Formulae, equations and amounts of substance

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

Topic 1.1: Formulae, Equations and Amounts of Substance

Welcome to Unit AS 1: Basic Concepts in Physical and Inorganic Chemistry! This chapter forms the mathematical foundation of your entire CCEA AS Chemistry course. While calculations can sometimes feel intimidating, the principles here work just like recipes in cooking. Once you learn the basic conversion steps and rules, you will be able to tackle any calculation question with confidence.

1. The Carbon-12 Scale and Relative Masses

Atoms are far too small to weigh on a standard laboratory balance. Instead, chemists compare the mass of every atom against a universal standard: an atom of the Carbon-12 ($^{12}\text{C}$) isotope, which is assigned a mass of exactly 12.00 units.

Core Definitions You Must Memorise

• Relative Atomic Mass ($A_r$): The weighted average mass of an atom of an element compared to $\frac{1}{12}\text{th}$ of the mass of an atom of Carbon-12.
• Relative Isotopic Mass: The mass of an atom of an isotope 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 compared to $\frac{1}{12}\text{th}$ of the mass of an atom of Carbon-12 (used for simple covalent molecules such as $\text{H}_2\text{O}$ and $\text{CO}_2$).
• Relative Formula Mass ($M_r$): The weighted mass of a formula unit of an ionic or giant covalent compound relative to $\frac{1}{12}\text{th}$ of the mass of an atom of Carbon-12 (used for substances like $\text{NaCl}$ or $\text{SiO}_2$).
• Molar Mass ($M$): The mass in grams of one mole of a substance, expressed in $\text{g mol}^{-1}$.

Calculating Relative Formula Mass ($M_r$)

To find the $M_r$ of any compound, add together the relative atomic masses ($A_r$) of all the individual atoms in its chemical formula.
Example: Calculate the $M_r$ of calcium hydroxide, $\text{Ca(OH)}_2$.
$M_r = 40.1 + (2 \times 16.0) + (2 \times 1.0) = 74.1$

Quick Key Takeaway: All atomic and molecular mass standards are defined relative to $\frac{1}{12}\text{th}$ of the mass of a $^{12}\text{C}$ atom.

2. The Mole and Avogadro's Constant

In everyday life, we use collective words for set quantities: a pair means 2, and a dozen means 12. In chemistry, our counting word is the mole (abbreviated as $\text{mol}$).

• The Mole: The amount of substance that contains the same number of stated particles (atoms, molecules, ions, or electrons) as there are atoms in exactly 12 g of the standard Carbon-12 isotope.
• The Avogadro Constant ($L$ or $N_A$): The number of particles in one mole of any substance, equal to $6.02 \times 10^{23}\text{ mol}^{-1}$.

Essential Formulae: Mass, Moles, and Particles

$$\text{Moles } (n) = \frac{\text{Mass in grams } (m)}{\text{Molar mass } (M)}$$

$$\text{Number of Particles} = n \times L = n \times (6.02 \times 10^{23})$$

Step-by-Step Example:

Question: How many molecules are present in $4.40\text{ g}$ of carbon dioxide, $\text{CO}_2$?
Step 1: Calculate the $M_r$ of $\text{CO}_2$:
$M_r = 12.0 + (2 \times 16.0) = 44.0\text{ g mol}^{-1}$
Step 2: Calculate the number of moles ($n$):
$n = \frac{4.40}{44.0} = 0.100\text{ mol}$
Step 3: Multiply by the Avogadro constant:
$\text{Number of molecules} = 0.100 \times (6.02 \times 10^{23}) = 6.02 \times 10^{22}\text{ molecules}$

Quick Key Takeaway: Mass in grams converts to moles by dividing by molar mass ($n = \frac{m}{M}$), and moles convert to total individual particles by multiplying by $6.02 \times 10^{23}$.

3. Formulae of Compounds & Water of Crystallisation

Empirical vs Molecular Formula

• 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 one molecule of a compound.

