Quantitative chemistry

Q: What is relative atomic mass (\(A_r\))?

The relative atomic mass (\(A_r\)) of an element is the average mass of an atom of that element compared to \(\frac{1}{12}\text{th}\) the mass of a carbon-12 atom.

Q: What is the relationship between one mole of a substance and its formula mass?

The mass of 1 mole of any substance in grams is numerically equal to its relative formula mass (\(M_r\)) or relative atomic mass (\(A_r\)).

Q: What is the difference between a limiting reactant and an excess reactant?

The limiting reactant is completely used up first and limits the amount of product formed, whereas the excess reactant is left over when the reaction stops.

Chemistry 1110 · CCEA GCSE · 9 min read · Updated: 31 Aug 2026

Welcome to Quantitative Chemistry (Unit 1.7)

Have you ever wondered how chemists measure out the exact right amounts of chemicals so nothing goes to waste? Because atoms and molecules are far too small to count individually with tweezers, chemists weigh them instead! Quantitative chemistry is simply the chemistry of measuring amounts, masses, and yields. Don't worry if maths isn't your favourite subject — every calculation here follows a clear, step-by-step recipe that you can master with practice.

1. Relative Masses and Percentage Composition

What is Relative Atomic Mass (\(A_r\))?

Atoms are unimaginably tiny. Rather than working with impossibly small numbers in grams, chemists compare the mass of every atom to a standard benchmark: an atom of carbon-12.

Official Definition: The relative atomic mass (\(A_r\)) of an element is the average mass of an atom of that element compared to \(\frac{1}{12}\text{th}\) the mass of a carbon-12 atom.

Memory Tip: You do not need to memorise \(A_r\) values — they are always provided on your CCEA Periodic Table in the exam!

What is Relative Formula Mass (\(M_r\))?

The relative formula mass (\(M_r\)) (sometimes called relative molecular mass) is simply the sum of all the relative atomic masses (\(A_r\)) of every atom shown in the chemical formula.

Step-by-Step Example: Calculating \(M_r\) for Water (\(\text{H}_2\text{O}\))
1. Identify the atoms: 2 Hydrogen atoms and 1 Oxygen atom.
2. Look up the \(A_r\) values: \(A_r(\text{H}) = 1\), \(A_r(\text{O}) = 16\).
3. Add them together: \(M_r = (2 \times 1) + 16 = 18\).

Step-by-Step Example: Calculating \(M_r\) with Brackets — \(\text{Ca(OH)}_2\)
1. The little '2' outside the bracket multiplies everything inside: 1 Calcium, 2 Oxygens, 2 Hydrogens.
2. Look up \(A_r\): \(A_r(\text{Ca}) = 40\), \(A_r(\text{O}) = 16\), \(A_r(\text{H}) = 1\).
3. Calculate: \(M_r = 40 + (2 \times 16) + (2 \times 1) = 40 + 32 + 2 = 74\).

Examiner Warning — Avoid this Common Mistake:
Never include the big balancing number when calculating \(M_r\)! If an equation shows \(2\text{MgO}\), the \(M_r\) of magnesium oxide is just \(24 + 16 = 40\), not 80. The big number 2 tells you how many moles react, not the formula mass itself.

Percentage Composition by Mass

This calculation tells you what percentage of a compound's total mass comes from one particular element.

The Formula:
\(\text{Percentage by mass} = \frac{\text{Total } A_r \text{ of the element in the formula}}{M_r \text{ of the compound}} \times 100\)

Example: What is the percentage by mass of oxygen in carbon dioxide (\(\text{CO}_2\))?
1. Total \(A_r\) of oxygen in \(\text{CO}_2 = 2 \times 16 = 32\).
2. \(M_r\) of \(\text{CO}_2 = 12 + (2 \times 16) = 44\).
3. \(\text{Percentage of Oxygen} = \frac{32}{44} \times 100 = 72.7\%\).

Key Takeaway: \(M_r\) is found by adding up all the atomic masses in the formula. Percentage composition shows the fraction of that mass contributed by one element.

2. The Mole and Mass Conversions

What is a Mole?

Just like a "baker's dozen" means 13 items and a "pair" means 2, the mole is the standard unit chemists use to measure chemical amounts.

