's mass comes from one particular element:
percentage mass of an element = Ar × number of atoms of that elementMr of the compound × 100
The relative atomic mass of every element is on the periodic table you are given — it is always the bigger of the two numbers. And don't forget the final × 100 when finding a percentage.
Find the relative formula mass of calcium carbonate, CaCO₃, and then the percentage by mass of calcium it contains. (Ar: C = 12, O = 16, Ca = 40.)
Step 1 — add up the Ar values to get Mr:
Mr(CaCO₃) = 40 + 12 + (3 × 16) = 40 + 12 + 48 = 100
Step 2 — put the total mass of calcium over the Mr, then × 100:
percentage of Ca = 40100 × 100 = 40%
🧪 Exam-style questions
Calculate the percentage by mass of magnesium in magnesium carbonate, MgCO₃.Use Ar: C = 12, O = 16, Mg = 24.
Show answer
- Mr(MgCO₃) = 24 + 12 + (3 × 16) = 84, and the mass of magnesium is 24.
- percentage of Mg = 2484 × 100 1 mark = 28.6% (to 3 s.f.) 1 mark
28.57%, 28.6% and 29% are all accepted — any correct rounding earns the mark when the question does not fix the number of figures.
What is the relative formula mass (Mr) of calcium hydroxide, Ca(OH)₂?Use Ar: H = 1, O = 16, Ca = 40. Tick (✓) one box.
5Chemical Measurements & Uncertainty
Whenever you take a measurement there is always some uncertainty — no reading is ever perfectly exact. To deal with this, scientists repeat their measurements, discard any anomalous results, and calculate a mean.
The range of the repeated readings (how spread out they are) is used as a simple measure of the uncertainty:
uncertainty = ± range2 where range = largest − smallest reading
A result is then written as the mean ± the uncertainty — for example, 25.40 ± 0.05 cm³.
Always leave out any obviously anomalous reading before you calculate the mean — the mean is taken only from the concordant (close-together) results.
A student measures the volume of acid used in a titration four times and records: 24.00, 24.10, 24.05 and 23.50 cm³. Find the mean volume and the uncertainty.
- Spot the anomaly — 23.50 cm³ lies well outside the tight cluster of the other three, so it is left out before averaging.
- Mean of the concordant results = (24.00 + 24.10 + 24.05) ÷ 3 = 72.15 ÷ 3 = 24.05 cm³.
- Uncertainty — range = 24.10 − 24.00 = 0.10, so uncertainty = ± 0.102 = ± 0.05 cm³.
The result is written as 24.05 ± 0.05 cm³.
🧪 Exam-style questions
A student records four titres: 12.30, 12.34, 12.32 and 11.90 cm³. Ignoring any anomalous result, calculate (a) the mean and (b) the uncertainty.
Show answer
- 11.90 cm³ is anomalous, so it is left out.
- (a) mean = (12.30 + 12.34 + 12.32) ÷ 3 = 36.96 ÷ 3 = 12.32 cm³ 1 mark
- (b) range = 12.34 − 12.30 = 0.04, so uncertainty = ± 0.04 ÷ 2 = ± 0.02 cm³ 1 mark
Final result: 12.32 ± 0.02 cm³.
A student repeats a measurement four times and records: 27.5, 27.4, 25.0 and 27.6 cm³. Which result is anomalous and should be left out before calculating the mean?Tick (✓) one box.
6The Mole H
Atoms are far too small and too numerous to count one by one, so chemists count them in huge, fixed-size groups called moles — in the same way you might count eggs in dozens.
The mole (mol) — the unit for the amount of a substance. One mole of any substance contains the same number of particles (atoms, molecules or ions) as one mole of any other substance.
That number is the Avogadro constant: 6.02 × 10²³ per mole.
The link between moles and mass is the key to the whole topic:
The mass of one mole of a substance, in grams, is numerically equal to its relative formula mass (Mr). This is called the molar mass (units: g/mol).
