IGCSE CIE Chemistry Calculation Question Types | IGCSE CIE 化学计算题型全面解析

📚 IGCSE CIE Chemistry Calculation Question Types | IGCSE CIE 化学计算题型全面解析

Calculation questions form a core part of the Cambridge IGCSE Chemistry examination, testing your ability to combine chemical concepts with mathematical skills. Based on the Cambridge IGCSE™ Chemistry Coursebook with Digital Access, this guide covers the most common calculation types you will encounter, from simple relative mass to more challenging titration and energy change problems. Mastering these techniques is essential for achieving a high grade.

计算题型是剑桥 IGCSE 化学考试的核心组成部分，考察你将化学概念与数学能力结合的能力。本文基于 Cambridge IGCSE™ Chemistry Coursebook with Digital Access，涵盖最常见的计算类型，从简单的相对质量到更具挑战性的滴定和能量变化问题。掌握这些技巧对于取得高分至关重要。

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1. Relative Atomic Mass and Relative Molecular Mass | 相对原子质量与相对分子质量

Relative atomic mass (Aᵣ) is the average mass of an atom of an element compared to 1/12 of the mass of a carbon‑12 atom. It has no units. Relative molecular mass (Mᵣ) is the sum of the Aᵣ values of all atoms in a molecule. For ionic compounds, the term relative formula mass is used, but the calculation is the same.

相对原子质量（Aᵣ）是一个元素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 Aᵣ 值的总和。对于离子化合物，使用相对式量这一术语，但计算方式相同。

r (H₂O): Mᵣ = (2 × 1) + 16 = 18. Calcium carbonate (CaCO₃): Mᵣ = 40 + 12 + (3 × 16) = 100.

计算 Mᵣ 时，将每种元素的 Aᵣ 乘以该元素的原子个数，再求和。例如，水（H₂O）：Mᵣ = (2 × 1) + 16 = 18。碳酸钙（CaCO₃）：Mᵣ = 40 + 12 + (3 × 16) = 100。

• Always use the Aᵣ values given in the Periodic Table, usually rounded to one decimal place in CIE examinations.
• 在 CIE 考试中，务必使用周期表给出的 Aᵣ 值，通常保留一位小数。
• Common mistake: forgetting to multiply when there is more than one atom of an element, e.g. in H₂O.
• 常见错误：当元素原子不止一个时忘记相乘，例如 H₂O 中的氢。

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2. The Mole and Avogadro's Number | 摩尔与阿伏伽德罗常数

ce that contains the same number of particles as there are atoms in exactly 12 g of carbon‑12. This number is Avogadro's constant, approximately 6.02 × 10²³.

一摩尔是物质的量，其所含的粒子数与恰好 12 g 碳‑12 中的原子数相同。这个数就是阿伏伽德罗常数，约为 6.02 × 10²³。

Number of particles = number of moles × 6.02 × 10²³

. For instance, 2 mol of water molecules contain 2 × 6.02 × 10²³ = 1.204 × 10²⁴ molecules.

你可以利用这个关系求算给定摩尔数的原子、分子或离子数目，或反向计算。例如，2 mol 水分子含有 2 × 6.02 × 10²³ = 1.204 × 10²⁴ 个水分子。

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3. Converting between Mass and Moles | 质量与摩尔转换

The link between mass of a substance and amount in moles is its molar mass (M), which has the same numerical value as Mᵣ but with units of g/mol.

物质的质量与摩尔数之间的桥梁是摩尔质量（M），其数值与 Mᵣ 相同，但单位为 g/mol。

n = m / M

calculate the mass of 0.5 mol of carbon dioxide (CO₂, M = 44 g/mol): m = 0.5 × 44 = 22 g.

其中 n = 摩尔数，m = 质量（克），M = 摩尔质量（g/mol）。要计算 0.5 mol 二氧化碳（CO₂，M = 44 g/mol）的质量：m = 0.5 × 44 = 22 g。

• Always ensure your mass is in grams, not kilograms.
• 务必确保质量单位为克，而非千克。
• If given a mass in tonnes or kg, convert to g first (1 kg = 1000 g).
• 若给出的质量是吨或千克，先换算成克（1 kg = 1000 g）。

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4. Empirical and Molecular Formulae | 经验式与分子式

The empirical formula gives the simplest whole‑number ratio of atoms in a compound. The molecular formula shows the actual number of each atom in a molecule and is a whole‑number multiple of the empirical formula.

