Mole Concept

Physics Chemistry  Biology  Mathematics
Science > Chemistry > Basic Concepts of ChemistryYou are Here

Molecule:

  • When two or more atoms are firmly held together by a chemical bond, a molecule is formed.
  • The molecule of an element may consist of one or more atoms of the same kind, while that of the chemical compound consists of different kinds of atoms.
  • The smallest particle of an element or compound which can exist in a free state and does not take part in a chemical reaction is called molecule.
  • Molecules are denoted by formula indicating the number of constituent elements in the compound.

Molecular Mass or Molar Mass:

  • The molecular mass or molar mass of a substance is defined as the ratio of the mass of one molecule of a substance to 1/12 th of the mass of 6C12 isotope taken as 12000 units.


Gram Molecular Mass or Molar Mass:

  • The molecular mass expressed in grams is called gram molecular mass (GMM)

Berzelius Hypothesis:

  • According to Dalton’s atomic theory atoms of different elements combine with each other in a simple whole number ratio, whereas Gay Lussac’s law of combining volumes, gases combine with each other in a simple whole ratio by their volumes.
  • A Swedish chemist Berzelius correlated the two laws and put forward is the concept called Berzelius hypothesis.
  • It states that equal volumes of all gases under similar conditions of temperature and pressure contain an equal number of atoms.
  • An Italian chemist Avogadro in 1811, modified the theory by differentiating between the ultimate particle of an element that takes part in reaction (atom) and the ultimate particle that has independent existence (molecule). He argued that the smallest particle of a substance which has independent existence is not an atom but molecule.

Avogadro’s Law:

  • Equal volumes of all gasses under the same conditions of temperature & pressure contain an equal number of molecules.
  • Explanation :
  • Let two gasses gas A and Gas B be taken in two containers having equal volume (V). Let the temperature of both the gasses be the same (T).Let the pressure of the two gases be same (P). By Avogadro’s law under such conditions of equal pressure, equal volume & equal temperature, the number of molecules of gas A in the container should be equal to the number of molecules of gas B in the container.

Validation of Avogadro’s Law:

  • Consider following chemical reactions

H2    +    Cl2   →    2HCl

1vol              1vol         2 vol

Thus the simple ratio of volumes is  1 : 1 : 2

  • Suppose 1 volume of hydrogen gas contains ‘n’ molecules. By Avogadro’s law at same temperature and pressure, there should be ‘n’ molecules of chlorine and ‘2n’ molecules of hydrogen chloride are involve.

Thus ‘n’ molecules of H2 +   ‘n’ molecules of Cl2 →  ‘2n’ molecules of HCl

Dividing this equation by n, we have

1 molecule of H2 +   1 molecule of Cl2 →  2 molecules of HCl

Hydrogen and chlorine are diatomic molecules.

2 atoms  of H2 +   2 atoms of Cl2 →  2 molecules of HCl

Dividing this equation by 2, we have

1 atom of H2 +   1 atom of Cl2 →  1 molecule of HCl

  • Now, number atoms of hydrogen, chlorine and hydrogen chloride are the whole numbers the Avogadro’s law is validated.

Importance of Avogadro’s Law:

  • It differentiates between atoms and molecules of gasses.
  • It modified Dalton’s atomic theory.
  • It explains Gay Lussac’s law of combining volume.
  • It helps in determination of the atomic weight of elements.
  • It established that at N.T.P.one gram mole of any gas occupies 22.4 dm3 by volume. one mole of a gas contains 6.023 × 1023 molecules of gas.
  • It gives the relation between the vapour density & the molecular mass. The relation is

Molecular weight = 2 × vapour density.

Vapour Density:

  • The vapour density of a gas is defined as the ratio of the weight of a certain volume of a gas to the weight of the same volume of hydrogen at the same temperature & pressure.

