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

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

This handbook brings together every key formula, equation, and quantitative relationship you need to master for OCR GCSE Chemistry (9–1). From moles and concentration to energy changes and reaction rates, all the essential tools are presented here with clear worked examples. Use it alongside your practical work and past papers to build confidence and speed.

本手册汇总了 OCR GCSE 化学（9–1）考试中你必须掌握的每一个关键公式、方程式和定量关系。从摩尔与浓度到能量变化与反应速率，所有核心工具都在这里，并配有清晰的示例。配合你的实验练习和历年真题使用，可以帮助你提升解题的信心与速度。

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

Relative atomic mass (Aᵣ) is the average mass of an atom of an element compared to 1/12th the mass of a carbon‑12 atom. It has no units. Aᵣ values are shown on the Periodic Table and account for the natural abundances of isotopes.

相对原子质量 (Aᵣ) 是指一个元素的原子平均质量与一个碳‑12 原子质量的 1/12 的比值，无单位。元素周期表上给出的 Aᵣ 值已考虑了同位素的天然丰度。

Relative formula mass (Mᵣ) applies to compounds. It is the sum of the Aᵣ values of all atoms in the formula unit. For ionic compounds we still use the term ‘relative formula mass’ even though they are not molecules.

相对式量 (Mᵣ) 用于化合物，它是化学式单元中所有原子的 Aᵣ 值之和。对于离子化合物，尽管它们不是分子，我们仍然使用“相对式量”这一术语。

Substance / 物质	Calculation / 计算	Mᵣ / 相对式量
H₂O	(2 × 1) + 16	18
NaCl	23 + 35.5	58.5
CaCO₃	40 + 12 + (3 × 16)	100

You must be able to calculate Mᵣ from a given formula and use it to work out the percentage by mass of an element in a compound: % mass = (total Aᵣ of element ÷ Mᵣ) × 100.

你必须能够根据化学式计算 Mᵣ，并用它来求算化合物中某元素的质量百分数：质量% = (元素的 Aᵣ 总和 ÷ Mᵣ) × 100。

2. The Mole & Molar Mass | 摩尔与摩尔质量

One mole of a substance contains 6.02 × 10²³ particles (Avogadro’s constant). The mass of one mole of a substance in grams is numerically equal to its Mᵣ. This quantity is called the molar mass, with units g/mol.

1 摩尔物质含有 6.02 × 10²³ 个微粒（阿伏伽德罗常数）。1 摩尔某物质的质量以克为单位时，数值上等于其相对式量 Mᵣ。这个量称为摩尔质量，单位为 g/mol。

For example, carbon has Aᵣ = 12, so 1 mol of carbon atoms weighs 12 g. Water has Mᵣ = 18, so 1 mol of H₂O molecules weighs 18 g. The molar mass of sodium chloride (NaCl) is 58.5 g/mol.

例如，碳的 Aᵣ = 12，因此 1 mol 碳原子的质量为 12 g。水的 Mᵣ = 18，因此 1 mol H₂O 分子的质量为 18 g。氯化钠 (NaCl) 的摩尔质量为 58.5 g/mol。

n = m / M

Where n = number of moles (mol), m = mass (g), M = molar mass (g/mol). This is the central formula for all quantitative chemistry at GCSE.

其中 n = 物质的量 (mol)，m = 质量 (g)，M = 摩尔质量 (g/mol)。这是 GCSE 阶段所有定量化学的中心公式。

3. Calculating Moles, Mass & Molar Mass | 摩尔、质量与摩尔质量的计算

To find the number of moles from a given mass, divide the mass by the molar mass. To find the mass, multiply the number of moles by the molar mass. Always write the formula first, substitute the numbers and check units.

