Amount of substance: AQA A-level Chemistry

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Atoms are too small for ordinary gram measurements to be convenient. Relative masses compare them with a shared reference: one-twelfth of the mass of one carbon-12 atom.

Relative isotopic mass describes an atom of one particular isotope. Relative atomic mass, Aᵣ, is a weighted mean for the element, allowing for its isotope abundances.

Because each relative mass is a ratio of two masses, it has no unit. Keep that separate from molar mass, which is normally expressed in g mol⁻¹. The definitions below give the precise exam wording.

Use the Aᵣ values on the Periodic Table supplied with the question. These are given to one decimal place, for example Cl = 35.5. Keep any more precise isotopic masses explicitly given in a mass-spectrum question.

Relative atomic mass, A_r

The weighted mean mass of an atom of an element relative to one-twelfth of the mass of one carbon-12 atom.

Relative molecular mass, M_r

The mean mass of a molecule relative to one-twelfth of the mass of one carbon-12 atom.

Relative formula mass

The sum of the relative atomic masses in the formula unit of an ionic compound. Ionic substances do not contain separate molecules.

Molar mass

The mass per mole of specified entities. Its unit is g mol⁻¹; its numerical value matches the corresponding relative mass.

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The formula tells you how many atoms of each element are present. To find a relative molecular or formula mass, add the Aᵣ contribution of every atom in the formula.

1. Write down how many atoms of each element are present.
2. Multiply the count of each element by its Aᵣ.
3. Add the contributions.

Brackets are the usual trap. In Al₂(SO₄)₃, the outside 3 multiplies both S and O, giving Al₂S₃O₁₂. It does not multiply the Al.

Use relative molecular mass for discrete molecules. For compounds such as ionic solids that do not contain molecules, use relative formula mass. Give no unit for a relative mass. The same numerical value used as a molar mass has units of g mol⁻¹.

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A mole is a unit for measuring amount of substance, rather like a dozen is a unit for counting items. You can have one mole, two moles or half a mole.

The mole and the Avogadro constant do different jobs. mol is the unit; L tells you how many specified particles there are per mole. Think of “dozen” and 12: the word names the counting unit, while the number tells you how many items it contains.

Use L = 6.022 × 10²³ mol⁻¹, rounded to four significant figures. So 1 mol contains approximately 6.022 × 10²³ specified particles. Use a different value if a question explicitly supplies one.

The amount of substance has symbol n and unit mol. Name what you are counting: atoms, molecules, formula units, ions or electrons. One mole of each contains the same number of entities, but their masses differ. One mole of O₂ molecules contains two moles of oxygen atoms.

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Before using a formula, decide what the question wants you to count. A molecule, an atom and an ion are not interchangeable labels.

Use molecules for substances made of separate molecules, such as water. For an ionic compound such as NaCl, use formula units rather than molecules.

Then look inside the formula. One H₂O molecule contains two H atoms and one O atom. One formula unit of an ionic compound may contain more than one of a particular ion. Apply that multiplier only when the question asks for those component particles.

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Use the relative molecular or formula mass, M_r, in mass-to-moles calculations. M_r has no unit. Its numerical value is the same as the mass of one mole in grams: M_r(H₂SO₄) = 98.1 means one mole of H₂SO₄ has a mass of 98.1 g.

1. Add the Aᵣ contributions in the formula to find M_r.
2. With mass in grams, divide by M_r to find amount in moles: n = m/M_r.
3. To find mass in grams, multiply the amount in moles by M_r: m = nM_r.

These are numerical calculation rules using mass in g and amount in moles. For atoms, use Aᵣ instead. Molar mass is the mass per mole and has unit g mol⁻¹; M_r itself remains unitless.

Compare with one mole to check your answer. If M_r(CaCO₃) = 100.1, then 0.350 mol must have a mass below 100.1 g.

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The Avogadro constant, L, connects amount in moles with particle number. Use L = 6.022 × 10²³ mol⁻¹ unless a question supplies another value.

• Moles to particles: multiply by L, using N = nL.
• Particles to moles: divide by L, using n = N/L.

