# Mole (unit)

The**mole**(symbol: mol) is one of the seven SI base units and is commonly used in chemistry. It measures the amount of substance of a system and is defined as the amount of substance that contains as many elementary entities as there are atoms in exactly 0.012 kilogram of carbon-12. This quantity is known as Avogadro's number and is approximately 6.02214199 × 10

^{23}.

When the mole is used, the elementary entities *must* be specified. Elementary entities may be:

A mole of atoms or molecules is also called a 'gram atom' or 'gram molecule'.

With a mole of a gas one means a mole of molecules.

A mole of any gas at a pressure of 100,000 Pa and a temperature of 300 K is 25 litres, see example.

Table of contents |

2 History 3 Utility of "moles" 4 Example calculation 5 See also |

## Background

Colloquially speaking, the mole is a convenient way of counting large numbers of particles. If you are dealing with this many molecules or atoms, then you have a mole of molecules or atoms. If you have half this number of these entities, then you have half a mole of them.

## History

The definition of a mole was originally chosen so that 14 grams (or 0.014 kg) of nitrogen constitute one mole; however, the definitions of the mole and the atomic mass unit are currently set so that one mole of carbon-12, which has a relative atomic mass of 12 u, is exactly 0.012 kg.

## Utility of "moles"

Number of particles is a more useful unit in chemistry than mass or weight, because reactions take place between atoms (for example, two hydrogen atoms and one oxygen atom make one molecule of water) that have very different weights (one oxygen atom weighs almost 16 times as much as a hydrogen atom). However, the raw numbers of atoms in a reaction are not convenient, because they are very large; for example, just one Millilitre of water contains over 3 × 10^{22} (or 30,000,000,000,000 billion) molecules.

## Example calculation

In this example, moles are used to calculate the mass of CO_{2} given off when 0.001 kg (or 1 g) of ethane is burnt. The formula involved is:

- 3.5 O
_{2}+ C_{2}H_{6}→ 2 CO_{2}+ 3 H_{2}O

_{2}and 3 moles of H

_{2}O. Notice that the amount of moles does not need to balance on either side of the equation. This is because a mole does not count mass or the number of atoms involved, simply the number of individual particles. In our calculation it is first necessary to work out the number of moles of ethane that has been burnt. The mass in grams of one mole of a substance is by definition its atomic or molecular mass. The atomic mass of hydrogen is 1, and the atomic mass of carbon is 12, so the molecular mass of C

_{2}H

_{6}is (2 × 12) + (6 × 1) = 30. One mole of ethane is 0.030kg (or 30 g). The amount burnt was 0.001 kg (or 1 g), or 1/30th of a mole. The molecular mass of CO

_{2}(the atomic mass of carbon is 12 and that of oxygen is 16) is 2 × 16 + 12 = 44, so one mole of carbon dioxide is 0.044 kg (or 44 g). From the formula we know that

- 1 mole of ethane gives off 2 moles of carbon dioxide

- 0.030 kg (30 g) of ethane gives off 2 × 0.044 kg (or 44 g) of carbon dioxide

*two*moles are produced. However, we also know that just 1/30th of a mole of ethane was burnt. Again:

- 1/30th of a mole of ethane gives off 2 × 1/30th of a mole of carbon dioxide.

- 30 × 1/30 g ethane gives off 44 × 2/30 g of carbon dioxide = 2.93 g = 0.00293 kg.

## See also