Skip to content

Course home

Sign up

3.1.5 The mole and the Avogadro constant

Chemists count particles in moles

Definition

Mole

The amount of a substance that contains 6.02 × 10²³ particles, such as atoms, molecules or ions.

Definition

Avogadro constant

The number of particles in one mole of a substance, 6.02 × 10²³.

  1. One drop of water contains more than 102110^{21}1021 molecules, so chemists count particles in very large, fixed groups.
  2. One mole of any substance contains 6.02×10236.02 \times 10^{23}6.02×1023 particles.
    1. This number is the Avogadro constant, written in standard form because it is so large.
  3. The amount of a substance in moles is given the unit mol.
  4. Always say which particles you are counting, because one mole of oxygen molecules, O2\text{O}_2O2​, contains two moles of oxygen atoms.
  5. For an ionic compound, one mole of formula units of NaCl\text{NaCl}NaCl contains one mole of Na+\text{Na}^{+}Na+ ions and one mole of Cl−\text{Cl}^{-}Cl− ions.
Example

A mole works like a dozen, except that it stands for 6.02×10236.02 \times 10^{23}6.02×1023 items instead of 121212.

The Avogadro constant converts moles into particles

  1. Multiply the amount in moles by the Avogadro constant to find the number of particles: number of particles=moles×6.02×1023\text{number of particles} = \text{moles} \times 6.02 \times 10^{23}number of particles=moles×6.02×1023
  2. Divide a number of particles by the Avogadro constant to find the amount in moles.
    1. 3.01×10233.01 \times 10^{23}3.01×1023 water molecules is 3.01×10236.02×1023=0.500 mol\dfrac{3.01 \times 10^{23}}{6.02 \times 10^{23}} = 0.500\,\text{mol}6.02×10233.01×1023​=0.500mol.
  3. To count the atoms in molecules, multiply by the number of atoms in each molecule.
    1. One mole of H2O\text{H}_2\text{O}H2​O contains two moles of hydrogen atoms and one mole of oxygen atoms.
Example
  • How many atoms are in 0.200 mol0.200\,\text{mol}0.200mol of carbon dioxide, CO2\text{CO}_2CO2​?
  • Number of molecules =0.200×6.02×1023=1.204×1023= 0.200 \times 6.02 \times 10^{23} = 1.204 \times 10^{23}=0.200×6.02×1023=1.204×1023.
  • Each molecule contains 333 atoms, so the number of atoms =3×1.204×1023=3.612×1023= 3 \times 1.204 \times 10^{23} = 3.612 \times 10^{23}=3×1.204×1023=3.612×1023.
  • To three significant figures, there are 3.61×10233.61 \times 10^{23}3.61×1023 atoms.

Mass of one mole equals relative formula mass

Definition

Relative formula mass

The mass of one formula unit of a substance compared with one-twelfth of the mass of a carbon-12 atom, given the symbol Mr.

  1. The mass of one mole of an element is its relative atomic mass in grams, so one mole of carbon atoms has a mass of 12.0 g12.0\,\text{g}12.0g.
  2. The mass of one mole of a compound is its relative formula mass in grams, so one mole of water, Mr=18.0M_r = 18.0Mr​=18.0, has a mass of 18.0 g18.0\,\text{g}18.0g.
  3. This works because relative masses compare particles on the same scale, and one mole always contains the same number of particles.
    1. A magnesium atom is about twice as heavy as a carbon atom, so 24.3 g24.3\,\text{g}24.3g of magnesium contains the same number of atoms as 12.0 g12.0\,\text{g}12.0g of carbon.
  4. The mass of one mole of a substance is called its molar mass, measured in grams per mole.
  5. Equal masses of different substances therefore contain different numbers of particles.
Common Mistake

Do not give MrM_rMr​ a unit, because it is a relative value, whereas the mass of one mole is measured in grams.

