📚 Typical Example Problems Explained in IGCSE CIE Chemistry | IGCSE CIE 化学：典型例题详解
IGCSE CIE Chemistry challenges students not only to recall facts but to apply concepts to solve structured problems. This article walks through a curated set of typical exam-style questions, unpacking the reasoning step by step. By working through these examples, learners can build confidence in calculations, chemical reasoning, and practical-based questions. Each section presents an original problem followed by a clear method and commentary in both English and Chinese.
IGCSE CIE 化学不仅要求学生记忆知识点，更要求他们运用概念来解决结构性问题。本文精选了一组典型的考题式例题，逐步拆解解题思路。通过练习这些例题，学生可以逐步建立起在计算、化学推理和实验类题目上的信心。每个小节均提供一道原创题目，并附上清晰的步骤和双语点评。
1. Mole Calculation Essentials | 摩尔计算基础
A sample of magnesium ribbon reacts completely with excess dilute hydrochloric acid to produce 240 cm³ of hydrogen gas at room temperature and pressure (r.t.p.). Calculate the mass of magnesium used. [Mg = 24.0; molar volume at r.t.p. = 24.0 dm³ mol⁻¹]
一段镁带与过量稀盐酸完全反应，在室温和常压下生成 240 cm³ 氢气。计算所用镁的质量。[Mg = 24.0；常温常压下气体摩尔体积 = 24.0 dm³ mol⁻¹]
Step 1: Convert the volume of hydrogen to dm³: 240 cm³ = 0.240 dm³. Step 2: Moles of H₂ = volume / molar volume = 0.240 / 24.0 = 0.0100 mol. Step 3: Balanced equation: Mg + 2HCl → MgCl₂ + H₂. Step 4: Mole ratio Mg : H₂ = 1 : 1, so moles of Mg = 0.0100 mol. Step 5: Mass of Mg = moles × Mᵣ = 0.0100 × 24.0 = 0.24 g.
步骤一：将氢气体积转换为 dm³：240 cm³ = 0.240 dm³。步骤二：H₂ 的物质的量 = 体积 / 摩尔体积 = 0.240 / 24.0 = 0.0100 mol。步骤三：配平方程式：Mg + 2HCl → MgCl₂ + H₂。步骤四：Mg 与 H₂ 的物质的量之比为 1 : 1，所以 Mg 的物质的量 = 0.0100 mol。步骤五：Mg 的质量 = 物质的量 × 相对原子质量 = 0.0100 × 24.0 = 0.24 g。
This classic single-replacement problem reinforces the direct link between gas volume, moles and reacting mass. Always check the mole ratio from the balanced equation and remember to convert cm³ to dm³ when using molar volume at r.t.p.
这道经典的置换反应题巩固了气体体积、物质的量和反应物质量之间的直接联系。务必要检查配平方程中的物质的量之比，并在使用常温常压下的摩尔体积时将 cm³ 转换为 dm³。
2. Empirical and Molecular Formulae from Percentage Composition | 由百分组成求经验式与分子式
A compound contains 40.0 % carbon, 6.7 % hydrogen and 53.3 % oxygen by mass. Its relative molecular mass is 180. Determine its empirical formula and its molecular formula.
某化合物含碳 40.0 %、氢 6.7 % 和氧 53.3 %（质量分数），其相对分子质量为 180。试确定其经验式和分子式。
Assume 100 g: C = 40.0 g, H = 6.7 g, O = 53.3 g. Moles: C = 40.0 / 12.0 = 3.33 mol; H = 6.7 / 1.0 = 6.7 mol; O = 53.3 / 16.0 = 3.33 mol. Divide by smallest (3.33): C = 1, H = 6.7/3.33 ≈ 2, O = 1. So empirical formula is CH₂O. Empirical formula mass = 12 + 2×1 + 16 = 30. Molecular mass / empirical mass = 180 / 30 = 6. Molecular formula = (CH₂O)₆ = C₆H₁₂O₆.
