Formulas and Equations

AS · 9 min

A correct formula and a balanced equation are the starting point of every chemical calculation, and most Paper 2 questions contain at least one "write an equation" mark. This note covers how to build formulas of ionic compounds from ionic charges, the ions you must know by heart, how to balance full and ionic equations with state symbols, and how to find empirical and molecular formulas from experimental data, including hydrated salts and combustion analysis.

Formulas of ionic compounds

An ionic compound is neutral overall, so the total positive charge equals the total negative charge. To write its formula you need the charge on each ion.

Predicting ionic charges from the Periodic Table

Atoms of main-group elements gain or lose electrons to reach the electronic configuration of the nearest noble gas.

group	1	2	13	15	16	17
typical ion	$\ce{M+}$	$\ce{M^2+}$	$\ce{M^3+}$	$\ce{X^3-}$	$\ce{X^2-}$	$\ce{X-}$
example	$\ce{Na+}$	$\ce{Mg^2+}$	$\ce{Al^3+}$	$\ce{N^3-}$	$\ce{O^2-}$	$\ce{Cl-}$

Transition metals can form ions with different charges, so the charge is shown by a Roman numeral in the name, which gives the oxidation number: iron(II) is $\ce{Fe^2+}$ and iron(III) is $\ce{Fe^3+}$ ; copper(I) is $\ce{Cu+}$ and copper(II) is $\ce{Cu^2+}$ .

Ions to learn

Key result

name	formula	name	formula
nitrate	$\ce{NO3-}$	ammonium	$\ce{NH4+}$
carbonate	$\ce{CO3^2-}$	zinc	$\ce{Zn^2+}$
sulfate	$\ce{SO4^2-}$	silver	$\ce{Ag+}$
hydroxide	$\ce{OH-}$	hydrogencarbonate	$\ce{HCO3-}$
phosphate	$\ce{PO4^3-}$

Method

Writing the formula of an ionic compound

Write the two ions with their charges, for example $\ce{Al^3+}$ and $\ce{SO4^2-}$ .
Find the lowest common multiple of the charges: here 6.
Use enough of each ion to reach that total: two $\ce{Al^3+}$ ( $+6$ ) and three $\ce{SO4^2-}$ ( $-6$ ).
Write the formula, using brackets around a polyatomic ion when there is more than one of it: $\ce{Al2(SO4)3}$ .

compound	ions	formula
calcium hydroxide	$\ce{Ca^2+}$ , $\ce{OH-}$	$\ce{Ca(OH)2}$
ammonium sulfate	$\ce{NH4+}$ , $\ce{SO4^2-}$	$\ce{(NH4)2SO4}$
iron(III) oxide	$\ce{Fe^3+}$ , $\ce{O^2-}$	$\ce{Fe2O3}$
sodium phosphate	$\ce{Na+}$ , $\ce{PO4^3-}$	$\ce{Na3PO4}$
calcium hydrogencarbonate	$\ce{Ca^2+}$ , $\ce{HCO3-}$	$\ce{Ca(HCO3)2}$
silver carbonate	$\ce{Ag+}$ , $\ce{CO3^2-}$	$\ce{Ag2CO3}$

Roman numerals also appear in the names of oxyanions, where they give the oxidation number of the central atom: sodium chlorate(I) is $\ce{NaClO}$ , potassium manganate(VII) is $\ce{KMnO4}$ and potassium dichromate(VI) is $\ce{K2Cr2O7}$ .

Watch out

Never change a subscript inside an ion to balance charges. Sulfate is always $\ce{SO4^2-}$ ; three of them are $\ce{(SO4)3}$ , not $\ce{SO12}$ . And do not put brackets around a single ion: $\ce{NaOH}$ , not $\ce{Na(OH)}$ .

Balancing equations

An equation shows the reactants and products and the amounts in which they react. A balanced equation has the same number of each type of atom on both sides (and, for ionic equations, the same total charge on both sides).

Method

Write correct formulas for every reactant and product. Never alter a formula to make it balance.
Balance one element at a time using coefficients in front of formulas. Leave elements that appear in several substances, and free elements like $\ce{O2}$ or $\ce{H2}$ , until last.
Check every element, and the total charge if ions appear.
Add state symbols.

State symbols

symbol	meaning
(s)	solid, including precipitates
(l)	liquid (pure liquid, for example water or molten salt)
(g)	gas
(aq)	aqueous: dissolved in water

Water formed in a reaction in solution is $\ce{H2O(l)}$ , not $\ce{H2O(aq)}$ .

