Amount of Substance & Solution Calculations

1. Relative Atomic and Molecular Mass

The masses of atoms and molecules are compared with the mass of an atom of carbon-12.

Relative Atomic Mass (\(A_r\))

The average mass of an atom of an element relative to 1/12 of the mass of an atom of carbon-12.

Relative Molecular Mass (\(M_r\))

The average mass of a molecule relative to 1/12 of the mass of an atom of carbon-12.

Ionic compounds do not contain molecules. For them, use the term relative formula mass. It is worked out in the same way, by adding up the A_r values in the formula.

Worked Example: Adding Up Relative Masses

M_r of H₂SO₄ = 2(1.0) + 32.1 + 4(16.0) = 98.1
Relative formula mass of CaCO₃ = 40.1 + 12.0 + 3(16.0) = 100.1

Relative masses have no units. Molar mass, M, has the same number but the unit g mol⁻¹.

A_r values are on the Periodic Table. To see how A_r is found from isotopes and mass spectra, go to Atomic Structure.

2. The Avogadro Constant and Particle Counting

The mole is the unit for amount of substance. One mole of any substance contains the same number of particles. This number is the Avogadro constant.

Avogadro Constant (\(L\))

The number of particles in one mole of a substance: \(L = 6.022 \times 10^{23}\text{ mol}^{-1}\). The value is given in the exam, so you do not need to learn it.

The mole can be used for any type of particle: atoms, molecules, ions, electrons or formula units. Always say which particle you mean. One mole of CO₂ molecules contains three moles of atoms.

Particle Number Relationship

\[ N = n \times L \]

Where \(N\) is the number of particles, \(n\) is the amount in moles and \(L\) is the Avogadro constant.

Worked Example: Counting Atoms

How many atoms are there in 0.500 mol of CO₂? (L = 6.02 × 10²³ mol⁻¹)

Step 1: each CO₂ molecule contains 3 atoms (1 C and 2 O).

Step 2: number of atoms = 0.500 × 6.02 × 10²³ × 3

Answer: 9.03 × 10²³ atoms

3. Core Mole Relationships

Solid Mass and Moles

Core Formula

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

Where \(n\) is amount of substance in moles (mol), \(m\) is mass in grams (g), and \(M\) is molar mass in g mol⁻¹. Relative molecular or formula mass, \(M_r\), is dimensionless. Its numerical value equals the molar mass when the latter is expressed in g mol⁻¹.

Solution Concentrations

For solutions, concentration represents the amount of solute dissolved in a specified volume of solution:

Concentration Formula

\[ n = c \times V \]

Where \(c\) is concentration in mol dm⁻³, and \(V\) is volume in dm³. To convert volume from cm³ to dm³, divide by 1000:

\[ V(\text{dm}^3) = \frac{V(\text{cm}^3)}{1000} \]

To convert molar concentration (mol dm⁻³) to mass concentration (g dm⁻³):

\[ \text{Concentration in g dm}^{-3} = \text{Concentration in mol dm}^{-3} \times M_r \]

Worked Example: Moles from Mass and from Concentration

4.00 g of NaOH (M_r = 40.0): n = 4.00 / 40.0 = 0.100 mol
25.0 cm³ of 0.100 mol dm⁻³ HCl: n = 0.100 × (25.0 / 1000) = 2.50 × 10⁻³ mol
0.100 mol dm⁻³ NaOH in g dm⁻³: 0.100 × 40.0 = 4.00 g dm⁻³

4. The Ideal Gas Equation

At low pressure and high temperature, real gas molecules occupy negligible volume compared to the container and experience negligible intermolecular forces. Under these conditions, gases obey the ideal gas equation:

Ideal Gas Law

\[ pV = nRT \]

Strict SI Unit Requirements

Variable	Physical Quantity	Strict SI Unit	Standard Conversion Rule
p	Pressure	Pascals (Pa)	1 kPa = 1000 Pa; 1 MPa = 1 × 10⁶ Pa
V	Volume	Cubic metres (m³)	1 dm³ = 1 × 10⁻³ m³; 1 cm³ = 1 × 10⁻⁶ m³
n	Amount	Moles (mol)	\(n = m / M_r\)
R	Gas Constant	J K⁻¹mol⁻¹	R is always given. Use the value in the question (8.31 or 8.314 J K⁻¹ mol⁻¹).
T	Temperature	Kelvin (K)	\(T(\text{K}) = \theta(^\circ\text{C}) + 273.15\)

Worked Example: Molar Mass from Gas Density

A sample of an unknown volatile liquid of mass 0.284 g is vaporised inside a gas syringe at 98.0 degrees C and 101.3 kPa. The resulting gas occupies a volume of 78.5 cm³. Calculate the relative molecular mass (M_r) of the substance.

Step 1: Convert all measurements to standard SI units:

\(p = 101.3\text{ kPa} = 101300\text{ Pa}\)
\(V = 78.5\text{ cm}^3 = 78.5 \times 10^{-6}\text{ m}^3\)
\(T = 98.0 + 273.15 = 371.15\text{ K}\)
\(R = 8.314\text{ J K}^{-1}\text{mol}^{-1}\)

Step 2: Rearrange ideal gas equation for moles (\(n\)):

\[ n = \frac{pV}{RT} = \frac{101300 \times 78.5 \times 10^{-6}}{8.314 \times 371.15} = \frac{7.95205}{3085.741} = 0.00257703\text{ mol} \]

Step 3: Calculate relative molecular mass (\(M_r\)):

\[ M = \frac{m}{n} = \frac{0.284}{0.00257703} = 110.204\text{ g mol}^{-1} \]

Reported to 3 significant figures: \(M_r = 110\).

