IGCSE Chemistry Calculations Mastery | IGCSE 化学计算题型全攻略

📚 IGCSE Chemistry Calculations Mastery | IGCSE 化学计算题型全攻略

Mastering calculations is essential for success in Cambridge IGCSE Chemistry. The ability to move confidently between mass, moles, volume and concentration unlocks almost every quantitative problem on the exam. This guide covers all major calculation types you will encounter, from foundation mole concepts to multi-step titration and yield problems, with step-by-step worked examples and clear bilingual explanations.

掌握计算是应对剑桥 IGCSE 化学考试的关键。能否熟练地在质量、摩尔、体积与浓度之间转换，决定了你能否解决考试中几乎所有的定量问题。本指南涵盖所有主要计算题型，从基础的摩尔概念到多步骤的滴定与产率问题，配以分步例题和清晰的双语解释。

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 1/12th of the mass of a carbon-12 atom. It has no units.

相对原子质量 (Aᵣ) 是元素一个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。

Relative molecular mass (Mᵣ) applies to covalent compounds and 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 identical.

相对分子质量 (Mᵣ) 适用于共价化合物，是分子中所有原子的 Aᵣ 之和。对于离子化合物，使用相对式量这一术语，但计算方法完全相同。

Many IGCSE questions provide a data table of Aᵣ values. A small set of common values is shown below:

许多 IGCSE 考题会提供 Aᵣ 数据表。以下是部分常见值：

Element	Aᵣ
Hydrogen (H)	1
Carbon (C)	12
Oxygen (O)	16
Sodium (Na)	23
Sulfur (S)	32

Worked example: Calculate the Mᵣ of sulfuric acid, H₂SO₄.

例题：计算硫酸 H₂SO₄ 的 Mᵣ。

Mᵣ = (2 × 1) + 32 + (4 × 16) = 2 + 32 + 64 = 98

2. The Mole Concept and Molar Mass | 摩尔概念与摩尔质量

One mole is the amount of substance that contains 6.02 × 10²³ particles (Avogadro constant). In IGCSE, the mole is almost always used to link mass and relative mass.

一摩尔是含有 6.02 × 10²³ 个微粒（阿伏伽德罗常数）的物质的量。在 IGCSE 中，摩尔几乎总是用来联系质量和相对质量。

Molar mass (M) is the mass of one mole of a substance. Its numerical value equals the Mᵣ, but it carries the unit g/mol.

摩尔质量 (M) 是一摩尔物质的质量。数值等于 Mᵣ，但带有单位 g/mol。

n = m / M

Where n = amount (mol), m = mass (g), M = molar mass (g/mol). This formula is the cornerstone of all quantitative chemistry.

式中 n = 物质的量 (mol)，m = 质量 (g)，M = 摩尔质量 (g/mol)。这个公式是所有定量化学的基石。

Example: How many moles are in 8.0 g of methane (CH₄)? (Aᵣ: H=1, C=12)

例题：8.0 g 甲烷 (CH₄) 是多少摩尔？

M(CH₄) = 12 + (4 × 1) = 16 g/mol. n = 8.0 / 16 = 0.50 mol.

M(CH₄) = 12 + (4 × 1) = 16 g/mol。n = 8.0 / 16 = 0.50 mol。

3. Converting Between Mass and Moles | 质量与摩尔数的换算

Once students can use n = m / M, they should practise both directions: mass to moles and moles to mass. Rearranging gives m = n × M.

学生掌握 n = m / M 后，应练习双向转换：由质量求摩尔数，以及由摩尔数求质量。变形后得 m = n × M。

Mass from moles: What mass of NaOH is needed to obtain 0.25 mol? (Na=23, O=16, H=1)

由摩尔数求质量：需要多少质量的 NaOH 才能得到 0.25 mol？

M(NaOH) = 23 + 16 + 1 = 40 g/mol. m = 0.25 × 40 = 10 g.

