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2.4 Mole Concept

🎯 Learning outcomes

After the completion of the chapter, the students should be able to:

  • Define a mole and explain its relation with mass, volume, and number of particles.
  • Interpret a balanced chemical equation in terms of interacting moles, representative particles, masses, and volume of gases (at STP).
  • Perform stoichiometric calculations involving the above quantities.
1Introduction & Building Blocks▼

Introduction

Imagine trying to count every grain of rice in a sack by picking them up one at a time. Impossible, right? Shopkeepers solve this by weighing rice instead of counting grains. Chemists face the same problem — atoms and molecules are far too small and far too numerous to count one by one. Their solution is a special counting unit called the mole.

Why do we even need a ‘mole’?

A single atom of hydrogen is so light that even a whole gram of hydrogen gas contains more atoms than there are grains of sand on Earth. We cannot weigh a single atom on a lab balance, so chemists needed a bridge between the invisible world of atoms and the visible world of grams that we can actually measure. That bridge is the mole.

Before defining the mole formally, get comfortable with these two building blocks: Gram Atomic Mass and Gram Molecular Mass

Gram Atomic Mass (Gram Atom)

The atomic mass of an element expressed in grams is called its gram atomic mass or gram atom.

Examples

  • Atomic mass of sodium (Na) = 23 amu
    Therefore, 23 g of Na = 1 gram atom of Na
    46 g of Na = 2 gram atoms of Na
  • Atomic mass of calcium (Ca) = 40 amu
    Therefore, 40 g of Ca = 1 gram atom of Ca

Gram molecular mass (gram molecule)

The molecular mass of a substance expressed in grams is called its gram molecular mass or gram molecule.

Examples

  • Molecular mass of CO₂ = 44 amu
    Therefore, 44 g of CO₂ = 1 gram molecule of CO₂
  • Molecular mass of H₂ = 2 amu
    Therefore, 4 g of H₂ = 2 gram molecules of H₂
  • Molecular mass of (water) H₂O = 1 x 2 + 16 = 18 amu
    18 g water = 1 gram molecule of water
    72 g water = 4 gram molecules of water
2The Mole Concept & Avogadro’s Number▼

Mole Concept

Scientists discovered that:

  • 1 gram atom of every element contains the same number of atoms.
  • 1 gram molecule of every substance contains the same number of molecules.

This fixed number is called Avogadro’s number.

Definition of Mole

A mole is the amount of substance that contains 6.022 × 10²³ particles.

Depending on the substance, a particle can be:

  • Atom: e.g., 1 mole of Na = 6.022 × 10²³ sodium atoms
  • Molecule: e.g., 1 mole of H₂O = 6.022 × 10²³ water molecules
  • Ion: e.g., 1 mole of Na⁺ = 6.022 × 10²³ sodium ions
  • Electron: e.g., 1 mole of e⁻ = 6.022 × 10²³ electrons
  • Formula unit: e.g., 1 mole of NaCl = 6.022 × 10²³ formula units

Think of it like a chemist’s ‘dozen’ — just as a dozen always means 12, a mole always means 6.022 × 10²³ particles.

💡 Did You Know?

6.022 × 10²³ is an unimaginably huge number! If every student in your class counted one particle every second without stopping, it would take far longer than the age of the universe to count one mole. That’s how big Avogadro’s number is!

Another definition states:

One mole is the amount of substance containing as many particles as there are atoms in exactly 12 g of the carbon-12 isotope.

1 mole = 1 gram atom / 1 gram molecule
        = gram atomic mass / gram molecular mass / gram formula mass
        = 6.022 × 10²³ particles (atoms/molecules/ions/formula units, electrons, etc.)
        = 22.4 litres of gas at NTP

Or,

no. of moles = mass in gramatomic or molecular mass = volume of gas at NTP22.4 = number of particles6.023 x 10²³
Mole concept diagram
3Worked Examples▼
📝 Worked Example 1

How many molecules are contained in 0.35 mole of N₂? (1)

Solution:

1 mole of N₂ molecule contains 6.022 x 10²³ molecules

0.35 moles of N₂ molecule contains 0.35 x 6.023 x 10²³ molecules

     = 2.1 x 10²³ molecules of N₂

📝 Worked Example 2

The absolute mass of 1 amu is

a. 1 g    b. 1 mg    c. 1.66 x 10⁻²⁴ g    d. 1.66 x 10⁻²⁷ g

Solution:

We know that 1 amu is the mass of 1/12th of the C¹² isotope.

We also know that, 6.023 x 10²³ C¹² isotopes = 12 g

1 C¹² isotope= 126.023 x 10²³ = 1.992 x 10⁻²³ g
1/12th of C¹² isotope= 1.992 x 10⁻²³12 = 1.660 x 10⁻²⁴ g

Therefore, 1 amu = 1.660 x 10⁻²⁴ g.

🗝️ Key Points to Remember

  • Mole is the SI unit of amount of substance.
  • One mole contains 6.022×10²³ particles.
  • One mole of gas occupies 22.4 L at STP.
  • Molar mass is numerically equal to atomic or molecular mass but expressed in grams.
  • Avogadro’s number connects microscopic particles with measurable laboratory quantities.

📄 Summary

The mole concept provides a bridge between the microscopic world of atoms and molecules and the macroscopic world of laboratory measurements. It allows chemists to calculate masses, volumes, and numbers of particles accurately. Understanding the mole concept is essential for solving numerical problems in chemistry and for studying chemical reactions quantitatively.

 

📥 Download 8.3 Mole Concepts- Notes

 

📥 Download 8.3 Mole Concpets- Important NEB questions

 

📥 Download 8.3 Mole Concpets- MCQs

 

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