Cambridge International AS & A Level Chemistry 9701 · Topic 2, with 4.1 The gaseous state

Atoms, molecules and stoichiometry

Chemistry is written in equations but done in grams, cubic centimetres and burette readings. Stoichiometry is the bridge: it converts what you can weigh and measure into what the equation actually counts, which is particles. Every calculation in this page is the same short journey — measurement to moles, moles across the equation, moles back to a measurement — and once you see that, the variety of questions collapses to a handful of steps in a different order.

The unified atomic mass unit2.1.1

Atoms are far too light to weigh one at a time in grams — a carbon atom comes in at about 2 × 10⁻²³ g, a number that is useless to work with and impossible to measure directly. So chemistry does what every measurement system does when absolute values are awkward: it picks a standard and compares everything to it.

Carbon-12 is the standard, and it is defined as weighing exactly 12 units — not approximately, exactly. Everything else is then measured against it, and because every mass on this page is a ratio of one mass to another, relative masses have no units. That is why you never write "the M_r of water is 18 g". It is 18. The grams arrive later, when you multiply by a mole.

Relative isotopic, atomic, molecular and formula mass2.1.2

Four terms, one standard. They are all "the mass of something compared with one twelfth of the mass of a carbon-12 atom"; the only difference is what the something is.

Why relative isotopic mass is almost a whole number

A proton and a neutron each weigh very close to one unit and an electron weighs almost nothing, so the mass of an isotope is close to its mass number — protons plus neutrons. At this level you may treat ³⁵Cl as 35 and ³⁷Cl as 37. The small discrepancies are real (they are the binding energy of the nucleus, showing up as mass) but they do not matter for any calculation in this syllabus.

Why relative atomic mass usually is not

Chlorine is 75% ³⁵Cl and 25% ³⁷Cl. Weighting the two isotopic masses by those abundances:

This is also why A_r values in the data booklet look untidy: 24.3 for magnesium, 63.5 for copper, 207.2 for lead. Each is a mixture average, and the mixture is set by how the element was made in stars and how it has decayed since.

Relative molecular against relative formula mass

Both are written M_r and both are worked out identically — add up the A_r values of every atom in the formula. The distinction is about what the formula means. Water really does exist as H₂O molecules, so 18.0 is a relative molecular mass. Sodium chloride does not exist as NaCl molecules at all; it is a giant lattice, and the formula records a 1 : 1 ratio. There is nothing there to call a molecule, so 58.5 is a relative formula mass.

Hydrates: where the arithmetic goes wrong

Water of crystallisation is written after a dot: CuSO₄·5H₂O. The dot is not a multiplication and not a bond — it means "and, built into the crystal, five waters per formula unit". For the M_r you treat those five waters as five complete H₂O units at 18.0 each.

The mole and the Avogadro constant2.2.1

An equation counts particles: 2H₂ + O₂ → 2H₂O says two hydrogen molecules for every oxygen molecule. A balance counts grams. The mole is the unit that lets one be turned into the other, and it is worth being precise about what it is.

Two phrases in that definition repay attention. "Amount of substance" is the formal name of the quantity whose unit is the mole, in the same way that "length" is the quantity whose unit is the metre; it is not a loose word for mass or volume. And "stated elementary units" means you must say what you have a mole of.

How big is 6.02 × 10²³?

Large enough that no everyday comparison helps much, which is the point: the number is set by how small atoms are, not chosen for convenience. It is defined so that one mole of carbon-12 weighs exactly 12 g. That single fact is what makes molar masses numerically equal to relative masses, and it is the reason the whole scheme works so smoothly.

Moles, mass and molar mass2.2.1

The molar mass, M, is the mass of one mole in grams per mole. Numerically it is the relative formula mass; the difference is that M_r has no units and M is in g mol⁻¹. M_r(H₂O) = 18.0; M(H₂O) = 18.0 g mol⁻¹.

Formulae of ionic compounds2.3.1

Everything downstream — equations, M_r values, reacting masses, titres — is built on getting the formula right first. A wrong formula does not produce a slightly wrong answer; it produces a confidently wrong one, and the method marks that follow are usually lost with it.

Step one: which ion is which

Step two: the charge

For a simple ion of a main-group element the charge comes from the group number: Group 1 gives 1+, Group 2 gives 2+, Group 13 gives 3+; Group 15 gives 3−, Group 16 gives 2−, Group 17 gives 1−. For a transition metal, or for lead and tin, the Roman numeral in the name is the charge — iron(III) is Fe³⁺, copper(II) is Cu²⁺. There is no shortcut for the compound ions; they have to be learnt.

