AS Chemistry Scheme of Work 4.2: Calculation Questions | AS 化学教学计划4.2：计算题型

📚 AS Chemistry Scheme of Work 4.2: Calculation Questions | AS 化学教学计划4.2：计算题型

Mastering calculation questions is essential for success in AS Chemistry. This article follows the structure of Scheme of Work 4.2, breaking down every major type of quantitative problem you will encounter — from mole concepts and empirical formulae to titrations and yield. The content is presented in linked pairs: each key point is given in English first, immediately followed by its Chinese version to support bilingual learning. Use this as your complete revision companion for calculation-based exam questions.

掌握计算题型是 AS 化学取得成功的关键。本文遵循教学计划 4.2 的结构，全面分解你将遇到的每一类定量问题——从摩尔概念、经验式到滴定和产率。所有内容以配对形式呈现：每个要点先给出英文，紧接着提供中文版本，以支持双语学习。将此作为你复习计算类考题的完整伴侣。

1. Relative Masses and Molar Mass | 相对原子质量与摩尔质量

The relative atomic mass (Aᵣ) of an element is the weighted average mass of its isotopes compared to 1/12th the mass of a carbon‑12 atom. It has no units.

元素的相对原子质量 (Aᵣ) 是其同位素质量的加权平均值，与碳‑12 原子质量的 1/12 相比较。它没有单位。

Relative molecular mass (Mᵣ) is the sum of the relative atomic masses of all atoms in a molecule. For ionic compounds, we use relative formula mass, calculated in the same way.

相对分子质量 (Mᵣ) 是分子中所有原子的相对原子质量之和。对于离子化合物，我们使用相对式量，计算方法相同。

Molar mass (M) is the mass of one mole of a substance, expressed in g mol⁻¹. Numerically it equals the relative atomic or formula mass.

摩尔质量 (M) 是一摩尔物质的质量，单位为 g mol⁻¹。数值上等于相对原子质量或相对式量。

M = m / n (g mol⁻¹)

Example: The molar mass of H₂O = (2 × 1.0) + 16.0 = 18.0 g mol⁻¹.

示例：H₂O 的摩尔质量 = (2 × 1.0) + 16.0 = 18.0 g mol⁻¹。

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 (Nₐ).

一摩尔任何物质恰好含有 6.02 × 10²³ 个微粒（原子、分子、离子或电子）。这个数被称为阿伏伽德罗常数 (Nₐ)。

The amount of substance n (mol) links mass and number of particles:

物质的量 n (mol) 将质量与粒子数联系起来：

n = m / M and N = n × Nₐ

To calculate the number of atoms in 0.50 mol of carbon: N = 0.50 × 6.02×10²³ = 3.01×10²³ atoms.

计算 0.50 mol 碳中的原子数：N = 0.50 × 6.02×10²³ = 3.01×10²³ 个原子。

Always check whether the question asks for atoms, molecules or ions — a diatomic molecule like O₂ contains two atoms per molecule.

务必核对题目要求的是原子、分子还是离子——像 O₂ 这样的双原子分子，每个分子含有两个原子。

3. Empirical and Molecular Formulae | 经验式与分子式

The empirical formula gives the simplest whole‑number ratio of atoms in a compound. The molecular formula shows the actual number of each atom in a molecule.

经验式给出化合物中原子的最简整数比。分子式表示分子中每种原子的实际个数。

To find an empirical formula:

求经验式的步骤：

Step 1: Divide the mass (or percentage) of each element by its Aᵣ to obtain moles.
步骤 1：将各元素的质量（或百分比）除以各自的 Aᵣ，得到摩尔数。
Step 2: Divide all mole values by the smallest number of moles.
步骤 2：将所有摩尔值除以最小的摩尔数。
Step 3: If necessary, multiply to get whole numbers.
步骤 3：如有必要，乘以整数得到最简比。

Example: A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Moles: C = 40.0/12.0 = 3.33; H = 6.7/1.0 = 6.7; O = 53.3/16.0 = 3.33. Divide by 3.33 gives ratio 1 : 2 : 1, so empirical formula is CH₂O.

示例：一种化合物含 40.0% 碳、6.7% 氢和 53.3% 氧（质量分数）。摩尔数：C = 40.0/12.0 = 3.33；H = 6.7/1.0 = 6.7；O = 53.3/16.0 = 3.33。除以 3.33 得到比例 1 : 2 : 1，因此经验式为 CH₂O。

To determine the molecular formula, divide the molar mass by the empirical formula mass: molecular formula = (empirical formula)ₙ where n = Mᵣ(compound) / Mᵣ(empirical).

