Formulas

1. Overview

Formulas and equations are the language of chemistry, and most questions in the course depend on writing them correctly. This section shows how to build the formula of an ionic compound from the charges on its ions, and how to write balanced equations and ionic equations with state symbols. It also explains the difference between empirical and molecular formulas, what hydrated and anhydrous salts are, and how to calculate a formula from experimental data.

Key Definitions

• Empirical formula: the simplest whole-number ratio of the atoms of each element present in a compound.
• Molecular formula: the actual number of atoms of each element present in one molecule of a compound; it is a whole-number multiple of the empirical formula.
• Polyatomic ion: an ion made of two or more atoms joined by covalent bonds, which carries an overall charge, such as SO₄²⁻ or NH₄⁺.
• Oxidation number: the charge an atom would have if all its bonds to atoms of other elements were ionic; in a name it is shown as a Roman numeral in brackets straight after the element, so iron(III) means iron with oxidation number +3, which as a simple ion is Fe³⁺.
• State symbols: symbols in brackets after each formula in an equation that show its physical state: (s) solid, (l) liquid, (g) gas and (aq) aqueous, meaning dissolved in water.
• Ionic equation: a balanced equation that shows only the ions and other species that take part in a reaction, leaving out spectator ions; both the atoms and the total charge must balance.
• Spectator ion: an ion that is present in a reaction mixture but takes no part in the reaction, so it appears unchanged, as the same aqueous ion, on both sides of the equation when every dissolved ionic compound is written as separate ions.
• Water of crystallisation: water molecules that form part of the crystal structure of a compound, usually a salt, in a fixed ratio to its other particles, as in CuSO₄·5H₂O.
• Hydrated: describes a crystalline compound that contains water of crystallisation, for example hydrated copper(II) sulfate, CuSO₄·5H₂O.
• Anhydrous: describes a substance that contains no water; an anhydrous salt is what remains when all its water of crystallisation has been removed, for example CuSO₄.

Content

Predicting the charge on an ion

In the main groups, atoms form ions by losing or gaining electrons until they have the electron arrangement of a noble gas. The group number (its last digit for Groups 13 to 17) gives the number of outer-shell electrons, and so the charge:

Group	Outer-shell electrons	Change	Charge	Examples
1	1	loses 1	1+	Li⁺, Na⁺, K⁺
2	2	loses 2	2+	Mg²⁺, Ca²⁺, Sr²⁺, Ba²⁺
13	3	loses 3	3+	Al³⁺
15	5	gains 3	3−	N³⁻ (nitride), P³⁻ (phosphide)
16	6	gains 2	2−	O²⁻ (oxide), S²⁻ (sulfide)
17	7	gains 1	1−	F⁻, Cl⁻, Br⁻, I⁻ (halides)

• Metals form positive ions. Non-metals form simple negative ions, whose names end in -ide.
• Boron, carbon and silicon normally form covalent bonds, not simple ions. Hydrogen forms H⁺ in acids.
• Transition elements, and the Group 14 metals tin and lead, can have more than one charge, such as Fe²⁺ and Fe³⁺. The name then gives the charge as a Roman numeral.

The ions you must know

Learn the names and formulas of these nine ions exactly. Seven of them are polyatomic ions: groups of covalently bonded atoms with an overall charge, which stay together as one unit. The other two, Zn²⁺ and Ag⁺, are simple metal ions.

Ion	Formula	Note
nitrate	NO₃⁻
carbonate	CO₃²⁻
hydrogencarbonate	HCO₃⁻	a carbonate ion with H⁺ added, so one negative charge
sulfate	SO₄²⁻
phosphate	PO₄³⁻
hydroxide	OH⁻	ends in -ide but contains two elements
ammonium	NH₄⁺	an ammonia molecule with H⁺ added; its salts, like those of sodium and potassium, are soluble
zinc	Zn²⁺
silver	Ag⁺

Zinc and silver are d-block metals, but in the compounds you meet each forms only one ion, so their names need no Roman numeral: zinc nitrate, silver nitrate.

