📚 Year 10 WJEC Chemistry: Formula and Theorem Quick Reference Handbook | Year 10 WJEC 化学：公式定理速查手册
This quick reference handbook gathers the key formulas, equations and principles you need to master for the Year 10 WJEC Chemistry course. Use it alongside your class notes and revision materials to strengthen your problem-solving skills and conceptual understanding. Every section gives the equation, the meaning of each symbol, the correct units and a short worked example so that you can apply it straight away in a test or an examination question.
本速查手册汇集了 Year 10 WJEC 化学课程中必须掌握的核心公式、方程式和原理。结合课堂笔记与复习资料使用，可帮助提升解题能力与概念理解。每一节都给出公式本身、各符号的含义、正确的单位以及简短例题，方便你在测验或考试题目中直接套用。
1. Atomic Structure and Relative Atomic Mass | 原子结构与相对原子质量
ss number (A) is the number of protons plus neutrons, so A = Z + N, where N is the number of neutrons. Atoms are electrically neutral, so Z is also equal to the number of electrons.
所有原子都由质子、中子和电子构成。原子序数 (Z) 是质子数，决定了元素的种类。质量数 (A) 为质子数与中子数之和：A = Z + N，其中 N 为中子数。原子不带电，因此 Z 也等于电子数。
A = Z + N（N = A − Z）
ment compared with 1/12 of the mass of a carbon-12 atom, and it has no units. For chlorine, about 75% of atoms have mass number 35 and 25% have mass number 37, so Ar(Cl) = (75 × 35 + 25 × 37) ÷ 100 = 35.5.
同位素是同一元素中 Z 相同但 N 不同的原子。相对原子质量 (Ar) 是元素各同位素质量的加权平均值，以碳-12 原子质量的 1/12 作标准比较，没有单位。对于氯，约 75% 的原子质量数为 35，25% 为 37，因此 Ar(Cl) = (75 × 35 + 25 × 37) ÷ 100 = 35.5。
Ar = Σ (abundance × isotope mass) ÷ 100
(2 × 1) + 16 = 18. For CaCO₃, Mr = 40 + 12 + (3 × 16) = 100. Remember that Mr is a ratio, so it never carries the unit g mol⁻¹; it is the molar mass in g mol⁻¹ that carries units.
相对分子质量 (Mr) 也称相对式量，是化学式中所有原子 Ar 值的总和。对于 H₂O，Mr = (2 × 1) + 16 = 18；对于 CaCO₃，Mr = 40 + 12 + (3 × 16) = 100。要注意 Mr 是一个比值，不含单位 g mol⁻¹；带单位的是摩尔质量。
Exam tip: mass spectra and percentage abundance data almost always ask for Ar. Multiply each abundance by its mass, add the products, then divide by the total abundance.
考试提示：质谱与丰度数据题几乎都在考查 Ar。将每个丰度乘以对应质量，求和后再除以丰度总和即可。
2. Mole and Molar Mass | 摩尔与摩尔质量
gadro constant). The molar mass (M) of a substance is the mass of one mole, with units of g mol⁻¹, and it is numerically equal to the relative atomic mass or the relative molecular mass.
摩尔是物质的量的国际单位。1 摩尔任何物质都含有 6.02 × 10²³ 个粒子（阿伏加德罗常数）。物质的摩尔质量 (M) 是 1 摩尔物质的质量，单位为 g mol⁻¹，数值上等于相对原子质量或相对分子质量。
n = m ÷ M
ns you will be given grams and asked for moles, but always check the unit before you substitute.
其中 n 为物质的量 (mol)，m 为质量 (g)，M 为摩尔质量 (g mol⁻¹)。该式可变形为 m = n × M 或 M = m ÷ n。WJEC 题目大多给出克数求摩尔数，但在代入前一定要先检查单位。
ions we usually use n = cV (see Section 5); for gases at room temperature and pressure, n = V (dm³) ÷ 24 (see Section 6); for particle counts, n = number of particles ÷ 6.02 × 10²³.
摩尔是定量化学的中心桥梁：它把质量与浓度、气体体积、粒子数目联系起来。处理溶液时常用 n = cV（见第 5 节）；处理室温和常压下的气体时，n = V (dm³) ÷ 24（见第 6 节）；计算粒子数时用 n = 粒子数 ÷ 6.02 × 10²³。
Worked example: 44, so n = 4.4 ÷ 44 = 0.10 mol, which contains 0.10 × 6.02 × 10²³ = 6.02 × 10²² molecules.
