📚 GCSE AQA Chemistry: Stoichiometry Explained | GCSE AQA 化学：化学计量考点精讲
Stoichiometry is the quantitative study of reactants and products in a chemical reaction. In GCSE AQA Chemistry, it involves calculations using relative atomic masses, the mole concept, balanced equations, and the relationships between mass, volume, and concentration. Mastering stoichiometry allows you to predict how much product will form or how much reactant is needed. This revision guide covers every key topic — from relative formula mass to atom economy — with clear step‑by‑step examples.
化学计量是对化学反应中反应物和产物的定量研究。在 GCSE AQA 化学中，它涉及运用相对原子质量、摩尔概念、配平方程式以及质量、体积和浓度之间的关系进行计算。掌握化学计量能让你预测产物的生成量或所需反应物的量。本复习指南涵盖所有核心主题——从相对式量到原子经济性——并配有清晰的逐步解析例题。
1. Relative Atomic Mass (Aᵣ) and Relative Formula Mass (Mᵣ) | 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 ½ the mass of a carbon‑12 atom. The relative formula mass (Mᵣ) of a compound is the sum of the Aᵣ values of all atoms in its formula. For example, water H₂O has Mᵣ = (2 × 1) + 16 = 18. There are no units because it is a relative measure.
元素的相对原子质量（Aᵣ）是其原子的平均质量与一个碳‑12 原子质量的 ½ 之比。化合物的相对式量（Mᵣ）是其化学式中所有原子的 Aᵣ 总和。例如，水 H₂O 的 Mᵣ = (2 × 1) + 16 = 18。该值没有单位，因为它是一个相对量度。
Always use the Aᵣ values given in the periodic table for your calculations. For ionic compounds, Mᵣ is often called relative formula mass; for molecules, it may be called relative molecular mass, but the calculation is the same.
在计算中始终使用元素周期表中给出的 Aᵣ 值。对于离子化合物，Mᵣ 通常称为相对式量；对于分子，则称为相对分子质量，但计算方法相同。
2. The Mole and Avogadro’s Constant | 摩尔与阿伏伽德罗常数
One mole of any substance contains 6.02 × 10²³ particles (atoms, molecules, ions, or formula units). This number is Avogadro’s constant (Nₐ). In GCSE Chemistry, you will mainly use the mole to connect mass and chemical equations rather than counting individual particles, but the concept underpins all stoichiometry.
物质的 1 摩尔含有 6.02 × 10²³ 个微粒（原子、分子、离子或式单元）。这个数字就是阿伏伽德罗常数（Nₐ）。在 GCSE 化学中，你主要使用摩尔来联系质量和化学方程式，而不是数单个微粒，但这一概念是所有化学计量的基础。
The mass of one mole of a substance in grams is numerically equal to its Mᵣ. For instance, 1 mol of carbon‑12 atoms has a mass of exactly 12 g, while 1 mol of H₂O molecules has a mass of 18 g.
物质 1 摩尔的质量（以克为单位）在数值上等于其 Mᵣ。例如，1 mol 碳‑12 原子的质量恰好是 12 g，而 1 mol 水分子（H₂O）的质量为 18 g。
3. Moles from Mass: Formula n = m / Mᵣ | 从质量求摩尔数：公式 n = m / Mᵣ
The key equation linking mass and moles is: number of moles (n) = mass (m) ÷ relative formula mass (Mᵣ). Always ensure the mass is in grams. If you know the mass and Mᵣ, you can find the number of moles.
连接质量和摩尔的关键方程是：摩尔数（n）= 质量（m）÷ 相对式量（Mᵣ）。务必确保质量以克为单位。如果已知质量和 Mᵣ，便可求出摩尔数。
n = m / Mᵣ
Example: Calculate the number of moles in 72 g of water (Mᵣ = 18). n = 72 / 18 = 4.0 mol.
例题：计算 72 g 水（Mᵣ = 18）中的摩尔数。n = 72 / 18 = 4.0 mol。
4. Balancing Equations and Mole Ratios | 配平方程与摩尔比例
A balanced symbol equation shows the exact numbers of moles of each reactant and product. The coefficients in front of each species give the mole ratio. For example, 2H₂ + O₂ → 2H₂O means 2 mol of hydrogen react with 1 mol of oxygen to produce 2 mol of water. Stoichiometric calculations rely on these ratios.