Mole Concept 02



Relation Between Molecular Mass and vapour Density:

  • The vapour density of a gas is defined as the ratio of the weight of a certain volume of a gas to the weight of the same volume of hydrogen at the same temperature & pressure.By definition of vapour density
  • Mole Concept 02
    By Avogadro’s law, Equal volumes of all gases under the same conditions of temp. & pressure, content equal number of molecules.

Mole Concept 03

This is the relation between the molecular weight and vapour density

Gram Molecular Volume (Molar volume) (GMV):

  • The volume occupied by one mole of a gas at STP (or NTP) is called as gram molar volume. It is 22.4 dm3 at NTP.
  • Proof: 

    Mathematically, Avogadro’s law is stated as “At constant temperature (T) and pressure (P) the volume (V) of a gas is directly proportional to the number of molecules (n).

Mole Concept 01

  • The molecular mass of gas corresponds to one mole of a gas hence we can say that one mole of a gas occupies 22.4 dm3 at STP.

Atomicity:

  • The number of atoms present in a molecule of a substance is called the atomicity of that substance.
  • Examples: The number in bracket indicates atomicity of that compound. Helium He (1), Dioxygen O2 (2),  Ozone O3 (3), 4 Phosphorous P4 (4), Sulphur S8 (8),  Carbon dioxide CO2 (3), Ammonia NH3 (4).
  • Atomicity indicates that how many atoms are present in the molecule of the element or the compound.


Atomicity of Element using Avogadro’s Law:

  • The number of molecules present in a molecule of a substance is called the atomicity of substance.
  • Let us consider the formation of ammonia. Experimentally it is found that 1 volume of nitrogen reacts with 3 volumes of hydrogen to form 2 volumes of ammonia. Thus,

N2       +   3H2      →      2NH3

1 vol             3 vol                2 vol

1/2  vol       3/2 vol             1vol

  • According to Avogadro ’s law “equal volumes of all gases under the identical condition of temperature and pressure contain the same number of molecules”. If 1 volume of ammonia contains n molecules  1/2 volume of nitrogen would contain n /2 molecules and 3/2  volume of hydrogen would contain 3n/2 molecules.

∴ n/2  molecules of Nitrogen + 3n/2 molecules of Hydrogen   → n molecules of Ammonia

∴  1/2 molecule of nitrogen + 3/2 molecule of hydrogen →   1molecule of ammonia.

  • As a molecule can be divided into the constituent atoms but the atoms cannot be divided further nitrogen molecule has to be diatomic, hydrogen diatomic and ammonia tetra-atomic.
  • Thus atomicity of Nitrogen =2, atomicity of hydrogen = 2 and atomicity of ammonia = 4.

Mole Concept:

  • The term mole was introduced by Wilhelm Ostwald. In Latin mole means heap or pile. He assumed substance as heap or pile of elementary entities like atoms, molecules, ions etc. and called it as mole.
  • When we are studying matter we require the actual number of molecules involved in the reaction. The atoms and molecules are discrete particles present in large number in the matter. To denote this number of atoms or molecules the term mole is used.
  • The amount of substance that contains as many elementary entities, like atoms, molecules, ions, photons as there are atoms in 0.012 kg of carbon 12 is called mole of a substance.
  • Example: 1 mole of hydrogen means 2 g of hydrogen. 1 mole of oxygen means 32 g of oxygen.
  • A mole is a fundamental unit of amount of substance in SI system of units. It is a collection of 6.022 × 1023 particles.   The number 6.022 × 1023  is also called as Avogadro’s number.
  • 1 mole also corresponds to the atomic or molecular mass of an element expressed in gram.  Thus 1 mole of a gas contains 6.022 × 1023 molecules of gas. 1-mole atom of a gas contains 6.022 × 1023 atoms of gas. 1-mole ions contain 6.022 × 1023 ions etc.

Mole and Avogadro’s Number:

  • The number of particles such as atoms, molecules, ions, in one mole of a substance is called Avogadro’s number or Avogadro’s constant.
  • It is denoted by ‘N’. Its value at STP is  6.022 × 1023 per mole. Thus 1 mole of a gas contains 6.022 × 1023 molecules of gas. 1-mole atom of a gas contains 6.022 × 1023 atoms of gas. 1-mole ions contain 6.022 × 1023 ions etc.

