📚 Cambridge IGCSE Chemistry Calculation Questions | 剑桥IGCSE化学计算题型
Calculation questions are a fundamental part of the Cambridge IGCSE Chemistry course and appear in both Paper 3 (Core) and Paper 4 (Extended). Mastering these problems requires a solid understanding of the mole concept, a methodical approach to using chemical equations, and confidence in unit conversions. This article reviews the full range of calculation types you may encounter, from basic relative masses to titration results and water of crystallisation. Each section presents key ideas in English, immediately followed by the same explanation in Chinese, helping bilingual learners build deep understanding.
计算题是剑桥 IGCSE 化学课程的重要组成部分，出现在 Paper 3（核心）和 Paper 4（拓展）中。要掌握这些题目，需要对摩尔概念有扎实的理解，能够有条理地运用化学方程式，并熟悉单位换算。本文回顾你可能遇到的所有计算题型，从基本的相对质量到滴定结果和结晶水计算。每个部分先用英文讲解关键思路，紧接着用中文给出相同的解释，帮助双语学习者建立深刻的理解。
1. Relative Atomic Mass and Relative Molecular Mass | 相对原子质量与相对分子质量
The relative atomic mass (Aᵣ) of an element is the average mass of its atoms compared to 1/12 of the mass of a carbon‑12 atom. It has no units. The relative molecular mass (Mᵣ) is the sum of the Aᵣ values of all the atoms in a molecule. For ionic compounds we often speak of relative formula mass, but the calculation method is identical.
元素的相对原子质量 (Aᵣ) 是指其原子的平均质量与一个碳‑12 原子质量的 1/12 的比值，没有单位。相对分子质量 (Mᵣ) 是分子中所有原子的 Aᵣ 值之和。对于离子化合物，我们常说相对式量，但计算方法完全相同。
For example, calculate Mᵣ of H₂SO₄: (2 × 1) + 32 + (4 × 16) = 98.
例如，计算 H₂SO₄ 的 Mᵣ：(2 × 1) + 32 + (4 × 16) = 98。
Having accurate Aᵣ and Mᵣ values is the first step in virtually every stoichiometric calculation in IGCSE.
获得准确的 Aᵣ 和 Mᵣ 数值是 IGCSE 几乎所有化学计量计算的第一步。
- Always use the Aᵣ values given in the Periodic Table provided in the exam.
- 始终使用试卷所提供周期表中的 Aᵣ 值。
- Mᵣ is calculated by adding the Aᵣ of each atom in the formula—remember to multiply when there are subscripts.
- Mᵣ 的计算是将化学式中各原子的 Aᵣ 相加——注意有下标时要乘以原子个数。
2. The Mole Concept and Avogadro’s Constant | 摩尔概念与阿伏伽德罗常数
One mole of any substance contains 6.02 × 10²³ particles (atoms, molecules, ions or electrons). This number is Avogadro’s constant. The mole allows chemists to count particles by weighing them, because the mass of one mole of a substance in grams is numerically equal to its Aᵣ or Mᵣ.
1 摩尔任何物质含有 6.02 × 10²³ 个粒子（原子、分子、离子或电子）。这个数字就是阿伏伽德罗常数。摩尔使化学家能够通过称重来计数粒子，因为 1 摩尔物质的质量（以克为单位）在数值上等于其 Aᵣ 或 Mᵣ。
For example, the molar mass of carbon (Aᵣ = 12) is 12 g/mol, so 12 g of carbon contains 6.02 × 10²³ carbon atoms.
例如，碳的摩尔质量 (Aᵣ = 12) 为 12 g/mol，因此 12 g 碳含有 6.02 × 10²³ 个碳原子。
The relationship between number of particles (N), number of moles (n) and Avogadro’s constant (L) is:
N = n × 6.02 × 10²³
粒子数 (N)、物质的量 (n) 和阿伏伽德罗常数 (L) 之间的关系为：
N = n × 6.02 × 10²³
3. Molar Mass and Converting Mass to Moles | 摩尔质量与质量和摩尔之间的换算
The central equation for most IGCSE calculations links mass, molar mass and amount in moles:
n = m / M
where n = amount in mol, m = mass in g, M = molar mass in g/mol.
IGCSE 大多数计算的核心公式将质量、摩尔质量和物质的量联系起来：
n = m / M
其中 n = 物质的量（mol），m = 质量（g），M = 摩尔质量（g/mol）。
For instance, to find the number of moles in 4.0 g of NaOH (Mᵣ = 40):
例如，求 4.0 g NaOH (Mᵣ = 40) 的物质的量：
n = 4.0 / 40 = 0.10 mol
Conversely, to find the mass of 0.25 mol of CO₂ (Mᵣ = 44):
反之，求 0.25 mol CO₂ (Mᵣ = 44) 的质量：
m = n × M = 0.25 × 44 = 11 g
Always check that your mass and molar mass are in the same units and that the formula is correctly rearranged.