The Golden Rule of Moles: The mass of 1 mole of any substance in grams is numerically equal to its relative formula mass (\(M_r\)) or relative atomic mass (\(A_r\)).
For example, the \(M_r\) of \(\text{H}_2\text{O}\) is 18, so 1 mole of water weighs exactly \(18\text{ g}\). The \(A_r\) of carbon is 12, so 1 mole of carbon weighs exactly \(12\text{ g}\).

The Core Mole Equation

\(\text{Number of moles} = \frac{\text{Mass (in g)}}{M_r \text{ (or } A_r \text{)}}\)

You can rearrange this equation using a formula triangle:
• \(\text{Mass (in g)} = \text{Moles} \times M_r\)
• \(M_r = \frac{\text{Mass (in g)}}{\text{Moles}}\)

Example: How many moles are present in \(20\text{ g}\) of sodium hydroxide (\(\text{NaOH}\))?
1. Calculate \(M_r(\text{NaOH}) = 23 + 16 + 1 = 40\).
2. \(\text{Moles} = \frac{\text{Mass}}{M_r} = \frac{20}{40} = 0.5\text{ moles}\).

Industrial Mass Conversions

In chemical manufacturing, reactions happen in kilograms (\(\text{kg}\)) or tonnes (\(\text{t}\)). Before putting numbers into the mole equation, always convert the mass to grams (\(\text{g}\)) first!

• \(1\text{ kilogram (kg)} = 1{,}000\text{ g} = 10^3\text{ g}\) (multiply by \(1{,}000\))
• \(1\text{ tonne (t)} = 1{,}000\text{ kg} = 1{,}000{,}000\text{ g} = 10^6\text{ g}\) (multiply by \(1{,}000{,}000\))

Example: Convert \(2.5\text{ tonnes}\) to grams:
\(2.5 \times 1{,}000{,}000 = 2{,}500{,}000\text{ g}\) (or \(2.5 \times 10^6\text{ g}\)).

Key Takeaway: Mass must always be in grams. Divide mass by \(M_r\) to find moles.

3. Balanced Equations, Reacting Masses, and Limiting Reactants

The Law of Conservation of Mass

During a chemical reaction, no atoms are created and no atoms are destroyed. The total mass of the reactants is always exactly equal to the total mass of the products.

Using Balanced Equations: Stoichiometric Mole Ratios

The big numbers in front of chemical formulas (coefficients) give you the mole ratio of the reactants and products.

Consider the reaction: \(2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}\)
This tells us: 2 moles of \(\text{Mg}\) react with 1 mole of \(\text{O}_2\) to produce 2 moles of \(\text{MgO}\).

The 3-Step Reacting Mass Calculation Method

Whenever you are asked to calculate the mass of a product from a given mass of reactant, follow these three reliable steps:

Step 1: Calculate moles of the known substance
\(\text{Moles} = \frac{\text{Mass}}{\text{M}_r}\)

Step 2: Use the balanced equation ratio to find moles of the target substance
Use the stoichiometric ratio between the known and unknown substances.

Step 3: Calculate the mass of the target substance
\(\text{Mass} = \text{Moles} \times M_r\)

Worked Example: What mass of magnesium oxide (\(\text{MgO}\)) is produced when \(12\text{ g}\) of magnesium (\(\text{Mg}\)) is completely burned in oxygen?
Equation: \(2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}\)
• Step 1: Moles of \(\text{Mg} = \frac{\text{Mass}}{A_r} = \frac{12}{24} = 0.5\text{ moles}\).
• Step 2: Ratio of \(\text{Mg} : \text{MgO}\) is \(2 : 2\) (which simplifies to \(1 : 1\)). Therefore, moles of \(\text{MgO} = 0.5\text{ moles}\).
• Step 3: \(M_r(\text{MgO}) = 24 + 16 = 40\). Mass of \(\text{MgO} = \text{Moles} \times M_r = 0.5 \times 40 = 20\text{ g}\).

Limiting Reactants

In many real-life experiments, reactants are not mixed in exact stoichiometric ratios.

• Limiting Reactant: The reactant that is completely used up first. It stops the reaction and limits the maximum amount of product formed.
• Excess Reactant: The reactant that is left over after the reaction has finished.

Everyday Analogy: If you have 10 burger patties and only 6 burger buns, you can only make 6 burgers. The buns are the limiting reactant, and the leftover patties are in excess.

Key Takeaway: Reacting mass calculations always follow: Mass \(\rightarrow\) Moles \(\rightarrow\) Ratio \(\rightarrow\) Target Mass. The limiting reactant controls how much product forms.

