📚 Mole Calculations and Stoichiometry for IGCSE Science | IGCSE 科学考点：摩尔计算与化学计量学
Mole calculations and stoichiometry form the backbone of quantitative chemistry at IGCSE level. Understanding the mole allows you to count particles by weighing them, predict product masses, and analyse chemical reactions with precision.
摩尔计算与化学计量学是 IGCSE 阶段定量化学的核心基础。理解“摩尔”这一概念，使你能够通过称量来“数”粒子、预测生成物的质量，并精确分析化学反应。
1. The Mole Concept | 摩尔的概念
The mole is the SI unit for the amount of substance. One mole of any substance contains the same number of particles, known as the Avogadro constant.
摩尔是物质的量的国际单位。任何物质的一摩尔都含有相同数量的粒子，这个数量称为阿伏伽德罗常数。
The Avogadro constant is approximately 6.02 × 10²³ per mole. This value applies to atoms, molecules, ions, or electrons, depending on the species you are counting.
阿伏伽德罗常数约为 6.02 × 10²³ 每摩尔。这个数值适用于原子、分子、离子或电子，具体取决于你所数的粒子种类。
For example, one mole of carbon atoms contains 6.02 × 10²³ carbon atoms. One mole of water molecules contains 6.02 × 10²³ water molecules.
例如，一摩尔碳原子含有 6.02 × 10²³ 个碳原子；一摩尔水分子则含有 6.02 × 10²³ 个水分子。
2. Molar Mass and Mass Calculations | 摩尔质量与质量换算
Molar mass is the mass of one mole of a substance, measured in grams per mole (g/mol). For elements, the molar mass is the relative atomic mass expressed in grams.
摩尔质量是指一摩尔物质的质量，单位为克每摩尔（g/mol）。对于元素而言，摩尔质量就是相对原子质量以克为单位表示的数值。
For compounds, add the relative atomic masses of all atoms in the formula. For example, the molar mass of CO₂ is 12 + 16 + 16 = 44 g/mol.
对于化合物，将化学式中所有原子的相对原子质量相加即可。例如，CO₂ 的摩尔质量为 12 + 16 + 16 = 44 g/mol。
The central conversion equation is: mass = moles × molar mass. You can rearrange it to find moles or molar mass when the other two quantities are known.
最核心的换算公式为：质量 = 物质的量 × 摩尔质量。已知其中任意两个量，即可通过公式变形求出第三个量。
m = n × M
3. Avogadro’s Constant and Particle Counts | 阿伏伽德罗常数与粒子数目
To find the number of particles in a sample, multiply the number of moles by the Avogadro constant.
要计算样品中的粒子数目，只需将物质的量乘以阿伏伽德罗常数即可。
Number of particles = moles × 6.02 × 10²³. This works for atoms, molecules, ions, or formula units.
粒子数目 = 物质的量 × 6.02 × 10²³。这一公式适用于原子、分子、离子或化学式单元。
Be careful whether the question asks for atoms or molecules. One mole of oxygen gas O₂ contains 6.02 × 10²³ molecules, but 2 × 6.02 × 10²³ oxygen atoms.
解题时要特别留意题目问的是原子还是分子。一摩尔氧气 O₂ 含有 6.02 × 10²³ 个分子，但含有 2 × 6.02 × 10²³ 个氧原子。
Use the following table to keep track of common conversions:
使用下表来梳理常见换算关系：
| Quantity | Formula | Units |
| Mass (m) | m = n × M | g |
| Moles (n) | n = m ÷ M | mol |
| Particles | particles = n × 6.02 × 10²³ | particles |
4. Gas Volume and Molar Volume | 气体体积与摩尔体积
At room temperature and pressure (RTP), one mole of any gas occupies the same volume: 24.0 dm³ (24 000 cm³). We call this the molar gas volume.
在室温常压（RTP）下，一摩尔任何气体都占据相同的体积：24.0 dm³（即 24 000 cm³）。这被称为气体摩尔体积。
To find the volume of a gas in dm³, multiply the number of moles by 24.0. To convert from cm³ to dm³, divide by 1000.
计算气体体积（以 dm³ 为单位）时，将物质的量乘以 24.0。若要将 cm³ 换算为 dm³，则需除以 1000。
Volume (dm³) = moles × 24.0
For example, 0.25 mol of carbon dioxide at RTP occupies 0.25 × 24.0 = 6.0 dm³. Always state the conditions because volume changes with temperature and pressure.
例如，0.25 mol 二氧化碳在室温常压下占据 0.25 × 24.0 = 6.0 dm³。作答时务必标明条件，因为体积会随温度和压强发生变化。
5. Concentration of Solutions | 溶液的浓度
Concentration measures how much solute is dissolved in a given volume of solution. The standard unit in IGCSE is mol/dm³, sometimes written as moldm⁻³.
浓度表示一定体积溶液中溶解了多少溶质。IGCSE 中常用的单位是 mol/dm³，也可写作 moldm⁻³。
Concentration = moles of solute ÷ volume of solution in dm³. A 1 mol/dm³ solution contains one mole of solute in every cubic decimetre of solution.
浓度 = 溶质的物质的量 ÷ 溶液体积（以 dm³ 为单位）。1 mol/dm³ 的溶液表示每立方分米溶液中含有 1 摩尔溶质。
c = n ÷ V
When using volumes in cm³ with concentration in mol/dm³, first convert cm³ to dm³ by dividing by 1000. For example, 250 cm³ = 0.250 dm³.
当体积以 cm³ 给出而浓度以 mol/dm³ 给出时，需先将 cm³ 除以 1000 换算为 dm³。例如，250 cm³ = 0.250 dm³。
6. Balancing Equations Using Moles | 用摩尔配平方程式
A balanced chemical equation shows the mole ratio in which substances react and are produced. The coefficients represent moles, not mass.
