A-Level OCR Chemistry: Mole Calculations | 摩尔计算考点精讲

📚 A-Level OCR Chemistry: Mole Calculations | 摩尔计算考点精讲

Mole calculations form the bedrock of quantitative chemistry, and they are assessed throughout OCR A-Level Chemistry papers. A solid grasp of the mole concept and its applications — from reacting masses and gas volumes to titrations and yield — is essential for success in both AS and A2 units. This guide walks you through every key calculation technique, with worked examples and tips to avoid common pitfalls.

摩尔计算是定量化学的基础，贯穿OCR A-Level化学的各个试卷。牢固掌握摩尔概念及其应用——从反应质量和气体体积到滴定和产率——对于在AS和A2单元中取得成功至关重要。本指南将带你学习每一项关键计算技巧，附有示例和避免常见错误的提示。

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

A mole is the amount of substance that contains exactly 6.022 × 10²³ specified elementary entities (atoms, molecules, ions, etc.). This number is Avogadro’s constant (Nₐ). In OCR, you must be comfortable using Nₐ to relate the number of particles to the amount in moles.

一摩尔是恰好含有 6.022 × 10²³ 个指定基本单元（原子、分子、离子等）的物质的量。这个数就是阿伏伽德罗常数（Nₐ）。在 OCR 考试中，你必须熟练运用 Nₐ 把粒子数目与摩尔量联系起来。

The relationship is:

关系式为：

Number of particles = n × Nₐ

Where n is the amount in mol. If you are asked how many atoms are in 0.450 mol of carbon, you multiply 0.450 by 6.022 × 10²³ to get 2.71 × 10²³ atoms. OCR often embeds these calculations within questions on redox titrations, percentage composition, or gas volumes. Remember to check whether you need the number of atoms or molecules; for diatomic gases like O₂, 1 mol of molecules contains 2 mol of atoms.

其中 n 为摩尔量。如果题目问 0.450 mol 碳中含有多少个原子，就用 0.450 乘以 6.022×10²³，得到 2.71×10²³ 个原子。OCR 经常把这种计算嵌入到氧化还原滴定、百分组成或气体体积的题目中。务必注意题目要求的是原子数还是分子数；对于 O₂ 等双原子气体，1 mol 分子含有 2 mol 原子。

2. Molar Mass and Mass-Mole Conversions | 摩尔质量与质量-摩尔转换

Molar mass (M) is the mass per mole of a substance, with units g mol⁻¹. It is numerically equal to the relative formula mass (M_r) or relative atomic mass (A_r). The core equation is:

摩尔质量（M）是每摩尔物质的质量，单位为 g mol⁻¹，数值上等于相对式量（M_r）或相对原子质量（A_r）。核心方程为：

n = m / M

To convert mass to moles, divide the mass in grams by the molar mass. For example, 8.55 g of aluminium sulfate, Al₂(SO₄)₃ (M_r = 342.3), gives n = 8.55 / 342.3 ≈ 0.0250 mol. OCR frequently tests this in the context of calculating reacting masses or preparing standard solutions. Be careful with hydrated salts — the water of crystallisation must be included in the molar mass (e.g., CuSO₄·5H₂O has M_r = 249.7).

将质量转换为摩尔数时，用质量（g）除以摩尔质量。例如，8.55 g 硫酸铝 Al₂(SO₄)₃（M_r = 342.3），n = 8.55 / 342.3 ≈ 0.0250 mol。OCR 经常在计算反应质量或配制标准溶液的背景下考察这一点。对含结晶水的盐要格外小心——结晶水必须计入摩尔质量（如 CuSO₄·5H₂O 的 M_r = 249.7）。

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

The empirical formula shows the simplest whole-number ratio of atoms in a compound. The molecular formula is a multiple of the empirical formula. OCR expects you to calculate these from percentage composition or combustion data.

经验式表示化合物中原子的最简整数比。分子式是经验式的整数倍。OCR 要求能根据百分组成或燃烧数据计算这些。

Worked example: A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Convert % to mass, then to moles: C: 40.0/12.0 = 3.33; H: 6.7/1.0 = 6.7; O: 53.3/16.0 = 3.33. Divide by smallest (3.33) to get ratio C: 1, H: 2, O: 1 → empirical formula CH₂O. If the relative molecular mass is 180, the molecular formula is C₆H₁₂O₆ (since 12+2+16 = 30, and 180/30 = 6).

计算示例：某化合物含碳 40.0%，氢 6.7%，氧 53.3%（质量分数）。将百分数转换为质量，再转换为摩尔数：C: 40.0/12.0 = 3.33；H: 6.7/1.0 = 6.7；O: 53.3/16.0 = 3.33。除以最小值（3.33）得比例 C:1, H:2, O:1 → 经验式 CH₂O。若相对分子质量为 180，则分子式为 C₆H₁

Published by TutorHao | A-Level Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）