始终检查质量和摩尔质量的单位是否一致，以及公式是否正确地变形。
4. Empirical Formula from Composition | 由组成求经验式
The empirical formula gives the simplest whole‑number ratio of atoms in a compound. It can be determined from percentage composition or from experimental mass data.
经验式表示化合物中各原子的最简整数比。它可以通过百分组成或实验质量数据来确定。
Steps: (1) Write the mass or percentage of each element. (2) Divide by the element’s Aᵣ to get moles. (3) Divide all mole values by the smallest mole value. (4) If necessary, multiply to obtain whole numbers.
步骤：(1) 写出各元素的质量或百分数。(2) 除以该元素的 Aᵣ，得到物质的量。(3) 将所有摩尔值除以最小的摩尔值。(4) 必要时，乘以适当的系数得到整数。
Example: a compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Assume 100 g sample:
例如：某化合物含 40.0% 碳，6.7% 氢和 53.3% 氧。假设样品质量为 100 g：
- C: 40.0 / 12 = 3.33 mol; H: 6.7 / 1 = 6.7 mol; O: 53.3 / 16 = 3.33 mol
- C: 40.0 / 12 = 3.33 mol; H: 6.7 / 1 = 6.7 mol; O: 53.3 / 16 = 3.33 mol
- Divide by 3.33 → C = 1, H = 2, O = 1 → empirical formula CH₂O.
- 除以 3.33 → C = 1, H = 2, O = 1 → 经验式 CH₂O。
5. Molecular Formula from Empirical Formula | 由经验式求分子式
The molecular formula is a multiple of the empirical formula. To find it you need the compound’s relative molecular mass (Mᵣ).
分子式是经验式的整数倍。要得到分子式，需要知道化合物的相对分子质量 (Mᵣ)。
First calculate the empirical formula mass. Then find the multiplier:
Multiplier = Mᵣ (given) / empirical formula mass
首先计算经验式的式量。然后求倍数：
倍数 = 给定的 Mᵣ / 经验式量
Using the previous example, CH₂O has empirical mass = 12 + (2×1) + 16 = 30. If the Mᵣ is found to be 180 by experiment, then multiplier = 180 / 30 = 6. The molecular formula is C₆H₁₂O₆.
用前例，CH₂O 的经验式量 = 12 + (2×1) + 16 = 30。如果通过实验测得 Mᵣ = 180，则倍数 = 180 / 30 = 6。分子式为 C₆H₁₂O₆。
Always check that the molecular formula makes chemical sense and that its Mᵣ matches the given value.
始终检查分子式在化学上是否合理，以及其 Mᵣ 是否与给定值相符。
6. Reacting Masses Using Moles in Equations | 利用化学方程式的反应质量计算
Balanced equations give the mole ratio in which reactants combine and products are formed. Once you know the moles of one substance, you can use the ratio to find the moles of another, then convert to mass.
配平的化学方程式给出了反应物化合和产物生成的摩尔比。一旦知道了某一种物质的物质的量，就可以利用比例求另一种物质的物质的量，然后换算为质量。
Example: 2Mg + O₂ → 2MgO. Calculate the mass of magnesium oxide produced when 6.0 g of magnesium burns completely in oxygen. (Aᵣ: Mg = 24, O = 16)
例如：2Mg + O₂ → 2MgO。计算 6.0 g 镁在氧气中完全燃烧生成的氧化镁质量。(Aᵣ: Mg = 24, O = 16)
Moles of Mg = 6.0 / 24 = 0.25 mol. From the equation, mole ratio Mg : MgO = 2 : 2 = 1 : 1, so moles of MgO = 0.25 mol. Mᵣ of MgO = 24 + 16 = 40. Mass of MgO = 0.25 × 40 = 10 g.
Mg 的物质的量 = 6.0 / 24 = 0.25 mol。根据方程式，摩尔比 Mg : MgO = 2 : 2 = 1 : 1，因此 MgO 的物质的量 = 0.25 mol。MgO 的 Mᵣ = 24 + 16 = 40。MgO 的质量 = 0.25 × 40 = 10 g。
The key is to move from mass → moles, use the mole ratio from the equation, then moles → mass.
关键在于从质量 → 物质的量，利用方程式中的摩尔比，再由物质的量 → 质量。
7. Limiting Reactants | 限量反应物
When two or more reactants are mixed in non‑stoichiometric amounts, the one that is completely consumed first is the limiting reactant. The other reactants are in excess. The amount of product is determined entirely by the limiting reactant.
