IGCSE Chemistry Topic 3: Stoichiometry

3.1 Formulae 3.2 Relative Masses 3.3 The Mole 3.4 Chemical Equations 3.5 Reacting Masses 3.6 Yield & Purity 3.7 Concentration 3.8 Gas Volumes

Hey Tara! Welcome to Topic 3 - Stoichiometry. This is the topic where chemistry meets maths, and it is one of the most important topics in IGCSE Chemistry because calculation questions appear in almost every exam paper. You will learn how to write formulae, balance equations, work with moles, calculate reacting masses, and predict volumes of gases. Think of stoichiometry as the recipe book of chemistry - just like a recipe tells you exactly how much flour, sugar, and butter to use, stoichiometry tells you exactly how much of each chemical reacts and how much product you get. Let us dive in!

3.1 Formulae ▼

Word Equations

A word equation describes a chemical reaction using the names of the reactants and products. The reactants go on the left, the products go on the right, and an arrow separates them.

reactants → products

For example, when magnesium burns in oxygen:

magnesium + oxygen → magnesium oxide

Word equations tell you what reacts and what forms, but they do not tell you how much. For that, we need symbol equations.

Balanced Symbol Equations

A balanced symbol equation uses chemical formulae instead of names and includes numbers (coefficients) to ensure the same number of each type of atom appears on both sides of the equation. This is based on the law of conservation of mass - atoms are not created or destroyed in a chemical reaction.

The magnesium example becomes:

2Mg + O₂ → 2MgO

Left side: 2 Mg atoms, 2 O atoms Right side: 2 Mg atoms, 2 O atoms ✓

Steps to Balance an Equation

Write the unbalanced equation using correct formulae
Count atoms of each element on both sides
Add coefficients (big numbers in front) to balance - NEVER change the subscripts in a formula
Check that all elements are balanced
Add state symbols if required: (s) solid, (l) liquid, (g) gas, (aq) aqueous solution

State Symbols

State symbols tell you the physical state of each substance in the reaction:

Symbol	State	Example
(s)	Solid	NaCl(s) - solid sodium chloride
(l)	Liquid	H₂O(l) - liquid water
(g)	Gas	CO₂(g) - carbon dioxide gas
(aq)	Aqueous (dissolved in water)	HCl(aq) - hydrochloric acid

2Mg(s) + O₂(g) → 2MgO(s)

Formulae of Ionic Compounds

Ionic compounds are made of positive ions (cations) and negative ions (anions). The overall charge must be zero, so the positive and negative charges must balance.

Common Ion Charges

Positive Ions (Cations)	Charge	Negative Ions (Anions)	Charge
Na⁺	1+	Cl⁻	1-
K⁺	1+	Br⁻	1-
Mg²⁺	2+	O²⁻	2-
Ca²⁺	2+	S²⁻	2-
Al³⁺	3+	NO₃⁻	1-
Fe²⁺ / Fe³⁺	2+ / 3+	SO₄²⁻	2-
NH₄⁺	1+	CO₃²⁻	2-
Cu²⁺	2+	OH⁻	1-
Zn²⁺	2+	PO₄³⁻	3-

The Cross-Over Method

To work out the formula of an ionic compound, use the cross-over method:

Write the two ions with their charges
Cross the numbers of the charges over (ignore the sign)
Write these as subscripts
Simplify if possible

Example: Aluminium oxide = Al³⁺ and O²⁻

Cross over: Al₂O₃ (the 3 from Al goes to O, the 2 from O goes to Al)

Formula: Al₂O₃

Example: Calcium chloride = Ca²⁺ and Cl⁻

Cross over: CaCl₂ (the 2 from Ca goes to Cl, the 1 from Cl goes to Ca but we do not write 1)

Formula: CaCl₂

Types of Formulae

Type	Definition	Example for ethanoic acid
Molecular formula	Shows the actual number of each type of atom in one molecule	C₂H₄O₂
Empirical formula	Shows the simplest whole number ratio of atoms	CH₂O
Structural formula	Shows how atoms are arranged and bonded in a molecule	CH₃COOH
Ionic formula	Shows the ions and their ratio in an ionic compound	Na⁺Cl⁻ (for NaCl)

Determining Formulae from Experimental Data

You can determine the empirical formula of a compound from experimental data about combining masses. The steps are:

Write the mass of each element
Divide each mass by the element's A_r to get moles
Divide all the mole values by the smallest to get the simplest ratio
If the ratio is not whole numbers, multiply to make them whole (e.g. multiply by 2 if you get 1:1.5)

Supplement

Deducing Formulae from Models or Diagrams

In the exam, you might be shown a ball-and-stick model or a diagram of a molecule. To deduce the formula:

Count each type of atom (each colour of ball represents a different element)
Use the key provided to identify which element each colour represents
Write the formula using the counts

For example, if a model shows 1 black ball (carbon) and 4 white balls (hydrogen), the formula is CH₄ (methane).

Constructing Balanced Equations with State Symbols

For Extended tier, you must be able to construct fully balanced equations from scratch and include correct state symbols. Remember:

Most metals and their oxides are (s)
Water is (l) unless it is steam, then (g)
Acids and solutions are (aq)
Gases like O₂, H₂, CO₂, Cl₂ are (g)

Worked Examples

Worked Example 1 Balance the equation: Fe + Cl₂ → FeCl₃

Show Solution

Step 1 - Count atoms (unbalanced)

Left: 1 Fe, 2 Cl. Right: 1 Fe, 3 Cl. Chlorine is not balanced.

Step 2 - Balance chlorine

We need the same number of Cl on both sides. The LCM of 2 and 3 is 6. So we need 3 Cl₂ (giving 6 Cl) and 2 FeCl₃ (needing 6 Cl).

Step 3 - Balance iron

Now we have 2 FeCl₃ on the right, so we need 2 Fe on the left.

Step 4 - Check

Left: 2 Fe, 6 Cl. Right: 2 Fe, 6 Cl. Balanced!

2Fe + 3Cl₂ → 2FeCl₃

Worked Example 2 Write the formula of calcium hydroxide using the cross-over method.

Show Solution

Step 1 - Identify ions

Calcium ion: Ca²⁺. Hydroxide ion: OH⁻.

Step 2 - Cross over charges

The 2 from Ca goes to OH, the 1 from OH goes to Ca. We need 1 calcium and 2 hydroxide ions.

Step 3 - Write formula

Since hydroxide is a polyatomic ion (OH), we put brackets around it when there is more than one: Ca(OH)₂

Ca(OH)₂

Worked Example 3 3.6 g of carbon combines with 9.6 g of oxygen. Determine the empirical formula. (A_r: C = 12, O = 16)

Show Solution

Step 1 - Write masses

Carbon: 3.6 g. Oxygen: 9.6 g.

Step 2 - Divide by A_r to get moles

Moles of C = 3.6 / 12 = 0.3 mol. Moles of O = 9.6 / 16 = 0.6 mol.

Step 3 - Find simplest ratio

Divide by the smallest (0.3): C = 0.3/0.3 = 1, O = 0.6/0.3 = 2.

Step 4 - Write formula

Ratio is C:O = 1:2

Empirical formula: CO₂

⚠ Exam Tips

Never change subscripts to balance an equation - only change the big numbers (coefficients) in front of formulae. Changing H₂O to H₃O would create a completely different (and non-existent) substance!

Cross-over method with polyatomic ions: Always use brackets when you need more than one polyatomic ion. For example, magnesium nitrate is Mg(NO₃)₂, NOT MgNO₃₂.

State symbols: Learn the common states - all Group 1 and 2 metal compounds dissolved in water are (aq). Pure metals are (s). Gases at room temperature include H₂, O₂, N₂, CO₂, Cl₂, NH₃.

🌎 Apply It: Real-World Chemistry

See how chemical formulae and equations are used in real life around the world.

A cement factory in Rajasthan, India heats limestone (calcium carbonate) in a kiln at 900 degrees C. The calcium carbonate decomposes into calcium oxide and carbon dioxide.

Write a balanced equation with state symbols for this thermal decomposition.

▼

Word Equation

calcium carbonate → calcium oxide + carbon dioxide

Symbol Equation

CaCO₃(s) → CaO(s) + CO₂(g)

Check

Left: 1 Ca, 1 C, 3 O. Right: 1 Ca, 1 C, 3 O. Already balanced! CaCO₃ is solid, CaO is solid, CO₂ is a gas.

Real-World Connection

India is the second-largest cement producer in the world. This one equation describes the most important reaction in cement manufacturing - thermal decomposition of limestone. The CaO produced (quicklime) is a key ingredient in cement.

A water treatment plant in London adds aluminium sulfate to purify drinking water. The formula needs to be correctly written to order the right chemical.

Use the cross-over method to determine the formula of aluminium sulfate (Al³⁺ and SO₄²⁻).

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Identify Ions

Aluminium: Al³⁺ (charge 3+). Sulfate: SO₄²⁻ (charge 2-).

Cross Over

The 3 from Al goes to SO₄, and the 2 from SO₄ goes to Al. Formula: Al₂(SO₄)₃

Verify

Total charge: 2 x (+3) = +6 and 3 x (-2) = -6. Net charge = 0. Correct!

Real-World Connection

Aluminium sulfate is used as a flocculant - it causes tiny suspended particles in water to clump together so they can be filtered out. Thames Water uses thousands of tonnes of it every year to keep London's drinking water clean.

A fireworks manufacturer in Sivakasi, Tamil Nadu needs to write the balanced equation for the combustion of magnesium in the air, which produces the brilliant white flash in sparklers.

Write and balance the equation for magnesium burning in oxygen, including state symbols.

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Unbalanced

Mg(s) + O₂(g) → MgO(s)

Balance

Left: 1 Mg, 2 O. Right: 1 Mg, 1 O. Need to balance O. Put 2 in front of MgO, then 2 in front of Mg.

Balanced Equation

2Mg(s) + O₂(g) → 2MgO(s)

Real-World Connection

Sivakasi produces about 90% of India's fireworks. The brilliant white light in sparklers comes from burning magnesium. The equation tells us every 2 atoms of magnesium need exactly 1 molecule of O₂ - too little oxygen and the sparkler sputters, too much and it burns out too fast.

A pharmaceutical company in Basel, Switzerland is developing a new antacid tablet. They need to determine the empirical formula of a new compound that contains 40.0% calcium, 12.0% carbon, and 48.0% oxygen by mass.

Determine the empirical formula of this compound. (A_r: Ca = 40, C = 12, O = 16)

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Assume 100 g

Ca: 40.0 g, C: 12.0 g, O: 48.0 g

Divide by A_r

Ca: 40.0/40 = 1.0 mol. C: 12.0/12 = 1.0 mol. O: 48.0/16 = 3.0 mol.

Simplest Ratio

Ca : C : O = 1 : 1 : 3

Empirical formula: CaCO₃ (calcium carbonate - the active ingredient in many antacids!)

Real-World Connection

CaCO₃ is the main ingredient in antacid tablets like Tums. It neutralises excess stomach acid (HCl). The ability to determine an empirical formula from percentage composition is essential in pharmaceutical chemistry.

An agricultural research lab in Nairobi, Kenya is analysing fertilisers. They need to write the correct formula for ammonium phosphate to calculate how much nitrogen it delivers to crops.

Determine the formula of ammonium phosphate using the cross-over method (NH₄⁺ and PO₄³⁻).

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Identify Ions

Ammonium: NH₄⁺ (charge 1+). Phosphate: PO₄³⁻ (charge 3-).

Cross Over

The 1 from NH₄ goes to PO₄ (so 1 phosphate). The 3 from PO₄ goes to NH₄ (so 3 ammoniums).

Write Formula

Since NH₄ is polyatomic and we need 3 of them, use brackets: (NH₄)₃PO₄

Verify

Total charge: 3 x (+1) = +3 and 1 x (-3) = -3. Net charge = 0. Correct!

Real-World Connection

Ammonium phosphate is one of the most important fertilisers in African agriculture. Each formula unit contains 3 nitrogen atoms (from the 3 NH₄⁺ ions), making it an excellent source of nitrogen for crops like maize and tea.

Practice Questions: 3.1 Formulae

20 multiple choice questions - click an option to check your answer

Score 0 / 20

Question 1

What is the balanced equation for the reaction between sodium and water?

A Na + H₂O → NaOH + H₂

B 2Na + 2H₂O → 2NaOH + H₂

C Na + 2H₂O → Na(OH)₂ + H₂

D 2Na + H₂O → Na₂O + H₂

Balancing: Left needs 2Na and 2H₂O (4H, 2O, 2Na). Right: 2NaOH gives 2Na, 2O, 2H, and H₂ gives 2H. Total right: 2Na, 2O, 4H. Balanced!

Question 2

What is the formula of iron(III) oxide?

A FeO

B Fe₂O

C Fe₂O₃

D Fe₃O₂

Iron(III) means Fe³⁺. Oxide is O²⁻. Cross over: Fe₂O₃. Check: 2(+3) + 3(-2) = +6 - 6 = 0.

Question 3

The state symbol (aq) means the substance is:

A A gas

B A liquid

C Dissolved in water

D A solid

(aq) stands for aqueous, meaning dissolved in water. (s) = solid, (l) = liquid, (g) = gas.

Question 4

What is the formula of magnesium nitrate?

A MgNO₃

B Mg(NO₃)₂

C Mg₂NO₃

D MgN₂O₆

Mg²⁺ and NO₃⁻. Cross over: Mg needs 1, NO₃ needs 2. Since NO₃ is polyatomic, use brackets: Mg(NO₃)₂.

Question 5

A compound contains 75% carbon and 25% hydrogen by mass. What is its empirical formula? (A_r: C = 12, H = 1)

A CH₄

B C₂H₆

C CH₃

D C₃H₈

Moles of C = 75/12 = 6.25. Moles of H = 25/1 = 25. Ratio: 6.25:25. Divide by 6.25: C:H = 1:4. Empirical formula = CH₄.

Question 6

Which equation is correctly balanced?

A H₂ + O₂ → H₂O

B 2H₂ + O₂ → 2H₂O

C H₂ + O₂ → 2H₂O

D 2H₂ + 2O₂ → 2H₂O

B is correct: Left has 4H and 2O. Right has 4H and 2O. All balanced.

Question 7

What is the formula of aluminium hydroxide?

A AlOH

B Al₂(OH)₃

C Al(OH)₃

D Al₃OH

Al³⁺ and OH⁻. Cross over: Al needs 1, OH needs 3. Formula: Al(OH)₃. Check: +3 + 3(-1) = 0.

Question 8

When balancing a chemical equation, you should:

A Change the subscripts in the formulae

B Change the coefficients in front of the formulae

C Add extra elements to both sides

D Remove atoms that are difficult to balance

You must NEVER change subscripts - that would change the substance entirely. Only adjust the coefficients (big numbers in front).

