Chapter 1: Some Basic Concepts of Chemistry
By Aarish Sir
๐น 1. Importance of Chemistry
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Chemistry is the central science – it links physics, biology, medicine, agriculture, and engineering.
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Everything around us (air, food, clothes, medicines, fuels, even our body) is matter → studied in chemistry.
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Applications:
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Medicine & Health Care – discovery of drugs, vaccines.
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Agriculture – fertilizers, pesticides.
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Industry – dyes, plastics, cement, glass, paints.
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Environment – ozone depletion, global warming.
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Daily life – soaps, detergents, cosmetics.
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๐น 2. Laws of Chemical Combination
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Law of Conservation of Mass (Lavoisier)
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Mass can neither be created nor destroyed in a chemical reaction.
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Example: H₂ + O₂ → H₂O
Mass of reactants = Mass of products.
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Law of Definite Proportion (Proust)
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A chemical compound always contains the same elements in a fixed ratio by mass.
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Example: Water (H₂O) → H:O = 2:16 = 1:8 by mass.
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Law of Multiple Proportion (Dalton)
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When two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in small whole-number ratios.
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Example:
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CO → 12:16 (C:O = 3:4)
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CO₂ → 12:32 (C:O = 3:8)
Ratio → 1:2
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Law of Gaseous Volumes (Gay-Lussac)
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When gases react, they do so in volumes that bear a simple ratio to each other and to the volume of products (at same T & P).
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Example: H₂ + Cl₂ → 2HCl
Volumes: 1 + 1 → 2
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Avogadro’s Law
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Equal volumes of gases at the same temperature and pressure contain equal number of molecules.
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This helped in finding atomic and molecular masses.
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๐น 3. Dalton’s Atomic Theory
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Matter is made up of tiny, indivisible particles called atoms.
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Atoms of a given element are identical in mass and properties.
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Compounds are formed by the chemical combination of atoms in small whole-number ratios.
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Atoms cannot be created or destroyed.
Limitations: Could not explain isotopes, isobars, and the discovery of subatomic particles.
๐น 4. Atomic and Molecular Mass
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Atomic Mass: Average mass of an atom compared to 1/12 of carbon-12 atom.
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Molecular Mass: Sum of atomic masses of all atoms in a molecule.
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Example: H₂O → 2(1) + 16 = 18 u.
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Molar Mass: Mass of 1 mole of a substance.
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H₂O = 18 g/mol.
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๐น 5. Mole Concept
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1 mole = 6.022 × 10²³ particles (Avogadro number).
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Relations:
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Moles = Given Mass / Molar Mass
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Moles = Number of Particles / Avogadro’s Number
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Moles of gas = Volume at STP / 22.4 L
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Example: Find moles in 9 g of H₂O.
M = 18 g/mol → n = 9/18 = 0.5 mol.
๐น 6. Percentage Composition
Example: % of H in H₂O
= (2/18) × 100 = 11.1%
% of O = 88.9%
๐น 7. Empirical and Molecular Formula
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Empirical Formula: Simplest whole-number ratio of atoms.
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Molecular Formula: Actual number of atoms in a molecule.
where
Example:
A compound has 40% C, 6.7% H, 53.3% O; Molar mass = 60 g.
→ Empirical formula = CH₂O (mass = 30).
→ n = 60/30 = 2.
→ Molecular formula = C₂H₄O₂.
๐น 8. Stoichiometry
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Quantitative relationship between reactants and products.
Example: How much O₂ is required to burn 2 g of H₂?
Reaction: 2H₂ + O₂ → 2H₂O
4 g H₂ reacts with 32 g O₂.
So, 2 g H₂ reacts with 16 g O₂.
๐น 9. Limiting Reagent
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The reactant which is completely consumed first, limiting the amount of product.
Example:
5 mol H₂ reacts with 2 mol O₂ → which is limiting?
Reaction: 2H₂ + O₂ → 2H₂O
Ratio = 2:1.
For 2 mol O₂ → 4 mol H₂ required.
We have 5 mol H₂ (excess), so O₂ is limiting reagent.
๐น 10. Concentration Terms
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Molarity (M): moles of solute / litre of solution.
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Molality (m): moles of solute / kg of solvent.
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Normality (N): gram equivalents / litre of solution.
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% by mass: (mass of solute / mass of solution) × 100.
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% by volume: (volume of solute / volume of solution) × 100.
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ppm: parts per million = mass of solute / mass of solution × 10⁶.
Q1. Define molar mass.
Ans: The mass of one mole of a substance (i.e., 6.022 × 10²³ entities) is called molar mass.
Unit: g mol⁻¹.
Q2. What is Avogadro’s number?
Ans: The number of particles (atoms, ions, or molecules) in one mole of a substance is called Avogadro’s number.
NA = 6.022 × 10²³ mol⁻¹.
Q3. Write SI unit of molarity.
Ans: Molarity = moles of solute / volume of solution (in L).
SI unit = mol L⁻¹.
๐น SA (2–3 Marks)
Q4. Define empirical formula and molecular formula.
Ans:
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Empirical formula: Shows the simplest whole-number ratio of atoms of each element present in a compound. Example: CH₂O (for glucose).
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Molecular formula: Shows actual number of atoms of each element present in a molecule of compound. Example: C₆H₁₂O₆ (glucose).
Q5. Calculate the number of molecules present in 11 g of CO₂.
Ans:
Molar mass of CO₂ = 44 g mol⁻¹
Moles of CO₂ = 11 / 44 = 0.25 mol
Number of molecules = 0.25 × 6.022 × 10²³
= 1.505 × 10²³ molecules
Q6. Define limiting reagent with an example.
Ans:
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The reactant which is completely consumed in a reaction and determines the extent of the reaction is called limiting reagent.
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Example: In the reaction 2H₂ + O₂ → 2H₂O, if 5 mol H₂ and 2 mol O₂ are present, O₂ is limiting reagent (because only 2 mol available instead of 2.5 required).
๐น LA (4–5 Marks)
Q7. A sample of Na₂CO₃·10H₂O weighing 106 g is heated strongly until constant mass is obtained. Find the mass of anhydrous Na₂CO₃ left.
Ans:
M(Na₂CO₃·10H₂O) = 286 g mol⁻¹
M(Na₂CO₃) = 106 g mol⁻¹
So, 286 g Na₂CO₃·10H₂O → 106 g Na₂CO₃
∴ 106 g Na₂CO₃·10H₂O → (106 × 106) / 286
= 39.3 g Na₂CO₃
Answer: 39.3 g
Q8. A mixture contains 5.3 g sodium carbonate and 4.2 g sodium bicarbonate. Calculate the volume of CO₂ liberated at STP on heating the mixture.
Ans:
Reaction:
2NaHCO₃ → Na₂CO₃ + CO₂ + H₂O
M(Na₂CO₃) = 106 g mol⁻¹
M(NaHCO₃) = 84 g mol⁻¹
Moles of Na₂CO₃ = 5.3 / 106 = 0.05 mol
Moles of NaHCO₃ = 4.2 / 84 = 0.05 mol
From reaction, 2 mol NaHCO₃ → 1 mol CO₂
∴ 0.05 mol NaHCO₃ → 0.025 mol CO₂
At STP, 1 mol CO₂ = 22.4 L
∴ Volume of CO₂ = 0.025 × 22.4 = 0.56 L
Answer: 0.56 L CO₂
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