Representing Chemical Reactions and Balancing Equations

From a reaction description to a word equation

A chemical equation records which substances react and which new substances form. The starting substances are reactants, written on the left. The substances formed are products, written on the right. An arrow means ‘react to form’; plus signs separate substances on the same side.

For example, sodium hydroxide reacts with hydrochloric acid to produce sodium chloride and water:

sodium hydroxide + hydrochloric acid → sodium chloride + water

The order in which substances appear in a sentence does not determine their position in the equation. If a question says that water and sodium chloride are formed when hydrochloric acid reacts with sodium hydroxide, the acids and alkalis still belong on the reactant side.

Sometimes you need to identify a reactant from the description. Carbon burning in air reacts with oxygen, so its word equation is:

carbon + oxygen → carbon dioxide

Replacing names with correct formulae

A symbol equation uses chemical formulae instead of names. Before balancing it, each formula must correctly represent the substance.

A subscript belongs to the formula itself: $\mathrm{H_2O}$ contains two hydrogen atoms for every oxygen atom. If there is no subscript, the number is one. Brackets allow a whole group to be repeated. In $\mathrm{Mg(NO_3)_2}$, the subscript 2 applies to everything inside the brackets, giving one magnesium, two nitrogen and six oxygen atoms.

For ionic compounds, the positive and negative charges must balance overall. For example, magnesium ions have charge $2+$ and nitrate ions, $\mathrm{NO_3^-}$, have charge $1-$. Two nitrate ions are therefore needed for each magnesium ion, giving $\mathrm{Mg(NO_3)_2}$. Brackets preserve the nitrate group while showing that there are two of them.

Some elements occur as molecules containing two atoms. Write hydrogen as $\mathrm{H_2}$, nitrogen as $\mathrm{N_2}$, oxygen as $\mathrm{O_2}$ and the halogens as $\mathrm{F_2}$, $\mathrm{Cl_2}$, $\mathrm{Br_2}$ and $\mathrm{I_2}$ when they are elemental substances. Carbon burning in oxygen is therefore represented by:

$$\mathrm{C + O_2 \rightarrow CO_2}$$

This equation already has one carbon atom and two oxygen atoms on each side.

Why equations must balance

During a chemical reaction, atoms are rearranged rather than created or destroyed. A balanced equation therefore has the same number of atoms of each element on both sides. This is the basis of conservation of mass in chemical reactions.

To balance an equation, change the coefficients: the numbers placed in front of whole formulae. A coefficient multiplies every atom in that formula. For example, $2\mathrm{H_2O}$ represents two water molecules, containing four hydrogen atoms and two oxygen atoms altogether.

Changing a subscript would change the substance, not its amount. $\mathrm{H_2O_2}$ is hydrogen peroxide, not water, so it cannot replace $\mathrm{H_2O}$ just to make an equation balance.

Balancing an equation step by step

Aluminium reacts with copper(II) oxide to form aluminium oxide and copper. Start by writing the correct formulae:

$$\mathrm{Al + CuO \rightarrow Al_2O_3 + Cu}$$

There are two aluminium atoms on the right, so put 2 before Al on the left. There are three oxygen atoms on the right, so put 3 before CuO on the left. This also gives three copper atoms on the left, so put 3 before Cu on the right:

$$\mathrm{2Al + 3CuO \rightarrow Al_2O_3 + 3Cu}$$

Check the result element by element:

The coefficients give the reacting ratio: two aluminium atoms react with three formula units of copper(II) oxide. Use the smallest whole-number ratio.

An unchanged group can make counting easier. Magnesium oxide reacts with nitric acid to form magnesium nitrate and water:

$$\mathrm{MgO + 2HNO_3 \rightarrow Mg(NO_3)_2 + H_2O}$$

There are two nitrate groups on the right, so two nitric acid units are needed on the left. This also supplies the two hydrogen atoms needed for water. Checking individual atoms gives one magnesium, two hydrogen, two nitrogen and seven oxygen atoms on each side.

Adding physical states

State symbols follow each formula: (s) means solid, (l) liquid, (g) gas and (aq) aqueous, meaning dissolved in water. Aqueous does not simply mean liquid: liquid water is $\mathrm{H_2O(l)}$, whereas sodium chloride dissolved in water is $\mathrm{NaCl(aq)}$.

The neutralisation equation with states is:

$$\mathrm{NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l)}$$

It is balanced without any written coefficients because a missing coefficient means one.

Higher Tier: balancing half-equations

An ion is a particle with an electrical charge. A half-equation follows the gain or loss of electrons by a particular particle. Both atoms and total charge must balance.

Copper(II) ions gain electrons to become copper atoms. Start with $\mathrm{Cu^{2+} \rightarrow Cu}$. The atoms already balance, but the charges do not. Adding two electrons, each with charge $1-$, to the left gives:

$$\mathrm{Cu^{2+} + 2e^- \rightarrow Cu}$$

The total charge is now zero on both sides. Gaining electrons is reduction.

Bromide ions lose electrons to form bromine. Because bromine is $\mathrm{Br_2}$, two bromide ions are needed:

$$\mathrm{2Br^- \rightarrow Br_2 + 2e^-}$$

There are two bromine atoms on each side, and the total charge on each side is $2-$. Losing electrons is oxidation. In a redox reaction, electrons lost by one particle are gained by another.

Higher Tier: removing spectator ions

An ionic equation focuses on the particles that actually change. Consider the neutralisation of hydrochloric acid by sodium hydroxide. In solution, these substances provide $\mathrm{H^+}$, $\mathrm{Cl^-}$, $\mathrm{Na^+}$ and $\mathrm{OH^-}$ ions. Writing the dissolved ions separately gives:

$$\mathrm{H^+(aq) + Cl^-(aq) + Na^+(aq) + OH^-(aq) \rightarrow Na^+(aq) + Cl^-(aq) + H_2O(l)}$$

The sodium and chloride ions appear unchanged on both sides. They are spectator ions, so cancel them to obtain:

$$\mathrm{H^+(aq) + OH^-(aq) \rightarrow H_2O(l)}$$

This shows the chemical change: hydrogen ions and hydroxide ions form water. The atoms balance, and the total charge is zero on both sides. Keep water as a molecule rather than splitting it into ions. Likewise, solids and gases remain intact when constructing an ionic equation.

Word and symbol equations

• Reactants go on the left; products go on the right.
• Word equations use names; symbol equations use correct formulae.
• Elemental diatomic molecules: $\mathrm{H_2, N_2, O_2, F_2, Cl_2, Br_2, I_2}$.

Balancing checklist

1. Write the correct formulae.
2. Count each element on both sides.
3. Adjust coefficients, not subscripts.
4. Recheck all atoms and use the smallest whole-number ratio.

A coefficient multiplies the whole formula; a subscript after brackets multiplies the whole bracketed group.

States: (s) solid, (l) liquid, (g) gas, (aq) dissolved in water.

Higher Tier: electrons and ions

Balance atoms and total charge.

• Reduction: gain of electrons; electrons on the left.
• Oxidation: loss of electrons; electrons on the right.

$$\mathrm{Cu^{2+} + 2e^- \rightarrow Cu}$$ $$\mathrm{2Br^- \rightarrow Br_2 + 2e^-}$$

For an ionic equation, write dissolved ionic substances as ions and cancel unchanged spectator ions. Do not split solids, liquids or gases into ions.

Neutralisation:

$$\mathrm{H^+(aq) + OH^-(aq) \rightarrow H_2O(l)}$$