经验式表示化合物中原子最简整数比。分子式显示分子中每种原子的实际个数，是经验式的整数倍。

Steps to find the empirical formula from percentage composition or mass data:

通过百分组成或质量数据求经验式的步骤：

• Divide the mass (or percentage) of each element by its Aᵣ to find the number of moles.
• 将各元素的质量（或百分比）除以其 Aᵣ，求得摩尔数。
• Divide each mole value by the smallest number of moles obtained.
• 将各摩尔值除以所得的最小摩尔数。
• If necessary, multiply all ratios by a simple factor to get whole numbers.
• 如有必要，将所有比值乘以一个简单的因子以获得整数。

Example: A compound contains 43.7% phosphorus and 56.3% oxygen. Find its empirical formula (Aᵣ: P = 31.0, O = 16.0).

示例：某化合物含 43.7% 磷和 56.3% 氧，求其经验式（Aᵣ: P = 31.0，O = 16.0）。

Element	P	O
% or mass	43.7	56.3
Moles = mass / Aᵣ	43.7 / 31.0 = 1.41	56.3 / 16.0 = 3.52
Divide by smallest (1.41)	1.41 / 1.41 = 1	3.52 / 1.41 = 2.5
Multiply to get whole numbers (×2)	2	5

Empirical formula = P₂O₅.

经验式 = P₂O₅。

To determine the molecular formula, divide the given Mᵣ of the compound by the empirical formula mass. Multiply the empirical formula subscripts by that factor.

要确定分子式，用给出的化合物 Mᵣ 除以经验式量，再将经验式的下标乘以该因子。

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5. Reacting Masses | 反应质量计算

Reacting mass calculations use the balanced chemical equation to predict the masses of substances consumed or produced.

反应质量计算利用配平的化学方程式来预测消耗或生成的物质质量。

Method:

方法：

• Write the balanced equation.
• 写出配平的化学方程式。
• Identify the mole ratio of the substances involved.
• 确定涉及物质的摩尔比。
• Convert the known mass into moles (n = m / M).
• 将已知质量转换为摩尔数 (n = m / M)。
• Use the mole ratio to find the moles of the unknown substance.
• 利用摩尔比求算出未知物质的摩尔数。
• Convert moles of the unknown substance into mass (m = n × M).
• 将未知物质的摩尔数换算为质量 (m = n × M)。

t mass of calcium oxide (CaO) is produced when 50 g of calcium carbonate (CaCO₃) is heated? Equation: CaCO₃ → CaO + CO₂. Mᵣ: CaCO₃ = 100, CaO = 56.

示例：加热 50 g 碳酸钙 (CaCO₃) 可生成多少质量的氧化钙 (CaO)？方程式：CaCO₃ → CaO + CO₂。Mᵣ: CaCO₃ = 100，CaO = 56。

ole ratio CaCO₃ : CaO = 1 : 1, so moles of CaO = 0.50 mol. Mass of CaO = 0.50 × 56 = 28 g.

CaCO₃ 的摩尔数 = 50 / 100 = 0.50 mol。摩尔比 CaCO₃ : CaO = 1 : 1，所以 CaO 的摩尔数 = 0.50 mol。CaO 的质量 = 0.50 × 56 = 28 g。

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6. Molar Gas Volume | 气体摩尔体积

At room temperature and pressure (RTP, about 20 °C and 1 atm), one mole of any gas occupies a volume of 24 dm³. This is the molar gas volume.

在室温和常压下（RTP，约 20 °C 和 1 atm），一摩尔任何气体所占体积为 24 dm³，这就是气体摩尔体积。

Volume of gas (dm³) = number of moles × 24

conditions are at RTP. For example, the volume of 3 mol of oxygen gas is 3 × 24 = 72 dm³.

若条件为室温常压，可直接使用此关系。例如，3 mol 氧气的体积为 3 × 24 = 72 dm³。

When a reaction produces a gas, first find the moles of the gas from the reacting masses or concentrations, then multiply by 24 to obtain the volume.

当反应产生气体时，先通过反应质量或浓度求出气体的摩尔数，再乘以 24 得到体积。

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7. Concentration Calculations | 浓度计算

Concentration is the amount of solute dissolved per unit volume of solution. It can be expressed in g/dm³ or mol/dm³.

浓度是单位体积溶液中溶解的溶质质量，可用 g/dm³ 或 mol/dm³ 表示。

Concentration (mol/dm³) = number of moles / volume (dm³)

or in mass units:

或用质量单位：

Concentration (g/dm³) = mass (g) / volume (dm³)

convert between the two: mass concentration (g/dm³) = molar concentration (mol/dm³) × Mᵣ.

二者换算：质量浓度 (g/dm³) = 摩尔浓度 (mol/dm³) × Mᵣ。

0.250 dm³. Moles NaOH = 4.0 / 40 = 0.10 mol. Concentration = 0.10 / 0.250 = 0.40 mol/dm³.

