IB Chemistry: Master Mole Calculations | IB 化学：摩尔计算 考点精讲

📚 IB Chemistry: Master Mole Calculations | IB 化学：摩尔计算 考点精讲

The mole is the cornerstone of quantitative chemistry in the IB Diploma Programme. From balancing equations to predicting product masses, mole calculations underpin nearly every stoichiometric problem you will encounter. Mastering these concepts is essential for success in Paper 1, Paper 2, and the Internal Assessment.

摩尔是IB文凭课程中定量化学的基石。从配平化学方程式到预测产物质量，摩尔计算几乎支撑着你将遇到的每一个化学计量问题。掌握这些概念对于在试卷1、试卷2和内部评估中取得成功至关重要。

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

A mole is defined as the amount of substance that contains exactly 6.02214076 × 10²³ elementary entities. This number is Avogadro’s constant (Nₐ). The entities can be atoms, molecules, ions, or electrons. In IB chemistry, we often use 6.02 × 10²³ for calculations unless more precision is required.

摩尔被定义为包含精确6.02214076 × 10²³个基本实体的物质的量。这个数就是阿伏伽德罗常数（Nₐ）。实体可以是原子、分子、离子或电子。在IB化学中，除非需要更高的精度，我们通常使用6.02 × 10²³进行计算。

The mole links the microscopic world of atoms to the macroscopic world of grams and litres. For any substance, one mole of particles weighs its relative atomic or formula mass in grams. This is the basis for all quantitative conversions.

摩尔将微观原子世界与宏观的克和升联系起来。对于任何物质，一摩尔的粒子质量以克计就是其相对原子质量或式量。这是所有定量换算的基础。

2. Molar Mass and Mass-Mole Conversions | 摩尔质量与质量-摩尔换算

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). For example, the molar mass of water (H₂O) is 18.02 g mol⁻¹. The key equation is: n = m / M, where n is amount in mol, m is mass in g, and M is molar mass.

摩尔质量（M）是一摩尔物质的质量，单位为 g mol⁻¹。它在数值上等于相对原子质量（Aᵣ）或相对式量（Mᵣ）。例如，水（H₂O）的摩尔质量是18.02 g mol⁻¹。关键公式是：n = m / M，其中n是物质的量（mol），m是质量（g），M是摩尔质量。

To convert from mass to moles, divide the mass by the molar mass. To convert from moles to mass, multiply the amount in moles by the molar mass. Always ensure your units cancel correctly. When using compound molar masses, sum the atomic masses of each element multiplied by its subscript.

要将质量转换为摩尔，用质量除以摩尔质量。要将摩尔转换为质量，用物质的量乘以摩尔质量。始终确保单位正确抵消。计算化合物摩尔质量时，加和每种元素的原子质量乘以其下标。

Example: How many moles are in 25.0 g of NaCl? (M = 58.44 g mol⁻¹). n = 25.0 / 58.44 = 0.428 mol. Conversely, 0.500 mol of NaCl has mass = 0.500 × 58.44 = 29.2 g.

举例：25.0 g NaCl中有多少摩尔？（M = 58.44 g mol⁻¹）。n = 25.0 / 58.44 = 0.428 mol。反过来，0.500 mol NaCl的质量 = 0.500 × 58.44 = 29.2 g。

3. Empirical and Molecular Formulae | 经验式与分子式

The empirical formula is 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. For example, benzene has molecular formula C₆H₆ and empirical formula CH. The relationship is: Molecular formula = n × (Empirical formula), where n is an integer.

经验式是化合物中原子的最简整数比。分子式显示分子中每种元素原子的实际数目。例如，苯的分子式是C₆H₆，经验式是CH。关系为：分子式 = n × (经验式)，其中n是整数。

To determine empirical formula from percentage composition: assume 100 g of sample, convert masses to moles, divide by the smallest number of moles to get the ratio. If a ratio is not whole, multiply all by a factor to clear fractions (e.g., 1:1.5 becomes 2:3). Then use molar mass to find n = (Molar mass) / (Empirical formula mass).

从百分组成确定经验式：假设样品100 g，将质量转换为摩尔，除以最小的摩尔数以得到比例。如果比例不是整数，将所有数乘以一个因子以消除分数（例如1:1.5变成2:3）。然后利用摩尔质量求出n = (摩尔质量) / (经验式式量)。

Example: A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Moles: C: 40.0/12.01 = 3.33, H: 6.7/1.01 = 6.63, O: 53.3/16.00 = 3.33. Divide by 3.33: C:1, H:1.99≈2, O:1 → empirical formula CH₂O. If molar mass is 180 g mol⁻¹, empirical mass = 30 g mol⁻¹, n = 6, molecular formula = C₆H₁₂O₆.

举例：一种化合物按质量含碳40.0%，氢6.7%，氧53.3%。摩尔：C: 40.0/12.01 = 3.33, H: 6.7/1.01 = 6.63, O: 53.3/16.00 = 3.33。除以3.33：C:1, H:1.99≈2, O:1 → 经验式 CH₂O。如果摩尔质量为180 g mol⁻¹，经验式式量 = 30 g mol⁻¹，n = 6，分子式 = C₆H₁₂O₆。

4. Reacting Masses and Stoichiometry | 反应质量与化学计量

Stoichiometry uses the mole ratios from a balanced chemical equation to relate the amounts of reactants and products. The coefficients in the equation give the mole ratio. To calculate reacting masses: convert given mass to moles, use the mole ratio to find moles of the required substance, then convert back to mass.

化学计量利用配平化学方程式中的摩尔比来关联反应物和产物的量。方程式中的系数给出摩尔比。计算反应质量：将给定质量转换为摩尔，利用摩尔比求出所需物质的摩尔，再转换回质量。

Crucial steps: 1) Write a balanced equation. 2) Convert known quantity to moles. 3) Use stoichiometric ratios to find moles of unknown. 4) Convert moles of unknown to required units (mass, concentration, volume). Always identify the limiting reactant when two masses are given.

关键步骤：1）写出配平方程式。2）将已知量转换为摩尔。3）利用化学计量比求出未知物的摩尔。4）将未知物的摩尔转换为所需单位（质量、浓度、体积）。当给出两种质量时，始终鉴定限量反应物。

Example: 2Al + 3Cl₂ → 2AlCl₃. How many grams of AlCl₃ form from 5.40 g of Al? Moles Al = 5.40/26.98 = 0.200 mol. Mole ratio Al:AlCl₃ = 2:2 = 1:1, so 0.200 mol AlCl₃. Mass AlCl₃ = 0.200 × 133.33 = 26.7 g.

举例：2Al + 3Cl₂ → 2AlCl₃。从5.40 g Al可以生成多少克AlCl₃？Al的摩尔数 = 5.40/26.98 = 0.200 mol。摩尔比 Al:AlCl₃ = 2:2 = 1:1，所以AlCl₃为0.200 mol。AlCl₃质量 = 0.200 × 133.33 = 26.7 g。

5. Molar Volume of a Gas | 气体摩尔体积

At standard temperature and pressure (STP: 273 K and 100 kPa), one mole of an ideal gas occupies 22.7 dm³ (or 22.7 L). The IB may also use the older definition of STP (0 °C, 1 atm) where molar volume is 22.4 dm³, but the 2023 syllabus uses 22.7 dm³ at 100 kPa. Always check the data booklet.

在标准状况（STP：273 K，100 kPa）下，一摩尔理想气体占22.7 dm³（或22.7 L）。IB也可能使用旧

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