**2.3 MOLE: **

(A practical chemical unit for handling atoms and
molecules) Since atoms and molecules are so small, it is
impossible to handle and
count atoms and molecules individually. Therefore, the
Chemists devised a special unit to describe very large
number of atoms, ions and molecules.

It is called mole and is abbreviated as mol. A mole can
be defined as “the molecular mass, atomic mass and
formula mass of a substance expressed in grams. Thus,
12g of carbon is equal to 1 mol of carbon atoms. 24g of
C is equal to 2 mol of carbon atoms" The mole concept
tries to give a practical meaning to the mass of
reactants and products in chemical reactions in terms of
the number of particles (atoms, molecules or ions)
involved. The actual countable number of particles in
one mole of a substance is 6.02 x 10^{23}
particles, which is referred as Avogadro number.

The (S.I.) definition of mole is the amount of
substance, containing as many elementary particles
(units) as there are atoms in exactly 12g of C-12 a.m.u.

It is also defined as the mass of any substance equal to
its atomic mass, molecular mass or formula mass in
grams.

Thus.

1 mole of C = 12 g

1 mole of Mg = 24 g

1 moleofH_{2}O = 18 g

1 mole of C0_{2} = 44 g

1 mole of CaCO_{3} = 100g

1 mole of Fe_{2}0_{3} = 160g

By formula, Number of moles = __
Give mass of substance__

Atomic mass or formula
mass

As, formula mass, represents the both covalent and
electrovalent compounds

2.3.1 Avogadro's Number (N_{4}):

A mole of substance always contains the same number of
particles (atoms, ions, molecules or formula units)
irrespective of its state, solid, liquid or gaseous,
that is 6.02x10^{23} particles. This constant
number has been determined by several methods, called
Avogadro's number (symbol N_{A}),in the honor of
Avogadro, the scientist who gave chemistry a method for
finding atomic and molecular masses.

Thus.

1 mole of C = 12g =
6.02x10^{23} atoms of carbon

1 mole of Mg = 24g = 6.02x10^{23}
atoms of magnesium

1 mole of H_{2}O = 18g
= 6.02x10^{23} molecules of water

1 mole of CO_{2} = 44g
= 6.02x10^{23} molecules of O_{2}

1 mole of NaCl = 58.5g = 6.02x10^{23}
F-units of NaCl

1 mole of CaC0_{3}= l00g
= 6.02x10^{23} F-units of CaCO_{3}

1 mole of Na+ = 23g = 6.02x10^{23}
ions of Na+

1 mole of CI- = 35.5g = 6.02x10^{23}
ions of CI-

Conversion of Mass into Moles and Moles into Mass of
Substance

Problem1.
Calculate the number of moles, in 50 g of each.

(a) Na (b) H_{2}O

Solution:

Number of moles __Given mass of substance__

Atomic mass or Formula mass

(a) Given,

i) Number of moles = ?

ii) Given mass of Na = 50g

iii) Atomic mass of Na = 23 a.m.u.

.'. Number of moles of Na = 50/33=2.173 moles of Na

(b) Given,

i) Number of moles of H_{2}O =
?

ii) Formula mass of H_{2}O = 18 a.m.u.

iii) Given mass of H_{2}O =
50g

Number of moles of H_{2}O = 50/18=2.777 moles of
H_{2}O

Problem 2.
What is the mass of 3 moles of each.

a) Al b) CO_{2}

Number of moles = Given mass of substance/Atomic mass
or Formula mass

.'. Mass of substance = number of moles x atomic mass or
formula mass in grams

(a) Given,

i) Number of moles of Al = 3 moles

ii) Atomic mass of Al = 27
grams

iii| Mass al Al =
Mole x atomic mass of Al

= 3x 27=81g

(b) Given,

i) Number of moles of CO_{2} =
3 moles

ii) Formula mass of CO_{2} = 44g

iii) Mass of CO_{2
}= Mole x Formula
mass of CO_{2}

= 3x42=132

Use of Avogadro’s number:

a) To calculate the number of atoms or molecules in a
given sample of substance.

b) To calculate the mass of single atom or molecule of
any substance.

Problem: 1.

Calculate the number of atoms in 9g of Al.

Solution:
According to Avogadro's number.

1 mole of Al= 27g = 6.02x10^{23} atoms.

This shows that:

27 g of Al contain 6.02x10^{23} atoms of
Al.

1g of Al will contain 6.02x10^{23}/27g

9g of Al will contain 6.02x10^{23} x9/27g
=2.006x10^{23}

This numerical can also be solved, by using the formula.

Number of atoms =N_{A}
x mass of substance/Atomic mass

Number of (Al) atoms =6.02x10^{23} x9/27g

=2.006x10^{23 }atoms of Al

Problem: 2.

Calculate the number of molecules in 9g of CO_{2}.

Solution:
According to Avogadro's number

1 mole of CO_{2} = 44g = 6.02x10^{23}
molecules

This shows that:

44 g of CO_{2} contain 6.02x10^{23}
molecules of CO_{2}

1 g of CO_{2} contain 6.02x10^{23}/44g
molecules

9 g of CO_{2} will contain 6.02x10^{23}x9g/44g

= 1.231x10^{23 }molecules of CO_{2}

By formula

Number of molecules =N_{A}
x mass of substance/Formula mass

Number of (CO_{2}) molecules = 6.02x10^{23}x9g/44g

=1.231x10^{23 }molecules of CO_{2}

Problem: 3.

Calculate the mass of one atom of carbon in grams

Solution:
According to Avogadro's number

1 mole of C = 12g = 6.02x10^{23 }atoms

This indicates that

6.02x10^{23 }atoms of C weigh 12 g

1 atoms of C =12g/6.02x10^{23 }= 12g x 10^{-23
}/6.02

= 1.993 x 10^{-23}g

Mass of one C-atom= 1.993 x 10^{-23}g

By formula

Mass of one atom=Atomic mass in g/N_{A}

Mass of one(C) atom=12g/6.02x10^{23 }
=1.993x02x10^{23}g

Problem: 4.

Calculate the mass of molecule of water (H_{2}O)
in gram.

Solution:

According to Avogadro’s number

1 mole of H_{2}O = 18g = 6.02x10^{23}
molecules

This indicates that

6.02x10^{23} molecules of (H_{2}O)
weight 18 g

1 molecules of (H_{2}O) weight 18g/6.02x10^{23}

= 18gx10^{-23}/6.02=2.90x10^{-23}g

By formula

Mass of one molecule = Formula mass in grams/NA

Mass of one (H_{2}O) molecules = 18g/6.02x10^{23}=2.90x10^{-23}g

Mass of one (H_{2}O)
molecule in grams = 2.90x10^{-23}g