IGCSE Chemistry: Calculation Practice | IGCSE 化学：计算题专项训练

📚 IGCSE Chemistry: Calculation Practice | IGCSE 化学：计算题专项训练

Mastering calculation questions is essential for achieving a top grade in IGCSE Chemistry. This article provides a systematic training guide covering the most common types of quantitative problems you will encounter — from moles and reacting masses to titrations and yields. Each section explains the underlying concept, presents key formulas, and walks you through worked examples to build both confidence and accuracy.

掌握计算题是 IGCSE 化学拿高分的关键。本文提供一个系统的专项训练指南，涵盖最常见的定量问题类型——从摩尔、反应质量到滴定和产率。每个小节解释核心概念，展示关键公式，并通过例题引导你逐步练习，帮助你建立信心并提高准确度。

1. Understanding Relative Atomic and Formula Mass | 理解相对原子质量和相对式量

Relative atomic mass (Aᵣ) is the weighted 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 formula mass (Mᵣ) is the sum of the relative atomic masses of all atoms in a formula unit. For a compound such as CaCO₃, you add Aᵣ(Ca) = 40, Aᵣ(C) = 12, and 3 × Aᵣ(O) = 3 × 16 = 48, giving Mᵣ = 40 + 12 + 48 = 100.

相对原子质量（Aᵣ）是一个元素的原子加权平均质量与一个碳‑12原子质量的 1/12 的比值，没有单位。相对式量（Mᵣ）是化学式中所有原子的相对原子质量之和。例如 CaCO₃，将 Aᵣ(Ca)=40，Aᵣ(C)=12，以及 3×Aᵣ(O)=3×16=48 相加，得到 Mᵣ = 40 + 12 + 48 = 100。

Always use the Aᵣ values given in the Periodic Table in your exam. For molecules like H₂O, Mᵣ = 2×1 + 16 = 18. For ionic compounds such as Na₂SO₄, Mᵣ = 2×23 + 32 + 4×16 = 142.

考试中务必使用周期表给出的 Aᵣ 值。对于分子如 H₂O，Mᵣ = 2×1 + 16 = 18；对于离子化合物如 Na₂SO₄，Mᵣ = 2×23 + 32 + 4×16 = 142。

2. Moles and Molar Mass | 摩尔与摩尔质量

The mole is the amount of substance that contains 6.02 × 10²³ particles (Avogadro’s number). Molar mass (M) is the mass of one mole of a substance, numerically equal to its relative formula mass but expressed in g mol⁻¹. For example, the molar mass of water is 18 g mol⁻¹.

摩尔是包含 6.02 × 10²³ 个微粒（阿伏伽德罗常数）的物质的量。摩尔质量（M）是一摩尔物质的质量，数值上等于其相对式量，但单位为 g mol⁻¹。例如水的摩尔质量是 18 g mol⁻¹。

number of moles (n) = mass (m) ÷ molar mass (M) n = m / M

This formula is the cornerstone of almost all stoichiometric calculations. Remember to convert mass to grams before using it.

这个公式是几乎所有化学计量计算的基础。使用前记得将质量换算为克。

3. Converting Mass to Moles and Vice Versa | 质量与摩尔的相互转换

To find the number of moles in a given mass, divide the mass by the molar mass. To find the mass from moles, multiply the number of moles by the molar mass.

计算一定质量中的摩尔数，用质量除以摩尔质量。已知摩尔数求质量，用摩尔数乘以摩尔质量。

Example: How many moles are present in 8.0 g of NaOH (Mᵣ = 40)?
n = m / M = 8.0 g ÷ 40 g mol⁻¹ = 0.20 mol.

例题：8.0 g NaOH（Mᵣ = 40）中含有多少摩尔？
n = m / M = 8.0 g ÷ 40 g mol⁻¹ = 0.20 mol。

Conversely, calculate the mass of 0.50 mol of CO₂ (Mᵣ = 44):
m = n × M = 0.50 mol × 44 g mol⁻¹ = 22 g.

反过来，计算 0.50 mol CO₂（Mᵣ = 44）的质量：
m = n × M = 0.50 mol × 44 g mol⁻¹ = 22 g。

4. Reacting Mass Calculations | 反应质量计算

Reacting mass problems require you to calculate the mass of a reactant needed or product formed from a given mass of another substance. The procedure always follows three steps:

反应质量计算要求你根据一种物质的质量，计算所需反应物或生成物的质量。步骤总是遵循三步：

Write the balanced chemical equation.
Convert the given mass to moles using n = m / M.
Use the mole ratio from the equation to find moles of the target substance, then convert to mass with m = n × M.

