Chemical equations and the laws of chemical combination: WASSCE Chemistry

A chemical equation is a recipe written in symbols: it shows what reacts, what forms, and in what proportions. Writing it correctly, balanced and with state symbols, is the first step in every calculation. This lesson covers writing and balancing equations, ionic equations, and the laws of chemical combination that equations obey.

Learning objectives

By the end of this lesson you should be able to:

write word equations and balanced symbol equations with state symbols
balance equations by inspection
write ionic equations and identify spectator ions
state and verify the laws of conservation of mass, constant composition and multiple proportions

From words to symbols

Word equation: magnesium + oxygen → magnesium oxide

Symbol equation (unbalanced): Mg + O₂ → MgO

Balanced, with state symbols:

2\text{Mg(s)} + \text{O}_2\text{(g)} \rightarrow 2\text{MgO(s)}

\begin{gathered}2\text{Mg(s)} + \text{O}_2\text{(g)} \\[6pt] \rightarrow 2\text{MgO(s)}\end{gathered}

State symbol	Meaning
(s)	solid
(l)	liquid
(g)	gas
(aq)	aqueous: dissolved in water

Balancing equations

A balanced equation has the same number of each kind of atom on both sides, because atoms are neither created nor destroyed. Balance by changing the coefficients (the numbers in front), never the subscripts in a formula.

Steps:

Write the correct formula of every reactant and product.
Count the atoms of each element on each side.
Balance one element at a time, leaving hydrogen and oxygen (or a free element) until last.
Check every element again.

Worked example

Balancing.

Aluminium burning in oxygen:

4\text{Al(s)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{Al}_2\text{O}_3\text{(s)}

\begin{gathered}4\text{Al(s)} + 3\text{O}_2\text{(g)} \\[6pt] \rightarrow 2\text{Al}_2\text{O}_3\text{(s)}\end{gathered}

Check: Al 4 = 4; O 6 = 6.

Methane burning in oxygen:

\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}

\begin{gathered}\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \\[6pt] \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}\end{gathered}

Check: C 1 = 1; H 4 = 4; O 4 = 2 + 2.

Propane (cooking gas) burning:

\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}

\begin{gathered}\text{C}_3\text{H}_8 + 5\text{O}_2 \\[6pt] \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}\end{gathered}

Balance C first (3 CO₂), then H (4 H₂O), then O: $6 + 4 = 10$ atoms, so 5 O₂.

Common mistake

Never change a formula to balance an equation. Writing H₂O₂ instead of 2H₂O changes water into hydrogen peroxide, a different substance.

Ionic equations

Many reactions in solution involve only some of the ions present. The others, which appear unchanged on both sides, are spectator ions and are left out of the ionic equation.

Worked example

Silver nitrate solution + sodium chloride solution.

Full equation:

\text{AgNO}_3\text{(aq)} + \text{NaCl(aq)} \rightarrow \text{AgCl(s)} + \text{NaNO}_3\text{(aq)}

\begin{gathered}\text{AgNO}_3\text{(aq)} + \text{NaCl(aq)} \\[6pt] \rightarrow \text{AgCl(s)} + \text{NaNO}_3\text{(aq)}\end{gathered}

Split the soluble ionic compounds into ions; Na⁺ and NO₃⁻ appear unchanged on both sides, so they are spectators. The ionic equation is:

\text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)} \rightarrow \text{AgCl(s)}

\begin{gathered}\text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)} \\[6pt] \rightarrow \text{AgCl(s)}\end{gathered}

A white precipitate of silver chloride forms. This is the test for chloride ions.

Reaction	Ionic equation
Any acid + alkali (neutralisation)	H⁺(aq) + OH⁻(aq) → H₂O(l)
Barium ions + sulphate ions	Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
Carbonate + acid	CO₃²⁻ + 2H⁺ → CO₂(g) + H₂O(l)
Zinc + copper(II) ions	Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

An ionic equation must balance both atoms and charges.

Laws of chemical combination

Law of conservation of mass

Definition

Matter is neither created nor destroyed in a chemical reaction: the total mass of the products equals the total mass of the reactants.