已知质量求物质的量：质量除以摩尔质量。已知物质的量求质量：物质的量乘以摩尔质量。解题时务必先写出公式，再代入数值，并检查单位。

Worked example: How many moles are present in 11.7 g of sodium chloride? M(NaCl) = 58.5 g/mol. n = 11.7 g ÷ 58.5 g/mol = 0.20 mol. The answer is given to two significant figures because the given mass is to three significant figures and the molar mass to three.

示例：11.7 g 氯化钠中含有多少摩尔？M(NaCl) = 58.5 g/mol。n = 11.7 g ÷ 58.5 g/mol = 0.20 mol。结果保留两位有效数字，因为已知质量与摩尔质量均为三位有效数字。

Another common task is to find the mass of a certain number of moles. For instance, to prepare 0.500 mol of glucose (C₆H₁₂O₆, Mᵣ = 180), you would need: m = 0.500 mol × 180 g/mol = 90.0 g.

另一种常见任务是求一定物质的量对应的质量。例如，配制 0.500 mol 葡萄糖 (C₆H₁₂O₆, Mᵣ = 180)，所需质量为：m = 0.500 mol × 180 g/mol = 90.0 g。

4. Concentration of Solutions | 溶液的浓度

Concentration is the amount of solute dissolved in a given volume of solution. The most common unit is mol/dm³, but g/dm³ is also used. Remember: 1 dm³ = 1000 cm³.

浓度是指一定体积溶液中溶解的溶质的量。最常用的单位是 mol/dm³，也使用 g/dm³。注意：1 dm³ = 1000 cm³。

c = n / V

c = concentration (mol/dm³), n = number of moles (mol), V = volume (dm³). To convert cm³ to dm³, divide by 1000.

c = 浓度 (mol/dm³)，n = 物质的量 (mol)，V = 体积 (dm³)。将 cm³ 转换为 dm³ 时，除以 1000。

Worked example: 2.0 g of sodium hydroxide (NaOH, M = 40 g/mol) is dissolved in water to make 250 cm³ of solution. Find the concentration in mol/dm³. First, n = 2.0 g ÷ 40 g/mol = 0.050 mol. V = 250 cm³ ÷ 1000 = 0.250 dm³. c = 0.050 mol ÷ 0.250 dm³ = 0.20 mol/dm³.

示例：将 2.0 g 氢氧化钠 (NaOH, M = 40 g/mol) 溶于水，配成 250 cm³ 溶液。求其浓度 (mol/dm³)。首先，n = 2.0 g ÷ 40 g/mol = 0.050 mol。V = 250 cm³ ÷ 1000 = 0.250 dm³。c = 0.050 mol ÷ 0.250 dm³ = 0.20 mol/dm³。

If a concentration is given in g/dm³, you can convert to mol/dm³ by dividing by the molar mass.

若浓度以 g/dm³ 给出，除以摩尔质量即可转换为 mol/dm³。

5. Gas Volumes (Molar Volume at RTP) | 气体体积（常温常压下的摩尔体积）

At room temperature and pressure (RTP, 20 °C and 1 atm), one mole of any gas occupies 24 dm³. This is the molar gas volume. The formula linking moles and volume is:

在常温常压下 (RTP, 20 °C 和 1 atm)，任何气体 1 摩尔所占体积均为 24 dm³。这就是气体摩尔体积。物质的量与体积的关系式为：

n = V / 24

where V is measured in dm³. If the volume is given in cm³, first convert to dm³ by dividing by 1000. The value 24 dm³/mol is only valid at RTP; if conditions change, the volume changes accordingly, but at GCSE you only need the RTP value.

其中 V 的单位为 dm³。若体积以 cm³ 给出，先除以 1000 转换为 dm³。24 dm³/mol 的值仅在 RTP 下有效；条件改变时体积也会改变，但 GCSE 阶段只要求掌握 RTP 下的数值。

Example: What volume does 0.40 mol of carbon dioxide occupy at RTP? V = n × 24 = 0.40 mol × 24 dm³/mol = 9.6 dm³.

