📚 Year 10 CIE Chemistry Formula & Theorem Quick-Reference Handbook | Year 10 CIE 化学：公式定理速查手册
This quick-reference handbook brings together the essential formulas, equations, and theorems that every Year 10 student needs for the CIE IGCSE Chemistry course. Understanding these principles is not just about memorisation; it is about building a toolkit that allows you to tackle quantitative problems, predict reaction outcomes, and explain chemical behaviour with confidence. Keep this guide close as you work through past papers and revision exercises. The formulas are organised by topic, so you can locate them instantly when you need to calculate moles, work out gas volumes, or balance energy changes.
这本速查手册汇集了 Year 10 学生在 CIE IGCSE 化学课程中必须掌握的核心公式、方程式和定理。理解这些原理不仅是为了记忆，更是为了构建一个工具包，让你能够自信地解决定量问题、预测反应结果并解释化学行为。在练习历年真题和复习题时，请将此指南放在手边。公式按主题分类，方便你在需要计算摩尔、求解气体体积或平衡能量变化时立即查阅。
1. Relative Atomic Mass and Relative Molecular Mass | 相对原子质量与相对分子质量
The relative atomic mass (Aᵣ) of an element is the average mass of its atoms compared to 1/12th the mass of a carbon-12 atom. It has no units because it is a ratio. You can find Aᵣ values on the Periodic Table; for example, oxygen has an Aᵣ of 16 and chlorine has an Aᵣ of 35.5, reflecting the natural abundance of its isotopes.
元素的相对原子质量 (Aᵣ) 是其原子的平均质量与碳-12 原子质量的 1/12 相比较的比值。由于是比值，它没有单位。你可以在元素周期表上查找 Aᵣ 值；例如，氧的 Aᵣ 为 16，氯的 Aᵣ 为 35.5，这反映了其同位素的天然丰度。
form the calculation: (2 × 1) + 32 + (4 × 16) = 98. This value is used directly in mole calculations when the substance exists as simple molecules. For giant ionic structures like sodium chloride, NaCl, we use the term relative formula mass, but it is calculated using the same method: Mᵣ of NaCl is 23 + 35.5 = 58.5.
化合物的相对分子质量 (Mᵣ) 是其化学式中所有原子的 Aᵣ 值之和。要计算硫酸 (H₂SO₄) 的 Mᵣ，计算过程为：(2 × 1) + 32 + (4 × 16) = 98。当物质以简单分子形式存在时，该值可直接用于摩尔计算。对于氯化钠 (NaCl) 等巨型离子结构，我们使用相对式量这一术语，但计算方法相同：NaCl 的 Mᵣ 为 23 + 35.5 = 58.5。
2. The Mole and Molar Mass | 摩尔与摩尔质量
t of substance. One mole of any substance contains exactly 6.02 × 10²³ particles (Avogadro' ar mass, expressed in grams per mole (g/mol), and it is numerically equal to its Aᵣ or Mᵣ.
摩尔是物质的量的基本单位。一摩尔任何物质恰好包含 6.02 × 10²³ 个粒子（阿伏伽德罗常数）。这些粒子可以是原子、分子、离子或式量单元。一摩尔物质的质量即为其摩尔质量，单位为克每摩尔 (g/mol)，数值上等于其 Aᵣ 或 Mᵣ。
Number of moles (n) = mass in grams (m) ÷ molar mass (M)
摩尔数 (n) = 质量（克, m）÷ 摩尔质量 (M)
ber of moles is therefore 8.0 ÷ 32 = 0.25 mol. Always check whether the substance is monatomic (e.g., Ar), diatomic (O₂, H₂, N₂, Cl₂), or polyatomic, as this affects the molar mass you use in calculations.
这个公式是进入定量化学的钥匙。如果你有 8.0 克氧气 (O₂)，首先计算其摩尔质量：2 × 16 = 32 g/mol。因此，摩尔数为 8.0 ÷ 32 = 0.25 mol。务必检查该物质是单原子（如 Ar）、双原子（O₂、H₂、N₂、Cl₂）还是多原子，因为这会直接影响计算中使用的摩尔质量。
3. Percentage Composition by Mass | 质量百分比组成
Percentage composition tells you the mass of a specific element present in 100 g of a compound. It is a critical concept for determining empirical formulas and checking the purity of samples. The formula applies to any element within a compound.