写出配平的化学方程式。
用 n = m / M 将已知质量转化为摩尔数。
利用方程式中的摩尔比求出目标物质的摩尔数，再用 m = n × M 转化为质量。

Worked example: What mass of magnesium oxide (MgO) is produced when 6.0 g of magnesium burns completely in oxygen? (Aᵣ: Mg = 24, O = 16)
Equation: 2Mg + O₂ → 2MgO
Moles of Mg = 6.0 g ÷ 24 g mol⁻¹ = 0.25 mol.
Mole ratio Mg : MgO = 2 : 2, so moles of MgO = 0.25 mol.
Mass of MgO = 0.25 mol × (24+16) g mol⁻¹ = 0.25 × 40 = 10 g.

例题：6.0 g 镁在氧气中完全燃烧，生成多少克氧化镁（MgO）？(Aᵣ: Mg=24, O=16)
方程式：2Mg + O₂ → 2MgO
Mg 的摩尔数 = 6.0 g ÷ 24 g mol⁻¹ = 0.25 mol。
摩尔比 Mg : MgO = 2 : 2，因此 MgO 的摩尔数 = 0.25 mol。
MgO 的质量 = 0.25 mol × (24+16) g mol⁻¹ = 0.25 × 40 = 10 g。

5. Gas Volume Calculations at RTP | 常温常压下气体体积计算

At room temperature and pressure (RTP, about 25 °C and 1 atm), one mole of any gas occupies a volume of 24 dm³ (or 24,000 cm³). This is called the molar gas volume. Use the formula:

在常温常压（RTP，约 25 °C 和 1 atm）下，一摩尔任何气体的体积为 24 dm³（或 24,000 cm³）。这称为摩尔气体体积。使用公式：

volume of gas (dm³) = number of moles × 24 dm³ mol⁻¹

If the volume is in cm³, remember 1 dm³ = 1000 cm³, so number of moles = volume in cm³ ÷ 24,000.

如果体积单位是 cm³，记住 1 dm³ = 1000 cm³，因此摩尔数 = 体积(cm³) ÷ 24,000。

Example: Calculate the volume of CO₂ produced at RTP when 10 g of CaCO₃ decomposes. (Mᵣ(CaCO₃)=100)
CaCO₃ → CaO + CO₂
Moles of CaCO₃ = 10 g ÷ 100 g mol⁻¹ = 0.10 mol.
Mole ratio CaCO₃ : CO₂ = 1 : 1, so 0.10 mol of CO₂ is formed.
Volume of CO₂ = 0.10 mol × 24 dm³ mol⁻¹ = 2.4 dm³.

例题：10 g CaCO₃ 分解时，在 RTP 下产生多少体积的 CO₂？(Mᵣ(CaCO₃)=100)
CaCO₃ → CaO + CO₂
CaCO₃ 的摩尔数 = 10 g ÷ 100 g mol⁻¹ = 0.10 mol。
摩尔比 CaCO₃ : CO₂ = 1 : 1，所以生成 0.10 mol CO₂。
CO₂ 体积 = 0.10 mol × 24 dm³ mol⁻¹ = 2.4 dm³。

6. Concentration of Solutions | 溶液的浓度

Concentration is usually expressed in mol dm⁻³ (molarity) or in g dm⁻³. The relationship is:

浓度通常用 mol dm⁻³（摩尔浓度）或 g dm⁻³ 表示。它们的关系为：

concentration (mol dm⁻³) = number of moles ÷ volume (dm³)
concentration (g dm⁻³) = mass (g) ÷ volume (dm³)

To convert between them, use the molar mass: concentration in mol dm⁻³ × molar mass = concentration in g dm⁻³.

两者之间的换算利用摩尔质量：mol dm⁻³ 浓度 × 摩尔质量 = g dm⁻³ 浓度。

Example: 5.85 g of NaCl is dissolved in water to make 250 cm³ of solution. Find its concentration in mol dm⁻³. (Mᵣ(NaCl)=58.5)
Volume in dm³ = 250 ÷ 1000 = 0.250 dm³.
Moles of NaCl = 5.85 g ÷ 58.5 g mol⁻¹ = 0.100 mol.
Concentration = 0.100 mol ÷ 0.250 dm³ = 0.400 mol dm⁻³.

例题：5.85 g NaCl 溶于水配成 250 cm³ 溶液，求其浓度（mol dm⁻³）。(Mᵣ(NaCl)=58.5)
体积 dm³ = 250 ÷ 1000 = 0.250 dm³。
NaCl 摩尔数 = 5.85 g ÷ 58.5 g mol⁻¹ = 0.100 mol。
浓度 = 0.100 mol ÷ 0.250 dm³ = 0.400 mol dm⁻³。

7. Titration Calculations | 滴定计算

Titration calculations rely on the mole ratio in the neutralisation equation. The core formula is:

滴定计算依赖于中和反应方程式中的摩尔比。核心公式为：

nₐ = cₐ × Vₐ and n_b = c_b × V_b

where c is concentration in mol dm⁻³ and V is volume in dm³ (so divide cm³ by 1000). Then use the mole ratio from the balanced equation to find the unknown.

