Essential AS Chemistry Formula Handbook | AS 化学：公式汇总手册

📚 Essential AS Chemistry Formula Handbook | AS 化学：公式汇总手册

This handbook collects the essential equations and formulas needed for AS-level Chemistry, covering mole calculations, energetics, equilibrium, acids and bases, electrolysis, and more. Save it for quick revision and reference.

这份手册汇总了AS化学所需的关键方程式和公式，涵盖摩尔计算、能量学、化学平衡、酸碱、电解等内容。可用于快速复习和查阅。

1. The Mole & Avogadro’s Number | 摩尔与阿伏加德罗常数

The mole is the central unit in chemistry. One mole of substance contains 6.02 × 10²³ entities (Avogadro’s number, L). The number of particles N is related to the amount of substance n by N = n × L.

摩尔是化学的核心单位。1摩尔物质包含6.02×10²³个基本单元（阿伏加德罗常数 L）。粒子数 N 与物质的量 n 的关系为 N = n × L。

N = n × L

Similarly, the mass m of a substance can be calculated from its molar mass M: m = n × M. Rearranging gives n = m / M and M = m / n. Remember to use consistent units: mass in grams, molar mass in g mol⁻¹.

类似地，物质的质量 m 可由其摩尔质量 M 计算：m = n × M。变形可得 n = m / M 和 M = m / n。注意使用一致的单位：质量以克为单位，摩尔质量以 g mol⁻¹ 为单位。

n = m / M

For gases at room temperature and pressure (RTP, 20°C, 1 atm), one mole occupies 24 dm³. The volume V of a gas is given by V = n × 24 dm³ mol⁻¹ (or 24,000 cm³).

对于室温常压（RTP，20°C，1 atm）下的气体，1摩尔占据24 dm³体积。气体体积 V = n × 24 dm³ mol⁻¹（或 24,000 cm³）。

V (gas) = n × 24 dm³

2. Empirical & Molecular Formulae | 经验式与分子式

The empirical formula shows the simplest whole-number ratio of atoms in a compound, while the molecular formula shows the actual number of atoms. Determine moles of each element; divide by smallest to get ratio. Molecular formula = (empirical formula)ₙ where n = relative molecular mass / empirical formula mass.

经验式表示化合物中原子最简整数比，而分子式表示实际原子数目。计算各元素物质的量；除以最小值得出整数比。分子式 = (经验式)ₙ，其中 n = 相对分子质量 / 经验式质量。

n = relative molecular mass / empirical formula mass

Example: If empirical formula is CH₂O and molar mass is 180 g mol⁻¹, empirical mass = 30 g mol⁻¹, so n = 180/30 = 6, molecular formula = C₆H₁₂O₆.

例如：经验式为CH₂O，摩尔质量为180 g mol⁻¹，经验式质量为 30 g mol⁻¹，则 n = 6，分子式为 C₆H₁₂O₆。

3. Stoichiometry & Reacting Masses | 化学计量学与反应质量

Balanced equations give mole ratios of reactants and products. Use n = m/M to convert masses to moles, apply the ratio, then convert back to mass. Limiting reagent calculations require identifying the reactant that runs out first.

配平方程式给出反应物和生成物的摩尔比。使用 n = m/M 将质量转换为物质的量，应用比例，再转换回质量。限量试剂计算需要先确定先消耗完的反应物。

Mass of product = (moles of limiting reagent) × (mole ratio) × M(product)

Gas volume calculations at RTP use V = n × 24 dm³. In titrations, moles of known reactant = c × V (in dm³), and then moles of unknown found via reaction stoichiometry.

在RTP下气体体积计算使用 V = n × 24 dm³。在滴定中，已知反应物的物质的量 = c × V (V以 dm³ 计)，然后通过化学计量比求出未知物的物质的量。

4. Concentration & Dilution | 浓度与稀释

Concentration c (mol dm⁻³) is the amount of solute per unit volume of solution: c = n / V. For dilution, the number of moles stays constant: c₁V₁ = c₂V₂. Always convert volumes to dm³ (1 dm³ = 1000 cm³).

浓度 c（mol dm⁻³）是单位体积溶液中溶质的物质的量：c = n / V。稀释时物质的量

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