Mole Calculations | 摩尔计算

📚 Mole Calculations | 摩尔计算

Mole calculations are the quantitative language of chemistry. They allow chemists to convert between particles, mass, gas volume, solution concentration and reacting amounts using one central unit: the mole.

摩尔计算是化学的定量语言。借助“摩尔”这一核心单位，化学家可以在微粒数、质量、气体体积、溶液浓度和反应量之间进行换算。

1. The Mole and Avogadro Constant | 摩尔与阿伏伽德罗常数

One mole of any substance contains exactly 6.02 × 10²³ elementary entities. This number is called the Avogadro constant, L, and its unit is mol⁻¹. In Cambridge A Level calculations, L = 6.02 × 10²³ mol⁻¹ is used unless a more exact value is specified.

一摩尔的任何物质都含有 6.02 × 10²³ 个基本单元。这个数称为阿伏伽德罗常数 L，单位为 mol⁻¹。在剑桥 A Level 计算中，除非另有说明，通常使用 L = 6.02 × 10²³ mol⁻¹。

N = n × L   or   n = N ÷ L

For example, 2.00 mol of sodium atoms contains 2.00 × 6.02 × 10²³ = 1.20 × 10²⁴ atoms. One molecule of CO₂ contains 3 atoms, so 1 mol of CO₂ molecules contains 3 mol of atoms.

例如，2.00 mol 钠原子含有 2.00 × 6.02 × 10²³ = 1.20 × 10²⁴ 个原子。一个 CO₂ 分子含有 3 个原子，因此 1 mol CO₂ 分子含有 3 mol 原子。

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. Its unit is g mol⁻¹ and it is numerically equal to the relative atomic mass Ar or relative formula mass Mr.

摩尔质量 M 是一摩尔物质的质量，单位为 g mol⁻¹，其数值等于相对原子质量 Ar 或相对式量 Mr。

n = m ÷ M   m = n × M   M = m ÷ n

Example: M(H₂O) = 18.0 g mol⁻¹, so 36.0 g of water is 36.0 ÷ 18.0 = 2.00 mol. Conversely, 0.250 mol of NaOH (M = 40.0 g mol⁻¹) has a mass of 0.250 × 40.0 = 10.0 g.

示例：M(H₂O) = 18.0 g mol⁻¹，所以 36.0 g 水为 36.0 ÷ 18.0 = 2.00 mol。反过来，0.250 mol NaOH（M = 40.0 g mol⁻¹）的质量为 0.250 × 40.0 = 10.0 g。

3. 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; the molecular formula gives the actual number of atoms in one molecule. To find the empirical formula, convert percentage composition by mass to moles, divide each by the smallest number of moles, and convert to whole numbers.

实验式给出化合物中各原子的最简整数比；分子式给出一个分子中原子的实际数目。求实验式的方法是：将质量百分组成换算为摩尔数，分别除以最小摩尔数，再化为整数。

• Assume 100 g of the compound, so percentages become masses in grams.
• Divide each mass by its relative atomic mass to get moles.
• Divide all mole values by the smallest value.
• Multiply by a small factor if necessary to get whole numbers.

假设取 100 g 化合物，因此质量分数可直接看作质量（克）。用每种元素的质量除以相对原子质量得到摩尔数，再用最小摩尔数去除所有值，必要时乘以一个小的整数以得到整比。

Example: a hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. n(C) = 85.7 ÷ 12.0 = 7.14 mol, n(H) = 14.3 ÷ 1.0 = 14.3 mol. Dividing by the smaller value gives CH₂ as the empirical formula. If the molar mass is 56.0 g mol⁻¹, the molecular formula is C₄H₈ because (12.0 + 2 × 1.0) × 4 = 56.0.

示例：某烃含碳 85.7%、氢 14.3%。n(C) = 85.7 ÷ 12.0 = 7.14

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