📚 IB CCEA Chemistry: Mole Calculations – Key Points Review | IB CCEA 化学：摩尔计算 考点精讲
The mole is the cornerstone of quantitative chemistry, underpinning everything from stoichiometry to solution concentrations. For both IB and CCEA specifications, a solid grasp of mole calculations is essential to succeed in exams and practical assessments. This article breaks down key concepts, formulas, and typical exam applications to help you master mole calculations with confidence.
摩尔是定量化学的基石，支撑着从化学计量到溶液浓度等各个领域。无论是 IB 还是 CCEA 课程大纲，扎实掌握摩尔计算对于考试和实验评估都至关重要。本文将深入解析关键概念、公式和常见考题应用，帮助你自信掌握摩尔计算。
1. The Mole and Avogadro's Constant | 摩尔与阿伏伽德罗常数
ary entities, known as the Avogadro constant, Nₐ. Entities can be atoms, molecules, ions, electrons, or formula units. This number allows chemists to count particles by weighing.
摩尔（符号 mol）是物质的量的国际单位。一摩尔精确包含 6.02214076 × 10²³ 个基本单元，这个数值被称为阿伏伽德罗常数 Nₐ。基本单元可以是原子、分子、离子、电子或化学式单元。这个数字使化学家能够通过称量来计数微粒。
Amount of substance, n, can be calculated from the number of particles, N, using the relationship below. Conversely, the number of particles can be found from the amount of substance.
物质的量 n 可由微粒数 N 通过下面的关系式计算得出；反之，也可由物质的量求出微粒数。
n = N / Nₐ and N = n × Nₐ
Because the entity must always be specified, the phrase "1 mol of chlorine" is ambiguous: it could mean 1 mol of chlorine atoms (Cl) or 1 mol of chlorine molecules (Cl₂). Examiners award marks for stating the entity explicitly.
由于必须明确基本单元，"1 mol 氯"这样的说法是有歧义的：它可能指 1 mol 氯原子（Cl），也可能指 1 mol 氯分子（Cl₂）。考试中明确写出基本单元才能得分。
- Nₐ = 6.022 × 10²³ mol⁻¹ (often rounded to 6.02 × 10²³ in calculations).
- Nₐ = 6.022 × 10²³ mol⁻¹（计算中常取 6.02 × 10²³）。
- The mole links the microscopic world of particles to the macroscopic world of the balance.
- 摩尔把微观的微粒世界与宏观的天平称量联系起来。
2. Particle Counts and Mole Ratios Inside a Formula | 微粒计数与化学式内的摩尔比
Avogadro's constant applies universally; for example, 1 mol of CO₂ contains 1 mol of CO₂ molecules, 1 mol of carbon atoms, and 2 mol of oxygen atoms. The subscripts in a formula are therefore mole ratios, not mass ratios.
阿伏伽德罗常数普遍适用；例如，1 mol CO₂ 含有 1 mol CO₂ 分子、1 mol 碳原子和 2 mol 氧原子。因此化学式中的下标是摩尔比，而不是质量比。
The same logic works for ions. One mole of NaCl produces 1 mol of Na⁺ ions and 1 mol of Cl⁻ ions when it dissolves, while 1 mol of MgCl₂ produces 1 mol of Mg²⁺ ions and 2 mol of Cl⁻ ions. This is why 1 mol dm⁻³ MgCl₂ has a chloride concentration of 2 mol dm⁻³.
同样的逻辑适用于离子。1 mol NaCl 溶解后产生 1 mol Na⁺ 离子和 1 mol Cl⁻ 离子，而 1 mol MgCl₂ 产生 1 mol Mg²⁺ 离子和 2 mol Cl⁻ 离子。这就是为什么 1 mol dm⁻³ 的 MgCl₂ 溶液中氯离子浓度为 2 mol dm⁻³。
| Substance | Amount | Atoms / ions present |
| H₂SO₄ | 1 mol | 2 mol H, 1 mol S, 4 mol O |
| H₂O | 0.50 mol | 1.0 mol H, 0.50 mol O |
| Al₂(SO₄)₃ | 1 mol | 2 mol Al³⁺, 3 mol SO₄²⁻ |
When a question asks for the number of atoms rather than moles, multiply the amount of each element by Nₐ. A common trap is forgetting to multiply by the subscript first.
当题目要求的是原子个数而非物质的量时，应把每种元素的物质的量乘以 Nₐ。常见的失分点是忘记先乘以化学式中的下标。
3. Molar Mass and Mass–Mole Conversions | 摩尔质量与质量–摩尔转换
The molar mass (M he relative atomic mass (Aᵣ) or relative formula mass (Mᵣ). For example, Aᵣ(Cu) = 63.5, so M(Cu) = 63.5 g mol⁻¹.
物质的摩尔质量（M）是一摩尔该物质基本单元的质量，单位为 g mol⁻¹。在数值上等于相对原子质量（Aᵣ）或相对化学式质量（Mᵣ）。例如 Aᵣ(Cu) = 63.5，因此 M(Cu) = 63.5 g mol⁻¹。
To convert a mass into an amount of substance, use the first relationship below, where m is mass in grams and M is molar mass. To find mass from moles, rearrange to the second form.
