📚 IGCSE WJEC Chemistry: Worked Examples Explained | IGCSE WJEC 化学：典型例题详解
Mastering IGCSE WJEC Chemistry requires not only understanding core concepts but also practising how to apply them in exam-style questions. This article presents ten carefully chosen worked examples that cover key topics: atomic structure, bonding, moles, electrolysis, energetics, rates, equilibrium, acids, and organic chemistry. Each example is broken down into clear steps, with English explanations immediately followed by Chinese translations, helping you strengthen both your subject knowledge and bilingual scientific literacy.
掌握 IGCSE WJEC 化学不仅需要理解核心概念，还需要练习如何在考试题型中运用它们。本文精选了十个典型例题，涵盖原子结构、化学键、摩尔计算、电解、能量变化、反应速率、平衡、酸碱和有机化学等关键主题。每个例题都配以清晰的步骤解析，英文解释后紧跟中文翻译，有助于你同时巩固学科知识与双语科学表达能力。
1. Isotopic Abundance and Relative Atomic Mass | 同位素丰度与相对原子质量
Example: A sample of chlorine contains 75% of chlorine‑35 atoms and 25% of chlorine‑37 atoms. Calculate the relative atomic mass (Aᵣ) of chlorine, giving your answer to one decimal place.
例题：某氯样品含有 75% 的氯‑35 原子和 25% 的氯‑37 原子。计算氯的相对原子质量 (Aᵣ)，结果保留一位小数。
In a real atomic sample, different isotopes exist in specific proportions. The Aᵣ is the weighted average mass of all isotopes compared with 1/12 of the mass of carbon‑12. Convert percentage abundance into decimal fractions first.
在实际原子样品中，不同同位素以特定比例存在。Aᵣ 是所有同位素相对于碳‑12 质量的 1/12 的加权平均质量。首先将丰度百分比转换为小数。
mass number by its fractional abundance, then add the results: (35 × 0.75) + (37 × 0.25).
用每种同位素的质量数乘以其小数丰度，然后将结果相加：(35 × 0.75) + (37 × 0.25)。
he relative atomic mass of chlorine in this sample is 35.5. Notice that this matches the value typically given in the Periodic Table.
计算：26.25 + 9.25 = 35.5。因此该样品中氯的相对原子质量为 35.5。注意该值与元素周期表中通常给出的数值一致。
2. Ionic Bonding and Properties | 离子键与性质
Example: Magnesium oxide (MgO) has a very high melting point of about 2850 °C. Explain, in terms of bonding and structure, why magnesium oxide has such a high melting point.
例题：氧化镁 (MgO) 的熔点非常高，约 2850 °C。请从化学键和结构的角度解释氧化镁为何具有如此高的熔点。
Magnesium oxide is an ionic compound. It consists of a giant lattice of Mg²⁺ and O²⁻ ions held together by strong electrostatic forces – ionic bonds – between the oppositely charged ions.
氧化镁是一种离子化合物。它由 Mg²⁺ 和 O²⁻ 离子构成的巨型晶格组成，相反电荷离子之间存在着强烈的静电引力，即离子键。
Melting requires overcoming these forces to allow ions to move freely. Because both ions carry double charges (Mg²⁺ and O²⁻), the ionic bonds are exceptionally strong compared with singly charged ions like Na⁺ and Cl⁻.
熔化需要克服这些作用力，使离子能够自由移动。由于两种离子都带双电荷 (Mg²⁺ 和 O²⁻)，相比 Na⁺ 和 Cl⁻ 等单电荷离子，离子键异常牢固。
A large amount of thermal energy is needed to break the many strong ionic bonds throughout the giant lattice, hence magnesium oxide has a very high melting point.
需要大量热能来破坏整个巨型晶格中大量牢固的离子键，因此氧化镁的熔点非常高。
3. Mole and Mass Calculations | 摩尔与质量计算
Example: methane (CH₄) burns completely in excess oxygen. (Aᵣ values: H = 1.0, C = 12.0, O = 16.0)
例题：计算 16 g 甲烷 (CH₄) 在过量氧气中完全燃烧时生成的二氧化碳 (CO₂) 的质量。(Aᵣ 值：H = 1.0，C = 12.0，O = 16.0)
he balanced equation. CH₄ + 2O₂ → CO₂ + 2H₂O. The mole ratio between CH₄ and CO₂ is 1 : 1.