So one mole of carbon (Ar = 12) has a mass of 12 g; one mole of water (Mr = 18) has a mass of 18 g.
number of moles = mass (g)Mr (g/mol)
Rearranging this one relationship lets you switch between mass and moles in either direction: mass = moles × Mr.
The measurement of amounts in moles applies to atoms, molecules, ions, electrons, formulae and equations. For example, one mole of carbon (C) contains the same number of atoms as the number of molecules in one mole of carbon dioxide (CO₂). Always show your working — you can pick up method marks even if the final number is wrong.
One mole of a compound contains 6.02 × 10²³ of that substance, but each compound holds more than one atom or ion — so you must scale each of these by its subscript.
For example, one mole of calcium chloride, CaCl₂, contains one mole of calcium ions (Ca²⁺) but two moles of chloride ions (Cl⁻), because every CaCl₂ unit contains two chlorides. So one mole of CaCl₂ gives 6.02 × 10²³ calcium ions and twice as many chloride ions (1.204 × 1024).
(a) Find the mass of 0.250 mol of zinc. (Ar Zn = 65.)
mass = moles × Mr = 0.250 × 65 = 16.25 g
(b) Find the number of moles in 2.64 g of carbon dioxide, CO₂. (Ar: C = 12, O = 16, so Mr = 44.)
moles = massMr = 2.6444 = 0.06 mol
That 0.06 mol contains 0.06 × (6.02 × 10²³) = 3.61 × 10²² molecules of CO₂.
🧪 Exam-style questions
Calculate the number of moles in 5.4 g of water, H₂O.Use Ar: H = 1, O = 16.
Show answer
- Mr(H₂O) = (2 × 1) + 16 = 18 1 mark
- moles = massMr = 5.418 = 0.3 mol 1 mark
Calculate the mass of 0.25 mol of sodium carbonate, Na₂CO₃.Use Ar: C = 12, O = 16, Na = 23.
Show answer
- Mr(Na₂CO₃) = (2 × 23) + 12 + (3 × 16) = 46 + 12 + 48 = 106 1 mark
- mass = moles × Mr = 0.25 × 106 = 26.5 g 1 mark
7Balancing Equations Using Masses H
The reverse is also possible: if you know the masses of every reactant and product, you can work out the balancing numbers (the coefficients) of the equation.
- Convert each mass into moles (mass ÷ Mr).
- Write the moles as a ratio.
- Simplify to the smallest whole numbers — these are the balancing numbers. (If you get values like 1 : 2 : 2.5, multiply them all by the same number to clear the decimal.)
You also need to be able to change the subject of the mole equation confidently, since these questions often ask you to rearrange before you substitute.
When methanol burns, 64 g of CH₃OH reacts with 96 g of O₂ to make 88 g of CO₂ and 72 g of H₂O. Deduce the balanced equation. (Mr: CH₃OH = 32, O₂ = 32, CO₂ = 44, H₂O = 18.)
Step 1 — convert every mass to moles:
| Substance | Mass ÷ Mr | Moles |
|---|---|---|
| CH₃OH | 64 ÷ 32 | 2 |
| O₂ | 96 ÷ 32 | 3 |
| CO₂ | 88 ÷ 44 | 2 |
| H₂O | 72 ÷ 18 | 4 |
Step 2 — write the mole ratio: 2 : 3 : 2 : 4 (already whole numbers).
Step 3 — these are the balancing numbers:
2CH₃OH + 3O₂ → 2CO₂ + 4H₂O
🧪 Exam-style questions
In the Haber process, 28 g of nitrogen (N₂) reacts with 6 g of hydrogen (H₂) to produce 34 g of ammonia (NH₃). Work out the moles of each substance, then complete the balanced equation.Use Mr: N₂ = 28, H₂ = 2, NH₃ = 17. Type the balancing numbers, then press Check.
