### The Mole & Avogadro’s Constant

**The mole**

- This is the mass of a substance containing the same number of fundamental units as there are atoms in exactly 12.000 g of
^{12}C - The mole is the unit representing the amount of atoms, ions, or molecules
- One mole is the amount of a substance that contains
**6.02 x 10**particles (Atoms, Molecules or Formulae) of a substance (6.02 x 10^{23}^{23 }is known as the**Avogadro Number)**

**Examples**

- 1 mole of Sodium (Na) contains 6.02 x 10
^{23}**Atoms**of Sodium - 1 mole of Hydrogen (H
_{2}) contains 6.02 x 10^{23}**Molecules**of Hydrogen - 1 mole of Sodium Chloride (NaCl) contains 6.02 x 10
^{23 }**Formula units**of Sodium Chloride

**Linking the mole and the atomic mass**

- One mole of any element is equal to the relative atomic mass of that element in grams
- For example one mole of carbon, that is if you had 6.02 x 10
^{23}atoms of carbon in your hand, it would have a mass of 12g - So one mole of helium atoms would have a mass of 4g, lithium 7g etc
- For a compound we add up the relative atomic masses
- So one mole of water would have a mass of 2 x 1 + 16 = 18g
- Hydrogen which has an atomic mass of 1 is therefore equal to
^{1}/_{12}the mass of a^{12}C atom - So one carbon atom has the same mass as 12 hydrogen atoms

**Extended Only**

### The Mole & the Volume of Gases

**Molar volume**

- This is the volume that one mole of any gas (be it molecular such as CO
_{2}or monoatomic such as helium) will occupy - It’s value is 24dm
^{3}or 24,000 cm^{3}at room temperature and pressure (r.t.p.)

**Calculations involving gases**

**General equation:**

Amount of gas (mol) = Volume of gas (dm^{3}) ÷ 24

or

Amount of gas (mol) = Volume of gas (cm^{3}) ÷ 24000

**1. Calculating the volume of gas that a particular amount of moles occupies**

**Equation:**

Volume of gas (dm^{3}) = Amount of gas (mol) x 24

or

Volume of gas (cm^{3}) = Amount of gas (mol) x 24000

**Example:**

**2. Calculating the moles in a particular volume of gas**

**Equation:**

Amount of gas (mol) = Volume of gas (dm^{3}) ÷ 24

or

Amount of gas (mol) = Volume of gas (cm^{3}) ÷ 24000

**Example:**