Formulae, equations and amounts of substance

Chemistry 1110 · CCEA A Level · 10 min read · Updated: 31 Aug 2026

Welcome to Formulae, Equations, and Amounts of Substance!

Welcome to one of the most fundamental chapters in AS Chemistry! If you have ever baked a cake, you know that measuring ingredients precisely is the secret to success. Chemistry works in exactly the same way. Atoms and molecules react in exact ratios, but because they are unimaginably tiny, we cannot count them individually. Instead, we "count by weighing".

In this chapter, you will master the language of chemical quantities: the mole, chemical formulae, reacting masses, solutions, and gas volumes. Don't worry if quantitative chemistry has felt intimidating in the past—we will break down every single calculation into simple, step-by-step recipes!

1. Relative Masses and the Mole Concept

Relative Atomic Mass (\(A_r\)) and Relative Molecular Mass (\(M_r\))

Individual atoms are far too light to weigh in grams on a standard laboratory balance. Instead, we compare the mass of every atom to a universally agreed standard: the carbon-12 isotope (\(^{12}\text{C}\)).

Relative Atomic Mass (\(A_r\)): The weighted average mass of an atom of an element relative to \(\frac{1}{12}\text{th}\) of the mass of an atom of carbon-12.

Relative Molecular Mass (\(M_r\)): The average mass of a molecule relative to \(\frac{1}{12}\text{th}\) of the mass of an atom of carbon-12. (For giant ionic structures like \(\text{NaCl}\), we often use the term Relative Formula Mass).

Example: To find the \(M_r\) of sulfuric acid (\(\text{H}_2\text{SO}_4\)):
\(M_r = (2 \times 1.0) + (1 \times 32.1) + (4 \times 16.0) = 98.1\)

What is a Mole?

A mole (abbreviated as \(\text{mol}\)) is simply a counting unit, just like a "pair" means \(2\) or a "dozen" means \(12\). One mole is the amount of substance that contains the same number of particles as there are atoms in exactly \(12\text{ g}\) of carbon-12.

This number is called the Avogadro Constant (\(L\) or \(N_A\)):

\(L = 6.02 \times 10^{23}\text{ mol}^{-1}\)

Did you know? If you had one mole of marbles, they would cover the entire surface of the Earth to a depth of several miles! That is how vast \(6.02 \times 10^{23}\) is.

Connecting Mass and Moles

The central equation connecting mass and moles is:

\(\text{Number of moles } (n) = \frac{\text{Mass in grams } (m)}{\text{Molar mass in g mol}^{-1} (M)}\)

Or simply: \(n = \frac{m}{M_r}\)

Memory Triangle: Picture \(m\) on top, with \(n\) and \(M_r\) on the bottom.
To find mass: \(m = n \times M_r\)
To find molar mass: \(M_r = \frac{m}{n}\)

Quick Review: Relative Masses & Moles

• All relative masses are measured against \(\frac{1}{12}\text{th}\) the mass of a \(^{12}\text{C}\) atom.
• \(1\text{ mol} = 6.02 \times 10^{23}\) particles.
• Key Formula: \(n = \frac{m}{M_r}\).

2. Empirical and Molecular Formulae

Definitions

Empirical Formula: The simplest whole-number ratio of atoms of each element present in a compound.

Molecular Formula: The actual number of atoms of each element present in a molecule of a compound.

Finding the Empirical Formula: A Step-by-Step Method

When given mass data or percentage compositions by mass, follow this foolproof \(4\)-step table:

1. List the elements present.
2. Write down the mass or percentage of each element.
3. Divide each mass by its relative atomic mass (\(A_r\)) to calculate moles.
4. Divide all values by the smallest mole value to find the simplest whole-number ratio.

Worked Example: A compound contains \(40.0\%\) carbon, \(6.7\%\) hydrogen, and \(53.3\%\) oxygen by mass. Find its empirical formula.

• Moles of \(\text{C} = \frac{40.0}{12.0} = 3.33\text{ mol}\)
• Moles of \(\text{H} = \frac{6.7}{1.0} = 6.70\text{ mol}\)
• Moles of \(\text{O} = \frac{53.3}{16.0} = 3.33\text{ mol}\)
• Divide by the smallest (\(3.33\)):
\(\text{C} = \frac{3.33}{3.33} = 1\), \(\text{H} = \frac{6.70}{3.33} = 2.01 \approx 2\), \(\text{O} = \frac{3.33}{3.33} = 1\)
• Empirical Formula = \(\text{CH}_2\text{O}\)

Finding the Molecular Formula from the Empirical Formula

To convert an empirical formula into a molecular formula, you need the compound's relative molecular mass (\(M_r\)):

\(\text{Multiplier } (k) = \frac{\text{Relative Molecular Mass } (M_r)}{\text{Empirical Formula Mass}}\)

If the compound above has an \(M_r\) of \(180.0\):
Mass of \(\text{CH}_2\text{O} = 12.0 + (2 \times 1.0) + 16.0 = 30.0\)
\(\text{Multiplier} = \frac{180.0}{30.0} = 6\)
Molecular Formula = \(\text{C}_6\text{H}_{12}\text{O}_6\)

Hydrated Salts and Water of Crystallisation

Many ionic compounds trap water molecules inside their crystal lattice. This trapped water is called the water of crystallisation (e.g., \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\)).

