📚 GCSE WJEC Chemistry: Calculation Bootcamp | GCSE WJEC 化学：计算题专项训练
Calculations are the backbone of GCSE WJEC Chemistry, turning qualitative ideas into quantitative predictions. This revision guide walks you through every major calculation type, from relative formula mass to titration results, equipping you with step-by-step methods and examiner tips. Work through the examples, then test yourself with the practice questions provided.
计算题是 GCSE WJEC 化学的支柱，它将定性概念转化为定量预测。这份复习指南将带你逐一攻克每一类主要计算题型——从相对式量到滴定结果，为你提供分步方法和考官建议。请仔细研读例题，然后用配套练习题自我检测。
1. Relative Atomic Mass and Relative Formula Mass | 相对原子质量与相对式量
Relative atomic mass (Aᵣ) is the average mass of an atom of an element compared to 1/12th the mass of a carbon‑12 atom. It has no units. On the WJEC data sheet, you will find Aᵣ values for all elements. Relative formula mass (Mᵣ) applies to compounds and is the sum of the Aᵣ values of all atoms in the formula.
相对原子质量（Aᵣ）是一个元素的原子平均质量，以碳‑12 原子质量的 1/12 作为比较标准。它没有单位。在 WJEC 数据表中，你可以查到所有元素的 Aᵣ 值。相对式量（Mᵣ）适用于化合物，是化学式中所有原子 Aᵣ 值的总和。
Example: Calculate the Mᵣ of ammonium sulfate, (NH₄)₂SO₄.
Aᵣ: N = 14, H = 1, S = 32, O = 16.
Mᵣ = 2 × [14 + (4 × 1)] + 32 + (4 × 16) = 2 × 18 + 32 + 64 = 36 + 32 + 64 = 132.
示例：计算硫酸铵 (NH₄)₂SO₄ 的 Mᵣ。
Aᵣ：N = 14，H = 1，S = 32，O = 16。
Mᵣ = 2 × [14 + (4×1)] + 32 + (4×16) = 2×18 + 32 + 64 = 36+32+64 = 132。
2. The Mole Concept | 摩尔概念
One mole of any substance contains 6.02 × 10²³ particles (Avogadro’s number). The mass of one mole of a substance is its Mᵣ in grams. This links the microscopic world to the laboratory balance. Number of moles = mass (g) ÷ Mᵣ (g/mol). Memorise this equation and always show units in your working.
1 摩尔任何物质都含有 6.02×10²³ 个微粒（阿伏伽德罗常数）。1 摩尔物质的质量等于其相对式量数值对应的克数。这便将微观世界与实验室天平联系了起来。摩尔数 = 质量(g) ÷ Mᵣ(g/mol)。请牢记这个公式，并始终在计算过程中标明单位。
Example: How many moles are present in 8.0 g of sulfur trioxide, SO₃? Mᵣ = 32 + (3×16) = 80. Moles = 8.0 ÷ 80 = 0.10 mol.
示例：8.0 g 三氧化硫 SO₃ 中含有多少摩尔？Mᵣ = 32 + (3×16) = 80。摩尔数 = 8.0 ÷ 80 = 0.10 mol。
3. Calculating Masses Using Moles | 利用摩尔计算质量
Once you know the number of moles, you can find the mass of any substance: mass = moles × Mᵣ. This is essential for predicting how much product will form or how much reactant is needed. Always begin by writing a balanced equation, then use the mole ratios.
一旦知道摩尔数，你就可以求出任意物质的质量：质量 = 摩尔数 × Mᵣ。这对于预测能生成多少产物或需要多少反应物至关重要。务必先写出配平的化学方程式，然后利用摩尔比进行计算。
Worked example: What mass of magnesium oxide, MgO, is produced when 4.8 g of magnesium burns completely in oxygen?
