GCSE CIE Chemistry: Formula Summary Handbook | GCSE CIE 化学：公式汇总手册

📚 GCSE CIE Chemistry: Formula Summary Handbook | GCSE CIE 化学：公式汇总手册

This article is a comprehensive revision guide covering all the essential formulae, mathematical relationships and quantitative concepts needed for the CIE GCSE Chemistry syllabus. Mastering these formulas is crucial for calculations involving the mole, reacting masses, concentrations, gas volumes, energy changes and more. Each section provides clear definitions, worked examples and practical tips to help you memorise and apply the equations confidently in your examinations.

本文是一份全面的复习指南，涵盖 CIE GCSE 化学大纲所需的全部重要公式、数学关系和定量概念。掌握这些公式对于进行摩尔计算、反应质量、浓度、气体体积、能量变化等至关重要。每个部分都提供清晰的定义、示例和实用技巧，帮助你记住这些方程并在考试中自信地应用。

1. Relative Atomic Mass and Relative Formula Mass | 相对原子质量与相对式量

The relative atomic mass (Aᵣ) of an element is the average mass of its atoms compared to 1/12ᵗʰ of the mass of a carbon‑12 atom. It has no units. When an element has several isotopes, Aᵣ is calculated from the mass and percentage abundance of each isotope using:

元素的相对原子质量 (Aᵣ) 是其原子的平均质量与一个碳‑12 原子质量的 1/12 相比较的数值，没有单位。当元素有多种同位素时，Aᵣ 根据每种同位素的质量和丰度百分比计算：

Aᵣ = Σ (isotopic mass × % abundance) / 100

For example, chlorine has two main isotopes: ³⁵Cl (75%) and ³⁷Cl (25%). Its Aᵣ = (35 × 75 + 37 × 25) / 100 = 35.5. The relative formula mass (Mᵣ) of a compound is the sum of the Aᵣ values of all the atoms shown in its formula. For magnesium chloride, MgCl₂: Mᵣ = 24 + (35.5 × 2) = 95.

例如，氯有两种主要同位素：³⁵Cl (75%) 和 ³⁷Cl (25%)。其 Aᵣ = (35 × 75 + 37 × 25) / 100 = 35.5。化合物的相对式量 (Mᵣ) 是其化学式中所有原子的 Aᵣ 总和。对于氯化镁 MgCl₂：Mᵣ = 24 + (35.5 × 2) = 95。

These values are obtained directly from the Periodic Table and are essential for all mole calculations. You must also be able to work out the percentage by mass of an element in a compound: % mass = (total Aᵣ of the element / Mᵣ of compound) × 100%.

这些数值直接从元素周期表获得，是所有摩尔计算的基础。你还必须能够计算化合物中某元素的质量百分比：质量% = (该元素的总 Aᵣ / 化合物的 Mᵣ) × 100%。

2. The Mole Concept | 摩尔的概念

The mole is the unit for amount of substance. One mole of any substance contains exactly 6.02 × 10²³ particles (Avogadro constant, Nₐ). The number of moles (n) is linked to mass (m in grams) by:

摩尔是物质的量的单位。一摩尔任何物质恰好包含 6.02 × 10²³ 个粒子（阿伏伽德罗常数，Nₐ）。摩尔数 (n) 与质量 (m，克) 的关系为：

n = m / Mᵣ

where Mᵣ is the molar mass in g/mol. To find the number of particles, use N = n × Nₐ. This formula allows you to convert between the mass of a substance, the number of moles and the number of atoms, molecules or ions.

其中 Mᵣ 是摩尔质量，单位为 g/mol。要计算粒子数，使用 N = n × Nₐ。这个公式可以使你在物质的质量、摩尔数以及原子、分子或离子数目之间进行转换。

For instance, how many moles are present in 8.0 g of sulfur (S₈)? Mᵣ of S₈ = 32 × 8 = 256, so n = 8.0 / 256 = 0.03125 mol. The number of S₈ molecules = 0.03125 × 6.02 × 10²³ = 1.88 × 10²². Moles are always at the heart of quantitative chemistry, so memorise n = m / Mᵣ and use it for all reacting mass problems.

例如，8.0 克硫 (S₈) 中有多少摩尔？S₈ 的 Mᵣ = 32 × 8 = 256，因此 n = 8.0 / 256 = 0.03125 mol。S₈ 分子数 = 0.03125 × 6.02 × 10²³ = 1.88 × 10²²。摩尔始终是定量化学的核心，务必牢记 n = m / Mᵣ，并将其用于所有反应质量问题。

3. Empirical and Molecular Formulae | 实验式与分子式

The empirical formula shows the simplest whole‑number ratio of atoms of each element in a compound. To find it from percentage or mass data, divide the mass (or percentage) by the Aᵣ of each element to obtain the mole ratio. Then divide by the smallest number of moles to get the simplest ratio.

