Quantitative Chemistry

In every chemical reaction, the atoms in the reactants are rearranged to make products. Because no atoms are created and no atoms are destroyed, the total mass stays the same.

This is the law of conservation of mass, and AQA mark schemes expect a one-sentence statement that mentions atoms — not just mass:

"No atoms are made or destroyed in a chemical reaction, so the total mass of the products equals the total mass of the reactants."

Think of a chemical reaction like rearranging Lego bricks. Take apart a Lego car and rebuild it as a Lego boat — you still have exactly the same bricks. The atoms are the bricks; the molecules are the models.

A simple worked example. When 2.0 g of hydrogen reacts completely with 16.0 g of oxygen in a sealed flask, the mass of water formed is:

$m(H_2O) = 2.0 + 16.0 = 18.0\,\text{g}$

Mass in = mass out. Always.

1. Word equation. Uses chemical names with a + between reactants and an arrow → between reactants and products.

magnesium + oxygen → magnesium oxide

2. Unbalanced symbol equation. Replace names with formulae:

$Mg + O_2 \rightarrow MgO$

Count atoms on each side:

Element	Left	Right
Mg	1	1
O	2	1

The oxygen count doesn't match — this is not yet balanced.

3. Balanced symbol equation. Insert big numbers (coefficients) in front of formulae so that every element has equal counts on both sides:

$2Mg + O_2 \rightarrow 2MgO$

Now Mg: 2 = 2 and O: 2 = 2. The equation is balanced.

Adding state symbols. AQA expects (s) solid, (l) liquid, (g) gas, (aq) aqueous (dissolved in water):

$2Mg(s) + O_2(g) \rightarrow 2MgO(s)$

Why coefficients, not subscripts? Subscripts are part of the formula — they tell you the recipe for that molecule. Changing $H_2O$ to $H_2O_2$ doesn't balance an equation; it changes water into hydrogen peroxide (a totally different substance).

Method (works every time):

1. Write the unbalanced equation with correct formulae.
2. List the atoms of each element on each side.
3. Pick the element appearing in fewest places; adjust a coefficient to match.
4. Re-count and continue. Save H and O for last in combustion problems.
5. Check every element has the same count on each side.

Worked example — propane combustion.

$C_3H_8 + O_2 \rightarrow CO_2 + H_2O$

Step	Action	Equation
1	Balance C (3 left → need 3 CO₂)	$C_3H_8 + O_2 \rightarrow 3CO_2 + H_2O$
2	Balance H (8 left → need 4 H₂O)	$C_3H_8 + O_2 \rightarrow 3CO_2 + 4H_2O$
3	Count O on right: $3 \times 2 + 4 \times 1 = 10$. Need 5 O₂ on left.	$C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$

Final check — C: 3 = 3 ✓; H: 8 = 8 ✓; O: 10 = 10 ✓. Balanced.

Worked example — iron + chlorine.

$Fe + Cl_2 \rightarrow FeCl_3$

Step	Action
1	Cl appears as 2 (left) and 3 (right). LCM = 6 → 3 Cl₂ and 2 FeCl₃
2	Now Fe: 1 left, 2 right → 2 Fe on left

$2Fe + 3Cl_2 \rightarrow 2FeCl_3$

Check — Fe: 2 = 2 ✓; Cl: 6 = 6 ✓.

The relative formula mass (Mᵣ) of a compound is the sum of the relative atomic masses (Aᵣ) of all the atoms shown in its formula.

For a molecule made of two non-metals (like water), Mᵣ is sometimes called the relative molecular mass — it means the same number. For ionic compounds (like NaCl) where there isn't really a 'molecule', AQA uses the term relative formula mass throughout.

Worked example — water, H₂O.

$M_r(H_2O) = 2 \times A_r(H) + 1 \times A_r(O) = 2 \times 1 + 16 = 18$

Worked example — sodium chloride, NaCl.

$M_r(NaCl) = 23 + 35.5 = 58.5$

Worked example — sulfuric acid, H₂SO₄.

$M_r(H_2SO_4) = 2 \times 1 + 32 + 4 \times 16 = 2 + 32 + 64 = 98$

You'll be given the periodic table in the exam, so you don't need to memorise Aᵣ values — but it speeds you up if you know the common ones (H = 1, C = 12, N = 14, O = 16, Na = 23, S = 32, Cl = 35.5, K = 39, Ca = 40, Fe = 56, Cu = 63.5).

Many ionic compounds have brackets to show that a whole group is repeated. The number outside the bracket multiplies every atom inside.

Worked example — calcium hydroxide, Ca(OH)₂.

The (OH)₂ means two hydroxide groups, so two O atoms and two H atoms.

$M_r(Ca(OH)_2) = 40 + 2 \times (16 + 1) = 40 + 2 \times 17 = 40 + 34 = 74$

Worked example — ammonium sulfate, (NH₄)₂SO₄.

Two ammonium groups means 2 N and 8 H. Then add SO₄ (1 S, 4 O).

