Compounds, formulae and equations

In a nutshell

Every calculation and every reaction in A-level chemistry starts from a correct formula and a balanced equation. Get one wrong and every mark that follows it is at risk.

This subtopic teaches you to work out the formula of an ionic compound from the charges on its ions, and to write balanced full and ionic equations, with state symbols, for reactions you know and for ones you have never met.

Assumed knowledge: Atomic structure and isotopes: protons, electrons and how an ion forms.

Core content

Predicting ionic charge from the periodic table

For many elements you can predict the charge on the ion from the element's group.

• Metals lose electrons to form positive ions. The charge equals the number of outer-shell electrons lost.
• Non-metals gain electrons to form negative ions. The charge equals the number of electrons needed to fill the outer shell.

Either way, the ion ends up with the same electron arrangement as the nearest noble gas.

A simple negative ion made from one non-metal atom takes the ending -ide.

The pattern does not work for Group 14 or Group 18: carbon and silicon bond covalently rather than forming ions, and the noble gases do not form ions. For most d-block metals (such as iron or copper) the charge cannot be predicted from the group, so the question will give it to you, often as a Roman numeral in the name: iron(III) means Fe³⁺, copper(II) means Cu²⁺.

Ions you must recall

Some ions cannot be predicted from the periodic table, so OCR expects you to know their names, formulae and charges by heart.

The first five are polyatomic ions: a group of atoms bonded together that carries an overall charge. The charge belongs to the whole group, not to one atom.

If a question just says "nitrate" or "sulfate", assume NO₃^- and SO₄^2-. The charge on any other ion (for example phosphate, PO₄^3-) will be given in the question.

Writing formulae of ionic compounds

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

The method:

1. Write down each ion with its charge (from the periodic table, from your recall list, or from the question).
2. Find the smallest whole-number ratio of ions that makes the charges cancel to zero.
3. Write the formula with the positive ion first. Put brackets round a polyatomic ion if you need more than one of it.
4. Leave the charges out of the final formula.

The brackets matter. Ca(OH)₂ has two O and two H; "CaOH₂" would mean one O and two H, which is a different (and wrong) formula.

Still don't get it? · making the charges cancel

Imagine a see-saw. Every positive ion is a weight on the left, every negative ion a weight on the right, and the number on each ion is how heavy it is. The compound only exists when the see-saw is perfectly level.

Now try aluminium oxide. Each aluminium weight is 3, each oxide weight is 2.

• One of each: 3 on the left, 2 on the right. Not level.
• You need a total that both 3 and 2 go into. The smallest is 6.
• 6 on the left needs two aluminium weights (2 × 3).
• 6 on the right needs three oxide weights (3 × 2).
• Level. So the compound has 2 aluminium and 3 oxide: Al₂O₃.

A handy check: the number of each ion in the formula often turns out to be the charge on the other ion (Al³⁺ gives the 3 to oxygen, O²⁻ gives the 2 to aluminium). But always simplify: Mg²⁺ and O²⁻ give MgO, not Mg₂O₂.

Exam version: write each ion's charge, find the smallest whole-number ratio so that the total positive charge equals the total negative charge, then write the formula with brackets round any polyatomic ion that appears more than once.

Balanced equations and state symbols

A balanced equation has the same number of atoms of each element on both sides. Atoms are never created or destroyed in a reaction.

To balance, change only the big numbers in front of formulae (the coefficients). Never change a formula to make it balance: turning H₂O into H₂O₂ makes a different substance.

The method:

1. Write the correct formula of every reactant and product.
2. Count the atoms of each element on each side.
3. Balance one element at a time, leaving elements that appear on their own (such as O₂ or H₂) until last.
4. Count again to check, then add state symbols.

Two shortcuts that save time:

• If a polyatomic ion such as SO₄ passes through the reaction unchanged, balance it as one unit.
• Remember the elements that exist as diatomic molecules: H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂.

State symbols show the physical state of each substance under the reaction conditions.

Water itself is always (l), never (aq): (aq) means "dissolved in water", and water cannot be dissolved in itself.

For example, a metal reacting with an acid, and a solid carbonate reacting with an acid:

• Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
• CaCO₃(s) + 2HNO₃(aq) → Ca(NO₃)₂(aq) + H₂O(l) + CO₂(g)

Ionic equations

An ionic equation shows only the species that take part in the reaction. The ions that are present but unchanged, the spectator ions, are left out.

The method:

1. Write the balanced full equation with state symbols.
2. Split every aqueous ionic compound (and every strong acid in solution, such as HCl(aq), HNO₃(aq) and H₂SO₄(aq)) into its separate ions.
3. Do not split solids (including precipitates), liquids such as water, gases, or covalent molecules.
4. Cancel the spectator ions: any ion that appears unchanged, in the same number, on both sides.
5. Check the equation balances for atoms and for charge.

