The Law of Conservation of Mass

This fundamental law states that during a chemical reaction, no atoms are created or destroyed—they are simply rearranged. This means the total mass of the reactants must always equal the total mass of the products. To show this, chemical reactions must be represented by balanced symbol equations, which have the same number of each type of atom on both sides of the reaction arrow.

Apparent Mass Changes in Reactions

In a non-enclosed system (like an open flask), mass can seem to increase or decrease. This is because a gas has either been added from the air or has escaped.

• Apparent Mass Increase: Occurs when a reactant combines with a gas from the air. For example, when magnesium is heated, it reacts with oxygen. 2Mg(s) + O₂(g) ⟶ 2MgO(s) The mass of the solid product is greater than the starting mass of magnesium because oxygen atoms from the air have been added.
• Apparent Mass Decrease: Occurs when a gas is produced and escapes. For example, the thermal decomposition of calcium carbonate produces carbon dioxide gas. CaCO₃(s) ⟶ CaO(s) + CO₂(g) If the container is unsealed, the carbon dioxide gas escapes, so the measured mass of the solid decreases. In a closed system, where no substances can enter or leave, the measured mass remains constant throughout the reaction.

Linking Conservation of Mass to Relative Formula Mass (Mr​)

For any balanced chemical equation, the sum of the relative formula masses (Mr​) of the reactants is equal to the sum of the relative formula masses of the products. Worked Example – Magnesium Oxide Formation (Relative atomic masses: Mg = 24, O = 16) 2Mg + O₂ ⟶ 2MgO

• Reactants Mr​: (2 × 24) + (2 × 16) = 48 + 32 = 80
• Products Mr​: 2 × (24 + 16) = 2 × 40 = 80 Because the totals match, mass is conserved.

How to Balance Equations

1. Write the correct chemical formulae for all reactants and products. Do not change these.
2. Count the atoms of each element on both sides of the equation.
3. Add large numbers (coefficients) in front of the formulae to balance the number of atoms.
4. Never change the small subscript numbers in a formula (e.g., the ‘2’ in H₂O).
5. Check that every element is balanced and that the coefficients are the smallest possible whole numbers.

Quick Tips for Balancing:

• Start by balancing elements that appear in the fewest formulae.
• If a polyatomic ion (e.g., sulfate, SO₄²⁻) appears unchanged on both sides, balance it as a single unit.
• In combustion reactions, it is often easiest to leave hydrogen and oxygen until last.

Exam tip if a question describes an experiment where mass seems to decrease, it is because a gaseous product has formed and escaped from the unsealed container. Conversely, an apparent mass increase is due to a gaseous reactant, usually oxygen from the air, combining with one of the starting materials.

What is Relative Formula Mass?

The relative formula mass of a substance is the sum of the relative atomic masses (Ar​) of all the atoms shown in its chemical formula. It is written as Mr​ and has no units. For simple molecules, you might see the term relative molecular mass, which is calculated in the same way. The Ar​ values you need will be on the periodic table provided in the exam.

How to Calculate Mr​

To calculate the Mr​, you must first identify the correct formula and count the number of atoms of each element. For each element, multiply its Ar​ by the number of atoms present, and then add all the totals together. Be careful with brackets; a subscript outside a bracket multiplies every atom inside it. For example, in Ca(OH)₂, there is one calcium atom, two oxygen atoms, and two hydrogen atoms.

Exam tip – Always show your full working when calculating Mr​. Marks can be awarded for the correct method, such as identifying the correct number of atoms, even if your final answer has an arithmetic error. This is especially important for compounds with brackets, which is where mistakes are most common.

