📚 IGCSE Chemistry: Worked Examples Explained | IGCSE 化学：典型例题详解
This article presents a collection of carefully worked examples covering the most common question types in IGCSE Chemistry. Each problem is broken down step by step to reinforce the core concepts and methods you need for exam success.
本文精选了 IGCSE 化学中最常见的典型例题，并逐步拆解分析，帮助你巩固核心概念与解题方法，从容应对考试。
1. Balancing Chemical Equations | 化学方程式的配平
Balancing equations is a fundamental skill. Always count the atoms of each element on both sides and adjust coefficients never change subscripts.
配平方程式是一项基本技能。必须始终保持反应前后的原子种类与数目相等，只能调整系数，不能改变化学式中的下标。
Example: Balance the equation Fe + Cl₂ → FeCl₃
例题：配平方程式 Fe + Cl₂ → FeCl₃
Step 1: List the number of atoms. Reactants: 1 Fe, 2 Cl. Products: 1 Fe, 3 Cl.
步骤1：列出原子个数。反应物：1个 Fe，2个 Cl。生成物：1个 Fe，3个 Cl。
Step 2: Balance chlorine first. To get 3 Cl on the left, we need 1.5 Cl₂ molecules, but we prefer whole numbers. Multiply the entire equation by 2 to clear fractions. This gives 2 Fe + 3 Cl₂ → 2 FeCl₃. Iron is now balanced (2 on each side).
步骤2：先配平氯原子。左侧需要 1.5 个 Cl₂ 分子才能得到 3 个 Cl，但系数最好为整数。将整个方程式乘以 2，得到 2 Fe + 3 Cl₂ → 2 FeCl₃。此时铁原子也配平（两侧各 2 个）。
Step 3: Check the final count: 2 Fe, 6 Cl on both sides. The equation is balanced.
步骤3：最终检查原子个数：两侧均有 2 个 Fe 和 6 个 Cl。方程式已配平。
Balanced equation: 2 Fe + 3 Cl₂ → 2 FeCl₃
配平后的方程式：2 Fe + 3 Cl₂ → 2 FeCl₃
2. Mole Calculations | 摩尔计算
The mole links mass to number of particles. The key equation is n = m / M where n is number of moles, m is mass in grams, and M is molar mass in g mol⁻¹.
摩尔把质量与微粒数联系起来。关键公式是 n = m / M，n 为物质的量（mol），m 为质量（g），M 为摩尔质量（g mol⁻¹）。
Example: Calculate the number of moles and the number of formula units in 10.0 g of calcium carbonate, CaCO₃. (Aᵣ: Ca = 40, C = 12, O = 16; Avogadro constant = 6.02 × 10²³)
例题：计算 10.0 g 碳酸钙 CaCO₃ 的物质的量及所含离子单元数目。（相对原子质量：Ca=40，C=12，O=16；阿伏伽德罗常数 = 6.02 × 10²³）
Step 1: Calculate the molar mass M of CaCO₃: 40 + 12 + (3 × 16) = 100 g mol⁻¹.
步骤1：计算 CaCO₃ 的摩尔质量 M：40 + 12 + (3 × 16) = 100 g mol⁻¹。
Number of moles = mass ÷ molar mass n = 10.0 g ÷ 100 g mol⁻¹ = 0.100 mol
Step 2: Find the number of formula units. Number of particles = n × Avogadro constant.
步骤2：求离子单元数目。微粒数 = n × 阿伏伽德罗常数。
Formula units = 0.100 × 6.02 × 10²³ = 6.02 × 10²²
Answer: 0.100 mol of CaCO₃ contains 6.02 × 10²² formula units.
答案：0.100 mol 的 CaCO₃ 含有 6.02 × 10²² 个 CaCO₃ 单元。
3. Empirical and Molecular Formula | 经验式与分子式
Empirical formula shows the simplest whole-number ratio of atoms in a compound. Molecular formula is a multiple of the empirical formula.
