📚 Year 10 AQA Chemistry: Formula & Theorem Quick Reference Handbook | AQA 化学：公式定理速查手册（Year 10）
This quick reference handbook collects all the essential formulas, equations and key relationships you will encounter in Year 10 AQA Chemistry. Whether you are revising for end‑of‑topic tests or preparing for mock exams, having these fundamental tools at your fingertips will help you solve numerical problems with confidence and understand the underlying chemical principles.
本速查手册汇集了 Year 10 AQA 化学课程中所有核心公式、方程式和关键定量关系。无论你是在为单元测验复习，还是在备战模拟考试，这些基础工具都能让你胸有成竹地解决计算题，并深入理解背后的化学原理。
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 compared to 1/12th of the mass of a carbon‑12 atom. The relative formula mass (Mᵣ) of a compound is the sum of the Aᵣ values of all the atoms shown in its formula.
元素的相对原子质量 (Aᵣ) 是其原子平均质量与碳‑12 原子质量的 1/12 之比。化合物的相对式量 (Mᵣ) 等于其化学式中所有原子的 Aᵣ 之和。
Examples | 示例:
- H₂O: Mᵣ = (2 × 1) + 16 = 18
- CaCO₃: Mᵣ = 40 + 12 + (3 × 16) = 100
- Mg(OH)₂: Mᵣ = 24 + (2 × 16) + (2 × 1) = 58
Mᵣ has no units; however, it represents the mass of one mole of the substance in grams.
Mᵣ 没有单位，但它表示一摩尔该物质的质量（克）。
2. The Mole and Avogadro’s Constant | 物质的量与阿伏伽德罗常数
ticles (atoms, molecules, ions or electrons). This number is called the Avogadro constant.
一摩尔任何物质都含有恰好 6.02 × 10²³ 个微粒（原子、分子、离子或电子），这个数值称为阿伏伽德罗常数。
Number of particles = number of moles × (6.02 × 10²³)
When a question gives the number of particles, you can find the amount in moles by dividing by Avogadro’s constant.
当题目给出微粒数时，可以通过除以阿伏伽德罗常数求出物质的量。
moles = number of particles ÷ (6.02 × 10²³)
3. Mass, Moles and Molar Mass | 质量、物质的量与摩尔质量
The most important equation in quantitative chemistry links mass, molar mass and amount in moles. Molar mass is numerically equal to the relative formula mass (Mᵣ) expressed in g/mol.
化学计量中最核心的公式将质量、摩尔质量和物质的量联系在一起。摩尔质量在数值上等于相对式量 (Mᵣ)，单位为 g/mol。
moles = mass (g) ÷ molar mass (g/mol)
You can rearrange this to find mass: mass = moles × molar mass.
可变形为：质量 = 物质的量 × 摩尔质量。
Worked example | 计算示例: How many moles are there in 8.0 g of NaOH? (Mᵣ = 40) → moles = 8.0 ÷ 40 = 0.20 mol.
8.0 g NaOH 的物质的量是多少？(Mᵣ = 40) → 物质的量 = 8.0 ÷ 40 = 0.20 mol。
4. Concentration of Solutions | 溶液浓度
The concentration of a solution is the amount of solute dissolved in a given volume of solvent. In Year 10, two units are commonly used: g/dm³ and mol/dm³.
溶液浓度表示单位体积溶剂中所含溶质的量。Year 10 阶段常用两种单位：g/dm³ 和 mol/dm³。
concentration (g/dm³) = mass of solute (g) ÷ volume of solution (dm³)
concentration (mol/dm³) = moles of solute ÷ volume of solution (dm³)
Remember: 1 dm³ = 1000 cm³. Always convert volumes to dm³ before using these formulas.
注意：1 dm³ = 1000 cm³，使用公式前务必将体积换算为 dm³。
Key conversion | 关键换算: concentration (mol/dm³) = concentration (g/dm³) ÷ Mᵣ.