例题：4.4 g CO₂ 的 Mr = 12 + (2 × 16) = 44，故 n = 4.4 ÷ 44 = 0.10 mol，含 0.10 × 6.02 × 10²³ = 6.02 × 10²² 个分子。
3. Empirical Formula, Molecular Formula and Percentage Composition | 实验式、分子式与百分组成
The empirical formula is the simplest whole-number ratio of atoms in a compound; the molecular formula is the actual number of atoms in one molecule. To find an empirical formula from masses, divide each mass by the Ar of that element, then divide every answer by the smallest one, and finally convert the ratio into small whole numbers.
实验式是化合物中各原子最简单整数比；分子式是分子中实际的原子数目。由质量求实验式：用每种元素的质量除以该元素的 Ar，再把所有结果除以其中最小值，最后把比值化为小整数。
Percentage composition tells you the mass of an element in 100 g of a compound. It is calculated by dividing the total mass of that element in the formula by Mr, then multiplying by 100. This is the routine used in fertiliser and alloy questions.
百分组成表示 100 g 化合物中某元素的质量。计算方法是用化学式中该元素的总质量除以 Mr，再乘以 100。肥料与合金类题目常用这一方法。
% element = (total mass of element in formula ÷ Mr) × 100
example, in ammonium nitrate NH₄NO₃, Mr = 80, and the mass of nitrogen is 2 × 14 = 28, so %N = (28 ÷ 80) × 100 = 35.0%. For percentage yield and atom economy, use the mass-based definitions in Section 4.
例如硝酸铵 NH₄NO₃ 的 Mr = 80，氮的总质量为 2 × 14 = 28，故 %N = (28 ÷ 80) × 100 = 35.0%。百分产率与原子经济性的计算见第 4 节，均基于质量。
Hydrated salts such as CuSO₄·5H₂O contain water of crystallisation. To find the value of x, subtract the mass after heating from the mass before heating to get the mass of water lost, then convert both masses into moles and take the ratio.
CuSO₄·5H₂O 这类水合盐含有结晶水。求 x 值时，用加热前的质量减去加热后的质量得到失去的水的质量，再分别把两种质量换算成摩尔，取比值即可。
4. Balanced Equations and Reacting Masses | 配平方程式与反应质量
A balanced equation shows the formulae of the reactants and products and the numbers of each particle involved. You may not change any subscript, because that changes the substance; you may only change the large number written in front of a formula. Balance one element at a time, leaving hydrogen and oxygen until last, and check every atom at the end.
配平方程式表示反应物与生成物的化学式以及各粒子的数目。绝对不能改动下标，因为那会改变物质本身，只能改变写在化学式前面的大数字。每次只配平一种元素，把氢和氧留到最后，最后逐一核对原子数。
2H₂ + O₂ → 2H₂O states: (s) (l) (g) (aq)
Use state symbols (s), (l), (g) and (aq) whenever a question asks for them, and remember that in an aqueous solution ionic compounds are shown as separate ions. Total mass is always conserved in a chemical reaction, which is why a balanced equation can be read as a recipe in moles.
题目要求时须写出状态符号 (s)、(l)、(g)、(aq)；在水溶液中，离子化合物要写成离子形式。化学反应中总质量守恒，因此配平方程式可以当作以摩尔为单位的配方来读。
Reacting mass calculations follow four steps: (1) write and balance the equation; (2) calculate the moles of the substance whose mass is given; (3) use the mole ratio from the equation to find the moles of the required substance; (4) convert those moles back into a mass, a volume or a concentration using the appropriate equation from this handbook.
反应质量计算分四步：(1) 写出并配平方程式；(2) 求出已知质量物质的摩尔数；(3) 用方程式中的摩尔比求出目标物质的摩尔数；(4) 用本手册中相应的公式把摩尔数换算回质量、体积或浓度。
100. A high yield does not guarantee a high atom economy, because a reaction may produce the wanted product efficiently but also create a large mass of by-product.
百分产率 = (实际产量 ÷ 理论产量) × 100；原子经济性 = (目标产物的 Mr ÷ 所有产物 Mr 之和) × 100。高产率并不等于高原子经济性，因为反应可能高效地生成目标产物，同时副产大量废物。
5. Solutions, Concentration and Titration | 溶液、浓度与滴定
Concentration is the amount of solute dissolved in a given volume of solution. The standard unit is mol dm⁻³, sometimes written as M. The key relationship links moles, concentration and volume in dm³, and it is the equation you will use most often in Year 10 practical work.
浓度是一定体积溶液中所含溶质的量，标准单位是 mol dm⁻³，有时写作 M。核心关系式把摩尔数、浓度与以 dm³ 为单位的体积联系在一起，是 Year 10 实验中最常用的公式。
n = c × V
cm³, divide by 1000 first: V (dm³) = V (cm³) ÷ 1000. Forgetting this conversion is the single most common error in this topic.
