📚 IGCSE OCR Chemistry: Formula Summary Handbook | IGCSE OCR 化学公式汇总手册
Mastering chemical calculations is the backbone of success in IGCSE OCR Chemistry. This handbook brings together every formula you must memorise and apply, from mole conversions to energy changes and atom economy. Each entry is paired with a clear explanation, making it your go-to quick reference for revision. Keep it handy as you tackle past papers and build confidence in quantitative analysis.
掌握化学计算是 IGCSE OCR 化学取得成功的基石。这本手册汇集了从摩尔换算到能量变化、原子经济性等所有必须记住和运用的公式。每个条目均配有清晰的解释，是您复习备考的首选快速参考。请随身携带，在做往年试题时反复查阅，增强对定量分析的信心。
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 weighted mean mass of an atom of an element compared to 1/12th the mass of a carbon-12 atom. It has no units and takes account of isotopic abundances.
相对原子质量 (Aᵣ) 是元素一个原子的加权平均质量与一个碳‑12 原子质量的 1/12 的比值，无单位，同时考虑了同位素丰度。
Relative formula mass (Mᵣ) applies to compounds and is simply the sum of the relative atomic masses of all the atoms present in one formula unit. For ionic substances we use the term ‘relative formula mass’ rather than ‘relative molecular mass’.
相对式量 (Mᵣ) 适用于化合物，即为一个式单元中所有原子的相对原子质量之和。对于离子型物质，我们使用“相对式量”而非“相对分子质量”。
Mᵣ = Σ (Aᵣ of each atom)
Example: Mᵣ of CaCO₃ = 40.1 + 12.0 + (3 × 16.0) = 100.1
示例：CaCO₃ 的 Mᵣ = 40.1 + 12.0 + (3 × 16.0) = 100.1
2. The Mole and Avogadro Constant | 摩尔与阿伏伽德罗常数
The mole is the SI base unit for amount of substance. One mole of any substance contains exactly 6.02 × 10²³ elementary particles (atoms, molecules, ions or electrons). This number is known as the Avogadro constant, Nₐ.
摩尔是物质的量的 SI 基本单位。1 摩尔任何物质恰好含有 6.02 × 10²³ 个基本微粒（原子、分子、离子或电子）。这个数称为阿伏伽德罗常数，Nₐ。
n = m / M
where n = amount of substance (mol), m = mass (g), M = molar mass (g mol⁻¹). Molar mass has the same numerical value as Mᵣ but carries units.
其中 n = 物质的量 (mol)，m = 质量 (g)，M = 摩尔质量 (g mol⁻¹)。摩尔质量的数值与 Mᵣ 相同，但具有单位。
n = N / Nₐ
N = actual number of particles, Nₐ = 6.02 × 10²³ mol⁻¹. Use this when you are given the number of atoms or molecules.
N = 实际粒子个数，Nₐ = 6.02 × 10²³ mol⁻¹。当给定原子或分子个数时，使用此公式。
3. Reacting Mass Calculations | 反应质量计算
Reacting mass problems connect the mass of reactants and products using the mole ratio from a balanced chemical equation. The calculation pathway always follows: mass → moles → mole ratio → moles → mass.
反应质量计算通过配平化学方程式中的摩尔比，将反应物和产物的质量关联起来。计算路径总是：质量 → 摩尔 → 摩尔比 → 摩尔 → 质量。
n(A) / coefficient(A) = n(B) / coefficient(B)
From the equation aA + bB → cC + dD, moles of A and B are related by: n(A) / a = n(B) / b. Rearrange to find unknown moles, then multiply by molar mass to obtain the target mass.
对于方程式 aA + bB → cC + dD，A 与 B 的摩尔关系为：n(A) / a = n(B) / b。变形求出未知摩尔数，再乘以摩尔质量即可得到目标质量。
4. Gas Volume Calculations (Molar Gas Volume) | 气体体积计算（摩尔气体体积）
At room temperature and pressure (rtp), typically 25 °C and 1 atmosphere, one mole of any gas occupies a fixed volume of 24 dm³ (or 24,000 cm³). This is called the molar gas volume, Vₘ.
在室温和常压下（rtp，通常为 25 °C、1 个大气压），1 摩尔任何气体的体积固定为 24 dm³（或 24,000 cm³）。此值称为摩尔气体体积，Vₘ。
V (dm³) = n × 24 dm³ mol⁻¹
If the volume is measured in cm³, convert to dm³ by dividing by 1000 first. For gases not at rtp, the ideal gas equation is not required at IGCSE.
如果测得的体积单位为 cm³，则先除以 1000 转换为 dm³。对于非 rtp 条件下的气体，IGCSE 不需要使用理想气体方程。
5. Concentration of Solutions | 溶液浓度
Concentration is a measure of how much solute is dissolved in a given volume of solvent. In IGCSE OCR Chemistry you work with both mol dm⁻³ and g dm⁻³.
浓度表示在一定体积溶剂中溶解的溶质多少。在 IGCSE OCR 化学中，你需要同时使用 mol dm⁻³ 和 g dm⁻³。
c (mol dm⁻³) = n (mol) / V (dm³)
Mass concentration (g dm⁻³) = m (g) / V (dm³)
To convert between the two: c (mol dm⁻³) = mass concentration (g dm⁻³) / M (g mol⁻¹).
两者之间的换算：c (mol dm⁻³) = 质量浓度 (g dm⁻³) / M (g mol⁻¹)。
6. Percentage Yield | 产率百分比
The percentage yield compares the actual mass of product obtained from an experiment to the theoretical mass predicted by stoichiometry. It indicates the efficiency of the reaction.