Worked Examples

(Relative atomic masses: H=1, O=16, Mg=24, S=32, Al=27)

Example 1: Magnesium Hydroxide, Mg(OH)₂

• Atoms: 1 x Mg, 2 x O, 2 x H

• Calculation: (1 × 24) + (2 × 16) + (2 × 1)

• = 24 + 32 + 2

• Mr​ = 58

Example 2: Aluminium Sulfate, Al₂(SO₄)₃

• Atoms: 2 x Al, 3 x S, 12 x O

• Calculation: (2 × 27) + (3 × 32) + (12 × 16)

• = 54 + 96 + 192

• Mr​ = 342

The Law of Conservation of Mass

The fundamental principle of conservation of mass states that during any chemical reaction, atoms are not created or destroyed, only rearranged. This means the total mass of all the reactants at the start of a reaction is always equal to the total mass of all the products at the end. Because of this law, all symbol equations must be balanced, showing the same number of each type of atom on both sides of the reaction arrow.

Apparent Mass Changes in Non-Enclosed Systems

In practice, if a reaction is carried out in an open container (a non-enclosed system), the mass measured on a balance might seem to change. This is always because a gas has either entered from the surrounding air or escaped from the reaction vessel.

• Apparent Mass Increase: This occurs when a reactant combines with a gas from the air, most commonly oxygen. When magnesium burns, it reacts with oxygen to form magnesium oxide. 2Mg(s) + O₂(g) ⟶ 2MgO(s) The mass of the solid product (magnesium oxide) is greater than the starting mass of the magnesium because oxygen atoms from the air, which were not initially on the balance, have been added to it. For example, if 2.4 g of magnesium reacts, it will form 4.0 g of magnesium oxide, meaning 1.6 g of oxygen was added.

• Apparent Mass Decrease: This happens when a gaseous product is formed and allowed to escape. For example, heating calcium carbonate causes it to decompose into calcium oxide and carbon dioxide gas. CaCO₃(s) ⟶ CaO(s) + CO₂(g) If the reaction is done in an open crucible, the carbon dioxide gas escapes and is not weighed. The mass of the solid remaining is therefore less than the starting mass. If 10.0 g of calcium carbonate is heated to produce 5.6 g of calcium oxide, the mass of carbon dioxide that escaped is 4.4 g (10.0 g – 5.6 g).

Reactions in Closed Systems

If a reaction is carried out in a closed system (e.g., a sealed flask where nothing can get in or out), the reading on a balance will remain constant. This is true even if a solid (a precipitate) forms from two solutions, or if a gas is produced but trapped within the container. For example, if a sealed flask containing 100.00 g of a solution and 3.00 g of sodium hydrogencarbonate is heated, the total mass will remain 103.00 g after the reaction, even though a gas is produced.

Linking Mass to Relative Formula Mass (Mr​)

For any balanced equation, the sum of the Mr​ values for the reactants equals the sum of the Mr​ values for the products.Check: 2Mg + O₂ ⟶ 2MgO (Ar​ values: Mg=24, O=16)

• Reactants Mr​: (2 × 24) + (2 × 16) = 48 + 32 = 80

• Products Mr​: 2 × (24 + 16) = 2 × 40 = 80 The totals match, proving that mass is conserved.

Exam tip – When answering questions about experiments, clearly distinguish between open and closed systems. Don’t just say “mass is lost”; specify that a gas has escaped. Similarly, state that mass appears to be gained because a gas from the air has reacted. To ensure accuracy in experiments involving heating a solid, you should heat it to a constant mass. This means heating, cooling, and reweighing the substance repeatedly until the mass reading no longer changes, which confirms the reaction is complete.

Key Measurement Terms

In quantitative chemistry, making careful measurements is crucial. It’s also important to understand the quality of those measurements using specific terms.

• Accuracy: This is how close a measurement is to the true value. Using properly calibrated equipment and avoiding systematic errors improves accuracy.

• Precision: This describes how close repeated measurements are to each other. Using apparatus with a finer scale (higher resolution) can improve precision.

• Repeatability: This is when the same person using the same method and equipment gets similar results.

• Reproducibility: This is when a different person or a different method produces similar results to the original experiment.

