📚 Year 9 OCR Chemistry: Quick Reference Handbook of Formulas and Theorems | Year 9 OCR 化学：公式定理速查手册
This quick reference handbook brings together the essential formulas, laws and theorems covered in the Year 9 OCR Chemistry course. Use it to revise key quantitative relationships, from conservation of mass to mole calculations and bond energies, all in one place.
这本速查手册汇集了 Year 9 OCR 化学课程中的重要公式、定律与定理。从质量守恒到摩尔计算以及键能，所有关键定量关系都在这里，方便你集中复习。
1. Law of Conservation of Mass | 质量守恒定律
In any chemical reaction, the total mass of the reactants is always equal to the total mass of the products. No atoms are created or destroyed; they are simply rearranged.
在任何化学反应中，反应物的总质量始终等于生成物的总质量。原子既不会凭空产生也不会凭空消失，只是被重新排列。
This law is the foundation of all chemical calculations. When balancing equations, the number of each type of atom must remain the same on both sides.
该定律是所有化学计算的基础。配平化学方程式时，两边每种原子的数目必须保持相等。
2. Relative Atomic Mass (Ar) | 相对原子质量
The relative atomic mass (Ar) of an element is the average mass of its atoms compared to 1/12th the mass of a carbon‑12 atom. It has no units.
相对原子质量 (Ar) 是指一个元素的原子平均质量与一个碳‑12 原子质量的 1/12 相比较所得的比值，没有单位。
If an element has several isotopes, Ar is the weighted mean of their mass numbers. The approximate formula is:
如果一种元素有几种同位素，Ar 就是它们质量数的加权平均值。近似计算公式为：
Ar = [(%abundance1 × mass1) + (%abundance2 × mass2) + ...] / 100
Ar = [(%丰度1 × 质量数1) + (%丰度2 × 质量数2) + ...] / 100
For example, chlorine has two main isotopes: chlorine‑35 (75%) and chlorine‑37 (25%). Its Ar is approximately (75×35 + 25×37) / 100 = 35.5.
例如，氯有两种主要同位素：氯‑35 (75%) 和氯‑37 (25%)。其 Ar 约为 (75×35 + 25×37) / 100 = 35.5。
3. Relative Formula Mass (Mr) | 相对式量
Relative formula mass (Mr) applies to compounds. It is the sum of the relative atomic masses of all the atoms shown in the chemical formula. Like Ar, Mr has no units.
相对式量 (Mr) 适用于化合物。它是化学式中所有原子的相对原子质量之和。与 Ar 一样，Mr 没有单位。
To calculate Mr for water, H₂O: Mr = (2 × 1) + (1 × 16) = 18.
计算水 (H₂O) 的 Mr：Mr = (2 × 1) + (1 × 16) = 18。
um nitrate, Mg(NO₃)₂, first count the atoms inside the brackets: Mg ×1, N ×2, O ×6. Then Mr = 24 + (2×14) + (6×16) = 148.
对于硝酸镁 Mg(NO₃)₂，先数括号内原子：Mg ×1, N ×2, O ×6。则 Mr = 24 + (2×14) + (6×16) = 148。
4. Percentage Composition | 百分组成
The percentage by mass of an element in a compound is the total mass of that element in one formula unit divided by the Mr of the compound, multiplied by 100.
化合物中某元素的质量百分数等于该元素在一个式量单位中的总质量除以化合物的 Mr，再乘以 100。
% element = (total mass of element in formula / Mr) × 100
元素% = (化学式中该元素总质量 / Mr) × 100
Example: find the percentage of oxygen in CO₂. Mr of CO₂ = 12 + (2×16) = 44. Mass of oxygen = 2×16 = 32. %O = (32 / 44) × 100 = 72.7%.
例题：求 CO₂ 中氧的百分含量。CO₂ 的 Mr = 12 + (2×16) = 44。氧的质量 = 2×16 = 32。O% = (32 / 44) × 100 = 72.7%。
5. The Mole and Molar Mass | 摩尔与摩尔质量
cles (atoms, molecules, ions or electrons). This number is called the Avogadro constant, Nₐ.
1 摩尔是物质中所含基本单元（原子、分子、离子或电子）数目恰好为 6.02 × 10²³ 个的量。这个数称为阿伏伽德罗常数，记作 Nₐ。
Molar mass (M) is the mass of one mole of a substance, expressed in g/mol. Numerically, it equals the relative atomic mass or relative formula mass.
