📚 WJEC Chemistry Year 11 Formula & Theorem Quick-Reference Handbook | WJEC 化学 11 年级公式定理速查手册
This quick-reference handbook covers all essential chemical formulas, equations, and principles required for the Year 11 WJEC GCSE Chemistry specification. Each section presents the core concept in English and Mandarin, followed by formula statements, worked applications, and key definitions. Use this guide to consolidate your understanding of quantitative chemistry, energy changes, reaction rates, and more.
本速查手册涵盖 WJEC 11 年级 GCSE 化学大纲中所有必备的化学公式、方程式和原理。每个小节均以中英双语介绍核心概念，随后给出公式表述、应用示例和关键定义。使用本手册可帮助你巩固对定量化学、能量变化、反应速率等内容的掌握。
1. Relative Atomic Mass and Relative Formula Mass | 相对原子质量与相对式量
The relative atomic mass (Ar) of an element is the average mass of its atoms, taking into account the abundances of its isotopes, compared to 1/12th the mass of a carbon‑12 atom. The relative formula mass (Mr) of a compound is the sum of the relative atomic masses of all atoms in its formula.
元素的相对原子质量（Ar）是其所有同位素原子质量的自然丰度平均值，以碳‑12 原子质量的 1/12 为标准。化合物的相对式量（Mr）是其化学式中所有原子的相对原子质量之和。
Calculation of Ar from isotopic abundances:
根据同位素丰度计算 Ar：
Ar = Σ (isotope mass × % abundance) / 100
Ar = Σ（同位素质量 × 丰度%）/ 100
Example: Chlorine has two isotopes — ³⁵Cl (75%) and ³⁷Cl (25%). Ar = (35 × 75 + 37 × 25) / 100 = 35.5.
示例：氯有两种同位素 — ³⁵Cl（75%）和 ³⁷Cl（25%）。Ar = (35×75 + 37×25) / 100 = 35.5。
Mr calculation: Mr of H₂SO₄ = (2×1) + 32 + (4×16) = 98.
Mr 计算： H₂SO₄ 的 Mr = (2×1) + 32 + (4×16) = 98。
2. The Mole and Avogadro’s Constant | 摩尔与阿伏伽德罗常数
s, ions, or electrons). This number is called Avogadro’s constant. The mole is the unit for amount of substance (symbol n).
1 摩尔任何物质均含有 6.02 × 10²³ 个微粒（原子、分子、离子或电子），该数值称为阿伏伽德罗常数。摩尔是物质的量的单位（符号 n）。
number of particles = n × 6.02 × 10²³
Use the mole to convert between mass, number of particles, and volume of gases.
利用摩尔可在质量、微粒数和气体体积之间进行换算。
3. Mass, Moles, and Molar Mass | 质量、摩尔与摩尔质量
The relationship between mass (m), molar mass (M), and amount in moles (n) is the most frequently used equation in quantitative chemistry.
质量（m）、摩尔质量（M）和物质的量（n）之间的关系是定量化学中最常用的方程。
n = m / M
l and is numerically equal to the relative formula mass (Mr). Rearranging gives m = n × M.
摩尔质量（M）的单位为 g/mol，数值上等于相对式量（Mr）。移项得 m = n × M。
Example: How many moles are in 10 g of CaCO₃? Mr = 100, so n = 10 / 100 = 0.10 mol.
示例：10 g CaCO₃ 是多少摩尔？Mr = 100，因此 n = 10 / 100 = 0.10 mol。
4. Concentration of Solutions | 溶液浓度
The concentration of a solution tells us how much solute is dissolved in a given volume of solvent. In WJEC, concentration is usually expressed in mol/dm³.
溶液浓度表示在一定体积溶剂中溶解了多少溶质。在 WJEC 中，浓度通常以 mol/dm³ 表示。
c = n / V
mol), V = volume (dm³). If volume is given in cm³, divide by 1000: V(dm³) = V(cm³) / 1000.
其中 c = 浓度（mol/dm³），n = 物质的量（mol），V = 体积（dm³）。若体积以 cm³ 给出，需除以 1000：V(dm³) = V(cm³) / 1000。
mple: 0.50 mol of NaCl dissolved in 250 cm³ of water. c = 0.50 / (250/1000) = 2.0 mol/dm³.
示例：0.50 mol NaCl 溶于 250 cm³ 水中。c = 0.50 / (250/1000) = 2.0 mol/dm³。
sed: mass concentration = mass (g) / volume (dm³). Conversion: mass concentration = c × M.