2Mg + O₂ → 2MgO
Moles of Mg = 4.8 ÷ 24 = 0.20 mol → moles of MgO = 0.20 mol (1:1 ratio) → mass MgO = 0.20 × (24+16) = 0.20 × 40 = 8.0 g.
完整示例：4.8 g 镁在氧气中完全燃烧，会产生多少克氧化镁 MgO？
2Mg + O₂ → 2MgO
Mg 的摩尔数 = 4.8 ÷ 24 = 0.20 mol → MgO 的摩尔数 = 0.20 mol（1:1 比）→ MgO 质量 = 0.20 × (24+16) = 0.20 × 40 = 8.0 g。
4. Empirical and Molecular Formulae | 经验式与分子式
Empirical formula is the simplest whole-number ratio of atoms in a compound. Molecular formula tells you the actual number of atoms of each element in a molecule. To find the empirical formula, divide the mass (or percentage) of each element by its Aᵣ, then find the simplest ratio.
经验式表示化合物中原子个数的最简整数比，分子式则表示一个分子中各元素原子的实际数目。求经验式时，用各元素的质量（或质量分数）除以其 Aᵣ，然后求出最简整数比。
Example: A compound contains 40% carbon, 6.7% hydrogen and 53.3% oxygen by mass. What is its empirical formula?
C: 40 ÷ 12 = 3.33; H: 6.7 ÷ 1 = 6.7; O: 53.3 ÷ 16 = 3.33. Divide by smallest (3.33): C = 1, H = 2.01 ≈ 2, O = 1. Empirical formula is CH₂O.
示例：某化合物含碳 40%、氢 6.7%、氧 53.3%（质量分数）。求经验式。
C: 40 ÷ 12 = 3.33；H: 6.7 ÷ 1 = 6.7；O: 53.3 ÷ 16 = 3.33。除以最小值 3.33：C = 1，H ≈ 2，O = 1。经验式为 CH₂O。
The molecular formula can be found if you know the relative molecular mass (Mᵣ) of the compound. Divide the Mᵣ by the empirical formula mass to get the multiplier.
若已知化合物的相对分子质量 (Mᵣ)，便可求得分子式。用 Mᵣ 除以经验式质量，得到倍数因子。
5. Conservation of Mass and Reacting Masses | 质量守恒与反应质量
In a chemical reaction, atoms are neither created nor destroyed, so total mass is conserved. This principle underpins reacting mass calculations. In a closed system, you can measure mass before and after to verify the law. If mass appears to decrease, a gas has escaped; if it increases, a gas from the air has reacted.
在化学反应中，原子既不能被创造也不能被消灭，因此总质量守恒。这一原理是反应质量计算的基础。在封闭系统中，你可以测量反应前后的质量来验证该定律。如果质量看似减少，说明有气体逸出；若质量增加，说明空气中的气体参与了反应。
Reacting mass check: 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂
Mᵣ of 2NaHCO₃ = 168, products total = 106 + 18 + 44 = 168. Mass is conserved.
反应质量检验：2NaHCO₃ → Na₂CO₃ + H₂O + CO₂
2NaHCO₃ 的 Mᵣ = 168，生成物总 Mᵣ = 106+18+44 = 168。质量守恒。
6. Concentration Calculations | 浓度计算
Concentration is usually measured in g/dm³ or mol/dm³. To convert between the two, use: concentration (mol/dm³) = concentration (g/dm³) ÷ Mᵣ. Remember that 1 dm³ = 1000 cm³, so always convert volumes into dm³ before calculating moles.
浓度通常以 g/dm³ 或 mol/dm³ 表示。两者之间的换算关系为：浓度 (mol/dm³) = 浓度 (g/dm³) ÷ Mᵣ。记住 1 dm³ = 1000 cm³，因此在计算摩尔数前，务必先将体积换算为 dm³。
Example: 5.85 g of NaCl is dissolved in 250 cm³ of water. What is the concentration in mol/dm³? Mᵣ NaCl = 58.5. Moles = 5.85 ÷ 58.5 = 0.100 mol. Volume = 250 ÷ 1000 = 0.250 dm³. Concentration = 0.100 ÷ 0.250 = 0.40 mol/dm³.