N is the number of particles and has no unit. The amount of substance, n, is measured in mol. Label the answer with the entity, such as molecules or chloride ions. L has unit mol⁻¹ because it gives the number of particles per mole.

If the target is an atom or ion within a formula, include its formula multiplier as well.

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The amount of a compound is not automatically the amount of every particle inside it. Use the formula to count how many of the requested particles occur in one molecule or formula unit.

1. Find the amount in moles of the stated substance.
2. Multiply by the number of target atoms or ions in each formula.
3. Use N = nL if the question asks for the actual particle number.

For electron counts, use atomic number and ionic charge first. Al³⁺ has 13 - 3 = 10 electrons; Cl⁻ has 17 + 1 = 18. Multiply the amount of those ions by their electron count, then by L.

Write the entity beside intermediate amounts. This makes it easier to spot an incorrect multiplier.

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For a mass-to-particles question, make two clear moves: mass to moles, then moles to particles.

1. Divide mass in grams by M_r to find moles of the substance. For atoms, use Aᵣ.
2. If needed, use the formula to find moles of the particular atom or ion being counted.
3. Multiply that mole amount by L.

For molecules or formula units, N = (m/M_r)L when m is in grams. If you need a component particle, include its formula multiplier. Writing both stages makes the target easier to track.

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A reverse calculation follows the same relationships in the opposite direction. Start with the target and the information you have.

1. Divide a particle count by L to find moles.
2. Undo any formula multiplier if you need moles of the whole compound.
3. To find mass in grams, use m = nM_r.
4. To find an unknown M_r, divide mass in grams by amount in moles: M_r = m/n. Give no unit for M_r.

Label the substance and entity at each step. Keep the full calculator value of the amount until the final answer.

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A good method can still lose marks through units, powers of ten or early rounding. Make a final check part of every calculation.

• Keep the calculator's full value during intermediate steps. Round only the final answer.
• Use sensible significant figures from the measured or supplied data. Exact subscripts and equation coefficients do not limit them.
• Check the unit and the entity: mol, g, atoms and molecules describe different answers.
• Check the direction. Dividing by a large molar mass should give fewer moles for a fixed mass.

In standard form, one non-zero digit comes before the decimal point. For example, 0.003045 mol = 3.045 × 10⁻³ mol. At three significant figures, 7.849 × 10²² becomes 7.85 × 10²².

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A reaction rearranges atoms; it does not create or destroy them. A balanced equation must therefore have the same number of atoms of each element on both sides.

1. Write the correct formulae before balancing.
2. Count the atoms on each side. Start with an element appearing in one reactant and one product if possible.
3. Change the coefficients in front of whole formulae. A coefficient multiplies every atom in that formula.
4. Recount every element, use the smallest whole-number coefficients and check the state symbols.

Never change a subscript to make the numbers balance. Changing H₂O to H₂O₂ would change the substance, not balance the original reaction.

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An ionic equation removes the ions that are only watching the reaction. These spectator ions appear unchanged on both sides.

1. Start with the balanced full equation, including state symbols.
2. Separate appropriate aqueous ionic substances, strong acids and strong alkalis into their ions. Keep the coefficients.
3. Keep solids, liquids, gases and undissociated molecular substances intact. An (aq) label alone does not mean every substance should be split.
4. Cancel spectator ions that are identical on both sides.
5. Check both atom numbers and total charge.

For a strong acid reacting with a strong alkali, the reacting particles are H⁺(aq) and OH⁻(aq), which form water. The remaining ions do not need to appear in the net equation.

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A reacting-mass calculation has one important bridge: the balanced equation connects moles of one substance to moles of another.

The coefficients in a balanced equation compare moles, not masses. In the grid, work across the amount row using these coefficients before converting back to mass.

1. Write the balanced equation above your grid.
2. Put in the known mass and the relevant Aᵣ or M_r values, with no units on relative masses.
3. Divide mass in grams by the relative mass to reach the moles row.
4. Use the equation coefficients to find moles of the other substance.
5. Multiply by its relative mass to find mass in grams.

The grid also works backwards from a product mass. Use M_r for molecules or formula units, and Aᵣ for atoms. Keep mass in grams and amount in moles.