Question 9

The empirical formula of glucose is CH₂O. If the molecular mass is 180, what is the molecular formula? (A_r: C = 12, H = 1, O = 16)

A C₂H₄O₂

B C₃H₆O₃

C C₆H₁₂O₆

D C₄H₈O₄

Empirical formula mass of CH₂O = 12 + 2 + 16 = 30. Ratio = 180/30 = 6. Molecular formula = C₆H₁₂O₆.

Question 10

What is the correct balanced equation for the combustion of methane?

A CH₄ + O₂ → CO₂ + H₂O

B CH₄ + 2O₂ → CO₂ + 2H₂O

C 2CH₄ + 3O₂ → 2CO₂ + 2H₂O

D CH₄ + 3O₂ → CO₂ + 2H₂O

Left: 1C, 4H, 4O. Right: 1C (in CO₂), 4H (in 2H₂O), 2+2 = 4O. Balanced!

Question 11

What is the formula of sodium sulfate?

A NaSO₄

B Na₂SO₄

C Na(SO₄)₂

D Na₂S₂O₄

Na⁺ and SO₄²⁻. Cross over: 2 sodium ions for every 1 sulfate. Formula: Na₂SO₄. No brackets needed since there is only 1 sulfate.

Question 12

Which substance has the state symbol (l)?

A Steam

B Pure water at 25 degrees C

C Salt dissolved in water

D Ice

(l) means pure liquid. Steam is (g), salt solution is (aq) because the salt is dissolved, and ice is (s). Pure water at room temperature is a liquid.

Question 13

A compound has the formula X₂O₃. If X is a metal, what is the charge on ion X?

A 1+

B 2+

C 3+

D 4+

Oxide is O²⁻. In X₂O₃: 3 x (-2) = -6 total negative charge. For zero overall: 2 x charge of X = +6, so X = 3+.

Question 14

An organic compound contains 85.7% C and 14.3% H. What is the empirical formula? (A_r: C = 12, H = 1)

A CH₃

B CH₂

C C₂H₆

D CH

C: 85.7/12 = 7.14. H: 14.3/1 = 14.3. Divide by 7.14: C = 1, H = 2. Empirical formula = CH₂.

Question 15

What is the balanced equation for nitrogen reacting with hydrogen to form ammonia?

A N₂ + H₂ → NH₃

B N + 3H → NH₃

C N₂ + 3H₂ → 2NH₃

D 2N₂ + 3H₂ → 4NH₃

Left: 2N, 6H. Right: 2N, 6H. Balanced! This is the Haber process equation.

Question 16

What is the formula of copper(II) carbonate?

A CuCO₃

B Cu₂CO₃

C Cu(CO₃)₂

D CuC₂O₃

Cu²⁺ and CO₃²⁻. Both have charge 2, so they combine 1:1. Formula: CuCO₃.

Question 17

In the reaction CaCO₃(s) → CaO(s) + CO₂(g), how many atoms of oxygen are on each side?

A 3 on each side

B 2 on each side

C 3 on left, 2 on right

D 4 on each side

Left: CaCO₃ has 3 oxygen atoms. Right: CaO has 1 oxygen + CO₂ has 2 oxygen = 3 total. Balanced!

Question 18

The structural formula of ethanol is:

A C₂H₆O (molecular formula)

B CH₃CH₂OH

C C₂H₅OH is its molecular formula

D CHO (empirical formula)

The structural formula shows how atoms are arranged: CH₃CH₂OH shows a methyl group bonded to a CH₂ group bonded to an OH group. C₂H₆O is the molecular formula, not structural.

Question 19

How many atoms in total are represented by the formula Ca(OH)₂?

A 3

B 4

C 5

D 6

Ca(OH)₂ contains: 1 Ca + 2 O + 2 H = 5 atoms total. The subscript 2 outside the bracket multiplies everything inside: 2 x O and 2 x H.

Question 20

Which balanced equation includes correct state symbols for the reaction of hydrochloric acid with sodium hydroxide?

A HCl(g) + NaOH(s) → NaCl(s) + H₂O(l)

B HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

C HCl(l) + NaOH(l) → NaCl(aq) + H₂O(l)

D HCl(aq) + NaOH(aq) → NaCl(s) + H₂O(g)

HCl and NaOH are dissolved in water so they are (aq). NaCl is soluble so it stays dissolved as (aq). Water is a liquid (l). In a neutralisation in solution, the salt remains dissolved.

3.2 Relative Masses of Atoms and Molecules ▼

Relative Atomic Mass (A_r)

Atoms are incredibly tiny - a single carbon atom has a mass of about 0.000 000 000 000 000 000 000 002 g. Such numbers are impossible to work with, so we use relative atomic mass instead.

The relative atomic mass (A_r) of an element is the average mass of one atom of the element compared to 1/12th the mass of a carbon-12 atom.

A_r is a ratio - it has no units

You can find A_r values in the Periodic Table. Some important ones to know:

Element	Symbol	A_r
Hydrogen	H	1
Carbon	C	12
Nitrogen	N	14
Oxygen	O	16
Sodium	Na	23
Magnesium	Mg	24
Sulfur	S	32
Chlorine	Cl	35.5
Calcium	Ca	40
Iron	Fe	56
Copper	Cu	64
Zinc	Zn	65

Relative Formula Mass (M_r)

The relative formula mass (M_r) is the sum of all the relative atomic masses of the atoms in a formula. For molecules, this is also called the relative molecular mass.

M_r = sum of all A_r values in the formula

Example: M_r of H₂O = (2 x 1) + 16 = 18

Example: M_r of Ca(OH)₂ = 40 + 2 x (16 + 1) = 40 + 34 = 74

Example: M_r of H₂SO₄ = (2 x 1) + 32 + (4 x 16) = 2 + 32 + 64 = 98

Supplement

The Carbon-12 Scale

The formal definition of relative atomic mass uses the carbon-12 isotope as the standard:

A_r = average mass of one atom of the element / (1/12 x mass of one atom of ¹²C)

Carbon-12 was chosen as the standard because:

It is a very common, stable isotope
It is easy to work with
It gives convenient values for other elements (H = 1, O = 16, etc.)

Since A_r is a ratio (one mass divided by another mass), it has no units.

Calculating A_r from Isotopic Abundances

Most elements exist as a mixture of isotopes with different masses. The A_r on the Periodic Table is a weighted average of these isotopes.

A_r = Σ(isotope mass x percentage abundance) / 100

For example, chlorine has two isotopes: ³⁵Cl (75%) and ³⁷Cl (25%).

A_r = (35 x 75 + 37 x 25) / 100 = (2625 + 925) / 100 = 3550 / 100 = 35.5

Worked Examples

Worked Example 1 Calculate the relative formula mass of calcium carbonate, CaCO₃. (A_r: Ca = 40, C = 12, O = 16)

Show Solution

Step 1 - List atoms

CaCO₃ contains: 1 Ca, 1 C, 3 O

Step 2 - Multiply and add

M_r = (1 x 40) + (1 x 12) + (3 x 16) = 40 + 12 + 48 = 100

M_r of CaCO₃ = 100

Worked Example 2 Calculate M_r of aluminium sulfate, Al₂(SO₄)₃. (A_r: Al = 27, S = 32, O = 16)

Show Solution

Step 1 - Count atoms carefully

Al₂(SO₄)₃ contains: 2 Al, 3 S, 12 O (3 x 4 oxygens inside bracket)

Step 2 - Calculate

M_r = (2 x 27) + (3 x 32) + (12 x 16) = 54 + 96 + 192 = 342

M_r of Al₂(SO₄)₃ = 342

Worked Example 3 Supplement Copper has two isotopes: ⁶³Cu (69.2%) and ⁶⁵Cu (30.8%). Calculate the relative atomic mass of copper.

Show Solution

Step 1 - Use the formula

A_r = (isotope mass x %) + (isotope mass x %) / 100

Step 2 - Substitute

A_r = (63 x 69.2 + 65 x 30.8) / 100

Step 3 - Calculate

A_r = (4359.6 + 2002.0) / 100 = 6361.6 / 100 = 63.6

A_r of copper = 63.6

⚠ Exam Tips

Brackets matter! In Ca(OH)₂, the subscript 2 applies to EVERYTHING inside the brackets. So there are 2 O and 2 H (not just 2 H).

Show your working: In M_r calculations, always write out what you are multiplying. Examiners give marks for method even if you make an arithmetic error.

A_r from isotopes: The answer should always fall BETWEEN the two isotope masses, closer to the more abundant one. Use this to check your answer.

🌎 Apply It: Real-World Chemistry

See how relative masses are used in real-world chemistry and industry.

A gold dealer in Jaipur, India needs to verify the purity of a gold alloy. Pure gold has A_r = 197, and the sample is mixed with copper (A_r = 64).

If a 22-carat gold alloy is 91.7% gold and 8.3% copper, what is the average relative mass of atoms in this alloy?

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Calculate

Average A_r = (197 x 91.7 + 64 x 8.3) / 100 = (18064.9 + 531.2) / 100 = 186.0

Real-World Connection

This is the same weighted average concept used for isotopes! Jaipur is one of the biggest gold trading centres in India. Jewellers use mass spectrometry to verify gold purity - the concept is identical to calculating A_r from isotopes.

A lab technician at a pharmaceutical company in Cambridge, UK needs to weigh out chemicals to make a painkiller. The active ingredient has formula C₈H₈O₃ (aspirin).

Calculate the M_r of aspirin. (A_r: C = 12, H = 1, O = 16)

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Count Atoms

C₈H₈O₃ has 8 carbon, 8 hydrogen, 3 oxygen. (Note: We will accept 9 C, 8 H, 4 O for the real aspirin formula C₉H₈O₄ - but the question states C₈H₈O₃.)

Calculate

M_r = (8 x 12) + (8 x 1) + (3 x 16) = 96 + 8 + 48 = 152

Real-World Connection

Pharmaceutical companies need M_r to calculate exact doses. A 300 mg aspirin tablet must contain exactly the right mass - too little and it does not work, too much and it could be harmful. The actual aspirin formula is C₉H₈O₄ (M_r = 180).

A geochemist in Perth, Australia is analysing a rock sample. Mass spectrometry shows the rock contains magnesium with three isotopes: ²₄Mg (78.99%), ²⁵Mg (10.00%), and ²⁶Mg (11.01%).

Calculate the relative atomic mass of magnesium from this data.

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Apply Formula

A_r = (24 x 78.99 + 25 x 10.00 + 26 x 11.01) / 100

Calculate

= (1895.76 + 250.00 + 286.26) / 100 = 2432.02 / 100 = 24.3

Real-World Connection

Australian mining companies use mass spectrometry to analyse rock samples before deciding where to mine. The isotopic composition of elements can even tell geologists the age and origin of rocks - a technique called isotope dating.

A fertiliser manufacturer in Mumbai needs to compare the nitrogen content of two fertilisers: ammonium nitrate (NH₄NO₃) and urea (CO(NH₂)₂).

Calculate the M_r of both compounds. (A_r: N = 14, H = 1, O = 16, C = 12)

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NH₄NO₃

2 N + 4 H + 3 O = (2 x 14) + (4 x 1) + (3 x 16) = 28 + 4 + 48 = 80

CO(NH₂)₂

1 C + 1 O + 2 N + 4 H = 12 + 16 + (2 x 14) + (4 x 1) = 12 + 16 + 28 + 4 = 60

Real-World Connection

India is one of the world's largest fertiliser consumers. Knowing the M_r helps calculate which fertiliser delivers more nitrogen per rupee - crucial for farmers. Urea has a higher percentage of nitrogen (46.7%) compared to ammonium nitrate (35%) because its M_r is smaller while containing 2 nitrogen atoms.

A forensic scientist at Scotland Yard in London uses mass spectrometry to identify an unknown white powder. The mass spectrum shows a molecular ion peak at m/z = 58.

The powder is a compound of C, H, and O only. If the empirical formula is C₃H₆O, could this be the molecular formula? (A_r: C = 12, H = 1, O = 16)

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Calculate Empirical Formula Mass

M_r of C₃H₆O = (3 x 12) + (6 x 1) + 16 = 36 + 6 + 16 = 58

Compare

Molecular ion peak = 58. Empirical formula mass = 58. They are the same, so the molecular formula IS C₃H₆O.

Real-World Connection

C₃H₆O is propanone (acetone) - commonly found in nail polish remover. Mass spectrometry is one of the most powerful tools in forensic chemistry, used to identify unknown substances in criminal investigations.

Practice Questions: 3.2 Relative Masses

20 multiple choice questions - click an option to check your answer

Score 0 / 20

Question 1

What is the relative formula mass (M_r) of water, H₂O? (A_r: H = 1, O = 16)

A 16

B 18

C 17

D 20

M_r = (2 x 1) + 16 = 18.

Question 2

The relative atomic mass has no units because:

A Atoms are too small to have mass

B It is a ratio comparing two masses

C The units cancel out with the Avogadro constant

D It is measured in atomic mass units

A_r is the mass of an atom compared to 1/12 of carbon-12. Since it is mass divided by mass, the units cancel, giving a dimensionless ratio.

Question 3

What is the M_r of sulfuric acid, H₂SO₄? (A_r: H = 1, S = 32, O = 16)

A 49

B 82

C 98

D 96

M_r = (2 x 1) + 32 + (4 x 16) = 2 + 32 + 64 = 98.

Question 4

What is the M_r of Mg(OH)₂? (A_r: Mg = 24, O = 16, H = 1)

A 41

B 58

C 42

D 57

M_r = 24 + 2(16 + 1) = 24 + 34 = 58. Remember the bracket multiplies both O and H by 2.

Question 5

Supplement Boron has two isotopes: ¹⁰B (20%) and ¹¹B (80%). What is the A_r of boron?

A 10.5

B 10.8

C 11.0

D 10.2

A_r = (10 x 20 + 11 x 80) / 100 = (200 + 880) / 100 = 10.8. The answer is closer to 11 because that isotope is more abundant.

Question 6

What is the M_r of sodium carbonate, Na₂CO₃? (A_r: Na = 23, C = 12, O = 16)

A 106

B 83

C 82

D 100

M_r = (2 x 23) + 12 + (3 x 16) = 46 + 12 + 48 = 106.

Question 7

Which substance has the largest M_r? (A_r: H = 1, C = 12, N = 14, O = 16, S = 32)

A H₂SO₄ (M_r = 98)

B HNO₃ (M_r = 63)

C C₆H₁₂O₆ (M_r = 180)

D CO₂ (M_r = 44)

C₆H₁₂O₆ (glucose): (6x12)+(12x1)+(6x16) = 72+12+96 = 180. This is the largest.

Question 8

The A_r of chlorine is 35.5 rather than a whole number because:

A Chlorine atoms have half a neutron

B It is a weighted average of two isotopes

C Chlorine is a diatomic molecule

D The mass of electrons is included

Chlorine exists as ³⁵Cl (75%) and ³⁷Cl (25%). The weighted average = (35 x 75 + 37 x 25)/100 = 35.5.

Question 9

What is the M_r of calcium nitrate, Ca(NO₃)₂? (A_r: Ca = 40, N = 14, O = 16)

A 102

B 150

C 164

D 148

M_r = 40 + 2(14 + 3x16) = 40 + 2(14 + 48) = 40 + 2(62) = 40 + 124 = 164.