假设 100 g 样品：C = 40.0 g，H = 6.7 g，O = 53.3 g。物质的量：C = 40.0 / 12.0 = 3.33 mol；H = 6.7 / 1.0 = 6.7 mol；O = 53.3 / 16.0 = 3.33 mol。除以最小量 (3.33)：C = 1，H ≈ 2，O = 1。因此经验式为 CH₂O。经验式质量 = 12 + 2×1 + 16 = 30。分子质量 / 经验式质量 = 180 / 30 = 6。分子式为 (CH₂O)₆ = C₆H₁₂O₆。
Percentage-to-formula questions often appear in CIE Paper 4. Keep a strict assumption of 100 g and round carefully. If the ratio gives a decimal close to 0.5 or 0.33, multiply all subscripts by a suitable integer.
由百分组成推断化学式的题目常出现在 CIE 试卷 4 中。严格假设 100 g 并仔细估算比例。若得到的原子个数比出现接近 0.5 或 0.33 的小数，需要将全部下标乘以适当的整数。
3. Equilibrium and Le Chatelier’s Principle | 平衡与勒夏特列原理
The reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) is exothermic in the forward direction. State and explain the effect on the equilibrium yield of SO₃ when (a) pressure is increased, (b) temperature is increased, (c) a catalyst is added.
反应 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) 正向放热。说明并解释以下改变对 SO₃ 平衡产率的影响：(a) 增大压强，(b) 升高温度，(c) 加入催化剂。
(a) Increasing pressure shifts the equilibrium to the side with fewer gas molecules. Reactants: 3 moles of gas; products: 2 moles of gas. Shift to right → higher yield of SO₃. (b) Forward reaction is exothermic. Increasing temperature favours the endothermic reverse reaction. Equilibrium shifts left → yield of SO₃ decreases. (c) A catalyst speeds up both forward and reverse reactions equally. Equilibrium position does NOT change; yield remains unchanged. Only rate of attainment is increased.
(a) 增大压强，平衡向气体分子数较少的一侧移动。反应物：3 mol 气体；生成物：2 mol 气体。平衡右移 → SO₃ 产率提高。(b) 正向反应放热。升高温度有利于吸热的逆反应。平衡左移 → SO₃ 产率降低。(c) 催化剂同等程度地加快正、逆反应速率。平衡位置不变；产率不变。仅加快达到平衡的速率。
CIE often asks to describe both the direction of shift and the reasoning in terms of moles of gas or energy change. The catalyst point is a classic pitfall: always state that yield does not change.
CIE 常要求学生既描述平衡移动方向，又根据气体物质的量或能量变化说明理由。关于催化剂这一点是经典易错点：一定要说明产率不变。
4. Titration Calculations and Concentration | 滴定计算与浓度
25.0 cm³ of sodium hydroxide solution required 20.0 cm³ of 0.100 mol dm⁻³ sulfuric acid for neutralisation. Calculate the concentration of the NaOH solution in mol dm⁻³ and g dm⁻³. [Na = 23.0, O = 16.0, H = 1.0]
25.0 cm³ 的氢氧化钠溶液恰好被 20.0 cm³ 的 0.100 mol dm⁻³ 硫酸中和。计算 NaOH 溶液的物质的量浓度（mol dm⁻³）和质量浓度（g dm⁻³）。
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Moles of H₂SO₄ = c × V = 0.100 × (20.0/1000) = 0.00200 mol. From equation, 1 mol H₂SO₄ reacts with 2 mol NaOH, so moles of NaOH = 2 × 0.00200 = 0.00400 mol. Concentration of NaOH = moles / volume (dm³) = 0.00400 / (25.0/1000) = 0.160 mol dm⁻³. Mᵣ of NaOH = 23.0 + 16.0 + 1.0 = 40.0. Mass concentration = mol dm⁻³ × Mᵣ = 0.160 × 40.0 = 6.40 g dm⁻³.
