Chapter 03

Stoichiometry

Cambridge O Level Chemistry 5070 Topic 3 revision chapter covering formulae, relative masses and the mole. The chapter teaches stoichiometry as one dependable route rather than a bag of formulae: identify the given quantity and its unit, convert to moles, apply the ratio taken from the balanced equation or the chemical formula, convert to the quantity the question asks for, then check the unit and the scale of the answer. Topic 3.1 covers stating the formulae of the elements and compounds named in the syllabus subject content, defining molecular formula as the number and type of different atoms in one molecule and empirical formula as the simplest whole-number ratio of the different atoms or ions in a compound, deducing a formula from a particle model or diagram, deducing the formula of an ionic compound from the charges on its ions, constructing word, symbol and ionic equations with state symbols, and deducing a symbol equation with state symbols from supplied information. Topic 3.2 covers relative atomic mass as the average mass of the isotopes of an element compared with one-twelfth of the mass of an atom of carbon-12, relative molecular mass as the sum of the relative atomic masses, relative formula mass for ionic compounds, and the difference between these unitless ratios and molar mass in grams per mole. Topic 3.3 covers the mole as the unit of amount of substance, the Avogadro constant of 6.02 x 10^23 particles per mole, the relationship between amount of substance, mass and molar mass, particle-number calculations, the molar gas volume taken as 24 dm3 at room temperature and pressure, concentration in g/dm3 and mol/dm3, reacting-mass and limiting-reactant calculations, volumes of gases and of solutions, conversion between cm3 and dm3, titration calculations from experimental data, empirical and molecular formulae from supplied data, and percentage yield, percentage composition by mass and percentage purity. Every worked example shows the relationship used, the substitution with units, the equation ratio where one applies, the final answer and a reasonableness check. Thirteen original diagrams make the quantities, ratios and unit gateways visible. All questions, datasets and figures are original Academiq Edu work.Show more

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A summary of this Chemistry chapter — open a section to read it. The full notes, worked examples and practice questions are in the study modules above.

What is Stoichiometry about?

Stoichiometry is the part of chemistry that uses the balanced equation to calculate how much of each substance reacts or is produced. In Cambridge O Level Chemistry 5070 almost every stoichiometry question follows one route: work out what you have been given, convert it to moles, apply the ratio of coefficients from the balanced equation, convert back to the quantity asked for, then check the unit and the size of the answer. Mass, number of particles, gas volume at r.t.p. and solution concentration are four different doors into moles, and moles are the only quantity a balanced equation can compare.

The mole (mol) is the unit of amount of substance. One mole contains \(6.02\times10^{23}\) particles — the Avogadro constant — and you must always name which particles: atoms, ions, molecules or formula units. Amount of substance is calculated as mass divided by molar mass, \(n = m/M\), where molar mass \(M\) is measured in \(\mathrm{g\,mol^{-1}}\). Relative atomic mass \(A_r\) and relative molecular mass \(M_r\) share the same numbers as molar mass but carry no unit, because each is a comparison with one-twelfth of the mass of a carbon-12 atom. One mole of any gas occupies \(24\,\mathrm{dm^3}\) at room temperature and pressure.

Key ideas to remember

Say the unit out loud and you can name the bridge. Name the bridge and you have already started the question.
In front of the formula counts it. Inside the formula defines it.
Smallest whole numbers whose charges cancel. Then check that they cancel.
Formulae first, states second, coefficients last — then count every element before you write the answer down.
Ionic and aqueous, or it stays whole. Cancel only what is identical in formula, charge and state — then check the charge.
\(A_r\) and \(M_r\) are counts of reference shares, so they are bare numbers. Molar mass is grams per mole, and it is the only one of the three that goes into \(n = m/M\).
A mole of what? Answer that before you multiply by \(6.02\times10^{23}\), and the entity trap can never catch you.
Write every substitution with its units attached. The unit that falls out tells you whether you rearranged correctly, before the answer can go anywhere else.