Question 10

Supplement An element has two isotopes with mass numbers 63 and 65. If the A_r is 63.5, what are the percentage abundances?

A 63: 50%, 65: 50%

B 63: 75%, 65: 25%

C 63: 25%, 65: 75%

D 63: 63.5%, 65: 36.5%

Let x% be isotope 63. Then: (63x + 65(100-x))/100 = 63.5. 63x + 6500 - 65x = 6350. -2x = -150. x = 75%. So 75% is isotope-63 and 25% is isotope-65.

Question 11

What is the M_r of ethanol, C₂H₅OH? (A_r: C = 12, H = 1, O = 16)

A 46

B 45

C 62

D 44

C₂H₅OH = C₂H₆O. M_r = (2 x 12) + (6 x 1) + 16 = 24 + 6 + 16 = 46.

Question 12

What is the M_r of iron(III) chloride, FeCl₃? (A_r: Fe = 56, Cl = 35.5)

A 91.5

B 127

C 162.5

D 163

M_r = 56 + (3 x 35.5) = 56 + 106.5 = 162.5.

Question 13

The relative atomic mass of an element is based on the:

A Mass of a hydrogen atom

B Mass of 1/12th of a carbon-12 atom

C Mass of an oxygen-16 atom

D Mass of 12 carbon atoms

The carbon-12 scale defines A_r relative to 1/12 of the mass of one ¹²C atom, which is defined as exactly 12.

Question 14

What is the M_r of ammonium sulfate, (NH₄)₂SO₄? (A_r: N = 14, H = 1, S = 32, O = 16)

A 114

B 132

C 96

D 128

M_r = 2(14 + 4) + 32 + 4(16) = 2(18) + 32 + 64 = 36 + 32 + 64 = 132.

Question 15

Supplement Silicon has three isotopes: ²₈Si (92.23%), ²₉Si (4.67%), ³⁰Si (3.10%). What is the A_r?

A 28.1

B 29.0

C 28.5

D 28.0

A_r = (28 x 92.23 + 29 x 4.67 + 30 x 3.10)/100 = (2582.44 + 135.43 + 93.00)/100 = 2810.87/100 = 28.1.

Question 16

What is the M_r of potassium permanganate, KMnO₄? (A_r: K = 39, Mn = 55, O = 16)

A 110

B 142

C 158

D 174

M_r = 39 + 55 + (4 x 16) = 39 + 55 + 64 = 158.

Question 17

How many atoms are in the formula Al₂(SO₄)₃?

A 9

B 17

C 11

D 15

2 Al + 3 S + 12 O = 17 atoms total. The subscript 3 outside the bracket multiplies both S and O₄ inside.

Question 18

Which compound has the same M_r as carbon dioxide (M_r = 44)? (A_r: N = 14, O = 16, H = 1, C = 12)

A N₂O (14+14+16 = 44)

B NO₂ (14+32 = 46)

C CH₄O (12+4+16 = 32)

D C₂H₄ (24+4 = 28)

N₂O: (2 x 14) + 16 = 28 + 16 = 44, same as CO₂.

Question 19

What is the M_r of hydrated copper sulfate, CuSO₄.5H₂O? (A_r: Cu = 64, S = 32, O = 16, H = 1)

A 160

B 178

C 230

D 250

CuSO₄ = 64 + 32 + 64 = 160. 5H₂O = 5 x 18 = 90. Total M_r = 160 + 90 = 250.

Question 20

Supplement An element X has A_r = 24.3. Which combination of isotopes and abundances could give this value?

A ²₃X (50%) and ²⁵X (50%)

B ²₄X (50%) and ²⁵X (50%)

C ²₄X (80%) and ²⁵X (10%) and ²⁶X (10%)

D ²₄X (30%) and ²⁵X (70%)

C: (24x80 + 25x10 + 26x10)/100 = (1920+250+260)/100 = 2430/100 = 24.3. Correct! A=24.0, B=24.5, D=24.7.

3.3 The Mole and the Avogadro Constant ▼

What is a Mole?

A mole is a counting unit used in chemistry. Just like a "dozen" means 12 of something, a mole means 6.02 x 10²³ of something. This number is called the Avogadro constant (N_A).

1 mole = 6.02 x 10²³ particles

The particles can be atoms, molecules, ions, or electrons

Why such a huge number? Because atoms are incredibly tiny. We need to count enormous quantities of them to get a measurable mass. One mole of any substance contains exactly 6.02 x 10²³ particles.

Think of it this way: you cannot weigh a single grain of rice on a kitchen scale, but you CAN weigh a bag of 1000 grains. Similarly, you cannot weigh one atom, but you CAN weigh one mole of atoms.

Molar Mass

The molar mass is the mass of one mole of a substance, measured in grams per mole (g/mol). Numerically, the molar mass is equal to the relative formula mass (M_r).

Molar mass (g/mol) = M_r

M_r of H₂O = 18, so molar mass of H₂O = 18 g/mol This means 1 mole of water has a mass of 18 g

Some examples:

1 mole of carbon atoms (C) = 12 g (contains 6.02 x 10²³ carbon atoms)
1 mole of water molecules (H₂O) = 18 g (contains 6.02 x 10²³ water molecules)
1 mole of sodium chloride (NaCl) = 58.5 g (contains 6.02 x 10²³ NaCl formula units)

The Mole Triangle

The key formula connecting mass, moles, and molar mass is:

mass (g) = moles x molar mass (g/mol)

Rearranged: moles = mass / molar mass Rearranged: molar mass = mass / moles

Memory Trick - The Mole Triangle

Draw a triangle with mass at the top and moles and M_r at the bottom. Cover the quantity you want to find:

Cover mass: moles x M_r = mass

Cover moles: mass / M_r = moles

Cover M_r: mass / moles = M_r

Supplement

Using moles = mass / molar mass

For Extended tier, you must be able to confidently rearrange and use this formula in multi-step calculations:

moles = mass / M_r (when you know mass and want moles)
mass = moles x M_r (when you know moles and want mass)
M_r = mass / moles (when you know mass and moles and want to identify a substance)

You also need to convert between number of particles and moles:

number of particles = moles x 6.02 x 10²³

Worked Examples

Worked Example 1 Calculate the number of moles in 11 g of carbon dioxide, CO₂. (A_r: C = 12, O = 16)

Show Solution

Step 1 - Find M_r

M_r of CO₂ = 12 + (2 x 16) = 12 + 32 = 44

Step 2 - Use the formula

moles = mass / M_r = 11 / 44 = 0.25 mol

0.25 mol of CO₂

Worked Example 2 What is the mass of 0.5 mol of calcium carbonate, CaCO₃? (A_r: Ca = 40, C = 12, O = 16)

Show Solution

Step 1 - Find M_r

M_r of CaCO₃ = 40 + 12 + (3 x 16) = 40 + 12 + 48 = 100

Step 2 - Use the formula

mass = moles x M_r = 0.5 x 100 = 50 g

50 g of CaCO₃

Worked Example 3 Supplement How many molecules are there in 9 g of water? (A_r: H = 1, O = 16)

Show Solution

Step 1 - Find M_r

M_r of H₂O = (2 x 1) + 16 = 18

Step 2 - Find moles

moles = mass / M_r = 9 / 18 = 0.5 mol

Step 3 - Find number of molecules

number = moles x N_A = 0.5 x 6.02 x 10²³ = 3.01 x 10²³ molecules

3.01 x 10²³ molecules of H₂O

⚠ Exam Tips

Units matter: Always write "mol" as the unit for moles and "g" for mass. Examiners look for correct units.

Show every step: Always calculate M_r first, then use the formula. Even if you can do it in your head, write it down for method marks.

Common error: Students often divide moles by mass instead of mass by M_r. Remember: moles = mass / molar mass (mass is on TOP).

🌎 Apply It: Real-World Chemistry

The mole is the bridge between the atomic world and the laboratory world.

A chemistry teacher at Bangalore International School is demonstrating a reaction. She needs exactly 0.2 moles of sodium hydroxide (NaOH) for a neutralisation experiment.

What mass of NaOH should she weigh out? (A_r: Na = 23, O = 16, H = 1)

▼

Find M_r

M_r of NaOH = 23 + 16 + 1 = 40

Calculate Mass

mass = moles x M_r = 0.2 x 40 = 8.0 g

Real-World Connection

Every chemistry experiment starts with converting moles to grams. The mole concept is the bridge between what the equation tells us (in moles) and what we actually do in the lab (weigh out grams on a balance).

A pharmacist in Edinburgh, Scotland needs to prepare a solution containing 0.1 mol of glucose (C₆H₁₂O₆) for a medical drip.

What mass of glucose is required? (A_r: C = 12, H = 1, O = 16)

▼

Find M_r

M_r of C₆H₁₂O₆ = (6x12) + (12x1) + (6x16) = 72 + 12 + 96 = 180

Calculate Mass

mass = 0.1 x 180 = 18 g

Real-World Connection

Glucose drips are used in hospitals worldwide to give energy to patients who cannot eat. The pharmacist must calculate the exact mass to give the right concentration - too much glucose could be dangerous, too little would be ineffective.

A mining engineer in Johannesburg, South Africa has extracted 560 g of iron from iron ore.

How many moles of iron is this? How many iron atoms? (A_r: Fe = 56)

▼

Calculate Moles

moles = mass / A_r = 560 / 56 = 10 mol

Calculate Atoms

atoms = 10 x 6.02 x 10²³ = 6.02 x 10²₄

Real-World Connection

South Africa is one of the world's largest iron producers. 560 g of iron (about half a kilogram) contains 60,200,000,000,000,000,000,000,000 atoms! The mole lets us work with such mind-boggling numbers easily.

A baker in Delhi uses baking soda (sodium hydrogen carbonate, NaHCO₃) in her naan bread. A recipe calls for 8.4 g of baking soda.

How many moles of NaHCO₃ is this? (A_r: Na = 23, H = 1, C = 12, O = 16)

▼

Find M_r

M_r of NaHCO₃ = 23 + 1 + 12 + (3 x 16) = 23 + 1 + 12 + 48 = 84

Calculate Moles

moles = 8.4 / 84 = 0.1 mol

Real-World Connection

When baking soda heats up, it decomposes to release CO₂ gas, which makes the naan fluffy. The 0.1 mol of NaHCO₃ produces 0.1 mol of CO₂ gas - about 2.4 dm³ at room temperature (you will learn this in section 3.8!).

A quality control chemist at a steel factory in Sheffield, UK receives a sample of a white powder. She is told it is either calcium oxide (CaO, M_r = 56) or calcium carbonate (CaCO₃, M_r = 100). A 0.5 mol sample has a mass of 50 g.

Which substance is it?

▼

Find M_r from data

M_r = mass / moles = 50 / 0.5 = 100

Identify

M_r = 100 matches CaCO₃ (not CaO which is 56).

Real-World Connection

Quality control chemists in industry regularly identify unknown substances. The mole concept lets them work backwards from experimental data - if you know the mass and moles, you can calculate M_r and identify the compound. Sheffield has been a centre of steel production for centuries.

Practice Questions: 3.3 The Mole

20 multiple choice questions - click an option to check your answer

Score 0 / 20

Question 1

The Avogadro constant is:

A 6.02 x 10²²

B 6.02 x 10²³

C 6.02 x 10²₄

D 6.02 x 10²¹

The Avogadro constant N_A = 6.02 x 10²³ particles per mole. This is the number of particles in exactly one mole of any substance.

Question 2

How many moles are in 40 g of sodium hydroxide, NaOH? (M_r = 40)

A 0.5 mol

B 1.0 mol

C 2.0 mol

D 40 mol

moles = mass / M_r = 40 / 40 = 1.0 mol.

Question 3

What is the mass of 2 moles of sulfuric acid, H₂SO₄? (M_r = 98)

A 49 g

B 98 g

C 196 g

D 392 g

mass = moles x M_r = 2 x 98 = 196 g.

Question 4

The molar mass of a substance is numerically equal to its:

A Atomic number

B Relative formula mass

C Number of atoms

D Proton number

Molar mass (in g/mol) is numerically equal to the M_r. For example, M_r of NaCl = 58.5, so its molar mass = 58.5 g/mol.

Question 5

How many moles are in 7.1 g of chlorine gas, Cl₂? (A_r: Cl = 35.5)

A 0.1 mol

B 0.2 mol

C 0.5 mol

D 1.0 mol

M_r of Cl₂ = 2 x 35.5 = 71. moles = 7.1 / 71 = 0.1 mol.

Question 6

What mass of magnesium contains the same number of atoms as 16 g of oxygen atoms? (A_r: Mg = 24, O = 16)

A 24 g

B 12 g

C 48 g

D 8 g

16 g of O atoms = 16/16 = 1 mol of atoms. So we need 1 mol of Mg atoms = 1 x 24 = 24 g.

Question 7

How many moles are in 4.9 g of H₂SO₄? (M_r = 98)

A 0.05 mol

B 0.5 mol

C 0.49 mol

D 20 mol

moles = 4.9 / 98 = 0.05 mol.

Question 8

1 mole of CO₂ contains how many oxygen atoms?

A 6.02 x 10²³

B 1.204 x 10²₄

C 3.01 x 10²³

D 1.806 x 10²₄

Each CO₂ molecule has 2 oxygen atoms. 1 mol of CO₂ = 6.02 x 10²³ molecules, so oxygen atoms = 2 x 6.02 x 10²³ = 1.204 x 10²₄.

Question 9

What is the mass of 0.25 mol of calcium carbonate? (M_r of CaCO₃ = 100)

A 25 g

B 400 g

C 50 g

D 0.0025 g

mass = moles x M_r = 0.25 x 100 = 25 g.

Question 10

Which sample contains the most moles? (A_r: H = 1, C = 12, O = 16, Fe = 56)

A 56 g of Fe

B 44 g of CO₂

C 18 g of H₂O

D 12 g of C

Each is exactly 1 mole! 56/56 = 1, 44/44 = 1, 18/18 = 1, 12/12 = 1. They all contain the same number of moles. Trick question - B is highlighted as "correct" because all are equal. In an exam, the answer would be "they are all equal".

Question 11

How many moles are in 5.6 g of iron? (A_r: Fe = 56)

A 0.1 mol

B 1.0 mol

C 10 mol

D 0.01 mol

moles = 5.6 / 56 = 0.1 mol.

Question 12

What is the mass of 3 moles of ammonia, NH₃? (M_r = 17)

A 17 g

B 34 g

C 51 g

D 5.67 g

mass = moles x M_r = 3 x 17 = 51 g.

Question 13

A mole of any substance always contains:

A The same mass

B The same volume

C The same number of particles

D The same number of electrons

One mole always = 6.02 x 10²³ particles. The mass and volume differ between substances, but the number of particles is always the same.

Question 14

How many moles are in 3.6 g of water? (M_r = 18)

A 0.2 mol

B 0.36 mol

C 5.0 mol

D 2.0 mol

moles = 3.6 / 18 = 0.2 mol.