Show answer
- moles: N₂ = 28 ÷ 28 = 1 · H₂ = 6 ÷ 2 = 3 · NH₃ = 34 ÷ 17 = 2 1 mark
- mole ratio N₂ : H₂ : NH₃ = 1 : 3 : 2 1 mark
- equation: N₂ + 3H₂ → 2NH₃ 1 mark (any equal multiple is accepted by the checker)
In a reaction, 0.5 mol of element A reacts with 1.5 mol of element B. What is the simplest whole-number mole ratio of A : B?Tick (✓) one box.
8Reacting Masses H
A balanced equation can be read in moles. For example:
Mg + 2HCl → MgCl₂ + H₂
tells you that 1 mole of magnesium reacts with 2 moles of hydrochloric acid to make 1 mole of magnesium chloride and 1 mole of hydrogen. These mole ratios let you calculate the mass of any product or reactant from the mass of another.
- Write the balanced equation (if it isn't already given).
- Convert the known mass into moles (mass ÷ Mr).
- Use the ratio in the equation to find the moles of the substance you want.
- Convert those moles back into a mass (moles × Mr).
Always work molar mass out from the formula, not from the big numbers in the equation. In 2MgO the molar mass of MgO is still 40 g/mol, not 80 — the 2 is a number of moles, not part of the formula.
Iron is extracted from iron(III) oxide in the blast furnace. Calculate the mass of iron produced from 160 g of iron(III) oxide, Fe₂O₃. (Ar: O = 16, Fe = 56.)
Draw a table under the equation. Put Ratio, Moles, Mass and Mr down the left, and the two substances you care about across the top. Write in what you already know, then work out the rest one box at a time.
Fe₂O₃ + 3CO → 2Fe + 3CO₂
| Fe₂O₃known | Fefind this | |
|---|---|---|
| Ratio | 11 | 12 |
| Moles | 41160 ÷ 160 | 521 × 2 |
| Mass | 2160 ggiven | 6112 g2 × 56 |
| Mr | 3160(2×56)+(3×16) | 356 |
Filling the table in order:
- Ratio — read the big numbers in the equation: Fe₂O₃ : Fe = 1 : 2.
- Mass you know — write 160 g under Fe₂O₃.
- Mr of each — work it out from the formula: Fe₂O₃ = (2×56)+(3×16) = 160; Fe = 56.
- Moles you know — mass ÷ Mr = 160 ÷ 160 = 1 mol.
- Moles you want — scale by the ratio (×2): 2 mol of Fe.
- Answer — moles × Mr = 2 × 56 = 112 g of iron.
Now work through a fresh example the same way. You can only fill one box at a time — each box unlocks once the one before it is correct. The prompt under the table tells you what to do next.
Calculate the mass of water produced when 4 g of methane, CH₄, burns completely in oxygen. (Ar: H = 1, C = 12, O = 16.)
CH₄ + 2O₂ → CO₂ + 2H₂O
| CH₄known | H₂Ofind this | |
|---|---|---|
| Ratio | ||
| Moles | ||
| Mass | 4 ggiven | g |
| Mr |
🧪 Exam-style questions
Calculate the mass of magnesium oxide formed when 6 g of magnesium burns completely in oxygen.
Show answer
- moles of Mg = 6 ÷ 24 = 0.25 mol 1 mark
- ratio Mg : MgO = 2 : 2 = 1 : 1, so moles of MgO = 0.25 mol 1 mark
- mass of MgO = 0.25 × 40 = 10 g 1 mark (Mr of MgO = 24 + 16 = 40)
Aluminium is extracted industrially by the electrolysis of aluminium oxide. Calculate the mass of aluminium produced from 51 tonnes of aluminium oxide, Al₂O₃.
Show answer
- Mr(Al₂O₃) = (2 × 27) + (3 × 16) = 54 + 48 = 102 1 mark
- moles of Al₂O₃ = 51 ÷ 102 = 0.5 1 mark (working in tonnes throughout)
- ratio Al₂O₃ : Al = 2 : 4 = 1 : 2, so moles of Al = 2 × 0.5 = 1.0 1 mark
- mass of Al = 1.0 × 27 = 27 tonnes 1 mark
The mole ratio works in any unit of mass, as long as you use the same unit on both sides.
