📚 Year 10 Cambridge IGCSE Chemistry: Formula & Theorem Quick Reference Handbook | Year 10 剑桥IGCSE化学：公式定理速查手册
This quick reference handbook brings together all the essential formulas, equations, and key principles you need to master for the Year 10 Cambridge IGCSE Chemistry course. Each section presents the core concepts with clear mathematical relationships, helping you revise stoichiometry, energetics, rates, and more efficiently.
本速查手册汇总了 Year 10 剑桥 IGCSE 化学课程必须掌握的所有核心公式、方程式和关键原理。每个小节都以清晰的数学关系呈现核心概念，帮助你高效复习化学计量、能量学、反应速率等内容。
1. Relative Atomic and Molecular Mass | 相对原子质量和相对分子质量
The relative atomic mass (Ar) of an element is the weighted average mass of its atoms relative to 1/12 the mass of a carbon-12 atom. It has no unit.
元素的相对原子质量 (Ar) 是其原子的加权平均质量与一个碳-12 原子质量的 1/12 之比，没有单位。
Relative molecular mass (Mr) is the sum of the Ar values of all the atoms present in a molecule. For ionic compounds, the term relative formula mass is often used.
相对分子质量 (Mr) 是分子中所有原子的 Ar 之和。对于离子化合物，通常使用相对式量一词。
Mr = Σ (Ar of each atom × number of atoms)
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Example: Mr of H2O = 2×1 + 16 = 18
示例：水的 Mr = 2×1 + 16 = 18
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Example: Mr of CaCO3 = 40 + 12 + (3×16) = 100
示例：碳酸钙的 Mr = 40 + 12 + (3×16) = 100
2. The Mole and Molar Mass | 摩尔与摩尔质量
One mole of any substance contains 6.02 × 1023 particles (Avogadro’s constant, NA). The molar mass (M) of a substance is the mass of one mole, expressed in g/mol, and is numerically equal to its Mr or Ar.
任何物质的一摩尔都含有 6.02 × 1023 个微粒（阿伏伽德罗常数, NA）。物质的摩尔质量 (M) 是一摩尔该物质的质量，单位为 g/mol，数值上等于其 Mr 或 Ar。
number of moles (n) = mass (m) / molar mass (M)
n = m / M
When mass is in grams and M in g/mol, n is in moles. Always convert mass to grams before using the formula.
当质量单位为克、M 单位为 g/mol 时，n 的单位为摩尔。使用公式前务必先将质量转换为克。
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Example: How many moles in 36 g of water? (M = 18 g/mol) n = 36/18 = 2.0 mol
示例：36 g 水中含多少摩尔？(M = 18 g/mol) n = 36/18 = 2.0 mol
3. Empirical and Molecular Formula | 经验式与分子式
The empirical formula gives the simplest whole-number ratio of atoms of each element in a compound. The molecular formula shows the actual number of atoms of each element in a molecule.
经验式给出化合物中各元素原子的最简整数比。分子式表示一个分子中各元素原子的实际数目。
To find empirical formula from percentage composition: divide the mass (or %) of each element by its Ar, then divide all results by the smallest value to obtain the simplest ratio.
从百分组成求经验式：将每种元素的质量（或百分比）除以其 Ar，再将所有结果除以最小值，得到最简整数比。
moles of element = mass / Ar ; ratio = divide by smallest mole
Molecular formula = (empirical formula)n, where n = Mr of compound / Mr of empirical formula.
分子式 = (经验式)n，其中 n = 化合物的 Mr / 经验式的 Mr。
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Example: A compound has 40% C, 6.7% H, 53.3% O. C: 40/12=3.33, H:6.7/1=6.7, O:53.3/16=3.33. Ratio 1:2:1 → CH2O. If Mr = 90, n = 90/30 = 3, molecular formula C3H6O3.
示例：化合物含40% C、6.7% H、53.3% O。C:40/12=3.33, H:6.7/1=6.7, O:53.3/16=3.33。比例1:2:1 → CH2O。若 Mr=90，n=90/30=3，分子式 C3H6O3。
4. Reacting Mass Calculations | 反应质量计算
Stoichiometry uses the balanced chemical equation to relate the number of moles of reactants and products. Coefficients in the equation give the mole ratio.