配平的化学方程式表明了每种反应物和产物的确切摩尔数。各物质前的系数给出了摩尔比。例如，2H₂ + O₂ → 2H₂O 表示 2 mol 氢气与 1 mol 氧气反应生成 2 mol 水。化学计量计算依赖这些比例。
When you perform calculations, always convert the given mass to moles first, then use the mole ratio to find moles of the unknown substance, and finally convert back to mass if required.
进行计算时，总是先将已知质量转换为摩尔，然后利用摩尔比求出未知物质的摩尔数，最后若需要再转换为质量。
5. Mass‑to‑Mass Calculations | 质量‑质量计算
In a typical reacting mass problem, you are given the mass of one substance and asked to find the mass of another. Step 1: write the balanced equation. Step 2: calculate the Mᵣ values for the two substances involved. Step 3: convert the given mass to moles using n = m / Mᵣ. Step 4: use the mole ratio to determine moles of the target substance. Step 5: convert moles back to mass using m = n × Mᵣ.
在典型的反应质量计算题中，会给出一种物质的质量，要求求出另一种物质的质量。第一步：写出配平的化学方程式。第二步：计算涉及两种物质的 Mᵣ。第三步：用 n = m / Mᵣ 将已知质量转换为摩尔。第四步：利用摩尔比求目标物质的摩尔数。第五步：用 m = n × Mᵣ 将摩尔数转换回质量。
Example: What mass of magnesium oxide (MgO) is produced when 6 g of magnesium burns completely in oxygen? 2Mg + O₂ → 2MgO. Mᵣ(Mg) = 24, Mᵣ(MgO) = 40. Moles of Mg = 6 / 24 = 0.25 mol. Mole ratio Mg : MgO = 1 : 1, so 0.25 mol MgO. Mass of MgO = 0.25 × 40 = 10 g.
例题：6 g 镁在氧气中完全燃烧可生成多少克氧化镁（MgO）？ 2Mg + O₂ → 2MgO。Mᵣ(Mg) = 24，Mᵣ(MgO) = 40。镁的摩尔数 = 6 / 24 = 0.25 mol。摩尔比 Mg : MgO = 1 : 1，故 MgO 为 0.25 mol。 MgO 质量 = 0.25 × 40 = 10 g。
6. Limiting Reactants | 限制反应物
In many reactions, one reactant is used up before the others; this is the limiting reactant. The amount of product formed depends entirely on the limiting reactant. To identify it, calculate the moles of each reactant, then compare the mole ratio required by the equation.
在许多反应中，一种反应物会先于其他反应物消耗完；这就是限制反应物。产物的生成量完全取决于限制反应物。要找出它，需计算各反应物的摩尔数，然后与方程式所需的摩尔比进行比较。
For example, if 2 mol of H₂ are needed for every 1 mol of O₂, and you have 3 mol of H₂ and 2 mol of O₂, H₂ is the limiting reactant because it requires only 1.5 mol of O₂, leaving excess O₂.
例如，若每 1 mol O₂ 需要 2 mol H₂，而你现有 3 mol H₂ 和 2 mol O₂，则 H₂ 是限制反应物，因为它只需要 1.5 mol O₂，O₂ 过量。
7. Percentage Yield | 百分产率
The percentage yield compares the actual mass of product obtained in an experiment to the theoretical mass predicted by stoichiometry. It is never 100% in practice due to incomplete reactions, side reactions, and loss during purification.
百分产率将实验中实际获得的产品质量与化学计量预测的理论质量进行比较。实际中产率从不会达到 100%，原因在于反应不完全、副反应以及提纯过程中的损失。
Percentage yield = (actual mass / theoretical mass) × 100%
If a reaction should produce 20 g of product but only 15 g is obtained, the yield is (15/20) × 100 = 75%.
若某反应理论应生成 20 g 产物，但仅获得 15 g，则产率为 (15/20) × 100 = 75%。
8. Molar Gas Volume (RTP) | 气体摩尔体积（常温常压）
At room temperature and pressure (RTP), one mole of any gas occupies a volume of 24 dm³. This applies to all gases regardless of their identity. The molar gas volume is used to link moles of a gas to its volume.
在常温常压（RTP）下，任何气体 1 摩尔所占的体积均为 24 dm³。这适用于所有气体，无论其种类。气体摩尔体积用于在气体摩尔数和体积之间建立联系。
Remember: 1 dm³ = 1000 cm³. You may need to convert volumes given in cm³ to dm³ by dividing by 1000 before using the molar volume.