Step three: balance the charges, then bracket carefully

The compound must be neutral overall, so you take whatever whole numbers of each ion make the charges cancel. Then the bracket rule: brackets go round a compound ion only when it is taken more than once. K₂SO₄ needs none because there is a single sulfate. Mg(NO₃)₂ needs them, because without them MgNO₃₂ would claim thirty-two oxygens.

Anhydrous, hydrated and water of crystallisation2.3.4

Anhydrous copper(II) sulfate is a white powder; add water and it becomes the familiar blue crystals CuSO₄·5H₂O, and the colour change is the standard test for the presence of water. The water is not contamination and not dampness: it sits at defined positions in the lattice, four of the five molecules bonded round the copper ion and the fifth held by hydrogen bonding to the sulfate.

Finding the water by experiment

Because the ratio is fixed, it can be measured. Weigh a crucible, weigh it with the hydrated salt, heat it, let it cool in a desiccator, and weigh again — repeating until two consecutive weighings agree. That is what heating to constant mass means, and it is the only way to be sure the dehydration is complete. The mass lost is the water; convert both masses to moles and the ratio is x.

Writing and balancing equations2.3.2

A balanced equation is a statement that atoms are conserved. Nothing is created or destroyed in a chemical change; the atoms are only rearranged, so whatever goes in must come out. Balancing is therefore not a puzzle with a trick — it is bookkeeping, and it always has an answer.

The procedure

1. Write the correct formulae first and then leave them alone. Remember the seven elements that exist as diatomic molecules: H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂.
2. Work left to right through the formulae, fixing one element at a time by changing the number in front.
3. Leave the element that appears in most places until last — usually oxygen, and usually in a combustion equation.
4. Go back to the start and check every element again, because fixing the last one often disturbs the first.
5. Clear any fractions by multiplying the whole equation through.

Ionic equations2.3.2

When ionic compounds dissolve, their ions separate and move independently. An equation written with the full formulae therefore pretends that pairs of ions travel together when they do not, and it hides what has actually changed. An ionic equation strips that away and shows only the species that take part.

How to get there

1. Write the full balanced equation, with state symbols.
2. Split every aqueous ionic compound and every strong acid into separate ions. Solids, liquids, gases and molecular substances such as water stay whole.
3. Cross out anything identical on both sides.
4. What is left is the ionic equation — check it still balances for both atoms and charge.

State symbols2.3.2

Four symbols, written in brackets after each formula: (s) solid, (l) liquid, (g) gas, (aq) aqueous — dissolved in water.

They are not decoration. They carry the information that distinguishes a precipitate from a dissolved product, tell you which substance can escape from the flask, and in some equations are the only difference between the two sides at all: I₂(s) → I₂(g) is sublimation, and without the symbols it says nothing.

Empirical and molecular formulae2.3.3

"Empirical" means found by experiment, and that is exactly what it is: the formula you can reach from measurements alone, without knowing anything about molecules. Weigh how much of each element is present, convert to moles, and the ratio falls out. Nothing in that procedure can tell you whether the real molecule is CH₂O or C₆H₁₂O₆ — both have the same composition by mass.

Calculating an empirical formula2.3.5

Four steps, always the same four, whether the data are masses or percentages.

1. Write down the mass of each element. If you are given percentages, assume 100 g of the compound — then the percentages are the masses in grams.
2. Divide each mass by that element's A_r to get moles.
3. Divide every answer by the smallest of them.
4. If the results are not yet whole numbers, multiply them all by the smallest factor that makes them so.

Percentages, and the 100 g trick

If a compound is 40.0% carbon by mass, then any sample contains 40.0% carbon; so you may as well pick the sample size that makes the arithmetic disappear. Choose 100 g and the percentage figure is the mass in grams. It works because a ratio does not care how much you started with.

Combustion data

A common variant gives the masses of CO₂ and H₂O produced when a compound burns. All the carbon ends up in the CO₂ and all the hydrogen in the H₂O, so:

• n(C) = n(CO₂) = mass of CO₂ ÷ 44.0
• n(H) = 2 × n(H₂O) = 2 × mass of H₂O ÷ 18.0
• If the compound might contain oxygen as well, find the masses of C and H from those moles, subtract both from the mass of compound burnt, and whatever is left is oxygen.

From empirical formula to molecular formula2.3.5

One extra step. Work out the mass of the empirical formula unit, divide the true M_r by it, and multiply every subscript by the whole number that comes out.

Topics 2.4 and 4.1

Reacting masses, volumes and gases

Everything so far has been about a single substance. From here on the balanced equation does the work, because its coefficients are the exchange rate between one substance's moles and another's.