9Limiting Reactants H
When two reactants are mixed, they are rarely present in the exact ratio the equation needs. Usually one is added in excess to make sure the other is completely used up.
The limiting reactant is the reactant that is completely used up in a reaction. It is called “limiting” because once it runs out the reaction stops, so it controls (limits) the amount of product that can form. The reactant left over is said to be in excess.
The amount of product is always proportional to the amount of the limiting reactant: use more of it (and keep the other in excess) and you make more product. To find which reactant is limiting, convert both to moles and compare them against the ratio in the balanced equation.
0.5 g of hydrogen reacts with 8 g of oxygen to form water. Identify the limiting reactant and calculate the mass of water formed. (Mr: H₂ = 2, O₂ = 32, H₂O = 18.)
2H₂ + O₂ → 2H₂O
- Convert both reactants to moles: H₂ = 0.5 ÷ 2 = 0.25 mol; O₂ = 8 ÷ 32 = 0.25 mol.
- Compare with the equation ratio (H₂ : O₂ = 2 : 1): 0.25 mol of H₂ needs only 0.125 mol of O₂. We have 0.25 mol of O₂ — more than enough — so O₂ is in excess and H₂ is the limiting reactant.
- Base the product on the limiting reactant: H₂ : H₂O = 2 : 2 = 1 : 1, so moles of H₂O = 0.25 mol. mass = 0.25 × 18 = 4.5 g.
Try It: Which Reactant Runs Out First?
Choose how many hydrogen (H₂) and oxygen (O₂) molecules to start with, then press React. They combine in the ratio 2H₂ + O₂ → 2H₂O — whatever is left over at the end is the reactant that was in excess, and the one that ran out is the limiting reactant.
These molecules drift freely. Choose how many H₂ and O₂ to start with, then press React.
🧪 Exam-style questions
Magnesium reacts with hydrochloric acid. 0.10 mol of magnesium is added to a solution containing 0.16 mol of HCl. Calculate the maximum mass of hydrogen gas that can be produced.
Show answer
- The ratio Mg : HCl is 1 : 2, so 0.10 mol of Mg would need 0.20 mol of HCl. Only 0.16 mol of HCl is present, so HCl is the limiting reactant. 1 mark
- ratio HCl : H₂ = 2 : 1, so moles of H₂ = 0.16 ÷ 2 = 0.08 mol 1 mark
- mass of H₂ = 0.08 × 2 = 0.16 g 1 mark (you must base the answer on the limiting reactant, not the magnesium)
0.3 mol of carbon is heated with 0.2 mol of oxygen. Which is the limiting reactant?
Tick (✓) one box.
10Concentration of Solutions
When a solid (the solute) dissolves in a liquid (the solvent) it forms a solution. The concentration tells you how much solute is packed into a given volume — the more solute in the same volume, the more concentrated the solution.
concentration (g/dm³) = mass of solute (g)volume of solution (dm³)
H Everyone needs to be able to calculate a concentration in g/dm³. Explaining how the concentration depends on the mass of solute and the volume of solution — for example, why dissolving the same mass in a larger volume gives a lower concentration — is a Higher Tier requirement.
Each unit of NaCl dissolves into one Na⁺ ion and one Cl⁻ ion.
Volumes in chemistry are usually measured in cm³, but concentration needs dm³. This conversion is one of the most common places students lose marks:
1 dm³ = 1000 cm³ = 1 litre. To go from cm³ → dm³, divide by 1000; to go from dm³ → cm³, multiply by 1000. Convert before you substitute, and sense-check: have you ended up with a bigger or smaller number than you expected?
Concentration in mol/dm³ T H
Concentration can also be measured in moles per dm³, which is far more useful for calculations because it links directly to the mole ratios in an equation:
concentration (mol/dm³) = amount of solute (mol)volume of solution (dm³)
The concentration–moles relationship is not printed on the exam paper, so you must learn it. It is a good idea to jot it down before you start a calculation so every part is in the right place.
