Some Basic Concepts of Chemistry

What this chapter is about

This chapter lays the foundation for all the chemistry you will study in senior secondary school. It begins with the nature of matter—how substances exist as elements, compounds and mixtures—and moves to the laws that govern chemical combinations. You learn to express amount of substance through the mole, connect laboratory masses to the number of atoms or molecules, and use balanced equations to predict how much product forms from given reactants.

A Class 11 student meets this chapter first because stoichiometry (the arithmetic of chemical reactions) underpins every calculation in physical, inorganic and organic chemistry. By the end, you should be able to convert between mass, moles and number of particles, write correct chemical formulas, balance equations, and solve problems on limiting reagent and percentage yield.

The chapter also introduces you to uncertainty in measurement, significant figures and dimensional analysis—skills you will use throughout physics and chemistry whenever you handle experimental data.

Key ideas

• Law of conservation of mass: In a chemical reaction, total mass of reactants equals total mass of products; matter is neither created nor destroyed.
• Law of definite proportions: A pure compound always contains its constituent elements in a fixed ratio by mass, regardless of source.
• Law of multiple proportions: When two elements form more than one compound, the different masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers.
• Mole concept: One mole of any substance contains exactly 6.022 × 10²³ entities (Avogadro constant, Nₐ). The molar mass (g mol⁻¹) equals the relative atomic or molecular mass expressed in grams.
• Molar mass and molecular mass: Molecular mass is the sum of atomic masses of all atoms in a formula; molar mass is the mass of one mole of that substance.
• Stoichiometry: A balanced equation gives the mole ratios in which reactants combine and products form, enabling mass and volume calculations.
• Limiting reagent: The reactant that is completely consumed first limits the amount of product; any other reactant present in excess remains unreacted.
• Empirical and molecular formulas: The empirical formula shows the simplest whole-number ratio of atoms; the molecular formula shows the actual number of atoms in one molecule.

Formulas and facts to remember

1. Number of moles: n = m / M, where m is mass in grams, M is molar mass in g mol⁻¹.
2. Number of particles: N = n × Nₐ, with Nₐ = 6.022 × 10²³ mol⁻¹.
3. Molar volume of an ideal gas at STP (0 °C, 1 bar): 22.7 L mol⁻¹ (or 22.4 L mol⁻¹ at older STP of 1 atm).
4. Percentage composition: (mass of element in 1 mol / molar mass of compound) × 100.
5. Molarity (M): moles of solute per litre of solution, unit mol L⁻¹.
6. Molality (m): moles of solute per kilogram of solvent, unit mol kg⁻¹.
7. Mass percent: (mass of solute / mass of solution) × 100.
8. Significant figures rule for multiplication/division: Result keeps as many significant figures as the quantity with the fewest.

Worked examples

Example 1 – Moles and particles

A sample of glucose (C₆H₁₂O₆) has a mass of 9.0 g. How many molecules does it contain?

Step 1: Calculate molar mass of glucose. M = (6 × 12) + (12 × 1) + (6 × 16) = 72 + 12 + 96 = 180 g mol⁻¹.

Step 2: Find number of moles. n = 9.0 g / 180 g mol⁻¹ = 0.050 mol.

Step 3: Find number of molecules. N = 0.050 mol × 6.022 × 10²³ mol⁻¹ = 3.01 × 10²² molecules.

---

Example 2 – Limiting reagent

Nitrogen and hydrogen react: N₂ + 3 H₂ → 2 NH₃. A cylinder contains 28 g of N₂ and 9.0 g of H₂. Which is the limiting reagent, and what mass of ammonia forms?

Step 1: Moles of each reactant. n(N₂) = 28 g / 28 g mol⁻¹ = 1.0 mol. n(H₂) = 9.0 g / 2.0 g mol⁻¹ = 4.5 mol.

Step 2: According to the equation, 1 mol N₂ needs 3 mol H₂. For 1.0 mol N₂, H₂ required = 3.0 mol; available = 4.5 mol (excess). So N₂ is the limiting reagent.

Step 3: Ammonia formed. From the equation, 1 mol N₂ gives 2 mol NH₃. n(NH₃) = 2.0 mol. Mass = 2.0 mol × 17 g mol⁻¹ = 34 g.

---

Example 3 – Empirical formula from percentage composition

An oxide of iron contains 70.0 % Fe and 30.0 % O by mass. Determine its empirical formula.

Step 1: Assume 100 g sample → 70.0 g Fe, 30.0 g O.

Step 2: Convert to moles. n(Fe) = 70.0 / 56 = 1.25 mol. n(O) = 30.0 / 16 = 1.875 mol.

Step 3: Divide by the smaller number. Fe: 1.25 / 1.25 = 1. O: 1.875 / 1.25 = 1.5.

Step 4: Multiply to get whole numbers (×2). Fe: 2, O: 3. Empirical formula: Fe₂O₃.

Common mistakes

• Adding atomic masses of only the unique elements and ignoring subscripts → count every atom, multiplied by its subscript.
• Using mass ratio directly as mole ratio in stoichiometry → convert masses to moles first, then apply the balanced equation's coefficients.
• Reporting an answer with more significant figures than the data justify → use the fewest significant figures among the measured quantities.
• Confusing molar mass (g mol⁻¹) with molecular mass (u) → numerically equal, but units differ; use g mol⁻¹ for lab calculations.
• Forgetting to identify the limiting reagent → always check which reactant runs out first before finding product mass.

Quick revision

• One mole = 6.022 × 10²³ particles; molar mass in grams equals relative atomic/molecular mass.
• n = m / M links mass to moles; N = n × Nₐ links moles to particle count.
• Balance the equation first; use mole ratios, not mass ratios, for stoichiometry.
• Limiting reagent decides maximum product; excess reagent remains.
• Empirical formula: simplest whole-number atom ratio; molecular formula may be a multiple of it.
• Keep significant figures consistent with the precision of your measurements.