Mole Concept and Stoichiometry

Chapter in a nutshell: This chapter connects the number, mass and volume of chemical particles through the mole (6.022 × 10²³ units) and Avogadro's number. Building on Gay-Lussac's Law of Combining Volumes and Avogadro's Law, it lets us deduce atomicity, link vapour density to molecular mass (M = 2 × V.D.), and perform stoichiometric mass–mass, mass–volume and volume–volume calculations from balanced chemical equations, plus derive empirical and molecular formulae from percentage composition.

1. Gas Laws — The Foundation

A gas responds to changes in pressure (P), temperature (T) and volume (V); changing any one affects the others. Three relationships are recalled from Class IX:

Absolute (Kelvin) scale: absolute zero = 0 K = −273 °C. To convert: T(K) = t(°C) + 273.

2. Gay-Lussac's Law of Combining Volumes

Statement: When gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the gaseous product(s), provided all volumes are measured at the same temperature and pressure.

Key points:

• Gay-Lussac published this in 1808 as the Law of Combining Volumes of Gases.
• It is valid only for gases — volumes of solids and liquids are treated as zero (negligible).
• It is essentially a volume-based statement; the coefficients of a balanced gaseous equation directly give the volume ratio.

Illustrative examples (volume ratios):

3. Avogadro's Law

Statement (Amedeo Avogadro, 1811): Equal volumes of all gases, under the same conditions of temperature and pressure, contain the same number of molecules.

So one litre of H₂ contains the same number of molecules as one litre of O₂, Cl₂ or any other gas at the same T and P. Avogadro distinguished atom (smallest particle of an element that takes part in a chemical reaction; may not exist independently) from molecule (smallest particle of an element or compound that can exist independently).

Applications of Avogadro's Law:

1. It explains Gay-Lussac's Law (since equal volumes hold equal molecules, the molecule ratio = volume ratio).
2. It determines the atomicity of elementary gases.
3. It helps determine the molecular formula of a gas.
4. It establishes the relation between vapour density and relative molecular mass (M = 2 × V.D.).
5. It gives the value of molar volume (22.4 dm³ at S.T.P.) and the relationship between gram molecular mass and gram molecular volume.

Deducing atomicity (worked logic): In H₂ + Cl₂ → 2HCl, x molecules of H₂ + x molecules of Cl₂ → 2x molecules of HCl. Dividing, ½ molecule H₂ + ½ molecule Cl₂ → 1 molecule HCl. Since atoms are indivisible, each molecule of H₂ and Cl₂ must contain at least two atoms — hence both are diatomic. Similarly N₂ + O₂ → 2NO proves nitrogen and oxygen are diatomic.

4. Atomicity

Atomicity = the number of atoms present in one molecule of an element.

• Molecules of the same type of atoms = homoatomic (e.g. P₄, O₃).
• Molecules of different types of atoms = heteroatomic (e.g. HCl, NH₃).
• Note distinction: 2H = two separate atoms of hydrogen; H₂ = one molecule of hydrogen (containing two atoms).

5. Relative Atomic Mass and Relative Molecular Mass

Atoms are too small to weigh directly, so masses are compared to a standard. Carbon-12 was adopted in 1961 as the standard (earlier H = 1 was used; either is accepted in ICSE).

Relative Atomic Mass (RAM / Atomic Weight): the number of times one atom of an element is heavier than 1/12th the mass of an atom of carbon-12.

$$\mathrm{Relative\ atomic\ mass = \dfrac{Mass\ of\ 1\ atom\ of\ the\ element}{\tfrac{1}{12} \times mass\ of\ one\ C\text{-}12\ atom}}$$

Fractional atomic masses & isotopes: Most atomic masses are not whole numbers because natural elements are a mixture of isotopes (same atomic number, different mass number). The RAM is the weighted average. For chlorine (isotopes of mass 35 and 37 in ratio 3 : 1):

$$\mathrm{Avg.\ atomic\ mass = \dfrac{(35\times3)+(37\times1)}{4} = \dfrac{142}{4} = 35.5}$$

Relative Molecular Mass (RMM / Molecular Weight): the number of times one molecule of a substance is heavier than 1/12th the mass of an atom of carbon-12. It is obtained by adding the relative atomic masses of all atoms in the molecule. Example: H₂SO₄ = (2×1) + 32 + (4×16) = 98 u.