• Uncertainty: This is the range within which the true value of a measurement is likely to lie. It is often written with a ‘±’ symbol (e.g., 25.0 ± 0.5 cm³).

Handling Data: Means and Anomalies

To ensure results are reliable, experiments should be repeated at least three times. Any result that does not fit the pattern of the others is called an anomaly and should be ignored when calculating the average. Example – Titration Results: A student obtains the following volumes: 24.80 cm³, 24.65 cm³, 28.10 cm³, 24.75 cm³. The result 28.10 cm³ is clearly an anomaly and should be excluded. The mean is calculated from the remaining concordant (closely agreeing) results: Mean = (24.80 + 24.65 + 24.75) ÷ 3 = 24.73 cm³

exam tip – When calculating a mean from experimental data, always state which result you are ignoring as an anomaly and give a reason (e.g., “it is not concordant with the other results”). For your final answer, give it to a number of significant figures that is consistent with the measurements you used.

Understanding Uncertainty

All measurements have some degree of uncertainty. The absolute uncertainty is the margin of error for a piece of equipment.

• For an analogue scale (like a ruler or measuring cylinder), it is usually ± half the smallest division. A measuring cylinder marked every 1 cm³ has an uncertainty of ± 0.5 cm³.

• For a digital instrument (like a balance), it is ± the smallest unit it can display. A balance that reads to two decimal places (e.g., 12.36 g) has an uncertainty of ± 0.01 g.

Percentage uncertainty shows how significant the uncertainty is relative to the measurement taken. Percentage Uncertainty = (Absolute Uncertainty ÷ Measured Value) × 100 For example, measuring 25.0 cm³ with a cylinder that has an uncertainty of ± 0.5 cm³ gives: Percentage Uncertainty = (0.5 ÷ 25.0) × 100 = 2.0%

Common Errors and How to Improve Measurements

Errors in experiments can be systematic (affecting all results in the same way) or random (causing results to scatter around a mean). Common sources of error include:

• Losing gas from a reaction before a bung is inserted.

• Evaporation of a solvent when heating.

• Judging a colour change by eye during a titration.

To improve the quality of measurements:

• Increase the quantity measured: Measuring a larger mass or volume reduces the percentage uncertainty. For example, the percentage uncertainty of weighing 20 g on a balance (± 0.01 g) is much lower than weighing 0.20 g.

• Use higher resolution apparatus: Use a pipette or burette instead of a measuring cylinder for measuring accurate volumes.

• Repeat and average: This reduces the effect of random errors.

What is a Mole?

In chemistry, a mole (unit symbol: mol) is a standard amount of a substance. It provides a way to count a very large number of particles. One mole of any substance contains 6.02 x 10²³ particles, a value known as the Avogadro Constant (NA​). These particles can be atoms, molecules, or ions, depending on what is being measured.

Calculating Moles from Mass

The most common calculation you will perform involves converting between the mass of a substance in grams (g) and the number of moles. This is done using the relative formula mass (Mr​).

The relationship is given by the formula: Moles (mol) = Mass (g) ÷ Relative Formula Mass (Mr​) n = m / Mr

You can rearrange this formula to find the mass if you know the number of moles: Mass (g) = Moles (mol) × Relative Formula Mass (Mr​) m = n × Mr

Exam tip – Pay close attention to the wording of questions involving elements that exist as molecules, like oxygen (O₂) or chlorine (Cl₂). If a question asks for the number of molecules, use the full formula (e.g., O₂) to calculate the Mr​. If it asks for the number of atoms, you will likely need an extra step, such as multiplying your final answer by two.

Calculating the Number of Particles

You can also calculate the exact number of atoms or molecules in a substance if you know the number of moles.

The formula for this is: Number of Particles = Moles (mol) × Avogadro Constant (NA​) Number of Particles = n × (6.02 x 10²³)

This can be rearranged to find the number of moles if you are given the number of particles.

