The Mole – Cambridge IGCSE/A‑Level Physics (9702)
Learning Objectives
- Understand that amount of substance is one of the seven SI base quantities.
- Know that its base unit is the mole (mol), defined by an exact number of elementary entities.
- Convert between mass, amount of substance and number of particles, and apply the mole in equations such as the ideal‑gas law.
Syllabus Context (Matter strand)
| Syllabus Item | Notes Section |
|---|
| 15.1 The mole – definition, Avogadro constant, molar mass | Definition of the mole; exact value of \(N_{\!A}\); molar‑mass conversion |
| 15.2 Using the mole in gas laws (ideal‑gas equation) | Ideal‑gas law with worked example; model‑limits box |
| AO‑3 practical skills – experimental determination of Avogadro’s number, vapour‑density method | Practical checklist, data‑sheet template, error‑analysis guide |
| Mathematical tools – algebra, proportional reasoning, scientific notation | Algebraic rearrangements; proportional‑reasoning exercise; scientific‑notation activity |
| Key concepts – models, testing, matter/energy | Conceptual links box |
Definition of the Mole
The mole is the amount of substance that contains exactly
\(N_{\!A}=6.02214076\times10^{23}\) elementary entities
where an “entity” may be an atom, molecule, ion, electron, etc. The value of \(N_{\!A}\) is exact by definition (it is not a measured quantity).
Why the Mole Is an SI Base Quantity
- It provides a direct bridge between the microscopic world (individual particles) and macroscopic quantities we can measure (mass, volume, pressure).
- It is defined independently of any other physical quantity – the numerical value of \(N_{\!A}\) is fixed.
- It is one of the seven SI base quantities (metre, kilogram, second, ampere, kelvin, candela, mole).
Relationship to Mass – Molar Mass
For a pure substance the mass of one mole is the molar mass \(M\) (units g mol⁻¹). The fundamental conversion is
\(n=\dfrac{m}{M}\qquad\text{or}\qquad m=nM\)
- \(n\) = amount of substance (mol)
- \(m\) = mass of the sample (g)
- \(M\) = molar mass (g mol⁻¹)
Common Molar Masses
| Substance | Formula | Molar mass \(M\) (g mol⁻¹) |
|---|
| Hydrogen gas | H2 | 2.016 |
| Oxygen gas | O2 | 31.998 |
| Carbon dioxide | CO2 | 44.009 |
| Water | H2O | 18.015 |
| Sodium chloride | NaCl | 58.44 |
Worked Example Calculations
- Number of molecules in 5.00 g of water
- Calculate moles:
\(n=\dfrac{5.00\ \text{g}}{18.015\ \text{g mol}^{-1}}=0.277\ \text{mol}\)
- Convert to particles:
\(N=n\,N_{\!A}=0.277\times6.02214076\times10^{23}=1.67\times10^{23}\) molecules
- Mass of 2.5 mol of carbon dioxide
\(m=nM=2.5\ \text{mol}\times44.009\ \text{g mol}^{-1}=110.0\ \text{g}\)
- Pressure of 2.0 mol of an ideal gas at 300 K occupying 10.0 L
Using \(pV=nRT\) with \(R=8.314\ \text{J mol}^{-1}\text{K}^{-1}=0.08206\ \text{L atm mol}^{-1}\text{K}^{-1}\):
\(p=\dfrac{nRT}{V}= \dfrac{2.0\times0.08206\times300}{10.0}=4.92\ \text{atm}\)
Using the Mole in Gas Laws
In the ideal‑gas equation
the variable \(n\) is the amount of substance (mol). This links the microscopic count of particles to macroscopic pressure, volume and temperature.
Model Limits (Key Concept: models of physical systems)
- The ideal‑gas law assumes point‑like particles with no intermolecular forces.
- It works well at low pressure and high temperature.
- At high pressure or low temperature real gases deviate; corrections are given by the Van der Waals equation.
Practical Skills Link (AO‑3)
Typical experiment – vapour‑density method
Goal: Determine the molar mass of a volatile liquid and, from it, an experimental value of \(N_{\!A}\).
Checklist for Planning & Evaluation
- Apparatus
- Analytical balance (±0.001 g)
- Heated bulb or evaporating dish
- Thermometer, barometer
- Glassware for collecting vapour (e.g., eudiometer)
- Procedure (summary)
- Weigh the empty bulb (mass \(m_0\)).