(a) 10 g of sodium hydroxide is dissolved to make 2 dm³ of solution. Find the concentration in g/dm³.
concentration = massvolume = 102 = 5 g/dm³
(b) T H 20 g of sodium hydroxide, NaOH, is dissolved to make 500 cm³ of solution. Find the concentration in mol/dm³. (Mr NaOH = 40.)
- Mass → moles: 20 ÷ 40 = 0.5 mol.
- cm³ → dm³: 500 ÷ 1000 = 0.5 dm³.
- Divide: concentration = 0.5 ÷ 0.5 = 1 mol/dm³.
🧪 Exam-style questions
12 g of sodium chloride is dissolved to make 250 cm³ of solution. Calculate the concentration of the solution in g/dm³.
Show answer
- convert the volume: 250 cm³ ÷ 1000 = 0.25 dm³ 1 mark
- concentration = massvolume = 120.25 = 48 g/dm³ 1 mark
Convert cm³ to dm³ before dividing — the single most common place to drop a mark here.
11Titration Calculations T H
A titration is a technique for finding the volumes of two solutions that react exactly together. If the concentration of one solution is known and the reacting volumes are measured, the concentration of the other solution can be calculated using the balanced equation — for example finding the concentration of an acid from a known alkali.
In practice, a measured volume of one solution is added to a conical flask with a pipette, together with a few drops of an indicator. The other solution is run in from a burette, swirling all the time, until the indicator just changes colour. This is the end point, where the two solutions have reacted exactly — the volume delivered is the titre. The titration is repeated until the titres agree closely, and the mean titre is used in the calculation.
The apparatus for a titration: acid is run from the burette into a measured volume of alkali plus indicator, drop by drop, until the colour just changes.
How to actually carry out a titration is the required practical covered in detail in C4: Chemical Changes. This topic focuses on the other half: how to analyse the results and turn a titre into a concentration.
Before any calculation you need a reliable titre. The first run is done quickly, just to find the approximate end point — this is the rough (trial) titre and is not used in the mean. The titration is then repeated carefully, adding solution drop by drop near the end point, until the titres are concordant — agreeing to within 0.10 cm³ of each other. Burette readings are taken to the nearest 0.05 cm³, and each titre is the final reading − initial reading:
| Rough | Run 1 | Run 2 | Run 3 | |
|---|---|---|---|---|
| Final reading / cm³ | 20.45 | 40.40 | 20.10 | 20.30 |
| Initial reading / cm³ | 0.00 | 20.45 | 0.05 | 0.30 |
| Titre / cm³ | 20.45 | 19.95 | 20.05 | 20.00 |
Runs 1–3 are concordant (all within 0.10 cm³), so they are used for the mean. The rough titre is too high — it is left out.
Just as in the uncertainty section, the mean is taken only from the concordant results — the rough titre is ignored:
mean titre = 19.95 + 20.05 + 20.003 = 60.003 = 20.00 cm³
This mean titre is the known burette volume that goes into the calculation — it is the 20.0 cm³ of acid used in the worked example below.
- Known solution → moles. moles = concentration × volume in dm³ (convert cm³ to dm³ by dividing by 1000).
- Use the equation ratio. Use the balancing numbers to convert to the moles of the unknown solution.
- Moles → concentration. concentration = moles ÷ volume in dm³ of the unknown solution.
25.0 cm³ of sodium hydroxide solution was exactly neutralised by 20.0 cm³ of dilute hydrochloric acid of concentration 0.100 mol/dm³. Calculate the concentration of the sodium hydroxide in mol/dm³.
Use the same table idea, but now the rows are Ratio, Moles, Concentration and Volume. Write in everything the question gives you, then bridge the two solutions through moles.
