The mole, molar mass, Avogadro's number

Almost every chemistry problem starts with one step: convert the given quantity into moles. The mole is chemistry’s "counting unit": atoms and molecules are far too small to count one by one, so they are measured in large "packets". The coefficients in an equation give the ratio of moles, not of masses or volumes. A student who masters this topic carries out the opening part of any calculation automatically: converting mass, gas volume, number of particles and solution concentration into moles and back. This topic underpins "Calculation problems" and "Solutions".

§1. The mole and the Avogadro constant

The mole is the unit of amount of substance: the amount containing as many particles as there are atoms in 12 g of carbon-12. This number is the Avogadro constant (named after A. Avogadro); since 2019 the mole has been defined by a fixed value of this number.
Molar mass M (g/mol) is the mass of 1 mol of a substance; numerically it equals the relative formula mass Mr: M(H₂O) = 18 g/mol, M(H₂SO₄) = 98 g/mol.
Key relationships:

When counting particles, check what is asked for: molecules, atoms, ions or electrons. For example, 1 mol of H₂O contains 6.02·10²³ molecules but 3·6.02·10²³ atoms and 10·6.02·10²³ electrons.

The mole is also the main tool for finding formulae: dividing the masses (or mass fractions) of the elements by their atomic masses gives the mole ratio, and reducing it to the smallest whole numbers gives the empirical formula. For example, a substance containing 2.4 g of carbon and 0.6 g of hydrogen has n(C) : n(H) = 0.2 : 0.6 = 1 : 3, empirical formula CH₃; if its molar mass is 30 g/mol, the molecular formula is C₂H₆. The number of ions in an ionic compound is found in the same way.

Worked example

Find the total number of atoms in 0.2 mol of Na[Al(OH)₄].

1. One formula unit contains Na 1 + Al 1 + O 4 + H 4 = 10 atoms.
2. Amount of atoms: 0.2·10 = 2 mol.
3. N = 2·6.02·10²³ = 1.204·10²⁴.

§2. Gases: molar volume and relative density

• Avogadro’s law: equal volumes of gases under the same conditions contain equal numbers of molecules, so the ratio of volumes equals the ratio of moles.
• Relative density: D_B(A) = M(A)/M(B); relative to air D = M/29, relative to hydrogen D = M/2. The average molar mass of air is 29 g/mol; gases with molar mass below 29 (H₂, He, NH₃, CH₄) are less dense than air.
• The average molar mass of a mixture:

• For non-standard conditions use the ideal gas equation pV = nRT (R = 8.314 J/(mol·K), p in kPa, V in L, T in K).

Worked example

A mixture of N₂ and H₂ has a density relative to hydrogen of 4.25. Find the volume fraction of N₂.

1. M̄ = 4.25·2 = 8.5 g/mol.
2. 28x + 2(1 − x) = 8.5 → 26x = 6.5 → x = 0.25.
3. N₂ 25 %, H₂ 75 %.

Worked example

What volume does 1 mol of gas occupy at 27 °C and 100 kPa?

1. T = 27 + 273 = 300 K.
2. V = nRT/p = 1·8.314·300/100.
3. V ≈ 24.9 L — more than at STP, because the temperature is higher and the pressure slightly lower.

§3. Calculating from an equation

1. Write and balance the equation.
2. Convert the given substance into moles.
3. Use the ratio of coefficients to find the moles of the unknown substance.
4. Convert to the required unit (g, L, number of particles).

For reactions between gases (under the same conditions) you can work directly with volumes: in 2H₂ + O₂ → 2H₂O(g), 2 L of hydrogen need 1 L of oxygen.

Worked example

What volume of H₂ (STP) is released when 13 g of zinc dissolves in excess hydrochloric acid?

1. Zn + 2HCl → ZnCl₂ + H₂.
2. n(Zn) = 13/65 = 0.2 mol.
3. n(H₂) = n(Zn) = 0.2 mol.
4. V = 0.2·22.4 = 4.48 L.

§4. Limiting and excess reagents; yield

If both reactants are given, divide each amount by its coefficient: the smaller value belongs to the limiting reagent, and the product is calculated from it.

In a multi-step process the overall yield is the product of the step yields: two steps of 80 % and 50 % give 40 % overall. The purity of a substance in a mixture or technical sample is handled similarly: first find the mass of pure substance, then convert it to moles.

Worked example

A mixture of 5.6 L of H₂ and 4.48 L of Cl₂ (STP) was exploded. What mass of HCl forms?

1. H₂ + Cl₂ → 2HCl; n(H₂) = 0.25 mol, n(Cl₂) = 0.2 mol — chlorine is limiting.
2. n(HCl) = 2·0.2 = 0.4 mol → m = 0.4·36.5 = 14.6 g (8.96 L).
3. H₂ left over: 0.05 mol.

§5. Equivalents and the law of equivalents

An equivalent is the amount of a substance that combines with or replaces 1 mol of hydrogen atoms. Equivalent mass:

Law of equivalents: substances react in equivalent amounts, m₁/E₁ = m₂/E₂.
The equivalent can depend on the reaction: in H₃PO₄ + NaOH → NaH₂PO₄ + H₂O only one hydrogen is replaced, so E(H₃PO₄) = 98/1 = 98, whereas for complete neutralisation it is 98/3 ≈ 32.7.

Worked example

1.2 g of a metal displaced 1.12 L of hydrogen (STP) from an acid. Find the metal’s equivalent mass and identify it (valency II).

1. n(H₂) = 1.12/22.4 = 0.05 mol → 0.1 mol of hydrogen atoms.
2. E = 1.2/0.1 = 12 g/mol.
3. A = E·valency = 12·2 = 24 — magnesium.

Check yourself

1. How many molecules are in 9 g of water?

2. What is the mass of 3.01·10²³ molecules of O₂?

3. A gas has a density relative to air of 2.2. What is its molar mass?

4. How many moles of ions form when 0.1 mol of Al₂(SO₄)₃ dissociates completely?

5. Heating 10 g of CaCO₃ released 2.24 L of CO₂ (STP). What is the yield?

6. What is the equivalent mass of K₂HPO₄ as a salt?

7. What is the mass of 11.2 L of CO₂ (STP)?

8. How many moles of atoms are in 1 mol of NH₃ molecules?

Common mistakes

• Confusing numbers of molecules and atoms: an O₂ molecule has two atoms.
• Using 22.4 L/mol under non-standard conditions: use pV = nRT.
• Calculating from the reagent in excess.
• Finding an equivalent from the formula alone, ignoring the reaction.
• Applying 22.4 to liquids and solids: it is for gases only.