Class 11 Chemistry Chapter 1: Some Basic Concepts of Chemistry Revision Notes

Key Definitions

Mole (mol)

The SI unit for amount of substance. One mole is the amount of substance that contains as many elementary entities (atoms, molecules, ions, etc.) as there are atoms in exactly 12 grams of carbon-12. This number is Avogadro's constant, approximately 6.022 × 10^23 entities.

Molar Mass

The mass of one mole of a substance, expressed in grams per mole (g/mol). Numerically, it is equal to the atomic or molecular mass expressed in atomic mass units (amu).

Molarity (M)

Defined as the number of moles of solute dissolved per litre of solution. It is temperature-dependent because volume changes with temperature. Molarity = (Moles of solute) / (Volume of solution in Litres).

Molality (m)

Defined as the number of moles of solute dissolved per kilogram of solvent. It is temperature-independent as mass does not change with temperature. Molality = (Moles of solute) / (Mass of solvent in Kilograms).

Limiting Reagent

The reactant that is completely consumed in a chemical reaction and thereby limits the amount of product formed. The other reactants are said to be in excess.

Empirical Formula

Represents the simplest whole-number ratio of various atoms present in a compound.

Molecular Formula

Shows the exact number of different types of atoms present in a molecule of a compound.

Significant Figures

Meaningful digits in a measured or calculated number. They represent the precision of a measurement.

Laws of Chemical Combination

Chemical reactions are governed by five fundamental laws, which form the bedrock of stoichiometry and chemical calculations. Understanding these laws is crucial for predicting the outcomes of reactions and determining quantities of reactants and products.

1. Law of Conservation of Mass (Antoine Lavoisier, 1789): This law states that matter can neither be created nor destroyed in a chemical reaction. This means the total mass of the reactants before a chemical change must equal the total mass of the products after the change. For example, if 10g of calcium carbonate decomposes, it will form 5.6g of calcium oxide and 4.4g of carbon dioxide, with the sum (10g) remaining constant.

1. Law of Definite Proportions (Joseph Proust, 1799): This law states that a given compound always contains exactly the same proportion of elements by weight, irrespective of its source or method of preparation. For instance, pure water (H₂O) will always contain hydrogen and oxygen in a 1:8 mass ratio, whether it's from a river or synthesized in a lab.

1. Law of Multiple Proportions (John Dalton, 1803): When two elements combine to form more than one compound, the weights of one element that combine with a fixed weight of the other are in a simple whole-number ratio. Consider carbon and oxygen forming CO and CO₂. In CO, 12g C combines with 16g O. In CO₂, 12g C combines with 32g O. The ratio of oxygen weights (16:32) is 1:2, a simple whole-number ratio.

1. Gay-Lussac's Law of Gaseous Volumes (1808): When gases combine or are produced in a chemical reaction, they do so in a simple whole-number ratio by volume, provided all measurements are made under the same conditions of temperature and pressure. For example, 1 volume of hydrogen combines with 1 volume of chlorine to form 2 volumes of hydrogen chloride gas (H₂ + Cl₂ → 2HCl).

1. Avogadro's Law (1811): This law states that equal volumes of all gases, under the same conditions of temperature and pressure, contain an equal number of moles (or molecules). This directly implies that at STP (Standard Temperature and Pressure: 0°C or 273.15 K and 1 atm), one mole of any gas occupies approximately 22.4 litres. These laws are foundational for quantitative analysis in chemistry.

Key Concepts & Formulas (Must Remember)

• Mole Concept: 1 mole = 6.022 × 10^23 entities = Gram Molecular Mass = 22.4 L at STP (for gases).
• Number of Moles (n): n = Given Mass (g) / Molar Mass (g/mol) OR n = Number of particles / Avogadro's Number.
• Percentage Composition: (Mass of element in compound / Molar mass of compound) × 100.
• Relationship between Empirical and Molecular Formula: Molecular Formula = (Empirical Formula)n, where n = Molar Mass / Empirical Formula Mass.
• Molarity (M): Moles of Solute / Volume of Solution (L). Units: mol/L.
• Molality (m): Moles of Solute / Mass of Solvent (kg). Units: mol/kg.
• Mole Fraction (χ): Moles of Component / Total Moles of all Components (sum of mole fractions for all components = 1).
• Density (d): Mass / Volume. Units: g/mL or g/cm³.
• Limiting Reagent: Identify by calculating moles of products formed from each reactant; the reactant giving the least product is limiting.

Worked Examples

• {"title":"Example 1: Calculating Molar Mass","bodyMarkdown":"Q: Calculate the molar mass of glucose (C₆H₁₂O₆).\nA: Atomic masses: C = 12.01 g/mol, H = 1.008 g/mol, O = 16.00 g/mol.\nMolar Mass = (6 × 12.01) + (12 × 1.008) + (6 × 16.00)\n= 72.06 + 12.096 + 96.00 = 180.156 g/mol."}
• {"title":"Example 2: Empirical Formula from Percentage Composition","bodyMarkdown":"Q: A compound contains 4.07% H, 24.27% C, and 71.65% Cl. Its molar mass is 98.96 g/mol. Find its empirical and molecular formula.\nA:\n1. Moles: H: 4.07/1.008 = 4.04 mol; C: 24.27/12.01 = 2.02 mol; Cl: 71.65/35.45 = 2.02 mol.\n2. Ratio (divide by smallest): H: 4.04/2.02 = 2; C: 2.02/2.02 = 1; Cl: 2.02/2.02 = 1.\n3. Empirical Formula = CH₂Cl.\n4. Empirical Formula Mass: 12.01 + (2 × 1.008) + 35.45 = 49.476 g/mol.\n5. n = Molar Mass / Empirical Formula Mass = 98.96 / 49.476 ≈ 2.\n6. Molecular Formula = (CH₂Cl)₂ = C₂H₄Cl₂."}

Rules for Determining Significant Figures

1. —
2. —
3. —
4. —
5. —

Exam Tip: Mastering Stoichiometry and Limiting Reagent

Stoichiometry and limiting reagent problems are frequent in exams. Always start by writing a balanced chemical equation. Convert all given masses to moles. For limiting reagent problems, calculate the moles of product formed from each reactant separately, assuming the other is in excess. The reactant that yields the least amount of product is your limiting reagent. The amount of product formed will be based on this limiting reagent. Pay close attention to units and significant figures throughout your calculations; silly errors here can cost valuable marks.

Practice Questions with Solutions

• Q: What is the volume occupied by 0.5 mole of CO₂ gas at STP? A: At STP, 1 mole of any gas occupies 22.4 L. So, 0.5 mole of CO₂ will occupy 0.5 × 22.4 L = 11.2 L.
• Q: Define molarity and molality. Which one is preferred for expressing concentration in experiments involving temperature changes? A: Molarity is moles of solute per liter of solution. Molality is moles of solute per kilogram of solvent. Molality is preferred for temperature-dependent experiments as it is based on mass, which doesn't change with temperature, unlike volume.
• Q: How many significant figures are there in 0.00250 g? A: There are 3 significant figures. Leading zeros (0.00) are not significant. Trailing zero after a decimal point (250) is significant.
• Q: If a reaction produces 10g of product based on reactant A, and 8g of product based on reactant B, which is the limiting reagent? A: Reactant B is the limiting reagent because it produces the least amount of product (8g), meaning it will be consumed completely first.

Frequently Asked Questions