请记住：1 dm³ = 1000 cm³。使用摩尔体积前，可能需要将 cm³ 单位的体积除以 1000 转换为 dm³。
9. Calculating Gas Volumes from Moles | 从摩尔数计算气体体积
To find the volume of a gas at RTP, use the formula: volume (dm³) = number of moles × 24. Conversely, if you know the volume, moles = volume (dm³) / 24.
要计算 RTP 下气体的体积，使用公式：体积 (dm³) = 摩尔数 × 24。反之，若已知体积，摩尔数 = 体积 (dm³) / 24。
Example: What volume does 2.5 mol of carbon dioxide occupy at RTP? V = 2.5 × 24 = 60 dm³. In cm³, this is 60,000 cm³.
例题：2.5 mol 二氧化碳在 RTP 下占多大体积？ V = 2.5 × 24 = 60 dm³。换算为 cm³ 即为 60,000 cm³。
10. Solution Concentration: mol/dm³ | 溶液浓度：摩尔每立方分米
Concentration measures how much solute is dissolved in a given volume of solution. In GCSE Chemistry, concentration is often expressed in mol/dm³. The equation is: concentration (c) = number of moles (n) / volume (V, in dm³).
浓度用于衡量一定体积溶液中所含溶质的量。在 GCSE 化学中，浓度常用 mol/dm³ 表示。公式为：浓度 (c) = 摩尔数 (n) / 体积 (V，单位 dm³)。
c = n / V
For a solution containing 0.5 mol of sodium hydroxide in 250 cm³ (0.25 dm³), c = 0.5 / 0.25 = 2.0 mol/dm³.
对于含有 0.5 mol 氢氧化钠的 250 cm³（0.25 dm³）溶液，c = 0.5 / 0.25 = 2.0 mol/dm³。
11. Converting Concentration to g/dm³ | 浓度单位换算：克每立方分米
Sometimes concentration is given in g/dm³. You can convert between mol/dm³ and g/dm³ using Mᵣ. Mass concentration (g/dm³) = molar concentration (mol/dm³) × Mᵣ. This is particularly useful for titration calculations or when preparing solutions.
有时浓度以 g/dm³ 给出。你可以利用 Mᵣ 在 mol/dm³ 和 g/dm³ 之间换算。质量浓度 (g/dm³) = 摩尔浓度 (mol/dm³) × Mᵣ。这在滴定计算或配制溶液时特别有用。
Example: A 0.4 mol/dm³ HCl solution has what concentration in g/dm³? Mᵣ(HCl) = 36.5. Mass concentration = 0.4 × 36.5 = 14.6 g/dm³.
例题：浓度为 0.4 mol/dm³ 的 HCl 溶液，其质量浓度是多少？ Mᵣ(HCl)=36.5。质量浓度 = 0.4 × 36.5 = 14.6 g/dm³。
12. Atom Economy | 原子经济性
Atom economy measures how efficiently atoms in the reactants end up in the desired product. It is a key concept in green chemistry. The formula is: Atom economy = (Mᵣ of desired product / sum of Mᵣ of all reactants) × 100%. A higher atom economy means less waste and more sustainable processes.
原子经济性衡量反应物中的原子有多少最终进入了目标产物。这是绿色化学的一个核心概念。公式为：原子经济性 =（目标产物的 Mᵣ / 所有反应物 Mᵣ 之和）× 100%。原子经济性越高，意味着废物越少，过程越可持续。
Example: In the reaction 2Mg + O₂ → 2MgO, the desired product is MgO. Atom economy = (Mᵣ(MgO) / [2 × Mᵣ(Mg) + Mᵣ(O₂)]) × 100% = (40 / [48 + 32]) × 100% = 50%.
例题：在反应 2Mg + O₂ → 2MgO 中，目标产物是 MgO。原子经济性 = (Mᵣ(MgO) / [2 × Mᵣ(Mg) + Mᵣ(O₂)]) × 100% = (40 / [48 + 32]) × 100% = 50%。
Reactions with low atom economy produce significant waste; achieving high atom economy is desirable in industrial synthesis.
原子经济性低的反应会产生大量废物；在工业合成中，实现高原子经济性是理想目标。
Published by TutorHao | Chemistry Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导