HCl + NaOH → NaCl + H₂O
| HClknown | NaOHfind this | |
|---|---|---|
| Ratio | 11 | 11 |
| Moles | 40.002000.100 × 0.0200 | 50.002001 : 1 |
| Conc. | 20.100mol/dm³ · given | 60.08000.00200 ÷ 0.0250 |
| Volume | 320.0 cm³= 0.0200 dm³ | 325.0 cm³= 0.0250 dm³ |
Filling the table in order:
- Ratio — read the big numbers in the equation: HCl : NaOH = 1 : 1.
- Known concentration — write 0.100 mol/dm³ under HCl.
- Volumes — write both in, converting cm³ → dm³ (÷ 1000): HCl = 0.0200 dm³, NaOH = 0.0250 dm³.
- Moles you know — concentration × volume = 0.100 × 0.0200 = 0.00200 mol.
- Moles you want — scale by the ratio (1 : 1): 0.00200 mol of NaOH.
- Answer — concentration = moles ÷ volume = 0.00200 ÷ 0.0250 = 0.0800 mol/dm³.
Try a fresh titration the same way. The volumes and the known concentration are already filled in — you fill the rest one box at a time, and each box unlocks once the one before it is correct. Watch the ratio: this acid is diprotic.
25.0 cm³ of sodium hydroxide of concentration 0.200 mol/dm³ was exactly neutralised by 20.0 cm³ of sulfuric acid, H₂SO₄. Calculate the concentration of the sulfuric acid in mol/dm³.
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
| NaOHknown | H₂SO₄find this | |
|---|---|---|
| Ratio | ||
| Moles | ||
| Conc. | 0.200mol/dm³ · given | mol/dm³ |
| Volume | 25.0 cm³= 0.0250 dm³ | 20.0 cm³= 0.0200 dm³ |
🧪 Exam-style questions
In a titration, 25.0 cm³ of sodium carbonate solution of concentration 0.0500 mol/dm³ was exactly neutralised by 25.0 cm³ of dilute hydrochloric acid. Calculate the concentration of the hydrochloric acid in mol/dm³.
Show answer
- moles of Na₂CO₃ = concentration × volume = 0.0500 × (25.0 ÷ 1000) = 0.00125 mol 1 mark
- ratio Na₂CO₃ : HCl = 1 : 2, so moles of HCl = 2 × 0.00125 = 0.00250 mol 1 mark
- concentration of HCl = moles ÷ volume = 0.00250 ÷ (25.0 ÷ 1000) 1 mark
- = 0.00250 ÷ 0.0250 = 0.100 mol/dm³ 1 mark
Convert the 25.0 cm³ volumes to dm³ (÷ 1000) before dividing — a very common place to drop a mark.
12Percentage Yield T
Even though no atoms are gained or lost, you can almost never collect the full calculated amount of product from a real reaction. The yield is the amount of product you actually obtain.
- the reaction may be reversible and not go to completion;
- some product is lost when it is separated and purified (e.g. during filtration);
- some reactants react in unexpected (side) reactions to give different products.
The percentage yield compares the actual yield with the maximum theoretical yield (the amount calculated from the balanced equation, assuming perfect conditions):
percentage yield = mass of product actually mademaximum theoretical mass of product × 100
H Higher Tier students must also be able to calculate the theoretical mass of a product from a given mass of reactant and the balanced equation (using the reacting-mass method from section 8).
(a) A reaction has a maximum theoretical yield of 2.0 g, but only 1.6 g of product is actually collected. Find the percentage yield.
percentage yield = 1.62.0 × 100 = 80%
(b) H 25 g of calcium carbonate is heated until it fully decomposes. 12.6 g of calcium oxide is collected. Calculate the percentage yield. (Mr: CaCO₃ = 100, CaO = 56.)
CaCO₃ → CaO + CO₂
- Theoretical mass first. moles CaCO₃ = 25 ÷ 100 = 0.25 mol. Ratio is 1 : 1, so 0.25 mol of CaO should form: mass = 0.25 × 56 = 14 g (the theoretical yield).
- Then compare with the actual yield. percentage yield = 12.614 × 100 = 90%.
