BITSAT Chemistry – Mole Concept, Stoichiometry & Redox Reactions
Study notes for quick revision and exam‑ready understanding
1. Mole Concept
| Concept | Simple Explanation | Formula / Key Value |
|---|---|---|
| Mole | Unit that counts entities (atoms, molecules, ions etc.) | 1 mol = 6.022 × 10²³ entities (Avogadro's number) |
| Molar Mass | Mass of 1 mol of a substance, expressed in g mol⁻¹ | M = Σ (atomic mass × number of atoms) |
| Conversion | mass (g) ↔ moles (mol) |
n = m / M and m = n × M |
Why it matters
The mole bridges the microscopic (atoms) and macroscopic (grams) worlds, allowing us to relate laboratory masses to the number of particles involved in a reaction.
2. Stoichiometry
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Mole Ratio – derived from the balanced chemical equation.
Example:
[ 2, \text{H}_2 + \text{O}_2 \rightarrow 2, \text{H}_2\text{O} ] → 2 mol H₂ : 1 mol O₂ : 2 mol H₂O -
Limiting Reagent – the reactant that is completely consumed first, limiting product formation.
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Theoretical Yield – maximum product obtainable from given reactants.
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Percent Yield – real yield compared to theoretical yield.
[ %,\text{Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100 ] -
Reaction Quotient (Q) – for dynamic systems (not usually BITSAT but useful).
Worked Example – Stoichiometry
Problem
From 10 g of iron(III) oxide (Fe₂O₃), how many grams of iron (Fe) can be produced in the reduction reaction with carbon?
[ \text{Fe}_2\text{O}_3 + 3,\text{C} \rightarrow 2,\text{Fe} + 3,\text{CO} ]
Solution
| Step | Calculation | Result |
|---|---|---|
| 1. Molar mass of Fe₂O₃ | (2\times55.85 + 3\times16.00 = 159.7,\text{g mol}^{-1}) | |
| 2. Moles of Fe₂O₃ | (n = \frac{10}{159.7} = 0.0626,\text{mol}) | |
| 3. Moles of Fe (from stoichiometry) | (n_{\text{Fe}} = 2 \times 0.0626 = 0.1252,\text{mol}) | |
| 4. Mass of Fe | (m = n_{\text{Fe}}\times55.85 = 0.1252\times55.85 = 6.99,\text{g}) | 6.99 g Fe |
Answer: 6.99 g of iron can be obtained theoretically.
3. Redox Reactions
| Concept | Simple Explanation | Key Point |
|---|---|---|
| Oxidation Number (ON) | Integer assigned to an atom in a compound. | Determined by rules (H +1, O –2, etc.) |
| Oxidation / Reduction | Loss / gain of electrons. | Oxidation = ↑ ON, Reduction = ↓ ON |
| Redox Pair | Oxidized form + Reduced form of the same element. | e.g., Fe³⁺ + e⁻ → Fe²⁺ |
| Half‑Reaction Method | Write separate oxidation and reduction equations, then balance electrons. | Used for balancing redox reactions in acidic or basic media |
| Standard Electrode Potential (E°) | Voltage for a half‑cell vs. SHE under standard conditions. | Positive E° → spontaneous reduction |
| Cell Potential (E°cell) | (E°{\text{cell}} = E°{\text{cathode}} - E°_{\text{anode}}) | Determines spontaneity in galvanic cells |
Identifying Oxidation Numbers
| Rule | Example |
|---|---|
| H in a compound is +1 unless in –H₃O⁺ | H₂O: H = +1 |
| O in most compounds is –2 | Fe₂O₃: each O = –2 |
| Sum |