2.1.3 Amount of substance

The mole is our chemical counting unit. One mole means the same number of specified particles, whatever those particles are, so I want you to say what species you are counting: 1 mol H₂O, 1 mol Na⁺ or 1 mol electrons are different statements.

That species label matters in equations too. If an equation tells you 2 mol H₂ reacts with 1 mol O₂, the coefficients are giving you a ratio of amounts of those exact species.

If you need to move between moles and individual particles, use the Avogadro constant. OCR gives Nₐ = 6.02 × 10²³ mol⁻¹ on the Data Sheet.

Multiply by Nₐ when you go from moles to particles; divide by Nₐ when you come back. Before you calculate, decide what the particles actually are — atoms, molecules, ions or electrons.

Mass and moles are linked by molar mass. In A-level calculations, the numerical value of the molar mass in g mol⁻¹ matches the relative formula or molecular mass, so once you have M you can move in either direction.

Write n = m/M before you substitute. It makes it much harder to accidentally multiply when the question needs a division.

At RTP, OCR gives you a molar gas volume of 24.0 dm³ mol⁻¹. So if the gas really is at RTP, you can convert directly between gas volume and amount without using pV = nRT.

Keep the units together: 24.0 dm³ is also 24 000 cm³. If a question supplies a different molar gas volume, use the value in the question.

Think of the empirical formula as the simplest whole-number ratio of atoms. The molecular formula tells you the actual number of each type of atom in one molecule.

So glucose has molecular formula C₆H₁₂O₆, but its empirical formula is CH₂O. OCR does not require formal definitions here, but you do need to use the two terms correctly.

For an empirical formula, turn each mass or percentage into moles first. Then divide every mole value by the smallest one and convert the resulting ratio to small whole numbers.

If the ratio comes out close to 1 : 1.5, for example, do not round 1.5 to 2. Multiply the whole ratio by 2 to get 2 : 3.

Once you have the empirical formula, find its empirical formula mass and compare that with the molecular Mᵣ. The ratio tells you how many empirical-formula units fit into one molecule.

We then multiply every subscript in the empirical formula by that same whole number.

A hydrated salt contains water molecules in a fixed ratio within its crystal lattice. We call those water molecules the water of crystallisation. An anhydrous salt has no water of crystallisation.

For example, CuSO₄·5H₂O contains five water molecules for every CuSO₄ formula unit. If you remove that water, you form anhydrous CuSO₄.

Treat the anhydrous salt and the water as two separate mole amounts, then compare them. If you heat a hydrate, the mass lost is the mass of water; the mass left is the mass of the anhydrous salt.

Label those two masses before you do any mole calculation. It stops the very common error of using the original hydrated mass as though it were the anhydrous salt.

In a reacting-mass calculation, the balanced equation sits in the middle of the route. Convert the mass you are given into moles, use the equation coefficients to move to the other substance, then convert those moles into the quantity the question asks for.

If you are unsure what to do next, write the species under the numbers. “0.20 mol Mg” is much safer than a bare 0.20.

• mass → moles
• use the balanced-equation ratio
• moles → required mass

For gases at the same temperature and pressure, the balanced equation gives you the mole ratio and therefore the reacting gas-volume ratio. If you need an actual amount of gas at RTP, OCR also lets you use 24.0 dm³ mol⁻¹.

Choose the shortest valid route. If the question gives two gas volumes at the same conditions, you often do not need to turn both into moles.

For a solution concentration in mol dm⁻³, use n = cV with V in dm³. If the question gives cm³, divide by 1000 before you substitute.

For concentration in g dm⁻³, you are dealing with mass per volume. You can convert between g dm⁻³ and mol dm⁻³ using the molar mass when you need to.

For pV = nRT, the chemistry is often easier than the units. With OCR’s R = 8.314 J mol⁻¹ K⁻¹, use pressure in Pa, volume in m³ and temperature in K.

Convert the units before you rearrange or substitute. That keeps powers-of-ten errors out of the algebra.

In a multi-step mole calculation, do not hunt for one giant formula. Work through the chemical route one link at a time and keep each number attached to its species.

Once you have moles of one substance, the balanced equation tells you the reacting ratio. Only after that ratio step should you convert to the requested mass, concentration, gas volume or particle number.

• get moles of the known species
• apply the equation ratio
• convert the new mole amount to the required quantity

Percentage yield compares what you actually obtained with the maximum amount the chemistry says you could obtain. So you need the theoretical yield before you can calculate the percentage.

If the question starts from reactant quantities, use the balanced equation to find that theoretical product amount first. Then compare the actual yield with it.

Atom economy asks where the atoms in your balanced equation end up. A high value means a larger fraction of the reactant atoms become the desired product, so less material becomes unwanted by-products.

Use the stoichiometric coefficients as well as the formula masses. If the equation forms two moles of the desired product, both moles belong in the numerator.

Choose apparatus from the precision the measurement needs. For mass, use an appropriate balance and avoid losing solid during transfer. For a fixed accurate solution volume, use volumetric apparatus; for a changing delivered volume, a burette is often the right tool. A gas syringe lets you measure gas volume directly.

In calculations, the apparatus is part of the chemistry. If you need an accurate 25.0 cm³ sample, a measuring cylinder is not a good substitute for a volumetric pipette.

If a process has high atom economy, more of the reactant atoms finish in the product you actually want. That means fewer unwanted by-products to separate, treat or dispose of.

When you explain the sustainability benefit, connect the chemistry to resource use: less waste can mean less raw material wasted and less energy or additional chemistry needed to handle by-products. Do not assume atom economy tells you everything about how green a process is; yield, energy use and hazards still matter.