Chapter 03
Stoichiometry
Syllabus 0620
Core Revision Module
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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.
Key ideas to remember
- Write it, weigh it, scale it. A balanced equation plus two relative masses is enough to answer every reacting-mass question on the Core route.
- 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. Add the shares in the formula and you have \(M_r\); call it a relative formula mass when the compound is ionic.
- Coefficients count particles. Multiply each by its relative mass, and the equation starts talking in grams. One scale factor, taken from the mass you were given, then carries it to your experiment.
- Solute per cubic decimetre of solution — counted in grams, or counted in moles. Bigger number, more concentrated; and "dilute" is a word, not a measurement.
What you need to be able to do
- Core 3.1.1 State the formulae of the elements and compounds named in the subject content. Taught in Lesson 3.1A.
- Core 3.1.2 Define the molecular formula of a compound as the number and type of different atoms in one molecule. Lesson 3.1A.
- Core 3.1.3 Deduce the formula of a simple compound from the relative numbers of atoms present in a model or a diagrammatic representation. Lesson 3.1A.
- Core 3.1.4 Construct word equations and symbol equations to show how reactants form products, including state symbols. Lesson 3.1C.
- Core 3.2.1 Describe relative atomic mass, \(A_r\), as the average mass of the isotopes of an element compared to \(\tfrac{1}{12}\)th of the mass of an atom of \(\mathrm{^{12}C}\). Lesson 3.2A.
- Core 3.2.2 Define relative molecular mass, \(M_r\), as the sum of the relative atomic masses. Relative formula mass, \(M_r\), will be used for ionic compounds. Lesson 3.2A.
- Core 3.2.3 Calculate reacting masses in simple proportions. Calculations will not involve the mole concept. Lesson 3.2B.
- Core 3.3.1 State that concentration can be measured in \(\mathrm{g/dm^3}\) or \(\mathrm{mol/dm^3}\). the Core concentration lesson.
- Supplement 3.1.5 Define the empirical formula of a compound as the simplest whole number ratio of the different atoms or ions in a compound. Lesson 3.1A, Supplement block.
- Supplement 3.1.6 Deduce the formula of an ionic compound from the relative numbers of the ions present in a model or a diagrammatic representation, or from the charges on the ions. Lesson 3.1B.
- Supplement 3.1.7 Construct symbol equations with state symbols, including ionic equations. Lesson 3.1D.
- Supplement 3.1.8 Deduce the symbol equation with state symbols for a chemical reaction, given relevant information. Lesson 3.1C, Supplement block.
- Supplement 3.3.2 State that the mole, mol, is the unit of amount of substance and that one mole contains \(6.02\times10^{23}\) particles, e.g. atoms, ions, molecules; this number is the Avogadro constant. Lesson 3.3A.
- Supplement 3.3.3 Use the relationship amount of substance (mol) = mass (g) ÷ molar mass (g/mol) to calculate: (a) amount of substance, (b) mass, (c) molar mass, (d) relative atomic mass or relative molecular / formula mass, (e) number of particles, using the value of the Avogadro constant. Lesson 3.3B; part (d) also in Lesson 3.2A, Supplement block.
- Supplement 3.3.4 Use the molar gas volume, taken as \(24\ \mathrm{dm^3}\) at room temperature and pressure, r.t.p., in calculations involving gases. Lesson 3.3E.
- Supplement 3.3.5 Calculate stoichiometric reacting masses, limiting reactants, volumes of gases at r.t.p., volumes of solutions and concentrations of solutions expressed in \(\mathrm{g/dm^3}\) and \(\mathrm{mol/dm^3}\), including conversion between \(\mathrm{cm^3}\) and \(\mathrm{dm^3}\). Lesson 3.3C, 3.3D, 3.3E and 3.3F.
- Supplement 3.3.6 Use experimental data from a titration to calculate the moles of solute, or the concentration or volume of a solution. Lesson 3.3G.
- Supplement 3.3.7 Calculate empirical formulae and molecular formulae, given appropriate data. Lesson 3.3H.
- Supplement 3.3.8 Calculate percentage yield, percentage composition by mass and percentage purity, given appropriate data. Lesson 3.3I.
