Amount of substance — OCR A-Level Chemistry

This row is a course and assessment guidance point, not an assessed answer target. Read it as the heading that introduces the mole as the chemist's counting unit for amount of substance. The mole links the number of particles to mass and to gas volume: one mole contains the Avogadro number of particles, approximately 6.02 × 10²³ mol⁻¹. Use the guidance to see how the specification builds from this heading into calculations such as n = m ÷ M, where n is amount in moles, m is mass in grams and M is molar mass in g mol⁻¹. Your task is to learn the definitions and the relationships, then practise applying them to given data. Treat any centre procedure or text-selection rule as guidance for teaching, not as examinable content.

This row opens the vocabulary of amount-of-substance calculations, so read it as a signpost rather than a stand-alone test item. The specification introduces a short list of terms, beginning with amount of substance, and expects you to explain each term and then use it correctly in calculations and written answers. Treat the list as a connected toolkit: the mole, the Avogadro constant, molar mass and concentration all depend on a secure meaning of amount of substance. When you read the course, build a glossary in your own words and attach one worked example to each term, such as converting a mass of 4.0 g of helium (molar mass 4.0 g mol⁻¹) to 1.0 mol. Check each definition against your textbook and class notes, and note where a term is a quantity, a unit or a constant, because examiners reward precise language.

Amount of substance is the quantity that counts how many specified entities a sample contains, and its unit is the mole. One mole contains the Avogadro constant of entities, approximately 6.02 × 10²³ mol⁻¹, so 0.50 mol of water contains about 3.01 × 10²³ water molecules. The entities must be stated, because a mole of chlorine atoms differs from a mole of chlorine molecules. Amount of substance is not the same as mass: 1.0 mol of helium has a mass of about 4.0 g, while 1.0 mol of carbon has a mass of about 12.0 g. In multiple-choice questions, decide whether the option names the correct quantity, the correct unit or the correct number of entities, and eliminate options that confuse mass with amount or that omit the entity.

The mole is the SI unit for amount of substance, symbol mol. Amount of substance counts how many specified particles a sample contains, not how heavy or how large it is. One mole contains the Avogadro number of particles, 6.02 × 10²³. So 2 mol of water contains 2 × 6.02 × 10²³ = 1.204 × 10²⁴ water molecules, while 0.5 mol of sodium ions contains 3.01 × 10²³ ions. Always state the particle type (atoms, molecules, ions, electrons) because the same amount can refer to different entities. The mole links mass and molar mass through n = m ÷ M, and links to gases and solutions through the molar gas volume and concentration. In multiple-choice questions, check the unit written after the number and confirm the particle named matches the question.

The Avogadro constant, symbol N_A, is the number of particles per mole: 6.02 × 10²³ mol⁻¹. The unit mol⁻¹ shows it is a count per mole, so multiplying an amount in moles by N_A gives the number of particles, and dividing a particle count by N_A gives the amount in moles. For example, 0.25 mol of oxygen molecules contains 0.25 × 6.02 × 10²³ = 1.505 × 10²³ molecules, and each molecule holds two atoms, so 3.01 × 10²³ atoms. Keep the value to three significant figures as given. In multiple-choice questions, check whether the answer should be a number of particles (no unit) or an amount (mol), and confirm which particle the question asks about.

Molar mass is the mass of one mole of a substance, quoted in g mol⁻¹. You find it by adding the relative atomic masses of every atom in the formula, so H₂O is (2 × 1.0) + 16.0 = 18.0 g mol⁻¹ and CO₂ is 12.0 + (2 × 16.0) = 44.0 g mol⁻¹. The number is the same as the relative molecular or formula mass, but the unit changes from none to g mol⁻¹. To convert between mass and moles use n = m ÷ M, rearranged as m = n × M or M = m ÷ n. A question may give a mass in grams and ask for moles, or give moles and ask for the mass in grams; always check the unit you are asked for. Keep the distinction clear: molar mass is a fixed property of the substance, whereas the actual mass in grams depends on how much you have.