M(NaOH) = 23 + 16 + 1 = 40 g/mol。m = 0.25 × 40 = 10 g。

Always check whether the question expects the answer in grams, kilograms or even tonnes; convert if needed.

始终注意题目要求的质量单位是克、千克还是吨，必要时要换算。

4. Molar Volume of Gases | 气体的摩尔体积

At room temperature and pressure (r.t.p., taken as 20 °C and 1 atm), one mole of any gas occupies a volume of 24 dm³. This is known as the molar gas volume.

在室温和常压下（r.t.p.，取 20 °C 和 1 atm），一摩尔任何气体的体积是 24 dm³，这称为气体摩尔体积。

V (dm³) = n × 24

If a volume is given in cm³, first convert to dm³ by dividing by 1000.

如果给出的体积单位是 cm³，先除以 1000 转换为 dm³。

Example: Calculate the volume of 2.0 mol of carbon dioxide at r.t.p.

例题：计算 2.0 mol 二氧化碳在 r.t.p. 下的体积。

V = 2.0 × 24 = 48 dm³.

V = 2.0 × 24 = 48 dm³。

Example 2: A sample of H₂ occupies 1200 cm³ at r.t.p. Find its amount in moles.

例题 2：某氢气样品在 r.t.p. 下体积为 1200 cm³，求其物质的量。

Convert to dm³: 1200 cm³ = 1.2 dm³. n = 1.2 / 24 = 0.050 mol.

转换为 dm³：1200 cm³ = 1.2 dm³。n = 1.2 / 24 = 0.050 mol。

5. Reacting Masses from Chemical Equations | 由化学方程式求反应质量

Reacting mass questions are solved by following a structured method:

解反应质量题需按固定的步骤进行：

Write the balanced equation → Find moles of the known substance → Use the mole ratio to find moles of the unknown → Convert moles back to mass.

书写配平的化学方程式 → 求出已知物质的量 → 利用摩尔比求未知物质的量 → 将摩尔数转换回质量。

Example: What mass of magnesium oxide is produced when 3.0 g of magnesium burns completely in oxygen? 2Mg + O₂ → 2MgO (Aᵣ: Mg=24, O=16)

例题：3.0 g 镁完全燃烧能生成多少质量的氧化镁？2Mg + O₂ → 2MgO

n(Mg) = 3.0 / 24 = 0.125 mol. Mole ratio Mg : MgO = 2 : 2 = 1 : 1, so n(MgO) = 0.125 mol. M(MgO) = 24 + 16 = 40 g/mol. m = 0.125 × 40 = 5.0 g.

n(Mg) = 3.0 / 24 = 0.125 mol。摩尔比 Mg : MgO = 2 : 2 = 1 : 1，所以 n(MgO) = 0.125 mol。M(MgO) = 24 + 16 = 40 g/mol。m = 0.125 × 40 = 5.0 g。

6. Limiting Reactant Problems | 限制反应物问题

When masses or moles of two reactants are given, one reactant may be used up first, limiting the amount of product. The limiting reactant is identified by comparing the mole ratio required by the equation with the moles actually present.

当给出两种反应物的质量或物质的量时，其中一种可能先耗尽，从而限制产物的量。通过比较方程所需摩尔比与实际摩尔数，可以确定限制反应物。

Method: For each reactant, divide the number of moles by its coefficient in the balanced equation. The smaller value indicates the limiting reactant.