步骤 1：写出配平的化学方程式。CH₄ + 2O₂ → CO₂ + 2H₂O。CH₄ 与 CO₂ 的摩尔比为 1 : 1。
Step 2: Calculate the molar mass of methane. Mᵣ(CH₄) = 12.0 + (4 × 1.0) = 16.0 g/mol.
步骤 2：计算甲烷的摩尔质量。Mᵣ(CH₄) = 12.0 + (4 × 1.0) = 16.0 g/mol。
number of moles of methane used. Moles = mass / molar mass = 16 g / 16.0 g/mol = 1.0 mol.
步骤 3：计算所用甲烷的物质的量。摩尔数 = 质量 / 摩尔质量 = 16 g / 16.0 g/mol = 1.0 mol。
ol. Mᵣ(CO₂) = 12.0 + (2 × 16.0) = 44.0 g/mol. Mass of CO₂ = 1.0 mol × 44.0 g/mol = 44.0 g.
步骤 4：利用 1:1 比例：生成的 CO₂ 的物质的量 = 1.0 mol。Mᵣ(CO₂) = 12.0 + (2 × 16.0) = 44.0 g/mol。CO₂ 的质量 = 1.0 mol × 44.0 g/mol = 44.0 g。
4. Determining Empirical Formula | 确定经验式
Example: A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Determine its empirical formula. (Aᵣ: C = 12.0, H = 1.0, O = 16.0)
例题：某化合物含有 40.0% 碳、6.7% 氢和 53.3% 氧（质量分数）。确定其经验式。(Aᵣ: C = 12.0，H = 1.0，O = 16.0)
Assume a 100 g sample so that the percentages convert directly into masses: 40.0 g C, 6.7 g H, 53.3 g O. Then convert each mass to moles.
假设样品为 100 g，于是百分数直接转换为质量：40.0 g C、6.7 g H、53.3 g O。然后将各质量换算为物质的量。
ol; Moles of H = 6.7 g ÷ 1.0 g/mol = 6.7 mol; Moles of O = 53.3 g ÷ 16.0 g/mol = 3.33 mol.
C 的物质的量 = 40.0 g ÷ 12.0 g/mol = 3.33 mol；H = 6.7 g ÷ 1.0 g/mol = 6.7 mol；O = 53.3 g ÷ 16.0 g/mol = 3.33 mol。
e smallest number of moles (3.33): C = 3.33/3.33 = 1, H = 6.7/3.33 ≈ 2, O = 3.33/3.33 = 1.
将各物质的量除以最小物质的量 3.33：C = 1，H ≈ 2，O = 1。
The simplest whole number ratio gives the empirical formula CH₂O.
最简整数比给出的经验式为 CH₂O。
5. Electrolysis of Aqueous Solutions | 水溶液的电解
Example: Predict the products formed at the anode and cathode during the electrolysis of concentrated aqueous sodium chloride (brine) using inert electrodes. Include relevant half‑equations.
例题：用惰性电极电解浓氯化钠水溶液（盐水），预测阳极和阴极的产物，并写出相关的半反应方程式。
ed to hydrogen gas rather than Na⁺ to sodium because Na⁺ is harder to reduce. Cathode half‑equation: 2H⁺ + 2e⁻ → H₂.
在浓盐水中，存在的离子有 Na⁺、Cl⁻、H⁺、OH⁻。在阴极，发生还原反应。借助金属活动性顺序：H⁺ 被还原为氢气，而不是 Na⁺ 被还原为钠，因为 Na⁺ 更难还原。阴极半反应：2H⁺ + 2e⁻ → H₂。
n. In concentrated NaCl, Cl⁻ is discharged preferentially to give chlorine gas. Anode half‑equation: 2Cl⁻ → Cl₂ + 2e⁻.
在阳极，Cl⁻ 和 OH⁻ 争夺氧化。在浓 NaCl 溶液中，Cl⁻ 优先放电生成氯气。阳极半反应：2Cl⁻ → Cl₂ + 2e⁻。
The solution left around the cathode becomes alkaline due to the production of NaOH as Na⁺ and OH⁻ remain. Overall products: hydrogen at the cathode, chlorine at the anode, and sodium hydroxide in solution.