也可使用质量浓度（g/dm³）：质量浓度 = 质量（g）/ 体积（dm³）。换算关系：质量浓度 = c × M。
5. Molar Volume of Gases | 气体摩尔体积
At room temperature and pressure (RTP, 20 °C and 1 atm), one mole of any gas occupies 24 dm³. This is known as the molar gas volume.
在常温常压下（RTP，20 °C，1 atm），1 摩尔任何气体的体积均为 24 dm³，称为气体摩尔体积。
V (dm³) = n × 24
This relationship is valid for all gases at RTP. For other conditions, the ideal gas equation is not required at this level.
该关系适用于 RTP 条件下的所有气体。本阶段不要求使用其他条件下的理想气体状态方程。
ol of CO₂ occupies 0.20 × 24 = 4.8 dm³. Conversely, 48 dm³ of O₂ equals 48 / 24 = 2.0 mol.
示例：0.20 mol CO₂ 的体积为 0.20 × 24 = 4.8 dm³。反过来，48 dm³ 的 O₂ 等于 48 / 24 = 2.0 mol。
6. Balanced Equations and Stoichiometry | 化学方程式与化学计量
A balanced chemical equation shows the ratio of moles of reactants and products. The numbers in front of substances are called stoichiometric coefficients.
配平的化学方程式表示反应物和生成物之间的摩尔比。物质前面的数字称为化学计量系数。
Example: 2H₂ + O₂ → 2H₂O means 2 mol of H₂ react with 1 mol of O₂ to produce 2 mol of H₂O.
示例：2H₂ + O₂ → 2H₂O 表示 2 mol H₂ 与 1 mol O₂ 反应生成 2 mol H₂O。
Use the mole ratio to calculate unknown quantities. Steps: (1) Convert given mass to moles; (2) Use mole ratio from equation; (3) Convert moles back to mass or volume.
利用摩尔比可计算未知量。步骤：（1）将已知质量转换为摩尔；（2）利用方程式中的摩尔比；（3）将摩尔转换回质量或体积。
Example: What mass of MgO is formed from 6 g of Mg? Mg = 6 / 24 = 0.25 mol. Mole ratio Mg:MgO = 1:1, so 0.25 mol MgO. Mass = 0.25 × 40 = 10 g.
示例：6 g Mg 能生成多少 MgO？2Mg + O₂ → 2MgO。Mg 的摩尔数 = 6 / 24 = 0.25 mol。Mg:MgO 摩尔比 = 1:1，因此生成 0.25 mol MgO。质量 = 0.25 × 40 = 10 g。
7. Percentage Yield | 百分比产率
The percentage yield compares the actual mass of product obtained from an experiment to the theoretical mass calculated from stoichiometry. It measures the efficiency of a reaction.
百分比产率将实验实际获得的产品质量与根据化学计量计算的理论质量进行比较，用于衡量反应的效率。
% yield = (actual yield / theoretical yield) × 100
Yield is always less than 100% due to incomplete reactions, side reactions, or product lost during purification.
由于反应不完全、副反应或纯化过程中产品损失，产率始终低于 100%。
Example: If the theoretical yield is 20.0 g and the actual yield is 15.5 g, % yield = (15.5 / 20.0) × 100 = 77.5%.
示例：若理论产量为 20.0 g，实际产量为 15.5 g，则 % 产率 = (15.5 / 20.0) × 100 = 77.5%。
8. Atom Economy | 原子经济性
Atom economy evaluates how much of the reactants’ mass ends up in the desired product. A higher atom economy means a more sustainable process with less waste.
原子经济性衡量反应物的质量中有多少进入了目标产物。原子经济性越高，过程越可持续，废弃物越少。
atom economy = (Mr of desired product / total Mr of all reactants) × 100
All reactants must be included, even if they are added in excess. The formula uses the Mr of the chemical equation as written.
所有反应物均需计入，即使它们过量。该公式使用化学方程式中各物质的 Mr。
60 + 84 = 244. Mr of desired product = 2×56 = 112. Atom economy = (112 / 244) × 100 = 45.9%.
示例：Fe₂O₃ + 3CO → 2Fe + 3CO₂。目标产物 = Fe。反应物总 Mr = (2×56+3×16) + 3×(12+16) = 160 + 84 = 244。目标产物 Mr = 2×56 = 112。原子经济性 = (112 / 244) × 100 = 45.9%。
9. Bond Energies and Enthalpy Change | 键能与焓变
Energy changes in a chemical reaction can be calculated using average bond energies. Bond breaking absorbs energy (endothermic); bond making releases energy (exothermic).