Worked example

10.0 g of calcium carbonate is heated strongly in an open crucible. It leaves 5.6 g of calcium oxide. How much carbon(IV) oxide escaped?

\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}

\begin{gathered}\text{CaCO}_3\text{(s)} \\[6pt] \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}\end{gathered}

By conservation of mass: $10.0 - 5.6 = 4.4$ g of CO₂ escaped. In a closed container the total mass would not change.

Law of constant (definite) composition

Definition

All pure samples of the same compound contain the same elements combined in the same proportion by mass, however they are made.

Worked example

Two samples of copper(II) oxide were made by different methods.

Sample A: 1.59 g of oxide gave 1.27 g of copper.
Sample B: 3.18 g of oxide gave 2.54 g of copper.

% copper in A $= \dfrac{1.27}{1.59} \times 100 = 79.9\%$ ; % copper in B $= \dfrac{2.54}{3.18} \times 100 = 79.9\%$ .

The percentages are equal, so the law is obeyed.

Law of multiple proportions

Definition

If two elements form more than one compound, the different masses of one element that combine with a fixed mass of the other are in a simple whole-number ratio.

Worked example

Carbon and oxygen.

In carbon(II) oxide, CO, 12 g of carbon combines with 16 g of oxygen.
In carbon(IV) oxide, CO₂, 12 g of carbon combines with 32 g of oxygen.

The masses of oxygen combining with a fixed 12 g of carbon are in the ratio $16 : 32 = 1 : 2$ , a simple whole-number ratio.

The law of reciprocal proportions and Gay-Lussac's law of combining volumes (gases) complete the laws of chemical combination. All of them are explained by Dalton's atomic theory: atoms combine in fixed whole numbers.

In Sierra Leone

Cooking gas sold in Sierra Leone is mainly propane and butane. Balancing their combustion equations shows that each molecule needs a lot of oxygen. Burning gas in a closed room without air flow gives incomplete combustion and poisonous carbon(II) oxide:

2\text{C}_3\text{H}_8 + 7\text{O}_2 \rightarrow 6\text{CO} + 8\text{H}_2\text{O}

\begin{gathered}2\text{C}_3\text{H}_8 + 7\text{O}_2 \\[6pt] \rightarrow 6\text{CO} + 8\text{H}_2\text{O}\end{gathered}

This is why kitchens need ventilation.

Key points

A balanced equation has equal numbers of each atom on both sides; balance with coefficients, never subscripts.
State symbols: (s), (l), (g), (aq).
Ionic equations leave out spectator ions and must balance atoms and charges; neutralisation is H⁺ + OH⁻ → H₂O.
Conservation of mass: total mass is unchanged. Constant composition: fixed proportions by mass. Multiple proportions: simple whole-number ratios.

WASSCE examination focus

WAEC tests this

Expect to balance equations (including combustion), to write ionic equations for precipitation and neutralisation, and to state and verify the laws of chemical combination from data. In verification questions, show the percentage or ratio calculation clearly before drawing the conclusion.

Quick check

Balance: Fe + Cl₂ → FeCl₃.
Write the ionic equation for the reaction of barium chloride solution with sodium sulphate solution.
State the law of constant composition.
Nitrogen forms N₂O and NO. Show that they obey the law of multiple proportions.

Answers:

2Fe + 3Cl₂ → 2FeCl₃.
Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s).
All pure samples of a compound contain the same elements combined in the same proportion by mass.
With a fixed 16 g of oxygen, N₂O has 28 g of N and NO has 14 g: ratio 28 : 14 = 2 : 1, a simple whole-number ratio.

Chemical equations and the laws of chemical combination

Learning objectives

From words to symbols

Balancing equations

Ionic equations

Laws of chemical combination

Law of conservation of mass

Law of constant (definite) composition

Law of multiple proportions

Key points

WASSCE examination focus

Quick check

Keep going with chemical equations and stoichiometry

More lessons in Chemical Equations and Stoichiometry