3.2 - Mole Concept

Mole and the Avogadro Constant

The mole, symbol mol, is the SI unit for the amount of substance.
One mole contains exactly $6.02214076 \times 1 0^{23}$ elementary entities. This fixed value is the Avogadro constant, $N_{A}$ , expressed in $mo l^{- 1}$ .
For calculations at this level:

N_{A} = 6.02 \times 1 0^{23} mo l^{- 1}

Type of substance	Entities in $1 mol$
Atomic substance, such as Cu	$6.02 \times 1 0^{23}$ atoms
Molecular substance, such as $H_{2} O$	$6.02 \times 1 0^{23}$ molecules
Ionic substance, such as $NaCl$	$6.02 \times 1 0^{23}$ formula units

n = \frac{N}{N _{A}}, N = n N_{A}

where $n$ is the number of moles and $N$ is the number of particles.

N = 0.2 (6.02 \times 1 0^{23}) = 1.204 \times 1 0^{23} Mg atoms .

n = \frac{3.01 \times 1 0 ^{24}}{6.02 \times 1 0 ^{23}} = 5 mol .

Distinguish molecules from their constituent atoms. Two moles of $O_{2}$ contain $1.204 \times 1 0^{24}$ molecules but twice as many oxygen atoms, or $2.408 \times 1 0^{24}$ atoms.

Molar mass, $M$ , is the mass of one mole of a substance. Its unit is $g mo l^{- 1}$ .
The numerical value of molar mass equals the substance’s relative mass:
- atomic substance: molar mass equals RAM;
- molecular substance: molar mass equals RMM;
- ionic substance: molar mass equals RFM.
Examples: Cu has RAM 64 and molar mass $64 g mo l^{- 1}$ ; $H_{2} O$ has RMM 18 and molar mass $18 g mo l^{- 1}$ ; $NaCl$ has RFM 58.5 and molar mass $58.5 g mo l^{- 1}$ .

n = \frac{m}{M}, m = n M

where $m$ is mass in grams.

Example: mass of $1.5 mol$ Al is $1.5 (27) = 40.5 g$ .
Example: $32 g$ $S O_{2}$ , for which $M = 64 g mo l^{- 1}$ , is $32/64 = 0.5 mol$ .
Example: $4.7 g$ $K_{2} O$ , with $M = 2 (39) + 16 = 94 g mo l^{- 1}$ , represents $4.7/94 = 0.05 mol$ .

Molar volume is the volume occupied by one mole of a gas.
At the same temperature and pressure, one mole of any gas occupies the same volume.
It applies to gases only and depends on temperature and pressure:

Condition	Molar volume, $V_{m}$
STP: $0^{\circ} C$ and $1 atm$	$22.4 d m^{3} mo l^{- 1}$
Room conditions: $25^{\circ} C$ and $1 atm$	$24 d m^{3} mo l^{- 1}$

n = \frac{V}{V _{m}}, V = n V_{m}

Use consistent volume units: $1 d m^{3} = 1000 c m^{3}$ .
Examples:
- $2.2 mol$ $H_{2}$ at STP occupies $2.2 (22.4) = 49.28 d m^{3}$ .
- $0.01 mol$ $N H_{3}$ at room conditions occupies $0.01 (24) = 0.24 d m^{3} = 240 c m^{3}$ .
- $600 c m^{3}$ $O_{2}$ at room conditions represents $0.600/24 = 0.025 mol$ .

The number of moles is the central quantity connecting particles, mass and gas volume:

N = n N_{A}, m = n M, V = n V_{m}

For a conversion between two non-mole quantities:
1. convert the given quantity to moles;
2. convert the moles to the required quantity.
Example: volume of $26.4 g$ $C O_{2}$ at room conditions:

M (C O_{2}) = 44 g mo l^{- 1}, n = \frac{26.4}{44} = 0.6 mol

V = 0.6 (24) = 14.4 d m^{3}

n = \frac{672}{22.4 \times 1000} = 0.03 mol

N = 0.03 (6.02 \times 1 0^{23}) = 1.806 \times 1 0^{22} molecules