Many different full equations share the same ionic equation, because the spectator ions differ but the change does not.

Every one balances for charge: in the carbonate equation, the left side is (2-) + 2(1+) = 0, and the right side has no ions, so it is also 0.

Still don't get it? · spectator ions and ionic equations

Think of a school disco. Before the music starts, everyone is milling around on their own. During the night two people pair up and leave the room together. Everyone else is still milling around exactly as before. If you wanted to describe what actually happened at the disco, you would only mention the pair who left; the rest were just watching.

Now the chemistry, one step at a time, for silver nitrate solution mixed with sodium chloride solution:

1. Both solutions are ionic and dissolved, so the beaker is full of separate Ag⁺, NO₃⁻, Na⁺ and Cl⁻ ions, all moving freely.
2. Ag⁺ and Cl⁻ meet and lock together as solid silver chloride, which drops out as a precipitate. They have changed: from dissolved ions to a solid.
3. Na⁺ and NO₃⁻ are still dissolved, free ions afterwards. Nothing about them has changed. They are the onlookers.
4. Write only what changed: Ag⁺(aq) + Cl⁻(aq) → AgCl(s).

Exam version: split the aqueous ionic compounds into ions, cancel the spectator ions (ions that appear unchanged on both sides), and check the ionic equation balances for atoms and for charge.

Equations for unfamiliar reactions

OCR can ask you to write an equation for a reaction you have never studied, using information given in the question. You use exactly the same skills.

1. Turn each name into a formula, using ion charges (predicted, recalled or given).
2. Use the description to decide on the products and their states: "fizzing" or "a gas is given off" means (g); "a precipitate forms" means (s); "a solution forms" means (aq).
3. Balance, add state symbols, and check atoms (and charge, for an ionic equation).

Worked examples

1. Formula from predicted and recalled charges

Write the formula of aluminium sulfate.

Aluminium is in Group 13, so it forms Al³⁺. Sulfate is SO₄^2- (recall).

$\text{charges: } \text{Al}^{3+} = +3 \qquad \text{SO}_4^{\,2-} = -2$ $\text{lowest common multiple of 3 and 2} = 6$ $2 \times (+3) = +6 \qquad 3 \times (-2) = -6$ $(+6) + (-6) = 0$ $\text{formula} = \text{Al}_2(\text{SO}_4)_3$

Check the atoms: 2 Al, 3 S, 12 O. The brackets are essential: "Al₂SO₄₃" is wrong because it does not show three sulfate ions.

2. Formula using a charge given in the question

Write the formula of ammonium phosphate. (Phosphate ion: PO₄^3-)

Ammonium is NH₄⁺ (recall). The phosphate charge is given.

$\text{charges: } \text{NH}_4^{\,+} = +1 \qquad \text{PO}_4^{\,3-} = -3$ $3 \times (+1) + 1 \times (-3) = 0$ $\text{formula} = (\text{NH}_4)_3\text{PO}_4$

Check the atoms: 3 N, 12 H, 1 P, 4 O. Brackets go round NH₄ because there are three of them; none round PO₄ because there is only one.

3. Balancing a full equation with state symbols

Write a balanced equation, with state symbols, for the complete combustion of propane, C₃H₈(g), to carbon dioxide and water at room temperature.

Start with the correct formulae:

C₃H₈ + O₂ → CO₂ + H₂O

1. Carbon: 3 on the left, so put 3 in front of CO₂.
2. Hydrogen: 8 on the left, so put 4 in front of H₂O (4 × 2 = 8).
3. Oxygen last (it appears on its own as O₂): right side has (3 × 2) + (4 × 1) = 10 O atoms, so put 5 in front of O₂.

$\text{O atoms on right} = (3 \times 2) + (4 \times 1) = 10 \;\Rightarrow\; \frac{10}{2} = 5\ \text{O}_2$

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

Water is (l) because the products are at room temperature.

4. Full equation to ionic equation

Aqueous zinc nitrate reacts with aqueous sodium hydroxide to give a white precipitate of zinc hydroxide and a solution of sodium nitrate. Write the full equation and the ionic equation, with state symbols.

Formulae from charges: zinc nitrate is Zn(NO₃)₂ (Zn²⁺, NO₃^-), zinc hydroxide is Zn(OH)₂, sodium nitrate is NaNO₃.

