GCSE AQA Chemistry: Calculation Masterclass | GCSE AQA 化学：计算题专项训练

📚 GCSE AQA Chemistry: Calculation Masterclass | GCSE AQA 化学：计算题专项训练

Quantitative chemistry is a core part of the AQA GCSE Chemistry specification. It tests your ability to use mathematical relationships to connect the masses of substances, the amounts in moles, and the volumes of gases and solutions. This masterclass will walk you through every type of calculation question you might face, from simple relative formula mass to more challenging titrations and atom economy. Each section includes worked examples written in clear steps, with paired English–Chinese explanations to reinforce understanding. By the end, you will have a complete toolkit to tackle any numerical problem confidently.

定量化学是 AQA GCSE 化学大纲的核心部分。它考查你运用数学关系将物质的质量、物质的量（摩尔）以及气体和溶液的体积联系起来的能力。本次专项训练将带你逐一攻克可能遇到的所有计算题型，从简单的相对式量到更具挑战性的滴定和原子经济性。每个部分都包含分步详细解答的例题，并配有中英对照解释以加深理解。学习结束后，你将拥有一整套工具，从容应对任何计算问题。

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1. Relative Atomic Mass and Relative Formula Mass | 相对原子质量与相对式量

Every element has a relative atomic mass (A_r), which is the average mass of its atoms compared to 1/12 of the mass of a carbon-12 atom. You can find A_r values on the Periodic Table; they are not always whole numbers because of isotopes. When atoms combine, we calculate the relative formula mass (M_r) by adding up the A_r values of all atoms in the formula.

每种元素都有一个相对原子质量（A_r），它是该元素原子的平均质量与一个碳-12原子质量的1/12的比值。你可以在周期表中找到A_r值；由于同位素的存在，这些数值不一定是整数。当原子结合时，我们将化学式中所有原子的A_r相加，计算出相对式量（M_r）。

For example, to find the M_r of sulfuric acid, H₂SO₄:
H: 2 × 1 = 2
S: 1 × 32 = 32
O: 4 × 16 = 64
Total M_r = 2 + 32 + 64 = 98.

例如，计算硫酸 H₂SO₄ 的 M_r：
H：2 × 1 = 2
S：1 × 32 = 32
O：4 × 16 = 64
总 M_r = 2 + 32 + 64 = 98。

Be careful with brackets, such as in Ca(OH)₂. Each bracket multiplies everything inside: one Ca (40), two O (2 × 16 = 32), and two H (2 × 1 = 2). So M_r = 40 + 32 + 2 = 74. Always work systematically to avoid missing any atom.

注意括号，例如 Ca(OH)₂。括号会将内部的原子数都乘以括号外的下标：一个 Ca (40)，两个 O (2 × 16 = 32)，两个 H (2 × 1 = 2)。因此 M_r = 40 + 32 + 2 = 74。要系统计算，避免遗漏任何原子。

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2. The Mole and Molar Mass | 摩尔与摩尔质量

One mole of any substance contains exactly 6.02 × 10²³ particles (Avogadro’s constant). The mass of one mole of a substance is its molar mass, which has the same numerical value as the relative formula mass M_r but is expressed in grams per mole (g mol⁻¹). For instance, the M_r of water is 18, so one mole of water has a mass of 18 g.

任何物质的一摩尔都恰好包含 6.02 × 10²³ 个微粒（阿伏伽德罗常数）。一摩尔物质的质量就是其摩尔质量，数值上与相对式量 M_r 相同，但单位是克每摩尔（g mol⁻¹）。例如，水的 M_r 是 18，因此一摩尔水的质量为 18 g。

The central equation linking moles (n), mass (m), and molar mass (M) is:

n = m / M or mass = moles × M_r

联系摩尔 (n)、质量 (m) 和摩尔质量 (M) 的核心方程为：

n = m / M 或 质量 = 摩尔 × M_r

This relationship is the key to almost every quantitative chemistry calculation. Memorise it and be prepared to rearrange it. Always check that your mass is in grams and your M_r is in g mol⁻¹.

这个关系式是几乎所有定量化学计算的钥匙。务必记住并能够熟练变形。始终确保质量单位为克，M_r 单位为 g mol⁻¹。

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3. Mass–Mole Conversions | 质量与摩尔的换算

Converting between mass and moles is the first step in many multi-step problems. To go from mass to moles, divide the mass by M_r. To go from moles to mass, multiply the number of moles by M_r.