6. Gram Atomic Mass and Gram Molecular Mass

• The quantity of an element that weighs equal to its GAM = one gram atom (e.g. 23 g Na = 1 g atom; 4 g He = 1 g atom).
• A sample whose mass equals its GMM = one gram molecule (e.g. 17 g NH₃ = 1 gram molecule).

7. The Mole Concept

It is impractical to count individual atoms/molecules, so a fixed collection — the mole — is used, just like a dozen (12) or a gross (144).

Definition: A mole is the amount of a pure substance containing the same number of chemical units (atoms, molecules or ions) as there are atoms in exactly 12 g of carbon-12, i.e. 6.022 × 10²³ units.

Avogadro's Number (Nₐ): the number of atoms present in 12 g (the gram atomic mass) of C-12 = 6.022 × 10²³. It is the number of elementary units in one mole of any substance.

8. Molar Volume

Molar volume = the volume occupied by one mole of a gas at S.T.P. = 22.4 dm³ (litres) = 22 400 cm³.

This follows from Avogadro's Law: the gram molecular mass of any gas contains Nₐ molecules and occupies 22.4 dm³ at S.T.P. (1 litre = 1 dm³ = 1000 cm³ = 1000 mL).

Caution: It is wrong to say "1 gram of any gas occupies 22.4 L at S.T.P." — it is 1 mole (1 GMM), not 1 gram, that occupies 22.4 L.

9. Key Mole Relationships (Formulae)

10. Vapour Density and Molecular Mass

Vapour Density (V.D.): the ratio of the mass of a certain volume of a gas (or vapour) to the mass of an equal volume of hydrogen, measured under the same conditions of temperature and pressure.

$$\mathrm{V.D. = \dfrac{Mass\ of\ a\ certain\ volume\ of\ gas}{Mass\ of\ an\ equal\ volume\ of\ hydrogen}}$$

• Vapour density is a ratio, so it has no units.
• Applying Avogadro's Law (equal volumes → equal molecules), V.D. = (mass of 1 molecule of gas) / (mass of 2 atoms of H). Multiplying both sides by 2 gives 2 × V.D. = mass of 1 molecule relative to 1 atom of H = relative molecular mass.

$$\boxed{\mathrm{Molecular\ mass = 2 \times Vapour\ Density}}$$

Example: V.D. of CO₂ = 22 → molecular mass = 2 × 22 = 44. V.D. of Cl₂ = 35.5 → M = 71.

11. Percentage Composition

The percentage composition of a compound is the percentage by mass of each element in one molecule.

$$\mathrm{\%\ of\ element = \dfrac{Mass\ of\ that\ element\ in\ the\ molecule}{Gram\ molecular\ mass\ of\ compound} \times 100}$$

Example — ammonium nitrate NH₄NO₃ (M = 80): %N = (28/80)×100 = 35 %, %O = (48/80)×100 = 60 %, %H = (4/80)×100 = 5 %.

12. Empirical and Molecular Formula

$$n = \dfrac{Molecular\ mass}{Empirical\ formula\ mass} = \dfrac{2 \times V.D.}{Empirical\ formula\ mass}$$

Steps to find empirical formula from % composition: (1) divide each element's % by its atomic mass → relative number of moles; (2) divide all by the smallest value → simplest ratio; (3) round to whole numbers → empirical formula. Then use V.D./molecular mass to find n and hence the molecular formula.

13. Information from a Chemical Equation & Stoichiometric Calculations

A balanced chemical equation conveys: (i) names of reactants and products; (ii) relative number of molecules; (iii) relative number of moles; (iv) relative masses; (v) for gases, relative volumes (Gay-Lussac). The equation must be balanced before any numerical work.

Three calculation types:

• Mass ↔ mass: use molar masses from the balanced equation.
• Mass ↔ volume: 1 mole of any gas = 22.4 dm³ at S.T.P.
• Volume ↔ volume: coefficients give the volume ratio directly (Gay-Lussac).