质量百分比组成告诉你 100 克化合物中某特定元素的质量。这是确定经验式和检查样品纯度的关键概念。该公式适用于化合物中的任何元素。
Percentage by mass = (total mass of element in formula ÷ Mᵣ of compound) × 100%
质量百分比 = (化学式中元素的总质量 ÷ 化合物的 Mᵣ) × 100%
= 100. The mass of calcium in one formula unit is 40. The percentage by mass of calcium is therefore (40 ÷ 100) × 100% = 40.0%.
让我们计算碳酸钙 (CaCO₃) 中钙的质量百分比。Aᵣ 值为：Ca = 40，C = 12，O = 16。CaCO₃ 的 Mᵣ 为 40 + 12 + (3 × 16) = 100。一个式量单元中钙的质量为 40。因此，钙的质量百分比为 (40 ÷ 100) × 100% = 40.0%。
This calculation is often reversed: given the percentage composition, you can deduce the empirical formula. Divide each percentage by the element's Aᵣ, then simplify the resulting ratios to the smallest whole numbers. This skill is tested frequently in CIE practical and theory papers.
这种计算经常被反过来使用：给定质量百分比组成，你可以推导出经验式。将每种元素的百分比除以其 Aᵣ，然后将所得比值简化为最小整数比。这项技能在 CIE 的实验卷和理论卷中经常考查。
4. Empirical and Molecular Formulae | 经验式与分子式
The empirical formula is the simplest whole-number ratio of atoms of each element in a compound. The molecular formula shows the actual number of atoms of each element in a molecule. The molecular formula is always a whole-number multiple of the empirical formula.
经验式是化合物中各元素原子的最简整数比。分子式显示一个分子中各元素原子的实际数量。分子式始终是经验式的整数倍。
ng by its Aᵣ; divide each mole value by the smallest number of moles obtained; and multiply if necessary to get whole numbers. For example, a compound contains 85.7% carbon and 14.3% 14.3 ÷ 1 = 14.3. Divide both by 7.14 to get a ratio of 1:2. The empirical formula is CH₂.
要通过实验数据确定经验式，请遵循以下步骤：将每种元素的质量（或百分比）除以其 Aᵣ，换算为摩尔数；将每个摩尔值除以所得的最小摩尔数；如有必要，乘以倍数得到整数。例如，一种化合物含有 85.7% 的碳和 14.3% 的氢（质量分数）。C 的摩尔数 = 85.7 ÷ 12 = 7.14；H 的摩尔数 = 14.3 ÷ 1 = 14.3。两者均除以 7.14，得到比值为 1:2。经验式为 CH₂。
12 + (2 × 1) = 14. The multiplier is 56 ÷ 14 = 4. The molecular formula is therefore C₄H₈.
如果后来发现该化合物的 Mᵣ 为 56，你就可以求出分子式。CH₂ 单元的质量为 12 + (2 × 1) = 14。倍数为 56 ÷ 14 = 4。因此分子式为 C₄H₈。
5. Molar Volume of Gases | 气体的摩尔体积
At room temperature and pressure (rtp), defined as 25 °C and 1 atm, one mole of any gas occupies a volume of 24 dm³. This is a simplifying theorem that allows you to link the number of moles of a gas directly to its volume, without needing to know its identity.
在室温常压 (rtp) 下，即 25 °C 和 1 atm 条件下，一摩尔任何气体占据的体积为 24 dm³。这是一条简化定理，让你能够将气体的摩尔数与其体积直接联系起来，而不需要知道气体的种类。
Volume of gas at rtp (dm³) = number of moles × 24
在 rtp 下的气体体积 (dm³) = 摩尔数 × 24
This relationship works in reverse: if you measure the volume of a gas produced in a reaction at rtp, you can calculate the number of moles by dividing the volume by 24. Remember that 1 dm³ equals 1000 cm³. If a volume is given in cm³, convert it to dm³ by dividing by 1000 before applying the formula.
这个关系式也可反向使用：如果你在 rtp 下测量了反应产生的气体体积，可以通过将体积除以 24 来计算摩尔数。请记住，1 dm³ 等于 1000 cm³。如果体积以 cm³ 给出，请先除以 1000 换算为 dm³，再应用公式。
This theorem is based on Avogadro's law, which states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. It explains why 24 dm³ of hydrogen contains the same number of molecules as 24 dm³ of carbon dioxide, even though their masses are very different.
