📚 GCSE Edexcel Chemistry: Formula Summary Handbook | GCSE Edexcel 化学：公式汇总手册
This comprehensive guide brings together every essential formula you need for GCSE Edexcel Chemistry. From moles to energy changes and titrations, mastering these equations will help you tackle quantitative problems with confidence.
这份全面的指南汇总了 GCSE Edexcel 化学所需的所有关键公式。从摩尔计算到能量变化和滴定，掌握这些方程式将使你自信地解决定量问题。
1. Relative Atomic Mass and Relative Formula Mass | 相对原子质量与相对式量
Relative atomic mass (Aᵣ) is the average mass of an atom of an element compared to 1/12th the mass of a carbon‑12 atom. It has no units.
相对原子质量（Aᵣ）是一个元素原子的平均质量与碳‑12 原子质量的 1/12 的比值，没有单位。
Relative formula mass (Mᵣ) applies to compounds. You calculate it by adding together all the Aᵣ values of the atoms shown in the chemical formula.
相对式量（Mᵣ）适用于化合物。计算方法是将化学式中各原子的 Aᵣ 相加。
Mᵣ = Σ (Aᵣ × number of atoms)
r H₂O: Mᵣ = (2 × 1) + 16 = 18. For calcium carbonate CaCO₃: Mᵣ = 40 + 12 + (3 × 16) = 100.
例如水 H₂O：Mᵣ = (2 × 1) + 16 = 18。碳酸钙 CaCO₃：Mᵣ = 40 + 12 + (3 × 16) = 100。
Always remember to multiply the Aᵣ by the number of atoms of that element present in the formula.
始终记得将该元素的 Aᵣ 乘以化学式中该原子的个数。
2. The Mole: n = m / Mᵣ | 摩尔：物质的量与质量的关系
f substance. One mole of any substance contains 6.02 × 10²³ particles (Avogadro constant).
摩尔（符号 n）是物质的量的单位。1 摩尔任何物质含有 6.02 × 10²³ 个微粒（阿伏伽德罗常数）。
The central formula linking mass and moles is:
联系质量与摩尔的中心公式是：
n = m / Mᵣ
where m is the mass in grams, Mᵣ is the relative formula mass, and n is the number of moles.
其中 m 为质量（克），Mᵣ 为相对式量，n 为摩尔数。
Example: How many moles in 8.0 g of sodium hydroxide NaOH (Mᵣ = 40)? n = 8.0 / 40 = 0.20 mol.
示例：8.0 g 氢氧化钠 NaOH（Mᵣ = 40）含多少摩尔？n = 8.0 / 40 = 0.20 mol。
o rearrange to find mass: m = n × Mᵣ. For 0.50 mol of CO₂ (Mᵣ = 44), m = 0.50 × 44 = 22 g.
你也可以变形公式求质量：m = n × Mᵣ。0.50 mol CO₂（Mᵣ = 44），m = 0.50 × 44 = 22 g。
3. Concentration: mol/dm³ and g/dm³ | 浓度：摩尔浓度与质量浓度
Concentration measures how much solute is dissolved in a given volume of solution. Two common units are used in GCSE Chemistry.
浓度表示单位体积溶液中溶质的含量。GCSE 化学中常用两种单位。
Molar concentration (c) in mol/dm³ is given by:
摩尔浓度（c，单位 mol/dm³）由下式给出：
c = n / V
where n is the number of moles and V is the volume in dm³.
其中 n 为摩尔数，V 为体积（dm³）。
Mass concentration in g/dm³ can be calculated directly from mass and volume:
质量浓度（g/dm³）可以直接由质量与体积计算：
Concentration (g/dm³) = mass (g) / volume (dm³)
To convert between the two, use: concentration (mol/dm³) = concentration (g/dm³) / Mᵣ.
两者之间的转换公式为：摩尔浓度 = 质量浓度 / Mᵣ。
entration of 5.85/0.25 = 23.4 g/dm³ and a molar concentration of 23.4/58.5 = 0.40 mol/dm³.
示例：5.85 g NaCl（Mᵣ = 58.5）溶于 250 cm³（0.25 dm³），质量浓度为 5.85/0.25 = 23.4 g/dm³，摩尔浓度为 23.4/58.5 = 0.40 mol/dm³。
4. Molar Gas Volume at RTP | 室温常压下的气体摩尔体积
At room temperature and pressure (RTP, 20°C and 1 atmosphere), one mole of any gas occupies 24 dm³. This is known as the molar gas volume.
在室温常压（RTP，20°C、1 个大气压）下，1 摩尔任何气体的体积为 24 dm³。这称为气体摩尔体积。
Volume (dm³) = n × 24
Example: The volume of 0.30 mol of hydrogen gas at RTP is 0.30 × 24 = 7.2 dm³.
示例：0.30 mol 氢气在 RTP 下的体积为 0.30 × 24 = 7.2 dm³。
Conversely, to find moles from a measured gas volume, use n = volume (dm³) / 24.
