Empirical Formulae & Formulae of Ionic Compounds
Section: 3. Stoichiometry | Syllabus: Cambridge AS Level Physics 9702
What is an Empirical Formula? Empirical Formula The simplest whole number ratio of atoms of each element in a compound. The empirical formula shows the simplest ratio, not necessarily the actual number of atoms in a molecule.
Shows the simplest ratio of atoms All numbers must be whole numbers The ratio must be in its simplest form (like fractions) May be the same as the molecular formula Compound Molecular Formula Empirical Formula Water H₂O H₂O (already simplest) Glucose C₆H₁₂O₆ CH₂O (÷ by 6) Ethane C₂H₆ CH₃ (÷ by 2) Hydrogen peroxide H₂O₂ HO (÷ by 2) Calculating Empirical Formula from Mass Step-by-Step Method Write down the mass of each element Divide by Aᵣ (relative atomic mass) to get moles Divide by the smallest number of moles to get ratio If needed, multiply all numbers to get whole numbers Write the empirical formula Remember The steps spell: MassDivideDivideMultiply - Mass → Divide by Aᵣ → Divide by smallest → Multiply to get whole numbers Worked Example 1: Simple Ratio Question: A compound contains 2.4 g of carbon and 0.8 g of hydrogen.
Find the empirical formula. (Aᵣ: C = 12, H = 1) Step 1: Write the masses C: 2.4 g H: 0.8 g Step 2: Divide by Aᵣ C: 2.4 ÷ 12 = 0.2 mol H: 0.8 ÷ 1 = 0.8 mol Step 3: Divide by smallest (0.2) C: 0.2 ÷ 0.2 = 1 H: 0.8 ÷ 0.2 = 4 Step 4: Already whole numbers!
Step 5: Write formula Empirical formula = CH₄ Check The ratio C:H is 1:4, which cannot be simplified further. CH₄ is the empirical formula (and also the molecular formula for methane). Worked Example 2: Need to Multiply Question: A compound contains 6.4 g of sulfur and 4.8 g of oxygen.
Find the empirical formula. (Aᵣ: S = 32, O = 16) Step 1: Write the masses S: 6.4 g O: 4.8 g Step 2: Divide by Aᵣ S: 6.4 ÷ 32 = 0.2 mol O: 4.8 ÷ 16 = 0.3 mol Step 3: Divide by smallest (0.2) S: 0.2 ÷ 0.2 = 1 O: 0.3 ÷ 0.2 = 1.5 Step 4: Multiply by 2 to get whole numbers S: 1 × 2 = 2 O: 1.5 × 2 = 3 Step 5: Write formula Empirical formula = S₂O₃ Important When you get decimals like 1.5, 2.5, or 1.33, you must multiply all numbers to get whole numbers.
Common multipliers: ×2 for .5, ×3 for .33 or .67, ×4 for .25 or .75 Calculating from Percentage Composition If given percentages instead of masses, treat the percentages as if they were grams . The method is exactly the same!
Example: A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Find empirical formula. (Aᵣ: C = 12, H = 1, O = 16) Treat % as grams: C: 40 g H: 6.7 g O: 53.3 g Divide by Aᵣ: C: 40 ÷ 12 = 3.33 mol H: 6.7 ÷ 1 = 6.7 mol O: 53.3 ÷ 16 = 3.33 mol Divide by smallest (3.33): C: 3.33 ÷ 3.33 = 1 H: 6.7 ÷ 3.33 = 2.01 ≈ 2 O: 3.33 ÷ 3.33 = 1 Empirical formula = CH₂O Formulae of Ionic Compounds Ionic Formula Shows the simplest ratio of ions in an ionic compound.
The total positive charge must equal the total negative charge. To write ionic formulae, you need to know the charges on the ions: Common Ion Charges +1 Ions +2 Ions +3 Ions -1 Ions -2 Ions Na⁺ (sodium) Mg²⁺ (magnesium) Al³⁺ (aluminium) Cl⁻ (chloride) O²⁻ (oxide) K⁺ (potassium) Ca²⁺ (calcium) Fe³⁺ (iron(III)) Br⁻ (bromide) S²⁻ (sulfide) H⁺ (hydrogen) Cu²⁺ (copper(II)) I⁻ (iodide) Ag⁺ (silver) Zn²⁺ (zinc) OH⁻ (hydroxide) NH₄⁺ (ammonium) Fe²⁺ (iron(II)) NO₃⁻ (nitrate) Polyatomic Ions (Groups of Atoms) NH₄⁺ - Ammonium OH⁻ - Hydroxide NO₃⁻ - Nitrate CO₃²⁻ - Carbonate SO₄²⁻ - Sulfate Writing Ionic Formulae - Method 1: Balancing Charges Rule: Total positive charges = Total negative charges Example 1: Sodium Chloride (Na⁺ and Cl⁻) Na⁺ has charge +1 Cl⁻ has charge -1 Need: +1 and -1 (already balanced) Formula: NaCl Example 2: Magnesium Oxide (Mg²⁺ and O²⁻) Mg²⁺ has charge +2 O²⁻ has charge -2 Need: +2 and -2 (already balanced) Formula: MgO Example 3: Magnesium Chloride (Mg²⁺ and Cl⁻) Mg²⁺ has charge +2 Cl⁻ has charge -1 Need: +2 total charge Need: 2 Cl⁻ ions to balance (+2 = -2) Formula: MgCl₂ Example 4: Aluminium Oxide (Al³⁺ and O²⁻) Al³⁺ has charge +3 O²⁻ has charge -2 Need charges to balance: 2 Al³⁺ = +6 charge 3 O²⁻ = -6 charge Formula: Al₂O₃ Writing Ionic Formulae - Method 2: Swap and Drop A quick method: Swap the charges and drop them as subscripts Steps Write the symbols with their charges Swap the numbers (ignore + and - signs) Drop them as subscripts Simplify if possible Example: Aluminium Oxide Al³⁺ and O²⁻ Swap the numbers: Al gets 2, O gets 3 Formula: Al₂O₃ Example: Calcium Chloride Ca²⁺ and Cl⁻ Swap: Ca gets 1, Cl gets 2 Ca₁Cl₂ = CaCl₂ (don't write the 1) Important After swapping, always simplify if possible.
For example, Mg²⁺ and O²⁻ would give Mg₂O₂, which simplifies to MgO. Ionic Formulae with Brackets When a polyatomic ion appears more than once , use brackets: Example: Calcium Hydroxide (Ca²⁺ and OH⁻) Ca²⁺ has charge +2 OH⁻ has charge -1 Need 2 OH⁻ ions to balance +2 charge Formula: Ca(OH)₂ (NOT CaOH₂!) Example: Aluminium Sulfate (Al³⁺ and SO₄²⁻) Al³⁺ has charge +3 SO₄²⁻ has charge -2 Swap and drop: Al gets 2, SO₄ gets 3 Formula: Al₂(SO₄)₃ When to Use Brackets Use brackets when a polyatomic ion …
Interactive revision notes, videos and practice questions load below.