1. Balanced full equation: Zn(NO₃)₂(aq) + 2NaOH(aq) → Zn(OH)₂(s) + 2NaNO₃(aq)
2. Split the aqueous ionic compounds (not the precipitate): Zn²⁺(aq) + 2NO₃^-(aq) + 2Na⁺(aq) + 2OH^-(aq) → Zn(OH)₂(s) + 2Na⁺(aq) + 2NO₃^-(aq)
3. Cancel the spectator ions (2Na⁺ and 2NO₃^- appear unchanged on both sides): Zn²⁺(aq) + 2OH^-(aq) → Zn(OH)₂(s)
4. Check the charge:

$\text{left: } (+2) + 2 \times (-1) = 0 \qquad \text{right: } 0$

Both sides are 0, and there is 1 Zn, 2 O and 2 H on each side, so the ionic equation is balanced.

5. An unfamiliar reaction from given information

When aluminium foil is added to dilute sulfuric acid, bubbles of hydrogen are given off and a solution of aluminium sulfate forms. Write a balanced equation with state symbols.

Formulae: aluminium is Al(s), sulfuric acid is H₂SO₄(aq), aluminium sulfate is Al₂(SO₄)₃(aq) (from Worked example 1), hydrogen is H₂(g) (diatomic).

1. Aluminium: 2 on the right, so put 2 in front of Al.
2. Sulfate as a unit: 3 SO₄ on the right, so put 3 in front of H₂SO₄.
3. Hydrogen: 3 × 2 = 6 H on the left, so put 3 in front of H₂.

2Al(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3H₂(g)

Common exam mistakes

• Ammonium written as ammonia. The ion is NH₄⁺; NH₃ is the neutral molecule ammonia. Ammonium sulfate is (NH₄)₂SO₄, not NH₄SO₄ or NH₃SO₄.
• Missing brackets round a polyatomic ion. "CaOH₂" and "MgNO₃2" are wrong; the formulae are Ca(OH)₂ and Mg(NO₃)₂.
• Mixing up -ate and -ide. Sulfate is SO₄^2- but sulfide is S^2-; nitrate is NO₃^- but nitride is N^3-. One letter changes the formula.
• Guessing the charge on zinc or silver. They are d-block metals, so students guess Zn⁺ or Ag²⁺. Zinc is always Zn²⁺ and silver Ag⁺ at this level: learn them.
• Changing a formula to balance an equation. Writing H₂O₂ instead of 2H₂O, or MgCl instead of balancing with 2HCl. Only the coefficients in front may change.
• Forgetting diatomic elements. Hydrogen gas is H₂(g) and oxygen is O₂(g), not H or O.
• Water as (aq). Water is H₂O(l). Likewise a precipitate is (s), not (aq).
• Leaving spectator ions in an ionic equation, or splitting a precipitate or water into ions. AgCl(s), BaSO₄(s) and H₂O(l) stay as whole formulae.
• An ionic equation that does not balance for charge, or ions written without their charges (for example "Ag + Cl → AgCl"). Always add up the charges on each side.
• Leaving out state symbols when the question asks for them. State symbols are often credited separately, so one missing or wrong symbol can cost a mark even when the equation is balanced.

Key definitions

OCR does not require formal definitions for this subtopic, but you must use these terms correctly.

• Formula of an ionic compound: the smallest whole-number ratio of ions in which the total positive charge equals the total negative charge.
• Balanced equation: an equation with the same number of atoms of each element on both sides. An ionic equation must also have the same total charge on both sides.
• Ionic equation: an equation showing only the ions and other species that take part in the reaction, with the spectator ions left out.
• Spectator ions: ions that are present in the reaction mixture but unchanged by the reaction, so they appear in the same form on both sides of the equation.
• State symbols: (s) solid, (l) liquid, (g) gas, (aq) aqueous, dissolved in water.
• Polyatomic ion: a group of atoms bonded together that carries an overall charge, for example NO₃^-, CO₃^2-, SO₄^2-, OH^- and NH₄⁺.

Specification

• I can predict the charge on an ion from the position of its element in the periodic table (Groups 1, 2, 13, 15, 16 and 17). (2.1.2(a)(i))
• I can recall the names and formulae of NO₃^-, CO₃^2-, SO₄^2-, OH^-, NH₄⁺, Zn²⁺ and Ag⁺, and I know "nitrate" and "sulfate" mean NO₃^- and SO₄^2-. (2.1.2(a)(ii))
• I can write the formula of an ionic compound from the charges of its ions, including charges given to me in a question, using brackets correctly. (2.1.2(a))
• I can construct balanced full equations, with state symbols, for reactions I have studied. (2.1.2(b))
• I can construct balanced ionic equations, with state symbols, by cancelling spectator ions, and check they balance for charge. (2.1.2(b))
• I can construct balanced equations for unfamiliar reactions from information given in a question, using ratios to balance them. (2.1.2(b), M0.2)

Related notes

• Atomic structure and isotopes
• Amount of substance
• Acids
• Redox
• Bonding and structure
• Qualitative analysis