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O。H₂SO₄ 的物质的量 = c × V = 0.100 × (20.0/1000) = 0.00200 mol。由方程式知 1 mol H₂SO₄ 与 2 mol NaOH 反应，所以 NaOH 的物质的量 = 2 × 0.00200 = 0.00400 mol。NaOH 的浓度 = 物质的量 / 体积 (dm³) = 0.00400 / (25.0/1000) = 0.160 mol dm⁻³。NaOH 的 Mᵣ = 23.0 + 16.0 + 1.0 = 40.0。质量浓度 = mol dm⁻³ × Mᵣ = 0.160 × 40.0 = 6.40 g dm⁻³。
Titration problems hinge on the mole ratio from the balanced equation. Always convert cm³ to dm³ by dividing by 1000. A common mistake is using the ratio 1:1 for diprotic acids like H₂SO₄ — remember it donates 2 H⁺ per molecule.
滴定计算的关键在于配平方程式中的物质的量之比。始终记住把 cm³ 除以 1000 转换为 dm³。对于硫酸这类二元酸，常见的错误是使用 1:1 的比——记住每个硫酸分子提供 2 个 H⁺。
5. Electrolysis of Aqueous Solutions | 水溶液电解
Describe the products formed at the anode and cathode during the electrolysis of concentrated aqueous sodium chloride using inert electrodes. Include ionic half-equations.
描述使用惰性电极电解浓氯化钠水溶液时阳极和阴极的产物。写出离子半方程式。
In concentrated NaCl(aq), the ions present are Na⁺, Cl⁻, H⁺, OH⁻. At the cathode (negative electrode): H⁺ ions discharge more easily than Na⁺. Reduction: 2H⁺ + 2e⁻ → H₂(g). Hydrogen gas is produced. At the anode (positive electrode): Cl⁻ ions are present in high concentration and discharge in preference to OH⁻. Oxidation: 2Cl⁻ → Cl₂(g) + 2e⁻. Chlorine gas is produced. Sodium hydroxide remains in solution.
在浓 NaCl 水溶液中，存在的离子有 Na⁺、Cl⁻、H⁺、OH⁻。在阴极（负电极）：H⁺ 比 Na⁺ 更容易放电。还原反应：2H⁺ + 2e⁻ → H₂(g)，生成氢气。在阳极（正电极）：Cl⁻ 浓度很高，优先于 OH⁻ 放电。氧化反应：2Cl⁻ → Cl₂(g) + 2e⁻，生成氯气。溶液中留下氢氧化钠。
This is a core electrolysis example where concentration overrides the normal preference for OH⁻ discharge. Students must be able to select the correct discharge based on the electrochemical series and concentration effects. For dilute NaCl, the anode product is oxygen due to OH⁻ discharge.
这是电解内容中的核心例子，其中浓度效应压过了通常的 OH⁻ 放电优先规则。学生必须能根据电化学序和浓度效应选择正确的放电物种。对于稀 NaCl，阳极产物是由于 OH⁻ 放电而生成的氧气。
6. Energy Profile and Enthalpy Change | 能量变化与焓变
When 50 cm³ of 1.0 mol dm⁻³ HCl is mixed with 50 cm³ of 1.0 mol dm⁻³ NaOH in a polystyrene cup, the temperature rises by 6.7 °C. Calculate the enthalpy change for the reaction HCl + NaOH → NaCl + H₂O. Assume the density of the solution is 1.0 g cm⁻³ and specific heat capacity is 4.2 J g⁻¹ °C⁻¹.
在聚苯乙烯杯中混合 50 cm³ 1.0 mol dm⁻³ HCl 和 50 cm³ 1.0 mol dm⁻³ NaOH，温度升高 6.7 °C。计算反应 HCl + NaOH → NaCl + H₂O 的焓变。假设溶液密度为 1.0 g cm⁻³，比热容为 4.2 J g⁻¹ °C⁻¹。
Total volume = 50 + 50 = 100 cm³ → mass = 100 g. Heat released q = mcΔT = 100 × 4.2 × 6.7 = 2814 J. Moles of HCl = 0.050 × 1.0 = 0.050 mol; same for NaOH. ΔH = −q / moles = −2814 / 0.050 = −56280 J mol⁻¹ ≈ −56.3 kJ mol⁻¹ (exothermic). The negative sign indicates heat is released to the surroundings.
