Chemistry Concept of Mole Lecture #1

CONCEPT OF MOLE

When relative mass of any substance represent in grams (g) is called one mole of substance or molar mass.

• The mole is related to mass,volume and number of particles of the given substance.

Examples:

There are four different chemical substances can be represent in moles.

i) Atom

Relative mass of Na atom = 23 amu

When it represent in grams it becomes

23 amu ----> 23 grams = One Mole or simply Mole = Molar atomic mass or Molar mass = One gram mole

ii) Molecule

Relative mass of H2O molecule = 18 amu

When it represent in grams it becomes

18 amu ----> 18 grams = One Mole or simply Mole = Molar molecular mass or Molar mass = One gram mole

iii) Formula Unit

Relative formula mass of NaCl = 58.5 amu

When it represent in grams it becomes

58.5 amu ----> 58.5 grams = One Mole or simply Mole = Formula unit mass or Molar mass = One gram mole

iv) Ion

Relative mass of H+ ion = 1.008 amu

When it represent in grams it becomes

1.008 amu ----> 1.008 grams = One Mole or simply Mole = Molar ionic mass or Molar mass = One gram mole

We know according to Avogadro's 

One mole of any substance = NA = 6.02 × 10^23 particles (atoms or molecules or formula units or ions)

In case of ideal gases one mole of ideal gase at STP (standard temperature and pressure) =Molar volume (Vm) = 22.414 dm3 = 6.02 × 10^23 atoms or molecules

Interconversion of different Quantities into Mole

i. Mass into Moles and vice versa

Moles (n) = Mass (m) / Molar mass (M)

ii. Number of particles into Moles

Moles(n) = Number of particles(N)/NA

iii. Volume of ideal gas into Moles

Moles (n) = Vol.(V)/Molar Vol.(Vm)
Relationship between different formulae

• m / Mr = N / NA
• m / Mr = V / Vm
• N / NA  = V / Vm

Q.1  The 4.4 g CO2 occupies ___________ volume at STP
A)  22.414 dm3
B)  11.207 dm3
C)  2.2414 dm3
D)  4.4000 dm3

Correct Ans: C
Explanation:
Use formula:
m / Mr = V / Vm
Or
V = m / Mr × Vm
V = 4.4 g / 44 g × 22.414 dm3
V = 0.1 × 22.414 = 2.2414 dm3
Q.2 How many total number of atoms and electrons are present in 1.8 g of NH4- ions
A) 3.01 × 10^23 and 1.2 NA
B) 6.01 × 10^22 and 12 NA
C) 1.20 × 10^23 and 1.1 NA
D) 6.01 × 10^23 and 11

Correct Ans: A
Explanation:
Use formula for calculating total no.of atoms
N / NA = m / Mr × atomicity
Or
N = m / Mr × NA × 5
N = 1.8 g / 18 gmol-1 × 6.02 × 10^23 × 5
N = 0.1 × 6.02 × 10^23 × 5
N = 3.01 × 10^23 (Total atoms)
Use formula for calculating total no.of atoms
N / NA = m / Mr × No.of es- in one ion
Or
N = m / Mr × NA × 12
N = 1.8 g / 18 gmol-1 × 6.02 × 10^23 × 12
N = 0.1 × 6.02 × 10^23 × 12
N = 1.2 ×  6.02 × 10^23 or 1.2 NA (Total electrons)


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