Year 11 Eduqas Chemistry Formula & Theorem Quick-Reference | Eduqas 化学公式速查手册

📚 Year 11 Eduqas Chemistry Formula & Theorem Quick-Reference | Eduqas 化学公式速查手册

This quick-reference guide brings together all of the essential formulae, equations and theorems that you need for Year 11 Eduqas GCSE Chemistry. Every formula is stated clearly with its units, followed by a concise explanation in both English and Chinese. Use it alongside your revision to memorise quantitative calculations, energy changes, rates, equilibrium and electrolysis.

本速查手册汇总了 Year 11 Eduqas 化学考试所需的所有关键公式、方程式与定理。每条公式均明确标注单位，并配有英中双语简要解释。可将它与日常复习结合使用，以便熟记定量计算、能量变化、速率、平衡和电解等核心内容。

1. Relative Masses and the Mole | 相对质量与摩尔

Aᵣ = average mass of one atom of an element ÷ 1/12 mass of a ¹²C atom

The relative atomic mass (Aᵣ) compares the average mass of an atom of an element to one-twelfth of the mass of a carbon‑12 atom. It has no units.

相对原子质量 (Aᵣ) 是某元素一个原子的平均质量与一个碳‑12 原子质量的 1/12 的比值，没有单位。

Mᵣ = sum of Aᵣ values of all atoms in a formula

The relative formula mass (Mᵣ) is used for both ionic and covalent substances. It equals the sum of the relative atomic masses of every atom shown in the formula.

相对分子质量 (Mᵣ) 适用于离子化合物和共价化合物，等于化学式中所有原子的相对原子质量之和。

n = m / M

Moles (n) equal mass (m) in grams divided by molar mass (M) in g mol⁻¹. Molar mass is numerically equal to the relative formula mass Mᵣ.

物质的量 n (mol) = 质量 m (g) ÷ 摩尔质量 M (g mol⁻¹)。摩尔质量的数值与相对分子质量 Mᵣ 相同。

N = n × Nₐ   (Nₐ = 6.02 × 10²³ mol⁻¹)

The number of particles (N) is obtained by multiplying the amount in moles by the Avogadro constant, Nₐ.

微粒数量 N = 物质的量 n × 阿伏加德罗常数 Nₐ (6.02 × 10²³ mol⁻¹)。

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2. Concentration of Solutions | 溶液浓度

c (mol dm⁻³) = n / V (dm³)

Concentration in mol dm⁻³ is the number of moles of solute dissolved in 1 dm³ of solution. Always convert volumes in cm³ to dm³ by dividing by 1000.

摩尔浓度 (mol dm⁻³) 是指每 1 dm³ 溶液中溶质的物质的量。务必先将体积从 cm³ 转换为 dm³，除以 1000。

concentration (g dm⁻³) = mass (g) / volume (dm³)

Mass concentration is often used when the molar mass is unknown. 1 g dm⁻³ means 1 gram of solute is dissolved in 1 dm³ of solution.

质量浓度 (g dm⁻³) 在摩尔质量未知时常用。1 g dm⁻³ 表示 1 dm³ 溶液中含有 1 克溶质。

c₁V₁ = c₂V₂   (dilution)

When a solution is diluted, the number of moles stays the same. The product of concentration and volume before dilution equals the product after dilution.

稀释溶液时，溶质的物质的量不变，稀释前后浓度与体积的乘积相等。

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3. Gases and Molar Volume | 气体与摩尔体积

V (dm³) = n × 24   (at RTP)

At room temperature and pressure (RTP, 20 °C and 1 atm), one mole of any gas occupies 24 dm³ (24 000 cm³). This is the molar gas volume.

在常温常压 (RTP, 20 °C, 1 atm) 下，1 摩尔任何气体的体积约为 24 dm³ (24 000 cm³)。这是气体摩尔体积。

The volume of gas produced or used in a reaction can be found directly from the moles of gas and the molar volume. If the volume is required in cm³, multiply by 24 000 instead of 24.

反应中产生或消耗的气体体积可直接由气体物质的量和摩尔体积求得。若体积需用 cm³ 表示，则乘以 24 000 而非 24。

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4. Reacting Masses from Equations | 根据方程式计算反应质量

work in moles → apply mole ratio from the balanced equation → convert to mass

Reacting mass calculations always follow three steps: convert given masses to moles, use the mole ratio from the balanced symbol equation, and convert the target moles back to mass using m = n × M.

反应质量计算始终遵循三步：将已知质量转换为物质的量，利用配平化学方程式的摩尔比，再将目标物质的量转换回质量 m = n × M。

The limiting reagent is the reactant that is completely consumed first. It determines the maximum amount of product formed.

限制反应物是先被完全消耗的反应物，它决定了产物的最大产量。

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5. Percentage Yield and Atom Economy | 百分产率与原子经济

% yield = (actual yield ÷ theoretical yield) × 100%

Percentage yield compares the mass of product actually obtained with the maximum theoretical mass calculated from the limiting reagent. It is always less than or equal to 100%.

百分产率将实际得到的产物质量与由限制反应物算出的最大理论质量进行比较，其值总是 ≤ 100%。

atom economy = (Mᵣ of desired product ÷ sum of Mᵣ of all reactants) × 100%

Atom economy measures how much of the starting mass ends up in the useful product. High atom economy reduces waste and makes a process more sustainable.

原子经济衡量起始物料中有多少最终进入目标产物。原子经济越高，废物越少，流程越可持续。

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6. Titration Calculations | 滴定计算

(c₁ × V₁) / n₁ = (c₂ × V₂) / n₂   (n₁, n₂ are coefficients in the balanced equation)

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