经验式表示化合物中各原子最简整数比，分子式则是经验式的整数倍。
Example: A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its molar mass is 56 g mol⁻¹. Determine the empirical and molecular formulae. (Aᵣ: C = 12, H = 1)
例题：某碳氢化合物含碳 85.7%、氢 14.3%（质量分数），其摩尔质量为 56 g mol⁻¹。求经验式和分子式。（相对原子质量：C=12，H=1）
Step 1: Assume 100 g of compound. Mass of C = 85.7 g, mass of H = 14.3 g.
步骤1：假设样品 100 g，则碳质量 85.7 g，氢质量 14.3 g。
Step 2: Convert to moles. n(C) = 85.7 ÷ 12 = 7.14 mol; n(H) = 14.3 ÷ 1 = 14.3 mol.
步骤2：换算为物质的量。n(C) = 85.7 ÷ 12 ≈ 7.14 mol；n(H) = 14.3 ÷ 1 = 14.3 mol。
Step 3: Divide by the smallest number of moles: C = 7.14 ÷ 7.14 = 1; H = 14.3 ÷ 7.14 = 2.00. The empirical formula is CH₂.
步骤3：除以最小摩尔数。C = 7.14 ÷ 7.14 = 1；H = 14.3 ÷ 7.14 ≈ 2.00。经验式为 CH₂。
Step 4: Find the molecular formula. Empirical formula mass = 12 + (2 × 1) = 14 g mol⁻¹. Factor = molar mass ÷ empirical mass = 56 ÷ 14 = 4.
步骤4：求分子式。经验式质量 = 12 + (2×1) = 14 g mol⁻¹。倍数 = 摩尔质量 ÷ 经验式质量 = 56 ÷ 14 = 4。
Step 5: Molecular formula = (CH₂)₄ = C₄H₈.
步骤5：分子式 = (CH₂)₄ = C₄H₈。
Answer: Empirical formula CH₂, molecular formula C₄H₈.
答案：经验式 CH₂，分子式 C₄H₈。
4. Concentration and Titration | 浓度与滴定
Titration calculations use the relationship n = c × V, where c is concentration in mol dm⁻³ and V is volume in dm³. Use the balanced equation to find reacting ratios.
滴定计算可用公式 n = c × V，c 为浓度（mol dm⁻³），V 为体积（dm³）。利用配平的方程式确定反应物比例。
Example: 25.0 cm³ of sodium hydroxide solution was neutralised by 20.0 cm³ of 0.100 mol dm⁻³ hydrochloric acid. Calculate the concentration of the NaOH solution.
例题：25.0 cm³ 的氢氧化钠溶液恰好被 20.0 cm³ 的 0.100 mol dm⁻³ 盐酸中和。求 NaOH 溶液的浓度。
Step 1: Write the balanced equation: NaOH + HCl → NaCl + H₂O. Mole ratio NaOH : HCl = 1 : 1.
步骤1：写出配平方程式：NaOH + HCl → NaCl + H₂O。物质的量之比 NaOH : HCl = 1 : 1。
Step 2: Calculate moles of HCl used. Convert volume to dm³: 20.0 cm³ = 0.0200 dm³. n(HCl) = 0.100 × 0.0200 = 0.00200 mol.
步骤2：计算 HCl 的物质的量。体积换算：20.0 cm³ = 0.0200 dm³。n(HCl) = 0.100 × 0.0200 = 0.00200 mol。
Step 3: Because the ratio is 1:1, n(NaOH) = 0.00200 mol. Volume of NaOH = 25.0 cm³ = 0.0250 dm³.
步骤3：由于 1:1，n(NaOH) = 0.00200 mol。NaOH 体积 = 25.0 cm³ = 0.0250 dm³。
c(NaOH) = n ÷ V = 0.00200 mol ÷ 0.0250 dm³ = 0.0800 mol dm⁻³
Answer: The concentration of NaOH is 0.0800 mol dm⁻³.
