Year 10 Eduqas Chemistry: Formula & Theorem Quick Reference Handbook | 公式定理速查手册

📚 Year 10 Eduqas Chemistry: Formula & Theorem Quick Reference Handbook | 公式定理速查手册

This handbook collects all the essential formulas and theorems you will encounter in Year 10 Eduqas Chemistry. Use it to quickly check how to convert between mass and moles, work out concentrations, predict gas volumes, calculate energy changes, and revise key chemical principles. Each formula is presented with a clear explanation and applied examples so you can master quantitative chemistry with confidence.

本手册汇集了 Year 10 Eduqas 化学课程中所有必备的公式与定理。你可以用它快速查阅质量与物质的量之间的换算、浓度计算、气体体积预测、能量变化计算，以及重要化学原理的复习。每一条公式都配有清晰的解释和应用示例，帮助你自信掌握定量化学。

1. Relative Atomic Mass and Formula Mass | 相对原子质量与化学式量

Relative atomic mass (A_r) is the average mass of an atom of an element compared to 1/12 of the mass of a carbon‑12 atom. It has no units. For compounds, we use relative formula mass (M_r), which is the sum of the A_r values of all atoms in the formula.

相对原子质量（A_r）是元素一个原子的平均质量与碳‑12 原子质量的 1/12 的比值，没有单位。对于化合物，我们使用相对化学式量（M_r），它是化学式中所有原子的 A_r 值之和。

The formula for M_r is simply:

M_r = Σ (A_r of each atom in the formula)

M_r 的计算公式为：

M_r = Σ（化学式中每个原子的 A_r 之和）

For example, the M_r of H₂ + 16 = 18. For ionic compounds like NaCl, we use relative formula mass in the same way: M_r(NaCl) = 23 + 35.5 = 58.5.

例如，H₂O 的 M_r 是 (2 × 1) + 16 = 18。对于离子化合物如 NaCl，我们同样使用相对化学式量：M_r(NaCl) = 23 + 35.5 = 58.5。

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

One mole of any substance contains exactly 6.02 × 10²³ particles (atoms, molecules, ions or electrons). This number is called Avogadro's constant, N_A.

一摩尔的任何物质都恰好含有 6.02 × 10²³ 个粒子（原子、分子、离子或电子）。这个数字称为阿伏伽德罗常数，记作 N_A。

The relationship between number of particles (N) and amount in moles (n) is:

N = n × N_A

粒子数（N）与物质的量（n）之间的关系为：

N = n × N_A

For instance, 0.5 mol of carbon atoms contains 0.5 × 6.02 × 10²³ = 3.01 × 10²³ atoms. The mole allows chemists to count particles by weighing, bridging the atomic world and the macroscopic world.

例如，0.5 mol 碳原子含有 0.5 × 6.02 × 10²³ = 3.01 × 10²³ 个原子。摩尔使化学家可以通过称量来“数”粒子，在原子世界与宏观世界之间架起桥梁。

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3. Mass-Mole Calculations | 质量-摩尔计算

The most frequently used equation in quantitative chemistry links mass, molar mass and moles:

n = m / M

定量化学中最常用的方程式将质量、摩尔质量与物质的量联系起来：

n = m / M

Here n is the amount in moles (mol), m is the mass in grams (g), and M is the molar mass in grams per mole (g mol⁻¹). Molar mass has the same numerical value as relative formula mass M_r.

其中 n 是物质的量（mol），m 是质量（g），M 是摩尔质量（g mol⁻¹）。摩尔质量的数值与相对化学式量 M_r 相同。

You can rearrange the equation to find mass or molar mass:

m = n × M

M = m / n

你可以重新排列公式以求出质量或摩尔质量：

m = n × M

M = m / n

For example, the mass of 0.2 mol of CO₂ (M = 44 g mol⁻¹) is 0.2 × 44 = 8.8 g. How many moles are in 5.85 g of NaCl? n = 5.85 / 58.5 = 0.10 mol.

例如，0.2 mol CO₂（M = 44 g mol⁻¹）的质量为 0.2 × 44 = 8.8 g。5.85 g NaCl 中含有多少摩尔？ n = 5.85 / 58.5 = 0.10 mol。

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4. Concentration Calculations | 浓度计算

Concentration measures how much solute is dissolved in a given volume of solution. In moles per cubic decimetre (mol dm⁻³), it is calculated by:

c = n / V

浓度衡量一定体积溶液中溶质的量，单位为摩尔每立方分米（mol dm⁻³），计算公式为：

c = n / V

c = concentration in mol dm⁻³, n = number of moles, V = volume of solution in dm³. Note: 1 dm³ = 1000 cm³. You may also see mass concentration in g dm⁻³, which is simply mass/volume.

c 为浓度（mol dm⁻³），n 为摩尔数，V 为溶液的体积（dm³）。注意：1 dm³ = 1000 cm³。你也可能会看到以 g dm⁻³ 为单位的质量浓度，它只是质量/体积。

If 0.25 mol of HCl is dissolved in 500 cm³ of water (0.500 dm³), the concentration is 0.25 / 0.50 = 0.50 mol dm⁻³. The equation can be rearranged to give n = c × V and V = n / c.

若将 0.25 mol HCl 溶于 500 cm³ 水（0.500 dm³）中，浓度为 0.25 / 0.50 = 0.50 mol dm⁻³。该公式可变形为 n = c × V 和 V = n / c。

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5. Gas Volume Calculations at RTP | 常温常压下气体体积计算

At room temperature and pressure (RTP, about 20 °C and 101 kPa), one mole of any gas occupies 24 dm³. This is the molar gas volume. The key formula is:

V = n × 24

在常温常压下（RTP，约 20 °C、101 kPa），一摩尔任何气体的体积均为 24 dm³。这就是气体摩尔体积。关键公式为：

V = n × 24

V = volume of gas at RTP in dm³, n = amount in moles. Conversely, n = V / 24.

V 为气体在 RTP 下的体积（dm³），n 为物质的量。反之，n = V / 24。

For example, 0.15 mol of O₂ has a volume of 0.15 × 24 = 3.6 dm³. If 12 dm³ of CO₂ 50 mol. Remember that this relationship holds for any gas, but only at the stated conditions.

例如，0.15 mol O₂ 的体积为 0.15 × 24 = 3.6 dm³。若收集到 12 dm³ CO₂，其摩尔数为 12 / 24 = 0.50 mol。请记住这一关系适用于任何气体，但仅限于指定条件下。

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6. Percentage Yield and Atom Economy | 产率与原子经济性

Percentage yield compares the actual mass of product obtained to the maximum theoretical mass predicted from the balanced equation.

% Yield = (actual yield / theoretical yield) × 100%

产率将实际获得的产品质量与根据配平方程式预测的最大理论质量进行比较。

产率 = (实际产量 / 理论产量) × 100%

Atom economy measures how much of the total mass of reactants ends up in the desired product. A higher atom economy means a more sustainable process with less waste.

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