实验式表示化合物中各元素原子的最简整数比。要从质量或百分比数据推求实验式，可将每种元素的质量（或百分比）除以其 Aᵣ 得到摩尔比，然后除以最小的摩尔数，得到最简比。

Example: A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Dividing by Aᵣ: 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 the empirical formula is CH₂O.

示例：某化合物含碳 40.0%，氢 6.7%，氧 53.3%。除以 Aᵣ：C → 40.0 / 12 = 3.33；H → 6.7 / 1 = 6.7；O → 53.3 / 16 = 3.33。除以 3.33 得 C:H:O = 1:2:1，所以实验式为 CH₂O。

The molecular formula gives the actual number of atoms in a molecule. It is a whole‑number multiple of the empirical formula. You need the Mᵣ of the compound: molecular formula = (empirical formula)ₓ where x = Mᵣ / empirical formula mass. If the Mᵣ of the above compound is 180, then x = 180 / (12+2+16) = 180 / 30 = 6, giving the molecular formula C₆H₁₂O₆.

分子式给出分子中实际的原子数目，它是实验式的整数倍。你需要知道化合物的 Mᵣ：分子式 = (实验式)ₓ，其中 x = Mᵣ / 实验式式量。如果上述化合物的 Mᵣ 为 180，则 x = 180 / (12+2+16) = 180 / 30 = 6，得到分子式 C₆H₁₂O₆。

4. Reacting Masses | 反应质量计算

In a chemical reaction, substances always react in fixed mole ratios, shown by the balanced equation. To calculate the mass of a reactant needed or product formed, follow three steps: convert the given mass to moles using n = m / Mᵣ; use the mole ratio from the equation to find moles of the unknown substance; convert moles back to mass using m = n × Mᵣ.

在化学反应中，物质总是按固定的摩尔比进行反应，这体现在配平的方程式中。要计算所需反应物或生成产物的质量，可以遵循三个步骤：使用 n = m / Mᵣ 将给定质量转换为摩尔数；利用方程式中的摩尔比求出未知物质的摩尔数；再用 m = n × Mᵣ 将摩尔数转换回质量。

For example, what mass of carbon dioxide is produced when 10.0 g of calcium carbonate decomposes? CaCO₃ → CaO + CO₂. Mᵣ of CaCO₃ = 100, so n(CaCO₃) = 10.0 / 100 = 0.10 mol. The mole ratio CaCO₃ : CO₂ is 1:1, so n(CO₂) = 0.10 mol. Mᵣ of CO₂ = 44, so mass of CO₂ = 0.10 × 44 = 4.4 g.

例如，10.0 克碳酸钙分解会产生多少质量的二氧化碳？CaCO₃ → CaO + CO₂。CaCO₃ 的 Mᵣ = 100，所以 n(CaCO₃) = 10.0 / 100 = 0.10 mol。摩尔比 CaCO₃ : CO₂ 为 1:1，因此 n(CO₂) = 0.10 mol。CO₂ 的 Mᵣ = 44，所以 CO₂ 的质量 = 0.10 × 44 = 4.4 g。

Always check that the equation is balanced and work carefully through the mole ratio. This method applies to any reacting mass calculation, including those involving limiting reactants.

务必确保方程式已配平，并仔细处理摩尔比。这种方法适用于任何反应质量计算，也包括涉及限量反应物的情况。

5. Concentration of Solutions | 溶液的浓度

The concentration of a solution can be expressed in mol/dm³ (molarity) or g/dm³. The key formula linking moles, concentration and volume is:

溶液的浓度可以用 mol/dm³（摩尔浓度）或 g/dm³ 表示。连接摩尔数、浓度和体积的关键公式为：

n = c × V

where n is the number of moles, c is the concentration in mol/dm³, and V is the volume in dm³. If the volume is given in cm³, convert it by dividing by 1000: V(dm³) = V(cm³) / 1000. You may also need the mass concentration: mass concentration (g/dm³) = mass (g) / volume (dm³), which is related to molarity by mass concentration = c × Mᵣ.

其中 n 是摩尔数，c 是浓度（mol/dm³），V 是体积（dm³）。如果体积以 cm³ 给出，需除以 1000 进行转换：V(dm³) = V(cm³) / 1000。你还可能需要质量浓度：质量浓度 (g/dm³) = 质量 (g) / 体积 (dm³)，它与摩尔浓度的关系为质量浓度 = c × Mᵣ。

For example, how many moles of HCl are in 25.0 cm³ of 0.500 mol/dm³ hydrochloric acid? V = 25.0 / 1000 = 0.0250 dm³; n = 0.500 × 0.0250 = 0.0125 mol. When diluting a solution, the number of moles stays the same, so c₁V₁ = c₂V₂. This dilution formula is especially useful in titration calculations.