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If a product mass is given, work back through the balanced equation to find how much reactant produced it. The route is the same as before, just travelled in the other direction.

For purity, distinguish the mass of pure reacting substance from the mass of the whole sample. Impurities contribute to the sample mass but, in the stated method, do not contribute to the measured reaction.

1. Use the product data to find moles of the pure reactant.
2. Convert those moles into the mass of pure reactant.
3. Divide by the total sample mass and multiply by 100.

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The limiting reactant is the one that runs out first. Once it is used up, no more product can form, even if some of the other reactant remains.

A coefficient is a big number written in front of an element or compound in a balanced equation. It multiplies the whole formula. If no number is written, the coefficient is 1. For example, in 2Al + 3Cl₂ → 2AlCl₃, the coefficients are 2, 3 and 2.

1. Convert each supplied reactant quantity into moles.
2. Divide each mole amount by its own coefficient in the balanced equation.
3. The smaller value identifies the limiting reactant and sets the maximum product amount.
4. To find excess left over, use the limiting amount to calculate how much of the other reactant reacts. Subtract that from its starting amount, then convert to mass if needed.

Compare n/coefficient, not the original masses or the mole amounts alone. Different coefficients mean different reaction requirements.

Dividing by a coefficient counts how many complete reaction groups a supply can support. For 2Al + 3Cl₂ → 2AlCl₃, each group needs 2 mol Al and 3 mol Cl₂. The supply supporting fewer complete groups runs out first.

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In a sequence of reactions, the product of one stage can become a reactant in the next. Carry its mole amount forward, then use the next balanced equation.

1. Finish the mole-ratio step for reaction 1.
2. Label the amount of intermediate formed.
3. Use that amount in reaction 2 and apply its own coefficient ratio.
4. If two reactant amounts are supplied in the second stage, check the limiting reactant again.

Do not carry one coefficient ratio across the whole chain unless you have first combined the equations correctly. Labelled intermediate amounts make the working easier to check.

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An empirical formula gives the simplest whole-number ratio of atoms. A mass ratio is not an atom ratio because different atoms have different masses.

1. Find the mass of each element. For percentages, assume a 100 g sample so each percentage becomes grams.
2. Divide each mass by its Aᵣ to find moles.
3. Divide every mole amount by the smallest one.
4. If a simple fraction remains, multiply the entire ratio to obtain whole numbers.
5. Use those whole numbers as the formula subscripts.

Do not round 1 : 1.5 to 1 : 2. Multiply both values by 2 to get 2 : 3. For a ratio of 1 : 1.33…, multiply both by 3 to get 3 : 4. This may look like 1 : 1.3 if rounded too early, so keep the full calculator values when choosing the multiplier.

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When a metal forms an oxide, the solid gains oxygen and its mass increases. First subtract the empty crucible mass so you are comparing the substances, not the apparatus.

The increase is the mass of oxygen gained, provided no solid is lost and the expected oxide is the only product. Convert the metal and oxygen masses to moles, then find their simplest ratio.

1. Heat, cool and weigh again until consecutive masses agree within the balance’s resolution.
2. Allow the apparatus to cool before weighing; hot apparatus gives unreliable readings.
3. Keep the lid slightly open so oxygen can enter while loss of solid is reduced.

Incomplete oxidation makes the apparent oxygen gain too small, so the calculated oxygen-to-metal ratio is too low. Keep heating, cooling and weighing until the mass stops increasing. This is evidence that all the metal has reacted with oxygen.

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The empirical formula gives the simplest atom ratio. The molecular formula gives the actual numbers of atoms in one molecule, so it is a whole-number multiple of the empirical formula.

1. Find the empirical formula first. If composition is given as percentages, begin with a 100 g sample.
2. Add the Aᵣ contributions to find the empirical-formula mass.
3. Divide the supplied Mᵣ by the empirical-formula mass.
4. Multiply every empirical-formula subscript by that whole-number multiplier.

Check that the final molecular formula adds up to the supplied Mᵣ. The multiplier applies to every element, not just the last one.