Finding the Empirical Formula (The 4-Step Table Method)

Example: A compound contains $40.0\%$ Carbon, $6.7\%$ Hydrogen, and $53.3\%$ Oxygen by mass. Find its empirical formula.
1. Mass or %: $\text{C} = 40.0\text{ g}$, $\text{H} = 6.7\text{ g}$, $\text{O} = 53.3\text{ g}$
2. Divide by $A_r$ to get moles:
$\text{C} = \frac{40.0}{12.0} = 3.33\text{ mol}$
$\text{H} = \frac{6.7}{1.0} = 6.70\text{ mol}$
$\text{O} = \frac{53.3}{16.0} = 3.33\text{ mol}$
3. Divide all numbers by the smallest mole value ($3.33$):
$\text{C} = \frac{3.33}{3.33} = 1$
$\text{H} = \frac{6.70}{3.33} = 2.01 \approx 2$
$\text{O} = \frac{3.33}{3.33} = 1$
4. State the empirical formula: $\text{CH}_2\text{O}$

To find the Molecular Formula: If the $M_r$ of this compound is given as $60.0\text{ g mol}^{-1}$:
• Mass of empirical unit $(\text{CH}_2\text{O}) = 12.0 + (2 \times 1.0) + 16.0 = 30.0$
• Multiplier $= \frac{M_r}{\text{Empirical mass}} = \frac{60.0}{30.0} = 2$
• Molecular formula $= 2 \times (\text{CH}_2\text{O}) = \text{C}_2\text{H}_4\text{O}_2$

Hydrated Salts and Water of Crystallisation

Many ionic salts contain water trapped inside their crystal lattice. This is called water of crystallisation.
• Hydrated Salt: A salt containing water of crystallisation (e.g., $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$).
• Anhydrous Salt: The salt remaining after all water of crystallisation has been removed (e.g., $\text{CuSO}_4$).

Exam Technique for Water of Crystallisation:
Treat water ($\text{H}_2\text{O}$, $M_r = 18.0$) like a single chemical unit in your empirical formula calculation.
1. Calculate the mass of the anhydrous salt and the mass of water lost.
2. Divide the anhydrous salt mass by its $M_r$, and the water mass by $18.0$.
3. Divide both values by the moles of anhydrous salt to find the value of $x$ in $\text{Salt} \cdot x\text{H}_2\text{O}$.

Quick Key Takeaway: Empirical formula gives the simplest ratio; molecular formula gives the true number of atoms. For hydrated salts, treat $\text{H}_2\text{O}$ as a single unit with $M_r = 18.0$.

4. Solutions and Concentrations

A solution is formed when a solute dissolves in a solvent. Concentration measures how much solute is dissolved per unit volume of solution.

Units and Core Formulae

• Volume conversion: In chemistry, standard volume is measured in cubic decimetres ($\text{dm}^3$).
$1\text{ dm}^3 = 1000\text{ cm}^3 = 1\text{ litre}$
To convert $\text{cm}^3$ to $\text{dm}^3$, divide by $1000$.

• Molar Concentration ($\text{mol dm}^{-3}$):

$$\text{Moles } (n) = c \times V\text{ (in dm}^3\text{)} = \frac{c \times V\text{ (in cm}^3\text{)}}{1000}$$

• Mass Concentration ($\text{g dm}^{-3}$):

$$\text{Mass concentration } (\text{g dm}^{-3}) = \text{Molar concentration } (\text{mol dm}^{-3}) \times M_r$$

Step-by-Step Example:

Question: Calculate the concentration in $\text{mol dm}^{-3}$ of a solution containing $5.30\text{ g}$ of $\text{Na}_2\text{CO}_3$ in $250\text{ cm}^3$ of water.
Step 1: Find $M_r$ of $\text{Na}_2\text{CO}_3 = (2 \times 23.0) + 12.0 + (3 \times 16.0) = 106.0\text{ g mol}^{-1}$
Step 2: Calculate moles of solute: $n = \frac{5.30}{106.0} = 0.0500\text{ mol}$
Step 3: Convert volume to $\text{dm}^3$: $V = \frac{250}{1000} = 0.250\text{ dm}^3$
Step 4: Calculate concentration: $c = \frac{n}{V} = \frac{0.0500}{0.250} = 0.200\text{ mol dm}^{-3}$

Quick Key Takeaway: Always ensure solution volumes are converted into $\text{dm}^3$ by dividing by $1000$ before using $n = c \times V$.

5. Gas Calculations

1. Molar Gas Volume at RTP

At Room Temperature and Pressure (RTP: $20^\circ\text{C} / 293\text{ K}$ and $101\text{ kPa}$), one mole of any gas occupies a volume of approximately $24.0\text{ dm}^3$ (or $24,000\text{ cm}^3$).