When heated strongly, the water evaporates, leaving the anhydrous salt behind:

\(\text{CuSO}_4 \cdot x\text{H}_2\text{O}\text{ (s)} \rightarrow \text{CuSO}_4\text{ (s)} + x\text{H}_2\text{O}\text{ (g)}\)

Common Exam Trick: Treat the anhydrous salt formula (e.g., \(\text{CuSO}_4\)) as one entity and \(\text{H}_2\text{O}\) as the other entity. Calculate the moles of each, then divide by the moles of the anhydrous salt to find \(x\).

Quick Review: Formulae

• Empirical = simplest whole number ratio; Molecular = actual number of atoms.
• Divide by \(A_r\) first, then divide by the smallest number.
• For hydrated salts, find the ratio of \(\text{moles of anhydrous salt} : \text{moles of }\text{H}_2\text{O}\).

3. Chemical Equations and Reacting Quantities

Balancing Equations and State Symbols

Matter cannot be created or destroyed. A balanced equation must have equal numbers of each type of atom on both sides.

Always remember to include state symbols when requested:
• \(\text{(s)}\) = solid
• \(\text{(l)}\) = pure liquid (e.g., liquid water \(\text{H}_2\text{O}\))
• \(\text{(g)}\) = gas
• \(\text{(aq)}\) = aqueous (dissolved in water)

Writing Ionic Equations

Ionic equations show only the reacting species and leave out spectator ions (ions that remain unchanged in solution on both sides of the arrow).

Step-by-step Guide to Ionic Equations:
1. Write the full balanced molecular equation with state symbols.
2. Split all soluble ionic compounds (indicated by \(\text{(aq)}\)) into their separate ions.
3. Leave solids \(\text{(s)}\), liquids \(\text{(l)}\), and gases \(\text{(g)}\) completely intact!
4. Cancel out all spectator ions that appear identically on both sides.

Example: Reaction between hydrochloric acid and sodium hydroxide:
Full equation: \(\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}\)
Split ions: \(\text{H}^+\text{(aq)} + \text{Cl}^-\text{(aq)} + \text{Na}^+\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{Na}^+\text{(aq)} + \text{Cl}^-\text{(aq)} + \text{H}_2\text{O(l)}\)
Cancel spectator ions (\(\text{Na}^+\) and \(\text{Cl}^-\)):
Net Ionic Equation: \(\text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{H}_2\text{O(l)}\)

Reacting Mass Calculations (Stoichiometry)

Use the 3-Step Mole Pipeline for reacting mass problems:

Step 1: Calculate the moles of the substance you know the mass of (\(n = \frac{m}{M_r}\)).
Step 2: Use the stoichiometric ratio from the balanced equation to find the moles of the desired substance.
Step 3: Convert those moles back into mass (\(m = n \times M_r\)).

Limiting Reagents

The limiting reagent is the reactant that is completely used up first, stopping the reaction and determining the maximum theoretical amount of product formed. Reactants present in quantities greater than required are in excess.

Tip: To identify the limiting reagent, calculate \(\frac{\text{moles available}}{\text{stoichiometric coefficient}}\) for each reactant. The one with the smallest value is your limiting reactant!

4. Solution Chemistry and Titrations

Concentration of Solutions

Concentration tells us how much solute is dissolved in a given volume of solution.

Key Units:
• Molar concentration: \(\text{mol dm}^{-3}\) (often written as \(\text{M}\))
• Mass concentration: \(\text{g dm}^{-3}\)

Vital Volume Conversion: Chemistry volumes are often measured in \(\text{cm}^3\), but concentration is per \(\text{dm}^3\) (\(1\text{ dm}^3 = 1000\text{ cm}^3 = 1\text{ litre}\)).
To convert \(\text{cm}^3\) to \(\text{dm}^3\), divide by \(1000\)!

Core Formulae:
\(\text{Moles } (n) = \text{Concentration } (c\text{ in mol dm}^{-3}) \times \text{Volume } (V\text{ in dm}^{-3})\)
\(n = \frac{c \times V\text{ (in cm}^3\text{)}}{1000}\)

To convert between \(\text{mol dm}^{-3}\) and \(\text{g dm}^{-3}\):
\(\text{Concentration in g dm}^{-3} = \text{Concentration in mol dm}^{-3} \times M_r\)

Volumetric Titration Calculations

Titration is an analytical technique used to find the unknown concentration of an acid or alkali by reacting it with a standard solution (a solution of accurately known concentration).