Writing the formula of an ionic compound

An ionic compound has no overall charge, so the total positive charge must equal the total negative charge.

1. Write the two ions with their charges.
2. Find the smallest numbers of each ion that make the charges cancel.
3. Write these numbers as subscripts after each ion (leave out a 1), and leave out the charges.
4. Put brackets round a polyatomic ion when you need more than one of it.

Compound	Ions used	Formula
magnesium oxide	Mg²⁺ and O²⁻	MgO (not Mg₂O₂)
calcium hydroxide	Ca²⁺ and 2 OH⁻	Ca(OH)₂
ammonium sulfate	2 NH₄⁺ and SO₄²⁻	(NH₄)₂SO₄
aluminium sulfate	2 Al³⁺ and 3 SO₄²⁻	Al₂(SO₄)₃
calcium phosphate	3 Ca²⁺ and 2 PO₄³⁻	Ca₃(PO₄)₂
calcium nitride	3 Ca²⁺ and 2 N³⁻	Ca₃N₂
calcium nitrate	Ca²⁺ and 2 NO₃⁻	Ca(NO₃)₂

The brackets matter. Ca(OH)₂ contains one Ca, two O and two H atoms; CaOH₂ would mean one O and two H.

Names with Roman numerals

When an element can have more than one oxidation number, the name shows it as a Roman numeral in brackets straight after the element. For a simple metal ion, the oxidation number is its charge.

• Name to formula. Iron(III) sulfate contains Fe³⁺ and SO₄²⁻. Two Fe³⁺ (+6) balance three SO₄²⁻ (−6), so the formula is Fe₂(SO₄)₃. Likewise copper(I) oxide is Cu₂O, copper(II) oxide is CuO and manganese(IV) oxide is MnO₂.
• Formula to name. Work back from the ion you know. In CrCl₃, three Cl⁻ give −3, so chromium is +3: chromium(III) chloride.
• In a negative ion, the numeral gives the oxidation number of the central atom. The oxidation numbers add up to the ion's charge, with oxygen as −2. In PO₄³⁻, P + 4(−2) = −3, so P is +5: phosphate(V). Likewise NO₃⁻ is nitrate(V), SO₄²⁻ is sulfate(VI) and SO₃²⁻ is sulfate(IV), which you will usually see under its traditional name, sulfite.

Balancing equations

A balanced equation has the same number of atoms of each element on both sides. Change only the numbers in front of formulas (the coefficients), never a subscript: H₂O₂ is hydrogen peroxide, not water.

1. Write the correct formula of every reactant and product.
2. Balance first the elements that appear in only one substance on each side; leave elements on their own, such as O₂, until last.
3. Treat a polyatomic ion that is unchanged, such as SO₄²⁻, as a single unit.
4. Use the smallest whole numbers, then count every element on both sides again.

For example, when octane, C₈H₁₈, burns completely, eight C atoms give 8CO₂ and eighteen H atoms give 9H₂O. That makes 8 × 2 + 9 = 25 O atoms on the right, which needs 12½O₂. Doubling clears the fraction:

2C₈H₁₈(l) + 25O₂(g) → 16CO₂(g) + 18H₂O(l)

Keep the fraction when an equation must refer to one mole of a substance, as for an enthalpy change of combustion: C₈H₁₈(l) + 12½O₂(g) → 8CO₂(g) + 9H₂O(l).

State symbols

Symbol	Meaning	Typical use
(s)	solid	metals, precipitates, solids that are heated
(l)	liquid	pure liquids, including water formed in a reaction
(g)	gas	gases given off, such as CO₂, H₂ and NH₃; steam
(aq)	aqueous, dissolved in water	solutions of salts, acids and alkalis; ions in solution

• Water formed in a reaction in solution is H₂O(l), not H₂O(aq): water is the solvent, not something dissolved in it.
• HCl(g) is hydrogen chloride gas; HCl(aq) is hydrochloric acid.
• To choose between (s) and (aq) for a product, you need to know whether it dissolves in water:

Usually soluble: (aq)	Usually insoluble: (s)
all nitrates; all common salts of Na⁺, K⁺ and NH₄⁺	most carbonates and phosphates (not those of Na⁺, K⁺ or NH₄⁺)
most chlorides, bromides and iodides	AgCl, AgBr and AgI
most sulfates	BaSO₄
Group 1 hydroxides and Ba(OH)₂	most other hydroxides, such as Mg(OH)₂, Cu(OH)₂ and Fe(OH)₃

Ionic equations

When an ionic compound dissolves, its ions separate. Ions that take no part in a reaction are spectator ions. An ionic equation leaves them out and shows only what changes.