其中 c 是浓度（mol dm⁻³），V 是体积（dm³，需将 cm³ 除以 1000）。再利用配平方程式中的摩尔比求出未知量。

Example: 25.0 cm³ of NaOH solution is neutralised by 20.0 cm³ of 0.100 mol dm⁻³ HCl. Calculate the concentration of NaOH.
HCl + NaOH → NaCl + H₂O
Moles of HCl = 0.100 mol dm⁻³ × (20.0 ÷ 1000) dm³ = 0.00200 mol.
Mole ratio HCl : NaOH = 1 : 1, so moles of NaOH = 0.00200 mol.
Concentration of NaOH = 0.00200 mol ÷ (25.0 ÷ 1000) dm³ = 0.0800 mol dm⁻³.

例题：25.0 cm³ NaOH 溶液被 20.0 cm³ 0.100 mol dm⁻³ HCl 中和。计算 NaOH 的浓度。
HCl + NaOH → NaCl + H₂O
HCl 的摩尔数 = 0.100 mol dm⁻³ × (20.0 ÷ 1000) dm³ = 0.00200 mol。
摩尔比 HCl : NaOH = 1 : 1，因此 NaOH 的摩尔数 = 0.00200 mol。
NaOH 浓度 = 0.00200 mol ÷ (25.0 ÷ 1000) dm³ = 0.0800 mol dm⁻³。

8. Percentage Yield | 产率计算

The percentage yield compares the actual mass of product obtained in an experiment to the theoretical mass calculated from the balanced equation. It is never greater than 100% due to incomplete reactions, side reactions, or product loss during purification.

产率是将实验实际得到的产品质量与根据配平方程式计算出的理论质量进行比较。由于反应不完全、副反应或纯化过程中的损耗，产率通常不会超过 100%。

percentage yield = (actual yield ÷ theoretical yield) × 100%

Example: In an experiment, 4.8 g of copper was heated with excess sulfur, and 5.6 g of copper(II) sulfide was obtained. Theoretical yield of CuS is 7.2 g. Calculate the percentage yield.
Percentage yield = (5.6 ÷ 7.2) × 100% = 77.8%.

例题：实验中，4.8 g 铜与过量硫加热反应，得到 5.6 g 硫化铜(II)。CuS 的理论产量为 7.2 g。计算产率。
产率 = (5.6 ÷ 7.2) × 100% = 77.8%。

Always use the same units for actual and theoretical yield; both must be masses (g or kg).

实际产量和理论产量的单位必须一致，通常都用质量（g 或 kg）。

9. 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 atoms 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。分子式则表示一个分子中原子的真实数目，是经验式的整数倍。

To find the empirical formula:

Divide the mass (or percentage) of each element by its Aᵣ.
Divide each result by the smallest number of moles to get the simplest ratio.
If necessary, multiply to obtain whole numbers.

求经验式的步骤：

用各元素的质量（或百分比）除以各自的 Aᵣ。
将每个结果除以最小的摩尔数，得到最简比。
如有需要，乘以倍数得到整数。

Example: A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Find its empirical formula. (Aᵣ: C=12, H=1, O=16)
C: 40.0 ÷ 12 = 3.33 H: 6.7 ÷ 1 = 6.70 O: 53.3 ÷ 16 = 3.33
Divide by smallest (3.33): C=1, H=2.01≈2, O=1. So empirical formula is CH₂O.
If the relative molecular mass is 60, molecular formula = (CH₂O) × n, n = 60 ÷ 30 = 2 → C₂H₄O₂.

例题：某化合物含碳 40.0%、氢 6.7%、氧 53.3%（质量分数），求其经验式。(Aᵣ: C=12, H=1, O=16)
C: 40.0 ÷ 12 = 3.33 H: 6.7 ÷ 1 = 6.70 O: 53.3 ÷ 16 = 3.33
除以最小值 (3.33)：C=1，H≈2，O=1。经验式为 CH₂O。
若其相对分子质量为 60，分子式 = (CH₂O) × n，n = 60 ÷ 30 = 2 → C₂H₄O₂。

10. Limiting Reactants | 限量反应物

In many reactions, one reactant is used up before the others. The reactant that is completely consumed is the limiting reactant; it determines the maximum amount of product. The other reactants are in excess.

