📚 IGCSE Chemistry: Formula Summary Handbook | IGCSE 化学：公式汇总手册
This handbook brings together every essential formula you will encounter in your IGCSE Chemistry course. Use it to consolidate your quantitative chemistry skills, from moles and mass calculations to energy changes, titrations, and electrolysis. Paired bilingual explanations help you master both the concepts and the terminology needed for the exam.
本手册汇集了IGCSE化学课程中所有重要的公式。你可以用它来巩固定量化学技能，涵盖摩尔、质量计算、能量变化、滴定以及电解等内容。中英对照的讲解能帮助你同时掌握概念和考试必备的术语。
1. Relative Atomic Mass & Molecular Mass | 相对原子质量与分子质量
The relative atomic mass (Ar) of an element is the average mass of its atoms compared to 1/12 of the mass of a carbon‑12 atom. It has no units. The relative molecular mass (Mr) of a compound is the sum of the Ar values of all the atoms in its formula.
相对原子质量（Ar）是一种元素原子平均质量与1个碳‑12原子质量的1/12相比较的比值，没有单位。相对分子质量（Mr）等于化学式中所有原子的Ar之和。
Ar = (average mass of one atom) / (1/12 × mass of one ¹²C atom)
The Ar values you use come directly from the Periodic Table. For example, the Mr of H₂SO₄ is calculated as: (2×1) + 32 + (4×16) = 98.
你使用的Ar值直接来自元素周期表。例如，H₂SO₄的Mr计算如下：(2×1) + 32 + (4×16) = 98。
2. The Mole and Avogadro’s Constant | 摩尔与阿伏伽德罗常数
One mole of any substance contains 6.022 × 10²³ particles (atoms, molecules, ions or electrons). This number is called Avogadro’s constant (L). The mole allows chemists to count particles by weighing.
1摩尔任何物质含有6.022 × 10²³个粒子（原子、分子、离子或电子）。这个数字称为阿伏伽德罗常数（L）。摩尔让化学家可以通过称重来“数”粒子。
number of particles = number of moles × L
If you have 2.0 mol of water, the number of molecules is 2.0 × 6.022 × 10²³ = 1.2044 × 10²⁴.
如果你有2.0 mol水，分子数目就是2.0 × 6.022 × 10²³ = 1.2044 × 10²⁴。
3. Mass-Mole-Particle Conversions | 质量-摩尔-粒子数换算
The link between mass and moles is the molar mass (M), which has the same numerical value as the relative atomic or molecular mass but is expressed in g mol⁻¹.
质量与摩尔之间的桥梁是摩尔质量（M），它在数值上与相对原子质量或相对分子质量相同，但单位为 g mol⁻¹。
number of moles (n) = mass (m) / molar mass (M)
For a given mass, first calculate moles, then multiply by Avogadro’s constant to find the number of particles. These two steps are the core of all quantitative chemistry.
对于给定质量，先计算摩尔数，再乘以阿伏伽德罗常数即可求得粒子数。这两步是所有定量化学的核心。
n = m / M and N = n × L
4. Molar Volume of Gases | 气体摩尔体积
At room temperature and pressure (RTP, about 25 °C and 1 atm), one mole of any gas occupies 24 dm³ (24 000 cm³). This is the molar gas volume and is incredibly useful for gas calculations.
在室温和常压（RTP，约25 °C、1 atm）下，1摩尔任何气体占据24 dm³（24 000 cm³）的体积。这就是气体摩尔体积，对气体计算非常有用。
volume of gas (dm³) = number of moles × 24 dm³
If the volume is given in cm³, use 24 000 cm³ instead. For example, 0.5 mol of carbon dioxide occupies 0.5 × 24 = 12 dm³ at RTP.
如果体积以 cm³ 给出，则用24 000 cm³。例如，0.5 mol二氧化碳在RTP下占据的体积为 0.5 × 24 = 12 dm³。
5. Concentration and Solution Volume | 浓度与溶液体积
The concentration of a solution tells you how much solute is dissolved in a given volume of solvent. In IGCSE, concentration is usually expressed in mol dm⁻³ or g dm⁻³.
溶液的浓度表示一定体积溶剂中溶解了多少溶质。在IGCSE中，浓度通常用 mol dm⁻³ 或 g dm⁻³ 表示。
concentration (c, mol dm⁻³) = moles of solute (n) / volume of solution (V, dm³)
This equation can be rearranged: n = c × V. Always ensure the volume is in dm³ before substituting. If you need concentration in g dm⁻³, simply use mass instead of moles: c (g dm⁻³) = m / V.
此公式可变形为：n = c × V。代入前请确保体积单位是dm³。如果需要以 g dm⁻³ 表示浓度，直接用质量替换摩尔数即可：c (g dm⁻³) = m / V。
6. Stoichiometry: Reacting Masses | 化学计量学：反应质量
Stoichiometry uses the balanced chemical equation to find the masses of reactants and products. The general steps are: write the equation, calculate moles of the known substance, use the mole ratio, then convert moles back to mass.
化学计量学利用配平后的化学方程式计算反应物与生成物的质量。基本步骤是：写出方程式，计算已知物质的摩尔数，利用摩尔比，再将摩尔数换算回质量。
m (wanted) = n (known) × (coefficient ratio) × M (wanted)
For example, in the reaction 2Mg + O₂ → 2MgO, 2 moles of Mg produce 2 moles of MgO. The mole ratio of Mg to MgO is 1:1. If you start with 2.4 g of Mg (Ar = 24), n = 0.1 mol, so mass of MgO = 0.1 × 40 = 4.0 g.
