What is the mole, and how does it link mass to number of particles?

The mole is the unit chemists use to count particles by weighing. One mole is the Avogadro constant of particles (about 6.02 times ten to the power 23), and the mass of one mole of a substance in grams is equal to its relative atomic or molecular mass. So the amount in moles equals the mass divided by the molar mass, and the number of particles equals the amount in moles multiplied by the Avogadro constant. The mole is the bridge between the laboratory masses we can weigh and the huge numbers of atoms and molecules taking part in a reaction.

How do you balance a chemical equation?

First write the correct formulae of all reactants and products, using the charges on the ions for ionic compounds; these formulae must not be changed. Then place numbers (coefficients) in front of the formulae so that the number of atoms of each element is the same on both sides. Count each element in turn and adjust the coefficients until they balance, then add state symbols. You balance by changing the big numbers in front, never by changing the small numbers inside a formula.

What is the limiting reagent, and why does it matter?

The limiting reagent is the reactant that is completely used up first, so it limits how much product can be made; the other reactant is in excess and some is left over. It matters because the amount of product is calculated from the limiting reagent, not from the reactant in excess. To find it, work out the amount in moles of each reactant, divide by its coefficient in the balanced equation, and the smallest result is the limiting reagent.

How do you carry out a titration calculation?

A titration calculation has three steps. First, use the known concentration and volume of one solution to find its amount in moles (moles equals concentration times volume in cubic decimetres). Second, use the mole ratio from the balanced equation to find the amount in moles of the other substance. Third, divide that amount by its volume in cubic decimetres to get its concentration, or multiply by its molar mass to get a mass. Careful unit conversion, especially cubic centimetres to cubic decimetres, is essential.

How is Stoichiometry assessed in O-Level Chemistry 6092?

It is examined across all three papers and is the most calculation-heavy topic. Paper 1 (multiple choice, 40 marks, 30 percent) tests quick mole and formula calculations. Paper 2 (structured and free response, 80 marks, 50 percent) rewards full reacting-mass, gas-volume, limiting-reagent, yield and titration calculations with correct working and units. Paper 3 (practical, 40 marks, 20 percent) includes a titration in which you record burette readings and use them in a calculation, so the practical and the arithmetic are tested together.

SingaporeChemistry

Stoichiometry and the Mole Concept (Singapore O-Level Chemistry 6092): relative masses and the mole, building and balancing formulae and equations, reacting masses and gas volumes with the limiting reagent and percentage yield, and concentration and titration

A Singapore O-Level Chemistry (SEAB 6092) overview of Stoichiometry and the Mole Concept. Relative atomic and molecular mass and the mole, building and balancing chemical formulae and equations, reacting masses and gas volumes with limiting reagent and percentage yield, and concentration and titration, with links to every dot point.

Generated by Claude Opus 4.87 min readSEAB-6092Updated 2026-06-13

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

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What this topic is really about
Relative masses and the mole
Formulae, equations and reacting quantities
Concentration and titration
How this topic is examined
Check your knowledge

What this topic is really about

Stoichiometry and the Mole Concept is the quantitative engine of chemistry: it lets you predict exactly how much of one substance reacts with or is produced from another. The unifying idea is the mole, which counts particles by weighing, and the balanced equation, which gives the ratio in which substances react. Almost every calculation in the course, from a reacting mass to a titration, starts by turning a mass or volume into moles, applying the mole ratio, and turning the answer back into a mass, volume or concentration. This guide ties the four dot points together and links to each one.

The complete set of dot-point pages for this topic, each with worked examples and questions, lives at /sg-o-level/chemistry/syllabus/stoichiometry-and-the-mole.

Relative masses and the mole

Relative masses and the mole defines relative atomic mass, relative molecular mass, the mole and the Avogadro constant, and shows how to convert between mass, amount in moles and number of particles. The central relationship is that amount in moles equals mass divided by molar mass.

Formulae, equations and reacting quantities

Chemical formulae and equations builds ionic formulae by balancing the charges on the ions and balances equations by adjusting the coefficients (never the formulae) so that atoms balance, then adds state symbols.

Mole calculations and reacting masses uses the mole ratio to find reacting masses and gas volumes (one mole of any gas occupies 24 cubic decimetres at room temperature and pressure), identifies the limiting reagent, and calculates percentage yield.

Reacting mass from a balanced equation

step Write the balanced equation

For the reaction of magnesium with oxygen:

2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}

step Find moles of the known substance

Mass of magnesium is $4.8\ \text{g}$ and its relative atomic mass is $24$ , so amount $= \dfrac{4.8}{24} = 0.20\ \text{mol}$ .

step Apply the mole ratio

From the equation, 2 mol of magnesium gives 2 mol of magnesium oxide, a 1 to 1 ratio, so $0.20\ \text{mol}$ of magnesium gives $0.20\ \text{mol}$ of magnesium oxide.

step Convert back to mass

Molar mass of magnesium oxide is $24 + 16 = 40$ , so mass $= 0.20 \times 40 = 8.0\ \text{g}$ of magnesium oxide.

