How do I write the formula of an ionic compound from the ion charges?

Write the symbols of the two ions with their charges, then balance the charges so the total positive charge equals the total negative charge. Use the smallest whole numbers of each ion that make the charges cancel, and write these as subscripts. For example, magnesium ions are 2+ and chloride ions are 1-, so you need two chloride ions for each magnesium ion, giving the formula magnesium chloride with the formula written as one magnesium to two chlorides.

How do I balance a chemical equation?

First write the correct formulae of all the reactants and products; never change a formula to balance. Then put numbers (coefficients) in front of the formulae so that there is the same number of each type of atom on both sides. Count the atoms of each element on each side, adjust the coefficients one element at a time, and check at the end. Finally add the state symbols (s), (l), (g) and (aq).

The mole is the chemist's unit for counting particles. One mole of any substance contains the Avogadro number of particles, which is the same very large number for every substance. Conveniently, one mole of a substance has a mass in grams equal to its relative atomic or molecular mass. So the mole links the mass you can weigh out to the number of particles taking part in a reaction, which is what makes calculations possible.

How do I use a balanced equation to find a reacting mass?

Work in three steps. First, convert the known mass into moles by dividing by the relative formula mass. Second, use the ratio of the coefficients in the balanced equation to find the moles of the substance you want. Third, convert those moles back to a mass by multiplying by its relative formula mass. The mole ratio from the balanced equation is the bridge between the two substances.

How is Stoichiometry and the Mole assessed in N(A)-Level Science 5107?

It is examined in the two Chemistry papers. Paper 3 (multiple choice, 20 marks) tests writing formulae, balancing simple equations, and one-step mole conversions. Paper 4 (structured, 30 marks) asks you to balance equations, calculate moles, and work out reacting masses or a simple percentage yield, with marks for clear working. A Periodic Table is provided in the multiple choice paper, and a calculator is allowed.

SingaporeChemistry

Stoichiometry and the Mole (Singapore N(A)-Level Science Chemistry 5107): writing formulae and balancing equations, the mole and relative masses, and using mole ratios to work out reacting masses

A Singapore N(A)-Level Science Chemistry (SEAB 5107) overview of Stoichiometry and the Mole. Writing formulae from ion charges and balancing symbol equations, the meaning of the mole and relative masses, and using the mole ratio from a balanced equation to calculate reacting masses, concentration and percentage yield, with links to every dot point.

Generated by Claude Opus 4.87 min readSEAB-5107Updated 2026-06-13

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is really about
Formulae and chemical equations
The mole and relative masses
Mole calculations and reacting masses
How this topic is examined
Check your knowledge

What this topic is really about

Stoichiometry and the Mole is the maths of chemistry: it lets you work out how much of each substance reacts and how much product you can make. Everything rests on the mole, the unit that links the mass you can weigh out to the number of particles that react. This guide ties the three dot points together and links to each one for the worked answers and practice.

The complete set of dot-point pages for this topic lives at /sg-n-level/chemistry/syllabus/stoichiometry-and-the-mole.

Formulae and chemical equations

Formulae and chemical equations shows how to write the formula of a compound by balancing ion charges, write word equations, and balance symbol equations with state symbols. The golden rule is never to change a formula to balance an equation, only to put numbers in front of the formulae.

The mole and relative masses

The mole and relative masses defines relative atomic and molecular mass, the mole and the Avogadro constant, and shows how to convert between mass, moles and number of particles. The key relationship is that moles equal mass divided by relative formula mass.

Mole calculations and reacting masses

Mole calculations and reacting masses uses the mole ratio from a balanced equation to find reacting masses, work with the concentration of solutions, and find a simple percentage yield. The method is always the same three steps: mass to moles, mole ratio, moles back to mass.

How this topic is examined

Show every step of working. Marks are given for the method (mass to moles, mole ratio, moles to mass), even if the final number slips.
Use the right relative formula mass. Add up the relative atomic masses correctly, including subscripts.
Read the mole ratio from the balanced equation. The coefficients are the bridge between reactant and product.

Check your knowledge

A mix of recall, formula and calculation questions covering Stoichiometry and the Mole. Use relative atomic masses H = 1, C = 12, O = 16, Mg = 24, Ca = 40. Attempt them under timed conditions, then check against the solutions.

Write the formula of calcium chloride, given calcium ions are 2+ and chloride ions are 1-. (1 mark)
Balance the equation: $\text{H}_2 + \text{O}_2 \rightarrow \text{H}_2\text{O}$ . (1 mark)
Calculate the relative formula mass of calcium carbonate, $\text{CaCO}_3$ . (1 mark)
Calculate the number of moles in 88 grams of carbon dioxide, $\text{CO}_2$ . (2 marks)
Magnesium reacts as $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$ . Calculate the mass of magnesium oxide formed from 12 grams of magnesium. (3 marks)

Solutions

The five questions cover writing a formula, balancing an equation, calculating relative formula mass, converting mass to moles, and a full three-step reacting-mass calculation.

Step 1: Write the formula of calcium chloride (Q1)

Calcium ions carry a 2+ charge and chloride ions carry a 1- charge. To make the total charge balance to zero, two chloride ions are needed for each calcium ion:

\text{CaCl}_2.

Final answer (Q1): $\text{CaCl}_2$ .

Step 2: Balance the equation for hydrogen and oxygen (Q2)

The unbalanced equation $\text{H}_2 + \text{O}_2 \rightarrow \text{H}_2\text{O}$ has two oxygen atoms on the left but only one on the right. Place a coefficient of 2 in front of $\text{H}_2\text{O}$ , then balance hydrogen by placing 2 in front of $\text{H}_2$ :

2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}.

Final answer (Q2): $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ .

Step 3: Calculate the relative formula mass of $\text{CaCO}_3$ (Q3)

Add up the relative atomic masses of every atom in the formula, counting the three oxygen atoms separately:

M_r(\text{CaCO}_3) = 40 + 12 + 3(16) = 40 + 12 + 48 = 100.

Final answer (Q3): $M_r = 100$ .

Step 4: Find the moles in 88 g of $\text{CO}_2$ (Q4)

First calculate the relative formula mass of $\text{CO}_2$ , then divide the mass by that value:

M_r(\text{CO}_2) = 12 + 2(16) = 44, \qquad n = \frac{88}{44} = 2 \text{ moles}.

Final answer (Q4): 2 moles of $\text{CO}_2$ .

Step 5: Find the mass of magnesium oxide formed from 12 g of magnesium (Q5)

Follow the three-step method: convert mass to moles, apply the mole ratio from the balanced equation, then convert moles back to mass.

n(\text{Mg}) = \frac{12}{24} = 0.5 \text{ moles.}

The balanced equation $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$ gives a Mg to MgO ratio of $2 : 2$ , so $0.5$ moles of Mg produce $0.5$ moles of MgO.

m(\text{MgO}) = 0.5 \times (24 + 16) = 0.5 \times 40 = 20 \text{ grams.}

Final answer (Q5): 20 grams of magnesium oxide.

Sources & how we know this

Singapore-Cambridge GCE Normal (Academic) Level Science (Physics, Chemistry) and Science (Chemistry, Biology) (Syllabus 5107) — Singapore Examinations and Assessment Board (2026)
Cambridge Assessment International Education, working with SEAB on the Singapore-Cambridge GCE N(A)-Level — Cambridge Assessment International Education (2026)