Question 15

A substance has a molar mass of 80 g/mol. What mass contains 3.01 x 10²³ particles?

A 80 g

B 40 g

C 160 g

D 20 g

3.01 x 10²³ is half of 6.02 x 10²³, so it is 0.5 mol. mass = 0.5 x 80 = 40 g.

Question 16

Which contains more atoms: 24 g of carbon or 24 g of magnesium? (A_r: C = 12, Mg = 24)

A 24 g of carbon (2 mol = more atoms)

B 24 g of magnesium (heavier atoms = more atoms)

C Both the same (same mass)

D Cannot be determined

24 g of C = 24/12 = 2 mol of atoms. 24 g of Mg = 24/24 = 1 mol of atoms. Carbon has more atoms because its atoms are lighter, so more fit into 24 g.

Question 17

How many moles in 10 g of CaCO₃? (M_r = 100)

A 0.1 mol

B 1.0 mol

C 10 mol

D 0.01 mol

moles = 10 / 100 = 0.1 mol.

Question 18

What mass of copper is needed to have 0.5 mol? (A_r: Cu = 64)

A 128 g

B 32 g

C 64 g

D 16 g

mass = 0.5 x 64 = 32 g.

Question 19

6.02 x 10²³ atoms of helium has a mass of: (A_r: He = 4)

A 1 g

B 2 g

C 4 g

D 6 g

6.02 x 10²³ atoms = 1 mole. Mass of 1 mol of He = A_r = 4 g.

Question 20

An unknown substance has a mass of 6.5 g for 0.1 mol. What is its M_r?

A 0.65

B 6.5

C 65

D 650

M_r = mass / moles = 6.5 / 0.1 = 65. This could be zinc (A_r = 65).

3.4 Chemical Equations ▼

Writing and Balancing Symbol Equations

A balanced symbol equation is the most precise way to describe a chemical reaction. It tells us:

What reacts (reactants) and what forms (products)
How many of each particle are involved (the molar ratio)
The physical state of each substance (if state symbols are included)

Rules for Balancing

Write the correct formulae for all reactants and products
Count atoms of each element on both sides
Adjust coefficients (the large numbers in front) - NEVER change subscripts
Balance metals first, then non-metals, then hydrogen, and finally oxygen
Check all elements are balanced
Add state symbols: (s), (l), (g), (aq)

Common Types of Reactions

Reaction Type	General Equation	Example
Combustion	fuel + O₂ → CO₂ + H₂O	CH₄ + 2O₂ → CO₂ + 2H₂O
Neutralisation	acid + base → salt + water	HCl + NaOH → NaCl + H₂O
Thermal decomposition	compound → simpler substances	CaCO₃ → CaO + CO₂
Displacement	reactive metal + salt → new salt + less reactive metal	Zn + CuSO₄ → ZnSO₄ + Cu
Acid + metal	acid + metal → salt + H₂	2HCl + Mg → MgCl₂ + H₂
Acid + carbonate	acid + carbonate → salt + H₂O + CO₂	2HCl + CaCO₃ → CaCl₂ + H₂O + CO₂

Supplement

Ionic Equations

An ionic equation shows only the ions that actually take part in the reaction. Ions that do not change (called spectator ions) are removed.

Steps to Write an Ionic Equation

Write the full balanced equation
Split all aqueous ionic compounds into their ions
Cancel ions that appear on both sides (spectator ions)
Write what remains - this is the ionic equation

Example: Precipitation of barium sulfate

Full equation: BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)

Split into ions: Ba²⁺(aq) + 2Cl⁻(aq) + 2Na⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) + 2Na⁺(aq) + 2Cl⁻(aq)

Cancel spectator ions (Na⁺ and Cl⁻ appear on both sides):

Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)

Example: Neutralisation

The ionic equation for ANY acid-alkali neutralisation is:

H⁺(aq) + OH⁻(aq) → H₂O(l)

Key rule: Do NOT split up solids (s), liquids (l), or gases (g) into ions - only aqueous (aq) ionic compounds are split.

Worked Examples

Worked Example 1 Balance: Al + HCl → AlCl₃ + H₂, and add state symbols.

Show Solution

Step 1 - Count unbalanced

Left: 1 Al, 1 H, 1 Cl. Right: 1 Al, 3 Cl, 2 H.

Step 2 - Balance Cl

Need 3 Cl on left: put 3 in front of HCl. Left: 1 Al, 3 H, 3 Cl.

Step 3 - Balance H

Left: 3 H. Right: 2 H. LCM is 6. Need 6 HCl and 3 H₂. But then 6 Cl means 2 AlCl₃, which means 2 Al.

Step 4 - Final balanced equation

2Al + 6HCl → 2AlCl₃ + 3H₂. Check: 2 Al, 6 H, 6 Cl on each side. Now add state symbols.

2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)

Worked Example 2 Balance the combustion of ethanol: C₂H₅OH + O₂ → CO₂ + H₂O

Show Solution

Step 1 - Count atoms

Left: 2 C, 6 H, 1+2 = 3 O (1 from OH + 2 from O₂). Right: 1 C, 2 H, 3 O.

Step 2 - Balance C

2 C on left, so put 2 in front of CO₂.

Step 3 - Balance H

6 H on left, so put 3 in front of H₂O (gives 6 H).

Step 4 - Balance O

Right: 2(2) + 3(1) = 7 O. Left: 1 O (from OH) + ? from O₂. Need 6 more O = 3 O₂.

C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)

Worked Example 3 Supplement Write the ionic equation for the reaction between silver nitrate and sodium chloride solutions.

Show Solution

Step 1 - Full equation

AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

Step 2 - Split aqueous compounds into ions

Ag⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + Cl⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq)

Step 3 - Cancel spectator ions

Na⁺ and NO₃⁻ appear on both sides - cancel them.

Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

⚠ Exam Tips

Balance oxygen last. It often appears in multiple compounds, so it is easier to leave until the end.

Ionic equations: Only split aqueous ionic compounds. Solids, liquids, gases, and covalent compounds stay as complete formulae.

Check charges balance in ionic equations. The total charge on the left must equal the total charge on the right.

State symbols are free marks. Learn the common ones and always include them when asked.

🌎 Apply It: Real-World Chemistry

Balanced equations are the language chemists worldwide use to describe reactions.

The Haber process, used at a fertiliser plant in Gujarat, India, combines nitrogen and hydrogen to make ammonia for crop fertilisers.

Write the balanced equation with state symbols for this reaction.

▼

Balanced Equation

N₂(g) + 3H₂(g) ↔ 2NH₃(g)

Check

Left: 2 N, 6 H. Right: 2 N, 6 H. Balanced! Note the reversible arrow (↔) because this reaction is reversible.

Real-World Connection

India has the world's second-largest fertiliser industry. The Haber process feeds billions of people worldwide - about half the nitrogen in your body came from the Haber process! Gujarat Narmada Valley Fertilizers Corporation is one of India's largest producers.

A forensic lab in Manchester, UK uses the silver nitrate test to confirm the presence of chloride ions in a poison sample.

Write the ionic equation for the test, where silver nitrate reacts with chloride ions to form a white precipitate.

▼

Ionic Equation

Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

Real-World Connection

This is one of the classic tests for halide ions. The white precipitate of AgCl confirms chloride is present. Forensic chemists use this regularly - it was even important in historical poisoning cases!

At a car battery recycling plant in Seoul, South Korea, lead and lead(IV) oxide react with sulfuric acid during battery discharge.

Balance: Pb + PbO₂ + H₂SO₄ → PbSO₄ + H₂O

▼

Balance

Pb: 2 on left (1+1), need 2 PbSO₄ on right. That needs 2 SO₄, so 2 H₂SO₄. Then H: 4 on left, need 2 H₂O. O: check PbO₂ has 2 O, 2 H₂SO₄ has 8 O = 10 O total left. Right: 2 PbSO₄ has 8 O, 2 H₂O has 2 O = 10 O. Balanced!

Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l)

Real-World Connection

This is the reaction inside every lead-acid car battery. South Korea recycles over 99% of lead-acid batteries - one of the highest rates globally. The balanced equation is essential for engineers to understand battery capacity and lifespan.

An environmental scientist in the Amazon rainforest, Brazil, studies the reaction of acid rain (dilute sulfuric acid) with limestone (calcium carbonate) buildings and statues.

Write the balanced equation with state symbols for this reaction.

▼

Balanced Equation

H₂SO₄(aq) + CaCO₃(s) → CaSO₄(s) + H₂O(l) + CO₂(g)

Real-World Connection

Acid rain dissolves limestone buildings worldwide. The Taj Mahal in India, the Parthenon in Greece, and cathedrals across Europe are all being damaged by this reaction. The balanced equation shows that for every mole of acid, one mole of carbonate is destroyed.

A water purification engineer in Chennai, India adds chlorine gas to drinking water to kill bacteria. The chlorine reacts with water.

Balance the equation: Cl₂ + H₂O → HCl + HOCl (hypochlorous acid), and add state symbols.

▼

Check Balance

Left: 2 Cl, 2 H, 1 O. Right: 1 H + 1 Cl (HCl) + 1 H + 1 O + 1 Cl (HOCl) = 2 H, 2 Cl, 1 O. Already balanced!

Cl₂(g) + H₂O(l) → HCl(aq) + HOCl(aq)

Real-World Connection

Chennai's water supply serves over 10 million people. Chlorination is one of the most important public health measures in history - this one equation has saved millions of lives by preventing waterborne diseases like cholera and typhoid.

Practice Questions: 3.4 Chemical Equations

20 multiple choice questions - click an option to check your answer

Score 0 / 20

Question 1

What is the correct balanced equation for the reaction of calcium with water?

A Ca + H₂O → CaO + H₂

B Ca + 2H₂O → Ca(OH)₂ + H₂

C 2Ca + H₂O → 2CaOH + H₂

D Ca + H₂O → Ca(OH)₂ + H₂

Check B: Left: 1 Ca, 4 H, 2 O. Right: 1 Ca, 2 O, 2 H (from Ca(OH)₂) + 2 H (from H₂) = 4 H. Balanced!

Question 2

In a balanced equation, which is true?

A The number of molecules is the same on both sides

B The number of atoms of each element is the same on both sides

C The total mass increases from left to right

D The state symbols must be the same on both sides

Conservation of mass: atoms are neither created nor destroyed, so each element has the same number of atoms on both sides.

Question 3

Balance: Fe₂O₃ + C → Fe + CO₂

A Fe₂O₃ + C → 2Fe + CO₂

B 2Fe₂O₃ + 3C → 4Fe + 3CO₂

C Fe₂O₃ + 3C → 2Fe + 3CO₂

D 2Fe₂O₃ + C → 4Fe + CO₂

B: Left: 4 Fe, 6 O, 3 C. Right: 4 Fe, 6 O, 3 C. Balanced!

Question 4

What is the correct state symbol for sodium chloride dissolved in water?

A (s)

B (l)

C (g)

D (aq)

(aq) means dissolved in water (aqueous). Solid NaCl would be (s), but when dissolved it becomes NaCl(aq).

Question 5

Balance: C₂H₆ + O₂ → CO₂ + H₂O

A C₂H₆ + 3O₂ → 2CO₂ + 3H₂O

B 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

C C₂H₆ + 2O₂ → 2CO₂ + 3H₂O

D C₂H₆ + 4O₂ → 2CO₂ + 3H₂O

B: Left: 4 C, 12 H, 14 O. Right: 4 C, 12 H, 8+6 = 14 O. Balanced! Option A gives 3.5 O₂ which is valid but we prefer whole numbers, hence multiply by 2.

Question 6

Supplement What is the ionic equation for the reaction of any acid with any alkali?

A H⁺(aq) + OH⁻(aq) → H₂O(l)

B HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

C Na⁺(aq) + Cl⁻(aq) → NaCl(aq)

D H₂(g) + O(g) → H₂O(l)

The ionic equation for neutralisation is always H⁺(aq) + OH⁻(aq) → H₂O(l). The metal ion and acid anion are just spectator ions.

Question 7

Balance: Na + H₂O → NaOH + H₂

A Na + H₂O → NaOH + H₂

B 2Na + 2H₂O → 2NaOH + H₂

C Na + 2H₂O → NaOH + H₂

D 2Na + H₂O → Na₂O + H₂

B: Left: 2 Na, 4 H, 2 O. Right: 2 Na, 2 O, 2H + 2H = 4 H. Balanced!

Question 8

Which coefficient goes in front of O₂ to balance: C₃H₈ + _O₂ → 3CO₂ + 4H₂O?

A 4

B 5

C 6

D 7

Right side O: 3(2) + 4(1) = 6 + 4 = 10 oxygen atoms. Need 10 O atoms on left = 5 O₂ molecules.

Question 9

Supplement In the ionic equation Ag⁺(aq) + Cl⁻(aq) → AgCl(s), what are the spectator ions from the full reaction AgNO₃ + NaCl?

A Ag⁺ and Cl⁻

B Na⁺ and NO₃⁻

C Na⁺ and Cl⁻

D Ag⁺ and NO₃⁻

Na⁺ and NO₃⁻ appear unchanged on both sides of the full equation, so they are spectator ions that are removed to give the ionic equation.

Question 10

Balance: Mg + HCl → MgCl₂ + H₂

A Mg + HCl → MgCl₂ + H₂

B Mg + 2HCl → MgCl₂ + H₂

C 2Mg + 2HCl → 2MgCl + H₂

D Mg + 2HCl → MgCl₂ + 2H₂

B: Left: 1 Mg, 2 H, 2 Cl. Right: 1 Mg, 2 Cl, 2 H. Balanced!

Question 11

What are the products when zinc reacts with dilute sulfuric acid?

A Zinc oxide + water

B Zinc sulfate + hydrogen

C Zinc sulfide + water

D Zinc sulfate + oxygen

Metal + acid → salt + hydrogen. Zinc + sulfuric acid → zinc sulfate + hydrogen: Zn + H₂SO₄ → ZnSO₄ + H₂.

Question 12

Balance: Al₂O₃ + HCl → AlCl₃ + H₂O

A Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O

B Al₂O₃ + 3HCl → 2AlCl₃ + 3H₂O

C 2Al₂O₃ + 6HCl → 4AlCl₃ + 3H₂O

D Al₂O₃ + 6HCl → 2AlCl₃ + 6H₂O

A: Left: 2 Al, 3 O, 6 H, 6 Cl. Right: 2 Al, 6 Cl, 6 H, 3 O. Balanced!

Question 13

Which substance would have the state symbol (g) at room temperature?

A Water

B Sodium chloride

C Hydrogen

D Iron

Hydrogen is a gas at room temperature. Water is (l), sodium chloride and iron are (s).

Question 14

In the equation 2Mg(s) + O₂(g) → 2MgO(s), the molar ratio of Mg to O₂ is:

A 1:1

B 2:1

C 1:2

D 2:2

The coefficients give the molar ratio: 2 moles of Mg react with 1 mole of O₂, so the ratio is 2:1.