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 Core 3.1.4
- 2. Reading the coefficients themselves 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. A coefficient counts particles. The equation only starts talking about mass once each coefficient has been multiplied by that substance's relative mass — and then it fixes a mass proportion you can scale. Lesson 3.2B Core 3.2.3
- 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. Lesson 3.2A Core 3.2.2
- 4. Ignoring a bracket subscript Repair The number outside a bracket multiplies everything inside it. \(\mathrm{Ca(NO_3)_2}\) holds 2 N and 6 O, not 1 and 3, and a missed bracket makes every relative mass built on that formula wrong. Lesson 3.1A Core 3.1.1
- 5. Taking the scale factor from the wrong substance Repair The factor always comes from the substance whose mass you were given, divided by its own number in the proportion. Divide by the other one and the answer is out by exactly the ratio — which usually still looks plausible. Lesson 3.2B Core 3.2.3
- 6. Leaving out state symbols Repair They are named inside the Core statement, so a symbol equation without them is an incomplete answer, not a stylistic choice. And they depend on the conditions the question describes, not on habit. Lesson 3.1C Core 3.1.4
- 7. 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 3.2A, Supplement block Supplement 3.3.3
- 8. 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 Supplement 3.3.2
- 9. 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 Supplement 3.3.4
- 10. 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 Supplement 3.3.5
- 11. 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 amount of substance instead — it costs one extra line and works every time. Lesson 3.3G Supplement 3.3.6
- 12. Rounding an empirical ratio too early, or inverting a percentage Repair A ratio of \(1 : 1.5\) is not \(1:2\); it is \(2:3\), reached by multiplying both numbers by 2. And 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 3.3H, Lesson 3.3I Supplement 3.3.7 Supplement 3.3.8
- "\(\mathrm{Ca(NO_3)_2}\) 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.
- "\(\mathrm{MgO}\) has a relative molecular mass." Fix The arithmetic is the same, the name is not. Magnesium oxide is ionic, so the sum of its relative atomic masses is a relative formula mass. Statement Core 3.2.2 makes that distinction part of the definition.
- "A formula can be written from the name alone, always." Fix Only for the substances the subject content names — that is exactly what statement Core 3.1.1 means by a defined list. For anything else the question supplies a model, a diagram or the formula itself; use what you were given rather than guessing.
- "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 in grams: it says an average chlorine atom would balance \(35.5\) shares of a carbon-12 atom. A single atom is far too light to weigh at all, which is exactly why the scale is a comparison in the first place.
- "\(A_r\) 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.
- "The coefficients are \(2 : 2\), so the masses are in the ratio \(1:1\)." Fix A coefficient counts particles. Multiply each one by its substance's relative mass first, then compare. \(2\times24 = 48\) against \(2\times40 = 80\).
- "\(M_r(\mathrm{MgO}) = 40\), so \(40\ \mathrm{g}\) of MgO is formed." Fix The relative mass is one rung of the ladder, not the answer. It has still to be multiplied by the coefficient and then scaled by the factor the given mass sets. Skipping step 4 answers a question nobody asked.
- "Scale factor \(=\) given mass \(\div\) the other substance's number." Fix The factor always comes from the substance you were given, divided by its own number in the proportion. Dividing by the wrong one gives an answer that is out by the ratio itself — often close enough to look plausible.
- "The equation was not balanced, but the relative masses were right." Fix Then the proportion is wrong, and so is everything after it. Balance first, every time; it costs ten seconds and it is the only step that cannot be recovered from later.
- "Concentration is measured in \(\mathrm{g/cm^3}\)." Fix That is the usual unit of density. The two units named for concentration are \(\mathrm{g/dm^3}\) and \(\mathrm{mol/dm^3}\), and both are per cubic decimetre.
- "The volume in the unit is the volume of water added." Fix It is the volume of the finished solution. That is why a solution is made up to a mark on a flask rather than by adding a measured volume of water to the solid.
- "A concentrated solution is one with a large mass." Fix Concentration says nothing about how much solution you have. A drop of \(2\ \mathrm{mol/dm^3}\) acid is more concentrated than a bucket of \(0.1\ \mathrm{mol/dm^3}\) acid, and the bucket contains far more solute in total.
- "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. On the Core route: the four questions above the Supplement box are your whole list. If you missed none of them, start at Lesson 3.1A and read at pace, skipping the labelled boxes; the chapter will still show you a route you can lean on when a question gets unfamiliar. On the Extended route: all eleven are yours. If several of the Supplement seven went wrong, do the unit and calculator clinic before Lesson 3.3A rather than after it — those failures compound.
- Three of these six are arithmetic, not chemistry Traps 2, 4 and 5 are all failures to carry a number through correctly, and all three are prevented by the same habit: write the multiplication out even when it looks trivial. Writing \(2 \times 24 = 48\) rather than jumping to "48" is what makes trap 2 impossible, and writing \(\dfrac{4.8\ \mathrm{g}}{48}\) with the substance named beside it is what makes trap 5 impossible.
