Formulae

3.1 Formulae

Chemical formula → a representation of the composition of a substance using chemical symbols and numerical subscripts.
Elements that exist as individual atoms → formula is simply the element’s symbol (e.g., Na, Fe, C).
Elements that exist as diatomic molecules (two atoms per molecule):
- Hydrogen → H₂
- Nitrogen → N₂
- Oxygen → O₂
- Fluorine → F₂
- Chlorine → Cl₂
- Bromine → Br₂
- Iodine → I₂
Other molecular elements:
- Phosphorus (P₄) – tetra-atomic molecules
- Sulfur (S₈) – octa-atomic molecules
Metals → generally represented as single atoms in formulas (e.g., Mg, Al, Zn) because they form giant metallic lattices.
Common compounds (must be memorised):
- Water → H₂O
- Ammonia → NH₃
- Methane → CH₄
- Carbon dioxide → CO₂
- Sulfur dioxide → SO₂
- Sodium chloride → NaCl
- Magnesium oxide → MgO
- Calcium carbonate → CaCO₃
- Hydrogen chloride → HCl
- Nitric acid → HNO₃
- Sulfuric acid → H₂SO₄
- Hydrochloric acid → HCl (aq)
- Sodium hydroxide → NaOH
- Potassium hydroxide → KOH
- Copper(II) sulfate → CuSO₄
- Zinc chloride → ZnCl₂

Molecular formula → shows the actual number and type of each atom in a single molecule.
Example:
- Ethene → C₂H₄
  - C = carbon, 2 atoms
  - H = hydrogen, 4 atoms
Example: Glucose → C₆H₁₂O₆ (contains 6 carbon atoms, 12 hydrogen atoms, 6 oxygen atoms per molecule)
The molecular formula does not show how atoms are bonded → for that, a structural formula is required.
Key point: Only applies to covalent substances (molecular substances). Ionic substances like NaCl do not have a molecular formula; they have a formula unit.

Empirical formula → simplest integer ratio of atoms/ions present in a compound.
For covalent compounds:
- Ethene (C₂H₄) → empirical formula = CH₂
- Glucose (C₆H₁₂O₆) → empirical formula = CH₂O
For ionic compounds:
- Sodium chloride → NaCl (already simplest ratio)
- Magnesium chloride → MgCl₂ (already simplest ratio)
Key difference:
- Molecular formula tells actual numbers.
- Empirical formula tells smallest whole number ratio.
Example: Butane has molecular formula C₄H₁₀ → empirical formula = C₂H₅.

Step-by-step approach:
1. Count the number of each atom type shown in the model/diagram.
2. Write element symbols in order (metal first in ionic compounds, carbon first in covalent organics).
3. Use subscripts to indicate number of atoms of each type.
4. If possible, simplify the ratio to the smallest whole numbers.
Example (molecular):
- Diagram shows 2 carbon atoms, 6 hydrogen atoms → formula = C₂H₆.
Example (ionic):
- Diagram shows twice as many chloride ions as magnesium ions → formula = MgCl₂.

Method for writing ionic formulas:
1. Write down the symbol of the cation (positive ion) with its charge.
2. Write down the symbol of the anion (negative ion) with its charge.
3. Swap and balance charges so the total positive and negative charges cancel out.
4. Write the formula without charges, with subscripts showing the number of ions.
Example:
- Na⁺ and Cl⁻ → charges are equal → NaCl.
- Mg²⁺ and Cl⁻ → need 2 chloride ions to balance one magnesium → MgCl₂.
- Al³⁺ and O²⁻ → need 2 aluminium and 3 oxygen → Al₂O₃.
For polyatomic ions:
- Use brackets if more than one is needed.
- Example: Ca²⁺ and SO₄²⁻ → CaSO₄ (no brackets needed, one sulfate ion).
- Example: Mg²⁺ and NO₃⁻ → Mg(NO₃)₂ (brackets used for nitrate).

Word equations: Describe the reaction in words.
- Example: Sodium + Chlorine → Sodium chloride.
Symbol equations: Use chemical formulas, balanced for atoms.
- Example: 2Na (s) + Cl₂ (g) → 2NaCl (s).
State symbols:
- (s) = solid
- (l) = liquid
- (g) = gas
- (aq) = aqueous (dissolved in water)
- Example: NaOH (aq) + HCl (aq) → NaCl (aq) + H₂O (l).
Ionic equations: Show only species that change during reaction.
- Example (neutralisation): H⁺ (aq) + OH⁻ (aq) → H₂O (l).
- Example (precipitation): Ag⁺ (aq) + Cl⁻ (aq) → AgCl (s).

Steps to deduce:
1. Identify reactants and products from description.
2. Determine physical states from context (solid metals, aqueous acids, gases, etc.).
3. Write chemical formulas for each substance.
4. Balance the equation so number of atoms of each element are the same on both sides.
Example 1: Calcium carbonate reacts with hydrochloric acid to produce calcium chloride, carbon dioxide and water.
- Formula: CaCO₃ (s) + HCl (aq) → CaCl₂ (aq) + CO₂ (g) + H₂O (l)
- Balance: CaCO₃ (s) + 2HCl (aq) → CaCl₂ (aq) + CO₂ (g) + H₂O (l).
Example 2: Magnesium burns in oxygen to form magnesium oxide.
- Formula: Mg (s) + O₂ (g) → MgO (s)
- Balance: 2Mg (s) + O₂ (g) → 2MgO (s).

Cations (positive ions):
- Group 1 metals: Li⁺, Na⁺, K⁺
- Group 2 metals: Mg²⁺, Ca²⁺, Ba²⁺
- Transition metals: Fe²⁺, Fe³⁺, Cu²⁺, Zn²⁺
- Ammonium: NH₄⁺
Anions (negative ions):
- Halides: F⁻, Cl⁻, Br⁻, I⁻
- Oxide: O²⁻
- Sulfide: S²⁻
- Nitrate: NO₃⁻
- Sulfate: SO₄²⁻
- Carbonate: CO₃²⁻
- Hydroxide: OH⁻

Example 1: Write formula for aluminium sulfate.
- Al³⁺ and SO₄²⁻ → swap charges → Al₂(SO₄)₃.
Example 2: Write ionic equation for precipitation of barium sulfate.
- Full: BaCl₂ (aq) + Na₂SO₄ (aq) → BaSO₄ (s) + 2NaCl (aq)
- Ionic: Ba²⁺ (aq) + SO₄²⁻ (aq) → BaSO₄ (s).
Example 3: Deduce empirical formula from % composition.
- A compound contains 40% C, 6.7% H, 53.3% O.
- Convert to moles: C = 40/12 ≈ 3.33, H = 6.7/1 = 6.7, O = 53.3/16 ≈ 3.33.
- Ratio = C: 3.33, H: 6.7, O: 3.33 → divide by 3.33 → C₁H₂O₁ → CH₂O.