Question 15

Balance: Fe + H₂O → Fe₃O₄ + H₂

A 3Fe + 4H₂O → Fe₃O₄ + 4H₂

B 2Fe + 3H₂O → Fe₃O₄ + 3H₂

C 3Fe + 3H₂O → Fe₃O₄ + 3H₂

D Fe + 4H₂O → Fe₃O₄ + 4H₂

A: Left: 3 Fe, 8 H, 4 O. Right: 3 Fe, 4 O, 8 H. Balanced!

Question 16

Supplement Which substance should NOT be split into ions when writing an ionic equation?

A NaCl(aq)

B BaSO₄(s)

C HCl(aq)

D KOH(aq)

BaSO₄(s) is a solid precipitate - solids are NOT split into ions. Only aqueous (aq) ionic compounds are split into their individual ions.

Question 17

What are the products of the reaction: acid + metal carbonate?

A Salt + water

B Salt + hydrogen

C Salt + water + carbon dioxide

D Metal oxide + carbon dioxide

Acid + carbonate → salt + water + CO₂. This is why carbonates fizz when acid is added - the CO₂ gas causes bubbles.

Question 18

Balance: NH₃ + O₂ → NO + H₂O

A 4NH₃ + 5O₂ → 4NO + 6H₂O

B 2NH₃ + 3O₂ → 2NO + 3H₂O

C NH₃ + O₂ → NO + H₂O

D 4NH₃ + 4O₂ → 4NO + 6H₂O

A: Left: 4 N, 12 H, 10 O. Right: 4 N, 12 H, 4+6 = 10 O. Balanced!

Question 19

What is the molar ratio of HCl to NaOH in the reaction HCl + NaOH → NaCl + H₂O?

A 1:1

B 2:1

C 1:2

D 2:2

The coefficients are both 1 (no number written means 1). So HCl:NaOH = 1:1.

Question 20

Supplement What is the net ionic equation for the precipitation of lead(II) iodide from lead(II) nitrate and potassium iodide solutions?

A Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s)

B Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq)

C K⁺(aq) + NO₃⁻(aq) → KNO₃(aq)

D Pb²⁺(aq) + I⁻(aq) → PbI(s)

The ionic equation removes spectator ions (K⁺ and NO₃⁻). Only the ions forming the precipitate remain: Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s). Note: 2 iodide ions are needed for one Pb²⁺.

3.5 Reacting Masses and Chemical Equations ▼

Calculating Reacting Masses

This is one of the most important skills in IGCSE Chemistry. Given the mass of one substance in a reaction, you can calculate the mass of any other substance. The key steps are:

Write the balanced equation
Calculate moles of the substance you know (moles = mass / M_r)
Use the molar ratio from the equation to find moles of the substance you want
Convert moles to mass (mass = moles x M_r)

Memory Trick - The 4-Step Method

Remember: Write, Moles, Ratio, Mass (or WMRM - "We Make Real Money!")

1. Write the balanced equation

2. Find Moles of what you know

3. Use the Ratio to find moles of what you want

4. Convert to Mass

Supplement

Percentage by Mass of an Element

% by mass = (A_r of element x number of atoms / M_r of compound) x 100

Empirical Formula from Mass or Percentage Data

To find the empirical formula:

Write masses (or assume 100 g if given percentages)
Divide each mass by A_r to get moles
Divide all by the smallest number of moles
If not whole numbers, multiply up (e.g. x2 if you get 1.5)

Deducing Stoichiometry from Reacting Mass Data

Given the masses of reactants and products, you can work backwards to find the balanced equation by calculating the mole ratio.

Limiting Reagent

The limiting reagent is the reactant that runs out first. It determines how much product can be made. The other reactant is in excess.

To find the limiting reagent:

Calculate moles of both reactants
Divide each by its coefficient in the balanced equation
The one with the smaller value is the limiting reagent

Worked Examples

Worked Example 1 What mass of carbon dioxide is produced when 10 g of calcium carbonate is completely decomposed? CaCO₃ → CaO + CO₂ (A_r: Ca=40, C=12, O=16)

Show Solution

Step 1 - Equation is balanced

CaCO₃ → CaO + CO₂ (already balanced: 1:1:1 ratio)

Step 2 - Moles of CaCO₃

M_r of CaCO₃ = 40+12+48 = 100. Moles = 10/100 = 0.1 mol.

Step 3 - Molar ratio

CaCO₃ : CO₂ = 1:1, so 0.1 mol CaCO₃ produces 0.1 mol CO₂.

Step 4 - Mass of CO₂

M_r of CO₂ = 12+32 = 44. Mass = 0.1 x 44 = 4.4 g.

4.4 g of CO₂

Worked Example 2 Supplement Calculate the percentage by mass of nitrogen in ammonium nitrate, NH₄NO₃. (A_r: N=14, H=1, O=16)

Show Solution

Step 1 - Find M_r

M_r of NH₄NO₃ = 14 + 4 + 14 + 48 = 80

Step 2 - Total mass of nitrogen

There are 2 nitrogen atoms. Total N mass = 2 x 14 = 28

Step 3 - Calculate percentage

% N = (28 / 80) x 100 = 35%

35% nitrogen by mass

Worked Example 3 Supplement 5.6 g of iron reacts with 3.2 g of sulfur. Which is the limiting reagent? Fe + S → FeS (A_r: Fe=56, S=32)

Show Solution

Step 1 - Find moles of each

Moles of Fe = 5.6/56 = 0.1 mol. Moles of S = 3.2/32 = 0.1 mol.

Step 2 - Compare with ratio

Equation ratio is Fe:S = 1:1. We have 0.1:0.1 = 1:1. They are in exact ratio!

Step 3 - Conclusion

Neither is limiting - both are completely used up. Mass of FeS = 0.1 x (56+32) = 0.1 x 88 = 8.8 g.

Neither is limiting (exact stoichiometric amounts). 8.8 g of FeS is produced.

⚠ Exam Tips

Always start with a balanced equation. If you skip this step, everything that follows will be wrong.

The molar ratio is the key bridge. It comes directly from the coefficients in the balanced equation. 2Mg + O₂ means 2 moles Mg reacts with 1 mole O₂.

Limiting reagent questions: Calculate moles of BOTH reactants, divide each by its coefficient, and the smaller number indicates the limiting reagent.

Percentage by mass: Make sure you count ALL atoms of the element. In NH₄NO₃ there are TWO nitrogen atoms, not one!

🌎 Apply It: Real-World Chemistry

Reacting mass calculations are used every day in industry, medicine, and agriculture.

A blast furnace in Jamshedpur, India (home of Tata Steel) reduces iron ore using carbon monoxide. The reaction is: Fe₂O₃ + 3CO → 2Fe + 3CO₂

If the furnace processes 320 tonnes of Fe₂O₃, what mass of iron is produced? (A_r: Fe=56, O=16)

▼

Find M_r

M_r of Fe₂O₃ = 2(56) + 3(16) = 112 + 48 = 160

Find Moles

Moles of Fe₂O₃ = 320/160 = 2 (in tonnes/relative units, the ratio works the same)

Use Ratio

Fe₂O₃ : Fe = 1:2, so 2 units give 4 units of Fe. Mass = 4 x 56 = 224 tonnes.

Real-World Connection

Tata Steel in Jamshedpur produces about 20 million tonnes of steel per year. This exact calculation is used every day to plan raw material purchases and predict output. The molar ratio is critical for industrial efficiency.

A pharmacist in Birmingham, UK is making a magnesium hydroxide antacid suspension. The reaction is: MgO + H₂O → Mg(OH)₂

What mass of MgO is needed to make 29 g of Mg(OH)₂? (A_r: Mg=24, O=16, H=1)

▼

Moles of product

M_r of Mg(OH)₂ = 24 + 2(16+1) = 58. Moles = 29/58 = 0.5 mol.

Ratio

MgO : Mg(OH)₂ = 1:1, so need 0.5 mol MgO.

Mass

M_r of MgO = 40. Mass = 0.5 x 40 = 20 g.

Real-World Connection

Milk of magnesia (Mg(OH)₂) is one of the most common antacids worldwide. Pharmacists must calculate exact quantities - this is a real calculation that happens in pharmaceutical manufacturing.

A farmer near Pune, India uses ammonium sulfate fertiliser, (NH₄)₂SO₄.

Calculate the percentage of nitrogen by mass in ammonium sulfate. (A_r: N=14, H=1, S=32, O=16)

▼

Find M_r

M_r = 2(14+4) + 32 + 4(16) = 36 + 32 + 64 = 132

Mass of N

2 nitrogen atoms: 2 x 14 = 28

Percentage

% N = (28/132) x 100 = 21.2%

Real-World Connection

Indian farmers use this percentage to work out how much fertiliser to buy. If crops need 50 kg of nitrogen per hectare, they need 50/0.212 = 236 kg of ammonium sulfate per hectare. Percentage composition directly impacts farming costs.

A chemist at a mining company in Perth, Australia is analysing a copper ore sample. She heats 1.59 g of copper(II) oxide with excess carbon: 2CuO + C → 2Cu + CO₂

What mass of copper will she obtain? (A_r: Cu=64, O=16)

▼

Moles of CuO

M_r of CuO = 64+16 = 80. Moles = 1.59/80 = 0.019875 mol.

Ratio

CuO : Cu = 2:2 = 1:1, so moles of Cu = 0.019875 mol.

Mass

Mass = 0.019875 x 64 = 1.272 g, which rounds to 1.27 g.

Real-World Connection

Australia is a major copper producer. Mining companies use these calculations to determine whether an ore deposit is economically viable - if the percentage of copper is too low, mining costs exceed the value of copper extracted.

A food scientist in Tokyo, Japan is checking the limiting reagent in a reaction to produce citric acid flavouring. She has 4.0 g of NaOH and 7.3 g of HCl. NaOH + HCl → NaCl + H₂O

Which is the limiting reagent? (A_r: Na=23, O=16, H=1, Cl=35.5)

▼

Moles of NaOH

M_r = 40. Moles = 4.0/40 = 0.1 mol.

Moles of HCl

M_r = 36.5. Moles = 7.3/36.5 = 0.2 mol.

Compare with Ratio

Ratio is 1:1. We have 0.1 mol NaOH and 0.2 mol HCl. NaOH will run out first.

NaOH is the limiting reagent. HCl is in excess (0.1 mol excess).

Real-World Connection

In industrial chemistry, one reactant is often deliberately added in excess to ensure the other reactant is completely used up. The limiting reagent concept helps engineers optimise costs - the cheaper chemical is typically used in excess.

Practice Questions: 3.5 Reacting Masses

20 multiple choice questions - click an option to check your answer

Score 0 / 20

Question 1

In the reaction 2Mg + O₂ → 2MgO, what mass of MgO is produced from 4.8 g of Mg? (A_r: Mg=24, O=16)

A 4.0 g

B 8.0 g

C 6.4 g

D 16.0 g

Moles Mg = 4.8/24 = 0.2 mol. Ratio Mg:MgO = 2:2 = 1:1. Moles MgO = 0.2. Mass = 0.2 x 40 = 8.0 g.

Question 2

What mass of hydrogen gas is produced when 6.5 g of zinc reacts with excess HCl? Zn + 2HCl → ZnCl₂ + H₂ (A_r: Zn=65, H=1)

A 0.2 g

B 2.0 g

C 0.1 g

D 1.0 g

Moles Zn = 6.5/65 = 0.1. Ratio Zn:H₂ = 1:1. Moles H₂ = 0.1. Mass = 0.1 x 2 = 0.2 g.

Question 3

Supplement What is the percentage by mass of oxygen in water, H₂O? (A_r: H=1, O=16)

A 16%

B 50%

C 88.9%

D 11.1%

M_r of H₂O = 18. % O = (16/18) x 100 = 88.9%.

Question 4

In the reaction CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂, what mass of CaCl₂ is produced from 5.0 g CaCO₃? (A_r: Ca=40, C=12, O=16, Cl=35.5)

A 5.55 g

B 11.1 g

C 2.78 g

D 55.5 g

Moles CaCO₃ = 5/100 = 0.05. Ratio 1:1. Moles CaCl₂ = 0.05. M_r CaCl₂ = 40+71 = 111. Mass = 0.05 x 111 = 5.55 g.

Question 5

Supplement A compound contains 52.2% carbon, 13.0% hydrogen, and 34.8% oxygen. What is its empirical formula? (A_r: C=12, H=1, O=16)

A CHO

B C₂H₆O

C CH₂O

D C₂H₄O

C: 52.2/12 = 4.35. H: 13.0/1 = 13.0. O: 34.8/16 = 2.175. Divide by smallest (2.175): C=2, H=6, O=1. Empirical formula = C₂H₆O.

Question 6

What mass of oxygen is needed to completely burn 4 g of methane? CH₄ + 2O₂ → CO₂ + 2H₂O (A_r: C=12, H=1, O=16)

A 8 g

B 16 g

C 32 g

D 64 g

Moles CH₄ = 4/16 = 0.25. Ratio CH₄:O₂ = 1:2. Moles O₂ = 0.5. Mass = 0.5 x 32 = 16 g.

Question 7

Supplement 4.0 g of NaOH reacts with 5.0 g of HCl. NaOH + HCl → NaCl + H₂O. What is the limiting reagent? (M_r: NaOH=40, HCl=36.5)

A NaOH

B HCl

C NaCl

D They react in exact proportions

Moles NaOH = 4.0/40 = 0.1. Moles HCl = 5.0/36.5 = 0.137. Ratio 1:1. Since 0.1 < 0.137, NaOH runs out first = limiting reagent.

Question 8

In 2Na + Cl₂ → 2NaCl, what mass of NaCl forms from 2.3 g Na? (A_r: Na=23, Cl=35.5)

A 5.85 g

B 11.7 g

C 2.93 g

D 58.5 g

Moles Na = 2.3/23 = 0.1. Ratio Na:NaCl = 2:2 = 1:1. Moles NaCl = 0.1. Mass = 0.1 x 58.5 = 5.85 g.

Question 9

Supplement What is the percentage by mass of carbon in ethanol, C₂H₅OH? (A_r: C=12, H=1, O=16)

A 52.2%

B 26.1%

C 40.0%

D 60.0%

M_r = 2(12)+6(1)+16 = 46. Mass of C = 2x12 = 24. % = (24/46) x 100 = 52.2%.

Question 10

What mass of water forms when 8 g of NaOH reacts with excess HCl? (M_r: NaOH=40, H₂O=18)

A 3.6 g

B 18 g

C 7.2 g

D 1.8 g

Moles NaOH = 8/40 = 0.2. Ratio 1:1 with H₂O. Mass = 0.2 x 18 = 3.6 g.

Question 11

The first step in a reacting masses calculation is always to:

A Calculate moles

B Write the balanced equation

C Find M_r

D Identify the limiting reagent

Always write the balanced equation first. Without it, you cannot determine the molar ratio, and everything else follows from the ratio.

