AS 2 Atoms, molecules and stoichiometry | A Level Chemistry 9701

Home › Learning Hub › A Level Chemistry › AS 2: Atoms, molecules and stoichiometry

🎯What you need to be able to do

Define the unified atomic mass unit, and use it to define relative atomic, isotopic, molecular and formula mass.
Define the mole in terms of the Avogadro constant and use it.
Write formulas of ionic compounds from ionic charges and oxidation numbers, predicting charges from the Periodic Table and recalling the common polyatomic ions.
Write balanced equations with state symbols, and ionic equations without spectator ions.
Define empirical and molecular formulas and calculate them from data; use the terms anhydrous, hydrated and water of crystallisation.
Carry out mole calculations for reacting masses and percentage yield, gas volumes, solution volumes and concentrations, and limiting and excess reagents — and deduce the stoichiometry of a reaction from the results.

📚The chemistry

This is the topic every other calculation in the course stands on. Rates, energetics, equilibria, electrolysis and titrations all start with “find the number of moles”, and a student who is fluent here has a head start on half of Paper 4.

2.1 Relative masses

Atoms are far too light to weigh in grams, so masses are compared on a scale fixed by carbon-12. The unified atomic mass unit, u, is defined as one twelfth of the mass of one atom of carbon-12. Every relative mass is a mass measured against that unit, so it is a ratio and has no units.

Relative isotopic mass — the mass of one atom of a particular isotope compared with 1/12 of the mass of a carbon-12 atom.
Relative atomic mass, A_r — the weighted average mass of the atoms of an element, taking account of the abundance of each isotope, compared with 1/12 of the mass of a carbon-12 atom.
Relative molecular mass, M_r — the weighted average mass of a molecule compared with 1/12 of the mass of a carbon-12 atom. For ionic compounds, which are not made of molecules, the same quantity is called relative formula mass, and is also written M_r.

A_r from isotopic abundances is a weighted mean. Chlorine is 75% ³⁵Cl and 25% ³⁷Cl, so

\[ A_\mathrm{r}(\mathrm{Cl}) = \frac{(35 \times 75) + (37 \times 25)}{100} = 35.5 \]

2.2 The mole

One mole is the amount of substance containing 6.022 × 10²³ particles — the Avogadro constant, L. The number was chosen so that one mole of a substance has a mass in grams equal to its M_r: 12.0 g of carbon, 18.0 g of water, 58.5 g of sodium chloride.

\[ n = \frac{m}{M} \qquad \text{number of particles} = n \times L \]

Say what the particles are. 1.00 mol of water contains 6.022 × 10²³ molecules but three times as many atoms.

2.3 Formulas and equations

Ionic formulas

Charges of simple ions follow from the group: Group 1 forms 1+, Group 2 forms 2+, Al forms 3+; Group 15 forms 3−, Group 16 2−, Group 17 1−. Transition metals have more than one charge, so their compounds carry a Roman numeral giving the oxidation number: iron(III) oxide contains Fe³⁺. Balance the charges to zero and you have the formula.

These polyatomic and other ions must be known by heart:

nitrate NO₃⁻

carbonate CO₃²⁻

sulfate SO₄²⁻

hydroxide OH⁻

ammonium NH₄⁺

hydrogencarbonate HCO₃⁻

phosphate PO₄³⁻

zinc Zn²⁺ and silver Ag⁺

So calcium phosphate is Ca₃(PO₄)₂ (3 × 2+ balances 2 × 3−) and ammonium sulfate is (NH₄)₂SO₄. Brackets go round a polyatomic ion whenever it is multiplied.

Equations and ionic equations

Balance by changing the numbers in front of formulas, never the formulas themselves, and add state symbols: (s), (l), (g), (aq). An ionic equation shows only the species that change. Write out the ions, then cross out the spectator ions that appear unchanged on both sides:

\[ \mathrm{AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)} \] \[ \mathrm{Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)} \]

Every neutralisation between a strong acid and a strong alkali reduces to H⁺(aq) + OH⁻(aq) → H₂O(l). An ionic equation must balance in charge as well as in atoms — check both.

Empirical and molecular formulas

The empirical formula is the simplest whole-number ratio of atoms of each element in a compound. The molecular formula is the actual number of atoms of each element in one molecule. Glucose has the molecular formula C₆H₁₂O₆ and the empirical formula CH₂O.