化学反应的能量变化可利用平均键能计算。断键吸收能量（吸热）；成键释放能量（放热）。
ΔH = Σ (bond energies of bonds broken) – Σ (bond energies of bonds formed)
ΔH = Σ（断裂键的键能总和） – Σ（形成键的键能总和）
A negative ΔH indicates an exothermic reaction; a positive ΔH indicates an endothermic reaction. The values are given in kJ/mol.
ΔH 为负值表示放热反应；ΔH 为正值表示吸热反应。数值单位为 kJ/mol。
ds broken: H-H (436) + Cl-Cl (243) = 679. Bonds formed: 2 × H-Cl (2 × 432 = 864). ΔH = 679 – 864 = –185 kJ/mol (exothermic).
示例：H₂ + Cl₂ → 2HCl。断裂键：H−H (436) + Cl−Cl (243) = 679。形成键：2 × H−Cl (2×432 = 864)。ΔH = 679 – 864 = –185 kJ/mol（放热）。
10. Titration Calculations | 滴定计算
Titration is a method used to determine the concentration of an unknown solution using a solution of known concentration. The key relationship uses the mole ratio from the balanced equation.
滴定是利用已知浓度溶液测定未知浓度溶液的方法。其关键关系来自配平方程式中的摩尔比。
cₐ × Vₐ / cb × Vb = nₐ / nb
irectly: n = c × V, and then apply mole ratio. Usually V is in cm³, so n = (c × V) / 1000.
或更直接地：n = c × V，然后应用摩尔比。通常 V 以 cm³ 为单位，因此 n = (c × V) / 1000。
tio 1:1. Moles of HCl = (0.10 × 30.0) / 1000 = 0.0030 mol. Moles of NaOH = 0.0030 mol. c(NaOH) = (0.0030 / 25.0) × 1000 = 0.12 mol/dm³.
示例：25.0 cm³ NaOH 溶液被 30.0 cm³ 0.10 mol/dm³ HCl 中和。计算 NaOH 的浓度。HCl + NaOH → NaCl + H₂O。摩尔比 1:1。HCl 的物质的量 = (0.10×30.0)/1000 = 0.0030 mol。NaOH 的物质的量 = 0.0030 mol。c(NaOH) = (0.0030 / 25.0)×1000 = 0.12 mol/dm³。
11. Rate of Reaction | 反应速率
The rate of a chemical reaction measures how quickly reactants are used up or products are formed. It can be determined by monitoring the change in amount or concentration over time.
化学反应速率衡量反应物消耗或产物生成的快慢，可通过监测量或浓度随时间的变化来确定。
mean rate = change in quantity / change in time
平均速率 = 变化量 / 变化时间
Units depend on the quantity measured — g/s for mass loss, cm³/s for gas volume produced, or mol/(dm³·s) for concentration changes.
单位取决于测量量 — 质量减少常用 g/s，气体体积用 cm³/s，浓度变化用 mol/(dm³·s)。
For a graph of volume of gas against time, the rate at any instant is the gradient (slope) of the tangent at that point.
对于气体体积-时间图，任意时刻的瞬时速率即为该点切线的斜率（梯度）。
12. pH and Hydrogen Ion Concentration | pH 值与氢离子浓度
The pH scale measures how acidic or alkaline a solution is. It is related to the concentration of hydrogen ions, [H⁺], in mol/dm³.
pH 标度用于衡量溶液的酸碱度，与溶液中的氢离子浓度 [H⁺]（mol/dm³）相关。
pH = –log₁₀[H⁺]
A solution of [H⁺] = 0.01 mol/dm³ has pH = –log₁₀(0.01) = 2. Decreasing pH by 1 means [H⁺] increases by a factor of 10.
[H⁺] = 0.01 mol/dm³ 的溶液 pH = –log₁₀(0.01) = 2。pH 值降低 1 意味着 [H⁺] 增大 10 倍。
centration. For dilution, use c₁V₁ = c₂V₂ to find new concentration before calculating pH.
对于强的一元酸，[H⁺] 等于酸的浓度。稀释时，先用 c₁V₁ = c₂V₂ 求出新浓度，再计算 pH。
Published by TutorHao | Chemistry Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图