What you need to be able to do

3.1.1 State the formulae of the elements and compounds named in the syllabus subject content.
3.1.2 Define the molecular formula of a compound as the number and type of different atoms in one molecule.
3.1.3 Define the empirical formula of a compound as the simplest whole-number ratio of the different atoms or ions in a compound.
3.1.4 Deduce the formula of a simple compound from the relative numbers of atoms or ions in a model or diagram.
3.1.5 Deduce the formula of an ionic compound from the charges on the ions.
3.1.6 Construct word equations, symbol equations and ionic equations showing how reactants form products, including state symbols.
3.1.7 Deduce the symbol equation with state symbols for a reaction, given relevant information.
3.2.1 Describe relative atomic mass, \(A_r\), as the average mass of the isotopes of an element compared with one-twelfth of the mass of an atom of \(\mathrm{^{12}C}\).
3.2.2 Define relative molecular mass, \(M_r\), as the sum of the relative atomic masses, and use relative formula mass, also \(M_r\), for ionic compounds.
3.3.1a State that the mole, mol, is the unit of amount of substance.
3.3.1b State that one mole contains \(6.02\times10^{23}\) particles — atoms, ions, molecules or other specified entities — and that this number is the Avogadro constant.
3.3.2a Use amount of substance = mass ÷ molar mass to calculate amount of substance.
3.3.2b Use the same relationship to calculate mass.
3.3.2c Use the same relationship to calculate molar mass.
3.3.2d Use mass and mole data to calculate a relative atomic mass or a relative molecular / formula mass.
3.3.2e Calculate a number of particles using the value of the Avogadro constant, naming the entity counted.
3.3.3 Use the molar gas volume, taken as \(24\,\mathrm{dm^3}\) at room temperature and pressure (r.t.p.), in calculations involving gases.
3.3.4a State and use concentration measured in \(\mathrm{g\,dm^{-3}}\).
3.3.4b State and use concentration measured in \(\mathrm{mol\,dm^{-3}}\).
3.3.5a Calculate stoichiometric reacting masses.
3.3.5b Identify and use limiting reactants.
3.3.5c Calculate volumes of gases at r.t.p.
3.3.5d Calculate volumes of solutions.
3.3.5e Calculate concentrations of solutions in \(\mathrm{g\,dm^{-3}}\).
3.3.5f Calculate concentrations of solutions in \(\mathrm{mol\,dm^{-3}}\).
3.3.5g Convert correctly between \(\mathrm{cm^3}\) and \(\mathrm{dm^3}\).
3.3.6 Use experimental data to calculate the concentration of a solution in a titration.
3.3.7a Calculate empirical formulae, given appropriate data.
3.3.7b Calculate molecular formulae, given appropriate data.
3.3.8a Calculate percentage yield.
3.3.8b Calculate percentage composition by mass.
3.3.8c Calculate percentage purity.