总体积 = 50 + 50 = 100 cm³ → 质量 = 100 g。释放的热量 q = mcΔT = 100 × 4.2 × 6.7 = 2814 J。HCl 的物质的量 = 0.050 × 1.0 = 0.050 mol；NaOH 相同。ΔH = −q / 物质的量 = −2814 / 0.050 = −56280 J mol⁻¹ ≈ −56.3 kJ mol⁻¹（放热）。负号表示热量释放到环境中。
Many marks are lost by forgetting to convert J to kJ or by omitting the negative sign for exothermic reactions. Also, look closely at the mole ratios: here both reactants are in 1:1 ratio, but if one is limiting, use the limiting amount.
许多考生因忘记将 J 转换为 kJ，或忽略了放热反应的负号而失分。另外，要仔细检查物质的量之比：此例中两种反应物为 1:1 比，但如果存在限制试剂，应使用限制试剂的物质的量。
7. Organic Chemistry: Naming and Isomerism | 有机化学：命名与同分异构
Draw the structural formula of 2-methylbutane and identify all the possible structural isomers of the molecular formula C₄H₁₀O that are alcohols. Name each isomer.
画出 2-甲基丁烷的结构简式，并写出分子式为 C₄H₁₀O 且属于醇类的所有可能的结构异构体。写出各异构体的名称。
2-methylbutane: a butane chain with a methyl branch on carbon 2: CH₃–CH(CH₃)–CH₂–CH₃. For C₄H₁₀O alcohols: four carbon chain possibilities. (1) butan-1-ol: CH₃–CH₂–CH₂–CH₂–OH. (2) butan-2-ol: CH₃–CH₂–CH(OH)–CH₃. (3) 2-methylpropan-1-ol: (CH₃)₂CH–CH₂–OH. (4) 2-methylpropan-2-ol: (CH₃)₃C–OH. These four are all structural isomers. Note: butan-2-ol has an optical isomer but in IGCSE only structural isomers are required.
2-甲基丁烷：丁烷主链上 2 位有一个甲基支链：CH₃–CH(CH₃)–CH₂–CH₃。对于 C₄H₁₀O 的醇类：可能有四种碳骨架。(1) 丁-1-醇：CH₃–CH₂–CH₂–CH₂–OH。(2) 丁-2-醇：CH₃–CH₂–CH(OH)–CH₃。(3) 2-甲基丙-1-醇：(CH₃)₂CH–CH₂–OH。(4) 2-甲基丙-2-醇：(CH₃)₃C–OH。以上四种均为结构异构体。注意：丁-2-醇存在对映异构体，但 IGCSE 阶段只要求结构异构。
Drawing organic structures clearly is vital — always show all bonds for the functional group. When identifying isomers, systematically change the position of the –OH group and the branching of the carbon chain.
清晰绘制有机结构极其重要——务必展示出官能团的所有键。在判断异构体时，要系统地改变 –OH 的位置和碳链的支链情况。
8. Rate of Reaction: Interpreting Graphs | 反应速率：图像解读
The volume of CO₂ produced when marble chips react with excess hydrochloric acid is recorded over time. The graph levels off after 120 seconds. Explain why the rate decreases during the reaction and how the final volume would change if the same mass of powdered marble is used instead.
记录大理石碎片与过量盐酸反应时产生 CO₂ 的体积随时间的变化。图像在 120 秒后趋于水平。解释为何反应过程中速率逐渐降低，并说明如果改用相同质量的大理石粉末，最终气体体积将如何变化。
The rate decreases because the concentration of acid falls as it is consumed. Also, the surface area of marble chips reduces as they dissolve. Both factors decrease collision frequency between reactant particles. When powdered marble is used, the surface area is much larger. The initial rate is faster, but the final volume of CO₂ remains exactly the same because the same mass of marble (same moles of CaCO₃) is used and acid is in excess.