$M_r = 2 \times (14 + 4) + 32 + 4 \times 16 = 2 \times 18 + 32 + 64 = 36 + 96 = 132$

Worked example — aluminium sulfate, Al₂(SO₄)₃.

$M_r = 2 \times 27 + 3 \times (32 + 4 \times 16) = 54 + 3 \times 96 = 54 + 288 = 342$

For a balanced equation, the total Mᵣ × coefficient on each side is equal. This is the algebraic way to see conservation of mass.

Worked example — formation of ammonia.

$N_2 + 3H_2 \rightarrow 2NH_3$

Side	Calculation	Total
Left	$1 \times 28 + 3 \times 2$	34
Right	$2 \times 17$	34

Equal — mass is conserved.

Worked example — combustion of methane.

$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$

Side	Calculation	Total
Left	$1 \times 16 + 2 \times 32$	80
Right	$1 \times 44 + 2 \times 18$	80

Match — equation correctly balanced and mass is conserved.

This 'check' is useful in the exam: if your totals don't match, you've miscounted atoms or miscalculated an Mᵣ.

In a chemical reaction, mass is always conserved overall. But on a balance you can only weigh what's inside the apparatus — so if a gas enters or leaves, the reading on the balance changes.

Closed system. A sealed flask, a stoppered conical flask or a sealed gas syringe. No gas can enter or leave. The balance reading stays exactly the same throughout the reaction.

Open system. A crucible without a lid, a beaker, or any apparatus connected to the air. Gases can enter (mass goes up) or leave (mass goes down).

Classic AQA exam example — burning magnesium in an open crucible.

$2Mg(s) + O_2(g) \rightarrow 2MgO(s)$

A piece of magnesium ribbon left in a crucible gains mass as it burns because oxygen from the air is absorbed into the solid product. The mass of MgO formed equals mass of Mg + mass of O₂ absorbed.

Classic AQA exam example — thermal decomposition of calcium carbonate.

$CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$

Heat a marble chip in an open boat and the balance reading falls — because CO₂ gas escapes into the air. The mass left behind (CaO) is less than the starting mass of CaCO₃.

Three useful rules:

1. All reactants and products are solids/liquids → no mass change.
2. A gas is a reactant in an open system → balance reading increases.
3. A gas is a product in an open system → balance reading decreases.

Worked example 1 — burning magnesium.

$2Mg(s) + O_2(g) \rightarrow 2MgO(s)$

Oxygen (gas) is a reactant. In an open crucible, the balance reading increases by the mass of O₂ absorbed.

Worked example 2 — neutralising HCl with NaOH (solutions).

$HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)$

No gas involved → mass stays the same even in a beaker.

Worked example 3 — adding acid to a carbonate (open beaker).

$CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)$

CO₂ gas is produced and escapes → balance reading decreases by the mass of CO₂ released.

Worked example 4 — same reaction, sealed conical flask.

The reaction still produces CO₂, but the gas is trapped inside the flask → balance reading is unchanged.

AQA mark schemes for these questions typically award 3 marks:

Mark	What the examiner wants
1	Name the gas involved (e.g. "carbon dioxide" or "oxygen")
2	State whether it enters or leaves the system, and why the balance changes
3	Conclude that mass is still conserved overall (no atoms are made or destroyed)

Model answer — why does the mass of a piece of magnesium increase when it burns in air?

"The magnesium reacts with oxygen from the air to form solid magnesium oxide. The oxygen atoms add to the magnesium, so the new product is heavier than the original magnesium. No atoms are made or destroyed — the mass that appears to be gained came from the oxygen in the air."

Model answer — why does the mass of marble chips decrease when they are heated?

"Heating calcium carbonate causes thermal decomposition into calcium oxide and carbon dioxide gas. The carbon dioxide escapes from the open boat into the air, so the balance reading falls. The total mass of CaO + CO₂ still equals the mass of the original CaCO₃ — mass is conserved overall."

When you measure the temperature of a reaction, the mass of a precipitate or the volume of acid added from a burette, you can never be exactly sure the value you've recorded is the true value. There are three big reasons:

1. Resolution of the instrument — a balance that displays only to the nearest 0.1 g cannot tell the difference between 24.31 g and 24.36 g; both round to 24.3 g.
2. Random error — small, unpredictable changes (a draught, a slight delay in stopping the timer, parallax when reading a burette) cause repeats to scatter around the true value.
3. Human judgement — when do you decide a colour change is 'complete'? When does the bubble just form?

AQA examiner reports highlight that students often quote a single reading as if it were exact. Top-band answers always:

• Take at least 3 repeats of every measurement.
• Identify and exclude anomalies before averaging.
• Quote the result as $\text{mean} \pm \text{uncertainty}$.

A note on terminology. Accuracy means how close your value is to the true value. Precision means how close your repeats are to each other. You can be precise but inaccurate (a consistent systematic offset) or accurate but imprecise (scattered around the right value). The spec separates these and AQA mark schemes reward correct usage.