化学计量利用配平的化学方程式将反应物和产物的摩尔数联系起来。方程式中的系数给出了摩尔比。
molar ratio = coefficient ratio from balanced equation
Steps for mass calculations: (i) convert known masses to moles using n = m/M; (ii) use the mole ratio to find moles of the unknown; (iii) convert moles back to mass using m = n × M.
质量计算步骤：(i) 用 n = m/M 将已知质量转换为摩尔；(ii) 利用摩尔比求出未知物的摩尔数；(iii) 用 m = n × M 将摩尔数转换回质量。
The limiting reactant is the one that is completely consumed first; it determines the maximum amount of product formed. Compare the available moles with the required ratio to identify the limiting reactant.
限量试剂是最先被完全消耗的反应物，它决定了最多能生成多少产物。比较现有摩尔数与所需比例，即可找出限量试剂。
5. Percentage Yield and Atom Economy | 百分产率与原子经济
Percentage yield compares the actual yield of a product with the theoretical yield calculated from the limiting reactant.
百分产率将实际获得的产物质量与根据限量试剂计算的理论产量进行比较。
% yield = (actual yield / theoretical yield) × 100%
Reasons for yield < 100% include incomplete reactions, side reactions, and loss during purification.
产率低于 100% 的原因包括反应不完全、副反应以及纯化过程中的损失。
Atom economy measures the proportion of reactant atoms that end up in the desired product. A higher atom economy reduces waste.
原子经济衡量的是反应物原子中最终进入目标产物的比例。原子经济越高，废物越少。
atom economy = (Mr of desired product / sum of Mr of all products) × 100%
Only one product may be desired; the sum of Mr of all products includes both desired and by-products as shown in the balanced equation.
可能只有一种目标产物；所有产物的 Mr 之和包括配平方程式中出现的所有产物（含副产物）。
6. Concentration of Solutions | 溶液浓度
Concentration expresses the amount of solute dissolved in a given volume of solvent. The SI unit is mol/dm3 (often written as M).
浓度表示一定体积溶剂中溶解的溶质数量，国际单位是 mol/dm3（常写作 M）。
concentration (mol/dm3) = number of moles / volume (dm3)
c = n / V
Volume must be in dm3. To convert cm3 to dm3, divide by 1000.
体积单位必须使用 dm3。若为 cm3，需除以 1000 进行换算。
Concentration can also be given in g/dm3: mass (g) / volume (dm3). To convert between g/dm3 and mol/dm3, divide or multiply by Mr.
浓度也可用 g/dm3 表示：质量 (g) / 体积 (dm3)。在 g/dm3 与 mol/dm3 之间换算时，除以或乘以 Mr。
Titration formula for reactions reaching equivalence point:
cA × VA / cB × VB = nA / nB
where nA and nB are the mole coefficients of the acid and base in the balanced equation.
其中 nA 和 nB 分别为配平方程式中酸和碱的摩尔系数。
7. Gas Volumes and Molar Volume | 气体体积与摩尔体积
At room temperature and pressure (rtp, 25 °C and 1 atm), one mole of any gas occupies 24 dm3 (or 24 000 cm3). This is the molar gas volume.
在常温常压下（rtp, 25 °C 和 1 atm），一摩尔任何气体占据 24 dm3（或 24 000 cm3）的体积。这就是气体的摩尔体积。
volume of gas (dm3) = number of moles × 24
V = n × 24 dm3/mol
To find moles from gas volume: n = V (dm3) / 24. Remember to convert cm3 to dm3 by dividing by 1000.
由气体体积求摩尔：n = V (dm3) / 24。注意将 cm3 除以 1000 换算为 dm3。
In reactions involving gases, the volume ratio is the same as the mole ratio from the balanced equation (Avogadro’s law).