配平的化学方程式展示了物质反应与生成时的摩尔比。系数代表的是物质的量（摩尔），而不是质量。
For example, 2H₂ + O₂ → 2H₂O means 2 moles of hydrogen react with 1 mole of oxygen to form 2 moles of water.
例如，2H₂ + O₂ → 2H₂O 表示 2 摩尔氢气与 1 摩尔氧气反应生成 2 摩尔水。
Use the mole ratio to convert from moles of one substance to moles of another. If 4 moles of hydrogen react fully, you need 2 moles of oxygen and you produce 4 moles of water.
利用摩尔比可将一种物质的物质的量换算为另一种物质的物质的量。若 4 摩尔氢气完全反应，需要 2 摩尔氧气，同时生成 4 摩尔水。
Always write the balanced equation before starting any calculation. The ratio between coefficients is your conversion factor.
开始任何计算之前，务必先写出配平的化学方程式。系数之间的比值就是你的换算因子。
7. Mass-Mass Stoichiometry | 质量-质量化学计量计算
Mass-mass calculations typically follow three steps: convert the given mass to moles, use the equation ratio to find moles of the target substance, then convert moles back to mass.
质量-质量计算通常分为三步：先将已知质量换算为物质的量，再通过方程式比例求出目标物质的物质的量，最后将物质的量换算回质量。
Consider the reaction: CaCO₃ → CaO + CO₂. If 50 g of CaCO₃ decomposes, first find moles: 50 ÷ 100 = 0.50 mol. The ratio is 1:1, so 0.50 mol CaO forms.
考虑反应：CaCO₃ → CaO + CO₂。若 50 g CaCO₃ 分解，先求物质的量：50 ÷ 100 = 0.50 mol。摩尔比为 1:1，因此生成 0.50 mol CaO。
Finally, convert moles of CaO to mass: 0.50 × 56 = 28 g. Always check molar masses using the periodic table provided in your exam.
最后，将 CaO 的物质的量换算为质量：0.50 × 56 = 28 g。切记使用考试所给元素周期表来核实摩尔质量。
mass → moles → ratio → moles → mass
8. Limiting Reagent | 限量反应物
In a chemical reaction, the limiting reagent is the reactant that runs out first. It determines the maximum amount of product that can form.
在化学反应中，限量反应物是最先消耗完毕的反应物。它决定了产物能够生成的最大量。
Any reactant that remains after the reaction has stopped is called the excess reagent. Identifying the limiting reagent is essential for accurate product predictions.
反应停止后剩余的反应物称为过量反应物。准确判断限量反应物是预测产物量的关键。
To find the limiting reagent, divide the number of moles of each reactant by its coefficient in the balanced equation. The reactant giving the smallest value is the limiting reagent.
判断限量反应物的方法是：将各反应物的物质的量除以它在配平方程式中的系数，所得数值最小者即为限量反应物。
For example, in the reaction N₂ + 3H₂ → 2NH₃, if you have 1 mol N₂ and 3 mol H₂, the ratio matches perfectly. But with 1 mol N₂ and 2 mol H₂, hydrogen is limiting.
例如，在反应 N₂ + 3H₂ → 2NH₃ 中，若你有 1 mol N₂ 和 3 mol H₂，比例完全匹配。但若只有 1 mol N₂ 和 2 mol H₂，则氢气是限量反应物。
9. Percentage Yield and Purity | 产率与纯度百分比
The theoretical yield is the maximum amount of product calculated from the balanced equation. The actual yield is what you obtain in the laboratory, which is often lower.
理论产率是根据配平方程式计算出的最大产物量；实际产率是实验室中实际获得的量，通常低于理论值。
Percentage yield = actual yield ÷ theoretical yield × 100%
Percentage purity compares the mass of the desired substance in a sample to the total sample mass.
纯度百分比用于比较样品中目标物质的质量与样品总质量。
Percentage purity = mass of pure substance ÷ total mass of sample × 100%
Common reasons for yield loss include incomplete reactions, side reactions, or product lost during separation. Exam questions often combine yield with mole calculations.
产率降低的常见原因包括反应不完全、发生副反应，或分离过程中产物损失。考试题目常将产率与摩尔计算结合起来考查。
10. Worked Examples and Exam Tips | 例题与考试技巧
Let us work through a mixed problem. Calculate the volume of carbon dioxide produced at RTP when 10 g of calcium carbonate reacts with excess hydrochloric acid.
我们完整解一道综合题。计算 10 g 碳酸钙与过量盐酸反应时，在室温常压下产生二氧化碳的体积。
Step 1: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. Moles of CaCO₃ = 10 ÷ 100 = 0.10 mol.
步骤一：CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂。CaCO₃ 的物质的量 = 10 ÷ 100 = 0.10 mol。
Step 2: The equation shows a 1:1 ratio between CaCO₃ and CO₂, so moles of CO₂ = 0.10 mol.
步骤二：方程式显示 CaCO₃ 与 CO₂ 的摩尔比为 1:1，因此 CO₂ 的物质的量 = 0.10 mol。
Step 3: Volume = 0.10 × 24.0 = 2.4 dm³. Always include units and round to the precision requested.
步骤三：体积 = 0.10 × 24.0 = 2.4 dm³。务必注明单位，并按题目要求保留适当精度。
Key exam tips: always show your working, write units at every stage, and check that your final answer is reasonable. A mass should never be negative, and a volume should make physical sense.
重要考试技巧：务必写出计算过程，每一步都要带单位，并检查最终答案是否合理。质量不应为负值，体积也应当符合物理实际。
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