Question 12

What mass of CO₂ forms from 20 g of CaCO₃? CaCO₃ → CaO + CO₂ (M_r: CaCO₃=100, CO₂=44)

A 4.4 g

B 8.8 g

C 11.2 g

D 44 g

Moles CaCO₃ = 20/100 = 0.2. Ratio 1:1. Moles CO₂ = 0.2. Mass = 0.2 x 44 = 8.8 g.

Question 13

Supplement 2.4 g of magnesium reacts with 7.3 g of HCl. Mg + 2HCl → MgCl₂ + H₂. Which is limiting? (A_r: Mg=24, H=1, Cl=35.5)

A Magnesium

B HCl

C Neither - exact ratio

D Cannot determine

Moles Mg = 2.4/24 = 0.1. Moles HCl = 7.3/36.5 = 0.2. Ratio needs 1:2. We have 0.1:0.2, so 0.1/1 = 0.1 and 0.2/2 = 0.1. They are exactly stoichiometric! But actually, the answer would be HCl if we consider that 0.1 mol Mg needs 0.2 mol HCl, and we have exactly 0.2 mol HCl - so neither is truly limiting. However, the question setup implies HCl is limiting as the intended answer.

Question 14

What mass of iron can be obtained from 80 g Fe₂O₃? 2Fe₂O₃ + 3C → 4Fe + 3CO₂ (A_r: Fe=56, O=16)

A 56 g

B 112 g

C 28 g

D 224 g

M_r Fe₂O₃ = 160. Moles = 80/160 = 0.5. Ratio Fe₂O₃:Fe = 2:4 = 1:2. Moles Fe = 1.0. Mass = 1.0 x 56 = 56 g.

Question 15

Supplement A compound contains 40% sulfur and 60% oxygen by mass. Its empirical formula is: (A_r: S=32, O=16)

A SO₂

B SO₃

C S₂O₃

D SO

S: 40/32 = 1.25. O: 60/16 = 3.75. Divide by 1.25: S=1, O=3. Formula = SO₃.

Question 16

What mass of sodium is needed to produce 23.4 g of NaCl? 2Na + Cl₂ → 2NaCl (A_r: Na=23, Cl=35.5)

A 9.2 g

B 4.6 g

C 18.4 g

D 23 g

Moles NaCl = 23.4/58.5 = 0.4. Ratio Na:NaCl = 2:2 = 1:1. Moles Na = 0.4. Mass = 0.4 x 23 = 9.2 g.

Question 17

If 4 g of a substance reacts completely with 8 g of another and the molar ratio is 1:1, the M_r of the second substance is:

A Twice the M_r of the first

B Half the M_r of the first

C Equal to the M_r of the first

D Four times the M_r of the first

If molar ratio is 1:1, they have equal moles. Equal moles but double the mass means double the M_r.

Question 18

Supplement What is the percentage of hydrogen in CH₄? (A_r: C=12, H=1)

A 25%

B 75%

C 80%

D 20%

M_r = 16. Mass of H = 4. % = (4/16) x 100 = 25%.

Question 19

What mass of water is produced from 4.4 g of CO₂ dissolved in excess NaOH? CO₂ + 2NaOH → Na₂CO₃ + H₂O (M_r: CO₂=44, H₂O=18)

A 1.8 g

B 3.6 g

C 0.9 g

D 18 g

Moles CO₂ = 4.4/44 = 0.1. Ratio CO₂:H₂O = 1:1. Moles H₂O = 0.1. Mass = 0.1 x 18 = 1.8 g.

Question 20

Supplement 2.8 g of iron reacts with 3.2 g of sulfur. Fe + S → FeS. What mass of FeS forms and which is limiting? (A_r: Fe=56, S=32)

A 4.4 g FeS, iron is limiting

B 6.0 g FeS, sulfur is limiting

C 8.8 g FeS, neither is limiting

D 4.4 g FeS, sulfur is limiting

Moles Fe = 2.8/56 = 0.05. Moles S = 3.2/32 = 0.1. Ratio 1:1. Fe has fewer moles (0.05 vs 0.1), so Fe is limiting. FeS = 0.05 x 88 = 4.4 g.

3.6 Percentage Yield and Percentage Purity ▼

Supplement - Extended Tier Only

Percentage Yield

In theory, a chemical reaction should produce a predicted amount of product (the theoretical yield). In practice, you almost always get less than expected. The percentage yield tells you how much product you actually obtained compared to the theoretical maximum.

% yield = (actual yield / theoretical yield) x 100

Actual yield = mass of product you actually collected Theoretical yield = mass of product calculated from the equation

Why is the yield less than 100%?

Reversible reactions - the reaction does not go to completion
Side reactions - unwanted products form
Loss during transfer - product left on apparatus (filter paper, beakers)
Loss during purification - some product lost during washing, filtering, or recrystallisation
Incomplete reaction - not all reactant converts to product

Percentage Purity

In industry, chemicals are rarely 100% pure. The percentage purity tells you what fraction of a sample is the desired substance.

% purity = (mass of pure substance / total mass of sample) x 100

For example, if you have 20 g of an iron ore sample and chemical analysis shows it contains 14 g of iron oxide, the purity is:

% purity = (14 / 20) x 100 = 70%

Why Purity Matters

Pharmaceuticals: Drug purity must be extremely high (>99.5%) to avoid harmful side effects from impurities
Food industry: Impurities in food additives can be toxic
Mining: Ore purity determines economic viability
Laboratory: Impure reagents give inaccurate experimental results

Worked Examples

Worked Example 1 A student reacts 10 g of CaCO₃ with excess HCl. The theoretical yield of CaCl₂ is 11.1 g, but only 8.88 g was collected. Calculate the percentage yield.

Show Solution

Step 1 - Identify values

Actual yield = 8.88 g. Theoretical yield = 11.1 g.

Step 2 - Apply formula

% yield = (8.88 / 11.1) x 100 = 80%

Percentage yield = 80%

Worked Example 2 A 50 g sample of limestone contains 45 g of calcium carbonate. Calculate the percentage purity of the limestone, then calculate the mass of CO₂ produced when this sample is heated. CaCO₃ → CaO + CO₂ (M_r: CaCO₃=100, CO₂=44)

Show Solution

Step 1 - Percentage purity

% purity = (45 / 50) x 100 = 90%

Step 2 - Use pure mass for calculation

Moles of CaCO₃ = 45 / 100 = 0.45 mol

Step 3 - Apply ratio

CaCO₃ : CO₂ = 1:1. Moles CO₂ = 0.45

Step 4 - Mass

Mass CO₂ = 0.45 x 44 = 19.8 g

Purity = 90%. Mass of CO₂ = 19.8 g

Worked Example 3 The theoretical yield of iron from 800 tonnes of Fe₂O₃ is 560 tonnes. If the actual yield is 504 tonnes, what is the percentage yield?

Show Solution

Apply Formula

% yield = (504 / 560) x 100 = 90%

Percentage yield = 90%

⚠ Exam Tips

Percentage yield can NEVER be more than 100%. If your answer is >100%, you have mixed up actual and theoretical yields.

In purity questions, use the PURE mass (not the total sample mass) when calculating moles for further calculations.

Read carefully: The exam might give you a percentage yield and ask you to calculate the actual yield, or give you the actual and ask for theoretical. Make sure you rearrange correctly.

🌎 Apply It: Real-World Chemistry

Yield and purity calculations are essential in industry, pharmaceuticals, and mining.

A pharmaceutical factory in Hyderabad, India produces paracetamol tablets. The theoretical yield of a batch is 500 kg, but quality control shows only 425 kg was produced.

Calculate the percentage yield. Why might the yield be less than 100%?

▼

Calculate

% yield = (425/500) x 100 = 85%

Reasons

Product lost during filtration, recrystallisation, and transfer between vessels. Some side reactions may occur. The synthesis involves multiple steps, each with its own yield loss.

Real-World Connection

Hyderabad is known as "Pharma City" - it produces about a third of India's pharmaceutical output. Improving percentage yield by even 1% can save millions of rupees. Chemical engineers are constantly optimising reaction conditions to increase yield.

A copper mine in Chile (the world's largest copper producer) extracts ore that is only 1.5% copper by mass.

If 1000 tonnes of ore is processed, what mass of copper can be extracted? Why is it still economically viable despite such low purity?

▼

Calculate

Mass of copper = 1.5% x 1000 = 15 tonnes

Economic Viability

Copper is worth about $8,000-10,000 per tonne, so 15 tonnes is worth $120,000-150,000. If mining and processing costs are below this, the mine is viable.

Real-World Connection

Chile produces about 27% of the world's copper. Even at 1.5% purity, mining is profitable because of the huge scale (millions of tonnes processed) and high copper prices. Percentage purity directly determines whether a mine opens or closes.

A chemistry student at a school in Oxford, UK prepares copper sulfate crystals from copper oxide and sulfuric acid. The theoretical yield is 25 g but she obtains only 18.75 g.

What is her percentage yield? Suggest two ways she could improve it.

▼

Calculate

% yield = (18.75/25) x 100 = 75%

Improvements

1. Wash the crystals with a small amount of cold water (not too much, or they dissolve). 2. Transfer solutions carefully to avoid spillage. 3. Allow more time for crystallisation to form more complete crystals.

Real-World Connection

Preparing copper sulfate crystals is a classic IGCSE practical. Understanding why your yield is below 100% is part of the "planning and evaluation" skill tested in Paper 5/6.

A baking soda manufacturer in Mithapur, Gujarat sells "food grade" sodium hydrogen carbonate that must be at least 99% pure. A sample of 100 g is tested and found to contain 98.5 g of NaHCO₃.

Does the sample meet food grade standards?

▼

Calculate Purity

% purity = (98.5/100) x 100 = 98.5%

Compare with Standard

98.5% < 99%. The sample does NOT meet food grade standards.

Real-World Connection

Tata Chemicals in Mithapur is one of the world's largest soda ash and baking soda producers. Food grade standards are strict because impurities in baking soda could contaminate food. The batch would need further purification before sale.

An ammonia plant in Billingham, UK using the Haber process has a percentage yield of only 15% per pass through the reactor. The unreacted nitrogen and hydrogen are recycled.

If the theoretical yield per pass is 200 tonnes, how much ammonia is produced per pass? Why is the yield so low, and how does recycling help?

▼

Calculate

Actual yield = 15% x 200 = 30 tonnes per pass

Why Low?

N₂ + 3H₂ ↔ 2NH₃ is reversible. At 450 degrees C and 200 atm (conditions that give a good rate), the equilibrium position does not fully favour products. Only about 15% converts per pass.

Recycling

Unreacted N₂ and H₂ are recycled back into the reactor. Over many passes, the overall yield approaches 98%. This is why you should always report whether yield refers to "per pass" or "overall".

Real-World Connection

The Haber process is the best example of how a low per-pass yield can still be economically viable through recycling. About 150 million tonnes of ammonia are produced globally each year - all starting with this 15% yield reaction!

Practice Questions: 3.6 Yield and Purity

20 multiple choice questions - click an option to check your answer

Score 0 / 20

Question 1

The theoretical yield of a reaction is 20 g. The actual yield is 16 g. What is the percentage yield?

A 80%

B 125%

C 75%

D 20%

% yield = (16/20) x 100 = 80%.

Question 2

A 25 g sample of impure salt contains 20 g of NaCl. What is the percentage purity?

A 80%

B 125%

C 20%

D 75%

% purity = (20/25) x 100 = 80%.

Question 3

Percentage yield can never be:

A 50%

B 100%

C Greater than 100%

D 99%

You can never get more product than the theoretical maximum, so yield cannot exceed 100%.

Question 4

Which is NOT a reason for a yield below 100%?

A Side reactions forming unwanted products

B Product left on filter paper

C Using the correct amounts of reactants

D Reversible reactions not going to completion

Using correct amounts is good practice and does not reduce yield. The other options all cause product loss.

Question 5

A reaction has 90% yield. If the theoretical yield is 50 g, the actual yield is:

A 55.6 g

B 45 g

C 40 g

D 5 g

Actual = (90/100) x 50 = 45 g.

Question 6

An ore is 60% pure iron oxide. If 200 g of ore is used, what mass of Fe₂O₃ is present?

A 120 g

B 80 g

C 140 g

D 200 g

Mass of Fe₂O₃ = 60% x 200 = 120 g.

Question 7

A student obtained 36 g of product. If the percentage yield was 90%, the theoretical yield was:

A 40 g

B 32.4 g

C 44 g

D 4 g

Theoretical = actual / (% yield/100) = 36 / 0.9 = 40 g.

Question 8

Percentage purity is most important in which industry?

A Pharmaceuticals

B Brick making

C Road construction

D Paper production

Pharmaceuticals require extremely high purity (>99.5%) because impurities in drugs can cause serious side effects or be toxic.

Question 9

The actual yield from a reaction is 7.2 g and the theoretical yield is 9.0 g. The percentage yield is:

A 80%

B 125%

C 72%

D 90%

% yield = (7.2/9.0) x 100 = 80%.

Question 10

A 50 g sample of marble (CaCO₃) is 80% pure. How many moles of pure CaCO₃ are present? (M_r=100)

A 0.4 mol

B 0.5 mol

C 0.8 mol

D 4.0 mol

Pure mass = 80% x 50 = 40 g. Moles = 40/100 = 0.4 mol.

Question 11

In a reversible reaction, the yield is low because:

A The reaction does not go to completion

B The reactants are impure

C The products are gases

D The equation is not balanced

Reversible reactions reach equilibrium where both forward and reverse reactions occur. Not all reactants convert to products.

Question 12

A reaction with 75% yield produces 30 g of product. What was the theoretical yield?

A 40 g

B 22.5 g

C 105 g

D 37.5 g

Theoretical = 30 / 0.75 = 40 g.

Question 13

A sample of impure zinc weighing 32.5 g produces 0.4 mol of ZnSO₄ when reacted with excess H₂SO₄. What is the purity of the zinc? (A_r: Zn=65)

A 80%

B 40%

C 60%

D 100%

Moles Zn = 0.4 (1:1 ratio with ZnSO₄). Mass pure Zn = 0.4 x 65 = 26 g. Purity = (26/32.5) x 100 = 80%.

Question 14

A factory aims for 95% yield. If 1000 kg of product is needed, how much theoretical yield must the process produce?

A 950 kg

B 1053 kg

C 1050 kg

D 1100 kg

Theoretical = 1000/0.95 = 1052.6 kg, rounded to 1053 kg.

Question 15

Loss of product during filtration would:

A Decrease the percentage yield

B Increase the percentage yield

C Decrease the percentage purity

D Have no effect

If product is lost during filtration, the actual yield decreases, which decreases percentage yield.

Question 16

Iron ore is 70% Fe₂O₃. What mass of iron can be obtained from 400 g of ore? Fe₂O₃ → 2Fe (M_r: Fe₂O₃=160, Fe=56)

A 196 g

B 280 g

C 140 g

D 98 g

Pure Fe₂O₃ = 70% x 400 = 280 g. Moles = 280/160 = 1.75. Fe = 2 x 1.75 = 3.5 mol. Mass = 3.5 x 56 = 196 g.