例如，25.0 cm³ 的 0.500 mol/dm³ 盐酸中含有多少摩尔 HCl？V = 25.0 / 1000 = 0.0250 dm³；n = 0.500 × 0.0250 = 0.0125 mol。稀释溶液时，摩尔数保持不变，因此 c₁V₁ = c₂V₂。这个稀释公式在滴定计算中特别有用。

6. Molar Volume of Gases | 气体摩尔体积

At room temperature and pressure (RTP, which is 20 °C and 1 atm), one mole of any gas occupies a volume of 24 dm³ (24 000 cm³). This is known as the molar gas volume. The relationship is:

在常温常压（RTP，即 20 °C 和 1 atm）下，一摩尔任何气体的体积都是 24 dm³ (24 000 cm³)，这被称为气体摩尔体积。数量关系为：

V (dm³) = n × 24

You can use this to find the volume of gas produced or the number of moles from a measured volume. If you are asked about other conditions, the question will provide the molar volume; otherwise, assume 24 dm³/mol.

你可以利用它来求算生成气体的体积，或从已测体积求算摩尔数。如果题目涉及其他条件，题干会给出摩尔体积；否则，假设为 24 dm³/mol。

Example: What volume of hydrogen (at RTP) is made when 0.50 g of magnesium reacts with excess acid? Mg + 2HCl → MgCl₂ + H₂. Mᵣ of Mg = 24, so n(Mg) = 0.50 / 24 = 0.0208 mol. Mole ratio Mg : H₂ = 1:1, so n(H₂) = 0.0208 mol. Volume = 0.0208 × 24 = 0.50 dm³ (500 cm³). Remember that the molar volume only applies to gases, not to solids or liquids.

示例：0.50 克镁与过量酸反应，在 RTP 下会生成多少体积的氢气？Mg + 2HCl → MgCl₂ + H₂。Mg 的 Mᵣ = 24，所以 n(Mg) = 0.50 / 24 = 0.0208 mol。摩尔比 Mg : H₂ = 1:1，故 n(H₂) = 0.0208 mol。体积 = 0.0208 × 24 = 0.50 dm³ (500 cm³)。切记摩尔体积仅适用于气体，不适用于固体或液体。

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

The percentage yield describes how much of the expected product was actually obtained. It is calculated using the formula:

产率表示实际获得预期产物的百分比，其计算公式为：

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

Theoretical yield is the maximum mass of product calculated from the limiting reactant using reacting mass steps. Actual yield is the mass measured in the experiment. A yield less than 100% may be due to incomplete reaction, side reactions or losses during purification.

理论产量是根据限量反应物通过反应质量计算得到的最大产物质量。实际产量是实验中测得的质量。产率低于 100% 可能由于反应不完全、副反应或纯化过程中的损失。

Atom economy measures how efficiently atoms in the reactants are converted into the desired product. It is a key concept in green chemistry:

原子经济性衡量反应物中的原子转化为目标产物的效率，这是绿色化学的核心概念：

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

For example, in the reaction 2Na + Cl₂ → 2NaCl, the desired product is NaCl. Mᵣ(NaCl) = 58.5, and sum of reactants Mᵣ = (2 × 23) + 71 = 46 + 71 = 117. Atom economy = (117 / 117) × 100% = 100% because there is only one product. A higher atom economy means fewer waste atoms and a more sustainable process.

例如，在反应 2Na + Cl₂ → 2NaCl 中，目标产物 NaCl 的 Mᵣ = 58.5，但计算原子经济性需用所有产物 Mᵣ 的总和还是目标产物？公式为：所需产物的 Mᵣ / 所有反应物 Mᵣ 总和。这里反应生成 2NaCl，所需产物 Mᵣ 为 2 × 58.5 = 117。反应物 Mᵣ 总和 = (2 × 23) + 71 = 117，原子经济性 = 100%。较高的原子经济性意味着废物原子更少，过程更具可持续性。

8. Titration Calculations | 滴定计算

Titration is a technique used to find the concentration of an unknown solution. The calculation uses the same n = c × V relationship, together with the mole ratio from the balanced equation. The typical procedure is:

滴定是用于测定未知溶液浓度的技术。计算时使用相同的 n = c × V 关系，并利用配平方程式中的摩尔比。典型步骤如下：

Write the balanced equation for the reaction.

写出反应的配平方程式。
Record the volumes used and convert to dm³.

记录所用的体积并转换为 dm³。
Calculate moles of the known solution using n = cV.