- Add a known mass of liquid (mass \(m_{\text{liq}}\)).
- Heat gently until the liquid vaporises completely and the vapour fills the bulb.
- Cool the bulb, then weigh the bulb with vapour (mass \(m_1\)).
- Record temperature \(T\) and pressure \(p\) of the vapour.
- Calculations
- Mass of vapour: \(m{\text{vap}} = m1 - m_0\).
- Density of vapour: \(\rho{\text{vap}} = \dfrac{m{\text{vap}}}{V_{\text{bulb}}}\) (bulb volume known).
- Use the ideal‑gas law to find the molar mass:
\[
M = \frac{\rho_{\text{vap}}\,RT}{p}
\]
- Experimental Avogadro number:
\[
N{\!A}^{\text{exp}} = \frac{M}{m{\text{u}}}
\]
where \(m_{\text{u}} = 1.66053906660\times10^{-24}\ \text{g}\) (atomic mass unit).
- Error analysis
- Random errors: balance reading, temperature fluctuations.
- Systematic errors: incomplete vapourisation, heat loss, leakage.
- Propagate uncertainties through the molar‑mass formula (use partial‑derivative method or percentage‑error approximation).
- Evaluation (Paper 5 style)
- Discuss how well the assumptions of the ideal‑gas law are satisfied.
- Identify the dominant source of error and suggest improvements (e.g., using a thermostatted bath, more accurate volume measurement).
Mathematical Toolbox
- Scientific notation & significant figures – e.g., \(6.02214076\times10^{23}\) (exact) vs. \(6.02\times10^{23}\) (3 sf).
- Algebraic rearrangement – from \(pV=nRT\) obtain \(p=\dfrac{nRT}{V}\), \(V=\dfrac{nRT}{p}\), \(T=\dfrac{pV}{nR}\).
- Proportional‑reasoning exercise
If 0.5 mol of an ideal gas occupies 12.0 L at 273 K and 1.00 atm, what pressure will 1.0 mol occupy at the same temperature if the volume is reduced to 6.0 L?
Solution:
\(p1V1=n1RT\) → \(p1 =\dfrac{n1RT}{V1}\).
Because \(n\) and \(T\) are unchanged, \(p\propto\frac{1}{V}\); halving the volume doubles the pressure: \(p_1 = 2.00\ \text{atm}\).
- Scientific‑notation manipulation activity
Write \(6.022\times10^{23}\) to 8 significant figures and then back to 3 significant figures.
Answer: \(6.0221408\times10^{23}\) → \(6.02\times10^{23}\).
- Error propagation (basic) – for \(M=\dfrac{\rho RT}{p}\), the relative uncertainty is
\[
\frac{\Delta M}{M}= \sqrt{\left(\frac{\Delta\rho}{\rho}\right)^2+\left(\frac{\Delta T}{T}\right)^2+\left(\frac{\Delta p}{p}\right)^2 }.
\]
Conceptual Links
- Models & testing: The mole is a model that quantifies a collection of particles; experiments such as the vapour‑density method test the model’s validity.
- Matter & energy: Molar mass links the mass of a macroscopic sample to the energy per particle in chemical reactions.
- Mathematics: Linear relationships in the ideal‑gas law demonstrate how mathematics translates a physical model into quantitative predictions.
Key Points to Remember
- 1 mol = \(6.02214076\times10^{23}\) elementary entities (exact by definition).
- Molar mass \(M\) (g mol⁻¹) enables the conversions: \(n=m/M\) and \(m=nM\).
- In the ideal‑gas equation \(pV=nRT\), \(n\) is the amount of substance in moles.
- The mole is an SI base unit, independent of other quantities, and underpins many A‑level calculations.
Syllabus‑to‑Notes Checklist (excerpt)
| Syllabus Item | Status |
|---|
| 15.1 The mole – definition, Avogadro constant, molar mass | ✓ Complete (exact \(N_{\!A}\) noted) |
| 15.2 Mole in gas laws (ideal‑gas equation) | ✓ Complete (worked example, model‑limits box) |
| AO‑3 practical – vapour‑density method | ✓ Expanded (checklist, data‑sheet ideas, error analysis) |
| Mathematical tools – algebra, proportional reasoning, scientific notation | ✓ Added exercises and propagation formula |
| Conceptual links (models, testing, matter/energy) | ✓ Included |