Mean of $n$ acceptable repeats:

$\bar{x} = \frac{x_1 + x_2 + \dots + x_n}{n}$

Range = largest − smallest acceptable value.

Uncertainty estimated as:

$\text{uncertainty} = \pm \frac{\text{range}}{2}$

Worked example. A student measures the temperature rise of a neutralisation four times: 6.4 °C, 6.2 °C, 6.5 °C and 9.1 °C.

• The 9.1 °C value is an obvious anomaly (much higher than the others) — exclude it.
• Mean: $(6.4 + 6.2 + 6.5)/3 = 6.37 \approx 6.4\,\text{°C}$.
• Range: $6.5 - 6.2 = 0.3\,\text{°C}$.
• Uncertainty: $\pm 0.15\,\text{°C}$ (round to match the data: $\pm 0.2$ °C).
• Report: temperature rise = 6.4 ± 0.2 °C.

Mark-scheme tip. AQA awards 1 mark for identifying the anomaly, 1 mark for the correct mean from the remaining repeats, and 1 mark for the uncertainty quoted as $\pm \tfrac{1}{2}$ × range. Three easy marks if you follow the routine.

Resolution is the smallest change a measuring instrument can detect or display. Common UK GCSE lab instruments:

Instrument	Typical resolution	Reading uncertainty
2 d.p. balance	0.01 g	± 0.005 g
3 d.p. balance	0.001 g	± 0.0005 g
50 cm³ burette	0.10 cm³	± 0.05 cm³ (per reading)
100 cm³ measuring cylinder	1 cm³	± 0.5 cm³
25 cm³ pipette	n/a (fixed)	± 0.06 cm³ (graded A)
Mercury thermometer (–10 → 110 °C)	1 °C	± 0.5 °C
Digital thermometer	0.1 °C	± 0.05 °C

A simple reading uncertainty for a single measurement is $\pm \tfrac{1}{2}$ of the resolution. For a burette you read twice (start and end), so the total uncertainty for the volume delivered is $2 \times 0.05 = \pm 0.10\,\text{cm}^3$.

Worked example — choosing the right instrument. A student wants to measure 25.0 cm³ of acid for a titration. Which is more appropriate, a 50 cm³ measuring cylinder or a 25 cm³ pipette?

• Measuring cylinder uncertainty: $\pm 0.5\,\text{cm}^3$ → percentage uncertainty = $0.5/25 \times 100 = 2.0\%$.
• Pipette uncertainty: $\pm 0.06\,\text{cm}^3$ → percentage uncertainty = $0.06/25 \times 100 = 0.24\%$.

The pipette is roughly ten times more precise — that's why titrations always use pipettes (or burettes) rather than measuring cylinders. Higher Tier mark schemes ask you to quantify the improvement with a percentage uncertainty calculation.

Percentage uncertainty:

$\%\,\text{uncertainty} = \frac{\text{uncertainty}}{\text{measured value}} \times 100$

The biggest percentage uncertainty in a multi-step measurement usually limits the overall accuracy — so focus your effort on improving the noisiest step (e.g. weighing tiny masses on a 2 d.p. balance is a common bottleneck).

Atoms and molecules are unimaginably small. A single drop of water contains roughly 1.5 × 10²¹ molecules — far more than the stars in the known universe. To make chemical accounting bearable, chemists count particles in moles.

Definition. One mole of any substance contains $6.02 \times 10^{23}$ particles. This number is the Avogadro constant, given the symbol $N_A$.

The 'particles' can be:

• Atoms — e.g. 1 mol of carbon atoms contains $6.02 \times 10^{23}$ C atoms.
• Molecules — e.g. 1 mol of water contains $6.02 \times 10^{23}$ H₂O molecules.
• Ions — e.g. 1 mol of Cl⁻ ions contains $6.02 \times 10^{23}$ Cl⁻ ions.
• Formula units — e.g. 1 mol of NaCl contains $6.02 \times 10^{23}$ Na⁺ ions and $6.02 \times 10^{23}$ Cl⁻ ions (2 × $N_A$ ions in total).

Why does the mole exist? Because chemical equations tell us how many particles react — but our lab apparatus can only weigh grams. The mole connects the two. The defining magic: 1 mole of a substance has a mass in grams equal to its Mᵣ value.

Substance	Mᵣ	Mass of 1 mole	Particles in 1 mole
H₂O	18	18 g	$6.02 \times 10^{23}$ H₂O molecules
CO₂	44	44 g	$6.02 \times 10^{23}$ CO₂ molecules
NaCl	58.5	58.5 g	$6.02 \times 10^{23}$ formula units
Mg	24	24 g	$6.02 \times 10^{23}$ Mg atoms

AQA exam note. $N_A = 6.02 \times 10^{23}\,\text{mol}^{-1}$ is given on the AQA Chemistry data sheet — you do not need to memorise it precisely, but you must use it confidently to convert moles ↔ particles.

Every mole calculation starts here:

$n = \frac{m}{M_r}$

• $n$ — number of moles (mol).
• $m$ — mass of the substance (g).
• $M_r$ — relative formula mass (g/mol, also called molar mass and given the symbol $M$).