关键转换: 物质的量浓度 (mol/dm³) = 质量浓度 (g/dm³) ÷ 相对式量。
5. Reacting Masses and Limiting Reactants | 反应质量与限量反应物
In a chemical reaction, the masses of reactants and products can be calculated using stoichiometry. The balanced equation gives the mole ratio.
在化学反应中，可利用化学计量关系计算反应物和产物的质量。配平的化学方程式给出了物质的量之比。
Steps | 步骤:
- Convert the known mass into moles: moles = mass ÷ Mᵣ.
- Use the mole ratio from the balanced equation to find moles of the unknown substance.
- Convert moles back to mass: mass = moles × Mᵣ.
- 将已知质量换算为物质的量：物质的量 = 质量 ÷ Mᵣ。
- 利用配平方程式中的摩尔比求出未知物的物质的量。
- 将物质的量换回质量：质量 = 物质的量 × Mᵣ。
The limiting reactant is the reagent that is completely used up first, determining the maximum yield of product. Compare the available moles to the required mole ratio to identify it.
限量反应物是最先被完全消耗的试剂，它决定了产物的最大产量。通过比较各反应物给出的物质的量与方程式所需的摩尔比，即可找出限量反应物。
6. Yield and Atom Economy | 产率与原子经济性
The percentage yield compares the actual mass of product obtained to the theoretical maximum mass predicted by stoichiometry.
产率用于比较实际获得的产物质量与根据化学计量关系算出的理论最大质量。
% yield = (actual yield ÷ theoretical yield) × 100
Atom economy measures how efficiently the atoms of the reactants are incorporated into the desired product. A higher atom economy means a greener process.
原子经济性衡量反应物中的原子有多少进入了目标产物。原子经济性越高，过程越绿色环保。
% atom economy = (Mᵣ of desired product ÷ sum of Mᵣ of all reactants) × 100
Both percentages are important for evaluating the sustainability and cost‑effectiveness of industrial processes.
这两个百分数对于评估工业过程的可持续性和经济性都非常重要。
7. Rate of Reaction | 反应速率
The rate of a chemical reaction can be measured by how quickly a reactant is used up or a product is formed. The average rate over a time interval is expressed as:
化学反应速率可通过反应物消耗或产物生成的快慢来衡量。某段时间内的平均速率可表示为：
mean rate = change in quantity ÷ time taken
The ‘quantity’ could be the mass of a reactant consumed (g), the volume of a gas produced (cm³), or the change in colour, pH or conductivity recorded with a sensor.
“变化量”可以是消耗的反应物质量 (g)、生成的气体体积 (cm³)，或传感器记录的溶液颜色、pH 或电导率的变化。
For graphical data, the rate at a specific point can be found by drawing a tangent and calculating its gradient. The units depend on what is measured, e.g. g/s or cm³/s.
对于图表数据，可通过绘制切线并计算其斜率求得某一时刻的瞬时速率。单位取决于测量对象，如 g/s 或 cm³/s。
8. Energy Changes and Bond Enthalpies | 能量变化与键焓
Chemical reactions either release energy (exothermic, ΔH negative) or absorb energy (endothermic, ΔH positive). The overall energy change can be estimated using average bond enthalpies.
化学反应要么放出能量（放热，ΔH 为负），要么吸收能量（吸热，ΔH 为正）。总能量变化可用平均键焓进行估算。
ΔH = total energy absorbed to break bonds − total energy released when forming bonds
Bond breaking is endothermic (positive value); bond making is exothermic (negative value). If more energy is released in bond making than is absorbed in bond breaking, the reaction is exothermic overall.
断裂化学键需要吸收能量（正值）；形成化学键则会释放能量（负值）。若形成键时释放的能量多于断裂键时吸收的能量，则该反应总体为放热。
You will be given a table of average bond enthalpies in the exam. Use the balanced equation to count the number and type of bonds involved.