示例：将 5.85 g NaCl 溶于 250 cm³ 水中。求物质的量浓度（mol/dm³）。Mᵣ NaCl = 58.5。摩尔数 = 5.85 ÷ 58.5 = 0.100 mol。体积 = 250 ÷ 1000 = 0.250 dm³。浓度 = 0.100 ÷ 0.250 = 0.40 mol/dm³。
7. Gas Volume Calculations | 气体体积计算
At room temperature and pressure (RTP), one mole of any gas occupies 24 dm³. Use this molar gas volume to link moles and volume. Volume (dm³) = moles × 24. This only applies at RTP (around 20 °C and 1 atm), which is specified in WJEC questions.
在室温和常压（RTP）下，1 摩尔任何气体的体积为 24 dm³。利用此摩尔气体体积可将摩尔数与体积联系起来。体积 (dm³) = 摩尔数 × 24。该关系仅适用于 RTP（约 20 °C，1 atm），这在 WJEC 试题中通常会注明。
Example: What volume of CO₂ is produced when 10.0 g of CaCO₃ reacts with excess acid? CaCO₃ → CaO + CO₂ (or with acid). Mᵣ CaCO₃ = 100. Moles = 10.0 ÷ 100 = 0.100 mol → moles CO₂ = 0.100 mol → volume = 0.100 × 24 = 2.4 dm³ (or 2400 cm³).
示例：10.0 g CaCO₃ 与过量酸反应，会产生多少体积的 CO₂？CaCO₃ → CaO + CO₂（或与酸反应）。Mᵣ CaCO₃ = 100。摩尔数 = 10.0 ÷ 100 = 0.100 mol → CO₂ 摩尔数 = 0.100 mol → 体积 = 0.100 × 24 = 2.4 dm³（即 2400 cm³）。
8. Percentage Yield and Atom Economy | 百分产率与原子经济性
Percentage yield compares the actual mass of product obtained to the theoretical maximum. % yield = (actual yield ÷ theoretical yield) × 100. Low yields can arise from side reactions, incomplete reactions, or product lost during purification.
百分产率将实际获得的产物质量与理论最大产量进行比较。% 产率 = (实际产量 ÷ 理论产量) × 100。产率偏低可能源于副反应、反应不完全或产物在提纯过程中损失。
Atom economy measures how much of the reactants end up in the desired product. Atom economy = (Mᵣ of desired product ÷ total Mᵣ of all reactants) × 100. High atom economy reduces waste and is a key green chemistry principle.
原子经济性衡量有多少反应物最终转化为目标产物。原子经济性 = (目标产物 Mᵣ ÷ 所有反应物总 Mᵣ) × 100。高原子经济性能减少废物，是绿色化学的核心原则。
Example: In the reaction CH₄ + 2O₂ → CO₂ + 2H₂O, if only CO₂ is desired, atom economy = 44 ÷ (16 + 64) × 100 = 55%.
示例：反应 CH₄ + 2O₂ → CO₂ + 2H₂O 中，若只有 CO₂ 为目标产物，原子经济性 = 44 ÷ (16 + 64) × 100 = 55%。
9. Titration Calculations | 滴定计算
Titrations are a classic practical in WJEC GCSE Chemistry, often involving neutralisation. Use the formula: moles = concentration (mol/dm³) × volume (dm³). At the endpoint, the mole ratio from the equation must be satisfied. Always record burette readings to 2 decimal places and average concordant titres (within 0.10 cm³).
