Year 11 Eduqas Chemistry: Formula & Theorem Quick Reference Handbook | 11年级Eduqas化学：公式定理速查手册

📚 Year 11 Eduqas Chemistry: Formula & Theorem Quick Reference Handbook | 11年级Eduqas化学：公式定理速查手册

This quick reference handbook brings together every essential formula, equation and theorem you need to master Year 11 Eduqas Chemistry. From moles and titrations to energy changes and equilibrium, these are the tools that will support your quantitative chemistry, explain reactions and secure marks in the exam. Keep it close, practise applying each relationship and you will build the confidence to tackle any calculation or explanation question.

这本速查手册汇总了11年级Eduqas化学必须掌握的每一个关键公式、方程和定理。从摩尔、滴定到能量变化和平衡，这些工具将支撑你的定量化学计算、解释反应现象并帮助你在考试中拿下分数。将它放在手边，反复练习应用每一个关系式，你就能建立起解答任何计算或解释题的信心。

1. Relative Atomic Mass and Relative Formula Mass | 相对原子质量与相对式量

Aᵣ) of an element is the average mass of its atoms compared to 1/12th the mass of a carbon‑ s of every atom in the formula. For example, Mᵣ of Mg(OH)₂ = 24.3 + 2×(16.0 + 1.0) = 58.3.

相对原子质量(Aᵣ)是某元素原子的平均质量与碳‑12原子质量的1/12相比较得到的数值，它考虑了同位素的丰度。计算化合物的相对式量(Mᵣ)时，只需将化学式中所有原子的Aᵣ相加。例如，Mg(OH)₂的Mᵣ = 24.3 + 2×(16.0 + 1.0) = 58.3。

You must always use the Aᵣ values from the Periodic Table provided in the exam, and remember that Mᵣ has no units because it is a relative measure.

考试时务必使用提供的周期表中的Aᵣ数值，并记住Mᵣ只是一个相对量度，因此没有单位。

2. The Mole and Molar Mass | 摩尔与摩尔质量

onstant). The mass of one mole of a substance in grams is numerically equal to its relative formula mass. This is called the molar mass, with units g/mol.

1摩尔物质所含的微粒数恰好是6.02 × 10²³（阿伏伽德罗常数）。1摩尔物质的质量以克为单位，在数值上等于其相对式量，称为摩尔质量，单位是g/mol。

oles. Always show units in calculations and convert mass into grams if given in kilograms.

核心三角关系式是：摩尔 = 质量(g) ÷ 摩尔质量(g/mol)。变形可得，质量 = 摩尔 × 摩尔质量，摩尔质量 = 质量 ÷ 摩尔。计算时务必标出单位，如果给出质量单位为千克，要先换算成克。

3. Concentration and Titration Core Formula | 浓度与滴定核心公式

volume is given in cm³, divide by 1000 first. For example, 0.050 mol dissolved in 250 cm³ gives c = 0.050 ÷ 0.250 = 0.20 mol/dm³.

溶液的浓度（单位mol/dm³）由下式给出：浓度 = 溶质的摩尔数 ÷ 溶液的体积(dm³)。常写作 c = n / V。如果体积以cm³给出，先除以1000。例如，0.050 mol溶质溶解在250 cm³中，c = 0.050 ÷ 0.250 = 0.20 mol/dm³。

re the stoichiometric coefficients from the balanced equation. This is derived from the fact that moles of reactants combine in the ratio of their coefficients. Use it to find an unknown concentration or volume.

在滴定中，两种反应溶液的关系为：(c₁ × V₁) / n₁ = (c₂ × V₂) / n₂，其中n₁和n₂是配平方程中的化学计量数。这是基于反应物的摩尔数以化学计量数之比结合。用它可求出未知浓度或体积。

4. Percentage Yield and Atom Economy | 产率百分比与原子经济性

Percentage yield compares the actual amount of product obtained to the maximum theoretical amount. The formula is:

Percentage yield = (actual mass ÷ theoretical mass) × 100

产率百分比将实际获得的产品量与最大理论量进行比较。公式为：

产率百分比 = (实际质量 ÷ 理论质量) × 100

You can also use moles rather than mass, as long as the same substance is used. Yield is always less than or equal to 100%. Reasons for lower yield include incomplete reactions, side reactions and product lost during purification.