使用 n = cV 计算已知溶液的摩尔数。
Use the mole ratio to find moles of the unknown solution.

利用摩尔比求出未知溶液的摩尔数。
Calculate its concentration using c = n / V.

用 c = n / V 计算其浓度。

For instance, 25.0 cm³ of NaOH is neutralised by 20.0 cm³ of 0.100 mol/dm³ H₂SO₄. Equation: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O. n(H₂SO₄) = 0.100 × 0.0200 = 0.00200 mol. Ratio NaOH : H₂SO₄ = 2:1, so n(NaOH) = 0.00400 mol. c(NaOH) = 0.00400 / 0.0250 = 0.160 mol/dm³.

例如，25.0 cm³ NaOH 被 20.0 cm³ 0.100 mol/dm³ H₂SO₄ 中和。方程式：2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O。n(H₂SO₄) = 0.100 × 0.0200 = 0.00200 mol。摩尔比 NaOH : H₂SO₄ = 2:1，所以 n(NaOH) = 0.00400 mol。c(NaOH) = 0.00400 / 0.0250 = 0.160 mol/dm³。

Always ensure you have identified the correct stoichiometric ratio. If the equation is not given, you must deduce it from the ions involved.

务必确保已识别出正确的化学计量比。如果题目未给方程式，你必须根据参与反应的离子推导。

9. Enthalpy Changes | 焓变

In CIE GCSE, you calculate the energy transferred in a reaction using the temperature change of a solution, usually water. The heat energy (q) absorbed or released is given by:

在 CIE GCSE 中，你利用溶液的温度变化（通常是水）来计算反应中传递的能量。吸收或放出的热量 (q) 计算公式为：

q = m × c × ΔT

where m is the mass of the solution (in g; for water 1 cm³ = 1 g), c is the specific heat capacity of water (4.18 J/g·°C) and ΔT is the temperature change (°C). The enthalpy change ΔH (in kJ/mol) is then:

其中 m 是溶液的质量（克；对于水，1 cm³ = 1 g），c 是水的比热容 (4.18 J/g·°C)，ΔT 是温度变化 (°C)。然后焓变 ΔH (kJ/mol) 为：

ΔH = –q / (1000 × n)

The negative sign indicates an exothermic reaction (temperature rises) gives a negative ΔH, while endothermic reactions have a positive ΔH.

负号表示放热反应（温度升高）给出负 ΔH，而吸热反应给出正 ΔH。

Example: 50 cm³ of 1.0 mol/dm³ HCl is neutralised by 50 cm³ of 1.0 mol/dm³ NaOH. The temperature rises by 6.8 °C. Total volume = 100 cm³, mass of solution ≈ 100 g. q = 100 × 4.18 × 6.8 = 2842 J. n(HCl) = 0.050 mol, so ΔH = –2842 / (1000 × 0.050) = –56.8 kJ/mol. This is the enthalpy of neutralisation.

示例：50 cm³ 1.0 mol/dm³ HCl 被 50 cm³ 1.0 mol/dm³ NaOH 中和，温度升高 6.8 °C。总体积 100 cm³，溶液质量 ≈ 100 g。q = 100 × 4.18 × 6.8 = 2842 J。n(HCl) = 0.050 mol，所以 ΔH = –2842 / (1000 × 0.050) = –56.8 kJ/mol。这就是中和焓。

10. Rate of Reaction | 反应速率

The rate of a chemical reaction tells us how quickly a reactant is used up or a product is formed. Two commonly used mathematical expressions are:

化学反应速率告诉我们反应物被消耗或产物生成的速度。常用的两种数学表达式为：

Mean rate = quantity of reactant used or product formed / time taken

Quantity can be expressed in mass (g), volume of gas (cm³) or concentration (mol/dm³). On a graph of amount versus time, the rate at any instant is given by the slope (gradient) of the tangent to the curve.

量可以用质量 (g)、气体体积 (cm³) 或浓度 (mol/dm³) 表示。在“量–时间”图上，任意时刻的速率等于曲线切线的斜率（梯度）。

For a graph of product volume vs time, the steeper the slope, the faster the reaction. Calculating the gradient: rate = change in y / change in x. Units will depend on the axes, e.g. cm³/s. Factors such as temperature, concentration, surface area and catalysts affect the rate, and you can compare rates using these gradient calculations.

对于产物体积随时间变化的图，斜率越大，反应越快。梯度计算：速率 = y 的变化 / x 的变化。单位取决于坐标轴，例如 cm³/s。温度、浓度、表面积和催化剂等因素会影响速率，你可以通过梯度计算来比较速率。

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GCSE CIE Chemistry: Formula Summary Handbook | GCSE CIE 化学：公式汇总手册