滴定是 WJEC GCSE 化学中的经典实验，通常涉及中和反应。使用公式：摩尔数 = 浓度 (mol/dm³) × 体积 (dm³)。到达终点时，必须满足化学方程式中的摩尔比。务必记录滴定管读数至小数点后两位，并取一致性滴定结果（差值不超过 0.10 cm³）的平均值。
Worked example: 25.0 cm³ of NaOH required 22.70 cm³ of 0.100 mol/dm³ HCl for neutralisation. NaOH + HCl → NaCl + H₂O. Find the concentration of NaOH.
Moles HCl = 0.100 × (22.70 ÷ 1000) = 0.00227 mol → moles NaOH = 0.00227 mol (1:1). Concentration NaOH = 0.00227 ÷ (25.0 ÷ 1000) = 0.0908 mol/dm³.
完整示例：25.0 cm³ NaOH 溶液需要 22.70 cm³ 0.100 mol/dm³ 的 HCl 才能中和。NaOH + HCl → NaCl + H₂O。求 NaOH 的浓度。
HCl 摩尔数 = 0.100 × (22.70 ÷ 1000) = 0.00227 mol → NaOH 摩尔数 = 0.00227 mol（1:1）。NaOH 浓度 = 0.00227 ÷ (25.0 ÷ 1000) = 0.0908 mol/dm³。
10. Common Mistakes and Tips | 常见错误与技巧
Even strong students lose marks by rushing. The top mistakes to avoid are: forgetting to balance equations before using mole ratios; mixing up cm³ and dm³; rounding too early in multi-step calculations; using the wrong Aᵣ or Mᵣ; and forgetting to convert percentages to masses before finding empirical formulae.
即使是优秀学生也会因匆忙而丢分。需要避免的主要错误有：在应用摩尔比之前忘记配平方程式；混淆 cm³ 和 dm³；在多步计算中过早四舍五入；使用错误的 Aᵣ 或 Mᵣ；在求经验式前忘记将百分数转化为质量。
Pro tip: Show all working clearly, even on simple steps. WJEC examiners award marks for the method even if the final answer is wrong. Label mole ratios underneath the balanced equation, and do a quick ‘sanity check’ – does your answer make sense chemically?
高手建议：清晰展示所有步骤，即便是简单的步骤。WJEC 考官会给计算方法分，即使最终答案错误。在配平方程式下方标注摩尔比，并快速进行“合理性检查”——你的答案从化学上讲得通吗？
11. Practice Question Set | 练习题集
Try these mixed questions to build confidence. Answers are provided so you can self-mark.
尝试以下混合题目来建立信心。附有答案，可自评。
- Calculate the Mᵣ of Ca(OH)₂. (Answer: 74) | 计算 Ca(OH)₂ 的 Mᵣ。（答案：74）
- How many moles in 3.2 g of O₂? (Answer: 0.10 mol) | 3.2 g O₂ 是多少摩尔？（答案：0.10 mol）
- What mass of CO₂ is produced when 2.0 g of carbon burns? (Answer: 7.33 g) | 2.0 g 碳燃烧生成多少克 CO₂？（答案：7.33 g）
- A solution contains 4.0 g NaOH in 500 cm³. Find concentration in mol/dm³. (Answer: 0.20 mol/dm³) | 某溶液在 500 cm³ 中含有 4.0 g NaOH。求物质的量浓度。（答案：0.20 mol/dm³）
- 15.0 cm³ of 0.20 mol/dm³ HCl reacts with excess NaOH. What volume of gas (at RTP) is produced? (Answer: 0.072 dm³ or 72 cm³) | 15.0 cm³ 0.20 mol/dm³ HCl 与过量 NaOH 反应。在 RTP 下生成的气体体积是多少？（答案：0.072 dm³ 或 72 cm³）
- In a titration, 25.0 cm³ H₂SO₄ neutralised 20.0 cm³ of 0.50 mol/dm³ NaOH. Find the concentration of the acid. (Answer: 0.20 mol/dm³) | 滴定中，25.0 cm³ H₂SO₄ 中和了 20.0 cm³ 0.50 mol/dm³ NaOH。求酸的浓度。（答案：0.20 mol/dm³）
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