## Monday, 29 December 2014

### 3.2 The Mole and the Number of Particles

The number of moles and the number of particles
Definition : Mole: One mole is the amount of any substance which contains the same number of particles as there are in 12g of Carbon-12, that is $6.02×{10}^{23}$ or the Avogrado constant.
■ Relationship between number of moles and number of particles.

1 mole of any substance contains $6.02×{10}^{23}$ particle of that substances (atoms, molecules or ions).

For atom :
 ○ 1 mole of lithium (Li) : $6.02×{10}^{23}$ atoms of lithium ○ 1 mole of carbon (C) : $6.02×{10}^{23}$ atoms of carbon

For molecule:
 ○ 1 mole of hydrogen ( ${\text{H}}_{2}$ ) : $2×6.02×{10}^{23}$ atoms of hydrogen or $6.02×{10}^{23}$ molecules of hydrogen. ○ 1 mole of carbon dioxide ( ${\text{CO}}_{2}$ ) : $3×6.02×{10}^{23}$ atom or $6.02×{10}^{23}$ molecules of carbon dioxide.

For ion:
 ○ 1 mole of sodium chloride (NaCl) : $6.02×{10}^{23}$ ions ${Na}^{+}$ and $6.02×{10}^{23}$ ions ${Cl}^{-}$ ○ 1 mole of zinc bromide ( ${\text{ZnBr}}_{2}$ ) $6.02×{10}^{23}$ ions ${Zn}^{+}$ and $2×6.02×{10}^{23}$ ions ${Br}^{-}$

■ Summary of relationship between number of moles and number of particles.
Worked-example 3.2(a)
Calculate the number of particle in
 (a) 0.5 mole of Copper
 Solution Number of particle in 0.5 mole of copper $\text{=}0.5×6.02×{10}^{23}$ $\text{=}3.01×{10}^{23}$ atom
 (b) 0.25 mole Lead(II) ions, ${Pb}^{2+}$
 Solution Number of particle in 0.25 mole of Lead(II) ions, $\text{=}0.25×6.02×{10}^{23}$ $\text{=}1.505×{10}^{23}$ ion ${Pb}^{2+}$
 (c) 3.0 moles Carbon Dioxide,
 Solution Number of particle in 3.0 moles of Carbon Dioxide $\text{=}3.0×6.02×{10}^{23}$ $\text{=}1.806×{10}^{24}$ molecule ${CO}_{2}$

Worked-example 3.2(b)
Calculate the number of moles of the following substances
 (a) $1.505×{10}^{24}$ atom of Silver, Ag
 Solution Number of moles $\begin{array}{c}\text{=}\frac{1.505×{10}^{24}}{6.02×{10}^{23}}\end{array}$ = 2.5 mol
 (b) $3.01×{10}^{22}$ Oxide ions, ${O}^{2-}$
 Solution Number of moles $\begin{array}{c}\text{=}\frac{3.01×{10}^{22}}{6.02×{10}^{23}}\end{array}$ = 0.5 mol
 (c) $4.515×{10}^{24}$ Nitrogen molecules, ${N}_{2}$
 Solution Number of moles $\begin{array}{c}\text{=}\frac{4.515×{10}^{24}}{6.02×{10}^{23}}\end{array}$ = 7.5 mol of molecules ${N}_{2}$

Worked-example 3.2(c)

(a) Find the number of atoms in 1.5 moles of ${Cl}_{2}$
 Solution $\begin{array}{c}\text{1.5 =}\frac{\text{Number of molecules}}{6.02×{10}^{23}}\end{array}$ Number of molecules = $1.5×6.02×{10}^{23}$ Number of molecules = $9.03×{10}^{23}$ Number of atom = $2×9.03×{10}^{23}$ Number of atom = $1.806×{10}^{24}$ atom

(b) Find the number of ions in 0.5 mole of Magnesium Chloride, ${MgCl}_{2}$
 Solution $\begin{array}{c}\text{0.5 =}\frac{\text{Number of particles}}{6.02×{10}^{23}}\end{array}$ Number of particles = $0.5×6.02×{10}^{23}$ Number of particles = $3.01×{10}^{23}$ Number of ions = $3×3.01×{10}^{23}$ Number of ions = $9.01×{10}^{23}$