示例：0.40 mol 二氧化碳在 RTP 下占据多大体积？V = n × 24 = 0.40 mol × 24 dm³/mol = 9.6 dm³。

6. Percentage Yield & Atom Economy | 百分产率与原子经济

Percentage yield compares the mass of product actually obtained to the theoretical mass predicted by the balanced equation.

百分产率比较实际获得的产品质量与由配平方程式计算出的理论质量。

% Yield = (Actual yield / Theoretical yield) × 100

Yields are never 100% in practice because of incomplete reactions, side reactions, and losses during purification. Understanding percentage yield helps chemists evaluate the efficiency of a process.

由于反应不完全、副反应以及提纯过程中的损失，实际产率永远不会达到 100%。理解百分产率有助于化学家评价工艺的效率。

Atom economy focuses on how much of the reactants end up in the desired product. It is a measure of green chemistry.

原子经济关注的是反应物中有多少最终进入了目标产物。它是衡量绿色化学的一项指标。

% Atom economy = (Mᵣ of desired product / Sum of Mᵣ of all reactants) × 100

A higher atom economy means fewer waste products and more sustainable processes. Reactions like addition polymerisation have 100% atom economy because all atoms are incorporated into the polymer.

原子经济越高，废物越少，过程越可持续。像加成聚合这样的反应具有 100% 的原子经济性，因为所有原子都进入了聚合物。

7. Empirical & Molecular Formulae | 经验式与分子式

The empirical formula gives the simplest whole‑number ratio of atoms in a compound. The molecular formula shows the actual numbers of atoms in a molecule.

经验式（最简式）表示化合物中原子个数的最简整数比。分子式则表示一个分子中原子的实际数目。

To find the empirical formula from experimental data: convert masses (or percentages) to moles by dividing by Aᵣ; then divide all mole values by the smallest number of moles to obtain the simplest ratio. If the ratios are not whole numbers, multiply to give whole numbers.

由实验数据确定经验式的方法：将质量（或百分含量）除以 Aᵣ 转换为物质的量；再将所有摩尔值除以其中最小的摩尔数，得到最简整数比。若比值不是整数，则乘以适当的系数使之变为整数。

Example: A compound contains 40% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Assume 100 g: C mol = 40/12 = 3.33; H = 6.7/1 = 6.7; O = 53.3/16 = 3.33. Divide by 3.33: C 1, H 2, O 1 → empirical formula CH₂O. The molecular formula is a multiple of the empirical formula, determined by comparing the Mᵣ of the compound with the empirical formula mass.

示例：某化合物含碳 40%、氢 6.7%、氧 53.3%（质量分数）。假定取 100 g：C 的摩尔数 = 40/12 = 3.33；H = 6.7/1 = 6.7；O = 53.3/16 = 3.33。除以 3.33 得：C 1, H 2, O 1 → 经验式为 CH₂O。分子式是经验式的整数倍，通过比较化合物的 Mᵣ 与经验式质量来确定。

8. Reacting Mass Calculations | 反应物质量计算

Reacting mass problems use the mole ratio from a balanced equation. Work through three steps: convert the known mass to moles, use the stoichiometric ratio to find moles of the unknown, then convert back to mass.

反应物质量计算运用配平方程式中的摩尔比。解题分三步：将已知物质的质量转换为物质的量；利用化学计量比求出未知物质的物质的量；再将其转换为质量。

Example: 2Mg + O₂ → 2MgO. What mass of magnesium oxide is produced when 6.0 g of magnesium burns? Aᵣ(Mg) = 24, Mᵣ(MgO) = 40. Step 1: n(Mg) = 6.0/24 = 0.25 mol. Step 2: mole ratio Mg : MgO = 2 : 2 = 1 : 1, so n(MgO) = 0.25 mol. Step 3: mass(MgO) = 0.25 × 40 = 10 g.