1. Write the balanced full equation with state symbols.
2. Split into ions every ionic compound that is (aq), and every strong acid, such as HCl(aq), HNO₃(aq) and H₂SO₄(aq).
3. Keep whole every solid, liquid and gas, covalent molecules such as H₂O and CO₂, and weak acids such as CH₃COOH.
4. Cross out the ions that appear unchanged on both sides.
5. Reduce to the smallest whole numbers, then check that both atoms and total charge balance.

For aluminium with dilute sulfuric acid, 2Al(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3H₂(g), the sulfate ions are spectators:

2Al(s) + 6H⁺(aq) → 2Al³⁺(aq) + 3H₂(g)

The charge is +6 on each side (six H⁺; two Al³⁺). It need not be zero, but it must be the same on both sides.

One ionic equation covers a whole type of reaction, whichever compounds supply the ions, as long as the other ions stay in solution. Here "acid" means a strong acid:

Reaction	Ionic equation
acid + alkali	H⁺(aq) + OH⁻(aq) → H₂O(l)
acid + a hydrogencarbonate solution	H⁺(aq) + HCO₃⁻(aq) → H₂O(l) + CO₂(g)
acid + solid calcium carbonate	CaCO₃(s) + 2H⁺(aq) → Ca²⁺(aq) + H₂O(l) + CO₂(g)
acid + zinc	Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)
ammonium salt warmed with alkali	NH₄⁺(aq) + OH⁻(aq) → NH₃(g) + H₂O(l)
silver nitrate + a chloride solution	Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

An insoluble solid keeps its full formula: calcium carbonate is written CaCO₃(s), not as separate Ca²⁺ and CO₃²⁻ ions.

Worked example 1: a precipitation reaction

Question: When a few drops of aqueous sodium phosphate are added to aqueous silver nitrate, a yellow precipitate of silver phosphate forms. Write the full equation and the ionic equation, with state symbols.

Step 1: build each formula from its ions. Silver nitrate is AgNO₃. Three Na⁺ balance one PO₄³⁻, so sodium phosphate is Na₃PO₄; likewise silver phosphate is Ag₃PO₄. The other product is sodium nitrate, NaNO₃.

Step 2: balance. Ag₃PO₄ needs 3AgNO₃, which brings three NO₃⁻, so the products need 3NaNO₃.

3AgNO₃(aq) + Na₃PO₄(aq) → Ag₃PO₄(s) + 3NaNO₃(aq)

Step 3: split the aqueous ionic compounds.

3Ag⁺(aq) + 3NO₃⁻(aq) + 3Na⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s) + 3Na⁺(aq) + 3NO₃⁻(aq)

Step 4: cross out the spectator ions, Na⁺ and NO₃⁻.

Step 5: check. Ag 3 = 3, P 1 = 1, O 4 = 4. Charge on the left: 3(+1) + (−3) = 0; on the right: 0.

Answer: Full equation: 3AgNO₃(aq) + Na₃PO₄(aq) → Ag₃PO₄(s) + 3NaNO₃(aq). Ionic equation: 3Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s).

Empirical and molecular formulas

The empirical formula is the simplest whole-number ratio of atoms; the molecular formula is the actual number of atoms of each element in one molecule. The molecular formula is a whole-number multiple of the empirical formula.