涉及气体的反应中，体积比等于配平方程式的摩尔比（阿伏伽德罗定律）。
8. Enthalpy Changes: Calorimetry | 焓变：量热法
An enthalpy change (ΔH) is the heat energy transferred in a reaction at constant pressure. It is measured in kJ/mol.
焓变 (ΔH) 是恒压条件下反应传递的热量，单位为 kJ/mol。
Simple calorimetry uses a metal or polystyrene cup; the heat exchanged is calculated using the specific heat capacity of water (or solution):
简单的量热法使用金属杯或聚苯乙烯杯；交换的热量利用水（或溶液）的比热容计算：
q = m × c × ΔT
where q = heat energy (J), m = mass of water/solution (g), c = specific heat capacity (for water, 4.18 J/g·°C), ΔT = temperature change (°C).
其中 q = 热量 (J)，m = 水/溶液的质量 (g)，c = 比热容（水为 4.18 J/g·°C），ΔT = 温度变化 (°C)。
To find the molar enthalpy change:
ΔH = −q / n
where n is the number of moles of the limiting reactant. The negative sign indicates that the reaction is exothermic (ΔH negative); if endothermic, q is absorbed and ΔH becomes positive.
其中 n 为限量试剂的摩尔数。负号表明反应放热（ΔH 为负）；若为吸热，q 被吸收，ΔH 为正。
Always convert the final answer to kJ/mol and apply the appropriate sign.
务必把最终结果换算为 kJ/mol，并加上正确的符号。
9. Bond Energy Calculations | 键能计算
Bond energy (bond dissociation energy) is the energy required to break one mole of a covalent bond in the gaseous state, in kJ/mol. Bond breaking is endothermic (positive), bond forming is exothermic (negative).
键能（键解离能）是指在气态下断裂一摩尔共价键所需的能量，单位为 kJ/mol。断键吸热（正值），成键放热（负值）。
The overall enthalpy change for a reaction can be estimated using average bond energies:
ΔH = Σ (bond energies of bonds broken) − Σ (bond energies of bonds formed)
Draw displayed formulae to count all the bonds present before and after the reaction, then apply the formula.
画出结构式以数清反应前后存在的所有键，然后套用公式。
Example: H2 + Cl2 → 2HCl. Bonds broken: H-H (436) + Cl-Cl (243) = 679 kJ; bonds formed: 2 × H-Cl (2 × 431 = 862 kJ). ΔH = 679 – 862 = -183 kJ/mol (exothermic).
示例：H2 + Cl2 → 2HCl。断开键：H-H (436) + Cl-Cl (243) = 679 kJ；生成键：2 × H-Cl (2 × 431 = 862 kJ)。ΔH = 679 – 862 = -183 kJ/mol（放热）。
10. Rates of Reaction | 反应速率
The rate of a chemical reaction measures how quickly reactants are used up or products are formed. Average rate can be expressed as:
化学反应速率衡量反应物消耗或产物生成的快慢。平均速率可表示为：
rate = change in quantity / change in time
The quantity may be mass of a reactant/product, volume of gas produced, or concentration of a species. Units include g/s, cm3/min, mol/dm3/s.
该物理量可以是反应物/产物的质量、产生气体的体积或某一物种的浓度。单位包括 g/s、cm3/min、mol/dm3/s 等。
For a graph of amount vs. time, the instantaneous rate at a point is the gradient of the tangent at that point. The initial rate is the gradient at t = 0.
对于数量-时间图，某时刻的瞬时速率是该点切线的斜率。初始速率是 t = 0 处的斜率。
Factors affecting reaction rate (collision theory): concentration of solutions, pressure of gases, surface area of solids, temperature, and presence of a catalyst. Increased frequency and/or energy of collisions leads to a higher rate.
影响反应速率的因素（碰撞理论）：溶液浓度、气体压强、固体表面积、温度以及催化剂。碰撞频率和/或能量的提高会使速率增大。
Catalysts provide an alternative pathway with lower activation energy (Ea), increasing the proportion of successful collisions without being consumed.
催化剂提供一条活化能 (Ea) 较低的反应路径，从而提高有效碰撞的比例，且自身不被消耗。
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