📚 Year 10 CCEA Chemistry: Formula & Theorem Quick Reference Handbook | CCEA 化学公式定理速查手册
This handbook pulls together the most important formulas, equations and key principles you need for Year 10 CCEA Chemistry. Whether you are tackling quantitative problems, rates of reaction or titration analysis, these tools will help you work confidently and accurately.
本手册汇集了 Year 10 CCEA 化学中最关键的公式、方程式和重要原理。无论你在处理定量计算、反应速率还是滴定分析，这些工具都能帮助你准确而自信地解题。
1. Relative Atomic Mass and Relative Formula Mass | 相对原子质量和相对式量
The relative atomic mass (Aᵣ) 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. The relative formula mass (Mᵣ) of a compound or molecule is the sum of the Aᵣ values of all the atoms in its formula. For simple molecular substances, we often call it relative molecular mass, but the calculation is the same.
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 之比，没有单位。化合物或分子的相对式量 (Mᵣ) 是其化学式中所有原子 Aᵣ 值的总和。对于简单分子物质，我们常称之为相对分子质量，但计算方式完全相同。
Example: For water, H₂O: Mᵣ = (2 × 1) + 16 = 18. For magnesium nitrate, Mg(NO₃)₂: Mᵣ = 24 + 2 × (14 + 3 × 16) = 148.
示例： 水 H₂O：Mᵣ = (2 × 1) + 16 = 18。硝酸镁 Mg(NO₃)₂：Mᵣ = 24 + 2 × (14 + 3 × 16) = 148。
2. The Mole Concept and Molar Mass | 摩尔概念与摩尔质量
One mole of any substance contains 6.02 × 10²³ particles (atoms, molecules, ions or formula units). This number is called the Avogadro constant. The molar mass (M) of a substance is the mass of one mole of that substance, with units grams per mole (g mol⁻¹). Numerically, the molar mass in g mol⁻¹ is equal to the relative formula mass Mᵣ.
1 摩尔任何物质包含 6.02 × 10²³ 个微粒（原子、分子、离子或式量单元），这个数值被称为阿伏伽德罗常数。物质的摩尔质量 (M) 是 1 摩尔该物质的质量，单位为克每摩尔 (g mol⁻¹)。数值上，以 g mol⁻¹ 为单位的摩尔质量等于相对式量 Mᵣ。
n = m / M
Here, n is the amount of substance in moles (mol), m is the mass in grams (g), and M is the molar mass in grams per mole (g mol⁻¹). This relationship is the foundation of all quantitative chemistry.
式中，n 是物质的量，单位为摩尔 (mol)；m 是质量，单位为克 (g)；M 是摩尔质量，单位为克每摩尔 (g mol⁻¹)。这一关系是所有定量化学的基础。
3. Reacting Mass Calculations | 反应质量计算
To find the mass of a reactant or product, follow these steps: (i) Write the balanced symbol equation. (ii) Convert the known mass to moles using n = m / M. (iii) Use the mole ratio from the balanced equation to determine the moles of the unknown substance. (iv) Convert those moles back to mass with m = n × M. This method works for solids, liquids and solutions.
要计算反应物或生成物的质量，可按照以下步骤进行：(i) 写出配平的化学方程式；(ii) 用 n = m / M 将已知质量换算为摩尔数；(iii) 利用配平方程式中的摩尔比，求出未知物质的摩尔数；(iv) 用 m = n × M 将摩尔数换算回质量。此方法适用于固体、液体和溶液。
Worked example: What mass of water is produced when 4.0 g of hydrogen burns completely? 2H₂ + O₂ → 2H₂O. M(H₂) = 2 g mol⁻¹, so n(H₂) = 4.0 / 2 = 2.0 mol. The equation shows a 1 : 1 mole ratio between H₂ and H₂O, so n(H₂O) = 2.0 mol. M(H₂O) = 18 g mol⁻¹, therefore m(H₂O) = 2.0 × 18 = 36 g.
计算示例：4.0 g 氢气完全燃烧可生成多少克水？2H₂ + O₂ → 2H₂O。 M(H₂) = 2 g mol⁻¹，故 n(H₂) = 4.0 / 2 = 2.0 mol。方程式显示 H₂ 与 H₂O 的摩尔比为 1 : 1，因此 n(H₂O) = 2.0 mol。M(H₂O) = 18 g mol⁻¹，所以 m(H₂O) = 2.0 × 18 = 36 g。
4. Concentration of Solutions | 溶液浓度
The concentration of a solution can be expressed in moles per cubic decimetre (mol dm⁻³) or grams per cubic decimetre (g dm⁻³). The key formula linking moles, volume and concentration is:
溶液的浓度可以用摩尔每立方分米 (mol dm⁻³) 或克每立方分米 (g dm⁻³) 表示。联系摩尔数、体积和浓度的核心公式为：
c = n / V or n = c × V
where c is concentration in mol dm⁻³, n is amount in mol, and V is volume in dm³. Always remember to convert cm³ to dm³ by dividing by 1000 before using this equation. For dilutions, the relationship c₁V₁ = c₂V₂ holds, where c₁ and V₁ are the stock solution concentration and volume, and c₂ and V₂ refer to the diluted solution.
