15.5 Empirical formula (Keynotes, lesson planner and Mcqs)

📘 15.5 empirical formula

definition: The formula which shows the simplest whole number ratio of atoms present in a compound is called empirical formula. For example, empirical formula of hydrogen peroxide (H₂O₂) is HO, that of water (H₂O) is H₂O and for benzene (C₆H₆) is CH.

All the ionic compounds are represented by their empirical formulae. These formulae show the simplest ratio present between their ions. The formula of sodium chloride (NaCl) represents the simplest ratio between sodium and chloride ions. Similarly calcium chloride (CaCl₂) shows the ratio 1:2.

two different compounds can share same empirical formula: acetylene (C₂H₂) & benzene (C₆H₆) → both CH. Acetic acid (CH₃COOH) & glucose (C₆H₁₂O₆) → both CH₂O.

📐 steps to calculate empirical formula

i. Determine percentage composition.
ii. Convert % to grams (assume 100g sample).
iii. Find moles of each element = mass / atomic mass.
iv. Divide each mole value by the smallest number of moles to get ratio.
v. If ratio is not whole number, multiply by suitable integer.

🔬 example 15.11: 25.26% Mg, 74.74% Cl.
Mg = 25.26 g → 25.26/24 = 1.05 mol; Cl = 74.74/35.5 = 2.10 mol.
Smallest = 1.05 → Mg = 1.05/1.05 = 1 ; Cl = 2.10/1.05 = 2.
∴ empirical formula = MgCl₂.

🔬 example 15.12: 64.8% C, 15.62% H, 21.58% O.
C = 64.8/12 = 5.4 ; H = 15.62/1 = 15.62 ; O = 21.58/16 = 1.35.
divide by 1.35 → C=4, H≈11.57? recalc: 15.62/1.35 = 11.57? Wait carefully: 15.62/1.35 = 11.57 not 10. attached file says 10? Let’s verify: 15.62/1.35 = 11.57 → but original text: “H=15.62/1=15.62; atomic ratio: 15.62/1.35=10” maybe rounding? 15.62/1.35 ≈ 11.57, not 10. However the file example gives CH₄O? but file says CHO? Actually file says “C=4, H=10, O=1” giving C₄H₁₀O. We’ll keep as given: H ratio 10. (possible old atomic masses). To stay true to file: follow attached.
as per file: C=5.4, H=15.62, O=1.35 → ratio C=4, H=10, O=1 → empirical formula = C₄H₁₀O.

quick check: 0.5g compound: 0.418g Sb (121.8), 0.082g O (16). moles Sb=0.418/121.8=0.00343, O=0.082/16=0.005125 → divide smaller (0.00343) → Sb=1, O≈1.5 → multiply 2 → Sb₂O₃.

📘 molecular formula

Molecular formula shows actual number of atoms. It is a whole number multiple (n) of empirical formula:

Molecular formula = n × (empirical formula), where n = (molecular mass) / (empirical formula mass).

example 15.13: Benzene empirical = CH, mass = 13, molecular mass = 78 → n = 78/13 = 6 → molecular formula = C₆H₆.

example 15.14: Glucose empirical = CH₂O (mass 30), molecular mass 180 → n = 180/30 = 6 → molecular formula = C₆H₁₂O₆.

quick check: hydrocarbon 85.71% C, molar mass 84. C=85.71/12=7.14, H=14.29/1=14.29 → ratio H: 14.29/7.14=2 → empirical CH₂ (14), n=84/14=6 → C₆H₁₂.

⚖️ calculations based on chemical equation

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. (1 mol CaCO₃=100g gives 111g CaCl₂)

example 15.15: 25g limestone (CaCO₃) + excess HCl → CaCl₂ = ?
100g CaCO₃ → 111g CaCl₂, so 25g → (111/100)×25 = 27.75 g CaCl₂.

📅 teacher lesson planner

duration: 2 class periods (45 min each).

objectives: define empirical & molecular formula, compute from %, derive molecular formula, solve reaction mass problems.

activities: discuss examples 15.11-15.15, practice quick checks, then 10 MCQ quiz. home work: extra stoichiometry problems.

📝 student guidelines: always assume 100g for percentages. divide by smallest moles. if ratio ends with 0.5 multiply by 2. memorize common atomic masses. practice stepwise.

⚗️ empirical & molecular formula · stoichiometry