Using Moles to Balance Equations

You can determine the balancing numbers in an equation if you know the masses of the reactants and products. The method involves converting masses to moles and then finding the simplest whole-number ratio.

Method:

1. Convert the known mass of each substance into moles using n = m / M_r.

2. Divide all the mole values by the smallest number of moles calculated.

3. The whole numbers you get are the balancing numbers (coefficients) for the equation.

Reacting Mass Calculations

A balanced chemical equation tells you the ratio of moles in which substances react. You can use this ratio to calculate the mass of a reactant or product.

Method:

1. Find the moles of the known substance: Use the mass you are given in the question and the formula n = m / M_r.

2. Use the balanced equation: Look at the mole ratio between the substance you know and the substance you want to find out about. Use this ratio to calculate the number of moles of the target substance.

3. Calculate the mass of the target substance: Rearrange the mole formula to m = n × M_r.

Worked Example: Calculate the mass of carbon dioxide produced when 10.0 g of calcium carbonate decomposes. (Ar​: Ca=40, C=12, O=16) CaCO₃(s) ⟶ CaO(s) + CO₂(g)

• Step 1 (Moles Known): Mr​ of CaCO₃ = 40 + 12 + (3×16) = 100. Moles of CaCO₃ = 10.0 g / 100 = 0.100 mol.

• Step 2 (Ratio): From the equation, the ratio of CaCO₃ : CO₂ is 1:1. Therefore, moles of CO₂ = 0.100 mol.

• Step 3 (Mass Target): Mr​ of CO₂ = 12 + (2×16) = 44. Mass of CO₂ = 0.100 mol × 44 = 4.40 g.

Exam tip – For any reacting mass calculation, your first step should always be to calculate the moles of the substance you have the most information about (usually the one you are given a mass for). Always show your working clearly, as marks are often given for each step.

Limiting Reactants

In a reaction, it is common for one reactant to be completely used up before the others. This is the limiting reactantbecause it stops the reaction and limits the amount of product that can be formed. The other reactants are said to be in excess.

Method to find the limiting reactant:

1. Calculate the moles of each reactant you are given a mass for.

2. Look at the mole ratio in the balanced equation.

3. Compare the moles you have with the ratio required. The reactant that runs out first is the limiting one. All further calculations about the amount of product formed must be based on the moles of this limiting reactant.

Why Use Moles for Balancing?

A balanced chemical equation must have the same number of atoms of each element on both sides. Using the concept of moles ensures the coefficients (the large numbers in front of formulae) correctly represent the reacting ratios of the substances.

Method for Balancing

1. Write the correct formulae for all reactants and products. Never change the small subscript numbers.

2. Balance the atoms, one element at a time, by adding coefficients in front of the formulae.

3. Count the atoms of each element on both sides to check they are equal.

4. Ensure the coefficients are the smallest possible whole numbers.

Worked Examples

Formation of Iron(III) Oxide (Rust)

• Unbalanced: Fe + O₂ ⟶ Fe₂O₃

• Logic: The product, Fe₂O₃, contains two iron atoms and three oxygen atoms.

  • To get two Fe atoms on the left, we place a 2 in front of Fe: 2Fe + O₂ ⟶ Fe₂O₃.

  • To get three O atoms on the left, we would need 1.5 O₂ molecules. Since we cannot have fractions, we must double all coefficients.

• Balanced: 4Fe + 3O₂ ⟶ 2Fe₂O₃

Combustion of Ethane

• Unbalanced: C₂H₆ + O₂ ⟶ CO₂ + H₂O

• Logic:

  1. Balance Carbon: There are two C atoms in C₂H₆, so place a 2 in front of CO₂. C₂H₆ + O₂ ⟶ 2CO₂ + H₂O

  2. Balance Hydrogen: There are six H atoms in C₂H₆, so place a 3 in front of H₂O. C₂H₆ + O₂ ⟶ 2CO₂ + 3H₂O

  3. Balance Oxygen: Count the oxygen atoms on the right side: (2 × 2) from CO₂ + (3 × 1) from H₂O = 7 oxygen atoms. To get 7 oxygen atoms on the left, we need 3.5 O₂ molecules.

