📚 IGCSE CCEA Chemistry: Calculation Practice | IGCSE CCEA 化学：计算题专项训练
Chemical calculations lie at the heart of IGCSE Chemistry. From determining the amount of a substance in moles to predicting yields and concentrations, a confident grasp of numerical problems is essential for success in the CCEA examination. This guide brings together the most common calculation types, with clear step-by-step examples that reinforce both understanding and exam technique.
化学计算是 IGCSE 化学的核心内容。从确定物质的量（摩尔）到预测产率和浓度，扎实掌握数值题目是在 CCEA 考试中取得高分的关键。本文汇集了最常出现的计算类型，并通过清晰的逐步示例巩固理解和应试技巧。
1. Relative Formula Mass (Mᵣ) Calculations | 相对式量 (Mᵣ) 的计算
Relative formula mass (Mᵣ) is the sum of the relative atomic masses (Aᵣ) of all atoms in a formula unit. It has no units because it compares the mass to one-twelfth of a carbon‑12 atom. For ionic compounds the term 'formula mass' is used, but the calculation is identical. Mᵣ is the foundation of every mole‑based problem.
相对式量（Mᵣ）是化学式单元中所有原子的相对原子质量（Aᵣ）的总和。它没有单位，因为它是与一个碳‑12 原子质量的十二分之一比较得出的。对于离子化合物使用“式量”这一术语，但计算方法完全相同。Mᵣ 是所有摩尔计算的基础。
Calculate Mᵣ by multiplying each element' Mg) = 24, Aᵣ(O) = 16, so Mᵣ = 24 + 16 = 40. For CaCO₃: Ca (40) + C (12) + 3×16 (48) = 100.
计算 Mᵣ 时，将每种元素的 Aᵣ 乘以其下标数，然后相加。对于 MgO：Aᵣ(Mg) = 24，Aᵣ(O) = 16，因此 Mᵣ = 24 + 16 = 40。对于 CaCO₃：Ca (40) + C (12) + 3×16 (48) = 100。
Mᵣ = Σ (Aᵣ of each element × number of atoms)
| Compound | Breakdown | Mᵣ |
|---|---|---|
| NaCl | 23 + 35.5 | 58.5 |
| H₂SO₄ | (2×1) + 32 + (4×16) | 98 |
| Al₂(SO₄)₃ | (2×27) + 3[32+(4×16)] | 342 |
2. Empirical Formula | 实验式（最简式）的计算
The empirical formula shows the simplest whole‑number ratio of atoms of each element in a compound. It is determined from percentage composition or directly from experimental mass data. The molecular formula is a whole‑number multiple of the empirical formula.
实验式表示化合物中各元素原子的最简整数比。它根据元素的质量百分比或直接的实验质量数据确定。分子式是实验式的整数倍。
3) if necessary, multiply through by a small integer to obtain whole numbers. Worked example: a compound contains 40.0% carbon, 6.7% hydrogen and 53.3% 6.7, O = 53.3÷16 = 3.33. Step 2: divide by 3.33 → C:1, H:2, O:1. Empirical formula = CH₂O.
计算实验式的步骤：(1) 用各元素的质量（或百分比）除以各自的 Aᵣ；(2) 将所得每个结果除以最小值；(3) 如果必要，乘以一个小整数以得到整数值。示例：某化合物含碳 40.0%、氢 6.7%、氧 53.3%。第一步：C = 40.0÷12 = 3.33，H = 6.7÷1 = 6.7，O = 53.3÷16 = 3.33。第二步：除以 3.33 → C:1，H:2，O:1。实验式 = CH₂O。
moles of element = mass (g) ÷ Aᵣ
3. Calculating Moles from Mass | 已知质量求物质的量
The mole is the SI unit for 'amount of substance' les (Avogadro constant). The formula linking mass, moles and molar mass (M, numerically equal to Mᵣ in g mol⁻¹) is used constantly in quantitative chemistry.
摩尔是“物质的量”的国际单位。1 摩尔含有 6.02 × 10²³ 个微粒（阿伏伽德罗常数）。质量、摩尔数与摩尔质量（M，以 g mol⁻¹ 为单位，数值等于 Mᵣ）之间的关系式在定量化学中不断被用到。
n = m ÷ M
mol), m = mass (g), M = molar mass (g mol⁻¹). Example: how many moles are in 8.0 g of NaOH? M = 40 g mol⁻¹. n = 8.0 ÷ 40 = 0.20 mol. For converting moles back to mass, use m = n × M.
其中 n = 物质的量 (mol)，m = 质量 (g)，M = 摩尔质量 (g mol⁻¹)。示例：8.0 g NaOH 中含有多少摩尔？Mᵣ(NaOH)=23+16+1=40，M=40 g mol⁻¹。n = 8.0 ÷ 40 = 0.20 mol。要将摩尔转换回质量，使用 m = n × M。
4. Mole Ratios from Equations | 根据化学方程式求物质的量之比
A balanced equation tells us the ratio in which substances react and form. The coefficients in front of each formula give the mole ratio. This allows us to calculate the mass of a product from a given mass of reactant, or vice versa.