Reacting masses from an equation2.4.1(a)

The coefficients in a balanced equation are ratios of moles, never of masses. That is the single idea this section rests on. 2Mg + O₂ → 2MgO says two moles of magnesium react with one mole of oxygen molecules — it does not say two grams with one gram, and the masses involved (48.6 g with 32.0 g) are not in a 2 : 1 ratio at all.

Limiting and excess reagents2.4.1(d)

When you are told how much of two reactants you have, one of them will usually run out first. That one is the limiting reagent, and it alone decides how much product can form. The other is in excess, and some of it will be left over at the end.

Percentage yield2.4.1(a)

The mass calculated from the equation is the theoretical yield — the most you could possibly get. What you actually collect is almost always less.

Why yields fall short

• The reaction is reversible and reaches equilibrium before it is complete.
• Side reactions use up reactant to make something else.
• Product is lost in handling — left on the filter paper, stuck to the glassware, lost in recrystallisation.
• The reactants were impure, so there was less there than the mass suggested.

Atom economy

Avogadro's law and the molar gas volume2.4.1(b)

This is a remarkable statement and it is worth pausing on. The molecules of carbon dioxide are much heavier and larger than those of hydrogen, yet a litre of each at the same conditions holds the same number of them. The reason is that a gas is mostly empty space: the molecules are so far apart that their own size and identity make almost no difference to how much room the sample takes up. What sets the volume is how many particles there are, how hard they are pushed together, and how fast they are moving.

The practical consequence is the molar gas volume, V_m: one mole of any gas occupies the same volume under the same conditions.

Avogadro's law also means that for gases, the coefficients of a balanced equation are ratios of volumes as well as of moles. In N₂(g) + 3H₂(g) → 2NH₃(g), 10 cm³ of nitrogen needs 30 cm³ of hydrogen and gives 20 cm³ of ammonia, with no conversion to moles needed at all — provided every substance is a gas and all volumes are at the same temperature and pressure.

The ideal gas equation4.1.1

The equation is the molar gas volume generalised. Rearranged as V/n = RT/p it says that the volume per mole depends only on temperature and pressure — not on which gas it is, which is Avogadro's law restated. Set T to 293 K and p to 101 kPa and it returns 24.1 dm³ mol⁻¹, the number you have been using as 24.0.

Units in pV = nRT4.1.1

The value 8.31 for R is tied to SI units. Put anything else into the equation and the answer is wrong by a factor that is rarely obvious. This is where most of the marks in this topic are lost, and it is entirely avoidable: convert everything before you start.

Finding M_r from pV = nRT4.1.1

This is the one route on the syllabus from bench measurements straight to a relative molecular mass, and it is the piece of evidence that turns an empirical formula into a molecular one.

Concentration of solutions2.4.1(c)

Two details in that definition matter. It is per cubic decimetre of solution, not of solvent — you make the solution up to the mark, you do not add a litre of water to the solid. And square brackets are the shorthand: [HCl] = 0.100 mol dm⁻³ means the concentration of hydrochloric acid is 0.100 mol dm⁻³.

The other unit

Concentration is sometimes quoted in g dm⁻³ — grams of solute per cubic decimetre. To convert, multiply or divide by the molar mass:

concentration in g dm⁻³ = concentration in mol dm⁻³ × M

Making a standard solution2.4.1(c)

To make one you need a solid that is a primary standard: pure, stable in air, not hygroscopic, of known formula including any water of crystallisation, and preferably with a large molar mass so that weighing errors matter less. Anhydrous sodium carbonate is the standard school example. Sodium hydroxide is not — it absorbs both water and carbon dioxide from the air, so a mass weighed out is never quite what the label says.

Carrying out a titration2.4.1(c)

A titration finds the volume of one solution that exactly reacts with a known volume of another. From that volume and one known concentration, the other concentration follows.

Titration calculations2.4.1(c)

The same three steps as reacting masses, with concentration and volume in place of mass at each end.

Deducing stoichiometry from experimental data2.4.1(e)

Every calculation so far has used the equation to predict a measurement. This one runs the other way: measure two amounts, and let them tell you the equation. It is the outcome most often left out of revision and it appears regularly on the papers.

Review

Review and data

Self-test30 questions

Written to the style of 9701 Paper 1 and the calculation parts of Paper 2. The explanations give the reasoning, not just the letter, and several of them name the specific error the wrong options are there to catch.

Definitions to learn word for word

Data used on this page

The relative atomic masses and constants below are the values the calculators on this page use, quoted to the precision of a standard tabulation. They are not a substitute for the Cambridge data booklet you are given in the examination, and any answer you submit should be worked out with the booklet's values. Small differences do occur — sulfur is 32.1 here and is sometimes tabulated as 32.0, which moves M_r(CuSO₄·5H₂O) between 249.6 and 249.5.