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Class 11 Chemistry Notes

Some Basic Concepts of Chemistry Class 11 Notes

The foundation chapter for all of chemistry: matter and its measurement, the mole concept and Avogadro's number, stoichiometry, percentage composition, empirical and molecular formulas, molarity and molality, and the laws of chemical combination — written for CBSE, JEE and NEET revision.

Class11SubjectChemistryCoversCBSE · JEE · NEET

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

What is the mole concept in one line?

A mole is the amount of substance containing Avogadro's number (6.022 × 10²³) of elementary particles; its mass in grams equals the substance's molar mass, linking the microscopic particle count to a measurable mass.

Laws of Chemical Combination

  • Law of conservation of mass — mass is neither created nor destroyed in a chemical reaction.
  • Law of constant proportions — a compound always contains the same elements in the same ratio by mass.
  • Law of multiple proportions — when two elements form more than one compound, the masses of one that combine with a fixed mass of the other are in a simple whole-number ratio.
  • Law of reciprocal proportions — relates combining ratios of elements.
  • Gay-Lussac's law — volumes of reacting gases (at constant T, P) are in simple whole-number ratios.

Atomic and Molecular Mass, Avogadro's Number

Mole

1 mol=6.022×1023 particles1\,\text{mol} = 6.022\times10^{23}\ \text{particles}

Avogadro's number NA=6.022×1023 mol−1N_A = 6.022\times10^{23}\ \text{mol}^{-1} counts particles (atoms, molecules or formula units). Number of moles n=mMn = \frac{m}{M} where m is mass (g) and M is molar mass (g/mol).

n=mM=NNAn = \frac{m}{M} = \frac{N}{N_A}
Mole relationships

Stoichiometry and the Mole

A balanced chemical equation gives the mole ratios of reactants and products. Stoichiometry uses these ratios to convert between amounts of substances — from moles to mass, volume or number of particles.

Reading a balanced equation

For N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \to 2\text{NH}_3 the mole ratio is 1 : 3 : 2 — meaning 1 mole of N₂ reacts with 3 moles of H₂ to give 2 moles of NH₃.

Percentage Composition, Empirical and Molecular Formulas

Empirical formula

EF=simplest whole-number ratioEF = \text{simplest whole-number ratio}

The empirical formula gives the simplest whole-number ratio of atoms. The molecular formula is a whole-number multiple of it: MF=n×EFMF = n \times EF where n=molar massEF massn = \frac{\text{molar mass}}{\text{EF mass}}.

%element=mass of elementmolar mass×100\% \text{element} = \frac{\text{mass of element}}{\text{molar mass}}\times100
Percentage composition

Water of crystallisation

In hydrated salts like CuSO₄·5H₂O, the formula mass includes the water molecules. Always count them when computing percentage composition by mass.

Stoichiometric Calculations and Limiting Reagent

Limiting reagent

The reactant that is completely consumed first, limiting the amount of product formed. Convert each reactant to moles of product; the one giving the least is limiting. The other is in excess.

Find the limiting reagent

For A+2B→C\text{A} + 2\text{B} \to \text{C} with 1 mol A and 1 mol B: B needs 2 mol per mol of A, so B (1 mol) limits — only 0.5 mol A reacts, giving 0.5 mol C.

Concentration Terms

  • Molarity (M):\text{Molarity (M)}:
  • moles per litre of solution.
  • Molality (m):\text{Molality (m)}:
  • moles per kg of solvent.
  • Mole fraction (x):\text{Mole fraction (x)}:
  • ratio of moles of one component to total moles.
  • Mass percent:\text{Mass percent:}
  • mass of solute per 100 g of solution.
  • ppm:\text{ppm:}
  • parts per million, for very dilute solutions.
M=nV(L),m=nw(kg),x1=n1n1+n2M = \frac{n}{V(L)},\quad m = \frac{n}{w(kg)},\quad x_1 = \frac{n_1}{n_1+n_2}
Concentration formulas

Dilution

On dilution the number of moles stays constant: M1V1=M2V2M_1V_1 = M_2V_2. Use this to find the new molarity after adding solvent.

Solved Examples

Example: What is the mass of 0.5 moles of CO2\text{CO}_2? (Molar masses: C = 12, O = 16)

Solution: M(CO2)=12+2×16=44 g/molM(\text{CO}_2) = 12 + 2\times16 = 44\,\text{g/mol}. So m=nM=0.5×44=22 gm = n M = 0.5\times44 = 22\,\text{g}.

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Mole

n=mM=NNAn = \frac{m}{M} = \frac{N}{N_A}

Avogadro's number

NA=6.022×1023 mol−1N_A = 6.022\times10^{23}\,\text{mol}^{-1}

Molarity

M=nV(L)M = \frac{n}{V(L)}

Molality

m=nw(kg)m = \frac{n}{w(kg)}

Molecular formula

MF=n×EFMF = n \times EF

Dilution

M1V1=M2V2M_1 V_1 = M_2 V_2

Mole fraction

x1=n1n1+n2x_1 = \frac{n_1}{n_1+n_2}

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

  • 1 mole = 6.022 × 10²³ particles = molar mass in grams.
  • Find the limiting reagent by comparing moles of product each reactant yields.
  • Molarity is per litre of solution; molality per kg of solvent.
  • Molecular formula = n × empirical formula.
  • Dilution: M₁V₁ = M₂V₂.

FAQ

Common questions

What is a mole?

A mole is the amount of a substance containing Avogadro's number (6.022 × 10²³) of elementary particles. Its mass in grams equals the substance's molar mass.

What is a limiting reagent?

The reactant that is consumed completely first, limiting the amount of product formed. The reaction stops when it runs out, regardless of other reactants in excess.

What is the difference between molarity and molality?

Molarity is moles of solute per litre of solution (temperature-dependent because volume changes); molality is moles per kilogram of solvent (temperature-independent).

How do I find an empirical formula?

Divide each element's mass by its atomic mass to get mole ratios, then divide by the smallest number and round to whole numbers to get the simplest ratio.

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