摩尔质量 (M) 是 1 摩尔物质的质量，单位为 g/mol。在数值上，它等于相对原子质量或相对式量。
The central relationship linking moles, mass and molar mass is:
联系物质的量、质量与摩尔质量的核心关系式为：
n = m / M
where n = amount in mol, m = mass in g, M = molar mass in g/mol.
式中 n = 物质的量 (mol)，m = 质量 (g)，M = 摩尔质量 (g/mol)。
Rearranging gives m = n × M, which is equally useful.
变形可得 m = n × M，同样常用。
6. Reacting Mass Calculations | 反应质量计算
To calculate the mass of a product or reactant, follow four steps:
要计算生成物或反应物的质量，请遵循以下四个步骤：
on to find moles of the unknown substance. Step 4: Convert moles of the unknown back to mass with m = n × M.
第一步：写出配平的化学方程式。第二步：用 n = m/M 将已知质量转化为物质的量。第三步：根据方程式中的化学计量数比，求出未知物的物质的量。第四步：用 m = n × M 将未知物的物质的量转换为质量。
Example: What mass of magnesium oxide forms when 6 g of magnesium burns completely? = 0.25 mol. Mole ratio Mg : MgO = 2 : 2, so n(MgO) = 0.25 mol. m(MgO) = 0.25 × 40 = 10 g.
例题：6 g 镁完全燃烧可生成多少克氧化镁？2Mg + O₂ → 2MgO。Mg 的物质的量 = 6 / 24 = 0.25 mol。计量比 Mg : MgO = 2 : 2，故 n(MgO) = 0.25 mol。m(MgO) = 0.25 × 40 = 10 g。
7. Concentration of Solutions | 溶液浓度
Concentration measured in grams per cubic decimetre (g/dm³) tells you the mass of solute dissolved in 1 dm³ of solution.
浓度若以克每立方分米 (g/dm³) 表示，则表示 1 dm³ 溶液中溶解的溶质质量。
Concentration (g/dm³) = mass of solute (g) / volume of solution (dm³)
浓度 (g/dm³) = 溶质质量 (g) / 溶液体积 (dm³)
When working with moles, the molar concentration (mol/dm³) is often used:
涉及物质的量时，常用摩尔浓度 (mol/dm³)：
c = n / V
ncentration in mol/dm³, n is amount in mol, V is volume in dm³. Remember 1 dm³ = 1000 cm³.
式中 c 为摩尔浓度 (mol/dm³)，n 为物质的量 (mol)，V 为溶液体积 (dm³)。记住 1 dm³ = 1000 cm³。
8. Bond Energy and Enthalpy Change | 键能与焓变
In a chemical reaction, bonds in the reactants are broken and new bonds in the products are formed. The overall energy change, ΔH, can be estimated using average bond energies.
在化学反应中，反应物中的化学键断裂，生成物中形成新的化学键。总能量变化 ΔH 可用平均键能来估算。
ΔH = Σ(bond energies of bonds broken) – Σ(bond energies of bonds formed)
ΔH = Σ(断裂键的键能) – Σ(形成键的键能)
Bond breaking is endothermic (absorbs energy), bond making is exothermic (releases energy).
断键吸热（吸热过程），成键放热（放热过程）。
If ΔH is negative the reaction is exothermic overall; if ΔH is positive it is endothermic.
若 ΔH 为负值，反应整体放热；若 ΔH 为正值，反应整体吸热。
Example: H₂ + Cl₂ → 2HCl. Bonds broken: H–H (436 kJ/mol) + Cl–Cl (243 kJ/mol) = 679 kJ. Bonds formed: 2 × H–Cl (2 × 431 = 862 kJ). ΔH = 679 – 862 = –183 kJ/mol, so the reaction is exothermic.
例题：H₂ + Cl₂ → 2HCl。断裂的键：H–H (436 kJ/mol) + Cl–Cl (243 kJ/mol) = 679 kJ。形成的键：2 × H–Cl (2 × 431 = 862 kJ)。ΔH = 679 – 862 = –183 kJ/mol，故该反应放热。
9. Factors Affecting Rate of Reaction | 反应速率影响因素
The rate of a chemical reaction can be increased by raising the temperature, increasing the concentration of reactants in solution, increasing the pressure of gaseous reactants, or increasing the surface area of solid reactants. Adding a suitable catalyst also speeds up the reaction without being used up.