考试中会提供平均键焓数据表。你需要根据配平的方程式统计涉及哪些化学键及其数目。
9. Electrolysis: Quantitative Relationships | 电解：定量关系
During electrolysis, the mass of a substance deposited or the volume of a gas released depends on the quantity of charge passed. Charge is related to current and time:
在电解过程中，析出物质的质量或释放气体的体积取决于通过的电量。电量与电流和时间的关系为：
charge (C) = current (A) × time (s)
At the Year 10 level, you mainly apply this qualitatively or through given ratios. The amount of product formed is proportional to the charge, but the exact mass also depends on the half‑equation and the ion’s Aᵣ.
在 Year 10 阶段，你主要运用这一关系的定性理解或按所给比例计算。生成物的量与电量成正比，但具体质量还取决于半反应方程式和离子的相对原子质量。
A simplified relationship for a metal deposit: mass ∝ (current × time). Higher current or longer time increases the yield.
对于金属沉积的简化关系：质量 ∝（电流 × 时间）。增大电流或延长通电时间均可提高产量。
10. Titration Calculations | 滴定计算
Titrations are used to determine the concentration of an unknown solution by reacting it with a solution of known concentration. The calculation uses the mole ratio from the balanced equation.
滴定实验通过将未知浓度溶液与已知浓度溶液反应，来测定前者的浓度。计算时需要使用配平方程式中的摩尔比。
General method | 一般步骤:
- Calculate moles of the known solution: moles = concentration (mol/dm³) × volume (dm³).
- Use the mole ratio to find moles of the unknown solution.
- Calculate its concentration: concentration = moles ÷ volume (dm³).
- 计算已知溶液的物质的量：物质的量 = 浓度 (mol/dm³) × 体积 (dm³)。
- 利用摩尔比求出未知溶液的物质的量。
- 计算其浓度：浓度 = 物质的量 ÷ 体积 (dm³)。
Always convert the titre volume from cm³ to dm³ by dividing by 1000. Reputable readings are averaged, discarding any anomalous results.
务必将滴定管读数（cm³）除以 1000 转换为 dm³。重复实验应取各次可信读数的平均值，剔除异常值。
11. Gas Volume and Molar Volume (Triple Science) | 气体体积与摩尔体积（Triple Science 适用）
At room temperature and pressure (RTP, 20 °C and 1 atmosphere), one mole of any gas occupies 24 dm³. This value helps convert between moles and volume of a gaseous product or reactant.
在常温常压下（RTP，20 °C 和 1 个大气压），一摩尔任何气体占据的体积为 24 dm³。该数值可用于气体产物或反应物的物质的量与体积之间的换算。
volume of gas (dm³) = moles of gas × 24
moles of gas = volume of gas (dm³) ÷ 24
duced in a reaction. If the question gives volume in cm³, first convert: dm³ = cm³ ÷ 1000.
该关系常与化学计量结合，用于计算反应生成的气体体积。若题目给出的体积单位是 cm³，需先换算：dm³ = cm³ ÷ 1000。
12. Percentage Composition and Empirical Formula | 百分组成与最简式
The percentage by mass of an element in a compound can be calculated from its formula:
化合物中某元素的质量分数可由其化学式计算得出：
% mass = (total mass of element in formula ÷ Mᵣ) × 100
The empirical formula gives the simplest whole‑number ratio of atoms in a compound. It is found by converting the measured masses or percentages of elements into moles and then determining the simplest ratio.
最简式表示化合物中各元素的原子个数最简整数比。求法是先将各元素的质量或百分数换算为物质的量，再求出最简整数比。
Steps | 步骤:
- Divide the mass (or percentage) of each element by its Aᵣ to get moles.
- Divide each mole value by the smallest mole value.
- If the ratios are not whole numbers, multiply by a suitable factor to obtain integers.
- 将各元素的质量（或百分比）除以各自的 Aᵣ，得到物质的量。
- 用最小的物质的量除以所有值。
- 若比值不是整数，则乘以合适的因数化为整数。
The molecular formula is a multiple of the empirical formula; it requires the Mᵣ of the compound.
分子式是最简式的整数倍，需要知道化合物的相对分子质量才能确定。
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