Balancing a combustion equation

Write a balanced equation, with state symbols, for the complete combustion of butane, $\ce{C4H10}$ .

Solution

Carbon first: $\ce{C4H10 + O2 -> 4CO2 + H2O}$ . Then hydrogen: 10 H gives $\ce{5H2O}$ . Oxygen on the right: $4 \times 2 + 5 = 13$ atoms, so $\tfrac{13}{2}\ \ce{O2}$ .

\ce{C4H10(g) + 13/2 O2(g) -> 4CO2(g) + 5H2O(l)}

or, doubling to remove the fraction:

\ce{2C4H10(g) + 13O2(g) -> 8CO2(g) + 10H2O(l)}

Fractions are acceptable in equations; they are used routinely in enthalpy definitions where one mole of a substance must appear.

Ionic equations

In aqueous solution, soluble ionic compounds, strong acids and strong alkalis exist as separate ions. Many of those ions take no part in the reaction: they are spectator ions. An ionic equation shows only the species that change, and must not include spectator ions.

Method

Writing an ionic equation

Write the full balanced equation with state symbols.
Split every aqueous ionic substance (and strong acid) into its ions. Do not split solids, liquids, gases, water, precipitates, or weak acids.
Cross out ions that appear unchanged on both sides.
Check atoms and charge balance.

Ionic equation for a precipitation

Write the ionic equation for the reaction between aqueous silver nitrate and aqueous sodium chloride.

Solution

Full equation: $\ce{AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)}$

Split aqueous species: $\ce{Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) -> AgCl(s) + Na+(aq) + NO3-(aq)}$

$\ce{Na+}$ and $\ce{NO3-}$ are spectators. Ionic equation:

\ce{Ag+(aq) + Cl-(aq) -> AgCl(s)}

Some ionic equations to know:

reaction	ionic equation
any acid with any alkali	$\ce{H+(aq) + OH-(aq) -> H2O(l)}$
aqueous carbonate with acid	$\ce{CO3^2-(aq) + 2H+(aq) -> CO2(g) + H2O(l)}$
test for sulfate	$\ce{Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)}$
metal displacement	$\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}$

For an insoluble carbonate such as $\ce{CaCO3}$ , the solid is not split into ions, so the ionic equation is $\ce{CaCO3(s) + 2H+(aq) -> Ca^2+(aq) + CO2(g) + H2O(l)}$ .

Exam tip

"Write an ionic equation" questions check that the charges balance as well as the atoms. In $\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}$ the charge is $+2$ on each side. Including a spectator ion, or omitting state symbols when asked for them, loses the mark.

Empirical and molecular formulas

Definition

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

Ethane has molecular formula $\ce{C2H6}$ and empirical formula $\ce{CH3}$ . Glucose, $\ce{C6H12O6}$ , and ethanoic acid, $\ce{C2H4O2}$ , share the empirical formula $\ce{CH2O}$ . Ionic compounds are described by empirical formulas only, because they do not form molecules.

Method

Finding an empirical formula

Write the mass (or percentage) of each element.
Divide each by its $A_r$ to get moles.
Divide every value by the smallest.
If the result is not close to whole numbers, multiply through by a small integer (values like 1.5 mean multiply by 2; 1.33 or 1.67 mean multiply by 3).
Molecular formula: divide $M_r$ by the empirical formula mass; multiply the empirical formula by the result.

From percentage composition to molecular formula

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its $M_r$ is 60. Find its empirical and molecular formulas.

Solution

	C	H	O
mass in 100 g	40.0	6.7	53.3
moles	$\tfrac{40.0}{12.0} = 3.33$	$\tfrac{6.7}{1.0} = 6.7$	$\tfrac{53.3}{16.0} = 3.33$
÷ smallest	1.00	2.01	1.00

Empirical formula $\ce{CH2O}$ , empirical formula mass $= 12.0 + 2.0 + 16.0 = 30.0$ .

$\dfrac{60}{30.0} = 2$ , so the molecular formula is $\ce{C2H4O2}$ .

Hydrated salts and water of crystallisation

Definition

Water of crystallisation is water that is chemically bonded within the crystal lattice of a salt.
A hydrated salt contains water of crystallisation, for example $\ce{CuSO4.5H2O}$ (blue crystals).
An anhydrous salt contains no water of crystallisation, for example $\ce{CuSO4}$ (a white powder).

Heating a hydrated salt drives off the water of crystallisation. The mass lost is the mass of water. Comparing moles of water with moles of anhydrous salt gives $x$ in $\ce{salt.xH2O}$ .