确定分子式时，用摩尔质量除以经验式质量：分子式 = (经验式)ₙ，其中 n = 化合物的 Mᵣ / 经验式的 Mᵣ。

4. Reacting Masses and Stoichiometry | 反应质量与化学计量

Chemical equations give the mole ratio of reactants and products. Use these ratios to calculate the masses of substances involved.

化学方程式给出了反应物和生成物的摩尔比。利用这些比值计算所涉及物质的质量。

mass → moles → mole ratio → moles of target → mass

For the reaction 2Mg + O₂ → 2MgO, how many grams of MgO are produced from 4.86 g of Mg?

对于反应 2Mg + O₂ → 2MgO，4.86 g Mg 能生成多少克 MgO？

Moles of Mg = 4.86 g / 24.3 g mol⁻¹ = 0.200 mol
Mole ratio Mg : MgO = 2 : 2 = 1 : 1, so moles of MgO = 0.200 mol
Mass of MgO = 0.200 mol × (24.3 + 16.0) g mol⁻¹ = 0.200 × 40.3 = 8.06 g

中文计算过程：Mg 的物质的量 = 4.86 / 24.3 = 0.200 mol，摩尔比 1:1，所以 MgO 物质的量 = 0.200 mol，质量 = 0.200 × 40.3 = 8.06 g。

In more complex cases, you may need to identify the limiting reactant first (see Section 9).

在更复杂的情况下，你可能需要先确定限量反应物（见第 9 节）。

5. Gas Volume Calculations at RTP | 室温常压下的气体体积计算

At room temperature and pressure (RTP, typically 25 °C and 1 atm), one mole of any gas occupies 24.0 dm³ (or 24 000 cm³). This is the molar gas volume (Vₘ).

在室温常压 (RTP，通常为 25 °C 和 1 atm) 下，一摩尔任何气体占据 24.0 dm³（或 24 000 cm³）。这就是气体摩尔体积 (Vₘ)。

n(gas) = V / 24.0 (dm³) or n = V / 24 000 (cm³)

If 0.500 mol of CO₂ is produced, the volume at RTP is 0.500 × 24.0 = 12.0 dm³.

若生成了 0.500 mol CO₂，在 RTP 下的体积为 0.500 × 24.0 = 12.0 dm³。

Many exam questions combine reacting masses with gas volumes: first calculate moles from the mass of a solid reactant, then use the mole ratio to find moles of gas, and finally convert to volume.

许多考题将反应质量与气体体积相结合：先由固体反应物的质量计算摩尔数，再利用摩尔比求得气体的物质的量，最后转换为体积。

Remember to check the units — volumes may be given in cm³ and must be converted to dm³ (÷1000) before using the 24.0 dm³ mol⁻¹ value, or use the 24 000 cm³ mol⁻¹ constant directly.

注意单位——体积可能以 cm³ 给出，必须转换为 dm³（除以 1000）后才能使用 24.0 dm³ mol⁻¹，或者直接使用常数 24 000 cm³ mol⁻¹。

6. Solution Concentrations and Molarity | 溶液浓度与摩尔浓度

Concentration measures the amount of solute dissolved in a given volume of solution. The most common unit is mol dm⁻³ (molarity, often symbol c).

浓度衡量的是溶解在一定体积溶液中的溶质的量。最常用的单位是 mol dm⁻³（摩尔浓度，常用符号 c）。

c = n / V (mol dm⁻³)

When mass of solute is given, first calculate n = m / M, then apply c = n / V (in dm³). If volume is in cm³, divide by 1000.

当给出溶质质量时，先计算 n = m / M，再代入 c = n / V（体积以 dm³ 为单位）。若体积为 cm³，需除以 1000。

A solution labelled ‘0.200 mol dm⁻³ NaOH’ means each dm³ of solution contains 0.200 mol of NaOH, i.e. 0.200 × 40.0 = 8.00 g per dm³.

标称“0.200 mol dm⁻³ NaOH”的溶液表示每 dm³ 溶液含有 0.200 mol NaOH，即每 dm³ 含 0.200 × 40.0 = 8.00 g。

Dilution calculations use the principle that moles of solute remain constant: c₁V₁ = c₂V₂. Always use consistent volume units.