To find an empirical formula from masses or percentages: divide each by A_r to get moles, divide every answer by the smallest, and multiply up if you are left with a ratio such as 1 : 1.5 or 1 : 1.33. To go on to the molecular formula, divide M_r by the empirical formula mass and multiply the empirical formula by that whole number.

Hydrated salts

A hydrated salt has water of crystallisation built into its crystal lattice in a fixed ratio: blue copper(II) sulfate is CuSO₄·5H₂O. Heating drives the water off, leaving the anhydrous salt, white CuSO₄. The number of water molecules is found by weighing before and after heating, exactly like an empirical formula: moles of anhydrous salt against moles of water lost.

2.4 Reacting masses and volumes

Every calculation here has the same three steps: convert what you know into moles, use the equation’s ratio, convert moles back into what you are asked for. The conversions:

solids: \( n = \dfrac{m}{M} \)

solutions: \( n = c \times V \) with V in dm³

gases at room conditions: \( n = \dfrac{V}{24.0\ \mathrm{dm^3}} \)

The molar volume, 24.0 dm³ mol⁻¹ at room conditions (22.4 at s.t.p.), is in the data section. It works for any gas, because equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. For other conditions use pV = nRT, which is in topic 4.

That same fact lets you do gas-volume problems without moles at all: for gases measured at the same temperature and pressure, volume ratios are mole ratios. 20 cm³ of propane needs 5 × 20 = 100 cm³ of oxygen, because C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.

Limiting reagent and percentage yield

When two reactants are mixed in amounts that do not match the equation, the one that runs out first is the limiting reagent and decides how much product can form; the other is in excess. Find it by working out how much of one reactant the other would need.

\[ \text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 \]

Theoretical yield is what the limiting reagent could produce if everything went perfectly. Actual yields are lower because of incomplete reaction, side reactions and product lost in transfer and purification.

Deducing stoichiometry

Running the calculation backwards tells you the equation. If 0.0200 mol of a metal reacts with 0.0400 mol of HCl, the ratio is 1 : 2 and the metal forms a 2+ ion. Titration questions in Paper 3 often end exactly this way.

✏️Worked example

5.00 g of calcium carbonate is added to 50.0 cm³ of 1.50 mol dm⁻³ hydrochloric acid. \[ \mathrm{CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)} \] (a) Identify the limiting reagent. (b) Calculate the maximum volume of carbon dioxide at room conditions. (c) 820 cm³ of gas was actually collected. Calculate the percentage yield. (d) Calculate the mass of the reagent left over. [A_r: Ca 40.1, C 12.0, O 16.0]

(a) Convert both to moles. M_r(CaCO₃) = 40.1 + 12.0 + 3(16.0) = 100.1.

\[ n(\mathrm{CaCO_3}) = \frac{5.00}{100.1} = 0.04995\ \mathrm{mol} \qquad n(\mathrm{HCl}) = 1.50 \times \frac{50.0}{1000} = 0.0750\ \mathrm{mol} \]

The equation needs 2 mol HCl per mol CaCO₃, so to use up all the carbonate would need 2 × 0.04995 = 0.0999 mol HCl. Only 0.0750 mol is present, so hydrochloric acid is the limiting reagent and calcium carbonate is in excess.

(b) Work from the limiting reagent: 2 HCl give 1 CO₂, so

\[ n(\mathrm{CO_2}) = \frac{0.0750}{2} = 0.0375\ \mathrm{mol} \qquad V = 0.0375 \times 24.0 = 0.900\ \mathrm{dm^3} = 900\ \mathrm{cm^3} \]

(c)

\[ \text{percentage yield} = \frac{820}{900} \times 100 = 91.1\% \]

(d) CaCO₃ used = n(CO₂) = 0.0375 mol, so the excess is 0.04995 − 0.0375 = 0.01245 mol, and its mass is 0.01245 × 100.1 = 1.25 g.

Check it. Two independent routes should agree. 0.0375 mol of CaCO₃ reacted, which is 0.0375 × 100.1 = 3.75 g; 5.00 − 3.75 = 1.25 g left over. And the gas volume is sensible: well under a litre from a few grams of solid. Every figure is quoted to 3 s.f., matching the data.