Step-by-step Titration Pipeline:
1. Calculate the mean titre using only concordant results (titres within \(\pm 0.10\text{ cm}^3\) of each other). Reject the initial rough titre.
2. Calculate the moles of the standard solution used (\(n = c \times V\)).
3. Use the balanced equation ratio to determine the moles of the unknown substance.
4. Calculate the unknown concentration: \(c = \frac{n}{V}\).

Quick Review: Solution Calculations

• Always convert \(\text{cm}^3\) to \(\text{dm}^3\) by dividing by \(1000\).
• Concordant titres must be within \(0.10\text{ cm}^3\).
• \(\text{Mass concentration (g dm}^{-3}\text{)} = \text{Molar concentration (mol dm}^{-3}\text{)} \times M_r\).

5. Gas Calculations

1. Molar Gas Volume at Room Temperature and Pressure (r.t.p.)

Avogadro’s Law: Equal volumes of all gases under the same conditions of temperature and pressure contain the same number of molecules.

At room temperature and pressure (\(20\ ^\circ\text{C}\) / \(293\text{ K}\) and \(1\text{ atm}\) / \(101.3\text{ kPa}\)), one mole of any gas occupies a volume of approximately \(24.0\text{ dm}^3\) (or \(24000\text{ cm}^3\)).

\(\text{Moles of gas } (n) = \frac{\text{Volume in dm}^3}{24.0} = \frac{\text{Volume in cm}^3}{24000}\)

2. The Ideal Gas Equation

When conditions are not at standard room temperature and pressure, we use the Ideal Gas Equation:

\(pV = nRT\)

Where:
• \(p\) = Pressure in Pascals (\(\text{Pa}\))
• \(V\) = Volume in cubic metres (\(\text{m}^3\))
• \(n\) = Amount of gas in moles (\(\text{mol}\))
• \(R\) = Molar gas constant \(= 8.314\text{ J K}^{-1}\text{ mol}^{-1}\)
• \(T\) = Temperature in Kelvin (\(\text{K}\))

WARNING: Unit Conversions are the #1 Pitfall!

Make sure you convert every quantity into SI units before plugging values into \(pV = nRT\):
• Pressure: \(\text{kPa} \times 1000 = \text{Pa}\)
• Temperature: \(^\circ\text{C} + 273 = \text{K}\)
• Volume: \(\text{dm}^3 \div 1000 = \text{m}^3\)
• Volume: \(\text{cm}^3 \div 1\,000\,000 = \text{m}^3\) (or \(\text{cm}^3 \times 10^{-6} = \text{m}^3\))

Example: Calculate the volume occupied by \(0.500\text{ mol}\) of nitrogen gas at \(150\text{ kPa}\) and \(25\ ^\circ\text{C}\).

• \(p = 150 \times 1000 = 150\,000\text{ Pa}\)
• \(T = 25 + 273 = 298\text{ K}\)
• \(n = 0.500\text{ mol}\)
• Rearrange: \(V = \frac{nRT}{p}\)
• \(V = \frac{0.500 \times 8.314 \times 298}{150\,000} = 0.00826\text{ m}^3\) (\(= 8.26\text{ dm}^3\))

Quick Review: Gas Calculations

• At r.t.p., use \(V = n \times 24.0\text{ dm}^3\).
• For all other conditions, use \(pV = nRT\).
• Always double-check your units: \(\text{Pa}\), \(\text{m}^3\), \(\text{K}\)!

6. Percentage Yield and Atom Economy

Chemical manufacturers need processes to be both efficient and sustainable. Two key metrics measure how well a reaction performs:

Percentage Yield

Percentage Yield compares the mass of product actually obtained in an experiment with the maximum theoretical mass predicted by stoichiometry.

\(\text{Percentage Yield} = \left( \frac{\text{Actual Mass of Product Obtained}}{\text{Theoretical Maximum Mass}} \right) \times 100\)

Why is percentage yield almost never \(100\%\)?
• The reaction may be reversible and reach equilibrium.
• Side reactions may produce unexpected by-products.
• Some product is lost during mechanical transfer, separation, or purification (e.g., sticking to glassware or filter paper).
• Reactants may not be \(100\%\) pure.

Atom Economy

Atom Economy is a measure of the proportion of starting materials that end up as useful products. It is a fundamental concept in Green Chemistry.

\(\text{Atom Economy} = \left( \frac{\text{Molecular mass of desired product}}{\text{Total molecular mass of all reactants}} \right) \times 100\)

Note: You must take stoichiometric balancing numbers into account when calculating the total molar masses!