The Unitary Method is used to scale: find the quantity for "1 mole / 1 volume," then multiply.

Worked Examples

Example 1 — Moles from mass. Find moles in 11 g of N₂ (N = 14). M(N₂) = 28. n = 11/28 = 0.39 mol.

Example 2 — Volume of a gas. Volume of 320 g SO₂ at S.T.P. (S = 32, O = 16). M(SO₂) = 64; n = 320/64 = 5 mol. Volume = 5 × 22.4 = 112 dm³.

Example 3 — Vapour density → molecular formula. A compound of C and H has V.D. = 13 and empirical formula CH. Empirical mass = 13. Molecular mass = 2 × V.D. = 26. n = 26/13 = 2. Molecular formula = (CH)₂ = C₂H₂ (acetylene).

Example 4 — Empirical formula from % composition. A compound has C = 14.4 %, H = 1.2 %, Cl = 84.5 % (H=1, C=12, Cl=35.5).

Example 5 — Volume–volume stoichiometry. Oxygen oxidises ethyne: 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O. What volume of ethyne at S.T.P. produces 8.4 dm³ of CO₂? 2 vol C₂H₂ → 4 vol CO₂, so 1 vol CO₂ needs ¼ vol C₂H₂. Volume of C₂H₂ = (2/4) × 8.4 = 4.2 dm³.

Example 6 — Mass–volume from an equation. O₂ is evolved by heating KClO₃ (catalyst MnO₂): 2KClO₃ →(MnO₂) 2KCl + 3O₂. Mass of KClO₃ needed to produce 6.72 L O₂ at S.T.P.? (K=39, Cl=35.5, O=16). 3 × 22.4 L O₂ comes from 2 × 122.5 g KClO₃. KClO₃ = (2 × 122.5)/(3 × 22.4) × 6.72 = 24.5 g.

Example 7 — Combustion calculation. Volume of O₂ for complete combustion of 8.8 g propane: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. 44 g C₃H₈ needs 5 × 22.4 dm³ O₂. For 8.8 g: (5 × 22.4 / 44) × 8.8 = 22.4 dm³ of O₂.

Example 8 — Molecules in a given mass. Number of molecules in 3.6 g of water. M(H₂O) = 18; n = 3.6/18 = 0.2 mol; molecules = 0.2 × 6.022 × 10²³ = 1.2 × 10²³.

Key Terms — Quick Glossary

Common Mistakes to Avoid

• Writing "1 gram of any gas = 22.4 L at S.T.P." — it is 1 mole (1 GMM), not 1 gram.
• Forgetting to balance the equation before doing mass/volume calculations — the ratios will be wrong.
• Confusing 2H (two atoms) with H₂ (one molecule of two atoms).
• Mixing up empirical (simplest ratio) and molecular (actual number) formula.
• Treating vapour density as having units — it is a dimensionless ratio; and forgetting M = 2 × V.D.
• Applying Gay-Lussac's / Avogadro's law to solids or liquids — they apply only to gases.
• Using 6.022 × 10²³ for molecules when the substance is ionic (count formula units / ions correctly, e.g. CaCl₂ gives 2 Cl⁻ ions per unit).
• Forgetting that volume–volume ratios hold only at the same temperature and pressure.

Likely Exam Questions (with crisp answers)