Rearranged:

$m = n \times M_r \qquad \text{and} \qquad M_r = \frac{m}{n}$

Worked example 1 — moles from mass. How many moles are in 36 g of water (H₂O, Mᵣ = 18)?

$n = \frac{m}{M_r} = \frac{36}{18} = 2\,\text{mol}$

Worked example 2 — mass from moles. What is the mass of 0.25 mol of sodium chloride (NaCl, Mᵣ = 58.5)?

$m = n \times M_r = 0.25 \times 58.5 = 14.625\,\text{g}$

Worked example 3 — moles to particles. How many molecules are in 0.50 mol of CO₂?

$N = n \times N_A = 0.50 \times 6.02 \times 10^{23} = 3.01 \times 10^{23}\,\text{molecules}$

Worked example 4 — particles to mass. Find the mass of $3.01 \times 10^{23}$ molecules of CO₂.

• $n = N / N_A = 3.01 \times 10^{23} / 6.02 \times 10^{23} = 0.50\,\text{mol}$
• $m = n \times M_r = 0.50 \times 44 = 22\,\text{g}$

These four conversions cover almost every AQA mole question on Paper 1.

Sometimes AQA asks you to find the number of particles in a given mass. The route is always the same:

mass → moles → particles

with $n = m/M_r$ first, then $N = n \times N_A$.

Worked example. How many oxygen atoms are in 8.0 g of O₂?

1. Mᵣ of O₂ = $2 \times 16 = 32$.
2. Moles of O₂: $n = 8.0 / 32 = 0.25\,\text{mol}$.
3. Molecules of O₂: $N = 0.25 \times 6.02 \times 10^{23} = 1.505 \times 10^{23}$.
4. Each O₂ molecule has 2 O atoms, so number of O atoms = $2 \times 1.505 \times 10^{23} = 3.01 \times 10^{23}$.

That last step trips up many candidates — always check the formula to see how many atoms (or ions) are in each particle.

Worked example — ions in NaCl. Calculate the number of chloride ions in 5.85 g of NaCl (Mᵣ = 58.5).

1. $n = 5.85 / 58.5 = 0.10\,\text{mol}$ of NaCl formula units.
2. Each formula unit contains 1 Cl⁻ ion. So number of Cl⁻ ions = $0.10 \times 6.02 \times 10^{23} = 6.02 \times 10^{22}\,\text{ions}$.

Worked example — atoms in a hydrocarbon. How many H atoms are in 2.0 g of CH₄ (Mᵣ = 16)?

1. $n = 2.0 / 16 = 0.125\,\text{mol}$ of CH₄.
2. Molecules: $0.125 \times 6.02 \times 10^{23} = 7.525 \times 10^{22}$.
3. Each molecule has 4 H atoms: $4 \times 7.525 \times 10^{22} = 3.01 \times 10^{23}\,\text{H atoms}$.

Stay disciplined — write the conversion chain (mass → moles → particles) every time.

A balanced equation works like a baking recipe: the big numbers in front (coefficients) tell you the mole ratio of reactants and products.

$2H_2 + O_2 \rightarrow 2H_2O$

Reads as: "2 mol of $H_2$ react with 1 mol of $O_2$ to give 2 mol of $H_2O$."

You can scale this freely. Half the recipe: 1 mol H₂ + 0.5 mol O₂ → 1 mol H₂O. Ten times: 20 mol H₂ + 10 mol O₂ → 20 mol H₂O. The ratios never change.

The universal recipe for any 'reaction mass' calculation:

mass A → moles A → moles B (multiply by ratio from equation) → mass B

Worked example. Calculate the mass of magnesium oxide formed when 4.8 g of magnesium burns in excess oxygen.

$2Mg + O_2 \rightarrow 2MgO$

1. Moles of Mg: $n = 4.8 / 24 = 0.20$ mol.
2. Mole ratio Mg : MgO = 2 : 2 = 1 : 1, so moles of MgO = 0.20 mol.
3. Mass of MgO: $m = n \times M_r = 0.20 \times (24 + 16) = 0.20 \times 40 = 8.0\,\text{g}$.

Worked example with different coefficients. What mass of $CO_2$ is produced when 8.0 g of methane combusts completely?

$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$

1. Moles of CH₄: $n = 8.0 / 16 = 0.50$ mol.
2. Mole ratio CH₄ : CO₂ = 1 : 1, so moles of CO₂ = 0.50 mol.
3. Mass of CO₂: $0.50 \times 44 = 22\,\text{g}$.

The big idea. Always go through moles. Mass-to-mass shortcuts (cross-multiply, etc.) only work when the ratio is 1 : 1 — they fail the moment the equation has coefficients other than 1.

When two reactants are mixed, only one runs out. That one is the limiting reactant — it determines how much product can form. The other is in excess.

Three-step routine.

1. Find the moles of each reactant.
2. Divide each by its coefficient in the balanced equation.
3. The smallest ratio identifies the limiting reactant.