To turn a fraction into a percentage you multiply by 100 — once. A frequent slip is to use the calculator’s % key (which already divides by 100) and then divide by 100 again, giving an answer 100× too small (e.g. 0.9% instead of 90%). Always sense-check: a percentage yield should sit between 0 and 100%.
🧪 Exam-style questions
A reaction has a maximum theoretical yield of 6.0 g. A student actually collects 4.2 g of product. Calculate the percentage yield.
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- percentage yield = actual yieldtheoretical yield × 100 = 4.26.0 × 100 1 mark = 70% 1 mark
The actual yield always goes on top, so the answer can never be more than 100%.
4.8 g of magnesium is burned completely in oxygen and 7.0 g of magnesium oxide is collected. Calculate the percentage yield of magnesium oxide.
Show answer
- moles of Mg = 4.8 ÷ 24 = 0.2 mol 1 mark
- ratio Mg : MgO = 1 : 1, so theoretical mass of MgO = 0.2 × 40 = 8.0 g 1 mark
- percentage yield = (7.0 ÷ 8.0) × 100 1 mark = 87.5% 1 mark
13Atom Economy T
Atom economy (atom utilisation) — a measure of the amount of the starting materials that end up as useful products. A high atom economy means little is wasted, which matters for sustainable development and for economic reasons.
percentage atom economy = Mr of the desired productsum of the Mr of all reactants × 100
Use the relative formula masses taken from the balanced equation, remembering to include the big numbers. A reaction that makes only the desired product has an atom economy of 100%; one that also makes by-products has a lower atom economy — although those by-products can sometimes be sold or used, which keeps the process economically worthwhile.
H Higher Tier students should also be able to explain why a particular reaction pathway is chosen, weighing up atom economy alongside percentage yield, rate, equilibrium position and the usefulness of any by-products.
Hydrogen is manufactured by reacting methane with steam (steam reforming). The desired product is the hydrogen. Calculate the atom economy. (Mr: CH₄ = 16, H₂O = 18, H₂ = 2.)
CH₄ + H₂O → CO + 3H₂
- Mr of the desired product: 3H₂ = 3 × 2 = 6.
- Sum of the Mr of all reactants: 16 + 18 = 34.
- Divide and × 100:
atom economy = 634 × 100 = 17.6%
This is a low atom economy — most of the starting mass ends up in the carbon monoxide by-product. Industrially this can still be worthwhile because the CO is itself useful and can be sold or used in other processes.
The most common mistake here is using Mr = 2 for the hydrogen instead of 3 × 2 = 6. The big number in 3H₂ tells you that three whole molecules of hydrogen are made, and every one of them counts towards the mass of useful product. Picture the actual particles on each side of the equation:
🧪 Exam-style questions
Quicklime (calcium oxide) is made by heating limestone. Calculate the atom economy for producing calcium oxide.
Show answer
- atom economy = Mr of desired productsum of Mr of reactants × 100 = 56100 × 100 1 mark = 56% 1 mark
There is only one reactant here (CaCO₃), so the denominator is just its Mr.
Iron is extracted from iron(III) oxide in the blast furnace. Calculate the atom economy for producing iron, where iron is the desired product.
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- Mr of desired product = 2 × 56 = 112 1 mark
- sum of Mr of reactants = 160 + (3 × 28) = 160 + 84 = 244 1 mark
- atom economy = (112 ÷ 244) × 100 = 45.9% 1 mark
Remember to multiply each formula mass by its big number before adding — here 3 × CO.
14Gas Volumes T H
Under the same conditions of temperature and pressure, a number of moles of any gas will occupy the same volume.
At room temperature and pressure (RTP) — 20 °C and 1 atmosphere — one mole of any gas occupies 24 dm³ (24,000 cm³). This is the molar gas volume and you need to learn this number.
number of moles of gas = volume of gas (dm³)24
If the volume is given in cm³ instead of dm³, divide by 24,000 instead of 24.
