A Level Chemistry

A Level Chemistry Revision – The Avogadro Constant

Welcome back to Beyond’s Science Blog! This A Level Chemistry post explores The Avogadro Constant.

You can also subscribe to Beyond Secondary Resources for access to thousands of worksheets and revision tools. Our site was created with teachers in mind and includes lots of teacher instructions, however, it also contains content for students that will be particularly useful when revising! You can sign up for a free account here and take a look around at our free resources before you subscribe too.

It is not possible to see or to weigh a single atom, so in order to count atoms, chemists compare the masses of large numbers of atoms.

The relative atomic mass (Ar) of hydrogen is 1.0. One mole of hydrogen atoms has a mass of 1g and contains a certain number of hydrogen atoms. This number of atoms is known as L.

The relative atomic mass of helium is 4.0. This means that a helium atom is four times heavier than a hydrogen atom. The same number, L, of helium atoms would have a mass of 4g. This is the mass of one mole of helium atoms.

In fact, one mole of any substance contains the same number of constituent particles. This number, L, is called the Avogadro constant.

The Avogadro constant, L = 6.022 × 1023mol-1

The Avogadro constant is the number of constituent particles in one mole of any substance.

The term ‘constituent particles’ could refer to atoms, molecules, ions or formula units depending on the structure of the given substance.

For example, neon, Ne, is a monoatomic element. Its constituent particles are single atoms. One mole of neon contains 6.022 × 1023 neon atoms.

Carbon dioxide, CO2, is a compound with a simple molecular structure. Its constituent particles are CO2 molecules. So, one mole of carbon dioxide will contain 6.022 × 1023 CO2 molecules.

We can use this information to calculate the number of atoms in one mole of carbon dioxide. Since each CO2 molecule contains three atoms (one carbon atom and two oxygen atoms), one mole of carbon dioxide contains (6.022 × 1023) × 3 = 1.8066 × 1024 atoms.

The constituent particle of an ionic compound or a substance with a macromolecular structure is one formula unit of that compound. For example, one mole of calcium oxide, CaO, contains 6.022 × 1023 formula units of CaO.

Since each formula unit of CaO contains two ions (one Ca2+ and one O2-), one mole of calcium oxide also contains (6.022 × 1023) × 2 = 1.2044 × 1024 ions.

The following equation can be used to calculate the number of constituent particles:

number of constituent particles = L × number of moles

You will always be given the value of the Avogadro constant if you need to use it in a calculation.

Example Question 1

Calculate the number of atoms in 23g of sodium, Na.

The constituent particles of sodium are Na atoms

Ar of Na = 23.0

moles of Na = \frac{\text{mass}}{\text{A}_\text{r}} = \frac{23}{23.0} = 1 mole

number of constituent particles = (6.022 × 1023) × 1 = 6.022 × 1023 sodium atoms

Example Question 2

a. Calculate the number of molecules in 56g of nitrogen gas, N2.

The constituent particles of nitrogen are N2 molecules

Mr of N2 = (14.0 × 2) = 28.0

moles of N2 = \frac{\text{mass}}{\text{A}_\text{r}} = \frac{56}{28.0} = 2 moles

number of constituent particles = (6.022 × 1023) × 2 = 1.2044 × 1024 N2 molecules

b. Calculate the number of nitrogen atoms in 56g of nitrogen gas, N2.

56g of N2 = 1.2044 × 1024 N2 molecules

Each N2 molecule contains two nitrogen atoms

(1.2044 × 1024) × 2 = 2.4088 × 1024 nitrogen atoms

Example Question 3

Calculate the number of ions in 29.25g of sodium chloride, NaCl.

The constituent particles of sodium chloride are NaCl formula units

Mr of NaCl = (23.0 × 1) + (35.5 × 1) = 58.5

moles of NaCl = \frac{\text{mass}}{\text{A}_\text{r}} = \frac{29.25}{58.5} = 0.5 moles

(6.022 × 1023) × 0.5 = 3.011 × 1023 formula units of NaCl

Each NaCl formula unit contains two ions (one Na+ ion and one Cl– ion).

(3.011 × 1023) × 2 = 6.022 × 1023 ions

Example Question 4

Calculate the number of electrons in 120.2g of silicon dioxide, SiO2.

The constituent particles of silicon dioxide are SiO2 formula units

Mr of SiO2 = (28.1 × 1) + (16.0 × 2) = 60.1

moles of SiO2 = \frac{\text{mass}}{\text{A}_\text{r}} = \frac{120.2}{60.1} = 2 moles

(6.022 × 1023) × 2 = 1.2044 × 1024 formula units of SiO2

Each SiO2 formula unit contains one silicon atom and two oxygen atoms

So, 2 moles of SiO2 contains 1.2044 × 1024 silicon atoms and 2.4088 × 1024 oxygen atoms.