式中 c 为浓度 (mol dm⁻³)，n 为物质的量 (mol)，V 为体积 (dm³)。使用此公式前，务必先将 cm³ 除以 1000 换算为 dm³。对于稀释过程，适用 c₁V₁ = c₂V₂ 这一关系，其中 c₁ 和 V₁ 为原溶液的浓度和体积，c₂ 和 V₂ 为稀释后溶液的浓度和体积。
If you are given mass instead of moles, you can combine m = n × M with the concentration formula: concentration (g dm⁻³) = mass (g) / volume (dm³).
如果已知的是质量而非摩尔数，可将 m = n × M 与浓度公式结合使用：浓度 (g dm⁻³) = 质量 (g) / 体积 (dm³)。
5. Gas Volumes at Room Temperature and Pressure | 室温常压下的气体体积
At room temperature and pressure (RTP, taken as 20 °C and 1 atmosphere), one mole of any gas occupies a volume of 24 dm³ (or 24,000 cm³). This means we can calculate the volume of a gas directly from its amount in moles.
在室温常压 (RTP，即 20 °C 和 1 个大气压) 下，1 摩尔任何气体所占的体积为 24 dm³（或 24,000 cm³）。这意味着我们可以直接由气体的摩尔数计算其体积。
V = n × 24 (volume in dm³)
Similarly, if you measure the volume of a gas produced in a reaction, you can find the moles present by n = V / 24. This applies only when the gas is collected at RTP.
类似地，如果测量了反应中产生的气体体积，可用 n = V / 24 计算出摩尔数。此换算仅适用于在 RTP 条件下收集的气体。
Example: 0.50 mol of CO₂ at RTP occupies 0.50 × 24 = 12 dm³. If 960 cm³ of hydrogen is collected at RTP, n(H₂) = 960 / 24000 = 0.040 mol.
示例：0.50 mol CO₂ 在 RTP 下占 0.50 × 24 = 12 dm³。如果在 RTP 下收集到 960 cm³ 氢气，则 n(H₂) = 960 / 24000 = 0.040 mol。
6. Empirical and Molecular Formulae | 实验式与分子式
The empirical formula shows the simplest whole‑number ratio of atoms of each element in a compound. To find it from mass or percentage data: (i) Divide the mass (or percentage) of each element by its relative atomic mass to get moles. (ii) Divide all the mole values by the smallest of them to obtain a ratio. (iii) Write the formula using these whole‑number ratios.
实验式表示化合物中各元素原子的最简整数比。由质量或百分含量推算实验式的方法为：(i) 将各元素的质量（或百分比）除以各自的相对原子质量，得到摩尔数；(ii) 将所有摩尔数值除以其中的最小值，得到整数比；(iii) 用这些整数比写出化学式。
The molecular formula gives the actual numbers of atoms in a molecule. It is often a whole‑number multiple of the empirical formula. You can determine the molecular formula if you know the relative molecular mass Mᵣ of the compound and the empirical formula mass.
分子式给出一个分子中原子的实际数目，通常是实验式的整数倍。若已知化合物的相对分子质量 Mᵣ 及实验式质量，即可确定分子式。
n = Mᵣ (molecule) / Mᵣ (empirical formula)
Here, n is the multiplier. Multiply the subscripts in the empirical formula by n to get the molecular formula.
式中 n 为倍数。将实验式中各原子的下标乘以 n，即得分子式。
7. Percentage Yield and Atom Economy | 产率与原子经济性
Percentage yield compares the mass of product actually obtained in an experiment with the theoretical maximum mass calculated from the balanced equation. It indicates the efficiency of a reaction.
产率用来比较实验中实际获得的产品质量与根据配平方程式算出的理论最大质量，反映了反应的效率。
% Yield = (actual mass / theoretical mass) × 100
Atom economy measures the proportion of reactant atoms that end up in the desired product. It is especially important in green chemistry and industrial processes.