The Significance of Avogadro’s Number:

  • Avogadro’s number is equal to the number of molecules in one gram mole or one gram molecular mass of any compound. Thus gram molecular mass of any substance is equal to the mass in grams of Avogadro’s number of 6.022 × 1023 molecules.
  • Avogadro’s number is equal to the number of atoms in one gram mole or one gram atomic mass of an element. Thus gram atomic mass of any element is equal to the weight in grams of Avogadro’s number of 6.022 × 1023 atoms.
  • Avogadro’s number is equal to the number of molecules in 22.4 dm3 of any gas at NTP.
  • The actual mass of an element or compound can be found using this number.

Concept of one gram atom:

  • one gram atom of an element is the amount of the element which is equal to the mass of 1 mole (or 6.023 × 1023 atoms) of the element or atomic mass of the element in grams.
  • Thus one gram atom of sodium is equal to one mole of sodium corresponds to 23 grams of sodium, because the atomic mass of sodium is 23 grams.


Schematic Diagram for Mole Calculations:

Number of moles

Where, m = Given mass, M = Molar mass

v = Given volume, V = Molar volume = 22.4 dm3

n = Number of moles

Number of atoms  = Number of molecules  × Atomicity

Conversions:

From To Factor
kg g × 103
mg g × 10-3
μg g × 10-6
metric ton kg × 103
metric ton g × 106
cm3 dm3 × 10-3
m3 dm3 × 103
litre dm3 × 1

Example – 1:

  • Calculate the number of moles of following.
  •  7.85 g of Fe (at. mass 56)

Given mass of Fe = 7.85 g

Fe is a monoatomic molecule. Hence molecular mass of Fe = Atomic mass of Fe = 56 g

Number of moles of Fe = Given mass of Fe / Molecular mass of Fe

Number of moles of Fe  =  7.85 g / 56 g = 0.1402

Number of moles of Fe = 0.1402

  • 7.9 mg of Ca (at. mass 40)

Given mass of Ca =7.9 mg = 7.9 × 10-3 g

Ca is a monoatomic molecule. Hence molecular mass of Ca = Atomic mass of Ca = 40 g

Number of moles of Ca = Given mass of Ca / Molecular mass of Ca

Number of moles of Ca = 7.9 × 10-3 g / 40 g = 1.975  × 10-4

Number of moles of Ca = 1.975  × 10-4

  • 1.46 metric tons of Al (at. mass 27)

Given mass of Al = 1.46 metric tons = 1.46 × 103 kg = 1.46 × 103 × 103  g = 1.46 × 106  g

Al is a monoatomic molecule. Hence molecular mass of Al = Atomic mass of Al = 27 g

Number of moles of Al = Given mass of Al / Molecular mass of Al

Number of moles of Al = 1.46 × 106  g / 27 g = 5.41  × 104

Number of moles of Al = 5.41  × 104

  • 65.5 mg of C (at. mass 12)

Given mass of C = 65.5 mg = 65.5 × 10-3 g

Al is a monoatomic molecule. Hence molecular mass of C = Atomic mass of C = 12 g

Number of moles of C = Given mass of C/ Molecular mass of C

Number of moles of C = 65.5 × 10-3 g / 12 g = 5.46 × 10-3

Number of moles of C  = 5.46 × 10-3

Example  – 2:

  • Calculate the number of moles and number of molecules of following.
  • 0.032 mg of methane