Each silicon atom has 14 electrons
(1.2044 × 1024) × 14 = 1.68616 × 1025

Each oxygen atom has 8 electrons.
(2.4088 × 1024) × 8 = 1.92704 × 1025

(1.68616 × 1025) + (1.92704 × 1025) = 3.6132 × 1025 electrons

Practice Questions

1. Which of the following statements is correct?
A. 0.5 moles of chlorine, Cl2, contains 3.001 × 1023 chlorine atoms.
B. 1.5 moles of magnesium oxide, MgO, contains 1.8066 × 1024 ions.
C. 12g of carbon, C, contains 6.022 × 1024 carbon atoms.
D. 36g of water, H2O, contains 1.2044 × 1023 molecules.

2. Which of the following contains the greatest number of atoms?
A. 27.9g of iron, Fe
B. 32.0g of oxygen, O2
C. 48.6g of magnesium, Mg
D. 111.1g of calcium chloride, CaCl2

3. Calculate the number of atoms in 1.70g of ammonia, NH3.

4. Calculate the number of protons in 10.0g of hydrogen, H2.

5. The relative atomic mass (Ar) of aluminium, Al, is 27.0. Calculate the mass, in kilograms, of one aluminium atom.

6. A sample of sodium oxide, Na2O, contains 3.011 × 1023 sodium ions, Na+. Calculate the mass, in grams, of the sodium oxide in the sample.

Answers

1. Which of the following statements is correct?

A. 0.5 moles of chlorine, Cl2, contains 3.001 × 1023 chlorine atoms.
0.5 moles of Cl2 = 6.022 × 1023 atoms

B. 1.5 moles of magnesium oxide, MgO, contains 1.8066 × 1024 ions.
Correct

C. 12g of carbon, C, contains 6.022 × 1024 carbon atoms.
12g of C = 6.022 × 1023 atoms

D. 36g of water, H2O, contains 1.2044 × 1023 molecules.
36g of H2O = 1.2044 × 1024 molecules

2. Which of the following contains the greatest number of atoms?

A. 27.9g of iron, Fe
27.9g of Fe = 0.5 moles = 3.011 × 1023 atoms

B. 32.0g of oxygen, O2
32.0g of O2 = 1 mole = 1.2044 × 1024 atoms

C. 48.6g of magnesium, Mg
48.6g of Mg = 2 moles = 1.2044 × 1024 atoms

D. 111.1g of calcium chloride, CaCl2
111.1g of CaCl2 = 1 mole = 1.8066 × 1024 atoms

111.1g of calcium chloride contains the greatest number of atoms.

3. Calculate the number of atoms in 1.70g of ammonia, NH3.

Mr of NH3 = (14.0 × 1) + (1.0 × 3) = 17.0

moles of NH3 = \frac{\textbf{mass}}{\textbf{A}_\textbf{r}} = \boldsymbol{\frac{1.70}{17.0}} = 0.1 moles

(6.022 × 1023) × 0.1 = 6.022 × 1022 NH3 molecules

Each NH3 molecule contains four atoms

(6.022 × 1022) × 4 = 2.4088 × 1023 atoms = 2.41 × 1023 (3s.f.)

4. Calculate the number of protons in 10.0g of hydrogen, H2.

Mr of H2 = (1.0 × 2) = 2.0

moles of H2 = \frac{\textbf{mass}}{\textbf{A}_\textbf{r}} = \boldsymbol{\frac{10}{2.0}} = 5 moles

(6.022 × 1023) × 5 = 3.011× 1024 H2 molecules

Each molecule contains two H atoms

(3.011 × 1024) × 2 = 6.022× 1024 H atoms, each with one proton

5. The relative atomic mass (Ar) of aluminium, Al, is 27.0. Calculate the mass, in kilograms, of one aluminium atom.

1 mole of Al = 27.0g = 6.022 × 1023 Al atoms

\boldsymbol{\frac{27.0}{6.022\,\times\,10^{23}}} = 4.483560279 × 10-23 g
= 4.483560279 × 10-26 kg
= 4.48 x 10-26 kg (3s.f.)

6. A sample of sodium oxide, Na2O, contains 3.011 × 1023 sodium ions, Na+. Calculate the mass, in grams, of the sodium oxide in the sample.

Each Na2O formula unit contains two Na+ ions.

\boldsymbol{\frac{3.011\,\times\,10^{23}}{2}} = 1.5055 x 1023 Na2O formula units

moles of Na2O = \boldsymbol{\frac{1.5055\,\times\,10^{23}}{6.022\,\times\,10^{23}}} = 0.25 moles

Mr of Na2O = (23.0 × 2) + (16.0 × 1) = 62.0

mass = 0.25 × 62.0 = 15.5g

Found this A Level Chemistry blog post on The Avogadro Constant useful? You can read our other Science revision blogs here.

Leave a Reply