Finding the water of crystallisation

A $2.46\ \text{g}$ sample of hydrated magnesium sulfate, $\ce{MgSO4.xH2O}$ , is heated to constant mass. The anhydrous $\ce{MgSO4}$ left has a mass of $1.20\ \text{g}$ . Find $x$ .

Solution

Mass of water $= 2.46 - 1.20 = 1.26\ \text{g}$ .

$n(\ce{MgSO4}) = \dfrac{1.20}{24.3 + 32.1 + 64.0} = \dfrac{1.20}{120.4} = 0.009967\ \text{mol}$

$n(\ce{H2O}) = \dfrac{1.26}{18.0} = 0.0700\ \text{mol}$

$x = \dfrac{0.0700}{0.009967} = 7.02$ , so $x = 7$ : the salt is $\ce{MgSO4.7H2O}$ .

Combustion analysis

When an organic compound containing C, H and possibly O burns completely, all its carbon ends up in $\ce{CO2}$ and all its hydrogen in $\ce{H2O}$ . Any mass not accounted for by C and H is oxygen.

Exam-style: combustion analysis

Complete combustion of $0.290\ \text{g}$ of a compound containing C, H and O produced $0.660\ \text{g}$ of $\ce{CO2}$ and $0.270\ \text{g}$ of $\ce{H2O}$ . Its $M_r$ is 58. Find its molecular formula.

Solution

Carbon: $n(\ce{CO2}) = \dfrac{0.660}{44.0} = 0.0150\ \text{mol}$ , so $n(\ce{C}) = 0.0150\ \text{mol}$ , mass $= 0.180\ \text{g}$ .

Hydrogen: $n(\ce{H2O}) = \dfrac{0.270}{18.0} = 0.0150\ \text{mol}$ , so $n(\ce{H}) = 2 \times 0.0150 = 0.0300\ \text{mol}$ , mass $= 0.0300\ \text{g}$ .

Oxygen: mass $= 0.290 - 0.180 - 0.0300 = 0.080\ \text{g}$ , so $n(\ce{O}) = \dfrac{0.080}{16.0} = 0.0050\ \text{mol}$ .

Ratio C : H : O $= 0.0150 : 0.0300 : 0.0050 = 3 : 6 : 1$ . Empirical formula $\ce{C3H6O}$ , mass 58.0.

$\dfrac{58}{58.0} = 1$ , so the molecular formula is $\ce{C3H6O}$ .

Watch out

Each $\ce{H2O}$ contains two hydrogen atoms. Forgetting to double the moles of water is the most common error in combustion analysis.

Summary

Ionic charges follow the group: $+1, +2, +3$ for Groups 1, 2, 13; $-3, -2, -1$ for Groups 15, 16, 17. Roman numerals give the charge on transition metal ions.
Learn $\ce{NO3-}$ , $\ce{CO3^2-}$ , $\ce{SO4^2-}$ , $\ce{OH-}$ , $\ce{NH4+}$ , $\ce{Zn^2+}$ , $\ce{Ag+}$ , $\ce{HCO3-}$ , $\ce{PO4^3-}$ .
Balance with coefficients only; check atoms and charges; add state symbols (s), (l), (g), (aq).
Ionic equations omit spectator ions; only aqueous ionic species are split.
Empirical formula: simplest whole-number ratio. Molecular formula: actual numbers of atoms in a molecule.
Water of crystallisation: compare moles of water lost with moles of anhydrous salt.

Practice

Question

Write formulas for: magnesium nitrate, ammonium sulfate, aluminium oxide, iron(II) phosphate, sodium hydrogencarbonate, silver sulfate, zinc hydroxide.
Balance: (a) $\ce{Fe2O3 + CO -> Fe + CO2}$ (b) $\ce{Al + HCl -> AlCl3 + H2}$ (c) $\ce{C2H5OH + O2 -> CO2 + H2O}$ .
Write ionic equations, with state symbols, for (a) aqueous barium chloride with aqueous sodium sulfate, (b) magnesium ribbon with aqueous copper(II) sulfate, (c) solid magnesium carbonate with dilute hydrochloric acid.
A hydrocarbon contains 85.7% carbon by mass and has $M_r = 56$ . Find its empirical and molecular formulas.
A compound contains 32.4% Na, 22.6% S and 45.0% O. Find its empirical formula and name it.
Heating $5.00\ \text{g}$ of $\ce{CuSO4.xH2O}$ to constant mass leaves $3.20\ \text{g}$ of anhydrous $\ce{CuSO4}$ . Find $x$ .
Complete combustion of $0.580\ \text{g}$ of a hydrocarbon gives $1.760\ \text{g}$ of $\ce{CO2}$ and $0.900\ \text{g}$ of $\ce{H2O}$ . $M_r = 58$ . Find the molecular formula.
Write the formulas of sodium chlorate(I), potassium manganate(VII) and chromium(III) sulfate.
Washing soda is $\ce{Na2CO3.xH2O}$ . Heating $1.430\ \text{g}$ of it to constant mass leaves $0.530\ \text{g}$ of $\ce{Na2CO3}$ . Find $x$ , and explain why the sample is heated to constant mass.
A $0.485\ \text{g}$ sample of a compound of carbon, hydrogen and chlorine burns to give $0.440\ \text{g}$ of $\ce{CO2}$ and $0.090\ \text{g}$ of $\ce{H2O}$ (all chlorine ends up in other products). $M_r = 97$ . Find its molecular formula.