反之，由气体体积求摩尔时，使用 n = 体积 (dm³) / 24。
This relationship is used when collecting gases by displacement or in combustion reactions.
该关系常用于排水集气或燃烧反应中收集气体时的计算。
5. Percentage Yield | 产率计算
Percentage yield compares the amount of product actually obtained (actual yield) to the amount expected from calculation (theoretical yield).
产率比较实际得到的产品量（实际产量）与根据计算预期得到的量（理论产量）。
% Yield = (actual yield / theoretical yield) × 100
Yields are usually less than 100% because of incomplete reactions, side reactions or product lost during purification.
产率通常低于 100%，因为反应不完全、发生副反应或纯化过程中产品损失。
ion where the theoretical yield was 15.0 g, the percentage yield is (12.0/15.0) × 100 = 80%.
示例：某反应理论产量为 15.0 g，实际获得 12.0 g 乙酸，则产率为 (12.0/15.0) × 100 = 80%。
6. Atom Economy | 原子经济性
Atom economy measures how efficiently atoms in the reactants are incorporated 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有多少进入了目标产物。这是绿色化学的核心概念。
% Atom economy = (Mᵣ of desired product / total Mᵣ of all reactants) × 100
Only the desired product appears in the numerator. By‑products and excess reactants are ignored.
分子中只包括目标产物。副产物和过量反应物不计入。
56). Total Mᵣ of reactants = 160 + (3×12) = 196. Atom economy = (2×56 / 196) × 100 = 57.1%.
示例：用碳还原氧化铁制铁：Fe₂O₃ + 3C → 2Fe + 3CO。目标产物 Fe（Mᵣ = 56），反应物总 Mᵣ = 160 + 36 = 196，原子经济性 = (112/196) × 100 = 57.1%。
7. Calorimetry and Heat Energy | 量热法与热能
When a reaction takes place in solution, the heat change can be determined using the specific heat capacity of water.
当反应在溶液中进行时，可以利用水的比热容来测定能量变化。
Q = m × c × ΔT
Q is the heat energy transferred (J), m is the mass of the solution (g), c is the specific heat capacity (4.18 J/g°C for water), and ΔT is the temperature change (°C).
Q 为转移的热量（J），m 为溶液的质量（g），c 是比热容（水的 c = 4.18 J/g°C），ΔT 是温度变化 (°C)。
Remember to use the mass of the whole solution, not just the solute, and to measure the temperature change accurately.
注意要用整个溶液的质量，而不仅仅是溶质，并准确测量温度变化。
lution increased in temperature by 6.5°C. Q = 50 × 4.18 × 6.5 = 1358.5 J (approx 1.36 kJ).
示例：中和反应中，50 g 溶液温度升高 6.5°C。Q = 50 × 4.18 × 6.5 = 1358.5 J（约 1.36 kJ）。
8. Bond Energy Calculations | 键能计算
Energy changes in a chemical reaction can be calculated using bond energies. Bond breaking is endothermic (energy taken in), bond making is exothermic (energy released).
化学反应的能量变化可利用键能计算。断键吸热（能量吸收），成键放热（能量释放）。
ΔH = Σ(bond energies of bonds broken) – Σ(bond energies of bonds formed)
If ΔH is negative, the reaction is exothermic overall; if positive, it is endothermic.
若 ΔH 为负值，总反应放热；若为正值，总反应吸热。
Example: For the combustion of hydrogen: 2H₂ + O₂ → 2H₂O. Bonds broken: 2×H–H (436) + 1×O=O (498) = 1370 kJ. Bonds formed: 4×O–H (464) = 1856 kJ. ΔH = 1370 – 1856 = -486 kJ.
示例：氢气燃烧：2H₂ + O₂ → 2H₂O。断裂键能总和：2×436 + 498 = 1370 kJ；形成键能总和：4×464 = 1856 kJ。ΔH = 1370 – 1856 = -486 kJ。
9. Titration Calculations | 滴定计算
Titration results allow you to find an unknown concentration by using a solution of known concentration. The key relationship uses the mole ratio from the balanced equation.
滴定结果可利用已知浓度的溶液求出未知浓度。关键关系是利用配平方程中的摩尔比。
c₁V₁ / n₁ = c₂V₂ / n₂
c₁ and c₂ are concentrations (mol/dm³), V₁ and V₂ are volumes (dm³ or consistent units), and n₁, n₂ are the stoichiometric coefficients for acid and base respectively.
c₁ 和 c₂ 为浓度（mol/dm³），V₁ 和 V₂ 为体积（dm³ 或统一单位），n₁、n₂ 为酸和碱在方程式中的系数。
. Then c₂ = (c₁ × V₁ × n₂) / (V₂ × n₁) = (0.100 × 20.0 × 1) / (25.0 × 1) = 0.0800 mol/dm³.
示例：25.0 cm³ NaOH（未知浓度）需要 20.0 cm³ 0
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