📚 GCSE Edexcel Chemistry: Mastering Calculations | GCSE Edexcel 化学计算题专项训练
Calculation questions are a core part of GCSE Edexcel Chemistry, linking abstract chemical ideas to measurable quantities. They appear across topics such as atomic structure, quantitative chemistry, energy changes and chemical analysis. Mastering the key formula, converting units correctly and showing clear working will help you secure up to 30% of your final grade.
计算题是 GCSE Edexcel 化学的核心组成部分，它将抽象的化学概念与可量化的数据联系起来。这类题目出现在原子结构、定量化学、能量变化和化学分析等多个主题中。熟练掌握关键公式、正确转换单位并展示清晰的解题步骤，将帮助你稳拿最多 30% 的总分。
This article breaks down the most common calculation types into bite-sized, bilingual explanations. For each section, the English explanation is immediately followed by its Chinese equivalent, so you can strengthen both your subject knowledge and your academic language skills.
本文把最常见的计算类型拆解成小模块，用双语逐一解释。每一节都是英文说明之后紧跟着中文配对，既能巩固化学知识，也能提升学术语言能力。
1. Relative Atomic & Formula Mass | 相对原子质量与相对分子质量
Relative atomic mass (Aᵣ) compares the average mass of an atom of an element to 1/12th the mass of a carbon-12 atom. It has no units and is found on the Periodic Table. For chlorine, Aᵣ = 35.5 because of the mixture of isotopes Cl-35 and Cl-37.
相对原子质量 (Aᵣ) 是元素的一个原子的平均质量与一个碳‑12 原子质量的 1/12 的比值。Aᵣ 没有单位，可在元素周期表中查找。例如氯的 Aᵣ = 35.5，因为天然氯是 Cl‑35 和 Cl‑37 两种同位素的混合物。
Relative formula mass (Mᵣ) is the sum of the Aᵣ values of all atoms shown in the formula. For Mg(OH)₂, Mᵣ = 24.3 + 2×(16.0+1.0) = 58.3. Always multiply the bracket contents before adding.
相对分子质量 (Mᵣ) 是化学式中各原子 Aᵣ 的总和。例如 Mg(OH)₂ 的 Mᵣ = 24.3 + 2×(16.0+1.0) = 58.3。计算时先乘以括号内原子的质量再相加。
2. The Mole and Molar Mass | 摩尔与摩尔质量
One mole of a substance contains exactly 6.02 × 10²³ particles (Avogadro constant). The molar mass of a substance is its Mᵣ expressed in grams. For example, the molar mass of water (H₂O) is 18 g/mol.
一摩尔任何物质含有 6.02 × 10²³ 个微粒（阿伏伽德罗常数）。物质的摩尔质量是把它的 Mᵣ 以克为单位表示出来的值。例如水 (H₂O) 的摩尔质量为 18 g/mol。
The key equation linking mass, moles and molar mass is:
mass (g) = moles × molar mass (g/mol)
连接质量、摩尔和摩尔质量的关键公式如下：
质量 (g) = 摩尔数 × 摩尔质量 (g/mol)
Rearrange it to find moles = mass / molar mass. Always check that your mass is in grams — convert from kg or mg if needed.
可变形为 摩尔数 = 质量 / 摩尔质量。计算前务必把质量单位转换成克，如果题目给出的单位是 kg 或 mg 要先行换算。
3. 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. To find it, convert the mass of each element to moles, divide by the smallest number of moles, and write the ratio as integers.
经验式表示化合物中各元素原子的最简整数比。求经验式的方法：将各元素的质量换算成摩尔数，除以最小的摩尔数，再把比例写成最简整数。
A worked example: a compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Assume 100 g, so you have 40.0 g C, 6.7 g H and 53.3 g O. Moles C = 40.0/12 = 3.33; H = 6.7/1 = 6.7; O = 53.3/16 = 3.33. Divide by 3.33 gives C:H:O = 1:2:1, so empirical formula is CH₂O.
例题：某化合物含碳 40.0%、氢 6.7%、氧 53.3%（质量分数）。假定样品 100 g，则含碳 40.0 g、氢 6.7 g、氧 53.3 g。碳的摩尔数 = 40.0/12 = 3.33；氢 = 6.7/1 = 6.7；氧 = 53.3/16 = 3.33。同除以 3.33 得 C:H:O = 1:2:1，因此经验式为 CH₂O。
The molecular formula is the actual number of atoms in a molecule. It is found by comparing the empirical formula mass with the given Mᵣ. If Mᵣ of the compound is 180, then n = 180 / 30 = 6, so molecular formula = C₆H₁₂O₆.
分子式是分子中原子的实际数目。将经验式的质量与题中给出的相对分子质量 Mᵣ 比较：若该化合物的 Mᵣ = 180，则 n = 180 / 30 = 6，分子式变为 C₆H₁₂O₆。
4. Reacting Masses | 反应质量计算
Reacting mass calculations use the balanced equation to find how much of a product is formed from a given mass of reactant. The steps are: write the balanced equation, find the moles of the known substance, use the mole ratio to find moles of the unknown, then convert to mass.