Substance	Molecular formula	Empirical formula
octane	C₈H₁₈	C₄H₉
ethanoic acid	C₂H₄O₂	CH₂O
glucose	C₆H₁₂O₆	CH₂O
phosphorus(V) oxide	P₄O₁₀	P₂O₅
sodium chloride	none: a giant ionic lattice	NaCl
silicon(IV) oxide	none: a giant covalent structure	SiO₂

• To get the empirical formula, divide every subscript in the molecular formula by their highest common factor. If that factor is 1, as in H₂O, the formulas are the same.
• Different compounds can share an empirical formula (ethanoic acid and glucose are both CH₂O), so you also need $M_{\text{r}}$ to find the molecular formula.
• Giant structures contain no molecules, so they have only an empirical formula.

Calculating an empirical formula

Step	What to do
1	Write the mass or percentage of each element. If one element is not given, find it by difference.
2	Divide each by its $A_{\text{r}}$ (from the Periodic Table, e.g. C 12.0) to get the amount of atoms in moles.
3	Divide every answer by the smallest one.
4	If a value is close to a half (1.5, 2.5), multiply all of them by 2; close to a third (1.33, 1.67), by 3; close to a quarter (1.25, 1.75), by 4. Round only values within about 0.1 of a whole number.
5	Write the formula from the whole-number ratio.

Percentages can be used as masses: 100 g of compound contains as many grams of each element as its percentage.

From empirical to molecular formula:

$$\text{multiplier} = \frac{M_{\text{r}}}{\text{empirical formula mass}}$$

Multiply every subscript by this whole number. $M_{\text{r}}$ may be given or found from gas data with $pV = nRT$. If the multiplier is not close to a whole number, recheck the ratio.

From combustion data. When a compound of C, H and O burns completely, all its carbon ends up in CO₂ and all its hydrogen in H₂O (two H atoms per H₂O). Find oxygen by difference, not from the products, because some of the oxygen in the CO₂ and H₂O came from the oxygen gas the compound burned in.

Worked example 2: from percentages by mass

Question: Compound Q contains carbon, hydrogen and oxygen only. By mass it is 49.3% carbon, 6.85% hydrogen and 43.8% oxygen. $M_{\text{r}}$ of Q is 146. Find the empirical and molecular formulas of Q. ($A_{\text{r}}$: C 12.0, H 1.0, O 16.0)

Step 1: amount of each element in 100 g, and the ratio.

	C	H	O
mass in 100 g / g	49.3	6.85	43.8
÷ $A_{\text{r}}$	49.3 ÷ 12.0 = 4.108	6.85 ÷ 1.0 = 6.85	43.8 ÷ 16.0 = 2.738
÷ smallest (2.738)	1.50	2.50	1.00
× 2	3.00	5.00	2.00

1.50 and 2.50 are halves, so double every value; do not round. The empirical formula is C₃H₅O₂.

Step 2: multiplier. Empirical formula mass = 3(12.0) + 5(1.0) + 2(16.0) = 73.0.

$$\text{multiplier} = \frac{146}{73.0} = 2.00$$

Answer: empirical formula C₃H₅O₂; molecular formula C₆H₁₀O₄.

Worked example 3: from combustion data

Question: 0.350 g of a liquid, R, which contains carbon, hydrogen and oxygen only, burns completely in excess oxygen. It forms 0.823 g of carbon dioxide and 0.226 g of water. $M_{\text{r}}$ of R is 112. Find the molecular formula of R. ($A_{\text{r}}$: C 12.0, H 1.0, O 16.0)

Step 1: mass of each element in R.

$$m(\text{C}) = 0.823 \times \frac{12.0}{44.0} = 0.2245\ \text{g} \qquad m(\text{H}) = 0.226 \times \frac{2.0}{18.0} = 0.02511\ \text{g}$$

$$m(\text{O}) = 0.350 - 0.2245 - 0.02511 = 0.1004\ \text{g}$$

Step 2: the ratio.

	C	H	O
mass / g	0.2245	0.02511	0.1004
÷ $A_{\text{r}}$ / mol	0.01871	0.02511	0.006275
÷ smallest	2.98	4.00	1.00

2.98 is within 0.1 of 3, so round it to 3; ratios from measured masses are rarely exact whole numbers. The empirical formula is C₃H₄O.

