Amount of substance: OCR A 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 your supplied Periodic Table, which gives values to one decimal place, for example Cl = 35.5. Use the exact data given in a question, including any more precise isotopic masses.

Relative isotopic mass

The mass of an atom of an isotope compared with one-twelfth of the mass of an atom of carbon-12.

Relative atomic mass, A_r

The weighted mean mass of an atom of an element compared with one-twelfth of the mass of an atom of carbon-12.

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A formula is an atom-counting instruction. 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 separate molecules and relative formula mass for a formula describing a giant structure. 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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Relative mass and molar mass often have the same numerical value, but they describe different quantities.

Relative molecular or formula mass is a comparison with the carbon-12 reference and has no unit. Molar mass, M, is the mass of one mole and is usually given in g mol⁻¹.

For water, Mᵣ = 18.0 and M = 18.0 g mol⁻¹. Use the molar mass when calculating n = m/M, with mass in grams. Do not put g mol⁻¹ on Mᵣ itself.

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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; N_A 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 N_A = 6.02 × 10²³ mol⁻¹, rounded to three significant figures. So 1 mol contains approximately 6.02 × 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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Molar mass tells you the mass of one mole. It is the bridge between the mass you can weigh and the amount in moles.

1. Find M by adding the Aᵣ contributions in the formula. Use g mol⁻¹ for molar mass.
2. For grams to moles, divide by M: n = m/M.
3. For moles to grams, multiply by M: m = nM.

Check the units before substituting. When M is in g mol⁻¹, mass must be in grams.

A quick check is to compare with one mole. If M = 100.1 g mol⁻¹, then 0.350 mol must have a mass below 100.1 g.

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The Avogadro constant tells you how many specified entities are in one mole. Use it to move between moles and particle number.

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

N is a count, so it has no unit. Label the answer with the entity, such as molecules or chloride ions. Use the supplied value of N_A; these questions normally use 6.02 × 10²³ mol⁻¹.

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 = nN_A 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 N_A.

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. The mole stage keeps the units and the target entity visible.

1. Divide the mass by the molar mass to find moles of the substance.
2. If needed, use the formula to find moles of the particular atom or ion being counted.
3. Multiply the appropriate mole amount by N_A.

The combined expression N = (m/M)N_A is valid when N counts the same basic entity described by M. If you need a component particle, remember its formula multiplier. Writing both stages can make that easier to see.

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A reverse calculation uses the same relationships in the opposite direction. Start by writing down the target and the information you have.

1. For a particle count to moles, divide by N_A.
2. Undo any formula multiplier if you need moles of the whole compound rather than its component particles.
3. For moles to mass, use m = nM.
4. If molar mass is the unknown, rearrange to M = m/n.

Label the substance and entity at every step. The worked examples show how the route changes when the question starts with a particle count or asks for an unfamiliar molar mass.

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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. It does not connect their gram values directly.

1. Write the balanced equation above your grid.
2. Put in the known mass and the relevant molar masses.
3. Use n = m/M to reach the moles row.
4. Move across that row using the equation's coefficient ratio.
5. Use m = nM to find the required mass.

The grid works in either direction. If the product mass is known, start there and follow the mole ratio backwards. Use g mol⁻¹ for M, which is numerically equal to the corresponding relative mass.

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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.

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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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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.

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The experiment measures a mass loss, so accurate weighing and controlled heating are essential.

1. Weigh the dry empty crucible, then weigh it with the hydrated salt.
2. Heat gently at first so the salt does not spit out of the crucible.
3. Cool safely and weigh. Repeat heating, cooling and weighing until the mass is constant.
4. Subtract the crucible mass to find the anhydrous salt mass.
5. Subtract the final sample mass from the initial sample mass to find water lost, then convert both masses to moles.

Repeat heating, cooling and weighing until consecutive readings agree. This constant mass is evidence that all the water has been removed. Heat enough to remove the water without decomposing the salt, and cool before each weighing. Losing solid as well as water would spoil the calculated mole ratio.

Allow the crucible to cool before each weighing: a hot object creates air currents and buoyancy effects that can make the balance reading unreliable.

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To explain an experimental error, follow its effect on both quantities in x = n(water lost)/n(salt left). Saying 'x is inaccurate' does not explain the direction.

For example, incomplete removal of water makes the measured water loss too small and leaves the residue too heavy. Use the ratio to explain what that does to x.

