Stoichiometry in IB Edexcel Chemistry: Key Points | IB Edexcel 化学：化学计量 考点精讲

📚 Stoichiometry in IB Edexcel Chemistry: Key Points | IB Edexcel 化学：化学计量 考点精讲

Stoichiometry is the quantitative study of reactants and products in chemical reactions. For IB and Edexcel Chemistry students, mastering stoichiometry means confidently converting between masses, moles, volumes and concentrations, while interpreting balanced equations. This revision guide covers the core principles, common pitfalls and the most examined calculation types to help you perform with precision in Paper 1 and Paper 2.

化学计量学是对化学反应中反应物与产物进行定量研究的学科。对 IB 和 Edexcel 化学考生而言，掌握化学计量就意味着能熟练地在质量、摩尔、体积和浓度之间进行换算，同时准确解读配平的化学方程式。本考点精讲覆盖核心原理、常见易错点以及最常考的计算类型，帮助你在卷一和卷二中精准作答。

1. The Mole Concept and Avogadro’s Constant | 摩尔概念与阿伏伽德罗常数

A mole is the amount of substance that contains exactly 6.02 × 10²³ elementary entities (atoms, molecules, ions, etc.). This number is Avogadro’s constant (Nₐ). Understanding the mole as a counting unit is the foundation of all stoichiometric calculations.

一摩尔物质恰好包含 6.02 × 10²³ 个基本单元（原子、分子、离子等），这个数值就是阿伏伽德罗常数（Nₐ）。将摩尔视为计数单位，是全部化学计量计算的基础。

Use the relationship: number of particles = amount (mol) × Nₐ. For example, 0.500 mol of CO₂ contains 0.500 × 6.02 × 10²³ CO₂ molecules, which means 3.01 × 10²³ molecules.

使用关系式：粒子数 = 物质的量（mol）× Nₐ。例如，0.500 mol CO₂ 含有 0.500 × 6.02 × 10²³ 个 CO₂ 分子，即 3.01 × 10²³ 个分子。

N = n × Nₐ

2. Molar Mass and Relative Masses | 摩尔质量与相对质量

Molar mass (M) is the mass of one mole of a substance, expressed in g mol⁻¹. It is numerically equal to the relative atomic mass (Aᵣ) or relative formula mass (Mᵣ) from the Periodic Table. Always link mass and moles via: mass = moles × molar mass.

摩尔质量（M）是一摩尔物质的质量，单位为 g mol⁻¹。它在数值上等于周期表中的相对原子质量（Aᵣ）或相对式量（Mᵣ）。始终通过质量 = 物质的量 × 摩尔质量将质量与摩尔联系起来。

m = n × M

For example, the Mᵣ of H₂SO₄ = 2(1.01) + 32.07 + 4(16.00) = 98.09, so the molar mass is 98.09 g mol⁻¹. Thus 0.100 mol of H₂SO₄ has a mass of 9.81 g.

例如，H₂SO₄ 的 Mᵣ = 2×1.01 + 32.07 + 4×16.00 = 98.09，因此摩尔质量为 98.09 g mol⁻¹。于是 0.100 mol H₂SO₄ 的质量为 9.81 g。

3. Empirical and 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 number of atoms of each element in a molecule. It is always an integer multiple of the empirical formula.

经验式表示化合物中各元素原子的最简整数比。分子式则表示一个分子中各元素原子的实际数目，它永远是经验式的整数倍。

To find an empirical formula: convert % composition or mass data to moles, then divide by the smallest number of moles to get the ratio. Multiply the empirical formula mass by an integer (n) to reach the given molar mass for the molecular formula.

求经验式的步骤：将百分组成或质量数据换算成摩尔，再除以最小摩尔数得到比例。用经验式质量乘以整数 n，使结果等于给定的摩尔质量，即可得到分子式。

Example: a hydrocarbon contains 85.7% C and 14.3% H by mass. Moles C = 85.7/12.01 = 7.14 mol; moles H = 14.3/1.01 = 14.16 mol. Divide by 7.14 → ratio C:H = 1:1.98 ≈ 1:2, so empirical formula = CH₂. If the molar mass is 56 g mol⁻¹, n = 56/14.03 = 4 → molecular formula = C₄H₈.

示例：某烃含碳 85.7%、氢 14.3%。碳的物质的量 = 85.7/12.01 = 7.14 mol；氢的物质的量 = 14.3/1.01 = 14.16 mol。除以 7.14 得比 C:H = 1:1.98 ≈ 1:2，经验式为 CH₂。若摩尔质量为 56 g mol⁻¹，n = 56/14.03 = 4，分子式为 C₄H₈。

4. Balancing Chemical Equations | 配平化学方程式

A balanced equation obeys the law of conservation of mass: the number of atoms of each element is the same on both sides. Never alter subscripts; only adjust coefficients. Balance elements that appear in the fewest compounds first, leaving O and H until last.