该定理基于阿伏伽德罗定律，即相同温度和压力下，等体积的气体含有相同数量的分子。这解释了为什么 24 dm³ 的氢气与 24 dm³ 的二氧化碳含有相同数量的分子，尽管它们的质量差别很大。
6. Concentration of Solutions | 溶液的浓度
Concentration is a measure of how much solute is dissolved in a given volume of solvent. The most common unit in Year 10 CIE chemistry is moles per cubic decimetre (mol/dm³), sometimes called molarity. This links the mole concept to solution chemistry and is essential for titration calculations.
浓度是衡量给定体积溶剂中溶解了多少溶质的量。Year 10 CIE 化学中最常见的单位是摩尔每立方分米 (mol/dm³)，有时称为摩尔浓度。这将摩尔概念与溶液化学联系起来，对于滴定计算至关重要。
Concentration (mol/dm³) = number of moles (n) ÷ volume of solution (dm³)
浓度 (mol/dm³) = 摩尔数 (n) ÷ 溶液的体积 (dm³)
in g/dm³ = concentration in mol/dm³ × molar mass. For example, a 0.5 mol/dm³ solution of NaOH (molar mass 40 g/mol) has a concentration of 0.5 × 40 = 20 g/dm³.
另一种单位是克每立方分米 (g/dm³)。你可以使用摩尔质量在两者间进行换算。关系式为：浓度 (g/dm³) = 浓度 (mol/dm³) × 摩尔质量。例如，浓度为 0.5 mol/dm³ 的 NaOH 溶液（摩尔质量 40 g/mol）的浓度为 0.5 × 40 = 20 g/dm³。
| Quantity | Formula |
| Concentration (mol/dm³) | n = c × V |
| Mass concentration (g/dm³) | mass = conc (g/dm³) × V (dm³) |
cm³ = 0.250 dm³. This is one of the most common sources of error in solution calculations.
在使用这些公式时，始终确保体积的单位是 dm³。如果给出的体积是 cm³，请先除以 1000：250 cm³ = 0.250 dm³。这是溶液计算中最常见的错误来源之一。
7. Reacting Masses and Limiting Reactants | 反应质量与限量反应物
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a balanced chemical equation. The coefficients in a balanced equation give the mole ratio in which substances react and are produced. This ratio is the central tool for all reacting mass calculations.
化学计量学是化学的一个分支，研究平衡化学方程式中反应物与产物之间的定量关系。平衡方程中的系数给出了物质反应和生成的摩尔比。这个比值是所有反应质量计算的核心工具。
he given mass to moles; (3) use the mole ratio to find moles of the unknown substance; (4) convert moles back to mass. For example, what mass of magnesium oxide forms when 6.0 g of magnesium burns in excess oxygen? g:MgO is 2:2, so moles of MgO = 0.25. Mass of MgO = 0.25 × (24 + 16) = 0.25 × 40 = 10.0 g.
要计算由给定质量的反应物生成的产物质量，请遵循这五步法：(1) 写出平衡方程式；(2) 将给定质量换算为摩尔数；(3) 利用摩尔比求出未知物质的摩尔数；(4) 将摩尔数换算回质量。例如，6.0 克镁在过量氧气中燃烧会生成多少质量的氧化镁？方程式：2Mg + O₂ → 2MgO。Mg 的摩尔数 = 6.0 ÷ 24 = 0.25。摩尔比 Mg:MgO 为 2:2，因此 MgO 的摩尔数 = 0.25。MgO 的质量 = 0.25 × (24 + 16) = 0.25 × 40 = 10.0 克。
The limiting reactant is the substance that is completely used up in a reaction, determining the maximum amount of product that can be formed. The other reactant is present in excess. To identify the limiting reactant, calculate the moles of each reactant, then compare the available mole ratio to the required mole ratio from the balanced equation.
限量反应物是反应中完全消耗掉的物质，它决定了能生成的最大产物量。另一种反应物是过量的。要识别限量反应物，计算每种反应物的摩尔数，然后将可用的摩尔比与平衡方程中所需的摩尔比进行比较。
8. Percentage Yield and Percentage Purity | 产率百分比与纯度百分比
In practical chemistry, the mass of product obtained (the actual yield) is often less than the mass predicted by stoichiometry (the theoretical yield). Percentage yield quantifies the efficiency of a reaction and is a key industrial and laboratory metric.
