2. Atoms, molecules and stoichiometry

Exam analysis

Most tested topics

2.1 Relative masses of atoms and molecules

One unified atomic mass unit, 1 u, is defined as one twelfth of the mass of one carbon-12 atom. It provides a common mass scale for atoms and other microscopic particles.

$\mathrm{1\ u = \frac{1}{12}\,mass\ of\ one\ ^{12}C\ atom}$

The standard is one twelfth of a carbon-12 atom's mass, not the mass of the whole atom. The unit u has dimensions of mass; by contrast, relative atomic, isotopic, molecular and formula masses are ratios and have no units.

Quantity	Definition relative to 1 u	Symbol
Relative isotopic mass	mass of one atom of a specified isotope ÷ 1 u	—
Relative atomic mass	weighted mean mass of an atom of an element ÷ 1 u	Aᵣ
Relative molecular mass	average mass of one molecule ÷ 1 u	Mᵣ
Relative formula mass	average mass of one formula unit ÷ 1 u	Mᵣ

Use relative molecular mass for a substance made of discrete molecules. Use relative formula mass for an ionic or giant structure represented by a formula unit. Both use Mᵣ and are found from the Aᵣ values of all atoms shown in the formula.

All four relative quantities are dimensionless ratios, so do not attach u, g or kg. Relative isotopic mass refers to one isotope; Aᵣ averages the naturally occurring isotopes of the element and therefore need not be a whole number.

2.2 The mole and the Avogadro constant

A mole is an amount of substance containing the Avogadro constant, L = 6.022 × 10²³ mol⁻¹, of specified entities. The entities may be atoms, molecules, ions, electrons or formula units, so always state what is being counted.

$\mathrm{N = nL}$

N is the number of entities, n is the amount in moles, and L is the Avogadro constant. For 0.250 mol of H₂O, N = 0.250 × 6.022 × 10²³ = 1.51 × 10²³ water molecules to three significant figures.

A chemical formula controls counts inside each entity: those water molecules contain twice as many H atoms, or 3.01 × 10²³ H atoms. One mole does not mean one mole of every atom shown in a formula.

The mole measures amount of substance, not mass or volume. Different one-mole samples contain the same number of specified entities but can have different masses and occupy different volumes.

2.3 Formulas

An ionic compound is electrically neutral: the total positive charge and total negative charge must be equal. Predict common monatomic-ion charges from Periodic Table position, then use the smallest whole-number ratio of ions that gives zero overall charge.

Source or name	Ion charge / formula
Group 1, Group 2, Group 13 metals	+1, +2, +3
Group 15, Group 16, Group 17 non-metals	−3, −2, −1
nitrate, carbonate, sulfate	NO₃⁻, CO₃²⁻, SO₄²⁻
hydroxide, ammonium	OH⁻, NH₄⁺
zinc, silver	Zn²⁺, Ag⁺
hydrogen carbonate, phosphate	HCO₃⁻, PO₄³⁻

$\mathrm{2(+3) + 3(-2)=0 \quad\Rightarrow\quad Fe_2(SO_4)_3}$

Iron(III) means Fe³⁺. Two Fe³⁺ ions give +6 and three sulfate ions give −6, so the formula is Fe₂(SO₄)₃. Put a polyatomic ion in brackets when more than one whole ion is required.

The Roman numeral states the metal's oxidation number, not the number of metal atoms. Do not carry ionic charge signs into the final neutral compound formula, and reduce ratios to their simplest whole numbers.

Write correct reactant and product formulas first. Change coefficients in front of formulas until every element is conserved; never change a subscript to balance an equation. For an ionic equation, split appropriate aqueous ionic substances into ions, cancel unchanged spectator ions, and check both atoms and total charge.

$\mathrm{AgNO_3(aq) + NaCl(aq) \longrightarrow AgCl(s) + NaNO_3(aq)}$

$\mathrm{Ag^+(aq) + Cl^-(aq) \longrightarrow AgCl(s)}$

Use (s) for solid, (l) for liquid, (g) for gas and (aq) for a species dissolved in water. In the example, Na⁺(aq) and NO₃⁻(aq) are spectators because they appear unchanged on both sides.

A balanced molecular equation conserves each element. A balanced ionic equation must also have the same total charge on both sides and must not contain spectator ions.

Formula	What it shows	Ethanoic acid example
Empirical	simplest whole-number ratio of atoms of each element	CH₂O
Molecular	actual number of atoms of each element in one molecule	C₂H₄O₂

$\mathrm{molecular\ formula = (empirical\ formula)_n}\qquad n=1,2,3,\ldots$

The empirical formula is obtained by simplifying all subscripts by their highest common factor. The molecular formula is a whole-number multiple of it; if the molecular subscripts are already in the simplest ratio, the two formulas are the same.

