use molar quantities where one mole of any substance is the amount containing a number of particles of that substance equal to the Avogadro constant NA

Published by Patrick Mutisya · 14 days ago

Cambridge A-Level Physics 9702 – The Mole

The Mole

Objective

Use molar quantities where one mole of any substance is the amount containing a number of particles equal to the Avogadro constant \$N_{\mathrm A}\$.

Key Definitions

  • Mole (mol): The SI base unit for amount of substance. One mole contains exactly \$N_{\mathrm A}=6.02214076\times10^{23}\$ elementary entities.

  • Avogadro constant \$N_{\mathrm A}\$: The number of particles in one mole. Fixed by definition.

  • Molar mass \$M\$: Mass of one mole of a substance, expressed in \$\mathrm{g\,mol^{-1}}\$ or \$\mathrm{kg\,mol^{-1}}\$.

Relationship Between Mass, Amount, and Number of Particles

The fundamental relation is

\$\$

n = \frac{m}{M} = \frac{N}{N_{\mathrm A}}

\$\$

where \$n\$ is the amount of substance (mol), \$m\$ the mass (kg or g), \$M\$ the molar mass, and \$N\$ the number of particles.

Converting Between Units

  1. Identify the quantity you have (mass, number of particles, or amount).

  2. Use the appropriate form of the equation \$n = m/M\$ or \$N = nN_{\mathrm A}\$.

  3. Solve for the desired quantity.

Common Molar Masses

SubstanceMolar mass \$M\$ (g mol\$^{-1}\$)
Hydrogen gas, \$\mathrm{H_2}\$2.016
Oxygen gas, \$\mathrm{O_2}\$31.998
Carbon dioxide, \$\mathrm{CO_2}\$44.009
Water, \$\mathrm{H_2O}\$18.015
Sodium chloride, \$\mathrm{NaCl}\$58.44

Worked Example

Problem: How many molecules are present in \$5.00\ \mathrm{g}\$ of \$\mathrm{CO_2}\$?

Solution:

  1. Calculate the amount of substance:

    \$\$

    n = \frac{m}{M} = \frac{5.00\ \mathrm{g}}{44.009\ \mathrm{g\,mol^{-1}}}=0.1136\ \mathrm{mol}

    \$\$

  2. Convert moles to number of molecules:

    \$\$

    N = nN_{\mathrm A}=0.1136\ \mathrm{mol}\times6.02214076\times10^{23}\ \mathrm{mol^{-1}}=6.84\times10^{22}\ \text{molecules}

    \$\$

Significance in Physics and Chemistry

  • Provides a bridge between macroscopic measurements (mass, volume) and microscopic entities (atoms, molecules, ions).

  • Essential for stoichiometric calculations in chemical reactions.

  • Used to define the mole in terms of a fixed number of particles, ensuring consistency across scientific disciplines.

Common Mistakes to Avoid

  • Confusing molar mass (g mol\$^{-1}\$) with molecular mass (atomic mass units, u). Remember \$1\ \text{u}=1\ \text{g mol}^{-1}\$.

  • Using \$N_{\mathrm A}=6.02\times10^{23}\$ as an approximation in high‑precision work; the defined value is \$6.02214076\times10^{23}\$.

  • Neglecting unit conversion when mass is given in kilograms but molar mass is in grams per mole.

Suggested diagram: A schematic showing the relationship between mass, moles, and number of particles, with arrows indicating the conversion using \$M\$ and \$N_{\mathrm A}\$.

Quick Reference Sheet

QuantitySymbolUnitRelation
Amount of substance\$n\$mol\$n = \dfrac{m}{M} = \dfrac{N}{N_{\mathrm A}}\$
Mass\$m\$g or kg\$m = nM\$
Number of particles\$N\$dimensionless\$N = nN_{\mathrm A}\$
Molar mass\$M\$g mol\$^{-1}\$\$M = \dfrac{m}{n}\$
Avogadro constant\$N{\mathrm A}\$mol\$^{-1}\$\$N{\mathrm A}=6.02214076\times10^{23}\$

Self‑Check Questions

  1. What is the mass of \$2.5\$ mol of \$\mathrm{NaCl}\$?

  2. How many atoms are there in \$0.250\ \mathrm{g}\$ of pure copper (Cu, \$M=63.546\ \mathrm{g\,mol^{-1}}\$)?

  3. If a gas occupies \$22.4\ \mathrm{L}\$ at STP, how many moles does it contain?

Answers to Self‑Check

  1. \$m = nM = 2.5\ \text{mol}\times58.44\ \text{g mol}^{-1}=146.1\ \text{g}\$

  2. \$n = \dfrac{0.250\ \text{g}}{63.546\ \text{g mol}^{-1}}=3.93\times10^{-3}\ \text{mol}\$

    \$N = nN_{\mathrm A}=3.93\times10^{-3}\times6.02214076\times10^{23}=2.37\times10^{21}\ \text{atoms}\$

  3. At STP, \$1\ \text{mol}\$ of an ideal gas occupies \$22.4\ \text{L}\$, so \$22.4\ \text{L}\$ corresponds to \$1\ \text{mol}\$.