Key terms in Stoichiometry

Empirical formula: The empirical formula of a compound is the simplest whole-number ratio of the different atoms or ions present in that compound, for example CH2 for ethene, whose molecular formula is C2H4.
Chemical equation: A chemical equation represents a reaction using formulae, with the same number of atoms of each element on both sides; the coefficients written in front of each formula give the ratio in which the substances react, and state symbols show the physical state of each species.
Ionic equation: An ionic equation shows only the species that actually change during a reaction: aqueous ionic compounds are written as separate ions, unchanged spectator ions are cancelled, and the equation must balance for atoms and for total charge.
Mole: The mole, symbol mol, is the unit of amount of substance; one mole of any substance contains 6.02 times 10 to the 23 specified particles, which may be atoms, ions, molecules or formula units.
Relative molecular mass: Relative molecular mass, Mr, is the sum of the relative atomic masses of all the atoms shown in the formula of a molecule; for an ionic compound the same sum is called the relative formula mass, and neither quantity has a unit.
Avogadro constant: The Avogadro constant is the number of specified particles in one mole of a substance, taken as 6.02 times 10 to the 23 per mole in Cambridge O Level Chemistry; it converts between amount of substance in moles and number of particles.
Amount of substance: Amount of substance is the quantity measured in moles that counts how many specified particles are present; it is calculated from mass divided by molar mass, and it is the quantity a balanced chemical equation compares between substances.
Limiting reactant: The limiting reactant is the reactant that is completely used up first in a chemical reaction and therefore sets the maximum amount of product that can form; it is identified by dividing each reactant's amount in moles by its coefficient in the balanced equation and taking the smaller result.
Molecular formula: The molecular formula of a compound states the number and type of different atoms present in one molecule of that compound, for example C2H4 for ethene, which contains two carbon atoms and four hydrogen atoms per molecule.
Molar gas volume: The molar gas volume is the volume occupied by one mole of any gas under stated conditions; for Cambridge O Level Chemistry it is taken as 24 cubic decimetres per mole at room temperature and pressure, equivalently 24 000 cubic centimetres per mole.
State symbol: A state symbol is the letter written in brackets after a formula in an equation to record the physical state of that species under the reaction conditions: (s) solid, (l) liquid, (g) gas and (aq) dissolved in water.
Formula unit: A formula unit is the simplest whole-number ratio of ions in an ionic compound, written so that the total positive charge exactly cancels the total negative charge; magnesium chloride, MgCl2, is one formula unit containing one magnesium ion and two chloride ions.
Empirical formula mass: The empirical formula mass is the sum of the relative atomic masses of the atoms shown in a compound's empirical formula; dividing the compound's relative molecular mass by it gives the whole-number multiplier that converts the empirical formula into the molecular formula.
Stoichiometric ratio: A stoichiometric ratio is the ratio of the coefficients in a balanced chemical equation; it compares amounts of substance in moles, so quantities must be converted to moles before the ratio is applied and converted back afterwards.
Molar mass: Molar mass, M, is the mass of one mole of a substance, measured in grams per mole; its numerical value matches the relative molecular or formula mass, but unlike those quantities it carries a unit and is the value used in the relationship amount of substance equals mass divided by molar mass.
Concentration: Concentration states how much solute is dissolved in a given volume of solution; it is measured either in moles of solute per cubic decimetre of solution or in grams of solute per cubic decimetre of solution, and the two are linked by the molar mass of the solute.
Percentage yield: Percentage yield is the mass of product actually obtained divided by the maximum mass the balanced equation predicts from the limiting reactant, multiplied by 100 per cent.
Titration: A titration is a procedure in which a measured volume of one solution is added from a burette to a known volume of another until the reaction is exactly complete; the volume delivered, called the titre, together with the known concentration allows the unknown concentration to be calculated.
Percentage purity: Percentage purity is the mass of the pure target substance present in a sample divided by the total mass of that impure sample, multiplied by 100 per cent; the mass of the pure substance is usually found from reaction data.
Percentage composition by mass: Percentage composition by mass is the mass contributed by a specified element within a compound's formula divided by the compound's relative molecular or formula mass, multiplied by 100 per cent.

