Stoichiometry Some Basic Concepts of Chemistry Class 11th Chapter 1 Science

Stoichiometry: Molar Calculations, Limiting Reagent, and Concentration Terms

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. The word "stoichiometry" comes from two Greek words — "stoicheion" meaning element and "metron" meaning measure — literally "element measuring." This chapter in CBSE Class 11 Chemistry builds on the mole concept and applies it to chemical equations, enabling students to calculate the amounts of reactants required and products formed in any given reaction. Stoichiometry is essential for industrial chemistry, pharmaceutical manufacturing, environmental science, and understanding the material balance in all chemical processes.

The mole concept is the foundation of all stoichiometric calculations. One mole of any substance contains exactly 6.022 × 10²³ particles (Avogadro's number), and its mass in grams equals its molar mass (molecular mass expressed in grams). For example, one mole of water (H₂O, molecular mass 18) has a mass of 18 grams and contains 6.022×10²³ water molecules. The number of moles n = given mass M / molar mass. The number of particles (atoms, molecules, ions) = n × NA = n × 6.022×10²³. For gases, one mole of any gas at STP (Standard Temperature and Pressure — 0°C and 1 atm) occupies 22.4 litres — this is the molar volume of a gas. At room temperature (25°C) and 1 atm, the molar volume is approximately 24.45 litres. For example, 11.2 litres of oxygen gas at STP contains 11.2/22.4 = 0.5 moles of O₂ molecules, which is 0.5 × 6.022×10²³ = 3.011×10²³ molecules of O₂. The percentage composition of an element in a compound is calculated as: Mass% = (Number of atoms × atomic mass / molecular mass) × 100. For methane CH₄: C mass% = (12/16)×100 = 75%, H mass% = (4/16)×100 = 25%.

The empirical formula represents the simplest whole-number ratio of atoms in a compound, while the molecular formula gives the actual number of atoms of each element per molecule. To determine the empirical formula from percentage composition: (1) Convert percentages to grams (assume 100 g of compound). (2) Convert grams to moles using atomic masses. (3) Divide each mole value by the smallest mole value to get the simplest ratio. (4) If necessary, multiply by suitable integers to obtain whole numbers. The molecular formula is found by dividing the molecular mass by the empirical formula mass: n = Molecular mass / Empirical formula mass. The actual molecular formula is n times the empirical formula. For example, a compound with empirical formula CH₂ (mass 14) has molecular mass 56. n = 56/14 = 4, so the molecular formula is C₄H₈. Balancing chemical equations is essential — it ensures that the number of atoms of each element is the same on both sides, obeying the law of conservation of mass. The mole ratios from the balanced equation serve as conversion factors between reactants and products. Given the number of moles of any one substance, the moles of any other can be found using the ratio of their coefficients. The limiting reagent is the reactant that is completely consumed first, determining the maximum amount of product that can be formed. The other reactants are in excess. To identify the limiting reagent: find the number of moles of each reactant, divide by its coefficient in the balanced equation — the smallest value indicates the limiting reagent. The theoretical yield is the amount of product calculated from the limiting reagent. The actual yield is always less than or equal to the theoretical yield due to losses, side reactions, and incomplete reactions. Percentage yield = (Actual yield / Theoretical yield) × 100.

• Mole concept: n = mass/molar mass; 1 mole = 6.022×10²³ particles; 1 mole of gas at STP = 22.4 L.
• Percentage composition: mass% of element = (total atomic mass of element / molecular mass) × 100.
• Empirical formula: smallest whole-number ratio from percentage composition; molecular formula = (empirical formula)×n where n = molecular mass/empirical mass.
• Stoichiometric calculations: use mole ratios from balanced equation to convert between amounts of different reactants and products.
• Limiting reagent: the reactant that produces the least product; identify by dividing moles by coefficient. Percentage yield = (actual/theoretical) × 100.