Worked Example: Volume of a Gas

Calculate the volume of 0.0100 mol of gas at 298 K and 100 kPa. Give your answer in cm³. (R = 8.31 J K⁻¹ mol⁻¹)

Step 1: p = 100 kPa = 100 000 Pa

Step 2: V = nRT / p = 0.0100 × 8.31 × 298 / 100 000 = 2.48 × 10⁻⁴ m³

Step 3: 1 m³ = 10⁶ cm³, so V = 248 cm³

5. Empirical and Molecular Formulas

Empirical Formula

The simplest whole-number ratio of atoms of each element present in a compound.

Molecular Formula

The actual number of atoms of each element present in one molecule of a compound.

The molecular formula is always an integer multiple (\(k\)) of the empirical formula:

\[ k = \frac{M_r\text{ of compound}}{\text{Empirical formula mass}} \]

Worked Example: Combustion Analysis

Combustion of 1.80 g of an organic compound containing only carbon, hydrogen, and oxygen produces 3.96 g of CO₂ and 2.16 g of H₂O. The relative molecular mass of the compound is 60.0. Determine its empirical and molecular formulas.

Step 1: Calculate masses of carbon and hydrogen:

Mass of C: 3.96 × (12.0/44.0) = 1.08 g
Mass of H: 2.16 × (2.0/18.0) = 0.24 g
Mass of O by difference: 1.80 − (1.08 + 0.24) = 0.48 g

Step 2: Find mole ratio:

Moles of C: 1.08/12.0 = 0.090 mol
Moles of H: 0.24/1.0 = 0.240 mol
Moles of O: 0.48/16.0 = 0.030 mol

Step 3: Divide by smallest value (0.030):

C: 0.090/0.030 = 3
H: 0.240/0.030 = 8
O: 0.030/0.030 = 1

Empirical formula = C₃H₈O.

Relative empirical formula mass = (3 × 12.0) + (8 × 1.0) + 16.0 = 60.0 (no units).

Since empirical mass equals \(M_r\) (60.0), integer multiplier \(k = 1\). The molecular formula is also C₃H₈O.

Worked Example: Empirical Formula from Percentage by Mass

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its M_r is 180. Find its empirical and molecular formulas.

Step 1: assume 100 g, so the percentages become masses. Divide each by A_r:

C: 40.0 / 12.0 = 3.33
H: 6.7 / 1.0 = 6.7
O: 53.3 / 16.0 = 3.33

Step 2: divide by the smallest (3.33): C 1, H 2.0, O 1. Empirical formula = CH₂O.

Step 3: empirical formula mass = 12.0 + 2(1.0) + 16.0 = 30.0. k = 180 / 30.0 = 6.

Molecular formula = C₆H₁₂O₆.

Go deeper

These UK AQA pages cover the same ideas in more detail. They may include content that is not in 9620.

6. Practice Questions

Practice Question 1 (2 Marks)

Define the term relative atomic mass.

Show Answer & Mark Scheme

The average mass of an atom of an element (1 mark)
relative to 1/12 of the mass of an atom of carbon-12 (1 mark)

Practice Question 2 (2 Marks)

Calculate the total number of ions in 0.0250 mol of magnesium chloride, MgCl₂. (L = 6.02 × 10²³ mol⁻¹)

Show Answer & Mark Scheme

Each MgCl₂ gives 3 ions (one Mg²⁺ and two Cl⁻), so 0.0250 × 3 = 0.0750 mol of ions (1 mark)
0.0750 × 6.02 × 10²³ = 4.52 × 10²² ions (1 mark)

Practice Question 3 (3 Marks)

A 0.320 g sample of a gas occupies 250 cm³ at 300 K and 100 kPa. Calculate the M_r of the gas. (R = 8.31 J K⁻¹ mol⁻¹)

Show Answer & Mark Scheme

p = 100 000 Pa and V = 250 × 10⁻⁶ m³ (1 mark)
n = pV / RT = (100 000 × 250 × 10⁻⁶) / (8.31 × 300) = 0.01003 mol (1 mark)
M_r = 0.320 / 0.01003 = 31.9 (1 mark)

Practice Question 4 (3 Marks)

2.43 g of magnesium burns in oxygen to form 4.03 g of magnesium oxide. Find the empirical formula of the oxide.

Show Answer & Mark Scheme

Mass of O = 4.03 - 2.43 = 1.60 g (1 mark)
Mg: 2.43 / 24.3 = 0.100 mol; O: 1.60 / 16.0 = 0.100 mol (1 mark)
Ratio 1 : 1, so the empirical formula is MgO (1 mark)

Next, use these mole formulas in reacting-mass, gas-volume and titration calculations on Balanced Equations and Calculations.

Required Practical 1

Make a volumetric solution and carry out a simple acid-base titration

Use moles and concentrations to make a standard solution and work out a titration result.

Open RP1 →

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