  4. Remove Fractions: Multiply all coefficients by 2 to get whole numbers.

• Balanced: 2C₂H₆ + 7O₂ ⟶ 4CO₂ + 6H₂O

Exam tip – When balancing combustion equations for hydrocarbons, always balance the elements in the order: Carbon, then Hydrogen, and finally Oxygen. This C-H-O order simplifies the process and helps avoid errors. If you end up with a fraction for oxygen, simply double every coefficient in the entire equation to resolve it.

What is Concentration?

Concentration tells you how much of a dissolved substance, called the solute, is in a specific volume of the solution. For GCSE Chemistry, concentration is typically expressed in grams per cubic decimetre (g/dm³) or moles per cubic decimetre (mol/dm³). Remember, one cubic decimetre (dm³) is the same as one litre, and 1 dm³ = 1000 cm³.

Key Formulas

To solve concentration problems, you need to be comfortable using and rearranging two main formulas. Always convert volumes to dm³ by dividing the cm³ value by 1000 before you start.

1. Concentration in g/dm³: Concentration (g/dm³) = Mass of solute (g) ÷ Volume of solution (dm³)

This can be rearranged to find the mass: Mass (g) = Concentration (g/dm³) × Volume (dm³)

2. Concentration in mol/dm³ (HT Only): Concentration (mol/dm³) = Moles of solute (mol) ÷ Volume of solution (dm³)

This can be rearranged to find the number of moles: Moles (mol) = Concentration (mol/dm³) × Volume (dm³)

Worked Examples

Example 1: Calculating Concentration in g/dm³ What is the concentration if 5.85 g of sodium chloride is dissolved in water to make a 0.250 dm³ solution? Concentration = 5.85 g / 0.250 dm³ = 23.4 g/dm³.

Example 2: Calculating Concentration in mol/dm³ (HT Only) Calculate the concentration in mol/dm³ for the solution in Example 1. (Mr​ of NaCl = 58.5)

• First, find the moles: n = m / M_r = 5.85 g / 58.5 = 0.100 mol.

• Then, calculate concentration: Concentration = 0.100 mol / 0.250 dm³ = 0.400 mol/dm³.

Example 3: Calculating Mass from Concentration (HT Only) How much hydrated copper(II) sulfate (CuSO₄·5H₂O, Mr​ = 249.5) is needed to make 500 cm³ of a 0.200 mol/dm³ solution?

• Convert volume: 500 cm³ = 0.500 dm³.

• Find moles: n = c × V = 0.200 mol/dm³ × 0.500 dm³ = 0.100 mol.

• Find mass: m = n × M_r = 0.100 mol × 249.5 = 24.95 g.

Exam tip – A common mistake is forgetting to convert volumes from cm³ to dm³. Before you do any calculation, check your volume unit and divide by 1000 if it’s in cm³. Clearly write down the formula you are using (c = n / V or c = m / V) and show your rearrangement if needed, as this can earn you marks even if the final calculation is wrong.

What is Percentage Yield?

In a chemical reaction, you often get less product than you theoretically calculated you would. Percentage yield is a calculation that compares the actual mass of product you collected from an experiment (actual yield) with the maximum mass you could have possibly made (theoretical yield). The theoretical yield is calculated from the balanced chemical equation.

The formula for percentage yield is: Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100

Calculating Percentage Yield

Worked Example: A student reacts 7.20 g of iron with excess oxygen to make iron(III) oxide. They collect 9.10 g of the product. Calculate the percentage yield. 4Fe + 3O₂ ⟶ 2Fe₂O₃ (Ar​: Fe = 56, O = 16. Mr​: Fe₂O₃ = 160)

1. Calculate moles of reactant: Moles of Fe = Mass / Ar​ = 7.20 g / 56 = 0.1286 mol.

2. Find theoretical moles of product using the ratio: The equation shows a 4 : 2 ratio between Fe and Fe₂O₃ (or 2 : 1). Theoretical moles of Fe₂O₃ = 0.1286 / 2 = 0.0643 mol.

