Mastering Calculation Questions from Complete Chemistry for Cambridge Secondary 1 Workbook | 掌握《完全剑桥初中化学1练习册》计算题型

📚 Mastering Calculation Questions from Complete Chemistry for Cambridge Secondary 1 Workbook | 掌握《完全剑桥初中化学1练习册》计算题型

The calculation questions in the Complete Chemistry for Cambridge Secondary 1 Workbook are designed to build your confidence in handling quantitative chemical problems. Mastering these skills – from relative atomic mass to concentration in g/dm³ – is essential for Cambridge Checkpoint Science and forms the backbone of IGCSE Chemistry.

《完全剑桥初中化学1练习册》中的计算题型旨在帮助你建立处理化学定量问题的信心。从相对原子质量到以 g/dm³ 为单位的浓度计算，掌握这些技能不仅是剑桥 Checkpoint 科学考试的关键，也为 IGCSE 化学打下坚实的基底。

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

Relative atomic mass (Aᵣ) is the average mass of an atom of an element compared to 1/12 the mass of a carbon-12 atom. It has no units. Relative formula mass (Mᵣ) is the sum of the Aᵣ values of all atoms in the chemical formula.

相对原子质量 (Aᵣ) 是元素一个原子的平均质量与碳-12原子质量的 1/12 比较得出的数值，无单位。化学式量 (Mᵣ) 是化学式中所有原子的 Aᵣ 值之和。

Example: Calculate Mᵣ of water (H₂O). Aᵣ of H = 1, O = 16. Mᵣ = (2 × 1) + 16 = 18.

示例：计算水 (H₂O) 的 Mᵣ。 Aᵣ(H) = 1，Aᵣ(O) = 16。 Mᵣ = (2 × 1) + 16 = 18。

Element	Symbol	Aᵣ
Hydrogen	H	1
Carbon	C	12
Oxygen	O	16
Sodium	Na	23
Sulfur	S	32

Common mistake: Forgetting to multiply the Aᵣ by the number of atoms shown in subscripts. In Ca(OH)₂, there are two OH groups, so O and H must each be counted twice.

常见错误：忘记将 Aᵣ 乘以下标所示的原子个数。在 Ca(OH)₂ 中，有两个 OH 基团，因此 O 和 H 各须计数两次。

---

2. Calculating Percentage Composition by Mass | 计算质量百分比组成

y mass of an element in a compound = (total mass of the element in the formula / Mᵣ) × 100%.

化合物中某元素的质量百分比 = (该元素在化学式中的总质量 / Mᵣ) × 100%。

ᵣ(N)=14, H=1, O=16. Mᵣ = (2×14) + (4×1) + (3×16) = 28+4+48=80. Mass of nitrogen = 2×14=28. %N = (28/80)×100% = 35%.

示例：求硝酸铵 NH₄NO₃ 中氮的质量百分比。Aᵣ(N)=14, H=1, O=16。 Mᵣ = (2×14)+(4×1)+(3×16)=28+4+48=80。氮的总质量 = 2×14=28。%N = (28/80)×100% = 35%。

Workbook questions often ask you to compare fertilisers by calculating the percentage of nutrient element they contain.

练习册中常要求通过计算营养元素的百分比来比较化肥。

---

3. Determining Empirical Formula from Mass Data | 从质量数据确定实验式

The empirical formula is the simplest whole-number ratio of atoms in a compound. It can be found from experimental mass or percentage data.

实验式（最简式）是化合物中原子最简整数比，可通过实验质量或质量百分比求得。

Steps: (1) Convert mass (or %) to moles by dividing by Aᵣ; (2) Divide all mole values by the smallest; (3) Multiply if necessary to get whole numbers.

步骤：(1) 用质量（或百分比）除以 Aᵣ 得到摩尔数；(2) 用最小值除所有摩尔值；(3) 必要时乘以整数得到最简单整数比。

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% .33, H=6.7/1=6.7, O=53.3/16=3.33. Divide by 3.33: C=1, H≈2, O=1. Empirical formula = CH₂O.