产率百分比将实验中实际得到的产物质量与根据化学计量学预测的理论质量进行比较，反映了反应的效率。
% yield = (actual yield / theoretical yield) × 100%
A yield less than 100% can be due to incomplete reaction, side reactions, or loss during purification. Yields greater than 100% suggest impure product or weighing errors.
产率低于 100% 可能源于反应不完全、副反应或纯化过程中的损失。产率大于 100% 则暗示产物不纯或称量误差。
7. Atom Economy | 原子经济性
Atom economy measures the proportion of reactant atoms that become part of 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进入到目标产物中的比例，是绿色化学和可持续工艺中的一个核心概念。
% atom economy = (Mᵣ of desired product / sum of Mᵣ of all reactants) × 100%
For the reaction A + B → C + D, where C is the desired product: sum of Mᵣ of reactants = Mᵣ(A) + Mᵣ(B). High atom economy reduces waste and is preferred industrially.
对于反应 A + B → C + D，C 为目标产物：反应物 Mᵣ 之和 = Mᵣ(A) + Mᵣ(B)。高原子经济性可减少废弃物，在工业上更受欢迎。
8. Titration Calculations | 滴定计算
Titration is used to find the unknown concentration of a solution by reacting it with a standard solution. The key relationship is derived from the balanced equation and the mole concept.
滴定通过将未知溶液与标准溶液发生反应来确定其未知浓度。核心关系式源自配平方程式和摩尔概念。
c₁V₁ / n₁ = c₂V₂ / n₂
where n₁ and n₂ are the stoichiometric coefficients in the equation for the two reactants. For a 1:1 reaction (e.g., HCl + NaOH → NaCl + H₂O), this simplifies to:
其中 n₁ 与 n₂ 是方程式中两种反应物的化学计量数。对于 1:1 的反应（例如 HCl + NaOH → NaCl + H₂O），可简化为：
c₁V₁ = c₂V₂
Remember V must be in dm³. If your values are in cm³, you can use the equation directly as long as both volumes share the same unit.
记住体积 V 的单位必须是 dm³。如果数据单位为 cm³，只要两个体积单位一致，也可直接代入方程。
9. Energy Changes (Bond Energy Calculations) | 能量变化（键能计算）
Energy changes in chemical reactions can be measured experimentally using calorimetry. The heat energy transferred, q, is calculated from the temperature change of a known mass of solution.
化学反应中的能量变化可通过量热法实验测定。传递的热量 q 根据已知质量溶液的温度变化计算。
q = m × c × ΔT
m = mass of solution (g), c = specific heat capacity (4.18 J g⁻¹ °C⁻¹ for water), ΔT = temperature change (°C).
m = 溶液的质量 (g)，c = 比热容（水的比热容为 4.18 J g⁻¹ °C⁻¹），ΔT = 温度变化 (°C)。
ΔH = – q / n or ΔH = – q / moles of limiting reactant
The negative sign ensures that exothermic reactions have a negative ΔH. For bond energy calculations, use:
负号确保放热反应的 ΔH 为负值。进行键能计算时，使用：
ΔH = Σ (bond energies of bonds broken) – Σ (bond energies of bonds formed)
10. Rate of Reaction Calculations | 反应速率计算
The rate of a chemical reaction tells us how quickly a reactant is used up or a product is formed. It can be expressed in different units depending on what is being monitored.
化学反应速率告诉我们反应物消耗或产物生成的快慢。根据监测量的不同，可用不同单位表示。
Average rate = change in amount or concentration / time taken
Common units include: cm³ of gas per second (cm³ s⁻¹), grams per second (g s⁻¹), or mol dm⁻³ s⁻¹ for concentration changes. For a graph of volume vs time, the instantaneous rate is given by the gradient of the tangent.
常见单位包括：每秒气体体积 (cm³ s⁻¹)、每秒质量 (g s⁻¹) 或浓度变化的 mol dm⁻³ s⁻¹。在体积-时间图中，瞬时速率由切线的斜率给出。
11. Empirical Formula and Molecular Formula | 实验式与分子式
The empirical formula is the simplest whole-number ratio of atoms of each element in a compound. The molecular formula gives the actual number of atoms of each element in a molecule.
实验式是化合物中各元素原子的最简整数比。分子式则给出一个分子中各元素的实际原子数目。
Steps to find empirical formula:
i) Convert mass or percentage composition to moles by dividing by Aᵣ. ii) Divide each mole value by the smallest number of moles. iii) Obtain the simplest integer ratio.
i) 将质量或百分比组成除以 Aᵣ 得到摩尔数。ii) 每个摩尔数除以其中最小的摩尔数。iii) 得到最简整数比。
Molecular formula = (Empirical formula) × n, where n = relative molecular mass / empirical formula mass.
分子式 = (实验式) × n，其中 n = 相对分子质量 / 实验式质量。
12. Percentage Composition and Water of Crystallisation | 百分比组成与结晶水
Percentage by mass of an element in a compound is found using the total mass of that element in one formula unit divided by Mᵣ, multiplied by 100.
化合物中某元素的质量百分比，等于一个式单元中该元素的总质量除以 Mᵣ，再乘以 100。
% element = (total Aᵣ of the element in formula / Mᵣ) × 100%
Water of crystallisation is the water trapped within the crystal structure of a hydrated salt. On gentle heating, the water is driven off, leaving the anhydrous salt. The number of moles of water per mole of salt (n) is calculated as:
结晶水是存在于水合盐晶体结构中的水。缓慢加热时水分逸出，留下无水盐。每摩尔盐对应的水摩尔数 (n) 计算如下：
n = (mass of water lost ÷ 18.0) / (mass of anhydrous salt ÷ Mᵣ of anhydrous salt)
The formula of the hydrated salt is then written as Salt·nH₂O.
水合盐的化学式写为 Salt·nH₂O。
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