📚 Unit 5: Chemical Formulae, Equations and Calculations (pp92-122) 9C Homework | 第5单元：化学式、方程式与计算（92-122页）9C作业
This unit brings together chemical formulae, balanced equations, the mole concept and quantitative calculations. You will use relative atomic mass, molar mass, molar gas volume, solution concentration and reacting mass ratios to answer numerical problems. The 9C Homework covering pages 92-122 is designed to test both your understanding of the key relationships and your ability to show clear working, with units, at every stage.
本单元综合了化学式、配平方程式、摩尔概念与定量计算。你将运用相对原子质量、摩尔质量、摩尔气体体积、溶液浓度以及反应质量比来解答计算题。92-122 页的 9C 作业旨在考查你对这些核心关系的理解，以及你在每一步都清晰写出计算过程与单位的能力。
1. Relative Atomic Mass and Relative Formula Mass | 相对原子质量与相对式量
Relative atomic mass (Aᵣ) is the weighted average mass of an atom compared with 1/12 of the mass of a carbon-12 atom. Relative formula mass (Mᵣ) is the sum of the Aᵣ values of all the atoms in a formula unit. Both Aᵣ and Mᵣ have no units because they are relative comparisons.
相对原子质量（Aᵣ）是原子质量与碳-12 原子质量的 1/12 相比所得的平均值。相对式量（Mᵣ）是化学式中所有原子的 Aᵣ 值之和。由于 Aᵣ 与 Mᵣ 都是相对比较值，因此没有单位。
ple, Mᵣ of H₂O = (2 × 1.0) + 16.0 = 18.0, and Mᵣ of CaCO₃ = 40.1 + 12.0 + (3 × 16.0) = 100.1.
Aᵣ 值通常可在周期表中查到。对于化合物，要把各原子的 Aᵣ 相加，并记得乘以下标数字。例如：H₂O 的 Mᵣ = (2 × 1.0) + 16.0 = 18.0；CaCO₃ 的 Mᵣ = 40.1 + 12.0 + (3 × 16.0) = 100.1。
Mᵣ(H₂O) = 18.0 Mᵣ(CaCO₃) = 100.1
For giant structures such as sodium chloride, the formula NaCl only shows the ratio of ions, so we use the term relative formula mass rather than relative molecular mass. This avoids implying that separate molecules exist.
对于氯化钠这类巨分子（离子型）结构，化学式 NaCl 只表示离子的比例，因此我们使用"相对式量"而不是"相对分子质量"，以免暗示存在独立分子。
| Element 元素 | H | C | N | O | Na | S | Cl | Ca | Cu |
|---|---|---|---|---|---|---|---|---|---|
| Aᵣ | 1.0 | 12.0 | 14.0 | 16.0 | 23.0 | 32.1 | 35.5 | 40.1 | 63.5 |
2. The Mole and the Avogadro Constant | 摩尔与阿伏伽德罗常数
particles, which may be atoms, molecules, ions or electrons. This number is called the Avogadro constant.
摩尔是物质的量的国际单位。1 摩尔含有 6.02 × 10²³ 个微粒，这些微粒可以是原子、分子、离子或电子。这个数值称为阿伏伽德罗常数。
number of particles = n × 6.02 × 10²³
of CO₂ contains 6.02 × 10²³ molecules but 3 × 6.02 × 10²³ atoms, because each molecule contains three atoms.