3. Calculate theoretical yield (mass): Theoretical mass = Moles × Mr​ = 0.0643 mol × 160 = 10.29 g.

4. Calculate percentage yield: Percentage Yield = (Actual Yield / Theoretical Yield) × 100 = (9.10 g / 10.29 g) × 100 = 88.4%

Why is Yield Less Than 100%?

It is very rare to achieve 100% yield in a real experiment. This can happen for several reasons:

• The reaction may be reversible and not go to completion.

• Unexpected side reactions might occur, forming different products.

• Some of the product can be lost during practical steps like filtration or when transferring between containers.

Exam tip – Do not confuse percentage yield with atom economy. Percentage yield is a measure of the practical efficiency of your experiment and tells you how much product you actually made. Atom economy is a theoretical measure, calculated from the equation, that tells you how many of the reactant atoms end up in the desired product. An experiment can have a high atom economy but a low percentage yield.

What is Atom Economy?

Atom economy is a measure of how efficiently the atoms from the reactants are converted into the desired product. It is a theoretical value calculated from the balanced equation and helps us see how ‘green’ or sustainable a reaction is. A high atom economy means less waste is produced.

The formula for atom economy is: Atom Economy (%) = (Mr​ of Desired Product ÷ Sum of Mr​ of ALL Reactants) × 100

Reactions that produce only one product have an atom economy of 100%.

Worked Examples

Example 1: Making Ethane (Addition Reaction) C₂H₄ + H₂ ⟶ C₂H₆ This reaction has only one product, ethane. Therefore, all the atoms from the reactants end up in the desired product. The atom economy is 100%.

Example 2: Making Chloromethane (Substitution Reaction) CH₄ + Cl₂ ⟶ CH₃Cl + HCl The desired product is chloromethane (CH₃Cl), but hydrogen chloride (HCl) is also produced as a waste by-product. (Mr​: CH₄=16, Cl₂=71, CH₃Cl=50.5, HCl=36.5)

• Sum of Mr​ of reactants = 16 + 71 = 87

• Mr​ of desired product (CH₃Cl) = 50.5

• Atom Economy = (50.5 ÷ 87) × 100 = 58.0% This means only 58% of the mass of the reactant atoms ends up in the useful product. The other 42% forms waste.

Why is Atom Economy Important?

Industrial processes with a high atom economy are better for the environment because they produce less waste material. They are also more economical because less money is spent separating the desired product from waste products and on treating that waste safely. Companies aim for processes with both high atom economy and high percentage yield to maximise profits and sustainability.

Exam tip – It is crucial not to mix up atom economy and percentage yield.

• Atom Economy is calculated before the experiment using only the balanced equation. It tells you the maximum possible efficiency.

• Percentage Yield is calculated after the experiment. It tells you how much product you actually made compared to the maximum possible amount. A reaction can have 100% atom economy but a very low percentage yield if, for example, the product was lost during the experiment.

Relating Moles, Concentration, and Volume

This topic builds on your understanding of moles and concentration. The key relationship you will use allows you to find the number of moles of a solute if you know its concentration and the volume of the solution.

The main formula is: Moles (mol) = Concentration (mol/dm³) × Volume (dm³) n = c × V

Always remember to convert any volumes given in cm³ into dm³ by dividing by 1000 before using this formula.

Titration Calculations

A titration is an experimental technique used to find the concentration of an unknown solution. By reacting it with a solution of known concentration (a standard solution), we can use the volumes and the mole ratio from the balanced equation to find the unknown value.