示例：将 4.0 g 氢氧化钠（NaOH，Mᵣ = 40）溶于水，配成 250 cm³ 溶液。计算其 mol/dm³ 浓度。体积 dm³ = 250 / 1000 = 0.250 dm³。NaOH 摩尔数 = 4.0 / 40 = 0.10 mol。浓度 = 0.10 / 0.250 = 0.40 mol/dm³。

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8. Titration Calculations | 滴定计算

Titration is used to find the unknown concentration of a solution by reacting it with a solution of known concentration. The key is the balanced equation, which gives the mole ratio.

滴定用于通过与已知浓度溶液反应来求未知溶液的浓度，关键是配平的化学方程式给出的摩尔比。

For a simple 1:1 reaction such as HCl + NaOH → NaCl + H₂O, the calculation simplifies to:

对于简单的 1:1 反应，如 HCl + NaOH → NaCl + H₂O，计算可简化为：

c₁V₁ = c₂V₂

For reactions with different mole ratios, e.g. H₂SO₄ + 2NaOH, you must include the ratio:

其中 c = 浓度 (mol/dm³)，V = 体积 (dm³)。对于摩尔比不同的反应，如 H₂SO₄ + 2NaOH，必须将比率纳入计算：

n(acid) / n(alkali) = (cₐVₐ) / (cₐₗₖₐₗᵢVₐₗₖₐₗᵢ)

A safer method is to always work through moles:

更稳妥的方法是始终通过摩尔数来求解：

• Calculate moles of the known reactant using its concentration and volume.
• 用已知反应物的浓度和体积计算其摩尔数。
• Use the mole ratio from the equation to find moles of the unknown.
• 利用方程式中的摩尔比求未知物的摩尔数。
• Divide the moles of the unknown by its volume to obtain its concentration.
• 将未知物的摩尔数除以其体积，得到其浓度。

so moles NaOH = 0.00250 mol. Concentration NaOH = 0.00250 / (30.0/1000) = 0.0833 mol/dm³.

示例：25.0 cm³ 的 0.100 mol/dm³ HCl 被 30.0 cm³ NaOH 溶液中和。求 NaOH 的浓度。HCl 摩尔数 = 0.100 × (25.0/1000) = 0.00250 mol。摩尔比 1:1，所以 NaOH 摩尔数 = 0.00250 mol。NaOH 浓度 = 0.00250 / (30.0/1000) = 0.0833 mol/dm³。

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9. Percentage Yield and Atom Economy | 百分产率与原子经济

Percentage yield compares the actual mass of product obtained to the theoretical mass predicted by the reacting mass calculation.

百分产率比较实际获得的产品质量与根据反应质量计算预测的理论质量。

Percentage yield = (actual yield / theoretical yield) × 100%

Yields are often less than 100% due to incomplete reactions, side reactions, or losses during purification.

由于反应不完全、副反应或提纯过程中的损失，产率通常低于 100%。

Atom economy measures the efficiency of a reaction in terms of how much of the reactants end up in the desired product.

原子经济衡量的是反应中反应物有多少最终进入目标产物中，以此反映反应的效率。

Atom economy = (Mᵣ of desired product / sum of Mᵣ of all reactants) × 100%

A higher atom economy means a more sustainable reaction with less waste.

较高的原子经济意味着反应更可持续，产生的废物更少。

These two calculations are distinct: yield tells you about the practical success, while atom economy tells you about the intrinsic 'greenness' of the reaction.

这两种计算截然不同：产率反映实际操作的成功程度，而原子经济反映反应本身的“绿色”程度。

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10. Simple Calorimetry Calculations | 简单量热计算

Calorimetry measures the heat energy released or absorbed during a chemical reaction. A common experiment involves burning a fuel to heat water and measuring the temperature rise.

量热法测量化学反应中释放或吸收的热能。常见实验是燃烧燃料加热水并测量温度升高。

q = m × c × ΔT

grams, c = specific heat capacity (4.2 J/g°C for water), ΔT = change in temperature (°C).

q = 能量变化（焦耳 J），m = 被加热的水（或溶液）的质量（克），c = 比热容（水为 4.2 J/g°C），ΔT = 温度变化 (°C)。

To express the energy per mole, divide q by the number of moles of fuel burned or reactant used. This gives the molar enthalpy change, often in kJ/mol (remember to convert J to kJ by dividing by 1000).

要表示每摩尔的能量，将 q 除以燃烧的燃料或使用的反应物的摩尔数，得到摩尔焓变，单位通常为 kJ/mol（记得将 J 除以 1000 换算为 kJ）。

Example: 0.92 g of ethanol (C₂H₅

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