配平的方程式遵循质量守恒定律：每种元素的原子数在两边的总数相等。绝不能改动下标，只能调整系数。从在最少化合物中出现的元素开始配平，最后配平氧和氢。

For ionic equations, balance atoms and overall charge. The net charge must be equal on both sides. State symbols (s, l, g, aq) should be included when writing full equations, especially in Edexcel papers.

对于离子方程式，需要配平原子和总电荷，两侧净电荷必须相等。在书写完整方程式时，尤其 Edexcel 试卷中，应注明状态符号（s、l、g、aq）。

5. Mass-to-Mass Calculations | 质量–质量计算

The core stoichiometric pathway is: mass → moles of known → moles of unknown (via mole ratio) → mass of unknown. The mole ratio comes from the coefficients in the balanced equation.

化学计量的核心计算路径为：质量 → 已知物的物质的量 → 通过摩尔比得到未知物的物质的量 → 未知物的质量。摩尔比来自配平方程式中的系数。

n(A) = mass(A)/M(A) → n(B) = n(A) × (coefficient B/coefficient A) → mass(B) = n(B) × M(B)

Worked example: 4.00 g of NaOH neutralise excess HCl. What mass of NaCl is produced? Mᵣ(NaOH) = 40.00, n(NaOH) = 4.00/40.00 = 0.100 mol. Equation: NaOH + HCl → NaCl + H₂O, ratio 1:1, so n(NaCl) = 0.100 mol. Mᵣ(NaCl) = 58.5, mass = 0.100 × 58.5 = 5.85 g.

示例：4.00 g NaOH 与过量 HCl 中和，生成的 NaCl 质量是多少？Mᵣ(NaOH)=40.00，n(NaOH)=4.00/40.00=0.100 mol。方程式：NaOH + HCl → NaCl + H₂O，摩尔比 1:1，故 n(NaCl)=0.100 mol。Mᵣ(NaCl)=58.5，质量=0.100×58.5=5.85 g。

6. Limiting Reactants and Excess | 限制试剂与过量

The limiting reactant is the one that is fully consumed, determining the maximum amount of product. Reactants still present after the reaction stops are said to be in excess. Always identify the limiting reactant before performing product calculations.

限制试剂是反应中完全消耗掉的那种反应物，它决定了产物能够生成的最大量。反应停止后仍有剩余的反应物称为过量试剂。在进行产物计算之前，务必先确定哪一个是限制试剂。

To find the limiting reactant, calculate the moles of each reactant and compare the mole ratio to the equation. Whichever reactant would be used up first based on the required ratio is the limiting one. Use its moles for all product calculations.

寻找限制试剂的方法是：先算出每种反应物的物质的量，并将摩尔比与方程式进行比较。根据方程式所需比例最先被用尽的那种反应物就是限制试剂。一切产物计算均基于该反应物的物质的量进行。

Example: 2.0 mol H₂ react with 1.5 mol O₂ to form H₂O. Equation: 2H₂ + O₂ → 2H₂O. 2.0 mol H₂ would need 1.0 mol O₂. O₂ present is 1.5 mol, so H₂ is limiting. Product H₂O = 2.0 mol (1:1 ratio with H₂).

示例：2.0 mol H₂ 与 1.5 mol O₂ 反应生成 H₂O。方程式：2H₂ + O₂ → 2H₂O。2.0 mol H₂ 需 1.0 mol O₂，而现有 1.5 mol O₂，故 H₂ 为限制试剂。产物 H₂O 为 2.0 mol（与 H₂ 1:1）。

7. Theoretical, Actual and Percentage Yield | 理论产量、实际产量与产率

The theoretical yield is the maximum product mass calculated from stoichiometry assuming complete reaction. The actual yield is the mass obtained experimentally. Percentage yield = (actual yield / theoretical yield) × 100%.

理论产量是基于化学计量关系、假设反应完全进行时计算出的最大产物质量。实际产量是实验中得到的产物质量。产率 =（实际产量 / 理论产量）× 100%。

Yields are rarely 100% due to incomplete reactions, side reactions, or losses during purification. When an IB or Edexcel question gives an actual yield, you must use it to find the percentage yield or work backwards to find the initial mass of a reactant.

由于反应不完全、副反应发生或纯化过程中的损失，产率很少能达到 100%。当 IB 或 Edexcel 题目给出实际产量时，你必须利用它求产率，或者反向推算反应物的初始质量。

8. Molar Volume of Gases and Gas Stoichiometry | 气体摩尔体积与气体计量

At standard temperature and pressure (STP: 0 °C, 100 kPa for IB; Edexcel often uses 20 °C and 1 atm or 100 kPa – check data book), one mole of any ideal gas occupies 22.7 dm³ (or 24.0 dm³ at 20 °C, 1 atm). Always note the conditions given in the question.