由于留下 Na⁺ 和 OH⁻，阴极附近溶液呈碱性，生成 NaOH。总产物：阴极产生氢气，阳极产生氯气，溶液中得到氢氧化钠。
6. Enthalpy Change Calculation | 焓变计算
Example: e the enthalpy change of neutralisation (ΔH). (Specific heat capacity of solution = 4.2 J/g·°C, density = 1.0 g/cm³)
例题：在量热实验中，将 50 cm³ 1.0 mol/dm³ HCl 与 50 cm³ 1.0 mol/dm³ NaOH 混合。温度从 20.0 °C 升高至 26.8 °C。计算中和焓变 (ΔH)。(溶液的比热容 = 4.2 J/g·°C，密度 = 1.0 g/cm³)
Total volume of solution = 100 cm³, so its mass m = 100 g. Temperature change ΔT = 26.8 – 20.0 = 6.8 °C. Heat energy released Q = m × c × ΔT.
溶液总体积 = 100 cm³，因此其质量 m = 100 g。温度变化 ΔT = 6.8 °C。释放的热量 Q = m × c × ΔT。
g species mix. Moles of HCl used = (50/1000) dm³ × 1.0 mol/dm³ = 0.050 mol. Same for NaOH.
Q = 100 g × 4.2 J/g·°C × 6.8 °C = 2856 J。这是反应物混合时释放的能量。所用 HCl 的物质的量 = (50/1000) dm³ × 1.0 mol/dm³ = 0.050 mol。NaOH 相同。
The enthalpy change of neutralisation is for the formation of 1 mole of water: ΔH = –Q / moles of water = –2856 J / 0.050 mol = –57120 J/mol, or –57.1 kJ/mol (to 3 s.f.). The negative sign shows the reaction is exothermic.
中和焓变对应生成 1 mol 水：ΔH = –Q / 水的物质的量 = –2856 J / 0.050 mol = –57120 J/mol，即 –57.1 kJ/mol（三位有效数字）。负号表示反应放热。
7. Rate of Reaction and Collision Theory | 反应速率与碰撞理论
Example: Explain, using collision theory, why increasing the surface area of a solid reactant increases the rate of reaction.
例题：用碰撞理论解释为什么增大固体反应物的表面积会提高反应速率。
For a reaction to occur, reactant particles must collide with sufficient energy (activation energy) and with the correct orientation. Only a fraction of collisions are successful.
发生反应需要反应物粒子发生碰撞，且碰撞能量不低于活化能，同时碰撞取向合适。只有一部分碰撞是有效碰撞。
In a solid reactant, only particles at the surface are exposed to collisions with the other reactant species. Breaking a large lump into smaller pieces increases the total surface area, exposing many more particles at the solid–fluid interface.
固体反应物中，只有表面的粒子暴露出来与其他反应物粒子碰撞。把大块破碎成小块会增加总表面积，在固–液/固–气界面上暴露出更多的粒子。
A greater surface area leads to a higher frequency of collisions between reactant particles per unit time. More frequent collisions mean more successful collisions per second, so the rate of reaction increases. Energy distribution and activation energy remain unchanged; the effect is purely on collision frequency.
更大的表面积导致单位时间内反应物粒子之间的碰撞频率更高。更频繁的碰撞意味着每秒发生更多的有效碰撞，因此反应速率提高。能量分布和活化能不变；这一效应完全源于碰撞频率的增加。
8. Reversible Reactions and Le Chatelier's Principle | 可逆反应与勒夏特列原理
Example: The Haber process is exothermic in the forward direction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = – Predict and explain the effect of increasing pressure on the equilibrium yield of ammonia.
例题：哈伯法正向放热：N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ/mol。预测并解释增大压强对氨的平衡产率的影响。
Le Chatelier's principle states that if a system at equilibrium is subjected to a change in pressure, temperature or concentration, the equilibrium position shifts to oppose the change.