很多反应中，一种反应物会先被耗尽。完全消耗的反应物称为限量反应物，它决定了产物的最大量。其他反应物则过量。

To identify the limiting reactant:

Convert the masses of both reactants to moles.
Use the mole ratio from the balanced equation to find which one produces less product.
The one producing the smaller amount of product is limiting.

确定限量反应物的方法：

把两种反应物的质量都转化为摩尔数。
利用配平方程式中的摩尔比，找出生成产物较少的一方。
生成较少产物的反应物即为限量反应物。

Example: 2.0 g of hydrogen reacts with 16.0 g of oxygen to form water. Which is limiting? (2H₂ + O₂ → 2H₂O)
Moles of H₂ = 2.0 ÷ 2 = 1.0 mol; moles of O₂ = 16.0 ÷ 32 = 0.50 mol.
From equation, 2 mol H₂ react with 1 mol O₂, so 1.0 mol H₂ would need 0.50 mol O₂.
We have exactly 0.50 mol O₂, so neither is in excess — they are in perfect stoichiometric ratio, but if H₂ were 1.2 mol, O₂ would be limiting.

例题：2.0 g 氢气与 16.0 g 氧气反应生成水。哪个是限量反应物？(2H₂ + O₂ → 2H₂O)
H₂ 的摩尔数 = 2.0 ÷ 2 = 1.0 mol；O₂ 的摩尔数 = 16.0 ÷ 32 = 0.50 mol。
根据方程式，2 mol H₂ 与 1 mol O₂ 反应，因此 1.0 mol H₂ 需要 0.50 mol O₂。
我们恰好有 0.50 mol O₂，两者恰好完全反应；但如果 H₂ 是 1.2 mol，O₂ 就会成为限量反应物。

Always base the calculation of theoretical yield on the limiting reactant.

理论产量的计算必须基于限量反应物。

11. Water of Crystallisation Calculations | 结晶水计算

Many salts contain water molecules as part of their crystal structure, e.g., CuSO₄·5H₂O. You may be asked to determine the value of x in a hydrated salt formula.

许多盐含有结晶水，例如 CuSO₄·5H₂O。题目可能要求你确定水合物化学式中 x 的值。

Method: Heat the hydrated salt to drive off the water and measure the mass loss. The mass of anhydrous salt remains. Convert both masses to moles and find their ratio.

方法：加热水合盐除去水分，称量损失的质量，得到无水盐的质量。将两者的质量转化为摩尔数，再求比值。

Example: 5.00 g of hydrated sodium carbonate, Na₂CO₃·xH₂O, was heated until all water was lost. The anhydrous salt weighed 1.85 g. Find x. (Mᵣ: Na₂CO₃=106, H₂O=18)
Mass of water lost = 5.00 − 1.85 = 3.15 g.
Moles of Na₂CO₃ = 1.85 ÷ 106 ≈ 0.01745 mol.
Moles of H₂O = 3.15 ÷ 18 = 0.175 mol.
Ratio H₂O : Na₂CO₃ = 0.175 ÷ 0.01745 ≈ 10. So x = 10, formula is Na₂CO₃·10H₂O.

例题：5.00 g 水合碳酸钠 Na₂CO₃·xH₂O 加热至完全失去水分，无水盐质量为 1.85 g。求 x。(Mᵣ: Na₂CO₃=106, H₂O=18)
失去水的质量 = 5.00 − 1.85 = 3.15 g。
Na₂CO₃ 摩尔数 = 1.85 ÷ 106 ≈ 0.01745 mol。
H₂O 摩尔数 = 3.15 ÷ 18 = 0.175 mol。
H₂O : Na₂CO₃ 的比值 = 0.175 ÷ 0.01745 ≈ 10。因此 x = 10，化学式为 Na₂CO₃·10H₂O。

12. Summary and Key Tips | 总结与关键技巧

Always show your working clearly, including units. Memorise the core formulas: n = m / M, volume of gas = n × 24 dm³ at RTP, concentration = n / V. Check that your answer is sensible — for example, a yield above 100% means you have made an error. Practice regularly with past paper questions to build speed and accuracy.

作答时务必清晰地展示计算过程并注明单位。牢记核心公式：n = m / M，气体体积 = n × 24 dm³ (RTP)，浓度 = n / V。检查答案是否合理——例如产率超过 100% 就说明有错误。定期练习历年真题，提高速度和准确度。

Mastering these calculation techniques will give you a solid foundation not only for IGCSE but also for further study in chemistry.

掌握这些计算技巧，不仅能为 IGCSE 打下坚实基础，也为后续化学学习铺平道路。

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IGCSE Chemistry: Calculation Practice | IGCSE 化学：计算题专项训练