速率降低是因为酸的浓度随着反应消耗而下降。同时，大理石碎块的表面积也随溶解而减小。两因素均降低了反应物粒子间的碰撞频率。若改用大理石粉末，其表面积显著增大。初始速率更快，但最终 CO₂ 体积完全不变，因为所用大理石质量相同（CaCO₃ 物质的量相同），且酸是过量的。
Rate-interpretation questions often combine the effect of concentration change during a reaction with particle size. The final yield depends on the limiting reactant, not on the particle size — a key point that separates surface area from total amount of product.
关于速率图像的题目常结合反应过程中浓度变化的影响与颗粒大小。最终产率取决于限制试剂，而非颗粒大小——这是区分表面积与产物总量的关键点。
9. Periodicity and Group Trends | 周期律与族趋势
Sodium, magnesium and aluminium all react with dilute hydrochloric acid. Describe the trend in reactivity across Period 3 and explain it in terms of atomic structure. Write balanced equations for the reactions of sodium and aluminium with dilute HCl.
钠、镁和铝都能与稀盐酸反应。描述第三周期反应性的变化趋势，并从原子结构角度加以解释。写出钠和铝与稀盐酸反应的配平方程式。
Reactivity decreases from sodium to aluminium for metals. Sodium reacts violently; magnesium reacts steadily; aluminium reacts slowly. Trend occurs because as we move across the period, nuclear charge increases (more protons) but shielding remains similar. Atomic radius decreases, and the outer electrons are held more tightly. More energy is required to remove the outer electrons, so metals become less reactive. Equations: 2Na + 2HCl → 2NaCl + H₂; 2Al + 6HCl → 2AlCl₃ + 3H₂.
对于金属，从钠到铝反应性降低。钠反应剧烈；镁反应平稳；铝反应缓慢。这一趋势的原因为：随着同周期从左到右，核电荷增加（质子数增加），但屏蔽效应相近。原子半径减小，外层电子被束缚得更紧。移去外层电子需更多能量，因此金属反应性减弱。方程式：2Na + 2HCl → 2NaCl + H₂；2Al + 6HCl → 2AlCl₃ + 3H₂。
Periodic trends are a staple of CIE Paper 2 and 4. When explaining reactivity, always link to nuclear charge, shielding, and the energy needed to lose electrons. Don’t forget to balance equations according to the valency of the metal.
元素周期律是 CIE 试卷 2 和 4 的必考点。解释反应性时，一定要联系核电荷、屏蔽效应和失去电子所需的能量。不要忘记根据金属的化合价来配平方程式。
10. Redox and Oxidation States | 氧化还原与氧化数
Identify the substance oxidised and the substance reduced in the reaction: CuO + H₂ → Cu + H₂O. Explain your answer in terms of oxidation states.
指明反应 CuO + H₂ → Cu + H₂O 中被氧化和被还原的物质，并从氧化数角度加以解释。
Assign oxidation states: Cu in CuO = +2, O in CuO = –2; H in H₂ = 0; Cu in Cu = 0; H in H₂O = +1, O in H₂O = –2. Cu changes from +2 to 0: gain of electrons → CuO is reduced. H changes from 0 to +1: loss of electrons → H₂ is oxidised. So CuO is the oxidising agent and H₂ is the reducing agent.
标出氧化数：CuO 中 Cu = +2，O = –2；H₂ 中 H = 0；Cu 中 Cu = 0；H₂O 中 H = +1，O = –2。Cu 从 +2 变为 0：得到电子 → CuO 被还原。H 从 0 变为 +1：失去电子 → H₂ 被氧化。因此 CuO 是氧化剂，H₂ 是还原剂。
Redox definitions in IGCSE still use ‘OIL RIG’ (Oxidation Is Loss, Reduction Is Gain of electrons), but oxidation states offer a consistent way to track electron transfer. When asked to identify the oxidising agent, always name the species that is itself reduced.