Worked example. Magnesium reacts with hydrochloric acid:

$Mg + 2HCl \rightarrow MgCl_2 + H_2$

If 2.4 g of Mg is added to a solution containing 0.10 mol of HCl, which is limiting?

1. Moles of Mg: $2.4 / 24 = 0.10$ mol; moles of HCl: 0.10 mol.
2. Divide by coefficients: Mg → $0.10/1 = 0.10$; HCl → $0.10/2 = 0.05$.
3. HCl has the smaller value → HCl is limiting, Mg is in excess.

Maximum mass of MgCl₂? Use the limiting reactant.

• Moles HCl → moles MgCl₂: ratio 2 : 1, so moles of MgCl₂ = $0.10/2 = 0.050$ mol.
• Mass MgCl₂ = $0.050 \times (24 + 2 \times 35.5) = 0.050 \times 95 = 4.75\,\text{g}$.

How much excess Mg is left?

• Moles of Mg used: ratio Mg : HCl = 1 : 2, so moles of Mg used = $0.10/2 = 0.05$ mol.
• Moles of Mg remaining: $0.10 - 0.05 = 0.05$ mol.
• Mass of excess Mg: $0.05 \times 24 = 1.2\,\text{g}$.

Why does it matter? In industry, the more expensive reactant is usually made limiting (so none is wasted) and the cheaper one is added in excess. AQA examiner reports praise candidates who explain the practical reasoning.

Quick-check. A common AQA trick is to give a small amount of one reactant with a much larger amount of another. The smaller mass is not always limiting — you must divide by the coefficient first.

Sometimes the question gives you the masses (or moles) of each substance and asks you to deduce the balancing numbers. The recipe:

1. Find the moles of each substance using $n = m/M_r$.
2. Divide every value by the smallest moles value to find the simplest ratio.
3. If you get non-integer ratios (e.g. 1 : 1.5 : 2.5), multiply all of them by a small integer to get whole numbers (here, multiply by 2 → 2 : 3 : 5).

Worked example. 4.6 g of sodium reacts with 7.1 g of chlorine to give 11.7 g of sodium chloride. Write the balanced equation. (Aᵣ: Na = 23, Cl = 35.5.)

1. Moles:
   • Na: $4.6/23 = 0.20$
   • Cl₂: $7.1/71 = 0.10$
   • NaCl: $11.7/58.5 = 0.20$
2. Divide each by the smallest (0.10):
   • Na: 0.20/0.10 = 2
   • Cl₂: 0.10/0.10 = 1
   • NaCl: 0.20/0.10 = 2
3. Equation: $2Na + Cl_2 \rightarrow 2NaCl$. ✓

Worked example with fractions. 0.40 g of hydrogen reacts with 3.20 g of oxygen to form 3.60 g of water.

1. Moles:
   • H₂: $0.40/2 = 0.20$
   • O₂: $3.20/32 = 0.10$
   • H₂O: $3.60/18 = 0.20$
2. Divide by smallest (0.10):
   • H₂: 2, O₂: 1, H₂O: 2.
3. Equation: $2H_2 + O_2 \rightarrow 2H_2O$. ✓

Sanity check. Mass conservation:

• Left: $0.40 + 3.20 = 3.60$ g.
• Right: 3.60 g. ✓ Always confirm.

A note on Higher Tier. This skill underpins empirical-formula calculations too (spec 5.3.1.2 follow-on). The technique is the same — convert masses to moles, divide by the smallest, round to integers.

Concentration tells you how much solute is dissolved in a given volume of solution. The simplest measure is grams per cubic decimetre:

$c = \dfrac{m}{V}$

• $c$ — concentration, g/dm³.
• $m$ — mass of solute, g.
• $V$ — volume of solution, dm³ (litres).

Critical unit conversion. Laboratory volumes are usually measured in cm³. Always convert before using the formula:

$1\,\text{dm}^3 = 1000\,\text{cm}^3 \quad \Rightarrow \quad V\,[\text{dm}^3] = V\,[\text{cm}^3]/1000$

Worked example 1. 5.85 g of NaCl is dissolved in 250 cm³ of water. Calculate the concentration in g/dm³.

1. Convert volume: $V = 250/1000 = 0.250\,\text{dm}^3$.
2. $c = m/V = 5.85/0.250 = 23.4\,\text{g/dm}^3$.

Worked example 2 — mass from concentration. What mass of solute is in 100 cm³ of a 12 g/dm³ solution?

1. $V = 0.100\,\text{dm}^3$.
2. $m = c \times V = 12 \times 0.100 = 1.2\,\text{g}$.

Mark-scheme tip. AQA usually awards 1 mark for the volume conversion and 1 mark for substituting into $c = m/V$. Write '$V = 250/1000 = 0.250\,\text{dm}^3$' explicitly — even if your final answer is wrong, you'll pick up the method mark.