由于这个数字极其庞大，化学家以摩尔而不是逐个计数的方式来计量微粒。答题时要清楚说明数的是哪种微粒：1 mol CO₂ 含 6.02 × 10²³ 个分子，但含 3 × 6.02 × 10²³ 个原子，因为每个分子中有三个原子。
- 0.50 mol of He contains 3.01 × 10²³ atoms. 0.50 mol He 含 3.01 × 10²³ 个原子。
- 2.0 mol of Na⁺ ions contains 1.20 × 10²⁴ ions. 2.0 mol Na⁺ 离子含 1.20 × 10²⁴ 个离子。
- 0.25 mol of H₂O contains 1.51 × 10²³ molecules. 0.25 mol H₂O 含 1.51 × 10²³ 个分子。
3. Molar Mass and Mass-Mole Conversions | 摩尔质量与质量-物质的量换算
The mass of one mole of a substance is its molar mass, with units g mol⁻¹. Numerically, the molar mass equals the relative formula mass. This gives the central equation for mass-mole conversions used throughout the whole unit.
1 摩尔物质的质量称为摩尔质量，单位为 g mol⁻¹。在数值上，摩尔质量等于相对式量。由此得到贯穿整个单元的质量-物质的量换算核心公式。
n = m ÷ M m = n × M
In the equation, n is the amount in moles, m is the mass in grams and M is the molar mass in g mol⁻¹. Always convert mass into grams and tidy up the units before substituting numbers.
式中 n 为物质的量（mol），m 为质量（g），M 为摩尔质量（g mol⁻¹）。代入数值前务必把质量换算成克，并先处理好单位。
Worked example: what is the amount, in moles, of 8.8 g of carbon dioxide? 0.20 mol. Reversing the calculation, 0.25 mol of NaOH has a mass of 0.25 × 40.0 = 10.0 g.
例题：8.8 g 二氧化碳的物质的量是多少？Mᵣ(CO₂) = 12.0 + (2 × 16.0) = 44.0，即 M = 44.0 g mol⁻¹。因此 n = 8.8 ÷ 44.0 = 0.20 mol。反向计算：0.25 mol NaOH 的质量为 0.25 × 40.0 = 10.0 g。
- Check the formula carefully before adding Aᵣ values. 相加 Aᵣ 之前先仔细核对化学式。
- Keep three significant figures unless the question states otherwise. 除题目另有要求，一般保留三位有效数字。
- Show n = m ÷ M, the substitution and the answer with unit. 写出 n = m ÷ M、代入数值与带单位的答案。
4. Percentage Composition by Mass | 质量百分比组成的计算
The percentage by mass of an element in a compound compares the total mass of that element in the formula with the relative formula mass of the whole compound. The equation is simple but the arithmetic must be careful.
化合物中某元素的质量百分比，是把化学式中该元素的总质量与整个化合物的相对式量进行比较。公式很简单，但运算必须仔细。
% element = (total Aᵣ of element ÷ Mᵣ of compound) × 100
For water, the percentage by mass of oxygen is (16.0 ÷ 18.0) × 100 = 88.9 %, and the percentage of hydrogen is (2.0 ÷ 18.0) × 100 = 11.1 %. The two values must add up to 100 %, which is a useful check.
以水为例，氧的质量百分比为 (16.0 ÷ 18.0) × 100 = 88.9 %，氢的质量百分比为 (2.0 ÷ 18.0) × 100 = 11.1 %。两者之和必须为 100 %，这是很好的检验方法。
f nitrogen is 2 × 14.0 = 28.0, so the percentage of nitrogen is (28.0 ÷ 80.0) × 100 = 35.0 %. This is why ammonium nitrate is a valuable fertiliser.
例题：硝酸铵 NH₄NO₃ 的 Mᵣ = 14.0 + (4 × 1.0) + 14.0 + (3 × 16.0) = 80.0；氮的总质量为 2 × 14.0 = 28.0，因此氮的百分比为 (28.0 ÷ 80.0) × 100 = 35.0 %。这正是硝酸铵是优质氮肥的原因。
5. Empirical and Molecular Formulae | 实验式与分子式
The empirical formula is the simplest whole-number ratio of atoms in a compound. The molecular formula gives the actual number of atoms in one molecule and is always a whole-number multiple of the empirical formula.