IGCSE 阶段氧化还原的定义仍使用 “OIL RIG”（氧化即失电子、还原即得电子），但氧化数提供了一种一致的方式来追踪电子转移。当被问到氧化剂时，一定要指出自身被还原的那种物质。
11. Preparing Soluble Salts by Titration and Filtration | 用滴定和过滤制备可溶性盐
Describe how a pure, dry sample of copper(II) sulfate-5-water crystals can be prepared starting from copper(II) oxide and dilute sulfuric acid. Include key apparatus and steps.
描述如何从氧化铜和稀硫酸出发，制备纯净干燥的硫酸铜-5-水晶体 (CuSO₄·5H₂O)。包括关键仪器和步骤。
Step 1: Warm dilute sulfuric acid in a beaker. Step 2: Add excess copper(II) oxide slowly while stirring until no more dissolves (unreacted solid remains). Step 3: Filter the mixture to remove excess CuO; the filtrate is copper(II) sulfate solution. Step 4: Heat the filtrate gently in an evaporating dish to concentrate it until crystallisation point (a drop forms crystals on a glass rod). Step 5: Leave to cool and crystallise. Step 6: Filter the crystals, wash with a little cold distilled water, and dry between filter papers.
步骤一：在烧杯中预热稀硫酸。步骤二：边搅拌边缓慢加入过量氧化铜，直至不再溶解（有未反应固体剩余）。步骤三：过滤混合物以除去过量 CuO；滤液为硫酸铜溶液。步骤四：将滤液在蒸发皿中缓缓加热浓缩，直至达到结晶点（用玻棒蘸取液滴可析出晶体）。步骤五：冷却结晶。步骤六：过滤晶体，用少量冷蒸馏水洗涤，在滤纸间压干。
This procedure uses filtration to remove excess insoluble base, followed by crystallisation. Questions often ask why excess CuO is used (to ensure all acid is neutralised) and why we do not simply evaporate to dryness (to obtain hydrated crystals, not anhydrous powder).
该过程通过过滤除去过量不溶性碱，然后结晶。考题常问为何使用过量 CuO（确保酸被完全中和）以及为何不直接蒸干（为获得水合晶体，而非无水粉末）。
12. Gas Tests and Identification | 气体检验与鉴定
A colourless gas is produced when a carbonate reacts with dilute acid. The gas gives a white precipitate with limewater. Name the gas and write the ionic equation for its reaction with limewater. Also state what would be observed if the gas is bubbled through bromine water.
某碳酸盐与稀酸反应产生一种无色气体。该气体使石灰水产生白色沉淀。说出该气体的名称，并写出其与石灰水反应的离子方程式。同时说明若将该气体通入溴水中会观察到什么。
The gas is carbon dioxide, CO₂. With limewater: Ca(OH)₂ + CO₂ → CaCO₃ + H₂O, ionic: Ca²⁺ + 2OH⁻ + CO₂ → CaCO₃↓ + H₂O. When CO₂ is bubbled through bromine water (which is orange), no reaction occurs and the colour remains unchanged because CO₂ is not an unsaturated hydrocarbon; it does not decolourise bromine water. This distinguishes CO₂ from alkenes like ethene which do decolourise bromine water.
该气体是二氧化碳 (CO₂)。与石灰水反应：Ca(OH)₂ + CO₂ → CaCO₃ + H₂O，离子方程式：Ca²⁺ + 2OH⁻ + CO₂ → CaCO₃↓ + H₂O。将 CO₂ 通入橙色的溴水中，不发生反应，颜色不变，因为 CO₂ 不是不饱和烃，它不会使溴水褪色。这一点可将 CO₂ 与乙烯等烯烃区分开来，后者能使溴水褪色。
Combining gas tests with explanations of reactivity is common in CIE. Students should know that only unsaturated organic compounds and a few reducing gases decolourise bromine water; CO₂ does not.
将气体检验与反应性解释相结合是 CIE 的常见考法。学生应知道只有不饱和有机物和少数还原性气体能使溴水褪色；CO₂ 不能。
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