- Four of these six are unit errors, not chemistry errors Traps 7, 9, 10 and 12 all come down to reading a unit — or the absence of one — correctly. That is why every Supplement 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 \(\dfrac{4.8}{24}\), most 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 Core selector on a question you have not seen “\(4.0\ \mathrm{g}\) of calcium reacts completely with chlorine. \(\mathrm{Ca} + \mathrm{Cl_2} \rightarrow \mathrm{CaCl_2}\). \(A_r\): Ca = 40, Cl = 35.5. Calculate the mass of calcium chloride formed.” I have: a mass, of calcium. I want: a mass, of calcium chloride. Balanced? Ca 1 : 1, Cl 2 : 2 — yes. So this is a Core reacting-mass question and the proportion route finishes it. Route: \(A_r(\mathrm{Ca}) = 40\) and \(M_r(\mathrm{CaCl_2}) = 40 + (2\times35.5) = 111\). Both coefficients are 1, so the proportion is \(40 \rightarrow 111\). Scale factor \(= 4.0 \div 40 = 0.10\). Answer: \(111 \times 0.10 = 11.1\ \mathrm{g}\). Check: the chloride must be heavier than the calcium alone, and the chlorine that joined it is \(71 \times 0.10 = 7.1\ \mathrm{g}\); \(4.0 + 7.1 = 11.1\). Consistent.
- 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 On every route. Wrong by \(\times 1000\)? A kilogram conversion. Wrong by exactly the ratio of two relative masses? The scale factor was taken from the wrong substance, or a coefficient was dropped. A few per cent out? Something was rounded too early. On the Extended route as well. Wrong by \(\times 1000\) can also be a \(\mathrm{cm^3}\) conversion or \(24\) used where \(24\,000\) was needed. Wrong by \(\times 100\)? A percentage that was not converted. Wrong by \(10^{23}\)? \(N\) and \(n\) were swapped. Knowing the signature turns a wrong answer into a diagnosis, and it is the reason a wrong answer is worth two minutes rather than a shrug.
- Marking yourself honestly Core paper, out of 25. Mark Q1–Q6 only. A score below about 18 usually means one of two things, and they need different repairs: if the marks went in Q1, Q2, Q3 or Q5 the problem is reading and writing formulae, so return to Lessons 3.1A and 3.1C. If they went in Q4, the problem is the proportion method, and Lesson 3.2B is the whole repair. Supplement paper, out of 57. Mark Q7–Q19 as well, and total out of 82. Sort every lost mark into one of three piles before you decide what to revise: a unit error, a ratio error, or a chemistry error (a wrong formula, a wrong state, an unbalanced equation). The largest pile names the section to return to, and it is almost never the one you expected. Do not average the two papers. They are different papers, and a strong Core score with a weak Supplement score is a clear and useful diagnosis: the foundations are sound and the mole route is not.
- Core route — you have finished Topic 3 Those eight rows are the whole of Topic 3 for a Core candidate. There is no ninth, and nothing in the box below is examinable on Papers 1 and 3. If all eight are ticked honestly, move on to Topic 4; if any is not, the link beside it is the shortest way back.
- Selected answers, if you want to check Core. 3.1.3: \(\mathrm{AB_2C_2}\). 3.2.2: \(M_r\bigl(\mathrm{(NH_4)_2CO_3}\bigr) = 96\), a relative formula mass, because the compound is ionic. 3.2.3: proportion \(80 \rightarrow 112\), scale factor \(10.0 \div 80 = 0.125\), so \(14.0\ \mathrm{g}\) of calcium oxide. Supplement. 3.1.5: \(\mathrm{CH_2O}\), and \(\mathrm{CO_2}\) is already simplest. 3.1.6: \(\mathrm{Fe_2(SO_4)_3}\) and \(\mathrm{(NH_4)_3PO_4}\). 3.3.3: \(0.0986\ \mathrm{mol}\); \(7.3\ \mathrm{g}\); \(44\ \mathrm{g\,mol^{-1}}\); \(1.204\times10^{23}\) chloride ions. 3.3.4: \(1.2\ \mathrm{dm^3}\), which is \(1200\ \mathrm{cm^3}\). 3.3.5: the acid is limiting (\(0.0500\ \mathrm{mol}\) of HCl allows only \(0.0250\ \mathrm{mol}\) of Zn, and \(3.25\ \mathrm{g}\) is \(0.0500\ \mathrm{mol}\)), giving \(0.0250\ \mathrm{mol}\) of hydrogen and \(0.60\ \mathrm{dm^3}\) at r.t.p. 3.3.6: \(0.160\ \mathrm{mol\,dm^{-3}}\) — the \(1:2\) ratio doubles what a \(1:1\) shortcut would give. 3.3.7: \(\mathrm{CH_2}\) and \(\mathrm{C_4H_8}\), multiplier 4. 3.3.8: \(35.0\%\) nitrogen.
- The five-minute version Core route. Write out one balanced equation, put the relative masses under it, multiply each by its coefficient, and write the sentence “x grams of A gives y grams of B”. Then write, underneath, “scale factor = given mass ÷ that substance's own number”. Those two lines are the whole of statement 3.2.3 on one piece of paper. Extended route. Do the above, then 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.
Syllabus reference and sources
Written against: Cambridge IGCSE Chemistry (0620) 2026–2028 Syllabus (Subject Content, Topic 3: Stoichiometry), Core and Supplement statements.
Written by: Academiq Instructor Panel
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