例如，对于反应 2Mg + O₂ → 2MgO，2 mol Mg 生成 2 mol MgO，Mg与MgO的摩尔比为1:1。若起始用2.4 g Mg（Ar = 24），则n = 0.1 mol，MgO的质量 = 0.1 × 40 = 4.0 g。
7. Percentage Yield | 产率
Percentage yield compares the actual amount of product obtained in an experiment to the theoretical amount predicted by stoichiometry. It indicates how efficient a reaction is.
产率将实验中实际得到的产物量与化学计量学预测的理论量进行比较，反映了反应的效率。
percentage yield = (actual yield / theoretical yield) × 100%
Yields are often less than 100% due to incomplete reactions, side reactions, or losses during purification. A high percentage yield is desirable in industry to reduce waste.
产率通常低于100%，原因包括反应不完全、副反应或提纯过程中的损失。工业上期望有高产量以减少浪费。
8. Atom Economy | 原子经济性
Atom economy measures the proportion of reactant atoms that end up in the desired product. It is a key concept in green chemistry and sustainable processes.
原子经济性衡量反应物原子中有多少最终进入期望产物。这是绿色化学和可持续工艺中的一个关键概念。
atom economy = (Mr of desired product / sum of Mr of all reactants) × 100%
If all reactants are incorporated into the single desired product, atom economy is 100%. Addition reactions typically have high atom economy, whereas substitution or elimination reactions often produce waste by‑products.
如果所有反应物都进入唯一的期望产物中，原子经济性为100%。加成反应通常原子经济性较高，而取代或消除反应常会产生副产物。
9. Empirical and Molecular Formulae | 实验式与分子式
The empirical formula is the simplest whole‑number ratio of atoms in a compound. The molecular formula shows the actual number of atoms in one molecule and is a whole‑number multiple of the empirical formula.
实验式是化合物中原子的最简整数比。分子式表示一个分子中各原子的实际数目，是实验式的整数倍。
n = Mr / empirical formula mass
To find the empirical formula: divide the mass (or percentage) of each element by its Ar, then divide each result by the smallest value to get the simplest ratio. To deduce the molecular formula, use the above equation.
求实验式的方法：用每种元素的质量（或百分含量）除以它的Ar，再将各结果除以最小值即得最简整数比。要推导分子式，使用上述公式。
10. Titration Calculations | 滴定计算
Titration is used to find the concentration of an unknown solution by reacting it with a standard solution of known concentration. The key formula relates the concentrations and volumes of the two solutions, taking the mole ratio into account.
滴定通过将未知浓度的溶液与已知浓度的标准溶液反应来测定其浓度。关键公式把两种溶液的浓度和体积联系起来，并计入摩尔比。
(c₁ × V₁) / n₁ = (c₂ × V₂) / n₂
where n₁ and n₂ are the coefficients from the balanced equation. For a 1:1 reaction (e.g. HCl + NaOH → NaCl + H₂O), the ratio simplifies to c₁V₁ = c₂V₂. Remember to use volumes in dm³ or change the concentration unit accordingly.
式中 n₁ 和 n₂ 是配平方程中的系数。对于 1:1 的反应（如 HCl + NaOH → NaCl + H₂O），该比可简化为 c₁V₁ = c₂V₂。请记住使用dm³为体积单位，或相应调整浓度单位。
11. Energy Changes in Reactions | 反应中的能量变化
In calorimetry experiments, the heat energy (q) absorbed or released by a reaction can be calculated using the specific heat capacity of water. The enthalpy change (ΔH) per mole is then found.
在量热实验中，反应吸收或放出的热量（q）可利用水的比热容进行计算。然后可求得每摩尔的焓变（ΔH）。
q = m × c × ΔT
Here, m is the mass of water (or solution) in grams, c is the specific heat capacity (4.18 J g⁻¹ °C⁻¹ for water), and ΔT is the temperature change in °C.
式中，m是水（或溶液）的质量（g），c是比热容（水的为4.18 J g⁻¹ °C⁻¹），ΔT是温度变化（°C）。
ΔH = -q / n
The negative sign indicates that exothermic reactions (temperature rise) have a negative ΔH, while endothermic reactions have a positive ΔH. Remember to convert q to kJ if required. This formula pair lets you link temperature data directly to the energy per mole of reactant.
公式前的负号表示放热反应（温度升高）的ΔH为负，吸热反应的ΔH为正。若需要，要将q转换为kJ。这一组公式能将温度数据直接与每摩尔反应物的能量联系起来。
12. Electrolysis Calculations (Charge and Mass) | 电解计算（电荷与质量）
Electrolysis calculations link the quantity of electric charge passed through a cell to the amount of substance produced at an electrode. This is based on the number of electrons involved in the electrode reaction.
电解计算将流过电解池的电量与电极上生成的物质量联系起来，其依据是电极反应中所涉及的电子数。
charge (Q, coulombs) = current (I, amperes) × time (t, seconds)
The number of moles of electrons transferred is then given by Q / F, where F is the Faraday constant (96 500 C mol⁻¹).
然后，转移电子的摩尔数可由 Q / F 求得，F是法拉第常数（96 500 C mol⁻¹）。
moles of substance produced = (I × t) / (n × F)
n is the number of electrons needed to produce one formula unit of the element (e.g., n = 2 for Cu²⁺ + 2e⁻ → Cu). To find the mass deposited, multiply moles by the molar mass of the substance. This powerful set of equations links electricity to chemistry.
n是生成一个化学式单位的元素所需的电子数（如 Cu²⁺ + 2e⁻ → Cu，n = 2）。要求出沉积的质量，将摩尔数乘以物质的摩尔质量即可。这一组方程式将电学与化学紧密相连。
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