也可以使用摩尔而非质量，只要针对同一种物质即可。产率总是小于或等于100%。产率偏低的原因包括反应不完全、发生副反应以及纯化过程中产品损失。

Atom economy measures the efficiency of a reaction in incorporating atoms from reactants into the desired product. The formula is:

Atom economy = (Mᵣ of desired product ÷ sum of Mᵣ of all reactants) × 100

原子经济性衡量反应中反应物的原子被结合进目标产物中的效率。公式为：

原子经济性 = (目标产物的Mᵣ ÷ 所有反应物Mᵣ之和) × 100

High atom economy means less waste and is a key principle of green chemistry. Use balanced equations to calculate the total Mᵣ of all reactants on the left‑hand side.

高原子经济性意味着更少的废弃物，是绿色化学的一个关键原则。需根据配平的方程计算左边所有反应物的Mᵣ总和。

5. Empirical and Molecular Formula | 实验式与分子式

The empirical formula gives the simplest whole‑ ivide each mole value by the smallest value obtained; (3) if the ratio is not whole numbers, multiply until all are integers. For example, a compound contains 40.0% carbon, 6.7% hydrogen and 53.3% 53.3/16 = 3.33. Divide by 3.33 gives ratio 1 C : 2 H : 1 O, so empirical formula is CH₂O.

实验式给出化合物中各元素原子的最简整数比。求法：(1) 将各元素的质量（或百分比）除以Aᵣ换算成摩尔数；(2) 将每个摩尔数除以其中的最小值；(3) 如果比率不是整数，进行倍乘直到全部为整数。例如，某化合物含碳40.0%、氢6.7%、氧53.3%。摩尔数：C 40.0/12 = 3.33，H 6.7/1 = 6.7，O 53.3/16 = 3.33。各除以3.33得比例1 C : 2 H : 1 O，所以实验式为CH₂O。

ular formula shows the actual number of atoms of each element in a molecule. It is a whole‑ compound is 180, empirical mass of CH₂O = 30, so 180/30 = 6; molecular formula is C₆H₁₂O₆.

分子式显示一个分子中各元素原子的实际数目。它是实验式的整数倍。将实验式质量乘以一个整数，使它与给定的相对分子质量(Mᵣ)相匹配。如果上述化合物的Mᵣ为180，CH₂O的实验式质量为30，则180/30 = 6；分子式为C₆H₁₂O₆。

6. Molar Volume of Gases | 气体的摩尔体积

At room temperature and pressure (RTP: 20°C and 1 atm), one mole of any gas occupies a volume of 24 dm³ (24 000 cm³). This is a fundamental conversion for all gas calculations at RTP. The relationship is:

Volume of gas (dm³) = moles of gas × 24

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（24 000 cm³）。这是RTP下所有气体计算的基础换算关系：

气体体积(dm³) = 气体摩尔数 × 24

lume of gas produced or reacted. For example, 0.40 mol of CO₂ occupies 0.40 × 24 = 9.6 dm³.

无论气体种类，只要条件是RTP，这一关系都适用。可以将它与根据质量计算摩尔的方法结合，从而求出产生或反应掉的气体体积。例如，0.40 mol CO₂所占体积为0.40 × 24 = 9.6 dm³。

7. Energy Changes and Bond Enthalpies | 能量变化与键能

In an exothermic reaction, energy is released to the surroundings and ΔH is negative. In an endothermic reaction, energy is absorbed and ΔH is positive. The overall enthalpy change for a reaction can be calculated using bond enthalpies with the formula:

ΔH = Σ(bond enthalpies broken) – Σ(bond enthalpies made)

在放热反应中，能量释放到周围环境中，ΔH为负值。在吸热反应中，能量被吸收，ΔH为正值。可以利用键能计算反应的总焓变，公式为：

ΔH = Σ(断裂键的键能总和) – Σ(形成键的键能总和)

Bond breaking is endothermic (positive) and bond making is exothermic (negative). When using given average bond enthalpies, remember they apply to gases, and you must draw the displayed formulae to count every bond broken and formed.

