📚 IGCSE WJEC Chemistry: Stoichiometry Revision | IGCSE WJEC 化学：化学计量考点精讲
Stoichiometry is the cornerstone of quantitative chemistry. It allows chemists to predict how much product will form, how much reactant is needed, and what happens at the particle level during a reaction. In the WJEC IGCSE Chemistry specification, stoichiometry ties together the mole concept, balanced equations, and essential calculations—from reacting masses to gas volumes and solution concentrations. This revision guide will walk you through every key concept, with examples and tips to help you solve problems confidently in your exam.
化学计量是定量化学的基石，它让化学家能够预测生成多少产物、需要多少反应物以及在反应中粒子层面发生了什么。在WJEC IGCSE化学大纲中，化学计量将摩尔概念、配平方程式以及反应质量、气体体积和溶液浓度等基本计算紧密结合在一起。这份复习指南将带你逐一掌握所有核心概念，结合例题和技巧，让你在考试中充满信心地解决计算题。
1. The Mole Concept | 摩尔的概念
it for amount of substance. One mole of any species contains exactly 6.02 × 10²³ (Avogadro's number) of its specified elementary particles— c world of grams and litres. The symbol for amount of substance is n, and its unit is mol.
摩尔是物质的基本单位。1摩尔任何物质恰好含有6.02 × 10²³（阿伏伽德罗常数）个指定的基本粒子——原子、分子、离子或式单位。这个巨大的数字为微观世界和可以用克、升测量的宏观世界之间搭建了一座桥梁。物质的量的符号是n，单位是mol。
You can think of the mole as a 'chemical dozen' arbon atoms. Since atoms and molecules are incredibly tiny, we cannot count them one by one; the mole gives us a practical way to measure them by weighing.
你可以把摩尔想象成“化学打”。就像一打鸡蛋表示12个鸡蛋一样，1摩尔碳原子表示6.02 × 10²³个碳原子。由于原子和分子极小，我们无法一个一个地数；摩尔为我们提供了一种通过称重来计量的实用方法。
2. Relative Atomic and Formula Mass | 相对原子质量与式量
he mass of a carbon-12 atom, and it has no units. You find Aᵣ values on the Periodic Table— Aᵣ values of all atoms in the formula. For magnesium oxide, MgO, Mᵣ = 24.3 + 16.0 = 40.3.
相对原子质量（Aᵣ）是指元素的原子平均质量与碳-12原子质量的1/12的比值，没有单位。你在周期表上可以找到Aᵣ值——例如，碳是12.0，氧是16.0，氯是35.5。相对式量（Mᵣ）用于化合物，是化学式中所有原子Aᵣ值的总和。对于氧化镁MgO，Mᵣ = 24.3 + 16.0 = 40.3。
ted copper sulfate CuSO₄·5H₂O, the Mᵣ = 63.5 + 32.1 + 4×16.0 + 5×(2×1.0 + 16.0) = 249.6. Always show your working step by step in exams to gain marks even if the final answer is slightly off.
计算Mᵣ时，要注意化学式中的括号和水合盐中的点。对于Ca(OH)₂，Mᵣ为40.1 + 2×(16.0 + 1.0) = 40.1 + 34.0 = 74.1。对于水合硫酸铜CuSO₄·5H₂O，Mᵣ = 63.5 + 32.1 + 4×16.0 + 5×(2×1.0 + 16.0) = 249.6。考试中务必逐步展示计算过程，即使最终答案略有偏差，也能拿到步骤分。
3. Molar Mass and Molar Calculations | 摩尔质量与摩尔计算
r instance, the molar mass of oxygen gas (O₂) is 2 × 16.0 = 32.0 g mol⁻¹. The central equation linking mass, moles and molar mass is:
n = m / M or moles = mass (g) ÷ molar mass (g mol⁻¹)
摩尔质量（M）是指1摩尔物质的质量，单位是g mol⁻¹。它的数值等于相对原子质量或相对式量。例如，氧气（O₂）的摩尔质量是2 × 16.0 = 32.0 g mol⁻¹。连接质量、摩尔数和摩尔质量的核心公式是：
n = m / M 即 摩尔数 = 质量 (g) ÷ 摩尔质量 (g mol⁻¹)
Example: How many moles are there in 5.00 g of sodium hydroxide, NaOH? mass are known: m = n × M. Practice rearranging the formula so you can solve for any unknown quickly.