勒夏特列原理指出，如果处于平衡状态的体系受到压强、温度或浓度的改变，平衡位置会向削弱该改变的方向移动。
oles of gas; products have 2 moles of gas. Increasing pressure favours the side with fewer gas molecules, because that reduces the total pressure.
数一下反应式中各边的气体分子数：反应物有 1 + 3 = 4 mol 气体；产物有 2 mol 气体。增大压强有利于气体分子数较少的一侧，因为这会降低总压强。
Therefore, the equilibrium shifts to the right (toward NH₃) to decrease the number of gaseous molecules, increasing the yield of ammonia. This is why the industrial process uses high pressure (around 200 atm).
因此，平衡向右（生成 NH₃ 的方向）移动以减少气体分子总数，从而提高氨的产率。这就是工业上使用高压（约 200 atm）的原因。
9. Acid–Base Titration Calculation | 酸碱滴定计算
Example: 25.0 cm³ of sulfuric acid (H₂SO₄) of unknown concentration is exactly neutralised by 30.0 cm³ of 0.100 mol/dm³ sodium hydroxide (NaOH). Calculate the concentration of the acid in mol/dm³.
例题：25.0 cm³ 未知浓度的硫酸 (H₂SO₄) 恰好被 30.0 cm³ 0.100 mol/dm³ 氢氧化钠 (NaOH) 中和。计算该酸的浓度（单位：mol/dm³）。
the balanced equation: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. The mole ratio H₂SO₄ : NaOH = 1 : 2.
写出配平的化学方程式：H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O。H₂SO₄ 与 NaOH 的物质的量之比为 1 : 2。
s of NaOH used = concentration × volume = 0.100 mol/dm³ × (30.0 / 1000) dm³ = 0.00300 mol.
所用 NaOH 的物质的量 = 浓度 × 体积 = 0.100 mol/dm³ × (30.0 / 1000) dm³ = 0.00300 mol。
:2 ratio, moles of H₂SO₄ that reacted = moles of NaOH ÷ 2 = 0.00300 mol ÷ 2 = 0.00150 mol.
由 1:2 的比例，反应的 H₂SO₄ 的物质的量 = NaOH 的物质的量 ÷ 2 = 0.00150 mol。
000) dm³ = 0.0600 mol/dm³. Always remember to convert cm³ to dm³ when using mol/dm³ units.
H₂SO₄ 的浓度 = 物质的量 / 体积 = 0.00150 mol / (25.0 / 1000) dm³ = 0.0600 mol/dm³。当使用 mol/dm³ 单位时，务必记得将 cm³ 转换为 dm³。
10. Organic Chemistry: Alkenes and Addition Reactions | 有机化学：烯烃与加成反应
Example: Ethene (C₂H₄) reacts with bromine (Br₂) at room temperature. Write the displayed formula for the product and explain why this reaction is described as an addition reaction.
例题：乙烯 (C₂H₄) 在室温下与溴 (Br₂) 反应。写出产物的完整结构式，并解释为何该反应被称为加成反应。
The carbon–carbon double bond in ethene contains a sigma bond and a pi bond; the pi bond is relatively weak and can break open to allow atoms to add across the double bond. No atoms are lost.
乙烯中的碳碳双键包含一个 σ 键和一个 π 键；π 键相对较弱，能够断裂打开，使原子加合到双键两侧。反应中没有原子的丢失。
When ethene is mixed with bromine, the reddish‑wn colour of bromine disappears. The reaction is: C₂H₄ + Br₂ → C₂H₄Br₂. The product is 1,2‑ la shows each carbon atom bonded to two hydrogen atoms and a bromine atom, with a single C–C bond.
乙烯与溴混合时，溴的红棕色会褪去。反应为 C₂H₄ + Br₂ → C₂H₄Br₂。产物为 1,2‑二溴乙烷，其完整结构式中每个碳原子连有两个氢原子和一个溴原子，碳碳之间为单键。
This is an addition reaction because two reactant molecules combine to form a single larger product, with the double bond becoming a single bond. No small molecule is eliminated. Unsaturated alkenes become saturated compounds.
这是一种加成反应，因为两个反应物分子结合生成单一更大的产物，双键转变为单键。没有小分子被消除。不饱和的烯烃转变为饱和化合物。
Published by TutorHao | Chemistry Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图