Common mistakes to avoid

1. Changing a subscript to balance an equation Repair A subscript is part of the substance's identity: turning \(\mathrm{H_2O}\) into \(\mathrm{H_2O_2}\) balances the oxygen and changes water into hydrogen peroxide. Only the number in front may change. Lesson 3.1C
2. Reading coefficients as a mass ratio Repair \(2\mathrm{H_2} + \mathrm{O_2} \rightarrow 2\mathrm{H_2O}\) does not mean 2 g of hydrogen with 1 g of oxygen. Coefficients count moles. Convert to moles before you touch the ratio. Lesson 3.3C
3. Giving \(A_r\) or \(M_r\) a unit Repair Both are comparisons against one-twelfth of a carbon-12 atom, so both are pure numbers. Writing \(M_r = 44\ \mathrm{g}\) is wrong; the quantity that carries \(\mathrm{g\,mol^{-1}}\) is molar mass. Lesson 3.2
4. Treating \(M_r\) and molar mass as the same quantity Repair They share a number and nothing else. \(M_r(\mathrm{CO_2}) = 44\); \(M(\mathrm{CO_2}) = 44\ \mathrm{g\,mol^{-1}}\). Only the second one can go into \(n = m/M\). Lesson 3.2
5. Not saying which particle you counted Repair "\(6.02\times10^{23}\) particles" is an incomplete answer. One mole of \(\mathrm{CO_2}\) is \(6.02\times10^{23}\) molecules, and \(1.806\times10^{24}\) atoms. Name the entity every time. Lesson 3.3A
6. Using 24 with a volume in \(\mathrm{cm^3}\) Repair \(24\) belongs to \(\mathrm{dm^3}\); \(24\,000\) belongs to \(\mathrm{cm^3}\). Mixing them is a factor-of-1000 error, which is exactly the size of gap that looks plausible on a calculator. Lesson 3.3E
7. Putting \(\mathrm{cm^3}\) straight into \(n = cV\) Repair \(c\) is per \(\mathrm{dm^3}\), so \(V\) must be in \(\mathrm{dm^3}\). Divide by 1000 first, on its own line, before you substitute. Lesson 3.3F
8. Applying a 1:1 shortcut to a titration that is not 1:1 Repair The relationship written as \(c_1V_1 = c_2V_2\) hides the equation ratio, and is only correct when that ratio happens to be \(1:1\). Go through moles instead — it costs one extra line and works every time. Lesson 3.3G
9. Rounding an empirical ratio too early Repair A ratio of \(1 : 1.5\) is not \(1:2\). It is \(2:3\), reached by multiplying both numbers by 2. Rounding here changes the compound. Lesson 3.3H
10. Inverting a percentage Repair Yield is actual over theoretical; purity is pure over total sample; composition is one element's mass over the whole formula mass. Write the fraction in words before you write it in numbers. Lesson 3.3I
"\(\mathrm{Ca(NO_3)_2}\) has 1 N and 3 O." Fix The subscript outside the bracket multiplies everything inside it: 2 N and 6 O. Missing this makes every downstream \(M_r\) wrong.
"The empirical formula of \(\mathrm{CO_2}\) is \(\mathrm{CO}\)." Fix \(1:2\) is already the simplest whole-number ratio, so the empirical formula of carbon dioxide is \(\mathrm{CO_2}\). Simplify only when both numbers share a factor.
"\(\mathrm{MgO}\) is a molecule of magnesium oxide." Fix Magnesium oxide is a giant ionic lattice with no discrete molecules. \(\mathrm{MgO}\) is a formula unit — the simplest ratio of ions — which is why its relative mass is called a relative formula mass.
"Calcium nitrate is \(\mathrm{CaNO_{32}}\)." Fix Without a bracket the 2 attaches to the oxygen subscript and creates a substance that does not exist. Write \(\mathrm{Ca(NO_3)_2}\).
"Sodium carbonate is \(\mathrm{Na_2(CO_3)}\)." Fix Brackets are only needed when the polyatomic ion is repeated. One carbonate ion needs none: \(\mathrm{Na_2CO_3}\).
"Iron oxide is \(\mathrm{FeO}\)." Fix "Iron oxide" is incomplete — iron has two common charges. Iron(II) oxide is \(\mathrm{FeO}\); iron(III) oxide is \(\mathrm{Fe_2O_3}\). Read the Roman numeral before writing anything.
"\(A_r\) of chlorine is 35.5, so a chlorine atom weighs 35.5 g." Fix \(35.5\) is a count of reference shares, not a mass. It is one mole of chlorine atoms that has a mass of \(35.5\ \mathrm{g}\).
"\(A_r\) is 35.5 because chlorine has half an extra neutron." Fix No individual atom has a fractional mass. \(A_r\) is an average over the isotopes present, so a non-whole value simply reflects the mixture.
"Magnesium oxide has a relative molecular mass of 40." Fix The number is right, the name is not. \(\mathrm{MgO}\) is ionic and has no molecules, so \(40\) is its relative formula mass.
"One mole of \(\mathrm{Cl_2}\) contains \(6.02\times10^{23}\) chlorine atoms." Fix It contains \(6.02\times10^{23}\) \(\mathrm{Cl_2}\) molecules, which is \(1.204\times10^{24}\) chlorine atoms.
"A mole is a very large mass." Fix A mole is an amount of substance, not a mass. One mole of hydrogen molecules has a mass of about \(2\ \mathrm{g}\); one mole of lead atoms has a mass of about \(207\ \mathrm{g}\). Same amount, very different masses.
"\(N\) and \(n\) are the same thing." Fix \(n\) is an amount in \(\mathrm{mol}\); \(N\) is a bare count of particles. They differ by a factor of \(N_A\), and mixing the symbols is how a calculation ends up \(10^{23}\) times wrong.
