Mole Concept Notes for Class 11, JEE and NEET: Formulas, Stoichiometry, Limiting Reagent and Concentration

The mole concept is the base of chemical calculations. It is part of NCERT Class 11 Chemistry Chapter 1, Some Basic Concepts of Chemistry, and it is used again in the chapters on solutions, equilibrium, electrochemistry and kinetics. These mole concept notes for Class 11 are written so that you can learn the ideas once and then use them in every numerical problem.

Where Mole Concept Notes for Class 11 Are Used Again

Tip from the eSaral Chemistry team: Mole concept questions appear on their own and inside other Physical Chemistry chapters. If the mole-mass-number conversions are automatic, you will make fewer arithmetic errors in Solutions, Electrochemistry and Equilibrium.

India's Best Exam Preparation for Class 11th - Download Now

Mole Concept  Notes

India's Best Exam Preparation for Class 11th - Download Now

Atomic Mass, Molecular Mass and Molar Mass in Mole Concept Notes for Class 11

Atomic Mass

The atomic mass of an element is the mass of one atom of the element, measured relative to 1/12th the mass of a carbon-12 atom. It is expressed in atomic mass units (u). 1 u = 1.66 × 10⁻²⁷ kg. The atomic mass of hydrogen is about 1 u, of carbon 12 u and of oxygen 16 u.

Average Atomic Mass

Most elements are mixtures of isotopes, so the atomic mass on the periodic table is an average, weighted by abundance. For chlorine (35Cl, 75%; 37Cl, 25%): average atomic mass = (35 × 0.75) + (37 × 0.25) = 35.5 u.

Molecular Mass

The molecular mass of a compound is the sum of the atomic masses of all the atoms in one molecule.

• H₂SO₄: 2(1) + 32 + 4(16) = 98 u
• Ca(OH)₂: 40 + 2(16 + 1) = 74 u

Molar Mass

The molar mass is the mass of one mole of a substance. It is numerically equal to the molecular mass but has the unit g/mol. For example, the molar mass of H₂O is 18 g/mol, of CO₂ is 44 g/mol and of NaCl is 58.5 g/mol. Atomic and molecular masses are in u, and molar mass is in g/mol.

The Mole and Avogadro’s Number in Mole Concept Notes for Class 11

Definition of the Mole

NCERT defines one mole as the amount of a substance that contains as many elementary entities as there are atoms in exactly 12 g of carbon-12. Since the 2019 revision of the SI, the mole is defined by fixing the number of entities: one mole contains exactly 6.02214076 × 10²³ elementary entities. In calculations, we use the rounded value:

Avogadro’s number: Nₐ = 6.022 × 10²³ mol⁻¹

Mole-Mass-Number-Volume Relationships in Mole Concept Notes for Class 11

1. Moles from mass: n = w/M, where w is the mass in grams and M is the molar mass in g/mol.

2. Number of particles: N = n × Nₐ, so N = (w/M) × Nₐ.

3. Moles of a gas from volume: at 273 K and 1 atm, one mole of an ideal gas occupies 22.4 L, so n = V/22.4 (V in litres). Under the IUPAC standard conditions of 273.15 K and 1 bar, the molar volume is 22.7 L. Many JEE and NEET problems use 22.4 L, so use the value that the question gives, or assumes. For other conditions, use PV = nRT.

Example: What is the volume of 4 g of O₂ at 273 K and 1 atm? n = 4/32 = 0.125 mol, so V = 0.125 × 22.4 = 2.8 L.

Empirical Formula and Molecular Formula in Mole Concept Notes for Class 11

The empirical formula gives the simplest whole-number ratio of atoms in a compound. The molecular formula gives the actual number of atoms in a molecule: molecular formula = n × empirical formula, where n = molecular mass/empirical formula mass.

Steps to Find the Empirical Formula from Percentage Composition

1. Treat the percentages as grams in a 100 g sample.
2. Divide each mass by the atomic mass to get the moles.
3. Divide all the mole values by the smallest one.
4. If the ratios are not whole numbers, multiply by a small integer (2, 3 and so on).
5. Write the formula.