Molar gas volume is the volume occupied by one mole of any gas, quoted in dm³ mol⁻¹. At room temperature and pressure it is taken as 24 dm³ mol⁻¹, so one mole of any gas occupies 24 dm³ under those conditions. Because the value is the same for all gases at a fixed temperature and pressure, you can convert between moles and gas volume using n = V ÷ 24, where V is in dm³, or V = n × 24. If a volume is given in cm³, divide by 1000 first, since 1 dm³ = 1000 cm³, so 240 cm³ is 0.240 dm³ and equals 0.0100 mol. The molar gas volume depends on temperature and pressure, so state the conditions when you use 24 dm³ mol⁻¹.

Determination of formulae means working out the simplest (empirical) formula and the molecular formula of a compound from data. For empirical formula, convert the mass or percentage of each element into moles by dividing by its molar mass, then divide every mole value by the smallest to obtain the simplest whole-number ratio; if a ratio is not whole, multiply all values by the same small integer. For a hydrated salt, treat the water of crystallisation as a component and find the ratio of anhydrous salt to water. For molecular formula, divide the relative molecular mass by the empirical formula mass and multiply the empirical formula by that whole number. Combustion data and gas volumes can supply the same mole information.

This row introduces a list of terms whose definitions you must know and use accurately within Amount of substance. The published statement is truncated after the colon, so the specific terms are not shown here; read them from your specification and use this note to learn how to handle defined terms. For each term, learn a precise definition, the quantity and unit it refers to, and a context where it applies. Distinguish closely related terms carefully, for example amount of substance in moles against mass in grams, and relative atomic mass against relative molecular mass. When you answer, use the term in its correct sense rather than a loose everyday meaning, and give units where a quantity is involved.

The empirical formula gives the simplest whole number ratio of atoms of each element in a compound. It is not the actual number of atoms in a molecule, so it is often a reduced version of the molecular formula. To find it, divide each element's amount in moles by the smallest amount, then scale to the lowest whole numbers. For example, a compound containing 0.20 mol C, 0.40 mol H and 0.20 mol O gives ratios 1 : 2 : 1, so the empirical formula is CH₂O. Glucose, C₆H₁₂O₆, has the same empirical formula, CH₂O. Water's empirical formula is H₂O because its ratio is already simplest. Ionic compounds are represented by their empirical formula, for example NaCl and MgCl₂.

The molecular formula shows the actual number and type of atoms of each element in a molecule. Unlike the empirical formula, it is not reduced to the simplest ratio. For example, ethane has the molecular formula C₂H₆, showing two carbon atoms and six hydrogen atoms in each molecule, while its empirical formula is CH₃. The molecular formula is a whole number multiple of the empirical formula, so molecular formula = (empirical formula)ₙ. You can find n by dividing the relative molecular mass by the empirical formula mass. For example, if the empirical formula is CH₂O (mass 30.0) and the relative molecular mass is 180, then n = 180 ÷ 30.0 = 6, giving C₆H₁₂O₆.

Empirical formula gives the simplest whole-number ratio of atoms in a compound; molecular formula gives the actual number of atoms per molecule. To find the empirical formula from percentage composition by mass, treat each percentage as a mass in grams (so 100 g of compound), divide each mass by the element's relative atomic mass to get moles, then divide every mole value by the smallest to obtain the ratio, multiplying to clear fractions. For example, 40.0% C, 6.7% H and 53.3% O gives 3.33 : 6.7 : 3.33, which simplifies to CH₂O. To find the molecular formula, divide the relative molecular mass by the empirical formula mass and multiply the empirical formula by that whole number; with Mr = 180 and CH₂O (mass 30), the factor is 6, giving C₆H₁₂O₆. Composition by mass works identically using the given masses instead of percentages.

An anhydrous salt contains no water of crystallisation; a hydrated salt contains water of crystallisation, the fixed amount of water built into its crystal lattice, written after a dot as in CuSO₄·5H₂O. To find the formula of a hydrated salt, determine the moles of anhydrous salt and the moles of water separately, then divide both by the smaller to obtain the ratio. From percentage composition, treat percentages as masses in 100 g: subtract the water percentage to get the anhydrous salt mass, divide each by its molar mass, and simplify. From experimental results, use the mass lost on heating as the mass of water and the residual mass as the anhydrous salt. For example, if heating leaves 0.80 g of anhydrous salt (Mr 160) and loses 0.90 g of water (Mr 18), the moles are 0.0050 and 0.050, giving a ratio of 1 : 10, so the formula is salt·10H₂O.