An empirical formula does not normally tell you the actual number of atoms in a molecule. For ionic compounds, the written formula already represents the simplest ratio of ions rather than a discrete molecule.

Water of crystallisation is water present in a fixed proportion within a crystalline compound. A hydrated compound contains this water; the corresponding anhydrous compound contains no water of crystallisation.

$\mathrm{CuSO_4\cdot 5H_2O\ \rightleftharpoons\ CuSO_4 + 5H_2O}$

CuSO₄·5H₂O is hydrated copper(II) sulfate and contains five water molecules per CuSO₄ formula unit. CuSO₄ is the anhydrous salt. The dot separates the salt formula from its fixed water ratio; it does not mean multiplication.

A wet solid or a salt merely dissolved in water is not automatically hydrated. Hydration describes water incorporated in the crystal in a definite stoichiometric ratio.

For each element, treat percentages as masses in a 100 g sample and divide mass by Aᵣ to obtain moles. Divide every mole value by the smallest, then multiply all ratios by the same small integer if needed to obtain whole numbers. These subscripts give the empirical formula.

Element	mass / g	divide by Aᵣ	simplest ratio
C	40.0	40.0 ÷ 12.0 = 3.33	1
H	6.7	6.7 ÷ 1.0 = 6.7	2
O	53.3	53.3 ÷ 16.0 = 3.33	1

The simplest ratio is C:H:O = 1:2:1, so the empirical formula is CH₂O and its empirical formula mass is 30.0.

$\mathrm{n=\frac{M_r}{empirical\ formula\ mass}=\frac{180}{30.0}=6}$

Multiply every empirical subscript by 6: the molecular formula is C₆H₁₂O₆. The multiplier must be a whole number; if it is not, recheck the mole ratio, rounding and supplied Mᵣ.

2.4 Reacting masses and volumes (of solutions and gases)

Start with a balanced equation. Convert each supplied quantity to moles, use the equation coefficients as a mole ratio, then convert the required moles to the requested mass, gas volume or solution quantity. Coefficients relate amounts in moles, not masses directly.

Quantity known	Convert to moles	Convert back
mass, m	n = m ÷ M	m = nM
solution concentration, c, and volume, V	n = cV, with V in dm³	c = n ÷ V
gas volume, V, at stated conditions	n = V ÷ Vₘ	V = nVₘ

Use the molar volume stated or justified by the conditions: Vₘ = 24.0 dm³ mol⁻¹ at room conditions, or 22.4 dm³ mol⁻¹ at s.t.p. (101 kPa and 273 K). Convert 1000 cm³ = 1 dm³ before substituting where necessary.

$\mathrm{CaCO_3 + 2HCl \longrightarrow CaCl_2 + CO_2 + H_2O}$

If 25.0 cm³ of 0.200 mol dm⁻³ HCl reacts with excess CaCO₃, n(HCl) = 0.200 × 0.0250 = 0.00500 mol. The 2:1 ratio gives 0.00250 mol CO₂, so at room conditions V(CO₂) = 0.00250 × 24.0 = 0.0600 dm³ = 60.0 cm³.

$\mathrm{percentage\ yield=\frac{actual\ yield}{theoretical\ yield}\times100\%}$

When two reactant amounts are supplied, compare n ÷ coefficient for each reactant. The smaller value identifies the limiting reagent and fixes the maximum product amount; the other reagent is in excess. Calculate unused excess only after subtracting the amount that reacts.

To deduce an unknown stoichiometric relationship, convert measured quantities to moles and reduce the mole amounts to the simplest justified ratio. Keep unrounded values through the working, then report the final answer to the significant figures given or requested. Check units, conditions, ratio direction and that percentage yield is actual ÷ theoretical.

2. Atoms, molecules and stoichiometry

Exam analysis

Most tested topics

2.1 Relative masses of atoms and molecules

The unified atomic mass unit is based on carbon-12

Relative masses compare particles with the unified atomic mass unit

2.2 The mole and the Avogadro constant

One mole contains the Avogadro constant of specified entities

2.3 Formulas

Balance ionic charges to write a neutral formula

Balance atoms and charge, then remove spectator ions

Empirical and molecular formulas describe different ratios

A hydrated salt contains a fixed ratio of water in its crystal

Convert composition to a mole ratio, then scale to the molecular formula

2.4 Reacting masses and volumes (of solutions and gases)

Use moles as the bridge in every stoichiometric calculation