例题：5.00 g氢氧化钠（NaOH）中含有多少摩尔？M(NaOH) = 23.0 + 16.0 + 1.0 = 40.0 g mol⁻¹。n = 5.00 / 40.0 = 0.125 mol。这个三变量关系式可以变形为已知摩尔数和摩尔质量求质量：m = n × M。多加练习公式变形，你就能快速解出任何一个未知量。
4. Avogadro's Number and Particles | 阿伏伽德罗常数与粒子数
or NA, is used to convert between moles and number of atoms, molecules, or ions. The relationship is:
Number of particles = n × (6.02 × 10²³)
1摩尔任何物质都含有6.02 × 10²³个粒子。这个常数，L或NA，用于摩尔与原子、分子或离子数目之间的转换。关系式为：
粒子数 = n × (6.02 × 10²³)
oles of a molecular substance, you must account for how many atoms are in each molecule. For 2.0 mol H₂O, the number of hydrogen atoms is 2.0 × 6.02 × 10²³ × 2 = 2.408 × 10²⁴ because each water molecule contains two hydrogen atoms. The WJEC exam often asks you to link moles, mass, and number of particles in multistep problems—so master each link separately first.
如果你有2.0摩尔水分子，分子数为2.0 × 6.02 × 10²³ = 1.204 × 10²⁴。要找出给定摩尔数的分子物质中的原子个数，必须考虑每个分子中含有多少个原子。对于2.0 mol H₂O，氢原子数为2.0 × 6.02 × 10²³ × 2 = 2.408 × 10²⁴，因为每个水分子含两个氢原子。WJEC考试常要求在多步计算中将摩尔、质量和粒子数联系起来——因此先分别掌握每个环节。
5. Molar Volume of Gases | 气体摩尔体积
At room temperature and pressure (RTP), which is 20 °C and 1 atmosphere, one mole of any gas occupies a volume of 24 dm³ (or 24 000 cm³). This is known as the molar gas volume, and it can be used to work out the volume of a gas from its amount in moles, and vice versa:
Volume of gas (dm³) = n × 24 or n = Volume / 24
在常温常压下，即20 °C和1个大气压，1摩尔任何气体的体积都是24 dm³（或24 000 cm³）。这被称为气体摩尔体积，可以用来根据摩尔数计算气体体积，反之亦然：
气体体积 (dm³) = n × 24 或 n = 体积 / 24
Example: What volume does 0.50 mol of carbon dioxide occupy at RTP? o dm³ by dividing by 1000 if you are using the 24 dm³ mol⁻¹ value; alternatively, you can use 24 000 cm³ mol⁻¹ directly. This concept is exclusive to gases—solids and liquids have negligible volumes compared to gases under these conditions.
例题：0.50 mol二氧化碳在常温常压下占有多大体积？体积 = 0.50 × 24 = 12 dm³。如果一个反应在常温常压下生成480 cm³氢气，那么H₂的摩尔数为 (480 / 1000) / 24 = 0.020 mol。记住，如果使用24 dm³ mol⁻¹，要把cm³除以1000转换为dm³；或者直接使用24 000 cm³ mol⁻¹。这个概念仅适用于气体——在这些条件下，固体和液体的体积与气体相比可以忽略不计。
6. Concentration of Solutions | 溶液浓度
The concentration of a solution is usually expressed in mol dm⁻³ or g dm⁻³. The key relationship for moles is:
n = c × V (where c is concentration in mol dm⁻³ and V is volume in dm³)
溶液的浓度通常以mol dm⁻³或g dm⁻³表示。与摩尔相关的关键关系是：
n = c × V （c为浓度，单位mol dm⁻³；V为体积，单位dm³）
0 = 0.0500 mol. If you are given concentration in g dm⁻³, you can convert to mol dm⁻³ by dividing by the molar mass. Many titration calculations in WJEC rely on this formula, so always ensure your volume is in dm³ before multiplying.
例如，求25.0 cm³浓度为2.00 mol dm⁻³的盐酸中HCl的摩尔数，先将25.0 cm³转换为0.0250 dm³，然后n = 2.00 × 0.0250 = 0.0500 mol。如果给定的浓度是g dm⁻³，你可以除以摩尔质量转化为mol dm⁻³。WJEC的许多滴定计算都依赖这个公式，所以务必在相乘前确保体积单位是dm³。
7. Using Balanced Equations | 利用平衡方程式进行化学计算
, the ratio is 2 moles of Mg react with 1 mole of O₂ to produce 2 moles of MgO. This ratio is the key to all stoichiometric calculations. You must first calculate the moles of the known substance, then use the ratio to find the moles of the unknown substance, and finally convert to the requested units.
一个配平的化学方程式提供了反应物和生成物的摩尔比。对于反应2Mg + O₂ → 2MgO，比例为2摩尔镁与1摩尔氧气反应生成2摩尔氧化镁。这个比值是所有化学计量计算的关键。你必须先计算出已知物质的摩尔数，然后利用比例求出未知物质的摩尔数，最后转换为所需的单位。
ven substance (using n = m/M, n = c×V, or n = V(gas)/24). 5) Use the mole ratio to find moles of the target. 6) Convert to mass, volume, or concentration as required. This systematic approach reduces mistakes.