Moles equal mass divided by relative formula mass

  1. Calculate the amount in moles from a mass using: moles=mass (g)Mr\text{moles} = \frac{\text{mass (g)}}{M_r}moles=Mr​mass (g)​
  2. Rearrange it to find the mass from an amount in moles: mass (g)=moles×Mr\text{mass (g)} = \text{moles} \times M_rmass (g)=moles×Mr​
  3. Always work out MrM_rMr​ first, using the relative atomic masses from the Periodic Table.
    1. For calcium carbonate, Mr(CaCO3)=40.1+12.0+(3×16.0)=100.1M_r(\text{CaCO}_3) = 40.1 + 12.0 + (3 \times 16.0) = 100.1Mr​(CaCO3​)=40.1+12.0+(3×16.0)=100.1.
  4. So 0.300 mol0.300\,\text{mol}0.300mol of calcium carbonate has a mass of 0.300×100.1=30.03 g0.300 \times 100.1 = 30.03\,\text{g}0.300×100.1=30.03g, which is 30.0 g30.0\,\text{g}30.0g to three significant figures.
Example
  • Calculate the amount in moles in 11.0 g11.0\,\text{g}11.0g of carbon dioxide (ArA_rAr​: C = 12.0, O = 16.0).
  • Mr(CO2)=12.0+(2×16.0)=44.0M_r(\text{CO}_2) = 12.0 + (2 \times 16.0) = 44.0Mr​(CO2​)=12.0+(2×16.0)=44.0.
  • moles=11.044.0=0.250 mol\text{moles} = \dfrac{11.0}{44.0} = 0.250\,\text{mol}moles=44.011.0​=0.250mol.

Linking mass and number of particles through moles

  1. To go from a mass to a number of particles, convert the mass into moles first and then multiply by the Avogadro constant.
    1. 4.86 g4.86\,\text{g}4.86g of magnesium is 4.8624.3=0.200 mol\dfrac{4.86}{24.3} = 0.200\,\text{mol}24.34.86​=0.200mol, which contains 0.200×6.02×1023=1.20×10230.200 \times 6.02 \times 10^{23} = 1.20 \times 10^{23}0.200×6.02×1023=1.20×1023 atoms.
  2. To go from a number of particles to a mass, divide by the Avogadro constant and then multiply by MrM_rMr​.
  3. The mass of one atom in grams is its relative atomic mass divided by the Avogadro constant.
    1. One magnesium atom has a mass of 24.36.02×1023=4.04×10−23 g\dfrac{24.3}{6.02 \times 10^{23}} = 4.04 \times 10^{-23}\,\text{g}6.02×102324.3​=4.04×10−23g.
  4. Because moles count particles, equal amounts in moles of two substances contain the same number of particles, whatever their masses.

Flowchart showing conversions between mass, moles, and particles using molar mass and the Avogadro constant.

Flow diagram showing conversions between mass, moles, and number of particles using molar mass and the Avogadro constant (6.02 × 10²³).

Exam technique
  • When a question gives a mass and asks for a number of particles, divide by MrM_rMr​ before you multiply by 6.02×10236.02 \times 10^{23}6.02×1023, because the Avogadro constant applies to an amount in moles.
  • Give the final answer in standard form, to the number of significant figures the question asks for.
Self review
  • What is the value of the Avogadro constant?
  • How many molecules are in 2.00 mol2.00\,\text{mol}2.00mol of water?
  • What is the mass of one mole of sodium chloride (ArA_rAr​: Na = 23.0, Cl = 35.5)?
  • How many moles are in 8.0 g8.0\,\text{g}8.0g of methane, CH4\text{CH}_4CH4​ (ArA_rAr​: C = 12.0, H = 1.0)?
  • Why do 12.0 g12.0\,\text{g}12.0g of carbon and 24.3 g24.3\,\text{g}24.3g of magnesium contain the same number of atoms?
PreviousNext

How was this guide?

Teach Genie

Review 3.1.5 The mole and the Avogadro constant by teaching Genie

Teach it back in your own words, spot gaps, and remember it better.

Start teaching
Genie and Baby Genie

Lesson

Recap your knowledge with an interactive lesson

8 minute activity

Start lesson

Atoms, molecules and ions are far too small to count individually, so chemists measure them in large fixed groups called moles. A mole is like a dozen, except that it represents 6.02×10236.02 \times 10^{23}6.02×1023 particles instead of 121212.

The number of particles in one mole is the Avogadro constant, NAN_{\text{A}}NA​. Its value is shown below:

NA=6.02×1023 mol−1 N_{\text{A}} = 6.02 \times 10^{23} \, \text{mol}^{-1} NA​=6.02×1023mol−1

The amount of substance has the symbol nnn and unit mol. Always identify the particles being counted, such as atoms, molecules, ions or formula units.

Flashcards

Remember key concepts with flashcards

20 flashcards

Practice flashcards

What is the Avogadro constant?

3.1.5 The mole and the Avogadro constant Revision Guide

  1. GCSE
  2. /Chemistry
  3. /3.1.5 The mole and the Avogadro constant

Revision notes for OCR GCSE Chemistry 3.1.5 The mole and the Avogadro constant: explanations and worked examples.

Revision guides