断键过程吸热（正值），成键过程放热（负值）。使用给出的平均键能时，要记住它们适用于气态，并且必须画出结构式以数清每一个断裂和形成的键。

For a simple calorimetry experiment, the heat energy transferred is calculated using:

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q = m × c × ΔT

where q is energy in J, m is mass of water (or solution) in g, c is specific heat capacity (4.2 J/g°C for water) and ΔT is the temperature change. This q can then be scaled to find ΔH per mole.

简单量热实验的热量计算为：

q = m × c × ΔT

其中q为能量(J)，m为水（或溶液）的质量(g)，c为比热容（水为4.2 J/g°C），ΔT为温度变化。然后可以将q换算成每摩尔的ΔH。

8. Rates of Reaction: Quantitative Expressions | 反应速率：定量表达式

The mean rate of a reaction can be expressed in terms of the quantity of reactant used or product formed over time:

Mean rate = change in quantity ÷ time taken

Quantity may be mass (g), volume of gas (cm³) or concentration (mol/dm³). The units of rate will then be g/s, cm³/s or mol/dm³/s respectively.

平均反应速率可以用反应物消耗量或生成物生成量随时间的变化来表示：

平均速率 = 量的变化 ÷ 所用时间

这个量可以是质量(g)、气体体积(cm³)或浓度(mol/dm³)，速率的单位则分别为g/s、cm³/s或mol/dm³/s。

For a reaction between marble chips and acid, you might measure the mass loss as CO₂ escapes, or collect the gas volume. Tangents to a curve on a graph allow you to calculate the rate at a specific instant.

对于大理石与酸的反应，可以测量CO₂逸出导致的质量损失，或收集气体体积。通过曲线的切线可以求出某一瞬间的速率。

nergy (greater than the activation energy) and correct orientation. Factors affecting rate—temperature, concentration/pressure, surface area and catalysts— lained by changes in collision frequency and the proportion of particles with energy ≥ Eₐ.

碰撞理论指出，要发生反应，粒子必须发生碰撞且能量足够（大于活化能）、取向正确。影响速率的因素——温度、浓度/压强、表面积和催化剂——都可以用碰撞频率和能量≥Eₐ的粒子比例的变化来解释。

9. Le Chatelier’s Principle and Equilibrium | 勒夏特列原理与平衡

Le Chatelier’s principle: If a system at dynamic equilibrium is subjected to a change in concentration, temperature or pressure, the position of equilibrium shifts to oppose the change.

勒夏特列原理：如果处于动态平衡的体系在浓度、温度或压强上发生变化，平衡位置将朝抵消该变化的方向移动。

For the Haber process (N₂ + 3H₂ ⇌ essure favours the side with fewer gas molecules (right, producing more NH₃). Increasing temperature favours the endothermic direction (left, reducing NH₃ yield), but is used to increase rate. Catalysts do not affect the position of equilibrium; they only speed up the attainment of equilibrium.

对于哈伯法（N₂ + 3H₂ ⇌ 2NH₃，ΔH = -92 kJ/mol），增大压强有利于气体分子数更少的一侧（向右，生成更多NH₃）。升高温度有利于吸热方向（向左，降低NH₃产率），但用于提高速率。催化剂不影响平衡位置，只加快达到平衡的速度。

Remember that for aqueous or gaseous systems, concentration changes shift equilibrium to use up the added substance or replace what was removed. Solids and pure liquids do not appear in the equilibrium constant expression, though Year 11 Eduqas does not require Kc calculations.

记住，对于水溶液或气体体系，浓度的变化会使平衡移向消耗掉加入的物质或补充被移除物质的方向。固体和纯液体不出现在平衡常数表达式中，但11年级Eduqas不要求进行Kc计算。

10. Electrolysis and Quantitative Half‑Equations | 电解与定量半反应

During electrolysis, the amount of substance produced at an electrode depends on the charge passed. The key relationship is:

Charge (C) = current (A) × time (s)

电解过程中，电极上生成物质的量取决于通过的电量。关键关系式为：

电量(C) = 电流(A) × 时间(s)

or volume of product, use the stoichiometric ratio of electrons to substance from the half‑equation.