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

2. The Ideal Gas Equation: $pV = nRT$

When conditions are not at RTP, we use the ideal gas equation. To get the correct answer, you must use standard SI units!

$$pV = nRT$$

• $p$ = Pressure in Pascals ($\text{Pa}$): If given in $\text{kPa}$, multiply by $1000$ ($1\text{ kPa} = 10^3\text{ Pa}$).
• $V$ = Volume in cubic metres ($\text{m}^3$):
From $\text{dm}^3$ to $\text{m}^3$: divide by $1000$ (or $\times 10^{-3}$)
From $\text{cm}^3$ to $\text{m}^3$: divide by $10^6$ (or $\times 10^{-6}$)
• $n$ = Amount of substance in moles ($\text{mol}$).
• $R$ = Molar gas constant: $8.31\text{ J K}^{-1}\text{ mol}^{-1}$ (provided on your data sheet).
• $T$ = Temperature in Kelvin ($\text{K}$): $T(\text{K}) = \theta(^\circ\text{C}) + 273$.

Ideal Gas Unit Conversion Memory Trick

Think P-V-T:
• Pressure: Needs to be in single Pa (not kilo-Pa).
• Volume: Needs to be in big $\text{m}^3$.
• Temperature: Must always be Kelvin (add $273$).

Quick Key Takeaway: At RTP, use $V = n \times 24.0\text{ dm}^3$. At other temperatures/pressures, use $pV = nRT$ with strict SI units ($\text{Pa}$, $\text{m}^3$, $\text{K}$).

6. Chemical Equations, Stoichiometry, and Ionic Equations

State Symbols

Always include state symbols when requested:
• $(s)$: Solid
• $(l)$: Liquid (e.g., pure $\text{H}_2\text{O}(l)$)
• $(g)$: Gas
• $(aq)$: Aqueous (dissolved in water)

Writing Balanced Ionic Equations

Ionic equations show only the reacting species, leaving out spectator ions (ions that do not change oxidation state or physical state during the reaction).

Step-by-Step Method:
1. Write the full balanced molecular equation:
$\text{HCl}(aq) + \text{NaOH}(aq) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l)$
2. Split all soluble aqueous ionic compounds into separate ions:
$\text{H}^+(aq) + \text{Cl}^-(aq) + \text{Na}^+(aq) + \text{OH}^-(aq) \rightarrow \text{Na}^+(aq) + \text{Cl}^-(aq) + \text{H}_2\text{O}(l)$
3. Cancel spectator ions present on both sides ($\text{Na}^+$ and $\text{Cl}^-$):
$\text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l)$

Stoichiometric Reacting Quantity Calculations

The Universal Calculation Pathway:
$$\text{Known Mass/Vol} \xrightarrow{\div M_r \text{ or } \div 24} \text{Moles of Known} \xrightarrow{\text{Molar Ratio}} \text{Moles of Unknown} \xrightarrow{\times M_r \text{ or } \times 24} \text{Target Mass/Vol}$$

Quick Key Takeaway: Stoichiometry uses the molar ratios from the balanced chemical equation to convert moles of a known substance into moles of an unknown substance.

7. Reaction Efficiency: Percentage Yield & Atom Economy

1. Percentage Yield

Percentage yield measures the practical efficiency of a laboratory reaction, comparing the amount of product actually collected to what could theoretically be made.

$$\% \text{ Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Maximum Yield}} \times 100$$

Why is actual yield often less than $100\%$?
• The reaction may be reversible or incomplete.
• Side reactions may produce unwanted alternative products.
• Product is lost during practical purification steps (e.g., left on filter paper or glassware).

2. Percentage Atom Economy

Atom economy is a theoretical measure of how sustainably a reaction is designed. It tells you what percentage of the mass of all reactants ends up in the desired useful product rather than waste products.

$$\% \text{ Atom Economy} = \frac{\text{Mass of Desired Product}}{\text{Total Mass of All Reactants}} \times 100 = \frac{M_r \text{ of Desired Product}}{\sum M_r \text{ of all Reactants}} \times 100$$

Key Differences Between Yield and Atom Economy:
• Percentage Yield: An experimental result based on practical measurements in the lab.
• Atom Economy: A theoretical value calculated directly from the stoichiometric equation, regardless of how carefully the experiment is carried out.
• Reactions with only one single product (addition reactions) always have an atom economy of $100\%$.