在实际化学实验中，获得的产物质量（实际产率）往往低于化学计量学预测的质量（理论产率）。产率百分比量化了反应的效率，是工业和实验室中的一项关键指标。
Percentage yield = (actual yield ÷ theoretical yield) × 100%
产率百分比 = (实际产率 ÷ 理论产率) × 100%
Reasons for a yield being less than 100% include: the reaction may be reversible and not go to completion; some product may be lost during filtration, transfer, or purification; unexpected side reactions may produce other substances. A yield greater than 100% usually indicates the product is impure or still contains solvent.
产率低于 100% 的原因包括：反应可能是可逆的，不能进行到底；在过滤、转移或纯化过程中可能会损失部分产物；意外的副反应可能产生其他物质。产率超过 100% 通常表明产物不纯或仍含有溶剂。
Percentage purity is used when a sample of a substance is not pure. It is the mass of the desired substance divided by the total mass of the sample, multiplied by 100%. This is particularly relevant when analysing minerals or when products contain unreacted starting materials.
Percentage purity = (mass of pure substance ÷ mass of impure sample) × 100%
纯度百分比 = (纯物质的质量 ÷ 不纯样品的质量) × 100%
limestone contains 9.2 g of calcium carbonate, the percentage purity is (9.2 ÷ 10.0) × 100% = 92.0% in subsequent stoichiometric calculations to find the actual amount of reactant available.
例如，如果一份 10.0 克的石灰石样品含有 9.2 克碳酸钙，则纯度百分比为 (9.2 ÷ 10.0) × 100% = 92.0%。然后在后续的化学计量计算中使用此值来确定实际可用的反应物量。
9. Energy Changes and Calorimetry | 能量变化与量热法
Chemical reactions either release energy to the surroundings (exothermic) or absorb energy from the surroundings (endothermic). The quantity of heat energy transferred can be calculated using calorimetry, generally by measuring the temperature change of a known mass of water or solution.
化学反应要么向周围环境释放能量（放热），要么从周围环境吸收能量（吸热）。传递的热能量可以通过量热法来计算，通常是测量已知质量的水或溶液的温度变化。
Heat energy transferred (Q) = m × c × ΔT
传递的热能 (Q) = m × c × ΔT
In this formula, Q is in joules (J), m is the mass of the substance being heated (usually water) in grams, c is the specific heat capacity (4.18 J/g°C for water), and ΔT is the temperature change in °C. This equation is the foundation of all simple calorimetry experiments, such as measuring the enthalpy change of neutralisation or combustion.
在该公式中，Q 的单位是焦耳 (J)，m 是被加热物质（通常是水）的质量，单位是克，c 是比热容（水为 4.18 J/g°C），ΔT 是温度变化，单位是 °C。该方程式是所有简单量热实验的基础，例如测量中和反应或燃烧的焓变。
To find the molar enthalpy change (the energy change per mole), divide Q by the number of moles of the limiting reactant. The unit then becomes J/mol or kJ/mol. Always pay attention to the sign: exothermic changes have negative ΔH values, endothermic changes have positive ΔH values. In CIE papers, you may be asked to convert between J and kJ (divide by 1000).
要求出摩尔焓变（每摩尔的能量变化），将 Q 除以限量反应物的摩尔数。此时单位变为 J/mol 或 kJ/mol。务必注意符号：放热变化的 ΔH 值为负，吸热变化的 ΔH 值为正。在 CIE 试卷中，你可能需要换算 J 和 kJ（除以 1000）。
10. Rate of Reaction | 反应速率
The rate of a chemical reaction measures how quickly reactants are converted into products. It can be expressed as the change in the amount of a reactant or product per unit time. In the CIE course, you must be able to calculate rates from graphs and from experimental data.
化学反应速率衡量反应物转化为产物的快慢。它可以表示为每单位时间内反应物或产物量的变化。在 CIE 课程中，你必须能够通过图表和实验数据来计算速率。
Average rate = change in quantity (mass, volume, or moles) ÷ time taken
平均速率 = 量的变化（质量、体积或摩尔数）÷ 所用时间
Common units for rate include g/s, cm³/s, or mol/s, depending on what quantity is being measured. A typical experiment tracks the volume of gas produced as a reaction proceeds. The rate at any given instant is found by drawing a tangent to the curve on a volume–time graph and calculating its gradient.