式中 n 为物质的量 (mol)，c 为浓度 (mol dm⁻³)，V 为体积 (dm³)。变形得 c = n ÷ V，V = n ÷ c。若体积以 cm³ 给出，必须先除以 1000：V (dm³) = V (cm³) ÷ 1000。忘掉这一步是本主题最常见的失分点。
applies. The volumes may be left in cm³ on both sides because the units cancel, but you must keep them consistent.
用水稀释溶液时溶质的量不变，因此可用稀释公式 c₁V₁ = c₂V₂。两边都用 cm³ 也可以，因为单位相互抵消，但必须保持一致。
Worked titration example: rom NaOH + HCl → NaCl + H₂O the ratio is 1 : 1, so n(HCl) = 2.50 × 10⁻³ mol and c(HCl) = 2.50 × 10⁻³ ÷ 0.0200 = 0.125 mol dm⁻³.
滴定例题：25.0 cm³、0.100 mol dm⁻³ 的 NaOH 被未知浓度的 HCl 中和，用去酸 20.0 cm³。n(NaOH) = 0.100 × 0.0250 = 2.50 × 10⁻³ mol。由 NaOH + HCl → NaCl + H₂O 可知摩尔比为 1 : 1，故 n(HCl) = 2.50 × 10⁻³ mol，c(HCl) = 2.50 × 10⁻³ ÷ 0.0200 = 0.125 mol dm⁻³。
6. Gases and Molar Volume | 气体与摩尔体积
At room temperature and pressure (RTP, about 20 °C and 1 atm), one mole of any gas occupies the same volume, 24 dm³, which equals 24 000 cm³. This is the molar volume of a gas, and it allows you to convert between the volume of a gas and the amount in moles without knowing which gas it is.
在室温常压（RTP，约 20 °C、1 atm）下，1 摩尔任何气体都占据相同的体积 24 dm³，即 24 000 cm³。这就是气体的摩尔体积，它让你无需知道是何种气体，就能在气体体积与摩尔数之间换算。
n = V (dm³) ÷ 24 V = n × 24
Remember to convert cm³ into dm³ before dividing, and always state that the volume refers to RTP. If a question specifies a temperature far from 20 °C, the value 24 dm³ no longer applies exactly and the ideal gas equation below is more appropriate.
除以 24 之前先要把 cm³ 换算成 dm³，并说明该体积是在 RTP 条件下测得。如果题目给出的温度与 20 °C 相差很远，24 dm³ 就不再精确适用，此时更适合用下面的理想气体方程。
pV = nRT
. To convert from °C to K, add 273. Remember that 1 dm³ = 1 × 10⁻³ m³ and 1 atm ≈ 101 kPa.
理想气体方程中，p 为压强 (Pa)，V 为体积 (m³)，n 为摩尔数，T 为热力学温度 (K)，R = 8.31 J mol⁻¹ K⁻¹。由 °C 换算 K 要加 273。注意 1 dm³ = 1 × 10⁻³ m³，1 atm ≈ 101 kPa。
Worked example: what is the mass of 4.8 dm³ of O₂ at RTP? n = 4.8 ÷ 24 = 0.20 mol, and M(O₂) = 32 g mol⁻¹, so m = 0.20 × 32 = 6.4 g.
例题：RTP 下 4.8 dm³ O₂ 的质量是多少？n = 4.8 ÷ 24 = 0.20 mol，M(O₂) = 32 g mol⁻¹，故 m = 0.20 × 32 = 6.4 g。
7. Bonding, Structure and Intermolecular Forces | 化学键、结构与分子间作用力
Ionic bonding transfers electrons from a metal to a non-metal, producing positive and negative ions held in a giant lattice by strong electrostatic attraction. The formula of an ionic compound is found by balancing the charges: the charge on one ion becomes the subscript of the other, then the ratio is simplified. For example Mg²⁺ with Cl⁻ gives MgCl₂.
离子键是金属把电子转移给非金属，生成正负离子，靠强静电引力构成巨型晶格。由离子电荷写化学式：一种离子的电荷数成为另一种离子的下标，再把比值化简。例如 Mg²⁺ 与 Cl⁻ 组合得 MgCl₂。
Covalent bonding shares pairs of electrons between non-metal atoms. A single bond shares two electrons, a double bond four and a triple bond six. Dot-and-cross diagrams must show only outer-shell electrons, and the total number of electrons shown must equal the sum of the outer electrons of all the atoms.