答案：NaOH 溶液的浓度为 0.0800 mol dm⁻³。
5. Electrolysis and Faraday’s Laws | 电解与法拉第定律
In electrolysis, the amount of substance produced is proportional to the quantity of electric charge. Use Q = I × t and then n(e⁻) = Q / F, where F = 96 500 C mol⁻¹.
电解中产物的量与通过的电量成正比。先用 Q = I × t 计算电量，再用 n(e⁻) = Q / F，F 为法拉第常数 96 500 C mol⁻¹。
Example: A current of 10.0 A is passed through molten aluminium oxide for 2.00 hours. Calculate the mass of aluminium produced. (Aᵣ: Al = 27; 1 F = 96 500 C mol⁻¹)
例题：用 10.0 A 的电流电解熔融氧化铝 2.00 小时。计算产生的铝的质量。（相对原子质量：Al=27；1 F = 96 500 C mol⁻¹）
Step 1: Calculate the total charge. Time = 2.00 × 3600 = 7200 s. Q = 10.0 A × 7200 s = 72 000 C.
步骤1：计算总电量。时间 = 2.00 × 3600 = 7200 s。Q = 10.0 A × 7200 s = 72 000 C。
Step 2: Calculate moles of electrons: n(e⁻) = Q / F = 72 000 ÷ 96 500 ≈ 0.746 mol.
步骤2：计算电子的物质的量：n(e⁻) = 72 000 ÷ 96 500 ≈ 0.746 mol。
Step 3: The cathode half-equation is Al³⁺ + 3e⁻ → Al. So 3 moles of electrons produce 1 mole of Al. n(Al) = 0.746 ÷ 3 = 0.2487 mol.
步骤3：阴极半反应为 Al³⁺ + 3e⁻ → Al，因此 3 mol 电子生成 1 mol Al。n(Al) = 0.746 ÷ 3 ≈ 0.2487 mol。
Step 4: Mass of Al = n(Al) × 27 = 0.2487 × 27 ≈ 6.71 g.
步骤4：铝的质量 = 0.2487 × 27 ≈ 6.71 g。
Answer: Approximately 6.71 g of aluminium is produced.
答案：约产生 6.71 g 铝。
6. Energy Changes and Bond Enthalpy | 能量变化与键焓
Enthalpy change of a reaction can be estimated using average bond energies: ΔH = Σ(bonds broken) – Σ(bonds formed).
可用平均键能估算反应焓变：ΔH = Σ(断裂键能) – Σ(形成键能)。
Example: Hydrogen reacts with chlorine: H₂ + Cl₂ → 2HCl. Calculate ΔH using the following bond energies (in kJ mol⁻¹): H–H 436, Cl–Cl 243, H–Cl 432.
例题：氢气与氯气反应：H₂ + Cl₂ → 2HCl。利用以下键能（kJ mol⁻¹）计算 ΔH：H–H 436，Cl–Cl 243，H–Cl 432。
Step 1: Identify bonds broken: 1 × H–H and 1 × Cl–Cl. Energy required = 436 + 243 = 679 kJ.
步骤1：确定断裂的键：1 个 H–H 和 1 个 Cl–Cl。所需能量 = 436 + 243 = 679 kJ。
Step 2: Identify bonds formed: 2 × H–Cl. Energy released = 2 × 432 = 864 kJ.
步骤2：确定生成的键：2 个 H–Cl。释放能量 = 2 × 432 = 864 kJ。
Step 3: ΔH = energy in – energy out = 679 – 864 = –185 kJ per mole of reaction as written.
步骤3：ΔH = 吸收能量 – 释放能量 = 679 – 864 = –185 kJ（按所写方程式反应一次）。
Answer: The reaction is exothermic, ΔH = –185 kJ mol⁻¹.
