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2.3 Formulas

Syllabus
9701–2028–2029
Topic
2.3
Level
AS

Learning objectives

2.3.1Ionic formulas and oxidation numbers• Write ionic formulas from ionic charges and oxidation numbers.- Predict ionic charge from Periodic Table position- Recall ions: NO3-, CO3^2-, SO4^2-, OH-, NH4+, Zn2+, Ag+, HCO3-, PO4^3-2.3.2Balanced and ionic equations• Write balanced equations and ionic equations without spectator ions; use appropriate state symbols.2.3.3Terms: empirical and molecular formula• Define/use: empirical and molecular formula2.3.4Terms: anhydrous, hydrated and water• Use terms: anhydrous, hydrated and water of crystallisation2.3.5Empirical and molecular formulas• Calculate empirical and molecular formulas, using given data

Balance ionic charges to write a neutral formula

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⁻

2(+3)+3(−2)=0⇒Fe2(SO4)3\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.

Balance atoms and charge, then remove spectator ions

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.

AgNO3(aq)+NaCl(aq)⟶AgCl(s)+NaNO3(aq)\mathrm{AgNO_3(aq) + NaCl(aq) \longrightarrow AgCl(s) + NaNO_3(aq)}

Ag+(aq)+Cl−(aq)⟶AgCl(s)\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.

Empirical and molecular formulas describe different ratios

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

molecular formula=(empirical formula)nn=1,2,3,…\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.

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

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.

CuSO4⋅5H2O ⇌ CuSO4+5H2O\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.

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

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.

n=Mrempirical formula mass=18030.0=6\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.