1. State Gay-Lussac's Law of Combining Volumes. When gases react they do so in volumes bearing a simple whole-number ratio to one another and to the gaseous products, all volumes measured at the same T and P.
2. State Avogadro's Law. Equal volumes of all gases under the same conditions of T and P contain the same number of molecules.
3. Define atomicity. Give the atomicity of H₂, P₄, S₈. Number of atoms in one molecule of an element; H₂ = 2, P₄ = 4, S₈ = 8.
4. Differentiate 2H and H₂. 2H = two separate hydrogen atoms; H₂ = one hydrogen molecule (two atoms bonded).
5. What is Avogadro's number? Nₐ = 6.022 × 10²³ — number of units in one mole.
6. Define molar volume; what is it for N₂ at S.T.P.? Volume occupied by 1 mole of a gas at S.T.P. = 22.4 dm³ (same for nitrogen).
7. Why is the atomic mass of chlorine 35.5? Cl has isotopes of mass 35 and 37 in ratio 3:1; weighted average = (35×3 + 37×1)/4 = 35.5.
8. Relate molecular mass and vapour density. Molecular mass = 2 × Vapour density.
9. State 5 applications of Avogadro's Law. Explains Gay-Lussac's law; finds atomicity; finds molecular formula; relates V.D. & molecular mass; gives molar volume / Avogadro's number.
10. Define empirical and molecular formula. Empirical = simplest whole-number ratio of atoms; molecular = actual number of atoms in a molecule.
11. Why is vapour density unitless? It is a ratio of two masses (gas to equal volume of H₂), so units cancel.
12. State the number of moles in 11 g of N₂. 11/28 = 0.39 mol.
13. What information does a chemical equation convey? Names, and relative numbers of molecules, moles, masses, and (for gases) volumes of reactants and products.
14. Why must the volume of a gas be stated with its T and P? Gas volume changes markedly with temperature and pressure, so the value is meaningless without them.
15. Define gram atomic mass with an example. Atomic mass expressed in grams; e.g. 16 g of oxygen = 1 gram atom.
16. What is the mass of 2.24 mL of propane at S.T.P.? 22 400 mL → 44 g; 2.24 mL → 44 × 2.24/22 400 = 0.0044 g ≈ 0.0001 g (per book scaling).
17. Calculate % of N in NH₄NO₃ (M = 80). (28/80) × 100 = 35 %.
18. What volume does 6.023 × 10²³ molecules of SO₂ occupy at S.T.P.? That is 1 mole → 22.4 L.
19. Empirical formula of butane (C₄H₁₀)? Divide by 2 → C₂H₅.
20. State the mass of 1 mole of NH₃. 17 g.

Chemical Equations & Formulas (quick reference)

Gas laws & mole relationships

• P₁V₁ = P₂V₂ (Boyle); V₁/T₁ = V₂/T₂ (Charles); P₁V₁/T₁ = P₂V₂/T₂ (combined)
• n = W/M ; n = V(dm³)/22.4 ; n = N/(6.022 × 10²³)
• N = n × 6.022 × 10²³ ; W = n × M ; V = n × 22.4 dm³
• Mass of one atom = GAM/(6.022 × 10²³) ; Mass of one molecule = GMM/(6.022 × 10²³)
• Molecular mass = 2 × Vapour Density
• % of element = (mass of element in molecule / molecular mass) × 100
• Molecular formula = n × Empirical formula ; n = molecular mass / empirical formula mass

Combining-volume / combustion equations (all balanced)

• H₂ + Cl₂ → 2HCl
• N₂ + 3H₂ → 2NH₃
• N₂ + O₂ → 2NO
• 2CO + O₂ → 2CO₂
• 2H₂ + O₂ → 2H₂O
• CH₄ + 2O₂ → CO₂ + 2H₂O
• 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
• C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
• 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
• C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
• 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
• CH₄ + 2Cl₂ → CH₂Cl₂ + 2HCl (substitution)
• C₂H₂ + 2Cl₂ → C₂H₂Cl₄ (addition)
• 2NO + O₂ → 2NO₂

Decomposition / thermal equations (all balanced)

• 2KClO₃ →(MnO₂, Δ) 2KCl + 3O₂
• CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
• 2Ca(NO₃)₂ →(Δ) 2CaO + 4NO₂ + O₂
• 4NH₃ + 5O₂ → 4NO + 6H₂O (catalytic oxidation of ammonia)

Useful molar masses (u / g mol⁻¹)

• H₂ = 2, O₂ = 32, N₂ = 28, Cl₂ = 71, CO₂ = 44, SO₂ = 64, NH₃ = 17, H₂O = 18, CH₄ = 16, NaOH = 40, CaCO₃ = 100, CaCl₂ = 111, NH₄NO₃ = 80, H₂SO₄ = 98, C₆H₁₂O₆ = 180.

Constants: Avogadro's number Nₐ = 6.022 × 10²³ ; Molar volume = 22.4 dm³ (22 400 cm³) at S.T.P. ; S.T.P. = 0 °C (273 K) and 760 mm Hg.