Working from the wrong reactant. Calculating CO₂ from the 0.04995 mol of CaCO₃ gives 1.20 dm³ and a “yield” of 68% — a plausible-looking answer that is simply wrong, because there is not enough acid to react with all that carbonate. Always identify the limiting reagent before you use a ratio. The second trap is forgetting the 2: the ratio of HCl to CO₂ is 2 : 1, not 1 : 1.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its M_r is 180. Find its empirical and molecular formulas.

Take 100 g. Moles: C = 40.0/12.0 = 3.33; H = 6.7/1.0 = 6.7; O = 53.3/16.0 = 3.33. Divide by the smallest (3.33): C 1, H 2.0, O 1. Empirical formula CH₂O, with empirical formula mass 12.0 + 2.0 + 16.0 = 30.0. 180 / 30.0 = 6, so the molecular formula is C₆H₁₂O₆.

2. 2.50 g of hydrated copper(II) sulfate, CuSO₄·xH₂O, is heated to constant mass, leaving 1.60 g of anhydrous CuSO₄. Find x. [M_r: CuSO₄ 159.6, H₂O 18.0]

Mass of water lost = 2.50 − 1.60 = 0.90 g. n(CuSO₄) = 1.60 / 159.6 = 0.01003 mol; n(H₂O) = 0.90 / 18.0 = 0.0500 mol. Ratio 0.0500 / 0.01003 = 4.99, so x = 5: CuSO₄·5H₂O. “Heated to constant mass” matters: it is how you know all the water has gone. If heating had stopped early, the water lost would be too small and x would come out low.

3. 25.0 cm³ of sodium hydroxide solution is neutralised by 20.0 cm³ of 0.100 mol dm⁻³ sulfuric acid. Write the equation and calculate the concentration of the sodium hydroxide.

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l). n(H₂SO₄) = 0.100 × 20.0/1000 = 0.00200 mol. The ratio is 1 : 2, so n(NaOH) = 0.00400 mol in 25.0 cm³. c = 0.00400 / (25.0/1000) = 0.160 mol dm⁻³. Sulfuric acid is diprotic; treating the ratio as 1 : 1 gives 0.0800, the most common wrong answer.

4. 20 cm³ of propane is burned in 150 cm³ of oxygen. All volumes are measured at room temperature and pressure. What is the final volume of gas, and what is it after passing through aqueous sodium hydroxide?

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l). Volume ratios equal mole ratios for gases at the same conditions, so 20 cm³ propane uses 100 cm³ O₂ and forms 60 cm³ CO₂. Oxygen left over: 150 − 100 = 50 cm³. At room temperature the water is a liquid and its volume is negligible. Final gas volume = 60 + 50 = 110 cm³. Sodium hydroxide absorbs the acidic CO₂, leaving 50 cm³ of oxygen.

5. Write ionic equations, with state symbols, for (a) aqueous barium chloride and aqueous sodium sulfate; (b) magnesium ribbon and dilute hydrochloric acid.

(a) Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s). Na⁺ and Cl⁻ are spectators. (b) Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g). Cl⁻ is the spectator. Check the charge: left 2+, right 2+. Magnesium stays as Mg(s) on the left, because a solid metal is not ionised.

6. 0.486 g of a Group 2 metal M reacts completely with excess dilute hydrochloric acid, giving 480 cm³ of hydrogen at room conditions. Identify M.

n(H₂) = 480 / 24 000 = 0.0200 mol. A Group 2 metal forms M²⁺: M + 2HCl → MCl₂ + H₂, a 1 : 1 ratio of metal to hydrogen, so n(M) = 0.0200 mol. A_r(M) = 0.486 / 0.0200 = 24.3: magnesium. The stoichiometry was deduced from the group; if the group were unknown you would try 1 : 1, 2 : 1 and 2 : 3 and see which gives a real A_r.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

Chemguide (Jim Clark) — the calculations section, including a set of worked problems on moles, concentrations and gas volumes
Royal Society of Chemistry — Starters for Ten and the mole-calculation worksheets, for quick daily practice
PhET — Reactants, Products and Leftovers, a simple visual of limiting reagents