Mole Calculations | 摩尔计算

📚 Mole Calculations | 摩尔计算

The mole is the central unit in quantitative chemistry. It enables chemists to count atoms, ions, and molecules by weighing them, bridging the atomic and macroscopic worlds. Mastering mole calculations is essential for solving stoichiometry problems, predicting yields, and interpreting analytical data across every topic from gas laws to titrations.

摩尔是定量化学的核心单位。它使化学家能够通过称量来计数原子、离子和分子，在原子世界与宏观世界之间架起桥梁。掌握摩尔计算对于解决化学计量问题、预测产率以及解释从气体定律到滴定等各主题的分析数据至关重要。

---

1. Introduction to the Mole | 摩尔简介

The mole (symbol mol) is the SI unit for amount of substance. One mole contains exactly 6.02214076 × 10²³ elementary entities. This number is called the Avogadro constant. A mole of carbon‑12 atoms has a mass of exactly 12 g, which originally defined the unit.

摩尔（符号 mol）是物质的量的国际单位。1摩尔恰好包含6.02214076 × 10²³个基本单元。这个数称为阿伏伽德罗常数。1摩尔碳‑12原子的质量恰好为12 g，这正是该单位最初的定义基础。

---

2. Avogadro’s Constant | 阿伏伽德罗常数

Avogadro’s constant, Nₐ = 6.02 × 10²³ mol⁻¹, is the number of atoms, molecules, or formula units per mole. It allows conversion between the number of particles and amount in moles: number of particles = n × Nₐ. For example, 2.00 mol of water contains 2.00 × 6.02 × 10²³ = 1.20 × 10²⁴ molecules.

阿伏伽德罗常数 Nₐ = 6.02 × 10²³ mol⁻¹ 是每摩尔的原子、分子或式单元的数目。它用于在粒子数与物质的量（摩尔）之间转换：粒子数 = n × Nₐ。例如，2.00 mol 水含有 2.00 × 6.02 × 10²³ = 1.20 × 10²⁴ 个分子。

---

3. Molar Mass | 摩尔质量

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). For H₂O, Mᵣ = (2 × 1.0) + 16.0 = 18.0, so M = 18.0 g mol⁻¹. Always use the atomic masses from the Periodic Table provided in the exam.

摩尔质量（M）是1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）。对于 H₂O，Mᵣ = (2 × 1.0) + 16.0 = 18.0，因此 M = 18.0 g mol⁻¹。务必使用考卷提供的周期表中的原子量。

---

4. Converting Mass to Moles | 质量与物质的量转换

The fundamental equation linking mass and moles is:

n = m ÷ M

where n = amount (mol), m = mass (g), M = molar mass (g mol⁻¹). Rearranging gives m = n × M. If a sample of NaOH (M = 40.0 g mol⁻¹) has a mass of 10.0 g, then n = 10.0 ÷ 40.0 = 0.250 mol.

连接质量与摩尔的基本方程为：n = m ÷ M，其中 n = 物质的量 (mol)，m = 质量 (g)，M = 摩尔质量 (g mol⁻¹)。移项得 m = n × M。若一份 NaOH (M = 40.0 g mol⁻¹) 的质量为 10.0 g，则 n = 10.0 ÷ 40.0 = 0.250 mol。

---

5. Moles of Gases – Molar Volume | 气体的摩尔体积

At room temperature and pressure (rtp, 20 °C and 1 atm), one mole of any gas occupies 24.0 dm³. The relationship is: n = V (dm³) ÷ 24.0 or V = n × 24.0. This applies to all ideal gases. For volumes in cm³, divide by 24 000. Do check the conditions given in a question—sometimes molar volume is quoted at 22.4 dm³ under standard temperature and pressure (stp, 0 °C and 1 atm).

在室温和常压（rtp，20 °C 和 1 atm）下，1摩尔任何气体的体积为24.0 dm³。关系式为：n = V (dm³) ÷ 24.0 或 V = n × 24.0。这适用于所有理想气体。若体积单位为 cm³，则除以24 000。注意题目给定的条件——有时标准状况 (stp, 0 °C 和 1 atm) 下摩尔体积为22.4 dm³。

---

6. Concentration and Moles in Solution | 溶液浓度与物质的量

Concentration (c) is the amount of solute per unit volume of solution, usually mol dm⁻³. The core equation is:

n = c × V (dm³)

If volume is in cm³, convert: V(dm³) = V(cm³) ÷ 1000. For example, 25.0 cm³ of 0.100 mol dm⁻³ HCl contains n = 0.100 × (25.0/1000) = 2.50 × 10⁻³ mol. This is fundamental in titration calculations.

浓度（c）是单位体积溶液中溶质的物质的量，通常单位为 mol dm⁻³。核心方程为：n = c × V (dm³)。若体积以 cm³ 给出，则转换：V(dm³) = V(cm³) ÷ 1000。例如，25.0 cm³ 的 0.100 mol dm⁻³ HCl 含 n = 0.100 × (25.0/1000) = 2.50 × 10⁻³ mol。这是滴定计算的基础。

---

7. Stoichiometry: Mole Ratios in Equations | 化学计量：方程式中的摩尔比

A balanced chemical equation gives the ratio of reacting moles. For the reaction 2H₂ + O₂ → 2H₂O, the mole ratio H₂ : O₂ : H₂O is 2 : 1 : 2. If 5.0 mol of H₂ react completely, 2.5 mol of O₂ are needed and 5.0 mol of H₂O are produced. Always start by writing the balanced equation, then calculate moles of a known substance, and finally use the mole ratio to find the required unknown.