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Complete combustion lets you count carbon and hydrogen through the products they form. For a compound containing only C, H and O, all its carbon enters CO₂ and all its hydrogen enters H₂O.

1. Find moles of CO₂. Each mole contains one mole of C atoms, so n(C) = n(CO₂).
2. Find moles of H₂O. Each mole contains two moles of H atoms, so n(H) = 2n(H₂O).
3. Convert the C and H amounts into masses. Subtract them from the original sample mass to find the oxygen originally in the compound.
4. Convert that oxygen mass to moles, then find the empirical atom ratio.

Oxygen in the combustion products also comes from the oxygen gas used in the reaction. That is why you cannot simply count it all as oxygen from the sample. If Mᵣ is supplied, use it to find the molecular formula.

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A hydrated salt has water molecules in a fixed ratio within its crystal structure. This is water of crystallisation; it is not simply a salt dissolved in water.

Hydrated salt

A salt containing water molecules as part of its crystal structure.

Water of crystallisation

Water molecules present in a fixed ratio within the crystal structure of a salt.

Anhydrous salt

A salt containing no water of crystallisation.

In salt·xH₂O, x tells you how many water molecules occur per formula unit of salt. Heating can remove that water and leave an anhydrous salt.

1. Find the mass of anhydrous salt left after heating.
2. Find the mass of water lost by subtracting the final sample mass from the initial sample mass.
3. Convert salt and water masses separately into moles.
4. Divide moles of water by moles of salt to find x.

Check the stated chemistry: this method assumes that the loss is water and that the residue is the intended anhydrous salt.

1. Weigh the empty crucible and then the crucible containing hydrated salt.
2. Heat gently at first to avoid spitting, then sufficiently to remove water without decomposing the salt.
3. Cool and weigh; repeat heating, cooling and weighing to constant mass.
4. Use the salt-left and water-lost mole amounts to calculate x.

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Concentration tells you how much solute is present in a given volume of solution. Use the final solution volume, not the volume of water added.

For concentration in mol dm⁻³, use c = n/V with V in dm³. Divide cm³ by 1000 to convert to dm³.

Mass concentration tells you grams of solute per dm³. Calculate mass divided by solution volume, or multiply concentration in mol dm⁻³ by M_r to get mass concentration in g dm⁻³.

Write the concentration unit before choosing the calculation. mol dm⁻³ and g dm⁻³ describe different quantities.

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Adding water spreads the same amount of solute through a larger volume. The concentration falls, but the solute's mole amount stays the same, assuming no reaction or loss.

That gives c₁V₁ = c₂V₂. The volumes must use matching units. If the stock volume is unknown, rearrange first to find the volume you need to measure.

1. Measure the required stock solution with suitable volumetric apparatus.
2. Transfer it completely to a volumetric flask.
3. Add deionised water to the calibration mark, using drops near the end.
4. Stopper and invert repeatedly to mix.

Make up to the required final volume. Do not simply add that volume of water to the stock solution.

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A standard solution has an accurately known concentration. To prepare one, both the mass of solute and the final solution volume must be accurate.

1. Calculate the required mass of a suitable pure, dry, stable solid. Weigh it accurately, using weighing by difference where appropriate.
2. Dissolve the solid fully in deionised water in a beaker, stirring with a glass rod.
3. Transfer the solution through a funnel into the volumetric flask.
4. Rinse the beaker, rod and funnel into the flask. This transfers solute that would otherwise be left behind.
5. Let a warm solution cool to room temperature. Add water nearly to the mark, then dropwise until the bottom of the meniscus is on the mark at eye level.
6. Stopper and invert repeatedly, then label the solution.

Quantitative transfer means transferring all the measured solute. A volumetric flask is calibrated to contain its stated volume at the calibration temperature.

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A titration finds the volume of one solution needed to react with a measured portion of another. The burette delivers the titrant; the conical flask holds the solution being analysed.