Answers

$\ce{Mg(NO3)2}$ , $\ce{(NH4)2SO4}$ , $\ce{Al2O3}$ , $\ce{Fe3(PO4)2}$ , $\ce{NaHCO3}$ , $\ce{Ag2SO4}$ , $\ce{Zn(OH)2}$ .
(a) $\ce{Fe2O3 + 3CO -> 2Fe + 3CO2}$ (b) $\ce{2Al + 6HCl -> 2AlCl3 + 3H2}$ (c) $\ce{C2H5OH + 3O2 -> 2CO2 + 3H2O}$ .
(a) $\ce{Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)}$ (b) $\ce{Mg(s) + Cu^2+(aq) -> Mg^2+(aq) + Cu(s)}$ (c) $\ce{MgCO3(s) + 2H+(aq) -> Mg^2+(aq) + CO2(g) + H2O(l)}$ .
C: $85.7 / 12.0 = 7.14$ ; H: $14.3 / 1.0 = 14.3$ ; ratio $1 : 2.00$ , so $\ce{CH2}$ (mass 14.0). $56 / 14.0 = 4$ : $\ce{C4H8}$ .
Na: $32.4 / 23.0 = 1.409$ ; S: $22.6 / 32.1 = 0.704$ ; O: $45.0 / 16.0 = 2.81$ . Divide by 0.704: $2.00 : 1 : 3.99$ , so $\ce{Na2SO4}$ , sodium sulfate.
Water $= 1.80\ \text{g} = 0.100\ \text{mol}$ . $n(\ce{CuSO4}) = 3.20 / 159.6 = 0.02005\ \text{mol}$ . $x = 0.100 / 0.02005 = 4.99$ , so $x = 5$ .
$n(\ce{C}) = 1.760 / 44.0 = 0.0400$ ; $n(\ce{H}) = 2 \times 0.900 / 18.0 = 0.100$ . Check: $0.0400 \times 12.0 + 0.100 \times 1.0 = 0.580\ \text{g}$ , so no oxygen. Ratio $1 : 2.5 = 2 : 5$ , empirical $\ce{C2H5}$ (mass 29.0). $58 / 29.0 = 2$ : $\ce{C4H10}$ .
$\ce{NaClO}$ , $\ce{KMnO4}$ , $\ce{Cr2(SO4)3}$ .
Water $= 0.900\ \text{g} = 0.0500\ \text{mol}$ . $n(\ce{Na2CO3}) = 0.530 / 106.0 = 0.00500\ \text{mol}$ . $x = 10$ . Heating to constant mass (heat, cool, weigh, repeat until two masses agree) ensures all the water of crystallisation has been driven off; otherwise the mass of water would be underestimated and $x$ would be too small.
$n(\ce{C}) = 0.440 / 44.0 = 0.0100$ , mass $0.120\ \text{g}$ . $n(\ce{H}) = 2 \times 0.090 / 18.0 = 0.0100$ , mass $0.010\ \text{g}$ . Mass of Cl $= 0.485 - 0.120 - 0.010 = 0.355\ \text{g}$ , $n = 0.355 / 35.5 = 0.0100$ . Ratio $1 : 1 : 1$ , empirical $\ce{CHCl}$ (mass 48.5). $97 / 48.5 = 2$ : $\ce{C2H2Cl2}$ .

Formulas of ionic compounds

Predicting ionic charges from the Periodic Table

Ions to learn

Balancing equations

State symbols

Ionic equations

Empirical and molecular formulas

Hydrated salts and water of crystallisation

Combustion analysis

Practice

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