当两种或多种反应物以非化学计量比混合时，首先完全消耗的反应物称为限量反应物。其他反应物为过量。产物的量完全取决于限量反应物。
To identify the limiting reactant, calculate the number of moles of each reactant. Divide each by its coefficient in the balanced equation. The smallest value indicates the limiting reactant.
要确定限量反应物，先计算各反应物的物质的量。将各物质的量除以配平方程式中各自的系数。最小的值对应的反应物即为限量反应物。
Example: 2H₂ + O₂ → 2H₂O. 4.0 g H₂ (Mᵣ = 2) is mixed with 32.0 g O₂ (Mᵣ = 32). Moles of H₂ = 4.0 / 2 = 2.0 mol; moles of O₂ = 1.0 mol. Ratio H₂ : 2.0/2 = 1.0; O₂ : 1.0/1 = 1.0. They are in exact stoichiometric ratio, but if 3.0 g H₂ were used, H₂ would be in excess and O₂ limiting.
例如：2H₂ + O₂ → 2H₂O。将 4.0 g H₂ (Mᵣ = 2) 与 32.0 g O₂ (Mᵣ = 32) 混合。H₂ 物质的量 = 4.0 / 2 = 2.0 mol；O₂ 物质的量 = 1.0 mol。比值 H₂ : 2.0/2 = 1.0；O₂ : 1.0/1 = 1.0，恰好为化学计量比。但如果使用 3.0 g H₂，则 H₂ 过量，O₂ 为限量反应物。
Use the moles of the limiting reactant to calculate the theoretical yield of product.
用限量反应物的物质的量计算产物的理论产量。
8. Percentage Yield | 产率
The theoretical yield is the maximum mass of product calculated from the limiting reactant. In practice, the actual yield is often lower due to incomplete reaction, side reactions or losses during purification. Percentage yield compares the actual yield to the theoretical yield.
理论产量是由限量反应物计算出的最大产品质量。在实际操作中，由于反应不完全、副反应或纯化过程中的损失，实际产量往往较低。产率是比较实际产量与理论产量。
Percentage yield = (actual yield / theoretical yield) × 100%
产率 = (实际产量 / 理论产量) × 100%
For instance, if the theoretical yield of MgO is 10 g but only 8.5 g is collected, then percentage yield = (8.5 / 10) × 100% = 85%.
例如，若 MgO 的理论产量为 10 g，但仅收集到 8.5 g，则产率 = (8.5 / 10) × 100% = 85%。
A high percentage yield is important for economic and environmental reasons. IGCSE questions often ask you to suggest reasons why the yield is less than 100%.
出于经济和环境原因，高产率非常重要。IGCSE 题目经常要求你解释产率低于 100% 的可能原因。
9. Volume of Gases and Molar Gas Volume | 气体体积与摩尔气体体积
At room temperature and pressure (r.t.p., about 20 °C and 1 atm), one mole of any gas occupies a volume of 24 dm³ (or 24,000 cm³). This is the molar gas volume. The relationship is:
n = volume (in dm³) / 24
在常温常压下（r.t.p.，约 20 °C，1 atm），1 摩尔任何气体的体积为 24 dm³（或 24,000 cm³）。这就是摩尔气体体积。关系式为：
n = 体积 (dm³) / 24
Example: Calculate the volume of carbon dioxide produced at r.t.p. when 5.0 g of CaCO₃ (Mᵣ = 100) decomposes completely: CaCO₃ → CaO + CO₂. Moles of CaCO₃ = 5.0 / 100 = 0.050 mol. From equation, moles of CO₂ = 0.050 mol. Volume of CO₂ = 0.050 × 24 = 1.2 dm³.
例题：计算 5.0 g CaCO₃ (Mᵣ = 100) 完全分解时在常温常压下产生的二氧化碳体积：CaCO₃ → CaO + CO₂。CaCO₃ 物质的量 = 5.0 / 100 = 0.050 mol。根据方程式，CO₂ 的物质的量 = 0.050 mol。CO₂ 体积 = 0.050 × 24 = 1.2 dm³。
Remember to convert volumes to dm³ if given in cm³ (1 dm³ = 1000 cm³).
如果给出的体积单位是 cm³，记得换算为 dm³（1 dm³ = 1000 cm³）。
10. Concentration of Solutions in mol/dm³ and g/dm³ | 溶液的浓度（mol/dm³ 和 g/dm³）
Concentration can be expressed in mol/dm³ or g/dm³. The two are linked by the molar mass:
concentration (mol/dm³) = concentration (g/dm³) / M
浓度可以用 mol/dm³ 或 g/dm³ 表示。两者通过摩尔质量联系起来：
浓度 (mol/dm³) = 浓度 (g/dm³) / M
Also, the number of moles in a solution of known volume is:
n = c × V
其中 c = 浓度（mol/dm³），V = 体积（dm³）。
同样，对于已知体积的溶液，物质的量为：
n = c × V
where c = concentration in mol/dm³, V = volume in dm³.