Percentage Yield vs. Atom Economy: Know the Difference

• Percentage Yield is practical: it measures how successfully you carried out a reaction in the lab.
• Atom Economy is theoretical: it measures how wasteful the balanced chemical reaction equation is by design.
• Addition reactions have \(100\%\) atom economy because only one product is formed.
• Substitution or elimination reactions have lower atom economies because waste by-products are produced alongside the desired product.

Quick Review: Reaction Efficiency

• \(\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\)
• \(\text{Atom Economy} = \frac{\text{Total } M_r \text{ of desired product}}{\text{Total } M_r \text{ of all reactants}} \times 100\)
• High atom economy \(\rightarrow\) less waste \(\rightarrow\) greener and more sustainable chemistry!

Master Summary of Core Formulae

• Mass & Moles: \(n = \frac{m}{M_r}\)
• Solutions: \(n = c \times V\text{ (in dm}^3\text{)} = \frac{c \times V\text{ (in cm}^3\text{)}}{1000}\)
• Concentration conversion: \(\text{g dm}^{-3} = \text{mol dm}^{-3} \times M_r\)
• Gases at r.t.p.: \(V\text{ (dm}^3\text{)} = n \times 24.0\)
• Ideal Gas: \(pV = nRT\) (with \(p\) in \(\text{Pa}\), \(V\) in \(\text{m}^3\), \(T\) in \(\text{K}\))
• Percentage Yield: \(\frac{\text{Actual}}{\text{Theoretical}} \times 100\)
• Atom Economy: \(\frac{\text{Mass of Desired Product}}{\text{Total Mass of Reactants}} \times 100\)

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the definition of relative atomic mass (\(A_r\))?

The weighted average mass of an atom of an element relative to \(\frac{1}{12}\text{th}\) of the mass of an atom of carbon-12.

How is a mole defined in chemistry, and what is the value of the Avogadro constant?

A mole is the amount of substance containing the same number of particles as there are atoms in exactly \(12\text{ g}\) of carbon-12; this number is the Avogadro constant, \(6.02 \times 10^{23}\text{ mol}^{-1}\).

What is the formula linking mass, moles, and molar mass?

\(n = \frac{m}{M_r}\), where \(n\) is the number of moles, \(m\) is mass in grams, and \(M_r\) is molar mass in \(\text{g mol}^{-1}\).

What is the difference between an empirical formula and a molecular formula?

An empirical formula gives the simplest whole-number ratio of atoms in a compound, while a molecular formula gives the actual number of atoms of each element in a molecule.

How is the formula of a hydrated salt containing water of crystallisation determined experimentally?

By heating the hydrated salt to remove water, then determining the mole ratio of anhydrous salt to water: \(\text{moles of anhydrous salt} : \text{moles of }\text{H}_2\text{O}\).

What are spectator ions, and how are they treated when writing an ionic equation?

Spectator ions are ions that remain unchanged in solution on both sides of a reaction equation; they are omitted from the net ionic equation.

What is a limiting reagent, and how is it identified in a chemical reaction?

The limiting reagent is the reactant that is completely consumed first; it is identified by finding which reactant provides the lowest value of \(\frac{\text{moles available}}{\text{stoichiometric coefficient}}\).

How are moles calculated from solution concentration and volume in \(\text{cm}^3\)?

\(n = \frac{c \times V}{1000}\), where \(c\) is concentration in \(\text{mol dm}^{-3}\) and \(V\) is volume in \(\text{cm}^3\).

What criteria must titration results meet to be considered concordant?

Titres must be within \(\pm 0.10\text{ cm}^3\) of each other, excluding the initial rough titre when calculating the mean.

What volume does one mole of any gas occupy at room temperature and pressure (r.t.p.)?

\(24.0\text{ dm}^3\) (or \(24000\text{ cm}^3\)), so \(n = \frac{V\text{ (dm}^3\text{)}}{24.0}\).

What are the correct SI units required for each variable in the Ideal Gas Equation \(pV = nRT\)?

Pressure (\(p\)) in \(\text{Pa}\), volume (\(V\)) in \(\text{m}^3\), moles (\(n\)) in \(\text{mol}\), and temperature (\(T\)) in \(\text{K}\).

How is percentage yield calculated, and why is it rarely \(100\%\)?

\(\text{Percentage Yield} = \left(\frac{\text{Actual Mass}}{\text{Theoretical Maximum Mass}}\right) \times 100\); it is below \(100\%\) due to incomplete reactions, side reactions, or mechanical product loss.

How is atom economy defined, and how does it differ from percentage yield?

Atom economy is \(\left(\frac{\text{Total } M_r \text{ of desired product}}{\text{Total } M_r \text{ of all reactants}}\right) \times 100\); it measures theoretical reaction waste rather than practical experimental efficiency.

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