步骤：1) 写出配平方程式。2) 在方程式下划出或突出相关的两种物质。3) 注明它们的摩尔比。4) 计算已知物质的摩尔数（用n = m/M、n = c×V或n = V(气体)/24）。5) 利用摩尔比求出目标物质的摩尔数。6) 根据需要转换为质量、体积或浓度。这个系统方法可以减少错误。
8. Mass-to-Mass Calculations | 质量-质量计算
equire you to find the mass of a product from a given mass of reactant, or vice versa. Let' mass of calcium oxide, CaO, can be produced by heating 25.0 g of calcium carbonate, CaCO₃? Equation: CaCO₃ → CaO + CO₂.
质量-质量计算要求你根据给定的反应物质量求生成物质量，或反之。我们来看一个例子：加热25.0 g碳酸钙（CaCO₃）能制得多少克氧化钙（CaO）？反应方程式：CaCO₃ → CaO + CO₂。
: CaO = 1 : 1, so moles of CaO = 0.2498 mol. Mass of CaO = 0.2498 × 56.1 = 14.0 g (3 s.f.). Always round your final answer to the appropriate number of significant figures based on the data given—normally three significant figures in this syllabus.
摩尔质量：M(CaCO₃) = 40.1 + 12.0 + (3×16.0) = 100.1 g mol⁻¹；M(CaO) = 40.1 + 16.0 = 56.1 g mol⁻¹。CaCO₃的摩尔数 = 25.0 / 100.1 = 0.2498 mol。CaCO₃与CaO的比例为1:1，因此CaO的摩尔数 = 0.2498 mol。CaO的质量 = 0.2498 × 56.1 = 14.0 g（三位有效数字）。务必根据所给数据修约最终答案至合适的有效数字——本大纲一般要求三位有效数字。
9. Limiting Reactants | 限制性反应物
In many reactions, one reactant is used up before the others. The substance that runs out first is the limiting reactant; it determines the maximum amount of product that can form. The other reactant(s) are said to be in excess. To identify the limiting reactant, calculate the moles of each reactant and compare the mole ratio from the balanced equation.
在许多反应中，一种反应物会在其他反应物之前耗尽。最先用完的反应物就是限制性反应物，它决定了最多能生成多少产物。其他反应物则被称为过量。要找出限制性反应物，需要计算每种反应物的摩尔数，并与平衡方程式中的摩尔比进行比较。
ol N₂ would need 6.0 mol H₂, but only 4.0 mol H₂ is available. Therefore, H₂ is the limiting reactant. The amount of NH₃ is determined by H₂: 3 mol H₂ yields 2 mol NH₃, so 4.0 mol H₂ yields (2/3) × 4.0 = 2.67 mol NH₃. All stoichiometry involving non-stoichiometric starting mixtures must first identify the limiting reactant.
例题：2.0 mol氮气与4.0 mol氢气反应生成氨：N₂ + 3H₂ → 2NH₃。根据比例，1 mol N₂需要3 mol H₂。2.0 mol N₂需要6.0 mol H₂，但实际只有4.0 mol H₂。因此，H₂是限制性反应物。NH₃的量由H₂决定：3 mol H₂生成2 mol NH₃，所以4.0 mol H₂生成(2/3) × 4.0 = 2.67 mol NH₃。所有涉及非化学计量比的起始混合物计算，都必须先确定限制性反应物。
10. Percentage Yield and Atom Economy | 产率百分数与原子经济性
Percentage yield compares the actual mass of product obtained in an experiment to the theoretical maximum mass calculated from stoichiometry. It is calculated as:
Percentage yield = (actual yield / theoretical yield) × 100%
产率百分数将实验实际获得的产品质量与根据化学计量计算出的理论最大质量进行比较。计算公式为：
产率百分数 = (实际产率 / 理论产率) × 100%
Yields are rarely 100% because of incomplete reactions, side reactions, and product lost during purification. Atom economy, on the other hand, measures the efficiency of a reaction in incorporating reactant atoms into the desired product. It is given by:
Atom economy = (Mᵣ of desired product / sum of Mᵣ of all reactants) × 100%
产率很少达到100%，原因包括反应不完全、发生副反应以及提纯时产品损失。另一方面，原子经济性衡量反应物原子转化为目标产物的效率，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 indicates a greener process with less chemical waste. WJEC questions may ask you to explain why a particular synthesis method is preferred based on its atom economy, or to suggest how a by-product affects the atom economy value. Always include the percentage sign in your final answer and show clear working.
高原子经济性意味着过程更绿色、化学废物更少。WJEC试题可能要求你根据原子经济性解释为何优选某种合成方法，或者说明副产物如何影响原子经济性数值。切记在最终答案中带上百分号，并展示清晰的计算过程。
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