示例：2Mg + O₂ → 2MgO。6.0 g 镁燃烧可制得多少克氧化镁？Aᵣ(Mg) = 24，Mᵣ(MgO) = 40。第一步：n(Mg) = 6.0/24 = 0.25 mol。第二步：摩尔比 Mg : MgO = 2 : 2 = 1 : 1，所以 n(MgO) = 0.25 mol。第三步：质量(MgO) = 0.25 × 40 = 10 g。

Always ensure the equation is correctly balanced before using the mole ratio. This technique applies to solids, solutions and gases alike, provided you use the appropriate conversion formulas.

使用摩尔比之前，务必确保方程式已正确配平。只要能运用相应的转换公式，这一方法同样适用于固体、溶液和气体。

9. Energy Changes & Bond Enthalpies | 能量变化与键焓

Chemical reactions involve energy transfers. Exothermic reactions release energy to the surroundings (temperature rises); endothermic reactions absorb energy (temperature falls).

化学反应伴随着能量的转移。放热反应向周围释放能量（温度升高）；吸热反应则从周围吸收能量（温度降低）。

For simple reactions, the overall energy change can be estimated using average bond enthalpies. The formula is:

对于简单反应，总能量变化可以通过平均键焓进行估算。公式为：

ΔH = Σ (bonds broken) – Σ (bonds formed)

Bond breaking is endothermic (positive values), bond making is exothermic (negative values). If the energy needed to break bonds is greater than the energy released forming new bonds, ΔH is positive and the reaction is endothermic. If it is less, ΔH is negative – exothermic.

断键吸热（正值），成键放热（负值）。如果断裂旧键所需的总能量大于形成新键所释放的总能量，ΔH 为正，反应吸热。如果小于，则 ΔH 为负——反应放热。

Example: H₂ + Cl₂ → 2HCl. Bond energies (kJ/mol): H–H 436, Cl–Cl 243, H–Cl 432. Bonds broken: 436 + 243 = 679 kJ. Bonds formed: 2 × 432 = 864 kJ. ΔH = 679 – 864 = –185 kJ/mol. The reaction is exothermic.

示例：H₂ + Cl₂ → 2HCl。键能 (kJ/mol)：H–H 436，Cl–Cl 243，H–Cl 432。断键吸收：436 + 243 = 679 kJ。成键释放：2 × 432 = 864 kJ。ΔH = 679 – 864 = –185 kJ/mol。该反应放热。

10. Rate of Reaction | 反应速率

The rate of a chemical reaction tells us how quickly a reactant is used up or a product is formed. It can be measured by monitoring change in mass, volume of gas evolved, colour change or pH over time.

化学反应速率告诉我们反应物消耗或产物生成的快慢。可通过监测质量变化、放出气体的体积、颜色变化或 pH 随时间的变化来测量。

Mean rate = Quantity of reactant used or product formed / Time

Units could be g/s, cm³/s, mol/s. For a changing rate, you may need to determine the rate at a specific point using the gradient of a tangent on a concentration–time graph.

单位可以是 g/s、cm³/s、mol/s。对于变化的速率，可能需要通过浓度–时间图上切线的斜率来确定某一点的瞬间速率。

Factors affecting rate include temperature, concentration (or pressure for gases), surface area of solids, and the presence of a catalyst. These factors can be explained using collision theory: reactions occur when particles collide with sufficient energy (activation energy) and correct orientation.

影响速率的因素有：温度、浓度（或气体压强）、固体的表面积以及催化剂的存在。这些因素可通过碰撞理论解释：只有当微粒以足够的能量（活化能）和正确的取向发生碰撞时，反应才会发生。

For OCR GCSE, you must be able to interpret graphs showing how much product is formed or reactant remaining against time, and explain how changing conditions alters the shape of such graphs.