Example: A compound contains 40% C, 6.67% H and 53.33% O. The moles are C: 40/12 = 3.33, H: 6.67/1 = 6.67 and O: 53.33/16 = 3.33. The ratio is 1 : 2 : 1, so the empirical formula is CH₂O. If the molecular mass is 180, n = 180/30 = 6, so the molecular formula is C₆H₁₂O₆.

Percentage Composition and Stoichiometry in Mole Concept Notes for Class 11

% of an element = (mass of the element in one mole of the compound ÷ molar mass of the compound) × 100. For example, the percentage of oxygen in H₂SO₄ is (64/98) × 100 = 65.3%.

Stoichiometry from a Balanced Equation

A balanced equation gives the mole ratios of the reactants and the products. Convert the mass to moles, use the mole ratio, and convert back to mass.

Example: For N₂ + 3H₂ → 2NH₃, 28 g of N₂ is 1 mol, which gives 2 mol of NH₃ = 34 g. The mass of H₂ needed is 3 mol × 2 g/mol = 6 g.

Limiting Reagent and Percentage Yield in Mole Concept Notes for Class 11

How to Find the Limiting Reagent

The limiting reagent is the reactant that is used up first. To find it:

1. Calculate the moles of each reactant.
2. Divide each by its stoichiometric coefficient.
3. The reactant with the smaller value is the limiting reagent.

Example: 2 g of H₂ reacts with 32 g of O₂ in 2H₂ + O₂ → 2H₂O. Moles of H₂ = 1, and 1/2 = 0.5. Moles of O₂ = 1, and 1/1 = 1. H₂ gives the smaller value, so H₂ is the limiting reagent. It forms 1 mol of H₂O = 18 g, and 0.5 mol (16 g) of O₂ is left over. If both ratios are equal, the reactants are used up together.

Percentage Yield

% yield = (actual yield ÷ theoretical yield) × 100. The theoretical yield is calculated from the limiting reagent, and the actual yield is never more than the theoretical yield. For example, a theoretical yield of 20 g and an actual yield of 15 g gives a yield of 75%.

Concentration Terms in Mole Concept Notes for Class 11

Molality is preferred for colligative properties, because it does not depend on temperature.

Solved Examples from Mole Concept Notes for Class 11

Example 1 (moles and particles): How many molecules and atoms are there in 11 g of CO₂?

n = 11/44 = 0.25 mol. Molecules = 0.25 × 6.022 × 10²³ ≈ 1.51 × 10²³. Each molecule has 3 atoms, so atoms ≈ 4.52 × 10²³.

Example 2 (empirical formula): An oxide of iron has 69.9% Fe and 30.1% O by mass. Find its empirical formula (Fe = 55.85, O = 16).

Fe: 69.9/55.85 = 1.25. O: 30.1/16 = 1.88. The ratio O/Fe = 1.5, so multiply by 2 to get 2 : 3. The empirical formula is Fe₂O₃.

Example 3 (limiting reagent): 5 mol of N₂ and 9 mol of H₂ react to form NH₃. Find the limiting reagent and the moles of NH₃ formed.

N₂: 5/1 = 5. H₂: 9/3 = 3. H₂ is the limiting reagent, and it gives 9 × (2/3) = 6 mol of NH₃.

Example 4 (molarity): 4 g of NaOH is dissolved in water to make 250 mL of solution. Find the molarity.

n = 4/40 = 0.1 mol. M = 0.1/0.25 = 0.4 M.

Example 5 (dilution): 100 mL of a 2 M solution is diluted to 500 mL. Find the new molarity.

M₂ = (2 × 100)/500 = 0.4 M.

Key Formulas in Mole Concept Notes for Class 11

Common Mistakes in Mole Concept Notes for Class 11 Problems

How to Study Mole Concept Notes for Class 11

1. Read the NCERT chapter first. Download the textbook from eSaral’s NCERT books for Class 11 page.
2. Solve the textbook questions with eSaral’s NCERT Solutions for Class 11 Chemistry Chapter 1.
3. Practise the three conversions (mass, moles, particles) until each takes under 30 seconds.
4. Test yourself with eSaral’s mole concept NEET previous year questions.