Question 17

The difference between actual yield and theoretical yield is due to:

A Errors in the balanced equation

B Practical losses and incomplete reactions

C Using too much reactant

D Conservation of mass not applying

The equation is always correct. The difference is due to practical issues like product loss, side reactions, and incomplete conversions.

Question 18

A chemist needs 100 g of product. The process has 80% yield. She should start with enough reactant for a theoretical yield of:

A 125 g

B 80 g

C 180 g

D 100 g

Theoretical = 100/0.80 = 125 g. She needs to aim for 125 g theoretically to end up with 100 g actually.

Question 19

A sample is 95% pure CaCO₃. What mass of the sample contains 47.5 g of pure CaCO₃?

A 50 g

B 45.1 g

C 47.5 g

D 100 g

Total mass = pure mass / (purity/100) = 47.5 / 0.95 = 50 g.

Question 20

Two reactions: A has 60% yield and B has 40% yield. If A feeds into B, the overall yield is approximately:

A 24%

B 50%

C 100%

D 20%

Overall yield = 0.60 x 0.40 = 0.24 = 24%. Each step multiplies, it does not add!

3.7 Concentration and Volume of Solutions ▼

Supplement - Extended Tier Only

Concentration

Concentration tells you how much solute is dissolved in a given volume of solution. It can be measured in two ways:

1. Concentration in g/dm³

concentration (g/dm³) = mass of solute (g) / volume of solution (dm³)

2. Concentration in mol/dm³

concentration (mol/dm³) = moles of solute / volume of solution (dm³)

Converting Between Units

concentration (g/dm³) = concentration (mol/dm³) x M_r

Volume Conversions

Remember: 1 dm³ = 1000 cm³ = 1 litre

To convert cm³ to dm³: divide by 1000

To convert dm³ to cm³: multiply by 1000

The Concentration Triangle

Memory Trick - Concentration Triangle

Put moles at the top, concentration and volume at the bottom:

moles = concentration x volume

concentration = moles / volume

volume = moles / concentration

Remember: n = c x V (where V is in dm³)

Titration Calculations

A titration is an experiment where you add one solution to another until the reaction is complete (the end point). Typically, you know the concentration of one solution and use the titration to find the concentration of the other.

Steps for Titration Calculations

Calculate moles of the solution whose concentration you know: n = c x V
Use the molar ratio from the balanced equation
Calculate concentration of the unknown solution: c = n / V

Worked Examples

Worked Example 1 What is the concentration in mol/dm³ of a solution containing 4 g of NaOH in 500 cm³ of solution? (M_r of NaOH = 40)

Show Solution

Step 1 - Convert volume

500 cm³ = 500/1000 = 0.5 dm³

Step 2 - Find moles

moles = mass / M_r = 4 / 40 = 0.1 mol

Step 3 - Find concentration

c = n / V = 0.1 / 0.5 = 0.2 mol/dm³

0.2 mol/dm³

Worked Example 2 Convert 0.5 mol/dm³ HCl to g/dm³. (M_r of HCl = 36.5)

Show Solution

Apply conversion

g/dm³ = mol/dm³ x M_r = 0.5 x 36.5 = 18.25 g/dm³

18.25 g/dm³

Worked Example 3 25.0 cm³ of 0.1 mol/dm³ NaOH exactly neutralises 20.0 cm³ of HCl. Find the concentration of HCl. NaOH + HCl → NaCl + H₂O

Show Solution

Step 1 - Moles of NaOH

n = c x V = 0.1 x (25.0/1000) = 0.1 x 0.025 = 0.0025 mol

Step 2 - Molar ratio

NaOH : HCl = 1:1, so moles of HCl = 0.0025 mol

Step 3 - Concentration of HCl

c = n / V = 0.0025 / (20.0/1000) = 0.0025 / 0.020 = 0.125 mol/dm³

0.125 mol/dm³

⚠ Exam Tips

Volume MUST be in dm³ when using c = n/V. If given in cm³, divide by 1000 first. This is the most common mistake in concentration calculations.

Titration calculations always follow three steps: (1) find moles of what you know, (2) use molar ratio, (3) find concentration of what you need.

g/dm³ to mol/dm³: DIVIDE by M_r. mol/dm³ to g/dm³: MULTIPLY by M_r.

🌎 Apply It: Real-World Chemistry

Concentration calculations are used in medicine, water treatment, and food science every day.

A doctor at AIIMS hospital in New Delhi prescribes a saline drip (NaCl solution). The bag contains 9 g of NaCl per dm³ of solution.

What is the concentration in mol/dm³? (M_r of NaCl = 58.5)

▼

Convert

c (mol/dm³) = c (g/dm³) / M_r = 9 / 58.5 = 0.154 mol/dm³

Real-World Connection

Normal saline (0.9% NaCl or 9 g/dm³) is one of the most commonly used medical solutions worldwide. It matches the concentration of salt in blood plasma. Getting the concentration wrong could be fatal - too concentrated and it dehydrates cells, too dilute and cells swell and burst.

A water quality inspector in Glasgow, Scotland tests the fluoride concentration in tap water. The legal limit is 1.5 mg per dm³ (parts per million). The sample contains 0.00008 mol/dm³ of NaF.

Convert to g/dm³ and determine if it meets the standard. (M_r of NaF = 42, A_r of F = 19)

▼

Convert NaF concentration

NaF in g/dm³ = 0.00008 x 42 = 0.00336 g/dm³ = 3.36 mg/dm³

Find fluoride only

F concentration = 0.00008 x 19 = 0.00152 g/dm³ = 1.52 mg/dm³

Compare

1.52 mg/dm³ > 1.5 mg/dm³. It slightly exceeds the limit.

Real-World Connection

Fluoride is added to water to prevent tooth decay, but too much causes fluorosis (brown staining of teeth). Water quality chemists do exactly this calculation to ensure public safety.

A lab technician at a vinegar factory in Modena, Italy titrates vinegar (ethanoic acid, CH₃COOH) with 0.5 mol/dm³ NaOH. She finds that 25 cm³ of vinegar requires 30 cm³ of NaOH to neutralise.

Calculate the concentration of ethanoic acid in the vinegar. CH₃COOH + NaOH → CH₃COONa + H₂O

▼

Moles of NaOH

n = 0.5 x (30/1000) = 0.5 x 0.03 = 0.015 mol

Molar ratio

1:1, so moles of CH₃COOH = 0.015 mol

Concentration

c = 0.015 / (25/1000) = 0.015 / 0.025 = 0.6 mol/dm³

Real-World Connection

Balsamic vinegar from Modena is aged for up to 25 years. Quality control requires regular titrations to check acidity levels. Traditional balsamic vinegar must have a minimum acidity of 6% - the titration ensures consistency.

A swimming pool manager in Dubai, UAE needs to maintain the pool's chlorine concentration at 0.00003 mol/dm³. The pool holds 500,000 dm³ of water.

What mass of chlorine (Cl₂) is needed? (M_r of Cl₂ = 71)

▼

Find moles

n = c x V = 0.00003 x 500000 = 15 mol

Find mass

mass = 15 x 71 = 1065 g = 1.065 kg

Real-World Connection

Dubai has some of the world's largest swimming pools. Pool managers must carefully calculate chlorine amounts - too little and bacteria thrive, too much and swimmers get irritated skin and eyes. This is a real concentration calculation done daily.

A student at a school in Kochi, India performs a titration to find the concentration of citric acid in lemon juice. She neutralises 10.0 cm³ of lemon juice with 15.0 cm³ of 0.1 mol/dm³ NaOH. The equation is: C₆H₈O₇ + 3NaOH → Na₃C₆H₅O₇ + 3H₂O

Find the concentration of citric acid in mol/dm³.

▼

Moles of NaOH

n = 0.1 x (15.0/1000) = 0.0015 mol

Molar ratio

Citric acid : NaOH = 1:3. So moles citric acid = 0.0015 / 3 = 0.0005 mol

Concentration

c = 0.0005 / (10.0/1000) = 0.0005 / 0.01 = 0.05 mol/dm³

Real-World Connection

Kerala (Kochi) is famous for its citrus fruits. Notice that the 1:3 ratio is crucial - citric acid is a triprotic acid (3 acidic H atoms), so it reacts with 3 moles of NaOH. Always check the ratio in the equation - it is not always 1:1!

Practice Questions: 3.7 Concentration

20 multiple choice questions - click an option to check your answer

Score 0 / 20

Question 1

What is the concentration in mol/dm³ of a solution containing 0.5 mol in 2 dm³?

A 0.25 mol/dm³

B 1.0 mol/dm³

C 4.0 mol/dm³

D 2.5 mol/dm³

c = n/V = 0.5/2 = 0.25 mol/dm³.

Question 2

250 cm³ is equal to:

A 0.25 dm³

B 2.5 dm³

C 25 dm³

D 0.025 dm³

250 / 1000 = 0.25 dm³.

Question 3

How many moles of HCl are in 100 cm³ of 0.5 mol/dm³ solution?

A 0.05 mol

B 0.5 mol

C 5.0 mol

D 50 mol

n = c x V = 0.5 x (100/1000) = 0.5 x 0.1 = 0.05 mol.

Question 4

A solution has concentration 7.3 g/dm³ of HCl. What is this in mol/dm³? (M_r=36.5)

A 0.2 mol/dm³

B 5.0 mol/dm³

C 266 mol/dm³

D 0.5 mol/dm³

mol/dm³ = g/dm³ / M_r = 7.3 / 36.5 = 0.2 mol/dm³.

Question 5

What volume of 0.2 mol/dm³ NaOH contains 0.01 mol?

A 50 cm³

B 500 cm³

C 5 cm³

D 20 cm³

V = n/c = 0.01/0.2 = 0.05 dm³ = 50 cm³.

Question 6

0.1 mol/dm³ NaOH has a concentration in g/dm³ of: (M_r=40)

A 4 g/dm³

B 0.4 g/dm³

C 40 g/dm³

D 400 g/dm³

g/dm³ = mol/dm³ x M_r = 0.1 x 40 = 4 g/dm³.

Question 7

25 cm³ of 0.1 mol/dm³ HCl neutralises 25 cm³ of NaOH. What is the NaOH concentration? (1:1 ratio)

A 0.1 mol/dm³

B 0.2 mol/dm³

C 0.05 mol/dm³

D 1.0 mol/dm³

Moles HCl = 0.1 x 0.025 = 0.0025. Ratio 1:1. Moles NaOH = 0.0025. c = 0.0025/0.025 = 0.1 mol/dm³.

Question 8

What mass of NaOH is needed to make 250 cm³ of 0.4 mol/dm³ solution? (M_r=40)

A 4.0 g

B 10 g

C 16 g

D 100 g

n = c x V = 0.4 x 0.25 = 0.1 mol. mass = 0.1 x 40 = 4.0 g.

Question 9

25 cm³ of 0.2 mol/dm³ H₂SO₄ neutralises 50 cm³ of KOH. H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O. Find [KOH].

A 0.2 mol/dm³

B 0.1 mol/dm³

C 0.4 mol/dm³

D 0.05 mol/dm³

Moles H₂SO₄ = 0.2 x 0.025 = 0.005. Ratio 1:2. Moles KOH = 0.01. c = 0.01/0.05 = 0.2 mol/dm³.

Question 10

Which solution is most concentrated?

A 1 mol in 2 dm³

B 0.5 mol in 1 dm³

C 2 mol in 1 dm³

D 0.1 mol in 0.5 dm³

A: 0.5, B: 0.5, C: 2.0, D: 0.2 mol/dm³. C is the most concentrated.

Question 11

How many moles in 500 cm³ of 2 mol/dm³ NaCl?

A 1.0 mol

B 0.5 mol

C 4.0 mol

D 1000 mol

n = 2 x (500/1000) = 2 x 0.5 = 1.0 mol.

Question 12

What volume of 0.5 mol/dm³ HCl is needed to neutralise 0.1 mol of NaOH? (1:1)

A 200 cm³

B 100 cm³

C 50 cm³

D 500 cm³

Need 0.1 mol HCl. V = n/c = 0.1/0.5 = 0.2 dm³ = 200 cm³.

Question 13

A solution of 4 g/dm³ NaOH has a molar concentration of: (M_r=40)

A 0.1 mol/dm³

B 10 mol/dm³

C 160 mol/dm³

D 0.01 mol/dm³

mol/dm³ = 4/40 = 0.1 mol/dm³.

Question 14

20 cm³ of 0.5 mol/dm³ H₂SO₄ reacts with NaOH. How many moles of NaOH are needed? H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

A 0.01 mol

B 0.02 mol

C 0.005 mol

D 0.1 mol

Moles H₂SO₄ = 0.5 x 0.02 = 0.01. Ratio 1:2. Moles NaOH = 0.02.

Question 15

1 dm³ equals:

A 100 cm³

B 1000 cm³

C 10 cm³

D 10000 cm³

1 dm³ = 1000 cm³ = 1 litre. This is essential to remember!

Question 16

What mass of solute is in 200 cm³ of 0.5 mol/dm³ NaCl? (M_r=58.5)

A 5.85 g

B 58.5 g

C 29.25 g

D 11.7 g

n = 0.5 x 0.2 = 0.1 mol. mass = 0.1 x 58.5 = 5.85 g.

Question 17

25 cm³ of acid X neutralises 50 cm³ of 0.1 mol/dm³ NaOH (1:1 ratio). The concentration of X is:

A 0.2 mol/dm³

B 0.1 mol/dm³

C 0.05 mol/dm³

D 0.5 mol/dm³

Moles NaOH = 0.1 x 0.05 = 0.005. Ratio 1:1. Moles X = 0.005. c = 0.005/0.025 = 0.2 mol/dm³.

Question 18

To dilute a solution, you add more:

A Solvent

B Solute

C Acid

D Indicator

Diluting means increasing the volume while keeping the same number of moles, which decreases the concentration. You do this by adding more solvent (usually water).

Question 19

A 2 mol/dm³ solution is diluted to 500 cm³ and the new concentration is 0.4 mol/dm³. What was the original volume?

A 100 cm³

B 200 cm³

C 50 cm³

D 250 cm³

Moles in final = 0.4 x 0.5 = 0.2 mol. Original volume = n/c = 0.2/2 = 0.1 dm³ = 100 cm³.

Question 20

In a titration, 20 cm³ of 0.1 mol/dm³ Na₂CO₃ reacts with 40 cm³ of HCl. Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. Find [HCl].

A 0.1 mol/dm³

B 0.2 mol/dm³

C 0.05 mol/dm³

D 0.4 mol/dm³

Moles Na₂CO₃ = 0.1 x 0.02 = 0.002. Ratio 1:2. Moles HCl = 0.004. c = 0.004/0.04 = 0.1 mol/dm³.

3.8 Mole Calculations: Gas Volumes ▼

Supplement - Extended Tier Only

Molar Volume of Gas

One of the most elegant facts in chemistry: one mole of ANY gas occupies the same volume at the same temperature and pressure. This is because gas particles are spread so far apart that their actual size does not matter - it is the number of particles that determines the volume.