在标准状况（STP：0 °C、100 kPa，IB 用此；Edexcel 常用 20 °C、1 atm 或 100 kPa，务必查数据手册）下，一摩尔任何理想气体体积为 22.7 dm³（或在 20 °C、1 atm 下为 24.0 dm³）。要始终注意题目给出的条件。

V = n × Vₘ

Gas stoichiometry links moles of gas to volume, allowing calculations without mass. For example, from the equation 2CO + O₂ → 2CO₂, 200 cm³ of CO at STP requires 100 cm³ O₂ and produces 200 cm³ CO₂, because volume ratio equals mole ratio for gases at the same T and P.

气体计量通过物质的量将气体体积联系起来，无需质量即可计算。例如，由方程式 2CO + O₂ → 2CO₂，在 STP 下 200 cm³ CO 需要 100 cm³ O₂ 并生成 200 cm³ CO₂，因为在同温同压下气体体积比等于摩尔比。

9. Concentration and Solution Stoichiometry | 浓度与溶液计量

Concentration (c) is the amount of solute per unit volume of solution, usually expressed in mol dm⁻³. The key equation is n = c × V, where V must be in dm³. For titration calculations, this relation is essential.

浓度（c）是单位体积溶液中溶质的物质的量，通常以 mol dm⁻³ 表示。核心公式为 n = c × V，其中 V 必须使用 dm³ 单位。对于滴定计算，这一关系至关重要。

n = c × V

When diluting a solution, the number of moles stays constant: c₁V₁ = c₂V₂. In titration, use the average titre volume and the mole ratio from the equation to find the unknown concentration. Always convert cm³ to dm³ by dividing by 1000.

稀释溶液时，溶质的物质的量保持不变：c₁V₁ = c₂V₂。在滴定中，利用平均滴定体积和方程式中的摩尔比来求算未知浓度。需始终将 cm³ 转换为 dm³（除以 1000）。

10. Percentage Composition and Purity | 百分组成与纯度

Percentage composition by mass of an element in a compound = (total mass of element in 1 mole / molar mass of compound) × 100%. This is tested frequently, especially for hydrated salts and minerals.

化合物中某元素的质量百分组成 =（1 摩尔化合物中该元素的总质量 / 化合物摩尔质量）× 100%。这一考点出现频繁，尤其见于水合盐和矿物相关题目。

Purity of a sample = (mass of pure substance / total mass of impure sample) × 100%. When an impure sample reacts, only the pure portion contributes to the product. Back‑titration is a common method for analysing purity.

样品纯度 =（纯物质质量 / 不纯样品总质量）× 100%。当不纯样品参与反应时，只有纯品部分产生产物。返滴定是分析纯度的常用方法。

11. Solving Stoichiometry Problems Efficiently | 高效解题策略

Develop a systematic approach: list all data with units, write the balanced equation, convert given quantities to moles, use the mole ratio, and finally convert to the required unit (mass, volume, concentration). Keep track of significant figures as per question data.

建立系统化解题流程：列出所有数据及单位，写出配平方程式，将已知量换算成摩尔，运用摩尔比，最后转换成所需单位（质量、体积、浓度）。严格按照题目数据的有效数字位数进行修约。

Common pitfalls include: forgetting to balance equations, mixing up limiting and excess reactants, using V in cm³ directly in n = cV without converting to dm³, and misreading gas molar volume conditions. Practise with a variety of past paper questions to build speed and accuracy.

常见错误包括：忘记配平方程式、混淆限制试剂与过量试剂、直接将 cm³ 代入 n = cV 而未转换为 dm³、读错气体摩尔体积的条件等。通过练习各类真题来提高解题速度与准确性。

12. Key Exam Tips for IB and Edexcel | IB 与 Edexcel 考试要点

In IB Paper 1 multiple choice, expect quick conversions and ratio‑based gas volume questions. In Paper 2, structured questions often combine moles, mass and concentration in a multi‑step problem. Edexcel International A‑Level features similar multi‑step calculations in Unit 1, often requiring clear method marks.

在 IB 卷一选择题中，会出现快速换算和基于比例的气体体积题。卷二结构化题目通常将摩尔、质量、浓度组合成多步计算题。Edexcel 国际 A‑Level 则在第一单元出现类似多步计算，常要求写出清晰解题步骤以获取步骤分。

Always show working clearly – even if the final answer is wrong, you can gain method marks. Memorise the core equations (n = m/M, n = cV, V = nVₘ) and unit conversions. For gases, check whether conditions are STP, RTP or something else stated in the question.

务必清晰地展示解题过程——即便最终答案错误，也能获得步骤分。熟记核心公式（n = m/M、n = cV、V = nVₘ）和单位换算。气体相关题要检查条件是 STP、RTP 还是题目指定的其他条件。

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