原子经济性衡量的是反应物原子进入目标产物中的比例，在绿色化学和工业生产中尤为重要。
% Atom economy = (Mᵣ of desired product / sum of Mᵣ of all reactants) × 100
Reactions with high atom economy produce less waste and make more sustainable use of materials.
原子经济性高的反应废物产生较少，能更可持续地利用原料。
8. Rate of Reaction | 反应速率
The rate of a chemical reaction measures how quickly reactants are used up or products are formed. It can be calculated as the change in amount or concentration of a substance per unit time.
化学反应速率衡量反应物消耗或产物生成的快慢程度，可以用单位时间内物质的变化量来计算。
Average rate = quantity of product formed / time taken
Common units are g s⁻¹, cm³ s⁻¹ or mol s⁻¹. The rate can be determined from the slope of a graph of volume of gas released or mass lost against time.
常用单位有 g s⁻¹、cm³ s⁻¹ 或 mol s⁻¹。可通过绘制气体释放体积或质量减少量对时间的关系图，由斜率求出反应速率。
To explain differences in rates, we use collision theory. For a reaction to occur, particles must collide with sufficient energy (activation energy) and with the correct orientation. Factors that increase the frequency of successful collisions — higher concentration, increased surface area of solids, higher temperature, and the presence of a catalyst — all increase the rate of reaction. A catalyst provides an alternative reaction pathway with a lower activation energy, but it is not used up.
解释速率差异需借助碰撞理论。反应要发生，微粒必须发生碰撞，且碰撞能量不低于活化能，同时取向合适。凡是能增加有效碰撞频率的因素——浓度提高、固体表面积增大、温度升高以及催化剂的存在——都会加快反应速率。催化剂能提供活化能较低的新反应路径，且自身在反应中不被消耗。
9. Titration Calculations | 滴定计算
Titration is a volumetric technique used to find the concentration of an unknown solution by reacting it with a solution of known concentration. The calculation relies on the balanced equation and the volumes delivered from the burette and pipette.
滴定是一种通过已知浓度的溶液与待测溶液反应以测定后者浓度的体积分析技术。相关计算需以配平方程式为基础，并利用滴定管和移液管量取的体积。
c₁V₁ / n₁ = c₂V₂ / n₂
where c₁ and V₁ are the concentration and volume of the first solution, c₂ and V₂ those of the second solution, and n₁ and n₂ are the stoichiometric coefficients (mole ratios) from the balanced equation. For a 1:1 acid–base reaction, the formula simplifies to cₐcᵢd Vₐcᵢd = cₐlkₐlᵢ Vₐlkₐlᵢ, provided both are measured in the same unit.
式中 c₁、V₁ 为第一种溶液的浓度和体积，c₂、V₂ 为第二种溶液的浓度和体积，n₁ 和 n₂ 为配平方程式中相应物质的化学计量系数（摩尔比）。对于 1:1 的酸碱反应，公式可简化为 cₐcᵢd Vₐcᵢd = cₐlkₐlᵢ Vₐlkₐlᵢ，前提是二者采用相同的体积单位。
Always record burette readings to ±0.05 cm³, perform several consistent titres, and discard any rough or anomalous values before calculating the mean titre volume.
记录滴定管读数应精确至 ±0.05 cm³，进行若干次重复滴定，舍去初测值或异常值后再计算平均滴定剂体积。
10. Conservation of Mass and Balancing Equations | 质量守恒与配平方程式
The law of conservation of mass states that no atoms are created or destroyed in a chemical reaction. The total mass of the reactants equals the total mass of the products. This is why we must balance chemical equations — the same number of each type of atom must appear on both sides of the arrow.
质量守恒定律指出，化学反应中原子的种类和数目不变，反应物的总质量等于生成物的总质量。正因如此，我们才必须配平化学方程式——箭头两边每种原子的数目必须相同。
When balancing equations, first write the correct formulae for all reactants and products, then adjust only the large coefficients in front of each formula until the atom counts match. Never change the subscripts inside a formula. State symbols (s), (l), (g) and (aq) can be added to show the physical states of the substances.
配平方程式时，先写出所有反应物和生成物的正确化学式，然后仅调整各化学式前的大系数，直至原子数目相符。切勿改动化学式内部的下标。可使用状态符号 (s)、(l)、(g) 和 (aq) 注明物质的物态。
In ionic equations, both mass and charge must be balanced. Spectator ions — ions that appear unchanged on both sides — are omitted to show only the species that actually change.
在离子方程式中，质量和电荷均需配平。未参与反应的旁观离子——在反应前后保持不变的离子——应被省略，仅展现实际发生变化的物种。
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