For Higher Tier work — especially titrations — concentration is most useful in moles per cubic decimetre:

$c = \dfrac{n}{V}$

• $c$ — concentration, mol/dm³.
• $n$ — moles of solute.
• $V$ — volume of solution, dm³.

Why moles? Reactions happen in fixed mole ratios. So mol/dm³ tells you how many particles are in the solution per unit volume — exactly what you need to compare with the balanced equation.

Converting between g/dm³ and mol/dm³. Use the molar mass:

$c\,[\text{g/dm}^3] = c\,[\text{mol/dm}^3] \times M_r$

Worked example. A 0.50 mol/dm³ NaOH solution. What is its concentration in g/dm³? (Mᵣ NaOH = 40.)

$c = 0.50 \times 40 = 20\,\text{g/dm}^3$

Worked example — finding mol/dm³ from a recipe. 4.0 g of NaOH is dissolved in 200 cm³ of water. Find the concentration in mol/dm³.

1. Moles: $n = m/M_r = 4.0/40 = 0.10$ mol.
2. Volume: $V = 200/1000 = 0.20$ dm³.
3. Concentration: $c = n/V = 0.10/0.20 = 0.50\,\text{mol/dm}^3$.

AQA exam note. Concentration is the centrepiece of any titration question. Once you know the mol/dm³ of two solutions and the volume reacted, you can find the unknown using the standard route: moles → mole ratio → moles → concentration.

When you add solvent to a solution, the moles of solute stay the same — only the volume increases. Hence the concentration drops.

For any dilution:

$c_1 V_1 = c_2 V_2$

• Subscript 1 = before dilution (stock).
• Subscript 2 = after dilution (final).
• Units must match on both sides.

Worked example 1. A 2.0 mol/dm³ HCl stock is diluted by taking 50 cm³ and adding water until the total volume is 250 cm³. Find the new concentration.

$c_2 = \dfrac{c_1 V_1}{V_2} = \dfrac{2.0 \times 50}{250} = 0.40\,\text{mol/dm}^3$

Worked example 2 — preparing a target concentration. You need 100 cm³ of 0.50 mol/dm³ NaCl. You have a 5.0 mol/dm³ stock. How much stock do you need?

$V_1 = \dfrac{c_2 V_2}{c_1} = \dfrac{0.50 \times 100}{5.0} = 10\,\text{cm}^3$

So take 10 cm³ of stock and dilute with water to a final volume of 100 cm³.

Practical note for AQA exam. When making a dilution, you add the stock to a volumetric flask and top up to the mark with distilled water (this is more accurate than mixing in a measuring cylinder).

Foundation-Tier dilution. Foundation candidates use the g/dm³ form: $c_1 V_1 = c_2 V_2$ works identically with grams instead of moles.

Percentage yield measures how successful a reaction was at producing the desired product:

$\%\,Y = \dfrac{\text{actual mass of product}}{\text{theoretical (maximum) mass}} \times 100$

• Actual mass — what you weighed at the end.
• Theoretical mass — what you would get if everything reacted and was collected with no loss. Calculated from the balanced equation and limiting reactant.

Real chemical reactions almost never give 100 %. The spec lists three reasons:

1. The reaction is reversible — products turn back into reactants, so equilibrium is reached before all reactants are converted.
2. Side reactions — the same reactants can produce unwanted by-products.
3. Practical losses — product sticks to filter paper, evaporates, or is lost when transferring between containers.

Worked example. A student heats 5.00 g of CaCO₃ and collects 2.50 g of CaO. Find the percentage yield. $CaCO_3 \rightarrow CaO + CO_2$. (Mᵣ: CaCO₃ = 100, CaO = 56.)

1. Moles of CaCO₃: $5.00/100 = 0.0500$ mol.
2. Mole ratio CaCO₃ : CaO = 1 : 1, so theoretical moles of CaO = 0.0500 mol.
3. Theoretical mass: $0.0500 \times 56 = 2.80\,\text{g}$.
4. $\%\,Y = (2.50/2.80) \times 100 = 89.3\%$.

Industry view. Higher yields = more product per batch = better economics. But pushing the yield up usually costs energy (higher temperature) or rare reagents. Industrial chemists balance yield against cost.

Common AQA marks: 1 mark for the theoretical mass, 1 mark for the calculation, 1 mark for the percentage (with units).

Atom economy asks a different question: of all the atoms that go into the reactants, what fraction end up in the useful product?

$\%\,AE = \dfrac{M_r\,\text{(desired product)}}{\sum M_r\,\text{(all products)}} \times 100$

Both numerator and denominator must include the coefficients from the balanced equation.

Why it matters. Reactions with low atom economy generate lots of waste — usually requiring expensive disposal, polluting the environment, or wasting raw materials. Green chemistry prioritises reactions with high atom economy.

A 100 % atom-economy reaction has ONE product. Examples:

• $N_2 + 3H_2 \rightarrow 2NH_3$ (Haber process) — every reactant atom ends up in ammonia.
• $C + O_2 \rightarrow CO_2$ — complete combustion.

