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NCERT Class 11
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1 Some Basic Concepts of Chemistry 2 Structure of Atom 3 Classification of Elements and Periodicity in Properties 4 Chemical Bonding and Molecular Structure 5 Thermodynamics 6 Equilibrium 7 Redox Reactions 8 Organic Chemistry – Some Basic Principles and Techniques 9 Hydrocarbons

1 Some Basic Concepts of Chemistry

1.1 Importance of Chemistry 1.2 Nature of Matter 1.3 Properties of Matter and Their Measurement 1.4 Uncertainty in Measurement 1.5 Laws of Chemical Combinations 1.6 Dalton’s Atomic Theory 1.7 Atomic and Molecular Masses 1.8 Mole Concept and Molar Masses 1.9 Percentage Composition 1.10 Stoichiometry and Stoichiometric Calculations

Stoichiometry and Stoichiometric Calculations

NCERT Reference: Chapter 1, Pages 20–25
Learning Objective: Understand how to calculate the quantities of reactants and products involved in chemical reactions, identify limiting reagents, and solve problems involving solutions.

Quick Notes:

  • Stoichiometry involves quantitative relationships between reactants and products in a balanced chemical equation.
  • We can use the mole ratio from the balanced equation to calculate required/reacted amounts.
  • Limiting reagent: The reactant that gets consumed first and limits the amount of product formed.
  • In solutions, concentration is expressed using:
    • Mass % = (Mass of solute / Mass of solution) × 100
    • Mole fraction = moles of component / total moles
    • Molarity (M) = moles of solute / volume of solution in litres
    • Molality (m) = moles of solute / mass of solvent in kg

Full Notes:

What is Stoichiometry?

Stoichiometry is the calculation of the quantities of substances involved in a chemical reaction, based on the balanced chemical equation.

A balanced equation tells us:

Example:

For the reaction: H2 + Cl2 → 2HCl

General Steps in Stoichiometric Calculations

  1. Write the balanced chemical equation
  2. Convert given data into moles
  3. Use mole ratios to relate substances
  4. Convert moles into mass, volume, or particles, as required

1.10.1 Limiting Reagent

In most reactions, one reactant may be in excess, while the other is completely consumed first — this is called the limiting reagent.

Definition: The limiting reagent is the reactant that gets consumed first and thus limits the amount of product formed. The other reactant is said to be in excess.

Example:2H2 + O2 → 2H2O

Steps to find limiting reagent:

  1. Convert mass to moles
  2. Compare the mole ratio of available reactants to the stoichiometric ratio
  3. The reactant that gives less product is the limiting reagent
Worked Example

Limiting reagent check: For 2H2 + O2 → 2H2O, which is limiting if we mix 4.0 mol H2 with 1.5 mol O2?

  1. Required stoichiometric ratio H2:O2 = 2:1.
  2. With 1.5 mol O2, maximum H2 that can react = 3.0 mol → leaves 1.0 mol H2 unreacted.
  3. Therefore O2 is limiting; water formed = 3.0 mol H2 → 3.0 mol H2O.

1.10.2 Reactions in Solutions

When reactions occur in solution, it is important to know how much solute is dissolved and how to express concentration.

Mass Percent (% by Mass)

NCERT 11 Chemistry diagram showing mass percent formula: mass of solute over mass of solution times 100.

Mass % = (Mass of solute / Mass of solution) × 100

Example: 10 g NaCl in 90 g water , solution = 100 g
Mass % of NaCl = (10 / 100) × 100 = 10%

Mole Fraction (χ)

NCERT 11 Chemistry diagram illustrating mole fraction definition and the identity χA + χB = 1 for a binary mixture.

Molarity (M)

NCERT 11 Chemistry formula diagram for molarity: moles of solute divided by volume of solution in litres.

Molarity (M) = moles of solute / volume of solution (L)

Example: 5.8 g NaCl (M = 58.44 g/mol) in 500 mL solution:
Moles = 5.8 / 58.44 = 0.0992 mol
Volume = 0.500 L
Molarity = 0.0992 / 0.500 = 0.198 M

Note: Molarity changes with temperature, because volume changes with temperature.

Molality (m)

NCERT 11 Chemistry formula diagram for molality: moles of solute per kilogram of solvent.

Molality (m) = moles of solute / mass of solvent (kg)

Example: 5.4 g glucose (M = 180 g/mol) in 100 g water
Moles = 5.4 / 180 = 0.03 mol
Solvent = 0.100 kg
Molality = 0.03 / 0.100 = 0.3 m

Note: Molality is independent of temperature, since mass does not change with temperature.

NCERT Summary of Units

Quantity Symbol Formula Depends on T?
Mass percent – (mass solute / mass solution) × 100 No
Mole fraction χ moles of component / total moles No
Molarity M moles of solute / volume of solution (L) Yes
Molality m moles of solute / mass of solvent (kg) No

Photo of Matt
Matt’s exam tip

If you're given volume of solution → use molarity. If you're given mass of solvent → use molality. Always watch for the limiting reagent in stoichiometry questions.

Summary