配平的化学方程式给出了反应的摩尔比。对于反应 2H₂ + O₂ → 2H₂O，摩尔比 H₂ : O₂ : H₂O 为 2 : 1 : 2。若5.0 mol H₂ 完全反应，则需要2.5 mol O₂ 并生成5.0 mol H₂O。务必从写出配平的方程式开始，然后计算已知物质的摩尔数，最后利用摩尔比求出未知量。

---

8. Calculating Reacting Masses | 计算反应质量

Combine mass–mole and stoichiometry steps. For example, what mass of CaO is produced from 10.0 g of CaCO₃? (Mᵣ: CaCO₃ = 100.1, CaO = 56.1)

Steps: n(CaCO₃) = 10.0 ÷ 100.1 = 0.0999 mol. Equation: CaCO₃ → CaO + CO₂ gives ratio 1:1, so n(CaO) = 0.0999 mol. Mass of CaO = 0.0999 × 56.1 = 5.61 g.

将质量‑摩尔转换与化学计量相结合。例如：10.0 g CaCO₃ 可生成多少克 CaO？（Mᵣ: CaCO₃ = 100.1，CaO = 56.1）

步骤：n(CaCO₃) = 10.0 ÷ 100.1 = 0.0999 mol。方程式 CaCO₃ → CaO + CO₂ 的摩尔比为1:1，因此 n(CaO) = 0.0999 mol。CaO的质量 = 0.0999 × 56.1 = 5.61 g。

---

9. Limiting Reactants | 限制反应物

In many reactions, one reactant is used up before the others; it is the limiting reactant. Calculate moles of all reactants, then divide each by its stoichiometric coefficient. The smallest value indicates the limiting reactant. The product amount is determined solely by this reactant. For instance, 2.0 mol of H₂ with 0.80 mol of O₂: H₂ ratio 2.0/2 = 1.0, O₂ ratio 0.80/1 = 0.80, so O₂ is limiting and will produce 1.6 mol of H₂O.

在许多反应中，一种反应物会在其他反应物之前被耗尽，它即为限制反应物。计算所有反应物的摩尔数，然后分别除以各自的化学计量系数。所得最小值表明限制反应物。产物的量完全由该反应物决定。例如，2.0 mol H₂ 和 0.80 mol O₂：H₂ 比值 2.0/2 = 1.0，O₂ 比值 0.80/1 = 0.80，因此 O₂ 是限制物，将生成 1.6 mol H₂O。

---

10. Percentage Yield | 产率百分比

The theoretical yield is the maximum mass of product from a given mass of reactant, calculated via stoichiometry. The actual yield is the mass obtained in the experiment. Percentage yield = (actual yield ÷ theoretical yield) × 100. If 4.00 g of CaO is collected in the CaCO₃ decomposition above, yield = (4.00/5.61) × 100 = 71.3%. Yields below 100% occur due to incomplete reactions, side reactions, or purification losses.

理论产率是由给定质量反应物通过化学计量计算得到的最大产物质量。实际产率是实验中得到的质量。产率百分比 = (实际产率 ÷ 理论产率) × 100。若上述 CaCO₃ 分解中收集到 4.00 g CaO，产率 = (4.00/5.61) × 100 = 71.3%。产率低于100%是由于反应不完全、副反应或纯化损失。

---

11. Empirical and Molecular Formulae | 实验式与分子式

The empirical formula gives the simplest whole‑number ratio of atoms in a compound. To find it, convert percentage composition or mass directly to moles of each element, then divide by the smallest number of moles. The molecular formula is a whole‑number multiple of the empirical formula. The multiplier n = Mᵣ (molecular) ÷ Mᵣ (empirical). For example, a compound with 40.0% C, 6.7% H, 53.3% O (by mass) and Mᵣ = 180: moles C:H:O = 40.0/12.0 : 6.7/1.0 : 53.3/16.0 = 3.33 : 6.7 : 3.33 → simplest ratio 1:2:1, empirical CH₂O. Multiplier = 180/30 = 6, so molecular formula is C₆H₁₂O₆.

实验式给出化合物中原子的最简整数比。要得出实验式，可将质量百分数或直接质量转换为各元素的摩尔数，再除以最小的摩尔数。分子式是实验式的整数倍。倍数 n = Mᵣ（分子）÷ Mᵣ（实验式）。例如，某化合物含 40.0% C、6.7% H、53.3% O（质量百分数），Mᵣ = 180：摩尔数 C:H:O = 40.0/12.0 : 6.7/1.0 : 53.3/16.0 = 3.33 : 6.7 : 3.33 → 最简比 1:2:1，实验式 CH₂O。倍数 = 180/30 = 6，故分子式为 C₆H₁₂O₆。

---

12. Water of Crystallisation | 结晶水

Many hydrated salts contain water molecules in their crystal lattice. The number of water molecules per formula unit can be calculated from mass loss on heating. If 4.92 g of hydrated MgSO₄·xH₂O loses 2.52 g of water on strong heating, leaving 2.40 g of anhydrous MgSO₄ (Mᵣ = 120.4), then n(MgSO₄) = 2.40/120.4 = 0.0199 mol. n(H₂O) = 2.52/18.0 = 0.140 mol. Ratio x = 0.140/0.0199 ≈ 7, so the formula is MgSO₄·7H₂O. This method is common in A‑Level practical exams.

许多水合盐的晶格中含有水分子。可利用加热失重计算每式单元中水分子的个数。若4.92 g 水合 MgSO₄·xH₂O 在强热下失水2.52 g，剩余2.40 g 无水 MgSO₄ (Mᵣ = 120.4)，则 n(MgSO₄) = 2.40/120.4 = 0.0199 mol，n(H₂O) = 2.52/18.0 = 0.140 mol。比值 x = 0.140/0.0199 ≈ 7，故该盐的化学式为 MgSO₄·7H₂O。此方法在A‑Level实验考试中十分常见。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）