Mole concept is arguably the broadest topic in chemistry, as it cuts across every other branch of chemistry, as far as reactions are concerned. It is the foundation of calculations in chemistry.
Chemical reactions, expressed as equations, are ways of confirming the Law of Conservation of Mass, which states that matter is neither created nor destroyed, but is transformed from one form to another. This is also what was paraphrased in one of the postulations of Dalton’s Atomic Theory, which is, atoms are neither created nor destroyed (during a chemical reaction), but are changed from one form to another.
The above implies that if two substances A and B combine to give another substance C, i.e,
A + B → C
then, the total masses of A and B will equal the mass of C. In other words, all of A and B will be converted to C, assuming there is no loss. Meaning that if we know the masses of A and B, we can easily calculate what we should be expecting as C.
For instance, in the reaction between sodium and chlorine to form sodium chloride;
2Na + Cl2 → 2NaCl
Before reacting (reactants):
Na = 2*1 = 2 atoms
Cl = 1*2 = 2 atoms
After reacting (products):
Na = 2*1 = 2 atoms
Cl = 2*1 = 2 atoms
we can see that 2 atoms of Na, combined with 2 atoms of Cl to form 2 moles of NaCl, made up of 2 atoms of Na and 2 atoms of Cl each; and if we carry out atom counts, we will observe that no atom was created (added)and none was destroyed (lost).
From the above example, it means that whenever Na and Cl combine to form NaCl, they MUST do so in the ratio of 2:1:2 That is to say, 2 moles of Na molecule will always combine with 1 mole of Cl molecule to form 2 moles of NaCl molecule. This is known as the stoichiometry of the reaction, (i.e, the relationship between the amounts of the reactants and the products, in terms of the mole ratio and mass ratio of the species involved).
Definition of Terms
Mole
Since the substances that react contain different amounts of elementary particles (atoms, molecules, ions, electrons etc), which directly or indirectly contribute to their masses; it is always difficult to measure the exact amount of a particular substance that may be required to react with another. Thus, the need to take their masses and volumes (if gaseous), to “the same base”. This “base” is known as the mole.
Hence, a mole can be defined as the amount of elementary entities or particles (atoms, molecules, ions, electrons etc) present in a substance, which is exactly as those present in 12 grammes of carbon-12.
Molar Mass
This is the mass of 1 mole of a substance expressed in grammes. Its is expressed in grammes per mole (g/mol).
Molar Volume
This is the volume occupied by 1 mole of a substance, usually gases, at standard temperature and pressure (s.t.p). It is also known as the molar gas volume (G.M.V), and has a constant value of 22.4dm^3 or 22400cm^3.
Avogadro Number
This is a constant value of 6.023 x 10^23, which represents the amount of elementary entities present in 1 mole of a substance. It can be likened to a dozen – 12, a score – 20, or a gross – 144.
Important Formulae Used in Mole Concept
1) Mole (n) = Reacting Mass (m)
——————
Molar Mass (M)
i.e, n = m/M
This implies that,
Reacting Mass = Mole x Molar Mass
Molar Mass = Reacting Mass/Mole
2) Mole (n) = Reacting Volume (v)
——————–
Molar Volume (V)
i.e, n = v/V
This implies that,
Reacting Volume = Mole x Molar Volume
Molar Volume = Reacting Vol/Mole
3) Mole (n) = Number of Particles
——————–
6.023 x 10^23
Relationship Between Mole & Mass (m), Volume (v) and Avogadro Number (L)
The relationships between the four quantities stated above can be expressed as shown below:
Molar Mass — Mole — Molar Volume
|
Avogadro Number
i.e, 1 mole = 6.023 x 10^23 = 22.4dm^3 = molar mass
1mole of CO2 = 44g of CO2 = 22.4dm^3 of CO2 = 6.023 x 10^23 molecules of CO2
1mole of O2 = 32g of O2 = 22.4dm^3 of O2 = 6.023 x 10^23 molecules of O2 = 2 x 6.023 x 10^23 atoms
1mole of Na = 23g of Na = 6.023 x 10^23 molecules of Na = 6.023 x 10^23 atoms of Na.
Example:
2Na(s) + Cl2(g) → 2NaCl(s)
You will observe that we have introduced the state symbols (s – solid, g – gas), which enables us to know the physical states at which the reacting species and products are at room temperature. Chemical calculation based on mass volume relationship
Example:
Calculate the volume of carbon dioxide formed at STP in ‘ml’ by the complete thermal decomposition of 3.125 g of pure calcium carbonate (Relative atomic mass of Ca=40, C=12, O=16)
Solution:
As per the equation,
CaCO3 → CaCO + CO2
40 + 12 + (16 x 3) 1 mole
100g 22.4 litres
100 g 24400 ml
Volume of carbon dioxide formed from 100 g of CaCO3 = 22400 ml
Volume of carbon dioxide from 3.125 g of CaCO3 = ?
CaCO3 : CO2
100 g : 22400 ml
3.125 g : x
Volume of carbon dioxide formed = 700 ml.
Example:
Calculate the volume of ammonia formed at STP in ‘ml’ by treating 2.675 g of ammonium chloride with excess of calcium hydroxide (Relative Atomic Mass of N=14, O=16, H=1, Cl=35.5, Ca=40).
Solution:
As per the equation,
NH4Cl + Ca(OH)2OH → CaCl2 + H2O + NH3
2 (14 + 4 + 35.5) 2 vols.(of NH3)
107 2 x 22.4 litres
107 g 44.8 litres or 44800 ml
NH4Cl : NH3
107 g : 44800 ml
2.675 g : x
x = (2.675 X 44800)/107
= 1120ml
Volume of ammonia formed at STP = 1120 ml.