某化合物含碳 40.0%、氢 6.7%、氧 53.3%。假设 100 g 样品：C 的摩尔数 40.0/12=3.33，H 6.7/1=6.7，O 53.3/16=3.33。均除以 3.33 得 C=1, H≈2, O=1。实验式为 CH₂O。

Note: If you are given the relative molecular mass, you can then find the molecular formula by comparing Mᵣ with empirical formula mass.

注意：若已知相对分子质量，便可比较 Mᵣ 与实验式量，求出分子式。

---

广告 / Sponsored

4. Reacting Mass Calculations Using Chemical Equations | 利用化学方程式计算反应质量

Chemical equations tell us the ratio in which moles react. We use the balanced equation and Mᵣ values to calculate reacting masses directly.

化学方程式揭示物质反应的摩尔比。我们依据配平的方程式和 Mᵣ 值直接计算反应质量。

O₂ → 2MgO. How many grams of magnesium oxide form when 12 g of magnesium burns completely? n, 2 mol Mg (2×24=48 g) produce 2 mol MgO (2×40=80 g). So 12 g Mg → (12/48)×80 = 20 g MgO.

示例：2Mg + O₂ → 2MgO。 12 g 镁完全燃烧生成多少克氧化镁？ Aᵣ(Mg)=24, O=16，Mᵣ(MgO)=40。 根据方程式，2 mol Mg (48 g) 生成 2 mol MgO (80 g)。 故 12 g Mg 生成 (12/48)×80 = 20 g MgO。

Mass product = (given mass reactant / Mᵣ reactant scaled) × Mᵣ product scaled

生成物质量 = (给定反应物质量 / 反应物的缩放质量) × 生成物的缩放质量

Always check that the equation is balanced before you start.

计算前务必确认方程式已配平。

---

5. Identifying the Limiting Reactant | 识别限量反应物

When two reactants are given, one will often be used up first – the limiting reactant – and it determines the amount of product formed.

当提供两种反应物的质量时，其中一种常会先消耗完——即限量反应物——它决定了产物的生成量。

1 mol O₂, so both are exactly used. If we had 4 g H₂ and 16 g O₂, O₂ = 0.5 mol, H₂ = 2 mol; O₂ is limiting. Water produced = 2×0.5=1 mol = 18 g.

示例：4 g H₂ 与 32 g O₂ 反应生成水。 2H₂ + O₂ → 2H₂O。 H₂ 的摩尔数 = 4/2=2 mol, O₂ 的摩尔数 = 32/32=1 mol。 方程式要求 2 mol H₂ 与 1 mol O₂ 恰好完全反应。 若是 4 g H₂ 与 16 g O₂，O₂=0.5 mol, H₂=2 mol，O₂ 是限量反应物，生成水 2×0.5=1 mol，即 18 g。

The limiting reactant can be identified by converting both given masses to moles, then comparing to the mole ratio from the equation.

辨别限量反应物的方法是：将两种物质的质量换算为摩尔数，再与方程式中的摩尔比进行比较。

---

6. Concentration Calculations in g/dm³ | 以 g/dm³ 为单位的浓度计算

ation (g/dm³) = mass of solute (g) / volume of solution (dm³). Remember: 1 dm³ = 1000 cm³.

浓度 (g/dm³) = 溶质质量 (g) / 溶液体积 (dm³)。 注意：1 dm³ = 1000 cm³。

e 200 cm³ of solution. Volume in dm³ = 200/1000=0.2 dm³. Concentration = 5/0.2 = 25 g/dm³.

示例：将 5 g 食盐溶于水配成 200 cm³ 溶液。体积 = 200/1000=0.2 dm³。 浓度 = 5/0.2 = 25 g/dm³。

Many workbook errors come from forgetting to convert cm³ to dm³. Always check the units given in the question.

练习册中的许多错误都源于忘记将 cm³ 换算为 dm³。 务必检查题目中的单位。

c (g/dm³) = m (g) / V (dm³)

浓度 c (g/dm³) = 溶质质量 m (g) / 溶液体积 V (dm³)

---

7. Making Solutions and Dilution Calculations | 配制溶液与稀释计算

n of a desired concentration, the mass of solute needed = concentration × volume (in dm³).

配制指定浓度的溶液时，所需溶质质量 = 浓度 × 体积 (dm³)。

Example: How many grams of sugar are needed to make 250 cm

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）