"\(n = M/m\)." Fix Inverted. Substitute the units and see: \(\mathrm{g\,mol^{-1}} \div \mathrm{g} = \mathrm{mol^{-1}}\), which is not an amount of substance. The correct arrangement, \(m/M\), gives \(\mathrm{mol}\).
"\(M_r\) can go straight into \(n = m/M\)." Fix Numerically you get away with it; as a statement it is wrong, because \(M_r\) has no unit. Write the molar mass with \(\mathrm{g\,mol^{-1}}\) and the substitution becomes self-checking.
"The number of particles has the unit mol." Fix \(N\) is a bare count and has no unit; what it needs instead is the name of the particle. \(N_A\) carries \(\mathrm{mol^{-1}}\), which is what cancels the \(\mathrm{mol}\) in \(n\).
Applying the ratio to grams. Fix The ratio belongs to the middle of the route, between two mole values. Cross into moles first, every time.
Using the wrong pair of coefficients. Fix Write the ratio out with both substances named — "\(1\ \mathrm{Fe_2O_3} : 2\ \mathrm{Fe}\)" — before you multiply. Naming them makes it almost impossible to pick up a coefficient belonging to a substance you were not asked about.
Using the reactant's molar mass to convert the product back. Fix Stage 4 uses B's molar mass, not A's. Writing \(M(\mathrm{CaO}) = 56\ \mathrm{g\,mol^{-1}}\) on its own line, with the substance named, keeps the two apart.
"The reactant with the smaller mass is limiting." Fix Mass is not a count. \(4.0\ \mathrm{g}\) of hydrogen is 2 mol of molecules while \(16.0\ \mathrm{g}\) of oxygen is only 0.5 mol. Convert first, then divide by the coefficient.
"Both reactants are used up completely." Fix Only when the supplied amounts happen to be in exactly the equation's ratio. Otherwise one is left over, and a question asking for the excess remaining is asking you to prove you noticed.
"Add the two product amounts together." Fix In method 2 you calculate the product twice to compare the answers, then keep the smaller one. The two figures are rival predictions, not contributions.
"Multiply the mass by 24." Fix The molar gas volume multiplies an amount, not a mass. Convert with \(n = m/M\) first, then multiply.
Dividing a volume in \(\mathrm{cm^3}\) by 24. Fix Convert to \(\mathrm{dm^3}\) first, or divide by \(24\,000\). Both give the same answer; mixing them is out by a factor of a thousand.
Quoting a gas volume without stating the conditions. Fix Write "at r.t.p." with the answer. The number \(24\) is only true under those conditions, and the phrase is what shows you know that.
Putting \(\mathrm{cm^3}\) straight into \(n = cV\). Fix Divide by \(1000\) first, on its own line. \(0.100 \times 25.0 = 2.5\) is not \(2.5\ \mathrm{mol}\) of anything; the real answer is \(2.5\times10^{-3}\ \mathrm{mol}\).
Using the volume of water added rather than the volume of solution. Fix Concentration is per \(\mathrm{dm^3}\) of finished solution. "Made up to \(250\ \mathrm{cm^3}\)" gives you the volume you need; "dissolved in \(250\ \mathrm{cm^3}\) of water" is a different statement, and a question will say which it means.
Reading \(\mathrm{g\,dm^{-3}}\) as if it were \(\mathrm{mol\,dm^{-3}}\). Fix Check the unit before choosing a route. A \(\mathrm{g\,dm^{-3}}\) value has to pass through the molar mass before it can meet an equation ratio, because ratios only speak in moles.
Dividing by the titre in step 5. Fix Divide by the volume of the solution whose concentration you are finding. Label both volumes with their solution's name as you write them down.
Using \(c_1V_1 = c_2V_2\) on a non-1 : 1 reaction. Fix Go through moles. The ratio line is the only place the chemistry enters the calculation, and a method without it cannot be right except by luck.
Averaging every titre including the rough. Fix Use the titres the question tells you to use. Do not invent your own selection rule, and do not silently drop a value the question kept.
Dividing the percentages by \(M_r\). Fix Each element's mass is divided by that element's own \(A_r\). \(M_r\) belongs to the last step, where it decides the multiplier.
Dividing by the first amount instead of the smallest. Fix Dividing by the smallest guarantees every result is at least 1, which is what makes the whole-number check readable. Circle the smallest before you divide anything.
Scaling only the awkward number. Fix Multiply every value in the ratio by the same integer. Scaling one alone changes the ratio, which is the one thing the whole method exists to preserve.
"Percentage yield = theoretical ÷ actual." Fix Actual on top. What you got is being compared with what was possible, so the possible amount is the whole.
Counting only one nitrogen in \(\mathrm{NH_4NO_3}\). Fix Read the whole formula. The percentages of every element must add to \(100\%\), which is the check that finds this.
Dividing the pure mass by the product mass instead of the sample mass. Fix Purity is about the sample you weighed out. The product only appears as the route to finding how much of that sample was the real substance.