Molecular mass of methane (CH4) = 12 × 1 + 1 × 4 = 12 + 4 = 16 g

Given mass of CH4 = 0.032 mg = 0.032 × 10-3 g = 3.2 × 10-5 g

Number of moles of CH4 =   n  =  Given mass of CH4 / Molecular mass of CH4

  n = 3.2 × 10-5 g / 16 g = 2 × 10-6

Now, Number of molecules = No. of moles × Avogadro’s number

Number of molecules = 2 × 10-6 × 6.022 × 1023

Number of molecules = 1.2044 × 1018

Number of moles of C  = 5.46 × 10-3 and Number of molecules = 1.2044 × 1018

  • 6.4 × 10-2 kg of sulphur dioxide

Molecular mass of methane (SO2) = 32 × 1 + 16 × 2 = 32 + 32 = 64 g

Given mass of SO2 = 6.4 × 10-2 kg = 6.4 × 10-2 × 103 g = 64 g

Number of moles of SO2 =   n  = Given mass of SO2 / Molecular mass of SO2

n = 64 g / 64 g = 1

Now, Number of molecules = No. of moles × Avogadro’s number

Number of molecules = 1 × 6.022 × 1023

Number of molecules = 6.022 × 1023

Number of moles of SO2 = 1 and Number of molecules =6.022 × 1023



  • 0.065 mg of water

Molecular mass of water (H2O) = 1 × 2 + 16 × 1 = 2 + 16 = 18 g

Given mass of H2O = 0.065 mg = 0.065 × 10-3 g = 6.5  × 10-5 g

Number of moles of H2O =   n  = Given mass of H2O / Molecular mass of H2O

n  = 6.5  × 10-5 g / 18 g = 3.611  × 10-6

Now, Number of molecules = No. of moles × Avogadro’s number

Number of molecules = 3.611  × 10-6 × 6.022 × 1023

Number of molecules = 2.174 × 1018

Number of moles of H2O = 3.611  × 10-6 and Number of molecules = 2.174 × 1018

  • 500 mg of carbon dioxide

Molecular mass of methane (CO2) = 12 × 1 + 16 × 2 = 12 + 32 = 44 g

Given mass of CO2 = 500 mg = 500 × 10-3  g = 0.5 g

Number of moles of CO2 =   n  = Given mass of CO2 / Molecular mass of CO2

n  = 0.5 g / 44 g = 1.136 × 10-2 

Now, Number of molecules = No. of moles × Avogadro’s number

Number of molecules = 1.136 × 10-2  × 6.022 × 1023

Number of molecules = 6.843 × 1021

Number of moles of CO2 = 1.136 × 10-2  and Number of molecules = 6.843 × 1021

Example – 3:

  • Calculate the number of moles, number of molecules and number of atoms of following.
  • 1.1. × 10-4 kg of carbon dioxide

Molecular mass of CO2 =12 × 1 + 16 × 2 = 12 + 32 = 44 g

Given mass of CO2 = 1.1. × 10-4 kg = 1.1. × 10-4 × 103 g = 0.11 g

Number of moles of CO2 = Given mass of CO2 / Molecular mass of CO2

Number of moles of CO2 = 0.11 g / 44 g = 2.5 × 10-3

Now, Number of molecules = No. of moles × Avogadro’s number

Number of molecules = 2.5 × 10-3 × 6.022 × 1023

Number of molecules of CO= 1.505 × 1021

Atomicity of CO2 is 3

Hence Number of atoms in CO2 = Number of molecules of CO2 × Atomicity of CO2

Hence Number of atoms of CO2 =  1.505 × 1021 × 3 = 4.515 × 1021

Number of moles  = 2.5 × 10-3 , Number of molecules = 1.505 × 1021  and Number of atoms =4.515 × 1021

  • 4.25. × 10-2 kg of ammonia

Molecular mass of NH3 =14× 1 + 1 × 3 = 14 + 3 = 17 g

Given mass of NH3 = 4.25. × 10-2 kg = 4.25. × 10-2 × 103 g = 42.5  g

Number of moles of NH3 = Given mass of NH3 / Molecular mass of NH3

Number of moles of NH3 = 42.5 g / 17 g = 2.5

Now, Number of molecules = No. of moles × Avogadro’s number

Number of molecules = 2.5  × 6.022 × 1023

Number of molecules of NH3 = 1.505 × 1024

Atomicity of NH3 is 4

Hence Number of atoms of NH3 = Number of molecules of NH3 × Atomicity of NH3

Hence Number of atoms in CO2 =  1.505 × 1024 × 4 = 6.020 × 1024

Number of moles  = 2.5, Number of molecules = 1.505 × 1024 and Number of atoms = 6.020 × 1024

  • 0.4 g of helium gas.

