Chemical formulae and equations

Section: 2 Atoms, Molecules and Stoichiometry  |  Syllabus: Cambridge AS Level Physics 9702

Types of Formula Empirical Formula The simplest whole-number ratio of atoms of each element in a compound. Molecular Formula The actual number of atoms of each element in one molecule of a compound. It is always a whole-number multiple of the empirical formula.

Compound Molecular Formula Empirical Formula Water H₂O H₂O Hydrogen peroxide H₂O₂ HO Glucose C₆H₁₂O₆ CH₂O Ethane C₂H₆ CH₃ Benzene C₆H₆ CH Key Point Ionic compounds (e.g. NaCl, MgO) are always expressed as empirical formulas - they have no discrete molecules, so a molecular formula is not applicable.

Calculating Empirical and Molecular Formulas To find the empirical formula from percentage composition or masses: Step 1: Divide the mass (or percentage) of each element by its Aᵣ → gives molar ratio.

Step 2: Divide all values by the smallest value → simplest ratio. Step 3: If necessary, multiply to give whole numbers. Empirical Formula from % Composition A compound contains 40.0% C, 6.7% H, 53.3% O by mass.

Molar ratios: C = 40.0/12 = 3.33 | H = 6.7/1 = 6.7 | O = 53.3/16 = 3.33 Divide by smallest (3.33): C = 1, H = 2, O = 1 Empirical formula: CH₂O Molecular Formula from Empirical Formula Empirical formula CH₂O, Mᵣ = 180 Empirical formula mass = 12 + 2 + 16 = 30 n = 180 ÷ 30 = 6 Molecular formula: C₆H₁₂O₆ (glucose) Hydrated Salts and Water of Crystallisation Anhydrous A salt that contains no water of crystallisation.

(e.g. anhydrous copper(II) sulfate, CuSO₄) Hydrated A salt that contains water molecules incorporated into its crystal structure. (e.g. hydrated copper(II) sulfate, CuSO₄·5H₂O) Water of Crystallisation Water molecules that are chemically bonded within the crystal lattice of a hydrated salt, shown after a dot in the formula (e.g.

CuSO₄·5H₂O - five water molecules per formula unit). Finding the Number of Waters of Crystallisation 4.90 g of hydrated barium chloride (BaCl₂·xH₂O) was heated to give 4.17 g of anhydrous BaCl₂. Mass of water lost = 4.90 − 4.17 = 0.73 g Moles of BaCl₂ = 4.17 ÷ 208 = 0.0201 mol Moles of H₂O = 0.73 ÷ 18 = 0.0406 mol Ratio H₂O : BaCl₂ = 0.0406 ÷ 0.0201 ≈ 2 Formula: BaCl₂·2H₂O Formulae of Ionic Compounds Ionic compounds are electrically neutral overall.

The formula is determined by balancing the charges of the ions present. Predicting Ionic Charge from the Periodic Table Group 1: 1+ (e.g. Na⁺, K⁺) Group 2: 2+ (e.g. Mg²⁺, Ca²⁺) Group 13: 3+ (e.g. Al³⁺) Group 15: 3− (e.g.

N³⁻, P³⁻) Group 16: 2− (e.g. O²⁻, S²⁻) Group 17: 1− (e.g. Cl⁻, Br⁻) Transition metals have variable oxidation states, shown by Roman numerals: Fe²⁺ (iron(II)), Fe³⁺ (iron(III)) Named Polyatomic Ions to Learn Ion Formula Charge Nitrate NO₃⁻ 1− Carbonate CO₃²⁻ 2− Sulfate SO₄²⁻ 2− Hydroxide OH⁻ 1− Ammonium NH₄⁺ 1+ Zinc Zn²⁺ 2+ Silver Ag⁺ 1+ Hydrogencarbonate HCO₃⁻ 1− Phosphate PO₄³⁻ 3− Writing an Ionic Formula: Aluminium Sulfate Al³⁺ and SO₄²⁻.

To balance: 2 × Al³⁺ (total +6) and 3 × SO₄²⁻ (total −6) Formula: Al₂(SO₄)₃ Writing and Balancing Equations A balanced chemical equation shows the correct formulae for all reactants and products, with coefficients adjusted so that atoms are conserved.

Never change a chemical formula to balance - only change the coefficients in front. Check that each element has the same number of atoms on both sides. Always include state symbols : (s) solid, (l) liquid, (g) gas, (aq) aqueous solution.

Balancing: Combustion of Propane Unbalanced: C₃H₈ + O₂ → CO₂ + H₂O Balanced: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l) Ionic Equations Ionic equations show only the species that actually change during the reaction - spectator ions (ions that appear unchanged on both sides) are omitted.

Step 1: Write the balanced full equation with state symbols. Step 2: Dissociate all soluble ionic compounds (aq) into their ions. Step 3: Cancel spectator ions that appear identically on both sides. Step 4: Write the net ionic equation.

Ionic Equation: Neutralisation Full: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) Dissociated: H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + Cl⁻(aq) + H₂O(l) Cancel spectators (Na⁺, Cl⁻): H⁺(aq) + OH⁻(aq) → H₂O(l) Ionic Equation: Precipitation Full: BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq) Ionic: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) (Na⁺ and Cl⁻ are spectator ions and are cancelled) Exam Tip Solids, liquids, and gases are never split into ions in ionic equations - only species labelled (aq) are dissociated.

BaSO₄(s) stays as BaSO₄(s).

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