Concentration and titration

Concentration and titration defines concentration in moles per cubic decimetre and grams per cubic decimetre and interconverts them, then applies the three-step titration calculation: moles of the known solution, mole ratio, then concentration or mass of the unknown. The key conversion is cubic centimetres to cubic decimetres by dividing by 1000.

How this topic is examined

Show clear, staged working. Moles of known, mole ratio, then the answer. Markers award method marks even if the final number slips.
Carry units and convert volumes. Convert cubic centimetres to cubic decimetres (divide by 1000) and use 24 cubic decimetres per mole for gases at room conditions.
Identify the limiting reagent before finding the product. The product amount comes from the limiting reagent, not the reactant in excess.

Stoichiometry and the Mole Concept is the quantitative engine of chemistry. The mole counts particles by weighing: one mole is the Avogadro constant of particles, and amount in moles equals mass divided by molar mass. Build ionic formulae from ionic charges and balance equations by adjusting coefficients, never the formulae. The universal method is to convert a mass, gas volume or solution volume into moles, apply the mole ratio from the balanced equation, then convert back; one mole of any gas occupies 24 cubic decimetres at room temperature and pressure. The limiting reagent is used up first and sets the amount of product, and percentage yield compares actual with theoretical yield. Concentration is in moles or grams per cubic decimetre, and a titration calculation has three steps: moles of the known, mole ratio, then the unknown.

Check your knowledge

A mix of recall, reasoning and calculation questions covering Stoichiometry and the Mole Concept. Attempt them under timed conditions, then check against the solutions. (Use relative atomic masses: H = 1, C = 12, O = 16, Na = 23, Cl = 35.5.)

Calculate the amount, in moles, of sodium hydroxide in 80 g of NaOH (molar mass 40). (2 marks)
Balance the equation: Na + Cl2 to NaCl. (1 mark)
Calculate the mass of carbon dioxide (molar mass 44) produced when 0.50 mol of carbon burns completely (C + O2 to CO2). (2 marks)
A 25.0 cm3 sample of hydrochloric acid is exactly neutralised by 20.0 cm3 of 0.100 mol/dm3 sodium hydroxide. Calculate the concentration of the acid (HCl + NaOH to NaCl + H2O). (3 marks)
Define the limiting reagent. (1 mark)

Solutions

Step 1: Use the mole formula to find the amount in moles (Q1)

The amount in moles is always mass divided by molar mass. This converts from the laboratory quantity you can weigh into the unit that the stoichiometry of the equation requires.

\text{amount} = \frac{\text{mass}}{\text{molar mass}} = \frac{80}{40} = 2.0\ \text{mol}.

Final answer: $2.0\ \text{mol}$

Step 2: Balance by adjusting coefficients so atoms on each side are equal (Q2)

The unbalanced equation $\text{Na} + \text{Cl}_2 \rightarrow \text{NaCl}$ has two chlorine atoms on the left but only one on the right. Placing a coefficient of $2$ in front of NaCl balances chlorine, and then placing $2$ in front of Na balances sodium.

2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl}.

Final answer: $2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl}$

Step 3: Use the mole ratio from the balanced equation, then convert back to mass (Q3)

The balanced equation $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$ shows that 1 mol of carbon produces exactly 1 mol of carbon dioxide. Multiplying the amount by the molar mass of carbon dioxide converts from moles back to grams.

Mole ratio is 1:1, so $0.50\ \text{mol}$ of carbon gives $0.50\ \text{mol}$ of $\text{CO}_2$ .

\text{mass} = 0.50 \times 44 = 22\ \text{g}.

Final answer: $22\ \text{g}$

Step 4: Three-step titration calculation (Q4)

A titration calculation always follows three stages: find the moles of the solution with the known concentration, use the mole ratio from the balanced equation to get the moles of the other substance, then divide by its volume (in dm $^3$ ) to find its concentration. The critical unit conversion is $\text{cm}^3 \div 1000 = \text{dm}^3$ .

Moles of $\text{NaOH} = 0.100 \times \dfrac{20.0}{1000} = 2.00 \times 10^{-3}\ \text{mol}$ .

The equation $\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$ has a 1:1 mole ratio, so moles of $\text{HCl} = 2.00 \times 10^{-3}\ \text{mol}$ .

\text{concentration of HCl} = \frac{2.00 \times 10^{-3}}{25.0/1000} = 0.0800\ \text{mol dm}^{-3}.

Final answer: $0.0800\ \text{mol dm}^{-3}$

Step 5: Define the limiting reagent in terms of its effect on the reaction (Q5)

The limiting reagent is the concept that explains why you cannot simply add more reactant and keep producing product. It is the reactant that is completely consumed first, so once it runs out the reaction stops regardless of how much of the other reactant remains.

Final answer: The limiting reagent is the reactant that is completely used up first in a reaction, so it determines the maximum amount of product that can be formed.

Sources & how we know this

Singapore-Cambridge GCE O-Level Chemistry (Syllabus 6092) — Singapore Examinations and Assessment Board (2026)
Cambridge Assessment International Education, working with SEAB on the Singapore-Cambridge GCE O-Level — Cambridge Assessment International Education (2026)