Worked example — production of $H_2$. Hydrogen can be made by two routes:

Route A — electrolysis of water:

$2H_2O \rightarrow 2H_2 + O_2$

Desired product: $H_2$. Mᵣ contributions:

• $2 \times M_r(H_2) = 2 \times 2 = 4$ (desired).
• $1 \times M_r(O_2) = 32$ (waste).
• Total $M_r$ products = $4 + 32 = 36$.

$\%\,AE = \dfrac{4}{36} \times 100 = 11.1\%$

Route B — reaction of methane with steam (steam reforming):

$CH_4 + 2H_2O \rightarrow CO_2 + 4H_2$

Desired product: $H_2$. Mᵣ contributions:

• $4 \times M_r(H_2) = 4 \times 2 = 8$ (desired).
• $1 \times M_r(CO_2) = 44$ (waste).
• Total Mᵣ products = $8 + 44 = 52$.

$\%\,AE = \dfrac{8}{52} \times 100 = 15.4\%$

Both routes have low atom economy because the unwanted product (O₂ or CO₂) is heavy compared with H₂. But route B has a higher atom economy than route A.

AQA exam tip. AQA examiner reports note that candidates often forget to multiply by the coefficient when calculating $M_r$ of each product. Always write '$2 \times 2 = 4$' rather than just '$M_r(H_2) = 2$'.

Yield and atom economy can be high or low independently — and industry must consider both.

Scenario	Example	What to do
High yield + high AE	Haber process (≈ 15 % yield per pass, but recycled; ~100 % AE)	Ideal — single waste-free product.
High yield + low AE	Combustion of fuels (≈ 100 % yield; AE depends on whether $CO_2$ counts as waste)	Hard to fix the AE; consider alternative routes.
Low yield + high AE	Reversible reactions at equilibrium	Use a catalyst, recycle unreacted reactants.
Low yield + low AE	Wasteful side reactions plus losses	Redesign the route — find a better catalyst or reaction.

Why AQA cares. Modern UK GCSE Chemistry explicitly links atom economy to sustainability, green chemistry and the economics of the chemical industry. Expect questions framed around 'evaluate this reaction's environmental impact'.

Worked example — comparing two routes. A student weighs both options for making ethanol:

• Hydration of ethene: $C_2H_4 + H_2O \rightarrow C_2H_5OH$ (AE = 100 %).
• Fermentation of glucose: $C_6H_{12}O_6 \rightarrow 2C_2H_5OH + 2CO_2$ (AE calculation below).

For fermentation:

• $M_r(C_2H_5OH) = 46$; $M_r(CO_2) = 44$.
• Desired product: $2 \times 46 = 92$.
• All products: $92 + (2 \times 44) = 92 + 88 = 180$.
• $\%\,AE = (92/180) \times 100 = 51.1\%$.

So hydration of ethene (100 % AE) is greener per molecule, but fermentation uses renewable feedstock (sugar from plants) — so the comparison is more nuanced than the AE number alone.

This kind of synoptic evaluation is exactly what AQA Higher Tier examiner reports highlight as distinguishing top-band candidates.

For any AQA titration calculation, follow exactly the same three-step routine. Practice it until it's automatic — examiners explicitly reward each step.

Step 1 — Moles of the known solution.

$n = c \times V$

• $n$ = moles.
• $c$ = concentration (mol/dm³).
• $V$ = volume (dm³ — convert from cm³ by ÷ 1000).

Step 2 — Apply the mole ratio. Read the balanced equation:

e.g. NaOH + HCl → NaCl + H₂O (ratio 1 : 1)

e.g. 2 NaOH + H₂SO₄ → Na₂SO₄ + 2 H₂O (ratio 2 : 1)

Multiply moles of the known by the ratio to get moles of the unknown.

Step 3 — Concentration of the unknown.

$c = \dfrac{n}{V}$

Use the volume of the unknown solution (again converted to dm³).

Worked example 1 — 1:1 ratio. 25.0 cm³ of 0.100 mol/dm³ NaOH neutralises 20.0 cm³ of HCl. Calculate the concentration of the HCl in mol/dm³.

Step 1. Moles of NaOH = $c \times V = 0.100 \times (25.0/1000) = 2.50 \times 10^{-3}\,\text{mol}$.

Step 2. Equation: NaOH + HCl → NaCl + H₂O. Ratio NaOH : HCl = 1 : 1. So moles of HCl = $2.50 \times 10^{-3}\,\text{mol}$.

Step 3. Concentration of HCl = $n/V = (2.50 \times 10^{-3}) / (20.0/1000) = 0.125\,\text{mol/dm}^3$.

Worked example 2 — 1:2 ratio. 30.0 cm³ of 0.200 mol/dm³ H₂SO₄ exactly neutralises 25.0 cm³ of NaOH. Find the concentration of NaOH.

Step 1. Moles of H₂SO₄ = $0.200 \times 0.0300 = 6.00 \times 10^{-3}\,\text{mol}$.