Worked Example: 25.0 cm³ of 0.200 mol/dm³ hydrochloric acid was needed to neutralise 40.0 cm³ of a sodium hydroxide solution. Calculate the concentration of the sodium hydroxide. HCl + NaOH ⟶ NaCl + H₂O

1. Calculate moles of the known substance (HCl):

   • Volume = 25.0 cm³ = 0.0250 dm³

   • Moles HCl = c × V = 0.200 mol/dm³ × 0.0250 dm³ = 0.00500 mol.

2. Use the mole ratio to find moles of the unknown (NaOH):

   • From the equation, the ratio HCl : NaOH is 1:1.

   • Therefore, moles of NaOH = 0.00500 mol.

3. Calculate the concentration of the unknown (NaOH):

   • Volume = 40.0 cm³ = 0.0400 dm³

   • Concentration NaOH = n / V = 0.00500 mol / 0.0400 dm³ = 0.125 mol/dm³.

Combining Mass and Concentration Calculations

Some questions may require you to calculate the mass of a substance needed to react with a solution.

Worked Example: Calculate the mass of sodium hydroxide needed to neutralise 50.0 cm³ of 0.500 mol/dm³ sulfuric acid.H₂SO₄ + 2NaOH ⟶ Na₂SO₄ + 2H₂O (Mr​ of NaOH = 40)

1. Calculate moles of sulfuric acid:

   • Volume = 50.0 cm³ = 0.0500 dm³

   • Moles H₂SO₄ = 0.500 mol/dm³ × 0.0500 dm³ = 0.0250 mol.

2. Use the mole ratio to find moles of sodium hydroxide:

   • The ratio H₂SO₄ : NaOH is 1:2.

   • Moles NaOH = 0.0250 mol × 2 = 0.0500 mol.

3. Calculate the mass of sodium hydroxide:

   • Mass NaOH = n × Mr = 0.0500 mol × 40 = 2.00 g.

Exam tip – For titration calculations, set your work out in clear, logical steps. First, calculate the moles of the substance you have both concentration and volume for. Second, use the mole ratio from the balanced equation to find the moles of the other substance. Finally, calculate the unknown concentration or mass. This structured approach helps prevent mistakes and can secure method marks.

Molar Volume of Gases

The core principle is that one mole of any gas occupies a volume of 24 dm³ (or 24,000 cm³) at room temperature and pressure (rtp). This constant, the molar gas volume, is used to convert between the amount of a gas in moles and the volume it occupies.

Key Formulas & Pitfalls

Use these essential formulas for your calculations:

• Volume(dm3)=moles×24

• Moles=24Volume(dm3)​ Remember to always convert volumes to dm³ before calculating (1 dm³ = 1000 cm³). Common mistakes include forgetting this conversion, ignoring the mole ratios from the balanced equation, and rounding numbers too early.

Worked Examples

Example 1: Calculating Gas Volume from Mass What volume of H₂ is produced at rtp when 0.200 g of magnesium reacts? (Mg+2HCl→MgCl2​+H2​) (Ar Mg = 24.3) First, find the moles of Mg: moles=24.30.200​=0.00823 mol. The mole ratio of Mg to H₂ is 1:1, so 0.00823 moles of H₂ are produced. The volume is $0.00823\times 24=0.1975$ dm³ (or 197.5 cm³).

Example 2: Calculating Gas Volume from another Gas Volume If 48 dm³ of hydrogen reacts at rtp, what volume of ammonia is formed? (N2​+3H2​→2NH3​) First, find the moles of hydrogen: moles=2448​=2.00 mol. The mole ratio of H₂ to NH₃ is 3:2. So, the moles of NH₃ produced is 32.00​×2=1.33 mol. The volume of ammonia is $1.33\times 24=31.9$dm³.

Exam tip – Always start with the balanced equation to find the mole ratio. Convert any given amounts to moles, use the ratio to find the moles of the gas you need, and then multiply by 24 to get its volume in dm³. Remember to show your working clearly.