反应质量计算依据配平的化学方程式，由已知反应物的质量求生成物的质量。步骤是：写出配平方程式，计算已知物质的摩尔数，利用化学计量数之比求出未知物的摩尔数，再转换为质量。
For example, 2Mg + O₂ → 2MgO. How many grams of MgO are made from 6.0 g of Mg? Moles of Mg = 6.0 / 24 = 0.25 mol. Ratio Mg:MgO = 1:1, so moles of MgO = 0.25 mol. Mᵣ of MgO = 40, so mass = 0.25 × 40 = 10 g.
例如反应 2Mg + O₂ → 2MgO，用 6.0 g 镁能生成多少克 MgO？镁的摩尔数 = 6.0 / 24 = 0.25 mol。Mg 与 MgO 的计量数之比为 1:1，故 MgO 的摩尔数 = 0.25 mol。MgO 的 Mᵣ = 40，质量 = 0.25 × 40 = 10 g。
When the ratio is not 1:1, always scale the moles carefully. Measure all atomic masses to one decimal place as in your data booklet.
当化学计量数比不是 1:1 时，一定要按比例仔细放大摩尔数。所有相对原子质量按数据手册取一位小数。
5. 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 maximum theoretical mass from the reacting mass calculation. It shows how efficient the reaction was in practice.
Percentage yield = (actual yield / theoretical yield) × 100
百分产率是将实际得到的产物质量与根据反应质量计算出的最大理论产量进行比较。它反映了反应实际操作中的效率。
百分产率 = (实际产量 / 理论产量) × 100
Atom economy measures how much of the total mass of reactants ends up in the desired product. It is a key concept in green chemistry.
Atom economy = (Mᵣ of desired product / sum of Mᵣ of all reactants) × 100
原子经济衡量的是总反应物质量中有多少进入了目标产物，这是绿色化学中的一个重要概念。
原子经济 = (目标产物的 Mᵣ / 所有反应物的 Mᵣ 总和) × 100
For instance, in the reaction CuO + H₂SO₄ → CuSO₄ + H₂O, desired product is CuSO₄ (Mᵣ = 159.6). Sum of reactant Mᵣ = 79.5 + 98.1 = 177.6. Atom economy = (159.6 / 177.6) × 100 = 89.9%.
例如反应 CuO + H₂SO₄ → CuSO₄ + H₂O，目标产物是 CuSO₄ (Mᵣ = 159.6)。反应物 Mᵣ 总和 = 79.5 + 98.1 = 177.6。原子经济 = (159.6 / 177.6) × 100 = 89.9%。
6. Concentration in mol/dm³ | 摩尔浓度
Concentration expresses how much solute is dissolved in a given volume of solution. The standard unit is mol/dm³ (often read as moles per cubic decimetre).
concentration (mol/dm³) = moles of solute / volume (dm³)
浓度表示在一定体积的溶液中溶有多少溶质，常用单位是 mol/dm³（读作摩尔每立方分米）。
浓度 (mol/dm³) = 溶质的摩尔数 / 体积 (dm³)
You may also see concentration given in g/dm³. To convert between them, use the molar mass: concentration in mol/dm³ = concentration in g/dm³ / molar mass.
有时浓度会以 g/dm³ 给出。两者换算时可借助摩尔质量：mol/dm³ 表示的浓度 = g/dm³ 表示的浓度 / 摩尔质量。
Remember that 1 dm³ = 1000 cm³. If a volume is given in cm³, divide by 1000 to convert to dm³ before using the formula.
注意 1 dm³ = 1000 cm³。如果题目给出的体积单位是 cm³，要先除以 1000 转换为 dm³ 后再代入公式。
7. Titration Calculations | 滴定计算
Titration calculations let you find an unknown concentration using a solution of known concentration. The key formula uses the balanced equation and the average titre volume.
滴定计算利用已知浓度的溶液来求未知溶液的浓度。关键是要结合配平的化学方程式和平均滴定体积。
For an acid‑base titration such as HCl + NaOH → NaCl + H₂O, the mole ratio is 1:1. If 25.0 cm³ of NaOH is neutralised by 30.0 cm³ of 0.100 mol/dm³ HCl, the concentration of NaOH is found by: moles of HCl = 0.100 × (30.0/1000) = 0.00300 mol. Thus moles of NaOH = 0.00300 mol, so [NaOH] = 0.00300 / (25.0/1000) = 0.120 mol/dm³.
以酸碱滴定 HCl + NaOH → NaCl + H₂O 为例，计量数比为 1:1。若 25.0 cm³ NaOH 恰好被 30.0 cm³ 0.100 mol/dm³ 的 HCl 中和，则 HCl 的摩尔数 = 0.100 × (30.0/1000) = 0.00300 mol。因此 NaOH 的摩尔数也为 0.00300 mol，[NaOH] = 0.00300 / (25.0/1000) = 0.120 mol/dm³。
Always convert cm³ to dm³. When the ratio is not 1:1, multiply or divide by the appropriate coefficients from the equation. Show your step‑by‑step working clearly to gain method marks.