This topic tests whether you can move between a measured quantity and the amount in moles, then use the balanced equation to find the amount of another substance, then convert back to the quantity asked for. For reacting masses, divide mass by molar mass to get moles, apply the mole ratio from the equation, then multiply by the molar mass of the target. For gas volumes at room temperature and pressure, one mole of any gas occupies 24 dm³, so volume in dm³ equals moles × 24. For solutions, concentration in mol dm⁻³ equals moles ÷ volume in dm³, so moles equals concentration × volume in dm³. Always convert cm³ to dm³ by dividing by 1000 before substituting. A typical MCQ gives a mass of one reactant and asks for the mass of product, or gives a gas volume and asks for concentration. Work in moles throughout and keep units consistent.

This specification point introduces the family of mole calculations you must be able to carry out: reacting masses, gas volumes and solution concentrations, all built on amount of substance in mol. Read it as a checklist of calculation types rather than a single fact. The core skill is converting between a measured quantity and moles, using the balanced equation to relate amounts, then converting to the requested quantity. For masses, use molar mass; for gases at room temperature and pressure, use 24 dm³ per mole; for solutions, use concentration in mol dm⁻³ and volume in dm³. When you revise, practise each type separately, then combine them in multi-step problems. Note that the statement is a heading with a colon, so the specific calculation types are the ones listed in the surrounding specification content.

This sub-statement asks you to handle the mass route through amount of substance calculations. You convert a mass in grams into moles by dividing by the molar mass, and you convert moles back into mass by multiplying by the molar mass, using n = m ÷ M and m = n × M. Molar mass is the sum of relative atomic masses in g mol⁻¹, so for water M = 18.0 g mol⁻¹ and 9.0 g is 0.500 mol. In a multiple-choice question you select the option matching your computed value and unit, checking significant figures and that the answer is chemically sensible. Always show the working: mass, molar mass, moles, then any further step such as concentration or gas volume.

This sub-statement asks you to handle the gas-volume route through amount of substance calculations. At room temperature and pressure, one mole of any gas occupies approximately 24 dm³, so moles = volume in dm³ ÷ 24, and volume = moles × 24 dm³. Convert cm³ to dm³ by dividing by 1000 before applying the molar volume, so 240 cm³ is 0.240 dm³ and equals 0.0100 mol. In a multiple-choice question you select the option matching your computed value and unit, checking that the temperature and pressure assumption is stated and that significant figures are sensible. Show the conversion step explicitly.

This sub-topic links the volume of a solution to its concentration and the amount of solute in moles. Concentration is the amount of solute per unit volume of solution, commonly expressed in mol dm⁻³, so a solution of concentration c and volume V contains n = c × V moles of solute, provided V is converted to dm³. Because 1 dm³ = 1000 cm³, a volume in cm³ is divided by 1000 before use, so 250 cm³ becomes 0.250 dm³. A 0.100 mol dm⁻³ solution in 250 cm³ therefore contains 0.100 × 0.250 = 0.0250 mol. Rearranging gives c = n ÷ V and V = n ÷ c. When a solution is diluted by adding solvent, the amount of solute stays the same, so c₁V₁ = c₂V₂. Questions test unit conversion, rearrangement and the distinction between amount in moles and concentration.

The ideal gas equation pV = nRT relates the pressure p, volume V, amount n and temperature T of a gas through the gas constant R. Pressure is in pascals (Pa), where 1 kPa = 1000 Pa and 1 atm ≈ 101 kPa; volume is in cubic metres (m³), where 1 dm³ = 1 × 10⁻³ m³ and 1 cm³ = 1 × 10⁻⁶ m³; temperature must be in kelvin, so add 273 to the Celsius value. R = 8.314 J K⁻¹ mol⁻¹. Rearranging gives n = pV ÷ RT, p = nRT ÷ V, V = nRT ÷ p and T = pV ÷ nR. A gas occupying 2.00 × 10⁻³ m³ at 100 kPa and 300 K contains n = (100000 × 2.00 × 10⁻³) ÷ (8.314 × 300) ≈ 0.0802 mol. Questions test unit conversion, rearrangement and correct substitution.