稀释计算基于溶质物质的量不变的原则：c₁V₁ = c₂V₂。始终使用一致的体积单位。

7. Titration Calculations | 滴定计算

Titration is used to determine an unknown concentration. The key is to write a balanced equation and identify the mole ratio between the reactants.

滴定用于测定未知浓度。关键是要写出配平的化学方程式，并确定反应物之间的摩尔比。

Typical procedure:

Record the volumes used (titre).
记录所用的体积（滴定值）。
Calculate moles of the known solution using n = cV.
利用 n = cV 计算已知溶液中溶质的物质的量。
Use mole ratio to find moles of the unknown.
利用摩尔比求得未知物的物质的量。
Finally calculate its concentration: c = n / V (using the pipette volume).
最后计算其浓度：c = n / V（使用移液管体积）。

Example: 25.0 cm³ of HCl is neutralised by 22.50 cm³ of 0.100 mol dm⁻³ NaOH. The equation: HCl + NaOH → NaCl + H₂O. Ratio 1:1. Moles NaOH = 0.100 × (22.50/1000) = 0.00225 mol. So moles HCl = 0.00225 mol. Concentration of HCl = 0.00225 / (25.0/1000) = 0.0900 mol dm⁻³.

示例：25.0 cm³ HCl 被 22.50 cm³ 0.100 mol dm⁻³ NaOH 中和。方程式：HCl + NaOH → NaCl + H₂O，摩尔比 1:1。NaOH 物质的量 = 0.100 × 0.02250 = 0.00225 mol。因此 HCl 物质的量 = 0.00225 mol。HCl 浓度 = 0.00225 / 0.0250 = 0.0900 mol dm⁻³。

For redox titrations (e.g. manganate(VII) with iron(II)), carefully deduce the mole ratio from the half‑equations.

对于氧化还原滴定（如高锰酸根与铁(II)），要通过半反应方程式细心推导摩尔比。

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

Percentage yield compares the actual mass of product obtained to the theoretical maximum mass predicted by stoichiometry.

产率将实际获得的产品质量与通过化学计量预测的理论最大质量进行比较。

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

Yields are often less than 100% due to incomplete reactions, side reactions, or losses during purification.

由于反应不完全、副反应或提纯过程中的损失，产率通常低于 100%。

Atom economy measures the efficiency of a reaction in converting starting materials into useful product:

原子经济性衡量反应将起始原料转化为有用产品的效率：

% atom economy = (Mᵣ of desired product / sum of Mᵣ of all reactants) × 100%

A higher atom economy means less waste and is a key principle of green chemistry.

较高的原子经济性意味着更少的废物，这是绿色化学的一个关键原则。

Example: The reaction C₂H₅OH + NaBr + H₂SO₄ → C₂H₅Br + NaHSO₄ + H₂O. Desired product C₂H₅Br has Mᵣ = 108.9. Sum of reactant Mᵣ: 46.0 + 102.9 + 98.1 = 247.0. Atom economy = (108.9 / 247.0) × 100% = 44.1%.

示例：反应 C₂H₅OH + NaBr + H₂SO₄ → C₂H₅Br + NaHSO₄ + H₂O。目标产物 C₂H₅Br 的 Mᵣ = 108.9，反应物的 Mᵣ 总和 = 46.0 + 102.9 + 98.1 = 247.0。原子经济性 = (108.9 / 247.0) × 100% = 44.1%。

9. Limiting and Excess Reactants | 限量与过量反应物

When two or more reactants are mixed in non‑stoichiometric amounts, the reactant that is completely consumed first is the limiting reactant. It determines the maximum amount of product formed.

当两种或多种反应物以非化学计量比混合时，首先被完全消耗的反应物称为限量反应物。它决定了生成物的最大量。

To identify the limiting reactant, calculate the number of moles of each reactant and divide by their respective coefficients in the balanced equation. The smallest value indicates the limiting species.

要找出限量反应物，计算每种反应物的物质的量，除以其在配平方程式中的系数。最小值对应的就是限量反应物。

Example: 10.0 g of Mg (Mᵣ = 24.3) react with 10.0 g of O₂ (Mᵣ = 32.0) to form MgO. Moles: Mg = 10.0/24.3 = 0.412 mol; O₂ = 10.0/32.0 = 0.313 mol. Equation: 2Mg + O₂ → 2MgO. Divide Mg by 2: 0.412/2 = 0.206; O₂ by 1: 0.313. The smaller value is 0.206, so Mg is the limiting reactant.