Example: How many moles of HCl are in 50.0 cm³ of 0.20 mol/dm³ hydrochloric acid? Volume in dm³ = 50.0 / 1000 = 0.050 dm³. n = 0.20 × 0.050 = 0.010 mol.
例题：50.0 cm³ 0.20 mol/dm³ 盐酸中含有多少摩尔 HCl？体积以 dm³ 计 = 50.0 / 1000 = 0.050 dm³。n = 0.20 × 0.050 = 0.010 mol。
To prepare a solution, you can dissolve a known mass in a solvent and make up to a known volume. These calculations often appear in practical‑based questions.
配制溶液时，可将已知质量溶于溶剂并定容至已知体积。这类计算常出现在与实验相关的题目中。
11. Titration Calculations | 滴定计算
Titrations are used to find the concentration of an unknown solution by reacting it with a solution of known concentration. The steps follow the same mass–mole pathway.
滴定法通过让未知浓度的溶液与已知浓度的溶液反应，来测定前者的浓度。其步骤遵循相同的质量–摩尔路径。
Typical IGCSE procedure: (1) Write the balanced equation. (2) Calculate moles of the known solution used (n = c × V). (3) Use the mole ratio to find moles of the unknown. (4) Calculate the unknown concentration or mass.
典型的 IGCSE 步骤：(1) 写出配平的化学方程式。(2) 计算所用已知溶液的物质的量 (n = c × V)。(3) 利用摩尔比求出未知物的物质的量。(4) 计算未知物的浓度或质量。
Example: 25.0 cm³ of NaOH solution is neutralised by 20.0 cm³ of 0.50 mol/dm³ H₂SO₄. Find the concentration of NaOH. Equation: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O. Moles of H₂SO₄ = 0.50 × (20.0/1000) = 0.010 mol. Mole ratio NaOH : H₂SO₄ = 2 : 1, so moles of NaOH = 0.020 mol. Volume of NaOH in dm³ = 25.0/1000 = 0.025 dm³. Concentration of NaOH = 0.020 / 0.025 = 0.80 mol/dm³.
例题：25.0 cm³ NaOH 溶液被 20.0 cm³ 0.50 mol/dm³ H₂SO₄ 中和。求 NaOH 的浓度。方程式：2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O。H₂SO₄ 物质的量 = 0.50 × (20.0/1000) = 0.010 mol。摩尔比 NaOH : H₂SO₄ = 2 : 1，因此 NaOH 物质的量 = 0.020 mol。NaOH 的体积以 dm³ 计 = 25.0/1000 = 0.025 dm³。NaOH 浓度 = 0.020 / 0.025 = 0.80 mol/dm³。
Always show the units at each step to avoid errors. If the question gives average titre, use the concordant results.
每一步都应标明单位以避免错误。如果题目给出平均滴定体积，应使用一致的滴定结果。
12. Water of Crystallisation | 结晶水计算
Many salts contain water molecules as part of their crystal structure, e.g. CuSO₄·5H₂O. The dot formula shows the number of water molecules per formula unit. Heating drives off the water, and the mass loss allows you to calculate ‘x’ in the formula.
许多盐在晶体结构中含有水分子，例如 CuSO₄·5H₂O。点式表示每个化学式单元所含的水分子数。加热可除去水分，通过质量减少可以计算式中的 ‘x’。
Method: (1) Find the mass of the hydrated salt before heating. (2) Heat to constant mass to remove all water. (3) Mass of anhydrous salt = final mass. (4) Mass of water lost = initial mass – final mass. (5) Convert both masses to moles. (6) Find the simplest ratio of anhydrous salt : water.
方法：(1) 称量加热前水合盐的质量。(2) 加热至恒重以除去所有水分。(3) 无水盐的质量 = 最终质量。(4) 失去的水的质量 = 初始质量 – 最终质量。(5) 将两个质量换算为物质的量。(6) 求无水盐与水的物质的量最简比。
Example: 2.50 g of hydrated magnesium sulfate (MgSO₄·xH₂O) is heated, leaving 1.22 g of anhydrous MgSO₄. (Mᵣ: MgSO₄ = 120, H₂O = 18). Mass of water = 2.50 – 1.22 = 1.28 g. Moles of MgSO₄ = 1.22 / 120 = 0.0102 mol. Moles of H₂O = 1.28 / 18 = 0.0711 mol. Ratio H
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