AQA A-Level Physical Chemistry 3.1.2 Amount of Substance Moles and Molar Mass
3.1.2

Moles and Molar Mass

Understand relative masses, the definition of the mole, and calculations involving mass and molar mass.

Atoms and molecules are far too small to be weighed or counted individually. To solve this problem, chemists use a standardized unit called the mole to bridge the gap between subatomic particles and laboratory scale masses.

🔑 Key Principle

The mole is a counting unit, much like a dozen represents 12 items. A mole represents exactly \( 6.022 \times 10^{23} \) particles. This number is carefully chosen so that the mass of one mole of an element in grams is numerically equal to its relative atomic mass.

Relative Atomic and Molecular Mass

Because atoms have extremely small masses, we measure their mass relative to a standard reference: the carbon-12 isotope.

Relative Atomic Mass (Ar)

The weighted mean mass of an atom of an element compared to one twelfth of the mass of an atom of carbon-12.

Relative Molecular Mass (Mr)

The weighted mean mass of a molecule compared to one twelfth of the mass of an atom of carbon-12. This is calculated by adding up the relative atomic masses of all the atoms in the molecular formula.

📝 AQA Examiner Tip

For ionic compounds (such as \( \text{NaCl} \)), which do not exist as simple molecules, we use the term relative formula mass instead of relative molecular mass. However, both use the symbol \( M_r \) and are calculated in the same way, by summing the \( A_r \) values of the constituent atoms.

What is the Mole?

In A-Level chemistry, you must know the official definition of the mole as defined by the carbon-12 standard.

The Mole

The amount of substance in grams that contains the same number of particles as there are atoms in exactly 12 grams of carbon-12.

This number of particles is known as Avogadro's constant (symbol \( L \) or \( N_A \)):

Avogadro's Constant (L)

\( L = 6.022 \times 10^{23} \text{ mol}^{-1} \)

This constant is provided on the AQA Periodic Table Data Sheet.

Molar Mass

The mass of one mole of a substance is called its molar mass (symbol \( M \)) and is measured in units of \( \text{g mol}^{-1} \).

✏️ Worked Example 1: Finding Molar Mass
Calculate the molar mass of sulfuric acid, \( \text{H}_2\text{SO}_4 \).

Use the relative atomic masses from the periodic table:

\( A_r(\text{H}) = 1.0 \), \( A_r(\text{S}) = 32.1 \), \( A_r(\text{O}) = 16.0 \)

Calculate the molar mass by summing the masses:

\[ M = (2 \times 1.0) + 32.1 + (4 \times 16.0) = 2.0 + 32.1 + 64.0 = 98.1 \text{ g mol}^{-1} \]

Mass-Mole Calculations

You can convert between the mass of a substance, the amount of substance in moles, and its molar mass using the following relationship:

\[ n = \frac{m}{M} \]

Where:

Mole Conversions Flowchart Mass (m in grams) divide by M multiply by M Moles (n in mol) multiply by L divide by L Mole Conversions Flowchart Mass (g) \( m = n \times M \) divide by M multiply by M Moles (mol) \( n = \frac{m}{M} \) multiply by L divide by L Particles (N) \( N = n \times L \)
✏️ Worked Example 2: Calculating Moles from Mass
Calculate the number of moles of sodium chloride, \( \text{NaCl} \), in \( 5.85 \text{ g} \) of the salt.

Step 1: Find the molar mass of \( \text{NaCl} \).

\( A_r(\text{Na}) = 23.0 \), \( A_r(\text{Cl}) = 35.5 \)

\[ M = 23.0 + 35.5 = 58.5 \text{ g mol}^{-1} \]

Step 2: Apply the equation \( n = \frac{m}{M} \).

\[ n = \frac{5.85}{58.5} = 0.100 \text{ mol} \]

✏️ Worked Example 3: Calculating Mass from Moles
Calculate the mass in grams of \( 0.250 \text{ moles of } \text{H}_2\text{O} \).

Step 1: Find the molar mass of \( \text{H}_2\text{O} \).

\( A_r(\text{H}) = 1.0 \), \( A_r(\text{O}) = 16.0 \)

\[ M = (2 \times 1.0) + 16.0 = 18.0 \text{ g mol}^{-1} \]

Step 2: Rearrange the equation to find mass: \( m = n \times M \).

\[ m = 0.250 \times 18.0 = 4.50 \text{ g} \]

Using Avogadro's Constant

You can calculate the total number of atoms, ions, or molecules in a given sample by multiplying the amount of substance in moles by Avogadro's constant.

\[ N = n \times L \]
✏️ Worked Example 4: Calculating Number of Atoms
Calculate the total number of oxygen atoms in \( 4.50 \text{ g of water, } \text{H}_2\text{O} \).

Step 1: Calculate the moles of \( \text{H}_2\text{O} \).

Molar mass of \( \text{H}_2\text{O} = 18.0 \text{ g mol}^{-1} \)

\[ n = \frac{4.50}{18.0} = 0.250 \text{ mol} \]

Step 2: Calculate the number of \( \text{H}_2\text{O} \) molecules.

\[ N_{\text{molecules}} = 0.250 \times 6.022 \times 10^{23} = 1.506 \times 10^{23} \text{ molecules} \]

Step 3: Calculate the number of oxygen atoms.

Each molecule of \( \text{H}_2\text{O} \) contains exactly 1 atom of oxygen, so the number of oxygen atoms is equal to the number of molecules:

\[ N_{\text{atoms of O}} = 1.506 \times 10^{23} \text{ atoms} \]

Study this topic on the go

Get flashcards and quizzes in ChemEasy, or plan your revision with ChemPlan IB.

See our apps →
← Topic Index Ideal Gas Equation →