答案：反应放热，ΔH = –185 kJ mol⁻¹。
7. Rate of Reaction and Graphs | 反应速率与图表
The rate of a reaction can be followed by measuring the volume of gas evolved. The instantaneous rate at a given time is found by drawing a tangent to the curve.
可通过测量放出气体体积追踪反应速率。某一时刻的瞬时速率由曲线上该点切线的斜率求得。
Example: The table shows the volume of CO₂ collected when calcium carbonate reacts with excess acid.
例题：下表给出了碳酸钙与过量酸反应时收集的 CO₂ 体积数据。
| Time / s | Volume of CO₂ / cm³ |
| 0 | 0 |
| 20 | 15 |
| 40 | 26 |
| 60 | 34 |
| 80 | 39 |
| 100 | 43 |
Estimate the rate of reaction at t = 30 s.
估算 t = 30 s 时的反应速率。
Step 1: Plot the curve of volume against time (a typical reaction curve). At 30 s, draw a tangent line that touches the curve only at that point.
步骤1：绘制体积-时间曲线（典型的反应曲线）。在 t=30 s 处画一条切线，仅在该点接触曲线。
Step 2: Take two convenient points on the tangent, for example (10 s, 8 cm³) and (50 s, 30 cm³).
步骤2：在切线上选两个便于读取的点，如 (10 s, 8 cm³) 和 (50 s, 30 cm³)。
Step 3: Calculate the slope, which equals the rate. Change in volume = 30 – 8 = 22 cm³; change in time = 50 – 10 = 40 s.
步骤3：计算斜率即速率。体积变化 = 30 – 8 = 22 cm³，时间变化 = 50 – 10 = 40 s。
Rate = 22 cm³ ÷ 40 s = 0.55 cm³ s⁻¹
Answer: The instantaneous rate at 30 s is approximately 0.55 cm³ s⁻¹.
答案：30 s 时的瞬时速率约为 0.55 cm³ s⁻¹。
8. Organic Chemistry: Naming and Isomers | 有机化学：命名与同分异构体
Isomers are compounds with the same molecular formula but different structural arrangements. Drawing all possible isomers tests your understanding of bonding and branching.
同分异构体是分子式相同但结构排列不同的化合物。画出所有可能的异构体能检验你对键合和支链的理解。
Example: The molecular formula C₄H₁₀ belongs to two alkanes. Draw their displayed structures and give their systematic IUPAC names.
例题：分子式 C₄H₁₀ 对应两种烷烃。画出它们的结构展示式并给出系统命名。
Step 1: Start with the straight-chain structure: four carbon atoms in a row.
步骤1：先画直链结构：四个碳原子排成一行。
Step 2: The continuous chain gives: CH₃–CH₂–CH₂–CH₃. This is butane.
步骤2：连续链得到 CH₃–CH₂–CH₂–CH₃，命名为丁烷（butane）。
Step 3: Introduce one branch by moving a carbon atom. The longest chain then becomes three carbons, with a methyl group on the second carbon. Structure: CH₃–CH(CH₃)–CH₃. This is methylpropane (common name isobutane). IUPAC name: 2-methylpropane.
步骤3：引入一个支链，将末端碳移到第二个碳上。最长链变为三个碳，第二个碳上连有一个甲基。结构为 CH₃–CH(CH₃)–CH₃，命名为 2-甲基丙烷（2-methylpropane）。
Step 4: Check that both have the formula C₄H₁₀ and no further distinct isomers exist.
步骤4：检查两者分子式均为 C₄H₁₀，且不存在其他不同的异构体。
Answer: The two isomers are butane (CH₃CH₂CH₂CH₃) and 2-methylpropane (CH₃CH(CH₃)CH₃).
答案：两种异构体是丁烷（CH₃CH₂CH₂CH₃）和 2-甲基丙烷（CH₃CH(CH₃)CH₃）。
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