How do you find the limiting reagent?

Divide the given moles of each reactant by its stoichiometric coefficient. The reactant with the smallest value is the limiting reagent — it is fully consumed first and determines the amount of product formed.

What is the difference between empirical and molecular formula?

Empirical formula gives the simplest whole-number ratio of atoms. Molecular formula gives the actual number of atoms. Molecular formula = n × Empirical formula, where n = Molar mass / Empirical formula mass.

What are the units of concentration?

Molarity (M) = moles of solute / litre of solution. Molality (m) = moles of solute / kg of solvent. Mole fraction = moles of component / total moles. Normality = equivalents / litre.

Class 11 Chemistry Basic Concepts and Chemical Calculations — Samacheer Kalvi

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Quick Summary

Basic Concepts of Chemistry builds the quantitative foundation for all of chemistry. The mole is the central unit connecting mass, number of particles, and volume (for gases). Stoichiometry — using balanced equations to calculate amounts — is the most tested skill in NEET chemistry.

Matter and Its Classification

Pure substances: Elements (one type of atom) and Compounds (chemically combined)
Mixtures: Homogeneous (uniform) or Heterogeneous (non-uniform)
Physical change: No new substance formed (e.g., melting)
Chemical change: New substance formed; irreversible (e.g., combustion)

Laws of Chemical Combination

Law	Statement
Conservation of mass	Mass of reactants = mass of products
Definite proportions	A compound always has the same ratio of elements by mass
Multiple proportions	Two elements forming multiple compounds — masses of one element that combine with fixed mass of other are in simple ratio
Reciprocal proportions (Equivalent proportions)	Equivalents of elements that react with each other
Gay-Lussac's law of combining volumes	Gases react in simple whole-number ratios by volume (at same T, P)

The Mole Concept

1 mole = 6.022 × 10²³ particles (Avogadro's number, Nₐ)

Answer

Quantity	Formula
Molar mass	Same as molecular mass in g/mol
Number of moles	n = mass (g) / molar mass
Number of particles	N = n × Nₐ
Volume at STP (gas)	V = n × 22.4 L

Stoichiometry

From a balanced equation, the mole ratio of reactants and products is fixed.

Steps: 1. Write and balance the equation. 2. Convert given mass to moles. 3. Use mole ratio from the equation. 4. Convert moles to mass/volume/particles as required.

Limiting Reagent

The reagent that is completely consumed first. Determines maximum product.

Finding limiting reagent: Divide moles of each reactant by its coefficient → smallest value = limiting reagent.

% yield = (actual yield / theoretical yield) × 100

Empirical and Molecular Formula

Empirical formula: Simplest ratio of atoms. Steps from % composition: 1. Assume 100 g → % = mass in g. 2. Divide each by atomic mass → moles. 3. Divide by smallest → ratio → empirical formula.

Molecular formula: n × empirical formula, where n = molar mass / empirical formula mass.

Concentration Units

Unit	Formula	Notes
Molarity (M)	mol solute / L solution	Changes with temperature
Molality (m)	mol solute / kg solvent	Independent of temperature
Mole fraction (χ)	nₐ / (nₐ + n_b)	Dimensionless; sum = 1
Normality (N)	equivalents / L	N = M × n-factor
% w/w	(mass solute / mass solution) × 100	—
% w/v	(mass solute / vol solution) × 100	—

Worked Examples

Example 1: Find the moles in 36 g of water (M = 18 g/mol). n = 36/18 = 2 mol. Molecules = 2 × 6.022×10²³ = 1.204 × 10²⁴ molecules.

Example 2: In the reaction N₂ + 3H₂ → 2NH₃, if 28 g N₂ and 6 g H₂ are used, find the limiting reagent. n(N₂) = 28/28 = 1 mol; n(H₂) = 6/2 = 3 mol. Ratio needed: 1:3. Ratio available: 1:3. Both are exactly stoichiometric — neither is limiting (both are used up). NH₃ formed = 2 mol = 34 g.

Example 3: A compound has 40% C, 6.67% H, 53.33% O. Find empirical formula. C: 40/12 = 3.33; H: 6.67/1 = 6.67; O: 53.33/16 = 3.33. Divide by 3.33: C:H:O = 1:2:1 → CH₂O (formaldehyde series). If molar mass = 180, n = 180/30 = 6 → molecular formula = C₆H₁₂O₆ (glucose).

Important Questions

One Mark

Define Avogadro's number.
How many moles are in 44 g of CO₂?
State the law of conservation of mass.
What is STP? What is the molar volume of a gas at STP?

Short Answer (3 marks)

Calculate the percentage composition of H₂SO₄.
Find the empirical formula of a compound containing 75% C and 25% H by mass.
In the reaction 2H₂ + O₂ → 2H₂O, if 4g H₂ and 32g O₂ react, find the mass of water formed.

Long Answer (5 marks)

Explain the mole concept. Calculate the number of atoms in 27 g of aluminium and the volume occupied by this many moles of any gas at STP.
A compound is 38.7% C, 9.7% H, and 51.6% O. Its molar mass is 62 g/mol. Find its empirical and molecular formula.

Common Mistakes

Using molecular mass in g instead of g/mol — the molar mass unit must be g/mol.
Stoichiometry: not balancing the equation before doing calculations.
Limiting reagent: comparing raw moles without dividing by stoichiometric coefficients.
Empirical formula: forgetting to divide all ratios by the smallest mole value.

Board Exam Tips

Mole concept numericals are the #1 most tested topic in Class 11 Chemistry board exams — do at least 20 practice problems.
Empirical and molecular formula problems appear as 3-mark or 5-mark questions.
NEET: Stoichiometry, limiting reagent, and % yield appear in 1–2 questions every year.
Show all working clearly — partial marks are awarded for correct method even if arithmetic is wrong.

Frequently Asked Questions

One mole is the amount of substance that contains 6.022 × 10²³ (Avogadro's number) particles (atoms, molecules, or ions). The mass of one mole of a substance in grams equals its molecular mass in amu.