速率的常见单位包括 g/s、cm³/s 或 mol/s，具体取决于测量的量。一个典型的实验是跟踪反应进行过程中产生的气体体积。任意给定时刻的速率可通过在体积–时间图上画出曲线的切线并计算其斜率来求得。
ce area of solids, and the presence of a catalyst. Each factor works by increasing the frequency of effective collisions, where particles meet with sufficient energy (≥ activation energy Eₐ) and the correct orientation.
根据碰撞理论，反应速率受多种因素影响：浓度（或气体的压力）、温度、固体表面积以及催化剂的存在。每个因素都是通过增加有效碰撞的频率来起作用的，有效碰撞是指粒子以足够的能量（≥ 活化能 Eₐ）和正确的取向相遇。
11. Formulas for Specific Chemical Processes | 特定化学过程的公式
Several core reaction types appear repeatedly in the Year 10 CIE syllabus. Being able to write balanced equations for these processes, and to interpret them in terms of moles, is an essential skill. Here are the key categories and their generalised formulas.
Year 10 CIE 教学大纲中有几类核心反应反复出现。能够为这些过程写出平衡方程式，并从摩尔的角度加以解读，是一项必备技能。以下是关键类别及其通用公式。
| Reaction Type | General Word Equation |
| Metal + Acid → Salt + Hydrogen | Zn + 2HCl → ZnCl₂ + H₂ |
| Metal Carbonate + Acid → Salt + Water + CO₂ | CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ |
| Neutralisation (Acid + Base) → Salt + Water | NaOH + HCl → NaCl + H₂O |
| Combustion of Hydrocarbon → CO₂ + H₂O | CH₄ + 2O₂ → CO₂ + 2H₂O |
| Thermal Decomposition of Carbonate → Oxide + CO₂ | CuCO₃ → CuO + CO₂ |
In each case, the balanced equation provides the mole ratio needed for quantitative predictions. For example, the neutralisation of sodium hydroxide by hydrochloric acid has a 1:1 ratio. If you know the moles of acid used, you immediately know the moles of base neutralised, which is fundamental to titration problems.
在每种情况下，平衡方程式都提供了定量预测所需的摩尔比。例如，氢氧化钠被盐酸中和的反应比例为 1:1。如果你知道所用酸的摩尔数，就立即知道被中和的碱的摩尔数，这是滴定问题的基础。
12. Graphs, Trends, and Mathematical Patterns | 图表、趋势与数学规律
Beyond plugging numbers into equations, the CIE course expects you to interpret graphical data. Common graphs include mass lost against time for reactions producing gases, temperature change against time for calorimetry, and volume of gas against time for reaction rate studies. The shape of each curve tells a chemical story: initial fast rate, gradual slowing, and finally a plateau when the limiting reactant is used up.
除了将数字代入方程式外，CIE 课程还要求你解读图表数据。常见的图表包括产生气体的反应中质量随时间减少的图、量热法中温度随时间变化的图，以及用于研究反应速率的气体体积随时间变化的图。每条曲线的形状都在讲述一个化学故事：初始快速，逐渐减慢，最终在限量反应物耗尽时达到平台。
In the Periodic Table, mathematical patterns underpin the group and period trends. For Group 1 elements, the reaction with water becomes more vigorous as you descend the group because the atomic radius increases, making it easier to lose the outer electron. The empirical formula of an oxide, such as Li₂O or Na₂O, reflects the Group 1 charge of +1. These trends are not given by single formulas but by the underlying atomic structure relationships: as you go down a group, the number of electron shells increases, which increases shielding and radius, affecting reactivity.
在元素周期表中，数学规律支撑着族和周期的趋势。对于第 1 族元素，从上到下与水反应的剧烈程度增加，因为原子半径增大，使得外层电子更容易失去。氧化物的经验式，如 Li₂O 或 Na₂O，反映了第 1 族元素的 +1 电荷。这些趋势并非由单一公式给出，而是由背后的原子结构关系决定的：同一族向下，电子层数增加，屏蔽效应和半径随之增大，从而影响反应性。
When tackling graph-based questions, always label your axes clearly, use appropriate scales, and draw lines of best fit. To find rates from a curved graph, construct a tangent at the specific time, then calculate the gradient: rise ÷ run. This mathematical skill is as important as any formula in the handbook.
在处理基于图表的问题时，务必清晰地标注坐标轴，使用合适的刻度，并画出最佳拟合线。要从曲线图中求速率，请在特定时间点作切线，然后计算斜率：纵坐标变化量 ÷ 横坐标变化量。这项数学技能与手册中任何公式同等重要。
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