提升温度、增大溶液中反应物浓度、增大气体反应物压强、增大固体反应物表面积，均可提高化学反应速率。加入合适的催化剂也能加快反应，而催化剂本身在反应中不被消耗。
Collision theory explains these effects: for a reaction to occur, particles must collide with sufficient energy (greater than or equal to the activation energy) and with the correct orientation.
碰撞理论解释了这些影响：发生反应需要粒子发生碰撞，并且碰撞能量须大于或等于活化能，同时要有正确的取向。
Increasing temperature gives particles more kinetic energy, so a greater proportion of collisions have enough energy to react. Increasing concentration, pressure or surface area increases the frequency of collisions.
升高温度使粒子动能增加，有足够能量发生反应的有效碰撞比例上升。增大浓度、压强或表面积则增大了碰撞频率。
10. pH Scale and Neutralisation | pH 值与中和反应
The pH scale ranges from 0 to 14. A pH less than 7 indicates an acidic solution, pH 7 is neutral, and a pH greater than 7 indicates an alkaline solution.
pH 值范围为 0 到 14。pH 小于 7 表示酸性溶液，pH 等于 7 为中性，pH 大于 7 表示碱性溶液。
Acids produce H⁺ ions in water; alkalis produce OH⁻ ions. During neutralisation, H⁺ and OH⁻ combine to form water:
酸在水中产生 H⁺ 离子；碱产生 OH⁻ 离子。中和反应中，H⁺ 与 OH⁻ 结合生成水：
H⁺ (aq) + OH⁻ (aq) → H₂O (l)
, hydrochloric acid + sodium hydroxide → sodium chloride + water: HCl + NaOH → NaCl + H₂O.
剩余的离子则形成盐。例如，盐酸 + 氢氧化钠 → 氯化钠 + 水：HCl + NaOH → NaCl + H₂O。
11. Balancing Chemical Equations | 化学方程式配平
Because atoms are conserved, a chemical equation must have the same number of each type of atom on both the reactant and product sides.
由于原子守恒，化学方程式两边每一种原子的数目必须相等。
To balance an equation, write the correct formulae first, then adjust the large numbers (coefficients) placed in front of each formula until the atom counts match. You must never change the small subscript numbers inside a formula.
配平方程式时，先写出正确的化学式，然后调整化学式前的大数字（化学计量数），直到每种原子总数相等。切勿更改化学式中的小下标数字。
t 2 in front of Na₂O: 2Na + O₂ → 2Na₂O. Now Na: 2 vs 4, so put 4Na on left. Final equation: 4Na + O₂ → 2Na₂O.
例题：配平 Na + O₂ → Na₂O。步骤 1：原子计数 (Na: 1 对 2, O: 2 对 1)。步骤 2：在左侧 Na 前加系数 2：2Na + O₂ → Na₂O。O 仍不平衡 (2 对 1)。在 Na₂O 前加系数 2：2Na + O₂ → 2Na₂O。此时 Na 为 2 对 4，于是左侧改为 4Na。最终方程式：4Na + O₂ → 2Na₂O。
12. Writing Chemical Formulae and State Symbols | 化学式书写与状态符号
For ionic compounds, the charges on the ions are used to deduce the simplest whole-number ratio. For example, magnesium ions Mg²⁺ and chloride ions Cl⁻ combine as MgCl₂ because two Cl⁻ ions are needed to balance one Mg²⁺.
书写离子化合物的化学式时，利用离子电荷推求最简整数比。例如，镁离子 Mg²⁺ 与氯离子 Cl⁻ 结合成 MgCl₂，因为需要两个 Cl⁻ 来平衡一个 Mg²⁺ 的电荷。
For molecular (covalent) compounds, prefixes often indicate the number of atoms: CO is carbon monoxide, CO₂ is carbon dioxide.
对于共价分子化合物，前缀常表示原子数目：CO 是一氧化碳，CO₂ 是二氧化碳。
State symbols must be added in equations: (s) for solid, (l) for liquid, (g) for gas, and (aq) for an aqueous solution (dissolved in water).
在方程式中必须标注状态符号：(s) 代表固体，(l) 代表液体，(g) 代表气体，(aq) 代表水溶液（溶于水中）。
(s) + H₂O(l). The state symbols show that a gas reacts with a solution to form a solid precipitate and a liquid.
例如，用石灰水检验二氧化碳：CO₂(g) + Ca(OH)₂(aq) → CaCO₃(s) + H₂O(l)。状态符号表明一种气体与一种溶液反应，生成固体沉淀和液体。
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