Examiner tips

What to do with your score There is no pass mark. Count only which questions you missed, and do those repairs first — in the order they are listed above, since each one is used by the next. If you missed nothing, start at Lesson 3.1A and read at pace; the chapter will still show you a route you can lean on when a question gets unfamiliar.
Five of these ten are unit errors, not chemistry errors Traps 2, 3, 4, 6 and 7 all come down to reading a unit — or the absence of one — correctly. Trap 2 belongs here because a coefficient counts moles, so reading it as grams is reading the wrong unit off the equation. That is why every worked example in this chapter carries its units through the substitution rather than adding them at the end. If you build the habit of writing \(\dfrac{4.8\ \mathrm{g}}{24\ \mathrm{g\,mol^{-1}}}\) instead of \(\dfrac{4.8}{24}\), half of this list stops being able to reach you.
The Roman numeral is the charge Iron(II) means \(\mathrm{Fe^{2+}}\); iron(III) means \(\mathrm{Fe^{3+}}\); copper(II) means \(\mathrm{Cu^{2+}}\). When a name gives you a Roman numeral you have been handed the charge and do not need to recall it. That is why iron(III) oxide is \(\mathrm{Fe_2O_3}\) while iron(II) oxide would be \(\mathrm{FeO}\).
Why percentages can be used as if they were masses A percentage by mass is the mass present in \(100\ \mathrm{g}\) of the compound. Dividing \(54.55\) by \(12\) is therefore genuinely "the moles of carbon in \(100\ \mathrm{g}\)". Since the method only ever uses the ratio of the amounts, the choice of \(100\ \mathrm{g}\) does not affect the answer — which is why you can start from percentages without converting them first.
Working the selector on a question you have not seen "\(50.0\ \mathrm{cm^3}\) of \(0.200\ \mathrm{mol\,dm^{-3}}\) hydrochloric acid is added to excess magnesium. Calculate the volume of hydrogen produced at r.t.p." I have: a concentration and a volume, of hydrochloric acid. I want: a gas volume in \(\mathrm{dm^3}\), of hydrogen. Different substances, so a ratio is needed. Route: \(n = cV\) (converting \(50.0\ \mathrm{cm^3}\) to \(0.0500\ \mathrm{dm^3}\)) → ratio from \(\mathrm{Mg} + 2\mathrm{HCl} \rightarrow \mathrm{MgCl_2} + \mathrm{H_2}\), which is \(2\ \mathrm{HCl} : 1\ \mathrm{H_2}\) → \(V = n \times 24\). Answer: \(n(\mathrm{HCl}) = 0.200 \times 0.0500 = 0.0100\ \mathrm{mol}\); \(n(\mathrm{H_2}) = 0.00500\ \mathrm{mol}\); \(V = 0.00500 \times 24 = 0.120\ \mathrm{dm^3}\), that is \(120\ \mathrm{cm^3}\) at r.t.p. The word "excess" told you the magnesium never needed checking.
The signature of each error Wrong by \(\times 1000\)? A \(\mathrm{cm^3}\) or a kg conversion. Wrong by \(\times 100\)? A percentage that was not converted. Wrong by \(\times 2\) or \(\times 3\)? A coefficient ratio that was dropped. Wrong by \(10^{23}\)? \(N\) and \(n\) were swapped. A few per cent out? Something was rounded too early. Knowing the signature turns a wrong answer into a diagnosis.
Marking yourself honestly Give yourself the mark only if the working would communicate the method to someone else — relationship, substitution with units, ratio where one applies, and a final answer with its unit. A right number with no visible route is worth less than a wrong number with a clear one, and it teaches you nothing about where you are weak.
Selected answers, if you want to check 3.3.2a: \(0.10\ \mathrm{mol}\). 3.3.2b: \(12.0\ \mathrm{g}\). 3.3.2c: \(36\ \mathrm{g\,mol^{-1}}\). 3.3.2d: \(M_r = 36\), no unit. 3.3.2e: \(1.204\times10^{23}\) chloride ions. 3.3.3: \(1.2\ \mathrm{dm^3}\). 3.3.4a: \(8.0\ \mathrm{g\,dm^{-3}}\). 3.3.5a: \(2.8\ \mathrm{g}\). 3.3.5d: \(0.0500\ \mathrm{dm^3} = 50.0\ \mathrm{cm^3}\). 3.3.5e: \(10.0\ \mathrm{g\,dm^{-3}}\). 3.3.5f: \(0.200\ \mathrm{mol\,dm^{-3}}\). 3.3.7b: \(\mathrm{C_4H_8}\), multiplier 4. 3.3.8b: \(40.0\%\). 3.1.4: \(\mathrm{A_2B_3}\).
The five-minute version If you have only a few minutes before an exam, do this and nothing else: draw the mole map, write the four bridge operations onto it with their unit conditions, and write the three percentage fractions in words. That is the part of Topic 3 which, once it is on paper in front of you, makes every other part findable.