4. Theoretical Yield and Percentage Yield

Theoretical vs Actual Yield

• Theoretical Yield: The maximum possible mass of product that could be formed, calculated assuming 100% of reactants are converted into products.
• Actual Yield: The real mass of product obtained when you carry out the experiment in the laboratory.

Percentage Yield Formula

\(\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\)

Worked Example: A student calculated that heating copper carbonate should produce a theoretical yield of \(5.0\text{ g}\) of copper oxide. After the experiment, they collected \(4.2\text{ g}\). What is the percentage yield?
\(\text{Percentage Yield} = \frac{4.2}{5.0} \times 100 = 84\%\).

Why is 100% Yield Rarely Achieved?

In chemistry exams, you are frequently asked why the actual yield is less than 100%. Learn these four key reasons:

1. Reversible reactions: The reaction may not go to completion because it reaches equilibrium.
2. Incomplete reactions: Some of the reactants may not fully react.
3. Side reactions: Unwanted side reactions may occur, creating different, unexpected products.
4. Loss during practical transfers: Some product is lost mechanically during filtration, pouring, or remaining stuck to glassware and filter paper.

Key Takeaway: Percentage yield compares what you actually made with the mathematical maximum. Real-life losses and side reactions keep it below 100%.

5. Hydrated Salts and Water of Crystallisation (Prescribed Practical C1)

Key Definitions

• Water of Crystallisation: Water chemically bonded into the crystal structure of a salt.
• Hydrated Substance: A solid compound that contains water of crystallisation (e.g., hydrated copper(II) sulfate, \(\text{CuSO}_4\cdot 5\text{H}_2\text{O}\)).
• Anhydrous Substance: A substance containing no water of crystallisation (e.g., anhydrous copper(II) sulfate, \(\text{CuSO}_4\)).

Calculating the \(M_r\) of a Hydrated Salt

The dot in a formula like \(\text{CuSO}_4\cdot 5\text{H}_2\text{O}\) means that for every 1 unit of \(\text{CuSO}_4\), there are 5 molecules of \(\text{H}_2\text{O}\) locked inside the crystal. To find total \(M_r\), add the mass of the water molecules!

Worked Calculation: Find the \(M_r\) of \(\text{CuSO}_4\cdot 5\text{H}_2\text{O}\)
(\(A_r\): \(\text{Cu} = 64\), \(\text{S} = 32\), \(\text{O} = 16\), \(\text{H} = 1\))
1. \(M_r \text{ of } \text{CuSO}_4 = 64 + 32 + (4 \times 16) = 160\).
2. \(M_r \text{ of } 5\text{H}_2\text{O} = 5 \times [ (2 \times 1) + 16 ] = 5 \times 18 = 90\).
3. \(\text{Total } M_r = 160 + 90 = 250\).

Common Error Alert: Do not multiply 160 by 90! The dot means you add the mass of the water to the mass of the salt.

Prescribed Practical C1: Determining Water of Crystallisation

In this required practical, you heat hydrated crystals to drive off all the water, leaving behind the anhydrous salt.

Method Outline:
1. Weigh an empty, dry crucible and lid.
2. Add a sample of hydrated crystals and weigh again (to find the initial mass of the hydrated salt).
3. Place the crucible on a pipeclay triangle on a tripod and heat gently with a Bunsen burner.
4. Cool in a desiccator and re-weigh.

Crucial Exam Term: "Heating to Constant Mass"
How do you know all the water has completely evaporated? You must heat, cool, re-weigh, and repeat until two consecutive mass readings are identical. If the mass is still decreasing, water is still being lost!

Finding the Degree of Hydration (\(x\))

To find the value of \(x\) in a formula like \(\text{Salt}\cdot x\text{H}_2\text{O}\), calculate the mole ratio:

Worked Example:
A sample of hydrated calcium sulfate, \(\text{CaSO}_4\cdot x\text{H}_2\text{O}\), has a mass of \(3.44\text{ g}\). When heated to constant mass, the anhydrous \(\text{CaSO}_4\) remaining has a mass of \(2.72\text{ g}\). Find the value of \(x\).
(\(M_r \text{ of } \text{CaSO}_4 = 136\); \(M_r \text{ of } \text{H}_2\text{O} = 18\))

• Step 1: Find mass of water lost:
\(\text{Mass of water} = 3.44\text{ g} - 2.72\text{ g} = 0.72\text{ g}\).