务必把 cm³ 换算成 dm³。当化学计量数比不是 1:1 时，需根据方程式中的系数进行乘除运算。清晰展示逐步计算能帮助你拿到过程分。
8. Gas Volumes and Molar Gas Volume | 气体体积与摩尔体积
At room temperature and pressure (rtp, about 20 °C and 1 atm), one mole of any gas occupies 24 dm³. This is the molar gas volume. You can use this fact to link moles and volume directly.
在常温常压下 (rtp, 大约 20 °C、1 atm)，一摩尔任何气体的体积均为 24 dm³，这就是气体摩尔体积。利用它可以直 接关联摩尔数与气体体积。
volume of gas (dm³) = moles of gas × 24
气体体积 (dm³) = 气体的摩尔数 × 24
For example, what volume of CO₂ is produced when 10 g of CaCO₃ decomposes? CaCO₃ → CaO + CO₂. Moles of CaCO₃ = 10 / 100 = 0.10 mol. Ratio 1:1, so moles of CO₂ = 0.10 mol. Volume = 0.10 × 24 = 2.4 dm³.
例如 10 g CaCO₃ 分解会生成多大体积的 CO₂？反应为 CaCO₃ → CaO + CO₂。CaCO₃ 的摩尔数 = 10 / 100 = 0.10 mol。化学计量数比 1:1，故 CO₂ 的摩尔数 = 0.10 mol。体积 = 0.10 × 24 = 2.4 dm³。
If the question asks for volume in cm³, multiply the result in dm³ by 1000. You may also need to use the ideal gas equation in more advanced questions, but at GCSE, the 24 dm³ rule is sufficient.
如果题目要求给出 cm³ 单位的体积，将 dm³ 的结果乘以 1000 即可。在更高层级的题目中可能需要理想气体方程，但在 GCSE 阶段使用 24 dm³ 规则已经足够。
9. Energy Changes Using Bond Enthalpies | 键焓计算能量变化
Bond enthalpy is the energy needed to break one mole of a covalent bond, measured in kJ/mol. In any reaction, bonds are broken in the reactants and new bonds are formed in the products.
键焓是断开一摩尔共价键所需要的能量，单位为 kJ/mol。任何化学反应中，旧键在反应物中断裂，新键在生成物中形成。
ΔH = total energy absorbed to break bonds − total energy released when forming bonds
ΔH = 断键吸收的总能量 − 成键释放的总能量
If more energy is released in bond making than is used in bond breaking, the reaction is exothermic (negative ΔH). If less is released, it is endothermic (positive ΔH).
若成键释放的能量大于断键所需的能量，反应为放热反应（ΔH 为负）；反之则为吸热反应（ΔH 为正）。
In an example, H₂ + Cl₂ → 2HCl. Bond breaking: 1 H–H (436 kJ) + 1 Cl–Cl (242 kJ) = 678 kJ absorbed. Bond forming: 2 H–Cl bonds, 2 × 431 = 862 kJ released. ΔH = 678 − 862 = −184 kJ/mol, so exothermic.
以 H₂ + Cl₂ → 2HCl 为例：断键吸收 1 个 H–H (436 kJ) + 1 个 Cl–Cl (242 kJ) = 678 kJ；成键形成 2 个 H–Cl，释放 2 × 431 = 862 kJ。ΔH = 678 − 862 = −184 kJ/mol，为放热反应。
Always draw the displayed formulae to count the exact number of each bond type. Triple or double bonds must have their correct bond enthalpies applied.
一定要画出结构式来准确统计各类键的数量。双键和三键必须代入对应的键焓数值，不能按单键处理。
10. Top Tips for Calculation Questions | 计算题高分技巧
Start by underlining the quantities given in the question and the unit required in the answer. Convert all units to the standard form — grams, dm³, or mol — before substituting into any formula.
首先圈划题目给出的数据和所求答案的单位。在代入任何公式之前，把所有单位转换为标准形式 — g、dm³ 或 mol。
Write the equation or formula at the beginning of your working. Even if you make a numerical mistake, an exam marker can award marks for a correct method.
在解题开头先把方程或公式写下。即使数字计算有误，阅卷老师也会根据正确的方法给予步骤分。
Check the stoichiometric ratio from the balanced equation carefully. A slip in the ratio is one of the most common errors in reacting mass and titration questions.
仔细核对配平方程中的化学计量数比。在反应质量和滴定计算中，弄错比例是最常见的错误之一。
Finally, reflect on whether your final answer is sensible. A yield above 100% or a negative mass should prompt you to re‑check your working.
最后，反思一下答案是否合理。如果产率超过 100% 或质量出现负值，就应该回头检查计算过程。
With regular practice of these calculation types, you will build speed and confidence. Use past paper questions to test yourself under timed conditions.
通过经常练习上述各种计算类型，你的解题速度和信心都会提升。可以用历年真题在限时条件下进行自我检测。
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