Frequently asked questions

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

The molecular formula gives the number and type of atoms in one molecule of a compound, for example \(\mathrm{C_6H_{12}O_6}\) for glucose. The empirical formula gives the simplest whole-number ratio of the different atoms or ions, so for glucose it is \(\mathrm{CH_2O}\). They describe the same substance but answer different questions. Simplify only when the numbers share a common factor: the empirical formula of \(\mathrm{CO_2}\) is still \(\mathrm{CO_2}\), because 1 : 2 is already the simplest ratio.

What is a mole, and why do you have to say which particle you are counting?

The mole (mol) is the unit of amount of substance. One mole contains \(6.02\times10^{23}\) particles, and that number is the Avogadro constant. The particles must be named because the count changes with the entity chosen: one mole of \(\mathrm{CO_2}\) is \(6.02\times10^{23}\) molecules but \(1.806\times10^{24}\) atoms. A mole is an amount, not a mass: one mole of hydrogen molecules has a mass of about 2 g, while one mole of lead atoms has a mass of about 207 g.

What is the difference between relative molecular mass and molar mass?

They share a number and nothing else. Relative molecular mass, \(M_r\), is the sum of the relative atomic masses in a formula; it is a comparison with one-twelfth of the mass of a carbon-12 atom, so it is a pure number with no unit. Molar mass, \(M\), is the mass of one mole and carries the unit \(\mathrm{g\,mol^{-1}}\). \(M_r(\mathrm{CO_2}) = 44\) but \(M(\mathrm{CO_2}) = 44\ \mathrm{g\,mol^{-1}}\), and only the molar mass can go into \(n = m/M\). Writing \(M_r = 44\ \mathrm{g}\) loses the mark.

How do you balance a chemical equation without changing the formulae?

Change only the coefficients, the large numbers written in front of a formula, because they count how many of that substance react. Never change a subscript: the small number inside a formula is part of the substance's identity, so turning \(\mathrm{H_2O}\) into \(\mathrm{H_2O_2}\) balances the oxygen but changes water into hydrogen peroxide. Write the correct formulae first, add state symbols second, adjust coefficients last, then count every element on both sides before you write the answer down.

How do you find the limiting reactant?

Convert the amount of each reactant to moles, then divide each amount by its coefficient in the balanced equation. The reactant with the smaller quotient is the limiting reactant: it is completely used up first, so it alone decides the maximum amount of product that can form. Calculate the product from the limiting reactant, never from the one in excess, and finish by checking that the masses balance, including the mass of the excess reactant left over.

When do you use 24 and when do you use 24 000 in a gas volume calculation?

The molar gas volume at room temperature and pressure (r.t.p.) is \(24\ \mathrm{dm^3}\), which is the same as \(24\,000\ \mathrm{cm^3}\). Read the unit on the volume you were given, then choose: 24 goes with \(\mathrm{dm^3}\) and 24 000 goes with \(\mathrm{cm^3}\). Mixing them is a factor-of-1000 error that looks plausible on a calculator. The same rule applies to concentration: \(n = cV\) needs \(V\) in \(\mathrm{dm^3}\), so divide a \(\mathrm{cm^3}\) volume by 1000 on its own line before you substitute. Always state at r.t.p. in the answer.

How do you calculate percentage yield, percentage purity and percentage composition?

Write the fraction in words before you write it in numbers, so it cannot go in upside down. Percentage yield is the mass of product actually obtained divided by the theoretical maximum predicted from the balanced equation, times 100. Percentage purity is the mass of the pure substance divided by the total mass of the impure sample, times 100. Percentage composition by mass is the mass of one element in the formula divided by the relative formula mass, times 100. Then check the unit and the scale of the answer.

Syllabus reference and sources

Written against: Cambridge O Level Chemistry (5070) 2026–2028 Syllabus (Subject Content, Topic 3: Stoichiometry).

Written by: Academiq Edu Instructor Panel

Source documents

Cambridge O Level Chemistry 5070 syllabus for examination in 2026, 2027 and 2028 (version 1)

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