Class 11 Chemistry Notes Stocihiometry – N factor calculation

N factor here we mean a conversion factor by which we divide molar mass of substance to get equivalent mass and it depends on nature of substance which varies from one condition to another condition. We can divide n-factor calculation is two category. 

In case of non-redox reaction.

(a) n factor of acid =Basicity of the acid
Basicity: Number of replaceable H⁺ ion.

Ex.1

n factor of HCl = 1
n factor of CH₃COOH=1
n factor of  H₂SO₄=2

(b)n factor of base= acidity of the base

Acidity: Number of replaceable OH^– ion.

Ex.2   

n factor of NaOH=1
n factor of Ca(OH)₂=2
n factor of Al(OH)₃=3
n factor of B(OH)₃=1 (because it is a mono basic acid)

(c) n factor Salt: Total number of positive or negative charge.

Ex.3

n factor of NaCl=1
n factor of Na₂SO₄=2
n factor of K₂SO₄. Al₂ (SO₄)₃. 24H₂O = 8

Ex.4   

Find the n factor of H₃PO₄ in the following reaction.
H₃PO₄ + Ca (OH)₃ → CaHPO₄ + 2H₂O

Sol.

Basicity of H₃PO₄ in the above reaction is 2
∴ the n factor of H₃PO₄ is 2

In case of redox reaction

(a) From oxidation number

n factor of oxidizing or reducing agent=change in oxidation number per molecule.

⇒ consider a salt A_xB_y in which the O.S. of A is +c. It changes to a compound A_dE in which the O.S. of A is +f. Here we are assuming that B does not undergo any change in O.S. A^+c_xB_y → A^+f_dE (Obviously A_xB_y must have reacted with some other substance to produce the product A_dE. That means other substance has the atom E in it.) The ‘n’ factor is = |xc-xf|.

Ex.5

Find the n factor of KMnO₄ in different medium.

Sol.

(i) In acidic medium
KMnO₄→Mn⁺²
Change in oxidation number of Mn=+7-2=5
∴ the n factor of KMnO₄=5

(ii) Basic medium
KMnO₄→K₂MnO₄
N factor of KMnO₄ = +7 – 6 = 1

(iii) Neutral medium
KMnO₄ → MnO₂
N factor of MnO₂ = +7 – 4 = 3

 (b)From ion electron method:

N factor = total number of electrons transferred per mole of the reactant

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