配平的化学方程式给出了各物质反应和生成的摩尔比。每个化学式前的系数表示了摩尔比例。据此，我们可以由已知反应物质量计算生成物质量，反之亦然。
ple: 2Mg + O₂ → 2MgO. How many grams of MgO are produced when 6.0 g of Mg burns completely? ss of MgO = 0.25 × 40 = 10 g. Always stick to these three stages: mass → moles → mole ratio → mass.
计算示例：2Mg + O₂ → 2MgO。当 6.0 g 镁完全燃烧时生成多少克氧化镁？Mᵣ(Mg)=24，Mᵣ(MgO)=40。步骤1：Mg 的物质的量 = 6.0 ÷ 24 = 0.25 mol。步骤2：根据方程式，Mg 与 MgO 的摩尔比为 2:2 = 1:1，因此 MgO 的物质的量 = 0.25 mol。步骤3：MgO 的质量 = 0.25 × 40 = 10 g。始终遵循这三个步骤：质量 → 物质的量 → 摩尔比 → 质量。
5. Concentration of Solutions | 溶液的浓度计算
and dm³ is vital: 1 dm³ = 1000 cm³. The key equation is n = c × V, where V must be in dm³.
溶液的浓度通常以 mol dm⁻³（摩尔每立方分米）表示。在 cm³ 和 dm³ 之间进行换算至关重要：1 dm³ = 1000 cm³。关键公式为 n = c × V，其中 V 必须以 dm³ 为单位。
n = c × V V(dm³) = V(cm³) ÷ 1000
Example: what mass of NaOH is needed to prepare 250 cm³ of 0.100 mol dm⁻³ solution? a titration, c₁V₁ = c₂V₂ (for a 1:1 reaction) can be used after adjusting for mole ratios.
示例：配制 250 cm³ 浓度为 0.100 mol dm⁻³ 的 NaOH 溶液需要多少质量的 NaOH？V = 250 ÷ 1000 = 0.250 dm³。n(NaOH) = 0.100 × 0.250 = 0.0250 mol。Mᵣ(NaOH)=40，因此质量 = 0.0250 × 40 = 1.0 g。另外，对于 1:1 反应，可通过滴定数据使用 c₁V₁ = c₂V₂，但需注意摩尔比。
6. Gas Volume Calculations (Molar Volume) | 气体体积计算（摩尔体积）
At room temperature and pressure (r.t.p.), one mole of any gas occupies 24 dm³ (24,000 cm³). This is the molar volume, Vₘ. Use it to interconvert moles and gas volumes. The relationship holds only for gases under the same conditions.
在室温和室压下（r.t.p.），任何气体 1 摩尔的体积为 24 dm³（24 000 cm³）。这就是摩尔体积 Vₘ。用它来进行摩尔与气体体积的相互换算。此关系只适用于相同条件下的气体。
volume (dm³) = n × 24 or n = volume (dm³) ÷ 24
Example: what volume of CO₂ is produced at r.t.p. when 10.0 g of CaCO₃ is heated? aCO₃ : CO₂ = 1:1, so n(CO₂) = 0.100 mol. Volume(CO₂) = 0.100 × 24 = 2.4 dm³ (or 2400 cm³).
示例：加热 10.0 g CaCO₃ 在 r.t.p. 下产生多大体积的 CO₂？CaCO₃ → CaO + CO₂。Mᵣ(CaCO₃)=100，n(CaCO₃) = 10.0 ÷ 100 = 0.100 mol。摩尔比 CaCO₃ : CO₂ = 1:1，因此 n(CO₂) = 0.100 mol。CO₂ 体积 = 0.100 × 24 = 2.4 dm³（或 2400 cm³）。
7. Percentage Yield and Atom Economy | 产率和原子经济性
Percentage yield compares the actual mass of product obtained to the theoretical mass predicted from the limiting reactant. It is always less than 100% in real experiments because of incomplete reactions, side reactions, or losses during separation.
产率是将实际获得的产品质量与根据限量试剂预测的理论质量进行比较。实际实验中由于反应不完全、副反应或分离过程中的损失，产率总是低于 100%。
% yield = (actual yield ÷ theoretical yield) × 100
Atom economy measures the efficiency of a reaction by considering how much of the reactants ends up in the desired product. Green chemistry aims for high atom economy.
原子经济性通过考虑有多少反应物最终进入目标产物来衡量反应的效率。绿色化学追求高原子经济性。
% atom economy = (Mᵣ of desired product ÷ sum of Mᵣ of all reactants) × 100
tom economy for Fe. Mᵣ(Fe₂O₃)=160, 3CO=84, total reactants=244. Desired product 2Fe = 112. % atom economy = (112 ÷ 244) × 100 ≈ 45.9%. The rest becomes waste CO₂.