Because equal moles mean equal volumes, the big numbers in a balanced equation also give the ratio of gas volumes. For example, in the complete combustion of propane the volumes are in the same 1 : 5 : 3 ratio as the moles:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
so 150 cm³ of propane needs 5 × 150 = 750 cm³ of oxygen, and makes 3 × 150 = 450 cm³ of carbon dioxide.
Read the units the question asks for. If you are given a mass, convert it to moles first (mass ÷ Mr), then multiply by 24 to get a volume in dm³.
(a) Volume from mass. Calculate the volume of 154 g of nitrogen gas, N₂, at RTP. (Mr N₂ = 28.)
- Mass → moles: 154 ÷ 28 = 5.5 mol.
- Moles → volume (× 24): 5.5 × 24 = 132 dm³.
(b) Volumes from an equation. 50 cm³ of methane is burned completely. What volume of oxygen reacts, and what volume of carbon dioxide is produced?
CH₄ + 2O₂ → CO₂ + 2H₂O
Equal volumes contain equal moles, so the big numbers give the volume ratio directly: O₂ = 2 × 50 = 100 cm³, and CO₂ = 1 × 50 = 50 cm³. (The water is liquid at RTP, so it is not counted as a gas volume.)
🧪 Exam-style questions
Calculate the volume occupied by 8 g of oxygen gas, O₂, at RTP.Mr(O₂) = 32. Molar gas volume = 24 dm³/mol.
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- moles of O₂ = mass ÷ Mr = 8 ÷ 32 = 0.25 mol 1 mark
- volume = moles × 24 = 0.25 × 24 = 6 dm³ 1 mark
100 cm³ of methane is burned completely. What volume of oxygen reacts with it? (All gas volumes measured at the same temperature and pressure.)
Show answer
- Equal volumes of gas contain equal moles, so the big numbers give the volume ratio directly: CH₄ : O₂ = 1 : 2.
- volume of O₂ = 2 × 100 = 200 cm³ 1 mark
- Conservation of mass: no atoms are created or destroyed in a reaction, so the total mass of products equals the total mass of reactants. This is why equations must be balanced.
- Balanced equation: same number of each type of atom on both sides. Balance by adding numbers in front of formulae — never change a formula.
- Mass change with a gas: mass appears to increase when a gas is absorbed from the air (e.g. a metal forming an oxide) and to decrease when a gas is released to the air (e.g. a carbonate giving off CO₂). In a sealed system the mass is unchanged.
- Relative formula mass (Mr) = sum of the Ar values of all the atoms in the formula. In a balanced equation the total Mr of the reactants equals the total Mr of the products.
- Uncertainty: report measurements to a sensible number of significant figures; the uncertainty of a mean can be estimated from the range of repeat results.
- The mole H: one mole contains the Avogadro number of particles, 6.02 × 10²³. Mass = Mr × moles, so moles = mass ÷ Mr.
- Balancing using masses H: convert masses to moles, divide by the smallest to get the whole-number ratio — this gives the balancing numbers.
- Reacting masses H: use the balanced equation to find the mole ratio, then convert between mass of one substance and mass of another.
- Limiting reactant H: the reactant fully used up; it limits the amount of product. The reactant in excess is left over. Amount of product is proportional to the moles of the limiting reactant.
- Concentration: in g/dm³, concentration = mass ÷ volume (dm³). The greater the mass dissolved in a given volume, the higher the concentration. (1 dm³ = 1000 cm³.)
- Titration T H: concentration in mol/dm³ = moles ÷ volume (dm³). Use the equation's mole ratio with one known concentration to find the other.
- Percentage yield T = (actual yield ÷ theoretical yield) × 100. Never 100% — losses occur, reactions may be incomplete or reversible, or there are side reactions.
- Atom economy T = (Mr of desired product ÷ sum of the Mr of all reactants) × 100. High atom economy means less waste and is more sustainable and economical.
- Gas volumes T H: equal volumes of any gases at the same temperature and pressure contain equal numbers of moles. One mole of any gas occupies 24 dm³ at RTP (20 °C, 1 atm), so volume (dm³) = moles × 24.