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Mole concept:

The mole is defined as the amount of a substance containing as many atoms, molecules, ions, electrons or other elementry entities as there are carbon atoms in exactly 12 g of ¹²C.

N_A = 6.022 X 10²³

The following are the definitions of 'mole' represented in the form of equations:

(1) no. of moles of molecules/atoms = weight in g

molecular/atomic wt

(2) no. of moles of gases = volume at NTP

standard molar volume (i.e. 22.4 L)

(3) no. of moles of entities = no. of entities

Avagadro constant (i.e. 6.022 X 10²³)

(4) no. of moles of solute = molarity X volume of solution in L

no. of millimoles = molarity X volume of solution in mL

(5) For a compound M_xN_y, x moles of N = y moles of M

Note: 1 mole is a fixed no. of particles but not a fixed weight.

Principle of Atom Conservation (POAC):

This principle states that the moles of atoms of an element are conserved throughout the reaction.

This means, moles of an element in reactants = moles of the element in products.

eg. CO + O₂ -----> CO₂

In this moles of C in CO = moles of C in CO₂

This means, moles of CO = moles of CO₂.

Similarly, for O,

moles of O in CO + moles of O in O₂ = moles of CO₂

Thus, moles of CO + 2(moles of O₂) = 2(moles of CO₂)

Advantages of Mole Method over other Methods:

(1) Balancing of chemical equations is not required in the majority of problems as the method of balancing the chemical equation is based on the principle of conservation of atom conservation.

(2) Number of reactions and their sequences, leading from reactants to products, need not be given.

Note: Whenever balanced chemical equation is given, mole method is very useful.

eg. 2KClO₃ -----> 2KCl + 3O₂

Thus, 2(moles of KClO₃) = 2(moles of KCl)

3(moles of KClO₃) = 2(moles of O₂)

3(moles of KCl) = 2(moles of O₂)

Concept of Limiting Reagent:

In single-reactant reactions, the calculations are carried out with only that amount of the reactant which has converted to the product.

In the reactions where more than on reactant is involved, one has to first identify the limiting reactant, i.e., the reactant which is completely consumed. All calculations are to be carried out with the amount of the limiting reactant only.

Initially ..... 5 moles 12 moles 0 moles

A + 2B -----> 4C

If A is limiting reactant: moles of C produced = 20

If B is limiting reactant: moles of C produced = 24

The reactant producing the least number of moles of the product is the limiting reactant and hence A is the limiting reactant. Thus,

Initially ..... 5 moles 12 moles 0 moles

A + 2B ------> 4C

Finally ..... 0 moles 4 moles 20 moles

The limiting reactant can also be ascertained by knowing the initial number of equivalents of each reactant. The reactant with the least number of equivalents is the limiting reactant.

Note: Generally, when the initial moles / equivalents of all the reactants is given, then the question is based on the concept of limiting reagent.

Eudiometry:

Reactants Absorbed gases

KOH solution CO₂ & SO₂

Alkaline Pyrogallol O₂

Soln of ammonical cuprous chloride CO

and if the mixture is cooled, H₂O vapour changes to liquid.

Hence, When temperature is above 100 ⁰C (373 K) then only volume of H₂O is considered.

C_xH_y + (x + (y/4)) O₂ -----> x CO₂ + (y/2) H₂O.

Eudiometry is mainly based on Avogadro's law,

A(g) + B(g) -----> C(g) + D(g)

a volume b volume c volume d volume

a moles b moles c moles d moles

Chemical Equivalence:

One equivalent of any substance is the amount of that substance that combines with 1 g H or displaces 1 g H.

no. of equivalents = weight

equivalent weight

And, in a compound, M_xN_y, no of equivalents of M = no of equivalents of N

Equivalent weights of some elements :

H - 1g; O - 8g, Cl - 35.5g .....

For metals in hydrides, oxides and chlorids,

eq. wt = atomic weight

valency (in the compd)

Equivalent weight = Molecular weight

n

Hence, no. of equivalents = no. of moles X n

Hence, Normality = Molarity X n

Law of chemical equivalence:

aA + bB -----> cC + dD

eq of A = eq of B = eq of C = eq of D

Neutralisation reactions:

For acids, n = no of replacable H⁺ ions

Thus, for HCl - n = 1

H₂SO₄ - n = 2

H₃PO₄ - n = 3

H₃PO₃ - n = 2

For bases, n = no of OH^- ions provided by the base.

Thus, for NaOH - n = 1

Ca(OH)₂ - n = 2

For salts, n = total charge on cations or anions

Thus, for NaCl - n = 1

CaCl₂ - n = 2

Ca₃(PO₄)₂ - n = 6

Note: When the reactions are provided, then take the n - factor according to the reaction.

eg. H₃PO₄ + NaOH -----> NaHPO₄ + H₂O

Here only 1 H is replaced, n = 1

When 2 or more indicators are mentioned in the question, then the reaction takes place in two phases.

Phenolphthalein shows colour change after the first step and Methyl Orange shows the colour change after the completion of the complete reaction.

2Na₂CO₃ + H₂SO₄ à 2NaHCO₃ + Na₂SO₄

n = 1 n = 2

2NaHCO₃ + H₂SO₄ à Na₂SO₄ + 2H₂CO₃

n = 1 n = 2

Redox Reactions:

n = total change in oxidation number for the molecule / ion

total number of electrons gained / lost by the molecule

K₂Cr₂O₇ (oxidizing agent) Cr ⁺⁶ à Cr ⁺³ n = 6

Na₂S₂O₃(reducing agent) S⁺² à S ^+2.5 n = 1

H₂C₂O₄ (reducing agent) à CO₂ n = 2

NaHC₂O₄ (reducing agent) à CO₂ n = 2

Na₂C₂O₄ (reducing agent) à CO₂n = 2

Note: H₂C₂O₄ NaHC₂O₄

For redox: n = 2 n = 2

For neutralistion n = 2 n = 1

KMnO₄ (oxidizing agent)

In acidic medium, KMnO₄ à Mn⁺² n = 5

In basic/neutral medium, KMnO₄ à MnO₂ n = 3

In limited and controlled amount of base, KMnO₄ à K₂MnO₄ n = 1

H₂O₂ (oxygen ? oxidation state = 1)

H₂O₂ à O₂ (as a reducing agent) n = 2

H₂O₂ à H₂O (as a oxidising agent) n = 2

Note: Oxidation number of s-block do not change.

For, FeC₂O₄, (in presence of oxidizing agent) n = 3 as

Fe⁺² à Fe ⁺³ + e^-

2C⁺³ à 2C⁺⁴ + 2e^-

Thus, total transfer of electrons = 3

Similarly, for H₂C₂O₄.NaHC₂O₄.xH₂O à CO₂ n = 4

And for H₂C₂O₄.FeC₂O₄.xH₂O à CO₂ n = 5

Disproportionation reaction:

Simultaneous oxidation & reduction of an element present in the molecule or ion in the same reaction.

In this,

Eq. wt. of molecule= (Eq. wt. due to oxidation) + (Eq. wt. due to reduction)

And ( 1 / n_overall) = ( 1 / n_oxi ) + ( 1 / n_red)

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