共价键是非金属原子之间共用电子对。单键共用 2 个电子，双键 4 个，三键 6 个。点叉图只需画出最外层电子，且所画电子总数必须等于各原子最外层电子数之和。
Between simple molecules there are weak intermolecular forces: London dispersion forces in all molecules, permanent dipole-dipole attraction in polar molecules, and hydrogen bonding when H is bonded to N, O or F. These forces are much weaker than covalent bonds, which is why simple molecular substances have low melting and boiling points while giant covalent substances such as diamond and silicon dioxide are extremely high melting.
简单分子之间存在较弱的分子间作用力：所有分子都有伦敦色散力，极性分子还有永久偶极–偶极作用，当 H 与 N、O、F 相连时还能形成氢键。这些作用力远弱于共价键，所以简单分子物质熔点沸点低，而金刚石、二氧化硅这类巨型共价结构熔点极高。
Metallic bonding consists of positive ions in a sea of delocalised electrons. This model explains why metals conduct electricity, are malleable and have high melting points. Dipole moments arise when two bonded atoms have different electronegativities, with fluorine being the most electronegative element.
金属键是正离子浸在离域电子海之中。该模型解释了金属为何导电、有延展性且熔点高。当成键两原子电负性不同时产生偶极，其中氟是电负性最大的元素。
8. Energetics: Enthalpy Change and Bond Energies | 热化学：焓变与键能
Enthalpy change (ΔH) is the heat energy transferred at constant pressure, usually measured in kJ mol⁻¹. An exothermic reaction releases heat to the surroundings and has a negative ΔH, while an endothermic reaction absorbs heat and has a positive ΔH. Combustion is exothermic; thermal decomposition is usually endothermic.
焓变 (ΔH) 是恒压下传递的热能，通常单位为 kJ mol⁻¹。放热反应向环境释放热量，ΔH 为负；吸热反应吸收热量，ΔH 为正。燃烧是放热反应，热分解通常是吸热反应。
In calorimetry experiments, the heat released is found from the temperature rise of the surrounding water using the specific heat capacity equation, then divided by the number of moles to give the molar enthalpy change.
在量热实验中，先用比热容公式由水温升高值求出放出的热量，再除以摩尔数，即得摩尔焓变。
Q = m × c × ΔT c(water) = 4.18 J g⁻¹ °C⁻¹
nthalpy, ΔH = −Q ÷ n ÷ 1000, with the sign negative for an exothermic reaction. Always add "insulation" and "use a lid" as improvements to reduce heat loss to the surroundings.
式中 Q 为热量 (J)，m 为被加热水的质量 (g)（数值上等于其体积的 cm³ 数），ΔT 为温度升高值 (°C)。求摩尔焓变时 ΔH = −Q ÷ n ÷ 1000，放热反应取负号。改进实验时应写"加隔热层""加盖子"以减少向环境的热损失。
Bond energy calculations use the idea that breaking bonds absorbs energy and making bonds releases energy. The overall enthalpy change is the energy needed to break all the bonds in the reactants minus the energy released when all the bonds in the products form.
键能计算依据：断键吸热，成键放热。总焓变等于断开会部反应物化学键所需能量减去生成全部生成物化学键时释放的能量。
ΔH = Σ(bonds broken) − Σ(bonds formed)
¹. Activation energy is the minimum energy that colliding particles need in order to react, and it is shown as the peak of the energy profile diagram.
例如 H₂ + Cl₂ → 2HCl 中，断键共 436 + 243 = 679 kJ mol⁻¹，成键共 2 × 431 = 862 kJ mol⁻¹，故 ΔH = 679 − 862 = −183 kJ mol⁻¹。活化能是碰撞粒子发生反应所需的最低能量，在能量变化图上表现为峰顶。
9. Rates of Reaction and Chemical Equilibrium | 反应速率与化学平衡
The rate of a reaction can be measured as the amount of a reactant used up, or the amount of a product formed, divided by the time taken. Typical units are g s⁻¹, cm³ s⁻¹ or mol dm⁻³ s⁻¹, and the gradient of a mass-time or volume-time graph gives the rate at that instant.
反应速率可以表示为单位时间内反应物的消耗量或生成物的生成量，常用单位为 g s⁻¹、cm³ s⁻¹ 或 mol dm⁻³ s⁻¹。质量–时间或体积–时间图上的斜率即为该时刻的瞬时速率。
According to collision theory, a reaction happens only when particles collide with energy at least equal to the activation energy and in the correct orientation. Increasing concentration or pressure, raising temperature, increasing surface area and adding a catalyst all increase the frequency or the success rate of collisions.
根据碰撞理论，只有当粒子碰撞能量不低于活化能且取向合适时反应才发生。增大浓度或压强、升高温度、增大接触面积以及加入催化剂，都能提高碰撞频率或有效碰撞的比率。