Step 2. Equation: H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O. So moles of NaOH = $2 \times 6.00 \times 10^{-3} = 1.20 \times 10^{-2}\,\text{mol}$.

Step 3. Concentration of NaOH = $(1.20 \times 10^{-2}) / 0.0250 = 0.480\,\text{mol/dm}^3$.

Common AQA reactions and their mole ratios:

Equation	Ratio (acid : alkali)
HCl + NaOH → NaCl + H₂O	1 : 1
HNO₃ + NaOH → NaNO₃ + H₂O	1 : 1
H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O	1 : 2
H₂SO₄ + 2 KOH → K₂SO₄ + 2 H₂O	1 : 2
2 HCl + Ca(OH)₂ → CaCl₂ + 2 H₂O	2 : 1
2 HCl + Na₂CO₃ → 2 NaCl + H₂O + CO₂	2 : 1

Rule of thumb. Always balance the equation first. Then read off the big numbers in front of the two substances you're working with.

Example: $H_2SO_4 + 2\,NaOH \to ...$ . Ratio of H₂SO₄ to NaOH is 1 : 2. So for every 1 mole of H₂SO₄, you need 2 moles of NaOH.

Multiplying / dividing by the ratio.

• If your known is the substance with the smaller coefficient → multiply to get the larger.
• If your known is the substance with the larger coefficient → divide to get the smaller.

Worked check (Worked Example 2 above): we knew moles of H₂SO₄ and needed moles of NaOH. Coefficient of H₂SO₄ = 1; coefficient of NaOH = 2. So multiply H₂SO₄ moles by 2 → NaOH moles. ✓

Volume conversion.

$1\,\text{dm}^3 = 1000\,\text{cm}^3 \quad \Rightarrow \quad V\,[\text{dm}^3] = V\,[\text{cm}^3] / 1000$

So 25.0 cm³ = 0.0250 dm³; 250 cm³ = 0.250 dm³; 1.00 dm³ = 1000 cm³.

Concentration unit conversion (mol/dm³ ↔ g/dm³):

$c\,[\text{g/dm}^3] = c\,[\text{mol/dm}^3] \times M_r$

So 0.100 mol/dm³ of NaOH (Mᵣ = 40) = 0.100 × 40 = 4.0 g/dm³.

Worked example. A NaOH solution has a concentration of 0.250 mol/dm³. Express this in g/dm³.

• Mᵣ(NaOH) = 23 + 16 + 1 = 40.
• $c = 0.250 \times 40 = 10.0\,\text{g/dm}^3$.

Worked example — going the other way. A solution of NaCl has 11.7 g/dm³. Express this in mol/dm³.

• Mᵣ(NaCl) = 23 + 35.5 = 58.5.
• $c = 11.7 / 58.5 = 0.200\,\text{mol/dm}^3$.

Why does AQA care about both units? Industrial chemists and pharmacists work in g/dm³ (easier to weigh out). Lab chemists work in mol/dm³ (matches the equation). UK GCSE Chemistry expects you to switch between the two using $M_r$.

Question. In a UK GCSE Chemistry titration, 25.0 cm³ of 0.150 mol/dm³ sulfuric acid (H₂SO₄) exactly neutralises 30.0 cm³ of sodium hydroxide solution (NaOH). The equation for the reaction is:

$H_2SO_4 + 2\,NaOH \to Na_2SO_4 + 2\,H_2O$

(a) Calculate the concentration of the NaOH in mol/dm³. (b) Express your answer in g/dm³.

Answer.

Step 1 — Moles of H₂SO₄ (the known).

• $V = 25.0/1000 = 0.0250\,\text{dm}^3$.
• $n = c \times V = 0.150 \times 0.0250 = 3.75 \times 10^{-3}\,\text{mol}$.

Step 2 — Mole ratio. From the equation: 1 mol H₂SO₄ reacts with 2 mol NaOH.

• Moles of NaOH = $2 \times 3.75 \times 10^{-3} = 7.50 \times 10^{-3}\,\text{mol}$.

Step 3 — Concentration of NaOH (the unknown).

• $V = 30.0/1000 = 0.0300\,\text{dm}^3$.
• $c = n/V = (7.50 \times 10^{-3}) / 0.0300 = 0.250\,\text{mol/dm}^3$. (a)

Part (b). Mᵣ(NaOH) = 23 + 16 + 1 = 40.

• $c\,[\text{g/dm}^3] = 0.250 \times 40 = 10.0\,\text{g/dm}^3$. (b)

Mark-scheme breakdown (5 marks typical for this style of question):

• Volume conversion 25.0/1000 = 0.0250 dm³ ✓ (1 mark)
• $n(H_2SO_4) = 0.150 \times 0.0250$ ✓ (1 mark)
• Use of mole ratio 1 : 2 ✓ (1 mark)
• $c(NaOH) = n/V = 0.250\,\text{mol/dm}^3$ ✓ (1 mark)
• Conversion to 10.0 g/dm³ ✓ (1 mark)

Always lay your working out in this structured way — even a wrong arithmetic answer can still score 4 out of 5.