1. Rinse the burette with deionised water, then its own solution. Fill it and run solution through the tip to remove air bubbles. Remove the filling funnel.
2. Rinse the pipette with deionised water, then the solution it will measure. Use a pipette filler to transfer a fixed aliquot (a measured portion of the solution) to the conical flask.
3. Let a standard transfer pipette drain. Do not blow out the drop designed to remain in it.
4. Add a few drops of a suitable indicator and place the flask on a white tile.
5. Read the burette meniscus at eye level and record the initial reading.
6. Do a rough titration to locate the endpoint. Repeat with fresh aliquots (measured portions), swirling and adding dropwise near the endpoint.
7. Wash splashes down the flask walls with deionised water. This changes volume but not the amount of solute being titrated.
8. Record the final reading and calculate final minus initial. Repeat until you obtain concordant titres (within 0.1 cm³ of each other).

Record burette readings to two decimal places, normally to the nearest 0.05 cm³ on a scale graduated every 0.10 cm³. Use eye protection, a pipette filler and the laboratory risk assessment.

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End point

The observed indicator colour change.

Equivalence point

The point at which the reactants have been mixed in their exact stoichiometric ratio.

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A burette scale increases downwards. The titre is the volume delivered, so subtract the initial reading from the final reading.

Read the bottom of a clear solution's meniscus at eye level. For a burette graduated every 0.10 cm³, estimate to the nearest 0.05 cm³ and record two decimal places.

Concordant titres (within 0.1 cm³ of each other) agree closely enough to use in the mean. Use a different tolerance if the question supplies one. Compare the titres themselves, not just the final readings.

Leave out the rough titre and justified outliers when finding the mean. If no set is concordant, more titrations are needed; do not choose an unsupported average.

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Use a grid with the reacting substances across the top and c (concentration), V (volume) and n (amount in moles) down the side. Add the coefficients from the balanced equation below the mole row.

1. Write the balanced equation and fill in the known concentrations and volumes. Convert each volume to dm³ by dividing its cm³ value by 1000.
2. Start in the column with both c and V known. Use n = cV to fill its mole row.
3. Work across the mole row using the equation ratio: divide by the known substance's coefficient, then multiply by the other substance's coefficient.
4. Work up the unknown substance's column: use c = n/V, dividing by that substance's own solution volume in dm³.
5. Give the concentration unit and round only the final answer.

Keep substance names beside the numbers. The burette volume and pipette volume belong to different solutions and cannot be swapped.

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A titration of a diluted sample first tells you about the diluted solution. You still need to work back to the original sample.

An aliquot is a measured portion taken from a well-mixed solution. To scale its mole amount to the full flask, multiply by whole-flask volume divided by aliquot volume.

For example, a 100.0 cm³ flask contains four 25.0 cm³ aliquots. If one contains 0.00200 mol acid, the whole flask contains 4 × 0.00200 = 0.00800 mol.

Those moles came from the original 0.0100 dm³ sample, so its concentration is 0.00800/0.0100 = 0.800 mol dm⁻³. Alternatively, multiply the diluted concentration by the dilution factor.

1 means before dilution; 2 means after dilution. Adding water changes the concentration and volume, but not the number of moles of solute. Use the same units for both volumes: either both cm³ or both dm³. The final volume is the total volume of the diluted solution, not the volume of water added.

First use the titration and balanced-equation mole ratio to find the diluted concentration. Then use the dilution equation to work back to the original concentration. The volume of the aliquot titrated is not the final dilution volume.

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A titre uses two measurements: an initial and a final burette reading. In this treatment, add their absolute uncertainties to obtain the titre uncertainty.

To find percentage uncertainty, divide the absolute uncertainty by the measured titre and multiply by 100. With the same apparatus, a larger titre gives a smaller percentage uncertainty.

A directional error is a mistake that makes a result too high or too low. For example, consistently adding acid past the end point makes the recorded titre too large. If that titre is used to calculate an unknown alkali concentration, the concentration comes out too high.

Explain the chain: what went wrong → whether the titre is too large or too small → how that changes the calculated result. This is different from uncertainty, which describes the range of values consistent with a measurement; it does not by itself tell you which way the result is wrong.

Use the uncertainty stated in the question. Two reading uncertainties alone give ±0.10 cm³ when each is ±0.05 cm³. A total including a further ±0.05 cm³ end-point judgement is ±0.15 cm³. Do not add the end-point term twice.

