Some Basic Concepts of Chemistry: Class 11 NCERT Chapter Guide

The Heart of Chemical Calculation: The Mole Concept

Atoms and molecules are incredibly tiny, so we can't count them individually. So, how do we connect the microscopic world of atoms to the macroscopic world of grams that we can weigh in a lab? The answer is the mole. A mole is just a specific number, much like a 'dozen' means 12. One mole of any substance contains 6.022 x 10²³ particles (atoms, molecules, or ions). This giant number is called Avogadro's Constant (Nₐ). The beauty of the mole is its direct link to the atomic or molecular mass. The mass of one mole of a substance in grams is numerically equal to its atomic/molecular mass in atomic mass units (u). For example, the atomic mass of Carbon is 12 u, so one mole of Carbon atoms weighs exactly 12 grams. This simple but powerful concept is the cornerstone of all quantitative chemistry.

Worked Examples: Molar Mass and Mole Calculations

• Example 1: Calculate the molar mass of Sulphuric Acid (H₂SO₄). Step 1: Identify the elements and the number of atoms of each in the formula. We have 2 atoms of Hydrogen (H), 1 atom of Sulphur (S), and 4 atoms of Oxygen (O). Step 2: Find the atomic mass of each element from the periodic table (approximate values are usually sufficient for school exams: H = 1 u, S = 32 u, O = 16 u). Step 3: Multiply the number of atoms of each element by its atomic mass and sum them up. Molar Mass = (2 × Atomic Mass of H) + (1 × Atomic Mass of S) + (4 × Atomic Mass of O) = (2 × 1) + (1 × 32) + (4 × 16) = 2 + 32 + 64 = 98 g/mol. Final Answer: The molar mass of H₂SO₄ is 98 g/mol.
• Example 2: How many moles are there in 49 g of Sulphuric Acid (H₂SO₄)? Step 1: Recall the formula relating moles, given mass, and molar mass: Number of moles (n) = Given Mass (m) / Molar Mass (M). Step 2: From the previous example, we know the molar mass (M) of H₂SO₄ is 98 g/mol. The given mass (m) is 49 g. Step 3: Substitute the values into the formula. n = 49 g / 98 g/mol n = 0.5 mol. Final Answer: There are 0.5 moles in 49 g of H₂SO₄.

Stoichiometry and the Limiting Reagent

A balanced chemical equation is like a recipe. Stoichiometry (from the Greek words stoicheion, meaning 'element', and metron, meaning 'measure') is the calculation of reactants and products in chemical reactions based on these balanced equations. For instance, the reaction to form water, 2H₂(g) + O₂(g) → 2H₂O(l), tells us that exactly 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water.

But what happens if the reactants are not mixed in this perfect 2:1 ratio? This brings us to the concept of the limiting reagent (or limiting reactant). The limiting reagent is the reactant that gets completely consumed first in a chemical reaction. Once it runs out, the reaction stops, no matter how much of the other reactants (called 'excess reagents') are left. The amount of product formed is always determined by the amount of the limiting reagent. Think of making sandwiches: if you have 10 slices of bread and only 3 slices of cheese, you can only make 3 sandwiches. The cheese is your limiting reagent.

Exam Trap: Always Identify the Limiting Reagent!

A very common mistake in stoichiometry problems is to assume reactants are present in the exact required ratio. Students often pick one of the given reactant masses and use it to calculate the product, ignoring the other reactant. This is incorrect! Rule #1: Always find the limiting reagent first. To do this, convert the given mass of all reactants into moles. Then, use the mole ratio from the balanced equation to see which reactant will be used up first. All further calculations for the amount of product formed or the amount of excess reagent left must be based on the initial amount of the limiting reagent.

Practice Questions with Solutions

• Q: Calculate the number of atoms in 8 g of Helium (He). (Atomic mass of He = 4 u) A: Step 1: Calculate the number of moles. Number of moles (n) = Given Mass / Molar Mass = 8 g / 4 g/mol = 2 moles. Step 2: Calculate the number of atoms using Avogadro's number (Nₐ = 6.022 x 10²³ atoms/mol). Number of atoms = Number of moles × Nₐ = 2 × (6.022 x 10²³). Final answer: The number of atoms is 12.044 x 10²³ or 1.2044 x 10²⁴ atoms.
• Q: For the reaction N₂(g) + 3H₂(g) → 2NH₃(g), how many grams of ammonia (NH₃) are produced if 28 g of nitrogen (N₂) reacts completely? A: Step 1: Convert the given mass of N₂ to moles. Molar mass of N₂ = 2 × 14 = 28 g/mol. Moles of N₂ = 28 g / 28 g/mol = 1 mole. Step 2: Use stoichiometry. The balanced equation shows 1 mole of N₂ produces 2 moles of NH₃. So, moles of NH₃ produced = 2 moles. Step 3: Convert moles of NH₃ to grams. Molar mass of NH₃ = 14 + (3 × 1) = 17 g/mol. Mass of NH₃ = Moles × Molar Mass = 2 mol × 17 g/mol = 34 g. Final answer: 34 g of ammonia are produced.
• Q: 50 g of N₂ and 10 g of H₂ are mixed to produce NH₃. Calculate the mass of NH₃ formed and identify the limiting reagent. (N₂ + 3H₂ → 2NH₃) A: Step 1: Calculate moles of each reactant. Moles of N₂ = 50 g / 28 g/mol ≈ 1.786 moles. Moles of H₂ = 10 g / 2 g/mol = 5 moles. Step 2: Identify the limiting reagent. From the equation, 1 mole N₂ needs 3 moles H₂. So, 1.786 moles of N₂ would need 1.786 × 3 = 5.358 moles of H₂. We only have 5 moles of H₂, which is less than what is needed. Therefore, H₂ is the limiting reagent. Step 3: Calculate the product based on the limiting reagent (H₂). From the equation, 3 moles of H₂ produce 2 moles of NH₃. So, 5 moles of H₂ will produce (2/3) × 5 = 3.33 moles of NH₃. Step 4: Convert moles of NH₃ to grams. Mass of NH₃ = 3.33 mol × 17 g/mol ≈ 56.6 g. Final answer: H₂ is the limiting reagent and approximately 56.6 g of NH₃ is formed.
• Q: Calculate the molarity of a solution prepared by dissolving 4 g of NaOH in enough water to form 250 mL of solution. A: Step 1: Calculate the molar mass of NaOH. Molar Mass = 23 (Na) + 16 (O) + 1 (H) = 40 g/mol. Step 2: Calculate the moles of NaOH. Moles = Given Mass / Molar Mass = 4 g / 40 g/mol = 0.1 moles. Step 3: Convert the volume of the solution to Liters. Volume (L) = 250 mL / 1000 mL/L = 0.25 L. Step 4: Calculate Molarity. Molarity (M) = Moles of solute / Volume of solution in L = 0.1 mol / 0.25 L = 0.4 M. Final answer: The molarity of the solution is 0.4 mol/L or 0.4 M.

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