Step 3: multiplier. Empirical formula mass = 3(12.0) + 4(1.0) + 16.0 = 56.0, and 112 ÷ 56.0 = 2.00.

Answer: the molecular formula of R is C₆H₈O₂.

Hydrated and anhydrous salts

Many salts crystallise from water with water molecules built into the crystal in a fixed ratio to the ions: water of crystallisation. A salt that contains it is hydrated; one with none is anhydrous.

• In CuSO₄·5H₂O, the dot means "combined with": five H₂O for every CuSO₄, so 1 mol of the hydrated salt contains 5 mol of water.
• $M_{\text{r}}$ (here a relative formula mass, as the salt is ionic) includes the water: 63.5 + 32.1 + 4(16.0) + 5(18.0) = 249.6 for CuSO₄·5H₂O. Use it whenever you weigh the hydrated salt.
• Heating drives the water off, CuSO₄·5H₂O(s) → CuSO₄(s) + 5H₂O(g); adding water to the anhydrous salt reverses the change.

Hydrated salt	Colour	Anhydrous salt	Colour
CuSO₄·5H₂O	blue	CuSO₄	white
CoCl₂·6H₂O	pink	CoCl₂	blue

Finding x in MX·xH₂O. Weigh an empty crucible with its lid, then the crucible and lid with the hydrated salt. Heat gently at first with the lid on, so that solid does not spit out, then strongly with the lid off. Replace the lid, cool and weigh. Repeat the heating, cooling and weighing until two readings agree: constant mass shows all the water has gone. The mass lost is water; the residue is the anhydrous salt.

$$x = \frac{n(\text{H}_2\text{O})}{n(\text{anhydrous salt})}$$

Each problem pushes x one way:

Problem	Mass lost	Value of x
not all the water driven off	too small	too low
solid spits out of the crucible	too large	too high
the anhydrous salt decomposes on very strong heating	too large	too high
crystals damp when first weighed	too large	too high

Worked example 4: water of crystallisation

Question: A sample of hydrated strontium chloride, SrCl₂·xH₂O, is heated in a crucible until its mass is constant. Find x, which is an integer. ($A_{\text{r}}$: Sr 87.6, Cl 35.5; $M_{\text{r}}$ of H₂O = 18.0)

Reading	Mass / g
crucible + lid	15.38
crucible + lid + hydrated salt	19.65
after first heating	18.01
after second heating	17.92
after third heating	17.92

Step 1: masses. The last two readings agree, so use 17.92 g.

• anhydrous SrCl₂ = 17.92 − 15.38 = 2.54 g
• water lost = 19.65 − 17.92 = 1.73 g

Step 2: amounts. Strontium is in Group 2, so Sr²⁺ needs two Cl⁻: $M_{\text{r}}$ of SrCl₂ = 87.6 + 2(35.5) = 158.6.

$$n(\text{SrCl}_2) = \frac{2.54}{158.6} = 0.01602\ \text{mol} \qquad n(\text{H}_2\text{O}) = \frac{1.73}{18.0} = 0.09611\ \text{mol}$$

Step 3: ratio.

$$x = \frac{0.09611}{0.01602} = 6.00$$

Answer: x = 6, so the formula is SrCl₂·6H₂O.

Stopping after the first heating would give water lost = 1.64 g and residue = 2.63 g, so x = 5.49. Too little water driven off gives a value of x that is too low.

Key Equations

Amount of atoms of an element ($n$ = amount in mol; $m$ = mass in g; $A_{\text{r}}$ from the Periodic Table): $$n = \frac{m}{A_{\text{r}}}$$

Empirical to molecular formula (the result is a whole number; multiply every subscript by it): $$\text{multiplier} = \frac{M_{\text{r}}}{\text{empirical formula mass}}$$

Combustion data (masses in g; CO₂ = 44.0 and H₂O = 18.0; oxygen by difference): $$m(\text{C}) = m(\text{CO}_2) \times \frac{12.0}{44.0} \qquad m(\text{H}) = m(\text{H}_2\text{O}) \times \frac{2.0}{18.0}$$