要把质量转换为物质的量，使用下面第一个关系式，其中 m 是质量（克），M 是摩尔质量。由物质的量求质量，则使用第二个变形公式。
n = m / M m = n × M M = m / n
Example: Calculate the moles in 10.0 g of CaCO₃ (M = 100.1 g mol⁻¹). n = 10.0 / 100.1 = 0.0999 mol, which rounds to 0.100 mol to three significant figures.
例题：计算 10.0 g CaCO₃ 中的物质的量（M = 100.1 g mol⁻¹）。n = 10.0 / 100.1 = 0.0999 mol，保留三位有效数字为 0.100 mol。
Working the other way: the mass of 0.250 mol NaOH (M = 40.0 g mol⁻¹) is m e unit and the correct number of significant figures, because CCEA and IB mark schemes both penalise unit and rounding errors.
反向计算：0.250 mol NaOH 的质量为 m = 0.250 × 40.0 = 10.0 g（M = 40.0 g mol⁻¹）。请务必写出单位并保留正确的有效数字，因为 CCEA 和 IB 的评分标准都会对单位错误和四舍五入错误扣分。
4. Empirical and Molecular Formulas | 实验式与分子式
The empirical formula is the simplest whole-number ratio of atoms in a compound; the molecular formula is the actual number of atoms in one molecule and is a whole-number multiple of the empirical formula. Both are found from experimental composition data.
实验式是化合物中各原子最简单的整数比；分子式是一个分子中实际的原子数目，是实验式的整数倍。两者都可以由实验组成数据求出。
- Assume 100 g of sample so that percentages become masses in grams.
- Divide each mass by the relative atomic mass to get moles of each element.
- Divide all mole values by the smallest one to get the simplest ratio.
- Multiply by a small integer if the ratio is not close to whole numbers (for example 1 : 1.5 becomes 2 : 3).
- 假设样品为 100 g，这样百分数就直接变成克数。
- 把每种质量除以相对原子质量，得到各元素的物质的量。
- 把所有物质的量除以其中最小值，得到最简单的比值。
- 若比值不是接近整数，则乘以小整数（例如 1 : 1.5 变为 2 : 3）。
| Element | % | ÷ Aᵣ | Ratio |
| C | 40.0 | 3.33 | 1 |
| H | 6.7 | 6.70 | 2 |
| O | 53.3 | 3.33 | 1 |
olecular mass is 180, the multiplier is 180 / 30 = 6, so the molecular formula is C₆H₁₂O₆.
实验式为 CH₂O，Mᵣ = 30.0。若测得相对分子质量为 180，则倍数为 180 / 30 = 6，因此分子式为 C₆H₁₂O₆。
5. Water of Crystallisation | 结晶水的计算
Many salts crystallise with a fixed number of water molecules per formula unit, written as a dot formula such as CuSO₄·5H₂O. Heating the hydrate drives off the water of crystallisation and leaves the anhydrous salt.
许多盐结晶时每个化学式单元带有固定数目的水分子，写作带点的形式，例如 CuSO₄·5H₂O。加热水合物会赶走结晶水，留下无水盐。
16.0 = 9.0 g. Then n(CuSO₄) = 16.0 / 159.6 = 0.100 mol and n(H₂O) = 9.0 / 18.0 = 0.500 mol, so x = 0.500 / 0.100 = 5, giving CuSO₄·5H₂O.
例题：25.0 g 水合硫酸铜样品加热至恒重，剩余 16.0 g 无水 CuSO₄。失去的水的质量为 25.0 − 16.0 = 9.0 g。则 n(CuSO₄) = 16.0 / 159.6 = 0.100 mol，n(H₂O) = 9.0 / 18.0 = 0.500 mol，因此 x = 0.500 / 0.100 = 5，即 CuSO₄·5H₂O。
- Heat to constant mass, not just once, so all water is removed.
- Cool in a desiccator before weighing to prevent reabsorbing moisture.
- Never heat strongly enough to decompose the salt or drive off SO₃.
- 要加热到恒重，而不是只加热一次，以确保水全部去除。
- 称量前在干燥器中冷却，防止重新吸收空气中的水分。
- 不要加热到盐分解或放出 SO₃ 的程度。
6. Solutions: Concentration, Volume and Dilution | 溶液：浓度、体积与稀释
Concentration (c , usually expressed in mol dm⁻³. Volume must be in dm³, so remember that 1 dm³ = 1000 cm³.
浓度（c）是单位体积溶液中所含溶质的物质的量，通常用 mol dm⁻³ 表示。体积必须以 dm³ 为单位，所以请记住 1 dm³ = 1000 cm³。
c = n / V n = c × V V = n / c
Example: how many moles of NaOH are in 250 cm³ of 0.100 mol dm⁻³ solution? Convert volume first: 250 / 1000 = 0.250 dm³. Then n = 0.100 × 0.250 = 0.0250 mol, which corresponds to 0.0250 × 40.0 = 1.00 g of NaOH.
例题：250 cm³ 的 0.100 mol dm⁻³ NaOH 溶液中含有多少摩尔 NaOH？先换算体积：250 / 1000 = 0.250 dm³。则 n = 0.100 × 0.250 = 0.0250 mol，对应 0.0250 × 40.0 = 1.00 g NaOH。
Dilution adds solvent but does not change the amount of solute, so the moles before and after dilution are equal. This gives the dilution equation, which is useful for preparing standard solutions from stock solutions.
稀释只是加入溶剂，不改变溶质的物质的量，所以稀释前后溶质的物质的量相等。由此得到稀释公式，用于由浓溶液配制标准溶液。
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