Quick Key Takeaway: High yield reduces practical material loss; high atom economy reduces unwanted chemical waste and promotes green, sustainable chemistry.

8. Top CCEA Examiner Pitfalls & Revision Checklist

Avoid these common mistakes reported by CCEA examiners in past AS 1 papers:

• Premature Rounding: Keep full unrounded figures in your calculator display throughout intermediate steps. Round only your final answer, typically to 3 significant figures unless specified otherwise.
• Formula Mass vs Stoichiometric Numbers: When finding molar mass ($M_r$), calculate the mass of one formula unit. Do not multiply the formula mass by the big balancing coefficient in the chemical equation at the initial conversion stage.
• Gas Equation Unit Trap: Forgetting to convert $\text{cm}^3$ to $\text{m}^3$ (divide by $10^6$) or $\text{kPa}$ to $\text{Pa}$ (multiply by $1000$) in $pV = nRT$.
• Inverting Ratios in Hydrated Salts: Always divide the moles of water by the moles of anhydrous salt ($\frac{n(\text{H}_2\text{O})}{n(\text{Salt})}$) to find $x$, never the other way around.
• Ionic Charges: Ensure that both the atoms and the total electrical charges balance on both sides of an ionic equation.

Quick check

Can you answer these now?

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

What is the standard standard isotope against which relative atomic and molecular masses are defined?

All relative mass values are defined compared to $\frac{1}{12}\text{th}$ of the mass of an atom of the Carbon-12 ($^{12}\text{C}$) isotope.

What is a mole and what is the value of the Avogadro constant?

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

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

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

How do you calculate the value of $x$ in a hydrated salt formula $\text{Salt} \cdot x\text{H}_2\text{O}$?

Divide the moles of water lost ($M_r = 18.0$) by the moles of the anhydrous salt: $x = \frac{n(\text{H}_2\text{O})}{n(\text{anhydrous salt})}$.

How is the molar concentration of a solution calculated from moles and volume?

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

What volume does one mole of any gas occupy at Room Temperature and Pressure (RTP)?

One mole of any gas occupies $24.0\text{ dm}^3$ (or $24,000\text{ cm}^3$) at RTP ($20^\circ\text{C}$ and $101\text{ kPa}$).

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 $\text{m}^3$, amount $n$ in $\text{mol}$, gas constant $R$ in $\text{J K}^{-1}\text{ mol}^{-1}$, and temperature $T$ in $\text{K}$.

How do you write a balanced ionic equation from a full molecular equation?

Split all soluble aqueous ionic compounds into separate ions and cancel out the spectator ions that remain unchanged on both sides.

What is the formula for calculating the percentage yield of a reaction?

$\% \text{ Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Maximum Yield}} \times 100$

How is percentage atom economy defined and calculated?

$\% \text{ Atom Economy} = \frac{M_r \text{ of Desired Product}}{\sum M_r \text{ of all Reactants}} \times 100$

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Formulae, equations and amounts of substance

Topic 1.1: Formulae, Equations and Amounts of Substance

1. The Carbon-12 Scale and Relative Masses

Core Definitions You Must Memorise

Calculating Relative Formula Mass (\(M_r\))

2. The Mole and Avogadro's Constant

Essential Formulae: Mass, Moles, and Particles

Step-by-Step Example:

3. Formulae of Compounds & Water of Crystallisation

Empirical vs Molecular Formula

Finding the Empirical Formula (The 4-Step Table Method)

Hydrated Salts and Water of Crystallisation

4. Solutions and Concentrations

Units and Core Formulae

Step-by-Step Example:

5. Gas Calculations

1. Molar Gas Volume at RTP

2. The Ideal Gas Equation: \(pV = nRT\)

Ideal Gas Unit Conversion Memory Trick

6. Chemical Equations, Stoichiometry, and Ionic Equations

State Symbols

Writing Balanced Ionic Equations

Stoichiometric Reacting Quantity Calculations

7. Reaction Efficiency: Percentage Yield & Atom Economy

1. Percentage Yield

2. Percentage Atom Economy

8. Top CCEA Examiner Pitfalls & Revision Checklist

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