示例：在炼铁反应 Fe₂O₃ + 3CO → 2Fe + 3CO₂ 中，计算铁（Fe）的原子经济性。Mᵣ(Fe₂O₃)=160，3CO=84，反应物总式量=244。目标产物 2Fe=112。原子经济性 = (112 ÷ 244) × 100 ≈ 45.9%。剩余部分成为废料 CO₂。
8. Titration Calculations | 滴定计算
In a titration, a solution of known concentration is used to determine the concentration of an unknown solution. The volume delivered at the equivalence point (end-point) is read from the burette. Multiple concordant titres are averaged for accuracy.
在滴定中，用一种已知浓度的溶液来确定未知溶液的浓度。在等当点（终点）时的滴定体积从滴定管读出。多个吻合的滴定读数取平均值以确保精度。
For an acid– oles of unknown; then c₂ = moles of unknown ÷ V₂ (dm³). If the ratio is not 1:1, adjust accordingly.
对于诸如 HCl + NaOH → NaCl + H₂O 的酸碱反应，摩尔比为 1:1。计算过程为：已知溶液的物质的量 = c₁ × V₁ (dm³)；利用摩尔比求出未知溶液的物质的量；然后 c₂ = 未知物的物质的量 ÷ V₂ (dm³)。如果不是 1:1 反应，需按比例调整。
× (23.40÷1000) = 0.00234 mol. Mole ratio 1:1, so n(NaOH) = 0.00234 mol. V(NaOH) = 25.0÷1000 = 0.0250 dm³. c(NaOH) = 0.00234 ÷ 0.0250 = 0.0936 mol dm⁻³. Always quote the final answer to an appropriate number of significant figures.
示例：25.0 cm³ NaOH 溶液需要用 23.40 cm³ 的 0.100 mol dm⁻³ HCl 恰好中和，计算 NaOH 的浓度。n(HCl) = 0.100 × (23.40÷1000) = 0.00234 mol。摩尔比为 1:1，因此 n(NaOH) = 0.00234 mol。V(NaOH) = 25.0÷1000 = 0.0250 dm³。c(NaOH) = 0.00234 ÷ 0.0250 = 0.0936 mol dm⁻³。最终答案应保留适当的有效数字。
9. Limiting Reactant and Excess | 限量试剂与过量试剂
When two or more reactants are mixed in non‑stoichiometric amounts, one will be used up first – the limiting reactant. The other is in excess. The amount of product formed is entirely determined by the limiting reactant. Identifying it correctly is essential.
当两种或两种以上反应物以非计量比混合时，其中一种会首先耗尽——即限量试剂。另一种则过量。生成物的量完全由限量试剂决定。正确识别限量试剂至关重要。
f Mg (Aᵣ=24) and 2.0 g of O₂ (Mᵣ=32) react: 2Mg + O₂ → 2MgO. Moles Mg = 2.4÷24 = 0.10 mol; moles O₂ = 2.0÷32 = 0.0625 mol. The equation needs 2 mol Mg per 1 mol O₂, so 0.10 mol Mg requires 0.050 mol O₂. Since 0.0625 mol O₂ is available, Mg is the limiting reactant and O₂ is in excess. The theoretical yield of MgO can then be calculated from the 0.10 mol of Mg.
方法：将每种反应物的质量转化为物质的量；将实际摩尔比与方程式的化学计量比进行比较。例如，2.4 g Mg (Aᵣ=24) 和 2.0 g O₂ (Mᵣ=32) 发生反应：2Mg + O₂ → 2MgO。Mg 的物质的量 = 2.4÷24 = 0.10 mol；O₂ 的物质的量 = 2.0÷32 = 0.0625 mol。方程式要求每 1 mol O₂ 需要 2 mol Mg，因此 0.10 mol Mg 需要 0.050 mol O₂。实际有 0.0625 mol O₂，所以 Mg 是限量试剂，O₂ 过量。然后可根据 0.10 mol Mg 计算 MgO 的理论产量。
10. Combined Calculations – A Step-by-Step Approach | 综合计算题 – 分步解析
Many exam questions combine several of the above skills. A typical problem might involve: using a titration to find the concentration of an acid, reacting it with a metal to collect a gas, and then calculating the percentage purity of the metal sample. The key is to break the problem into manageable stages and write down each conversion clearly.
许多考题会将多项上述技能综合在一起。一个典型题目可能包括：通过滴定确定一种酸的浓度，让酸与某种金属反应收集气体，然后计算该金属样品的纯度百分比。关键在于将题目分解为可操作的步骤，并清晰写出每一步的换算。
³ of H₂ at r.t.p. Calculate the percentage purity of the magnesium sample. Mg + 2HCl → MgCl₂ + H₂. Step 1: n(H₂) = volume ÷ 24000 = 480 ÷ 24000 = 0.0200 mol. Step 2: mole ratio Mg : H₂ = 1:1, so n(Mg) that reacted = 0.0200 mol. Step 3: mass of pure Mg = 0.0200 × 24.3 = 0.486 g. Step 4: % purity = (0.486 ÷ 0.500) × 100 = 97.2%. Notice that the HCl information was extra – you only need the gas volume if the acid is in excess.
综合示例：0.500 g 不纯的镁与 30.0 cm³ 的 2.00 mol dm⁻³ HCl（过量）反应。产生的气体在
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