实验式是化合物中各原子数的最简整数比。分子式表示一个分子中实际的原子数目，它总是实验式的整数倍。
To find an empirical formula: divide the mass or percentage of each element by its Aᵣ, divide all the answers by the smallest value, then multiply if necessary to obtain whole numbers.
求实验式的步骤：把各元素的质量或百分比除以其 Aᵣ；再把所有结果除以其中最小的数值；必要时再乘以一个整数以得到整比。
Example: a compound contains 40.0 % calcium, 12.0 % carbon and 48.0 % ) : (12.0 ÷ 12.0) : (48.0 ÷ 16.0) = 1.00 : 1.00 : 3.00, so the empirical formula is CaCO₃.
例题：某化合物按质量含 40.0 % 钙、12.0 % 碳、48.0 % 氧。Ca : C : O = (40.0 ÷ 40.1) : (12.0 ÷ 12.0) : (48.0 ÷ 16.0) = 1.00 : 1.00 : 3.00，故实验式为 CaCO₃。
n = Mᵣ ÷ empirical formula mass
(mass 30.0) and Mᵣ = 180.0, so n = 180.0 ÷ 30.0 = 6 and the molecular formula is C₆H₁₂O₆.
例题：葡萄糖的实验式为 CH₂O（式量 30.0），Mᵣ = 180.0，因此 n = 180.0 ÷ 30.0 = 6，分子式为 C₆H₁₂O₆。
6. Writing and Balancing Chemical Equations | 化学方程式的书写与配平
A balanced symbol equation obeys the law of conservation of mass: the number of atoms of each element must be the same on both sides. Reactants are written on the left and products on the right, separated by an arrow.
配平的符号方程式遵循质量守恒定律：两边各元素的原子数目必须相等。反应物写在左边，生成物写在右边，中间用箭头隔开。
2H₂ + O₂ → 2H₂O
Balance by inspection, starting with the element that appears in the fewest formulae. Never change a subscript to balance an equation, because that changes the identity of the substance; only change the number in front of a formula.
用观察法配平，从出现在最少化学式中的元素开始。切勿通过改动下标来配平，因为那会改变物质的种类；只能改动化学式前面的系数。
Example: the complete combustion of methane. Carbon appears once on each side, so start with hydrogen: four hydrogen atoms on the left need two water molecules on the right. That gives four oxygen atoms on the right, so two oxygen molecules are needed on the left.
例题：甲烷的完全燃烧。两边各有 1 个碳原子，因此先从氢入手：左边 4 个氢原子需要右边 2 个水分子；这样右边共有 4 个氧原子，于是左边需要 2 个氧分子。
CH₄ + 2O₂ → CO₂ + 2H₂O
The coefficients in a balanced equation give the reacting mole ratio. In the equation above, 1 mol of methane reacts with 2 mol of oxygen to produce 1 mol of carbon dioxide and 2 mol of water. This ratio is the bridge between the mole concept and all reacting mass calculations.
配平方程式中的系数给出反应的摩尔比。上式表示 1 mol 甲烷与 2 mol 氧气反应，生成 1 mol 二氧化碳和 2 mol 水。这个比例正是摩尔概念与所有反应质量计算之间的桥梁。
7. State Symbols and Ionic Equations | 状态符号与离子方程式
State symbols show the physical state of each substance under the reaction conditions and are required in many exam answers.
状态符号表示各物质在反应条件下的物理状态，很多考题的答案中都要求写出。
| Symbol 符号 | Meaning 含义 | Example 示例 |
|---|---|---|
| (s) | solid 固体 | CaCO₃(s) |
| (l) | liquid 液体 | H₂O(l) |
| (g) | gas 气体 | CO₂(g) |
| (aq) | aqueous solution 水溶液 | NaCl(aq) |
. For the precipitation of silver chloride, the full equation is AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq), and the ionic equation is shown below.
离子方程式只表示真正发生变化的离子。两边都没有变化的离子称为旁观离子，需要消去。以氯化银沉淀为例，完整方程式为 AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq
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