Precision

How closely repeated measurements agree.

Accuracy

How close a result is to the true or accepted value.

Random error

A varying effect that causes scatter.

Systematic error

A consistent effect that shifts results in one direction.

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A titration can find how much pure reacting substance was present in an impure sample. Keep track of whether each amount refers to an aliquot or the whole prepared solution.

1. Use the titre and known concentration to find moles of titrant.
2. Apply the equation ratio to find moles of the sample substance in the aliquot.
3. Scale up to the whole flask using the flask-volume/aliquot-volume ratio.
4. Convert to mass of pure substance, then divide by the original sample mass and multiply by 100.

This method assumes that the impurities do not react with the titrant or interfere with the endpoint.

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A long calculation becomes more manageable when every amount has a label: which substance, and which part of the sample? Distinguish an aliquot, the whole flask, the original sample and a production batch.

Start with the requested quantity and write an equation containing it. Work backwards through any missing quantities, then carry full calculator values through the forward calculation.

In a back titration, find the reagent left over and subtract it from the known amount originally added. Only the difference reacted with the sample. Apply the sample-reaction ratio after that subtraction.

In a standardisation chain, the concentration found in the first titration becomes data for the second. Do not round it prematurely.

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The ideal gas equation connects pressure, volume, temperature and amount. Write pV = nRT, then rearrange for the quantity you need.

1. Use pressure in Pa, volume in m³ and temperature in K when using R = 8.31 J mol⁻¹ K⁻¹.
2. Convert before substituting: add 273 to temperature in °C for these school calculations, and use 1 dm³ = 10⁻³ m³ or 1 cm³ = 10⁻⁶ m³.
3. Calculate the amount in moles. If mass is supplied, use M_r = m/n to find relative molecular mass, using mass in grams.

The ideal-gas model is a better approximation at low pressure and high temperature. The particles then occupy a smaller fraction of the volume, and attractions have less effect.

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For gases at the same temperature and pressure, volume is proportional to amount in moles. This lets you use balanced-equation coefficients as gas-volume ratios.

Check both conditions before using the shortcut. It does not turn the equation coefficients into volume ratios for liquids or solids.

If a molar gas volume Vₘ is supplied, use n = V/Vₘ with matching volume units. For example, with Vₘ = 24.0 dm³ mol⁻¹, first convert 120 cm³ to 0.120 dm³. Then n = 0.120/24.0 = 0.00500 mol.

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A reaction involving a solid, a solution and a gas still uses the same mole-ratio bridge. Each starting quantity just has a different route to moles.

1. Use n = m/M_r for the solid and n = cV for the solution, with V in dm³.
2. If both reactant quantities are supplied, check which is limiting.
3. Use the limiting amount and balanced equation to find moles of gas.
4. Convert the gas amount to volume using pV = nRT or a supplied molar gas volume.

Do not calculate a gas volume from a reactant that is actually in excess. The reactant that runs out sets the maximum amount of gas.

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The theoretical yield is the maximum amount of desired product predicted by the balanced equation. The actual yield is what you recover from the experiment.

1. Calculate the theoretical yield from the limiting reactant.
2. Divide actual yield by theoretical yield and multiply by 100.
3. Compare matching quantities: mass with mass, amount with amount, or suitable gas volumes measured under matching conditions.

A reaction may be incomplete, competing reactions may form other products, or product may be lost during transfer and purification.

A result above 100% suggests an apparent product mass that is too high, for example because the product is wet or impure. It does not mean more pure product formed than the theoretical maximum.

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Atom economy asks how much of the reactant mass is directed into the chosen desired product by the balanced equation. Percentage yield asks how much of that product is actually recovered.

Use the formula-mass contribution of the desired product, including its coefficient, divided by the total reactant formula-mass contributions, also including coefficients. Multiply by 100.

Conservation of mass means the denominator can also be the total formula-mass contribution of all products. Do not include catalysts, solvents or arbitrary excess quantities.

A high atom economy can reduce waste, but choosing a sustainable process also needs energy use, hazards, raw-material cost and supply, renewability, rate, yield, separation and recycling to be considered. A useful by-product may also affect the choice.