在 OCR GCSE 中，你必须能够解读产物生成量或反应物剩余量随时间变化的曲线，并解释改变条件如何改变曲线的形状。

11. Titration Calculations | 滴定计算

Titration is used to find the concentration of an unknown solution by reacting it with a solution of known concentration. The key formula when both solutions are expressed in mol/dm³ and the reaction ratio is 1:1 is:

滴定法通过让未知浓度溶液与已知浓度溶液反应，来求出未知溶液的浓度。当两种溶液均以 mol/dm³ 表示且反应比为 1:1 时，关键公式为：

c₁V₁ = c₂V₂

However, you must always account for the stoichiometry. For a reaction aA + bB → products, the correct relationship is:

但你必须始终考虑化学计量比。对于反应 aA + bB → 产物，正确的关系式为：

nₐ / n_b = a / b

You can first find moles of the known solution (n = cV), then use the mole ratio to find moles of the unknown, and finally its concentration.

你可以先求已知溶液的物质的量 (n = cV)，然后利用摩尔比求出未知物的物质的量，最后求得其浓度。

Worked example: 25.0 cm³ of NaOH solution is neutralised by 20.0 cm³ of 0.100 mol/dm³ H₂SO₄. The equation is H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Moles of H₂SO₄ = 0.100 × 0.0200 = 0.00200 mol. Ratio H₂SO₄ : NaOH = 1 : 2, so moles NaOH = 0.00400 mol. Concentration NaOH = 0.00400 / 0.0250 = 0.160 mol/dm³.

示例：25.0 cm³ NaOH 溶液被 20.0 cm³ 0.100 mol/dm³ 的 H₂SO₄ 中和。方程式为 H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O。H₂SO₄ 的物质的量 = 0.100 × 0.0200 = 0.00200 mol。H₂SO₄ : NaOH 摩尔比 = 1 : 2，故 NaOH 的物质的量 = 0.00400 mol。NaOH 浓度 = 0.00400 / 0.0250 = 0.160 mol/dm³。

12. Writing & Balancing Equations | 化学方程式的书写与配平

Every chemical equation must balance – the same number and type of atoms on each side. This reflects the Law of Conservation of Mass. Start by writing the correct formulae for reactants and products, then balance by placing numbers (coefficients) in front of formulae. Never change the subscripts inside a formula.

每个化学方程式都必须配平——两侧原子种类和数目相同，这体现了质量守恒定律。首先写出反应物和产物的正确化学式，然后在化学式前面添加数字（系数）进行配平。切勿改动化学式内部的下标。

State symbols add important information: (s) solid, (l) liquid, (g) gas, (aq) aqueous solution. OCR expects you to include them where possible, particularly in ionic equations.

状态符号提供重要信息：(s) 固体，(l) 液体，(g) 气体，(aq) 水溶液。OCR 期望你在可能的情况下添加状态符号，尤其是在离子方程式中。

Ionic equations show only the particles that actually change. For example, the neutralisation reaction: H⁺(aq) + OH⁻(aq) → H₂O(l). Spectator ions (Na⁺ and Cl⁻ in the reaction of HCl with NaOH) are omitted. To write a net ionic equation, first write the full balanced equation, separate aqueous ionic substances into their ions, then cancel species that appear on both sides.

离子方程式只表示实际发生变化的微粒。例如，中和反应：H⁺(aq) + OH⁻(aq) → H₂O(l)。旁观离子（如 HCl 与 NaOH 反应中的 Na⁺ 和 Cl⁻）被省略。书写净离子方程式时，先写出完整配平的化学方程式，将可溶离子化合物拆成离子，然后消去两边均出现的物种。

Memorise the common polyatomic ions – carbonate (CO₃²⁻), sulfate (SO₄²⁻), nitrate (NO₃⁻), ammonium (NH₄⁺), hydroxide (OH⁻) – as they appear frequently in OCR GCSE papers.

记住常见的多原子离子——碳酸根 (CO₃²⁻)、硫酸根 (SO₄²⁻)、硝酸根 (NO₃⁻)、铵根 (NH₄⁺)、氢氧根 (OH⁻)，它们经常出现在 OCR GCSE 考卷中。

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