示例：10.0 g Mg (Mᵣ = 24.3) 与 10.0 g O₂ (Mᵣ = 32.0) 反应生成 MgO。物质的量：Mg = 0.412 mol；O₂ = 0.313 mol。方程式：2Mg + O₂ → 2MgO。Mg 除以 2 得 0.206，O₂ 除以 1 得 0.313。较小值是 0.206，因此 Mg 是限量反应物。

The mass of MgO produced is based on Mg: moles MgO = 0.412 mol, mass = 0.412 × 40.3 = 16.6 g. O₂ is in excess.

生成 MgO 的质量基于 Mg：MgO 物质的量 = 0.412 mol，质量 = 0.412 × 40.3 = 16.6 g。O₂ 过量。

10. Common Errors and Exam Strategies | 常见错误与应试策略

Many marks are lost on calculation questions through careless unit conversions. Always convert cm³ to dm³ by dividing by 1000 before using concentration or molar volume formulas.

计算题中许多失分是由于粗心的单位换算造成的。在应用浓度或摩尔体积公式前，务必先将 cm³ 除以 1000 转换为 dm³。

Ignoring the mole ratio from the balanced equation is another frequent mistake. Never assume a 1:1 ratio unless the equation clearly shows it.

忽略配平方程式中的摩尔比是另一个常见错误。除非化学方程式明确显示 1:1 的比例，否则绝不要默认这个比值。

Rounding errors accumulate if you round intermediate answers too early. Keep all figures in your calculator and round only the final answer to the appropriate number of significant figures (usually 3 s.f.).

如果过早对中间结果进行舍入，舍入误差会累积。将计算器中的所有数字保留，只将最终结果舍入到适当的小数位数（通常 3 位有效数字）。

For multi‑step problems, organise your working clearly: write down the balanced equation, list the given data, convert to moles, apply the mole ratio, and finally convert to the required quantity (mass, volume, concentration).

对于多步问题，清晰地组织你的解题过程：写出配平的方程式，列出已知数据，转换为物质的量，应用摩尔比，最后转换为所需物理量（质量、体积、浓度）。

Practise past‑paper questions under timed conditions. Calculation questions often follow predictable patterns, so familiarity with the method will boost your speed and confidence.

在限时条件下练习历年真题。计算题通常遵循可预测的模式，因此熟悉解题方法会提高你的速度和信心。

11. Worked Example: Multi‑Step Calculation | 例题：多步计算

A 2.50 g sample of impure calcium carbonate reacts with excess hydrochloric acid. The volume of carbon dioxide collected at RTP is 480 cm³. Calculate the percentage purity of the sample.

一块 2.50 g 的不纯碳酸钙样品与过量盐酸反应。在室温常压下收集到 480 cm³ 二氧化碳。计算样品的纯度。

Step 1: Write the equation: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. Mole ratio CaCO₃ : CO₂ = 1 : 1.

步骤 1：写出方程式：CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂。CaCO₃ 与 CO₂ 的摩尔比 = 1 : 1。

Step 2: Moles of CO₂ = volume / 24 000 = 480 / 24 000 = 0.0200 mol.

步骤 2：CO₂ 的物质的量 = 体积 / 24 000 = 480 / 24 000 = 0.0200 mol。

Step 3: Moles of CaCO₃ that reacted = 0.0200 mol (1:1 ratio).

步骤 3：反应的 CaCO₃ 物质的量 = 0.0200 mol（1:1 比率）。

Step 4: Mass of pure CaCO₃ = n × M = 0.0200 × 100.1 = 2.002 g.

步骤 4：纯 CaCO₃ 的质量 = 0.0200 × 100.1 = 2.002 g。

Step 5: Percentage purity = (mass of pure / mass of sample) × 100 = (2.002 / 2.50) × 100 = 80.1%.

步骤 5：纯度百分比 = (纯物质质量 / 样品质量) × 100 = (2.002 / 2.50) × 100 = 80.1%。

This example combines gas volumes, stoichiometry and purity — a classic AS calculation.

这个例题综合了气体体积、化学计量和纯度——一道典型的 AS 计算题。

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AS Chemistry Scheme of Work 4.2: Calculation Questions | AS 化学教学计划4.2：计算题型