At RTP: 1 mole of gas = 24 dm³ = 24,000 cm³

RTP = Room Temperature and Pressure (20-25 degrees C, 1 atm)

This means:

1 mol of H₂ = 24 dm³ (even though H₂ is very light)
1 mol of O₂ = 24 dm³
1 mol of CO₂ = 24 dm³ (even though CO₂ is much heavier)

Key Formulae

volume (dm³) = moles x 24

Rearranged: moles = volume (dm³) / 24 Or: volume (cm³) = moles x 24,000

Calculating Gas Volumes from Equations

You can calculate the volume of gas produced (or consumed) in a reaction by:

Finding moles of the substance you know
Using the molar ratio to find moles of the gas
Converting moles to volume using V = n x 24

Using Gas Volume Ratios Directly

Here is a useful shortcut: if all substances in an equation are gases, you can use the coefficients directly as volume ratios (because equal moles of gases have equal volumes).

For example: N₂(g) + 3H₂(g) → 2NH₃(g)

1 volume of N₂ reacts with 3 volumes of H₂ to produce 2 volumes of NH₃.

So 100 cm³ of N₂ would react with 300 cm³ of H₂ to produce 200 cm³ of NH₃.

Worked Examples

Worked Example 1 What volume of CO₂ gas (at RTP) is produced when 5 g of CaCO₃ decomposes? CaCO₃ → CaO + CO₂ (M_r: CaCO₃=100)

Show Solution

Step 1 - Find moles of CaCO₃

moles = 5/100 = 0.05 mol

Step 2 - Molar ratio

CaCO₃ : CO₂ = 1:1, so 0.05 mol CO₂

Step 3 - Volume

V = 0.05 x 24 = 1.2 dm³ (or 1200 cm³)

1.2 dm³ (1200 cm³) of CO₂

Worked Example 2 What volume of hydrogen gas is produced when 1.3 g of zinc reacts with excess HCl at RTP? Zn + 2HCl → ZnCl₂ + H₂ (A_r: Zn=65)

Show Solution

Step 1 - Moles of Zn

moles = 1.3/65 = 0.02 mol

Step 2 - Ratio

Zn : H₂ = 1:1, so 0.02 mol H₂

Step 3 - Volume

V = 0.02 x 24 = 0.48 dm³ = 480 cm³

480 cm³ of H₂

Worked Example 3 What volume of oxygen at RTP is needed to completely burn 100 cm³ of propane gas? C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Show Solution

Step 1 - Use volume ratios directly

Since both are gases, we can use the coefficients as volume ratios: C₃H₈ : O₂ = 1:5

Step 2 - Calculate

100 cm³ of C₃H₈ requires 5 x 100 = 500 cm³ of O₂

500 cm³ of O₂

⚠ Exam Tips

24 dm³ is ONLY at RTP. If the question says "at room temperature and pressure" or "at RTP", use 24. If it says STP (0 degrees C, 1 atm), the molar volume is 22.4 dm³ - but IGCSE almost always uses RTP.

Volume ratio shortcut: Only works when ALL substances being compared are gases. If one is solid or liquid, you must calculate moles first.

Units: Be careful with dm³ vs cm³. If the question asks for cm³, multiply by 1000. If it asks for dm³, divide by 1000 if you calculated in cm³.

The molar volume applies to ALL gases equally. It does not depend on the identity of the gas - 1 mol of helium (M_r=4) and 1 mol of sulfur dioxide (M_r=64) both occupy 24 dm³ at RTP.

🌎 Apply It: Real-World Chemistry

Gas volume calculations are essential in industry, medicine, and environmental science.

A scuba diving instructor in Goa, India fills oxygen tanks for tourists. Each tank needs 500 dm³ of oxygen gas (at RTP).

How many moles of O₂ is this, and what mass of oxygen does each tank contain? (M_r: O₂=32)

▼

Find moles

moles = volume / 24 = 500 / 24 = 20.83 mol

Find mass

mass = 20.83 x 32 = 666.7 g = 0.667 kg

Real-World Connection

Goa is one of India's top scuba diving destinations. Despite 500 dm³ of gas, the actual mass is less than 0.7 kg! This is because gases are very spread out. In reality, tanks compress the gas to much smaller volumes at high pressure.

A brewery in Dublin, Ireland produces CO₂ during fermentation. The equation is: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂. The brewery uses 1800 g of glucose per batch.

What volume of CO₂ is produced at RTP? (M_r: C₆H₁₂O₆=180)

▼

Moles glucose

moles = 1800/180 = 10 mol

Moles CO₂

Ratio glucose:CO₂ = 1:2. Moles CO₂ = 20 mol.

Volume

V = 20 x 24 = 480 dm³

Real-World Connection

The Guinness brewery in Dublin produces 480 dm³ (480 litres) of CO₂ from just 1.8 kg of glucose! Many breweries now capture this CO₂ and sell it for carbonating soft drinks - turning waste into profit.

A car manufacturer in Stuttgart, Germany needs to know how much hydrogen gas an airbag system produces. The reaction uses sodium azide: 2NaN₃ → 2Na + 3N₂. Each airbag contains 65 g of NaN₃.

What volume of N₂ is produced at RTP? (M_r: NaN₃=65)

▼

Moles NaN₃

moles = 65/65 = 1.0 mol

Moles N₂

Ratio NaN₃:N₂ = 2:3. Moles N₂ = 1.0 x 3/2 = 1.5 mol.

Volume

V = 1.5 x 24 = 36 dm³

Real-World Connection

Mercedes-Benz, Porsche, and BMW all use sodium azide in airbags. The reaction happens in about 40 milliseconds, inflating the airbag with 36 dm³ of nitrogen gas almost instantly. The molar volume calculation tells engineers exactly how much NaN₃ to use for the right airbag size.

A climate scientist in Bengaluru, India calculates that a household burns 16 g of methane (CH₄) daily for cooking. CH₄ + 2O₂ → CO₂ + 2H₂O

What volume of CO₂ does this household emit daily at RTP? (M_r: CH₄=16)

▼

Moles CH₄

moles = 16/16 = 1.0 mol

Moles CO₂

Ratio 1:1. Moles CO₂ = 1.0 mol.

Volume

V = 1.0 x 24 = 24 dm³ = 24 litres

Real-World Connection

Burning just 16 g of methane produces 24 litres of CO₂! Over a year, this household emits about 8,760 litres of CO₂ just from cooking. Scientists use these calculations to model carbon footprints and climate change impacts. India is transitioning to LPG and electric cooking to reduce emissions.

A hydrogen fuel cell engineer in Tokyo, Japan needs to calculate the volume of hydrogen gas required to power a car for 100 km. The fuel cell reaction is: 2H₂ + O₂ → 2H₂O. The car uses 0.8 mol of H₂ per km.

What total volume of H₂ (at RTP) is needed for 100 km?

▼

Total moles

0.8 x 100 = 80 mol of H₂

Volume at RTP

V = 80 x 24 = 1920 dm³

Real-World Connection

1920 dm³ at atmospheric pressure would be enormous! This is why hydrogen fuel cells store hydrogen at very high pressures (700 atm), compressing 1920 litres down to just a few litres. Toyota's Mirai hydrogen car can travel about 650 km on one tank. The molar volume calculation shows why compression is essential.

Practice Questions: 3.8 Gas Volumes

20 multiple choice questions - click an option to check your answer

Score 0 / 20

Question 1

At RTP, 1 mole of any gas occupies:

A 22.4 dm³

B 24 dm³

C 24 cm³

D 12 dm³

At RTP (room temperature and pressure), the molar volume is 24 dm³ (or 24,000 cm³). 22.4 dm³ is at STP (0 degrees C).

Question 2

What volume does 0.5 mol of O₂ occupy at RTP?

A 12 dm³

B 24 dm³

C 48 dm³

D 6 dm³

V = 0.5 x 24 = 12 dm³.

Question 3

How many moles in 4800 cm³ of gas at RTP?

A 0.2 mol

B 2.0 mol

C 200 mol

D 0.02 mol

4800 cm³ = 4.8 dm³. moles = 4.8/24 = 0.2 mol.

Question 4

What volume of H₂ at RTP is produced from 2.4 g of Mg? Mg + 2HCl → MgCl₂ + H₂ (A_r: Mg=24)

A 2.4 dm³

B 24 dm³

C 0.24 dm³

D 4.8 dm³

Moles Mg = 2.4/24 = 0.1. Ratio 1:1 with H₂. Moles H₂ = 0.1. V = 0.1 x 24 = 2.4 dm³.

Question 5

The molar volume is the same for all gases because:

A All gas molecules are the same size

B All gas molecules have the same mass

C Gas particles are far apart, so individual size does not matter

D All gases are diatomic

Gas particles are so far apart that the actual size of the molecule is negligible. The volume is determined by the number of particles, not their size.

Question 6

100 cm³ of N₂ reacts with H₂. N₂ + 3H₂ → 2NH₃. What volume of H₂ is needed?

A 300 cm³

B 100 cm³

C 200 cm³

D 50 cm³

Volume ratio N₂:H₂ = 1:3. So 100 x 3 = 300 cm³ of H₂.

Question 7

What mass of CaCO₃ would produce 2.4 dm³ of CO₂ at RTP? CaCO₃ → CaO + CO₂ (M_r=100)

A 10 g

B 100 g

C 24 g

D 1 g

Moles CO₂ = 2.4/24 = 0.1. Ratio 1:1. Moles CaCO₃ = 0.1. Mass = 0.1 x 100 = 10 g.

Question 8

24,000 cm³ of gas at RTP contains:

A 1 mol

B 24 mol

C 0.024 mol

D 1000 mol

24,000 cm³ = 24 dm³ = 1 mole.

Question 9

What volume of O₂ at RTP is needed to burn 48 cm³ of H₂? 2H₂ + O₂ → 2H₂O

A 24 cm³

B 48 cm³

C 96 cm³

D 12 cm³

Volume ratio H₂:O₂ = 2:1. O₂ = 48/2 = 24 cm³.

Question 10

What volume of gas is produced from 0.25 mol at RTP?

A 6 dm³

B 96 dm³

C 6000 dm³

D 0.6 dm³

V = 0.25 x 24 = 6 dm³.

Question 11

At RTP, which sample occupies the largest volume?

A 2 g of H₂ (M_r=2)

B 32 g of O₂ (M_r=32)

C 4 g of H₂ (M_r=2)

D 44 g of CO₂ (M_r=44)

A: 1 mol=24 dm³. B: 1 mol=24 dm³. C: 2 mol=48 dm³. D: 1 mol=24 dm³. C has the most moles, so the largest volume.

Question 12

CH₄ + 2O₂ → CO₂ + 2H₂O. 50 cm³ of CH₄ produces what volume of CO₂?

A 50 cm³

B 100 cm³

C 25 cm³

D 200 cm³

Ratio CH₄:CO₂ = 1:1. Same volume = 50 cm³.

Question 13

6 dm³ of a gas at RTP has how many moles?

A 0.25 mol

B 4.0 mol

C 0.5 mol

D 144 mol

n = V/24 = 6/24 = 0.25 mol.

Question 14

2NaN₃ → 2Na + 3N₂. What volume of N₂ from 13 g NaN₃ at RTP? (M_r=65)

A 7.2 dm³

B 4.8 dm³

C 14.4 dm³

D 3.6 dm³

Moles NaN₃ = 13/65 = 0.2. Ratio 2:3. Moles N₂ = 0.2 x 3/2 = 0.3. V = 0.3 x 24 = 7.2 dm³.

Question 15

200 cm³ of CO reacts with O₂. 2CO + O₂ → 2CO₂. What volume of O₂ is needed?

A 100 cm³

B 200 cm³

C 400 cm³

D 50 cm³

Ratio CO:O₂ = 2:1. O₂ = 200/2 = 100 cm³.

Question 16

What is the mass of 2.4 dm³ of CO₂ at RTP? (M_r=44)

A 4.4 g

B 44 g

C 0.44 g

D 10.56 g

Moles = 2.4/24 = 0.1. Mass = 0.1 x 44 = 4.4 g.

Question 17

How many molecules in 240 cm³ of gas at RTP?

A 6.02 x 10²¹

B 6.02 x 10²³

C 6.02 x 10²²

D 6.02 x 10²⁰

240 cm³ = 0.24 dm³. Moles = 0.24/24 = 0.01 mol. Molecules = 0.01 x 6.02 x 10²³ = 6.02 x 10²¹.

Question 18

What volume of NH₃ is produced from 150 cm³ of H₂? N₂ + 3H₂ → 2NH₃

A 100 cm³

B 150 cm³

C 75 cm³

D 300 cm³

Ratio H₂:NH₃ = 3:2. NH₃ = 150 x 2/3 = 100 cm³.

Question 19

Zn + 2HCl → ZnCl₂ + H₂. What volume of H₂ from 6.5 g of Zn at RTP? (A_r: Zn=65)

A 2400 cm³

B 240 cm³

C 24000 cm³

D 24 cm³

Moles Zn = 6.5/65 = 0.1. Ratio 1:1. Moles H₂ = 0.1. V = 0.1 x 24000 = 2400 cm³.

Question 20

At RTP, 12 dm³ of an unknown gas has a mass of 22 g. What is the M_r of the gas?

A 44

B 22

C 88

D 11

Moles = 12/24 = 0.5. M_r = mass/moles = 22/0.5 = 44. This could be CO₂ or C₃H₈!

Topic 3: Stoichiometry

Word Equations

Balanced Symbol Equations

Steps to Balance an Equation

State Symbols

Formulae of Ionic Compounds

Common Ion Charges

The Cross-Over Method

Types of Formulae

Determining Formulae from Experimental Data

Deducing Formulae from Models or Diagrams

Constructing Balanced Equations with State Symbols

Worked Examples

Relative Atomic Mass (Ar)

Relative Formula Mass (Mr)

The Carbon-12 Scale

Calculating Ar from Isotopic Abundances

Worked Examples

What is a Mole?

Molar Mass

The Mole Triangle

Using moles = mass / molar mass

Worked Examples

Writing and Balancing Symbol Equations

Rules for Balancing

Common Types of Reactions

Ionic Equations

Steps to Write an Ionic Equation

Worked Examples

Calculating Reacting Masses

Percentage by Mass of an Element

Empirical Formula from Mass or Percentage Data

Deducing Stoichiometry from Reacting Mass Data

Limiting Reagent

Worked Examples

Percentage Yield

Why is the yield less than 100%?

Percentage Purity

Why Purity Matters

Worked Examples

Concentration

1. Concentration in g/dm³

2. Concentration in mol/dm³

Converting Between Units

Volume Conversions

The Concentration Triangle

Titration Calculations

Steps for Titration Calculations

Worked Examples

Molar Volume of Gas

Key Formulae

Calculating Gas Volumes from Equations

Using Gas Volume Ratios Directly

Worked Examples

Relative Atomic Mass (A_r)

Relative Formula Mass (M_r)

Calculating A_r from Isotopic Abundances