Molecular mass of He =4 g

Given mass of He = 0.4 g

Number of moles of He = Given mass of He / Molecular mass of He

Number of moles of He = 0.4 g / 4 g = 0.1

Now, Number of molecules = No. of moles × Avogadro’s number

Number of molecules = 0.1  × 6.022 × 1023

Number of molecules of He = 6.022 × 1022

Atomicity of He is 1

Hence Number of atoms of He = Number of molecules of He × Atomicity of He

Hence Number of atoms in He = 6.022 × 1022 × 1 = 6.022 × 1022

Number of moles  = 0.1, Number of molecules = 6.022 × 1022 and Number of atoms = 6.022 × 1022

  • 5.6 cm3 of ammonia at STP.

Given volume of ammonia = 5.6 cm3 = 5.6 × 10-3 dm3

1 mole of ammonia at STP occupies 22.4 dm3 by volume.

Number of moles of NH3 = Given volume of NH3 / 22.4 dm3

Number of moles of NH3 = 5.6 × 10-3 dm3 / 22.4 dm3 = 2.5 × 10-4 

Now, Number of molecules = No. of moles × Avogadro’s number

Number of molecules = 2.5 × 10-4  × 6.022 × 1023

Number of molecules of NH3 = 1.5055  × 1020

Atomicity of NH3 is 4

Hence Number of atoms of NH3 = Number of molecules of NH3 × Atomicity of NH3

Hence Number of atoms of NH3 = 1.5055  × 1020 × 4 = 6.022 × 1021

Number of moles  = 2.5 × 10-4 , Number of molecules = 1.5055  × 1020 and Number of atoms =6.022 × 1021

  • 5.6 dm3 of ammonia at STP.

Given volume of ammonia = 5.6  dm3

1 mole of ammonia at STP occupies 22.4 dm3 by volume.

Number of moles of NH3 = Given volume of NH3 / 22.4 dm3

Number of moles of NH3 = 5.6  dm3 / 22.4 dm3 = 0.25

Now, Number of molecules = No. of moles × Avogadro’s number

Number of molecules = 0.25  × 6.022 × 1023

Number of molecules of NH3 = 1.5055  × 1023

Atomicity of NH3 is 4

Hence Number of atoms of NH3 = Number of molecules of NH3 × Atomicity of NH3

Hence Number of atoms of NH3 = 1.5055  × 1023 × 4 = 6.022 × 1023

Number of moles  = 0.25, Number of molecules = 1.5055  × 1023 and Number of atoms =6.022 × 1023



  • 7.6 dm3 of hydrogen at STP.

Given volume of hydrogen = 7.6 dm3

1 mole of hydrogen at STP occupies 22.4 dm3 by volume.

Number of moles of H2 = Given volume of H2 / 22.4 dm3

Number of moles of H2 = 7.6 dm3 / 22.4 dm3 =0.34

Now, Number of molecules = No. of moles × Avogadro’s number

Number of molecules = 0.34  × 6.022 × 1023

Number of molecules of H2  = 2.047  × 1023

Atomicity of H2 is 2

Hence Number of atoms of H2 = Number of molecules of H2  × Atomicity of H2

Hence Number of atoms of NH3 = 2.047  × 1023 × 2 = 4.094  × 1023 

Number of moles  = 0.34 , Number of molecules = 2.047  × 1023 and Number of atoms = 4.094  × 1023 

Science > Chemistry > Basic Concepts of ChemistryYou are Here
Physics Chemistry  Biology  Mathematics

Leave a Comment

Your email address will not be published. Required fields are marked *