方法：对每种反应物，用其摩尔数除以配平方程中的系数。数值较小者即限制反应物。

Example: 2H₂ + O₂ → 2H₂O. 4.0 g of H₂ reacts with 32.0 g of O₂. Find the limiting reactant and the mass of water formed. (H=1, O=16)

例题：2H₂ + O₂ → 2H₂O。4.0 g H₂ 与 32.0 g O₂ 反应。找出限制反应物和生成水的质量。

n(H₂) = 4.0 / 2 = 2.0 mol. n(O₂) = 32.0 / 32 = 1.0 mol. Check ratio: H₂ 2.0/2 = 1.0; O₂ 1.0/1 = 1.0. Neither is in excess; both are exactly consumed. Water formed: n(H₂O) = 2.0 mol, m = 2.0 × 18 = 36 g.

n(H₂) = 4.0 / 2 = 2.0 mol。n(O₂) = 32.0 / 32 = 1.0 mol。检查比值：H₂ 2.0/2 = 1.0；O₂ 1.0/1 = 1.0。两者均恰好完全反应，无过量。生成水：n(H₂O) = 2.0 mol，m = 2.0 × 18 = 36 g。

If instead 2.0 g H₂ (1.0 mol) had been used with 32.0 g O₂ (1.0 mol), H₂ would give 1.0/2 = 0.5, O₂ 1.0/1 = 1.0, so H₂ is limiting and the theoretical water yield = 1.0 mol → 18 g.

若改用 2.0 g H₂ (1.0 mol) 与 32.0 g O₂ (1.0 mol)，H₂ 的比值为 0.5，O₂ 为 1.0，H₂ 限制反应，理论产水量 = 1.0 mol → 18 g。

7. Percentage Yield and Atom Economy | 产率与原子经济性

Percentage yield compares the actual mass of product obtained with the theoretical mass predicted by stoichiometry.

产率（百分率）将实际获得的产物质量与化学计量学预测的理论质量进行比较。

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

% 产率 = (实际产量 / 理论产量) × 100%

Yields are often less than 100% because of incomplete reactions, side reactions, or loss during purification.

产率常低于 100%，原因包括反应不完全、副反应或纯化过程中的损失。

Atom economy measures how efficiently reactants are converted into the desired product. It is a concept emphasised in ‘green chemistry’.

原子经济性衡量反应物转化为目标产物的效率，这是 ‘绿色化学’ 强调的概念。

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

原子经济性 = (目标产物的 Mᵣ / 所有反应物的 Mᵣ 之和) × 100%

While IGCSE questions on atom economy are less frequent, they do appear and require careful identification of the ‘desired’ product from the equation.

虽然 IGCSE 中原子经济性的考题较少，但仍有出现，需要仔细从方程中识别 ‘目标’ 产物。

8. Concentration Calculations | 浓度计算

Concentration can be expressed in mol/dm³ (molarity) or g/dm³. The two are linked by the molar mass:

浓度可以用 mol/dm³（摩尔浓度）或 g/dm³ 表示。两者通过摩尔质量联系：

c (mol/dm³) = c (g/dm³) / M

The key relationship when using molarity is:

使用摩尔浓度时的关键关系是：

n = c × V (dm³)

Example: A student dissolves 5.85 g of NaCl in water to make 250 cm³ of solution. Calculate the concentration in mol/dm³. (Na=23, Cl=35.5)

例题：某学生将 5.85 g NaCl 溶于水，配成 250 cm³ 溶液。计算摩尔浓度。

M(NaCl) = 23 + 35.5 = 58.5 g/mol. n = 5.85 / 58.5 = 0.10 mol. V = 250 / 1000 = 0.25 dm³. c = 0.10 / 0.25 = 0.40 mol/dm³.

M(NaCl) = 58.5 g/mol。n = 5.85 / 58.5 = 0.10 mol。V = 0.25 dm³。c = 0.10 / 0.25 = 0.40 mol/dm³。

If the question asks for concentration in g/dm³, simply divide mass by volume in dm³: 5.85 / 0.25 = 23.4 g/dm³.

若题目要求 g/dm³ 浓度，直接用质量除以体积 (dm³)：5.85 / 0.25 = 23.4 g/dm³。

9. Titration Calculations | 滴定计算

Titration questions require us to use

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IGCSE Chemistry Calculations Mastery | IGCSE 化学计算题型全攻略