Percentage by mass of an element ($N$ = number of its atoms in the formula): $$\%\ \text{by mass} = \frac{N \times A_{\text{r}}}{M_{\text{r}}} \times 100$$

Water of crystallisation ($n$ in mol; $x$ = water molecules per formula unit): $$x = \frac{n(\text{H}_2\text{O})}{n(\text{anhydrous salt})} \qquad M_{\text{r}}(\text{hydrated salt}) = M_{\text{r}}(\text{anhydrous salt}) + 18.0x$$

Charge balance: in an ionic compound, total positive charge = total negative charge; in an ionic equation, total charge on the left = total charge on the right.

Common Mistakes to Avoid

• Mistake: missing or misplaced brackets. Fix: put brackets round a polyatomic ion whenever there is more than one of it: Ca(OH)₂, not CaOH₂ or CaOH; (NH₄)₂SO₄, not NH₄₂SO₄.
• Mistake: a wrong ion formula, or confusing -ide with -ate. Fix: check that the charges cancel. Zinc carbonate is ZnCO₃, not Zn(CO₃)₂ or ZnCO₂. Calcium nitride (N³⁻) is Ca₃N₂, but calcium nitrate (NO₃⁻) is Ca(NO₃)₂.
• Mistake: leaving spectator ions in an ionic equation, or giving the full equation instead. Fix: in Worked example 1, Na⁺ and NO₃⁻ appear on both sides, so the ionic equation is just 3Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s).
• Mistake: splitting an insoluble solid into ions, or using the wrong state symbols. Fix: write CaCO₃(s) + 2H⁺(aq) → Ca²⁺(aq) + H₂O(l) + CO₂(g). The solid carbonate stays whole and the water is (l), not (aq).
• Mistake: balancing the atoms but not the charge. Fix: Al(s) + 2H⁺(aq) → Al³⁺(aq) + H₂(g) has equal atoms but +2 on the left and +3 on the right. The balanced version is 2Al(s) + 6H⁺(aq) → 2Al³⁺(aq) + 3H₂(g).
• Mistake: rounding a ratio such as 1.50 to 2, or dividing by atomic number instead of $A_{\text{r}}$. Fix: in Worked example 2, double 1.50 : 2.50 : 1 to 3 : 5 : 2 (C₃H₅O₂). Rounding gives C₂H₃O, and 146 ÷ 43.0 = 3.40 is not a whole number, a sign the ratio is wrong.
• Mistake: giving the empirical formula when the molecular formula is asked for. Fix: use $M_{\text{r}}$ to find the multiplier. Phosphorus(V) oxide exists as P₄O₁₀ molecules, so P₂O₅ is only its empirical formula; Q in Worked example 2 is C₆H₁₀O₄, not C₃H₅O₂.
• Mistake: ignoring water of crystallisation in $M_{\text{r}}$. Fix: 2.50 g of CuSO₄·5H₂O is 2.50 ÷ 249.6 = 0.0100 mol. Dividing by 159.6, the $M_{\text{r}}$ of CuSO₄, gives 0.0157 mol, which is wrong.

Exam Tips

• Write the formula of an ion with its charge (SO₄²⁻) and of a compound without charges (Na₂SO₄). Name a metal that forms more than one ion with its Roman numeral: iron(III) chloride, not iron chloride. Give a formula or a name, whichever is asked for.
• Construct an equation by writing every formula first, then balancing with coefficients only; count each element, and the charge, on both sides. Add state symbols whenever asked, and always for enthalpy changes and ionisation energies.
• Write an ionic equation with only the species that change, the smallest whole-number coefficients and a state symbol on every species.
• Define the two formulas precisely: the empirical formula is the simplest whole-number ratio of atoms of each element; the molecular formula is the actual number of atoms of each element in one molecule.
• Calculate with a table showing the ÷ $A_{\text{r}}$ and ÷ smallest rows, keeping 3 significant figures in the ratios so that 1.50 is not mistaken for 2. When a question says x is an integer, give a whole number.