然后，可利用法拉第常数求出电子的摩尔数：1摩尔电子携带96 500 C的电量。因此 e⁻ 的摩尔数 = 电量 ÷ 96 500。要计算产物的质量或体积，需利用半反应方程中电子与物质的化学计量比。

roduced = 0.300 ÷ 3 = 0.100 mol. Multiply by molar mass (27.0 g/mol) to get mass = 2.70 g.

例如：在电解熔融氧化铝时，Al³⁺ + 3e⁻ → Al。如果通过了0.300 mol电子，产生的Al的摩尔数 = 0.300 ÷ 3 = 0.100 mol。乘以摩尔质量(27.0 g/mol)得到质量为2.70 g。

11. Conservation of Mass and Balancing Equations | 质量守恒与配平方程

The Law of Conservation of Mass states that no atoms are lost or made during a chemical reaction; the total mass of reactants equals the total mass of products. This is why we must balance chemical equations so that the number of each type of atom is the same on both sides.

质量守恒定律指出，在化学反应中没有原子产生或消失；反应物的总质量等于生成物的总质量。这就是我们必须配平化学方程，使各种原子的数目在两边相等的原因。

In a closed system, mass stays constant even if a gas is produced, because nothing escapes. In an open system, a gas may leave, causing an apparent mass decrease. This concept is vital when explaining mass changes in precipitation or effervescence experiments.

在封闭系统中，即使产生气体，质量也保持不变，因为没有物质外逸。在开放系统中，气体可能逸散，导致表观质量减少。在沉淀反应或产生气泡实验中，解释质量变化时这一概念至关重要。

A balanced ionic equation shows only the ions that change, while half‑ → Zn²⁺(aq) + Cu(s). Zn loses 2 electrons (oxidation), Cu²⁺ gains 2 electrons (reduction).

离子方程式仅显示发生变化的离子，而半反应式则将氧化和还原过程分开。例如，锌与硫酸铜(II)反应：Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)。Zn失去2个电子（氧化），Cu²⁺得到2个电子（还原）。

12. Key Ions and Solubility Rules (Quick Reference) | 重要离子与溶解性规则（速查）

You must know the formulas and charges of common ions: sulfate SO₄²⁻, nitrate NO₃⁻, carbonate CO₃²⁻, hydroxide OH⁻, ammonium NH₄⁺, and metal ions such as Na⁺, Ca²⁺, Al³⁺, Fe²⁺/Fe³⁺. Recognising these helps in writing balanced equations and predicting precipitates.

你必须掌握常见离子的化学式和电荷：硫酸根SO₄²⁻、硝酸根NO₃⁻、碳酸根CO₃²⁻、氢氧根OH⁻、铵根NH₄⁺，以及金属离子如Na⁺、Ca²⁺、Al³⁺、Fe²⁺/Fe³⁺。认识它们有助于书写配平方程和预测沉淀。

General solubility rules for Year 11: all sodium, potassium and ammonium salts are soluble; all nitrates are soluble; most chlorides, bromides and iodides are soluble (except lead and silver halides); most sulfates are soluble (except lead, barium and calcium sulfate); most carbonates and hydroxides are insoluble (except those of Group 1 and ammonium). Use these to deduce precipitation reactions.

11年级需掌握的一般溶解性规则：所有钠盐、钾盐和铵盐均可溶；所有硝酸盐均可溶；大多数氯化物、溴化物和碘化物可溶（铅和银的卤化物除外）；大多数硫酸盐可溶（铅、钡和钙的硫酸盐除外）；大多数碳酸盐和氢氧化物不可溶（第一主族和铵的化合物除外）。利用这些规则推断沉淀反应。

Testing for anions: chloride (white precipitate with acidified silver nitrate), sulfate (white precipitate with acidified barium chloride), carbonate (bubbling with dilute acid, gas turns limewater cloudy). These tests are based on solubility or gas evolution.

阴离子检验：氯离子（与酸化硝酸银溶液产生白色沉淀），硫酸根离子（与酸化氯化钡溶液产生白色沉淀），碳酸根离子（与稀酸反应产生气泡，气体使石灰水变浑浊）。这些检验基于溶解性或气体生成。

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