Stoichiometry links a balanced equation's mole ratio to measurable quantities. Read the equation, confirm it is balanced, then convert what you are given into moles: mass ÷ molar mass, concentration × volume in dm³, or gas volume ÷ 24.0 dm³ at room temperature and pressure. Apply the mole ratio from the coefficients to find moles of the target substance, then convert back to the requested unit. For example, in 2H₂ + O₂ → 2H₂O, burning 4.0 g of H₂ (2 mol) needs 1 mol O₂ and gives 2 mol H₂O. Keep units consistent and use the limiting reagent when two reactants are given.

Percentage yield compares the amount actually obtained with the theoretical maximum from the balanced equation: percentage yield = (actual yield ÷ theoretical yield) × 100. Atom economy measures how much of the reactant mass ends up in the desired product: atom economy = (molar mass of desired product ÷ total molar mass of all reactants) × 100, using the balanced equation. A reaction can have high atom economy yet low percentage yield if it is slow or incomplete, and vice versa. For example, addition reactions typically have 100% atom economy because all reactant atoms appear in the single product, whereas substitution reactions produce by-products and a lower atom economy.

This row is a lead-in to a family of calculation skills in amount-of-substance work, so read it as a heading that introduces the calculation types listed beneath it rather than as a single task. Your job is to work through the course material that follows and practise each calculation type it names, choosing the right relationship, converting units consistently and keeping track of significant figures. Treat the sub-items as a checklist: for each one, find a worked example, then attempt a fresh problem without looking at the solution. Record which step you find hardest, because that is what to revise. The heading itself carries no marks; the assessed skill is performing the calculations it introduces, so build fluency with moles, mass, concentration, gas volumes and equations as the material directs.

Percentage yield compares the amount of product actually obtained with the maximum amount predicted from the balanced equation, expressed as a percentage. Calculate the theoretical amount in moles from the limiting reactant, convert to the expected mass or volume, then divide the actual yield by the theoretical yield and multiply by 100. Related quantities work the same way: you may be given the percentage yield and asked for the actual or theoretical amount, so rearrange the relationship rather than recalculating from scratch. Always identify the limiting reactant first, since the theoretical yield depends on it, and keep units consistent. A yield above 100% signals an error such as an impure or wet product, not a genuine result.

Atom economy measures how much of the total mass of reactants ends up in the desired product, expressed as a percentage. Calculate it as (molar mass of desired product × 100) ÷ total molar mass of all reactants, using the balanced equation and relative formula masses. For example, in the reaction N₂ + 3H₂ → 2NH₃, the desired product is ammonia: (2 × 17.0 × 100) ÷ (28.0 + 3 × 2.0) = 3400 ÷ 34.0 = 100%, because all reactant atoms appear in the product. A reaction producing a useful product alongside a waste by-product has a lower atom economy. Atom economy depends only on the equation, not on the actual yield obtained, so a reaction can have high atom economy but low percentage yield.

Practical measurement underpins amount-of-substance calculations. For mass, use a balance of appropriate precision, tare or zero it, and record to the balance's resolution. For volumes of solutions, use a volumetric pipette for a fixed accurate volume and a burette for variable delivered volumes, reading the bottom of the meniscus at eye level; a measuring cylinder gives only approximate volumes. For gas volumes, collect gas over water in an inverted measuring cylinder or use a gas syringe, reading the volume at room temperature and pressure and noting the syringe scale. Record all readings with units and correct significant figures, and repeat titrations to obtain concordant results.

Atom economy compares the mass of atoms in the desired product with the total mass of atoms in all products, expressed as a percentage. A process with high atom economy converts most reactant atoms into the wanted product, so fewer atoms end up in unwanted by-products. This supports sustainability because fewer raw materials are wasted per tonne of product, less separation and disposal of by-products is needed, and the process makes more efficient use of finite resources. For example, an addition reaction that gives a single product has 100% atom economy, whereas a substitution that also produces a large unwanted co-product has a lower value. A multiple-choice question may ask you to identify which statement is a genuine sustainability benefit, so link the high atom economy to reduced waste and efficient use of resources rather than to reaction rate or yield.