• Step 2: Calculate moles of anhydrous salt and water:
\(\text{Moles of } \text{CaSO}_4 = \frac{2.72}{136} = 0.02\text{ moles}\)
\(\text{Moles of } \text{H}_2\text{O} = \frac{0.72}{18} = 0.04\text{ moles}\)

• Step 3: Determine the simplest whole-number ratio:
Divide both numbers by the smallest mole value (\(0.02\)):
\(\text{CaSO}_4 = \frac{0.02}{0.02} = 1\)
\(\text{H}_2\text{O} = \frac{0.04}{0.02} = 2\)

• Result: The formula is \(\text{CaSO}_4\cdot 2\text{H}_2\text{O}\), so \(x = 2\).

Key Takeaway: "Heating to constant mass" ensures all water is driven off. The mole ratio of anhydrous salt to water gives the value of \(x\).

Quick Chapter Summary & Checklist

Before sitting your Unit 1 examination, make sure you can:
• Calculate \(M_r\) and percentage by mass of an element.
• Convert masses between tonnes, kilograms, and grams.
• Use the mole formula: \(\text{Moles} = \frac{\text{Mass}}{M_r}\).
• Use balanced equations to carry out 3-step reacting mass calculations.
• Define theoretical and actual yield, and calculate percentage yield.
• State the 4 reasons why percentage yield is less than 100%.
• Describe how to heat to constant mass in Prescribed Practical C1 to find \(x\).

Quick check

Can you answer these now?

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

What is relative atomic mass (\(A_r\))?

The relative atomic mass (\(A_r\)) of an element is the average mass of an atom of that element compared to \(\frac{1}{12}\text{th}\) the mass of a carbon-12 atom.

How is relative formula mass (\(M_r\)) calculated?

\(M_r\) is calculated by adding together the relative atomic masses (\(A_r\)) of all the atoms present in the chemical formula.

What formula is used to calculate the percentage composition by mass of an element in a compound?

\(\text{Percentage by mass} = \frac{\text{Total } A_r \text{ of the element in the formula}}{M_r \text{ of the compound}} \times 100\)

What is the relationship between one mole of a substance and its formula mass?

The mass of 1 mole of any substance in grams is numerically equal to its relative formula mass (\(M_r\)) or relative atomic mass (\(A_r\)).

What is the core formula linking moles, mass, and relative formula mass?

\(\text{Number of moles} = \frac{\text{Mass (in g)}}{M_r \text{ (or } A_r\text{)}}\)

How do you convert kilograms and tonnes into grams for quantitative chemistry calculations?

Multiply kilograms (\(\text{kg}\)) by \(1{,}000\) to get grams (\(\text{g}\)), and multiply tonnes (\(\text{t}\)) by \(1{,}000{,}000\) (or \(10^6\)) to get grams.

What is the 3-step method for calculating reacting masses from a balanced equation?

1. Calculate moles of the known substance (\(\text{Moles} = \frac{\text{Mass}}{M_r}\)).
2. Use the stoichiometric ratio from the balanced equation to find moles of the target substance.
3. Calculate the mass of the target substance (\(\text{Mass} = \text{Moles} \times M_r\)).

What is the difference between a limiting reactant and an excess reactant?

The limiting reactant is completely used up first and limits the amount of product formed, whereas the excess reactant is left over when the reaction stops.

How is percentage yield calculated?

\(\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\)

What are the four main reasons why the percentage yield of a chemical reaction is usually less than 100%?

1. Reversible reactions (reaching equilibrium).
2. Incomplete reactions.
3. Unwanted side reactions.
4. Mechanical loss during practical transfers (e.g., filtration or remaining stuck to glassware).

What is the difference between a hydrated substance and an anhydrous substance?

A hydrated substance contains water chemically bonded into its crystal structure (water of crystallisation), whereas an anhydrous substance contains no water of crystallisation.

In Prescribed Practical C1, what does the phrase "heating to constant mass" mean and why is it done?

It means repeatedly heating, cooling, and re-weighing a sample until two consecutive mass readings are identical, ensuring all water of crystallisation has been driven off.

How is the degree of hydration (\(x\)) calculated for a hydrated salt formula \(\text{Salt}\cdot x\text{H}_2\text{O}\)?

Find the mass of water lost and mass of anhydrous salt remaining, convert both masses into moles, and find their simplest whole-number mole ratio (\(\text{Moles of } \text{H}_2\text{O} \div \text{Moles of salt}\)).

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