Heat, cool and weigh the sample, then repeat until consecutive masses agree. This is evidence that no more water is being lost. Also check that the salt has not decomposed: a steady mass alone cannot identify the solid left behind.

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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 molar concentration by molar mass: mass concentration = cM.

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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An acid releases H⁺(aq) in aqueous solution. An alkali is a water-soluble base that produces OH⁻(aq). Familiar examples are listed in the table.

Strength and concentration are different ideas. A strong acid ionises essentially completely in water; a weak acid ionises only partly. Concentration tells you how much acid is present per unit volume.

Diluting a strong acid makes it less concentrated, not weak. Use the ionisation equations below to see the difference between essentially complete and partial ionisation.

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Salt

An ionic compound formed when the replaceable hydrogen ions of an acid are replaced by metal ions or ammonium ions (NH₄⁺).

Neutralisation is the reaction of an acid with a base. When H⁺(aq) reacts with OH⁻(aq), water forms. The full equation depends on the base: acid plus aqueous ammonia forms an ammonium salt without a water product.

The other ions form the salt. Use the acid to identify the negative ion, then balance charges with the positive ion from the base.

Different bases give different additional products. Metal oxides and hydroxides form salt and water; carbonates also produce carbon dioxide. Use the table and balanced equations to choose the correct mole ratio.

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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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For the burette diagram, the delivered volume is the final scale reading minus the initial scale reading. A single scale reading is not necessarily the delivered titre.

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The titre is final burette reading minus initial reading. Select concordant titres (within 0.1 cm³ of each other) before calculating a mean, and show which results you used.

Leave out the rough titre and justified non-concordant results. Use a different concordance tolerance if the question supplies one.

A titre depends on two burette readings, so add their absolute uncertainties in this treatment. Percentage uncertainty is absolute titre uncertainty divided by the titre, multiplied by 100.

For the same burette uncertainty, a larger titre gives a smaller percentage uncertainty. Use a different apparatus uncertainty if the question supplies one.

A directional error makes the result too high or too low. For example, consistently overshooting the end point gives too large a titre. When known acid is in the burette and unknown alkali is in the flask, this makes the calculated alkali concentration too high. Explain what changes, how the titre changes and how this changes the answer. Uncertainty alone does not tell you the direction of an error.

This example includes the two reading uncertainties only. Include any additional contribution specified in a question; do not assume that one uncertainty value applies to every burette or experiment.

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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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An aliquot is a measured portion of a well-mixed solution. A titration of that portion gives the amount in the aliquot, not automatically the amount in the whole flask.

Multiply the aliquot mole amount by whole-flask volume divided by aliquot volume to find the total. Use matching volume units in that ratio.

If the flask was made by diluting an original sample, those total solute moles came from that original sample. Divide by the original sample volume to recover its concentration. Label each stage so the dilution factor is applied only once.

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A back titration measures what is left over. You add a known excess of reagent to the sample, then titrate the unreacted excess.

1. Calculate the amount in moles of reagent originally added.
2. Use the titration and its equation ratio to find the amount in moles of that reagent left over.
3. Subtract: amount that reacted with the sample = initial amount - amount remaining.
4. Use the sample-reaction equation to find the amount in moles of sample substance, then its mass or purity.

The two reactions may have different coefficient ratios. Apply each ratio at the stage where it belongs.

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Molar gas volume, Vₘ

The gas volume per mole, V/n, at a stated temperature and pressure.

At room temperature and pressure, use a molar gas volume of 24.0 dm³ mol⁻¹, unless the question supplies a different value.

To find moles, divide the gas volume by the molar gas volume. To find volume, multiply moles by it. Make the units match first: 24.0 dm³ is 24,000 cm³.

At the same temperature and pressure, equal volumes of ideal gases contain equal mole amounts. You can therefore use balanced coefficients as volume ratios when all the substances being compared are gases under those conditions.

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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 = 8.314 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 = m/n to find molar mass.

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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A gas calculation is only as good as the volume measurement. Use a gas-tight apparatus with a syringe plunger that moves freely, and begin collection promptly so gas is not lost.

Measure temperature and pressure consistently. Collection over water is suitable only if the gas is sufficiently insoluble and does not react with the water.

To find molar mass, first calculate moles from the gas data, then use M = m/n.

If gas escapes, measured volume is too small and the calculated mole amount is too small. With the gas mass used in the calculation unchanged, dividing by that smaller n makes the calculated M too high.

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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.