Empirical & Molecular Formula
Section: 3. Stoichiometry | Syllabus: Cambridge AS Level Physics 9702
Empirical vs Molecular Formula - Quick Recap 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 Example: Glucose: Molecular formula = C₆H₁₂O₆ Glucose: Empirical formula = CH₂O (simplest ratio 1:2:1) Finding Empirical Formula from Masses Step 1: Write down the masses (or percentage by mass) of each element Step 2: Divide each mass by the Ar to get moles Step 3: Divide all moles by the smallest number of moles Step 4: If needed, multiply to get whole numbers Step 5: Write the empirical formula Example 1: From Mass Data 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: Masses given ✓ C = 2.4 g H = 0.8 g Step 2: Convert to moles Moles of C = 2.4 ÷ 12 = 0.2 mol Moles of 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 Example 2: From Percentage Composition Question: A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.
Find the empirical formula. (Ar: C=12, H=1, O=16) Step 1: Percentages (treat as grams) C = 40 g H = 6.7 g O = 53.3 g Step 2: Convert to moles Moles of C = 40 ÷ 12 = 3.33 mol Moles of H = 6.7 ÷ 1 = 6.7 mol Moles of O = 53.3 ÷ 16 = 3.33 mol Step 3: Divide by smallest (3.33) C: 3.33 ÷ 3.33 = 1 H: 6.7 ÷ 3.33 = 2 O: 3.33 ÷ 3.33 = 1 Step 4: Already whole numbers ✓ Step 5: Write formula Empirical formula = CH₂O Dealing with Decimals Sometimes step 3 gives decimals that need converting to whole numbers: Common patterns: If you get 0.5 → multiply all by 2 If you get 0.33 or 0.67 → multiply all by 3 If you get 0.25 or 0.75 → multiply all by 4 If you get 1.5 → multiply all by 2 If you get 1.33 or 2.67 → multiply all by 3 Example 3: With Decimal Ratios Question: A compound contains 7.2 g of carbon and 2.4 g of hydrogen.
Find the empirical formula. (Ar: C=12, H=1) Step 2: Convert to moles Moles of C = 7.2 ÷ 12 = 0.6 mol Moles of H = 2.4 ÷ 1 = 2.4 mol Step 3: Divide by smallest (0.6) C: 0.6 ÷ 0.6 = 1 H: 2.4 ÷ 0.6 = 4 Step 4: Already whole numbers ✓ Step 5: Empirical formula = CH₄ Example 4: Needing Multiplication Question: Mole ratio is C:H:O = 1:1.5:1.
Find the empirical formula. Step 4: Multiply to remove 0.5 C: 1 × 2 = 2 H: 1.5 × 2 = 3 O: 1 × 2 = 2 Step 5: Empirical formula = C₂H₃O₂ Finding Molecular Formula from Empirical Formula Step 1: Find the empirical formula Step 2: Calculate the Mr of the empirical formula Step 3: Divide the Mr of the molecule by the Mr of the empirical formula Step 4: Multiply the empirical formula by this number Example 5: Molecular Formula Calculation Question: A compound has the empirical formula CH₂O and a relative molecular mass of 180.
Find the molecular formula. (Ar: C=12, H=1, O=16) Step 1: Empirical formula given: CH₂O ✓ Step 2: Mr of CH₂O Mr = 12 + (2 × 1) + 16 = 30 Step 3: Divide molecular Mr by empirical Mr 180 ÷ 30 = 6 Step 4: Multiply empirical formula by 6 CH₂O × 6 = C₆H₁₂O₆ Answer: Molecular formula = C₆H₁₂O₆ (glucose) Example 6: Complete Calculation Question: A hydrocarbon contains 85.7% carbon and 14.3% hydrogen.
Its Mr is 56. Find both the empirical and molecular formulas. (Ar: C=12, H=1) Part A: Find empirical formula Step 2: Moles C: 85.7 ÷ 12 = 7.14 mol H: 14.3 ÷ 1 = 14.3 mol Step 3: Divide by 7.14 C: 7.14 ÷ 7.14 = 1 H: 14.3 ÷ 7.14 = 2 Empirical formula = CH₂ Part B: Find molecular formula Mr of CH₂ = 12 + 2 = 14 Multiplier: 56 ÷ 14 = 4 Molecular formula = CH₂ × 4 = C₄H₈ Using a Table Method Many students find a table helpful: Element Mass (g) Ar Moles (÷Ar) ÷ smallest ×2 if needed C 2.4 12 0.2 1 1 H 0.8 1 0.8 4 4 Result: CH₄ From Combustion Data You can find empirical formulas from combustion experiments: Key principle: All carbon goes into CO₂, all hydrogen goes into H₂O Strategy: Find moles of CO₂ → gives moles of C Find moles of H₂O → gives moles of H (multiply by 2) If compound also contains O, calculate by subtraction Example 7: From Combustion Question: Complete combustion of 2.3 g of a hydrocarbon produces 6.6 g of CO₂ and 4.5 g of H₂O.
Find the empirical formula. (Ar: C=12, H=1, O=16) Step 1: Find moles of CO₂ Mr of CO₂ = 44 Moles = 6.6 ÷ 44 = 0.15 mol Therefore moles of C = 0.15 mol Step 2: Find moles of H₂O Mr of H₂O = 18 Moles = 4.5 ÷ 18 = 0.25 mol Each H₂O has 2 H, so moles of H = 0.25 × 2 = 0.5 mol Step 3: Find ratio C : H = 0.15 : 0.5 Divide by 0.15: 1 : 3.33 Multiply by 3: 3 : 10 Answer: Empirical formula = C₃H₁₀ When Empirical = Molecular For some compounds, the empirical and molecular formulas are the same: Water: Both H₂O (can't simplify further) Carbon dioxide: Both CO₂ Sulfuric acid: Both H₂SO₄ This happens when the formula is already in its simplest ratio.
Common Mistakes to Avoid Not dividing by Ar: Must convert mass to moles first! Dividing by the wrong number: Always divide by the smallest number of moles Forgettin…
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