质量与摩尔之间的换算是许多多步计算的第一步。从质量换为摩尔，用质量除以 M_r。从摩尔换为质量，用摩尔数乘以 M_r。

Worked example: How many moles are present in 5.00 g of calcium carbonate, CaCO₃? (A_r: Ca = 40, C = 12, O = 16)
M_r of CaCO₃ = 40 + 12 + (3 × 16) = 100
Moles = mass / M_r = 5.00 / 100 = 0.0500 mol

例题：5.00 g 碳酸钙 CaCO₃ 中含有多少摩尔？（A_r：Ca = 40，C = 12，O = 16）
CaCO₃ 的 M_r = 40 + 12 + (3 × 16) = 100
摩尔 = 质量 / M_r = 5.00 / 100 = 0.0500 mol

When dealing with large or small numbers, use standard form and give your answer to an appropriate number of significant figures, usually 3 significant figures unless the question states otherwise.

当处理较大或较小的数值时，使用科学计数法，并将答案保留适当的有效数字，通常为3位有效数字，除非题目另有要求。

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4. Reacting Mass Calculations from Equations | 根据化学方程式的反应质量计算

A balanced chemical equation tells you the mole ratio in which substances react and are produced. To calculate the mass of a product or reactant, follow these steps:

配平的化学方程式告诉你物质反应和生成的摩尔比例。要计算产物或反应物的质量，请遵循以下步骤：

Step 1: Write the balanced equation.
Step 2: Calculate the M_r of the substance whose mass is given and the substance whose mass is asked for.
Step 3: Convert the given mass to moles using n = m / M_r.
Step 4: Use the mole ratio from the equation to find the moles of the target substance.
Step 5: Convert moles of the target substance back to mass using mass = n × M_r.

步骤 1：写出配平的化学方程式。
步骤 2：计算已知质量物质和所求质量物质的 M_r。
步骤 3：用 n = m / M_r 将已知质量换算为摩尔。
步骤 4：根据方程式中的摩尔比，计算目标物质的摩尔数。
步骤 5：用 质量 = n × M_r 将目标物质的摩尔数换算回质量。

Example: What mass of magnesium oxide (MgO) is produced when 6.00 g of magnesium burns completely in oxygen? (A_r: Mg = 24, O = 16)
Equation: 2Mg + O₂ → 2MgO
M_r of Mg = 24, M_r of MgO = 24 + 16 = 40
Moles of Mg = 6.00 / 24 = 0.250 mol
Mole ratio Mg : MgO = 2 : 2 = 1 : 1, so moles of MgO = 0.250 mol
Mass of MgO = 0.250 × 40 = 10.0 g

例题：6.00 g 镁在氧气中完全燃烧，生成多少克氧化镁 (MgO)？（A_r：Mg = 24，O = 16）
方程式：2Mg + O₂ → 2MgO
Mg 的 M_r = 24，MgO 的 M_r = 24 + 16 = 40
Mg 的摩尔数 = 6.00 / 24 = 0.250 mol
摩尔比 Mg : MgO = 2 : 2 = 1 : 1，所以 MgO 的摩尔数 = 0.250 mol
MgO 的质量 = 0.250 × 40 = 10.0 g

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5. Limiting Reactants | 限制反应物

In many reactions, one reactant is used up first, stopping the reaction and leaving the other reactant in excess. The limiting reactant determines the maximum amount of product that can form. To identify it, calculate the moles of each reactant, then divide by its stoichiometric coefficient. The smallest value indicates the limiting reactant.

在许多反应中，一种反应物会先消耗完，使反应停止，另一种则过量剩余。限制反应物决定了能生成的产物的最大量。要找出限制反应物，先计算每种反应物的摩尔数，再除以其化学计量系数，最小值即为限制反应物。

Example: 3.0 g of hydrogen (H₂) reacts with 16.0 g of oxygen (O₂) to make water. Identify the limiting reactant. (A_r: H = 1, O = 16)
Equation: 2H₂ + O₂ → 2H₂O
Moles of H₂ = 3.0 / 2 = 1.5 mol; divide by coefficient 2 → 0.75
Moles of O₂ = 16.0 / 32 = 0.50 mol; divide by coefficient 1 → 0.50
0.50 < 0.75, so O₂ is the limiting reactant.

例题：3.0 g 氢气 (H₂) 与 16.0 g 氧气 (O₂) 反应生成水。找出限制反应物。（A_r：H = 1，O = 16）
方程式：2H₂ + O₂ → 2H₂O
H₂ 的摩尔 = 3.0 / 2 = 1.5 mol；除以系数 2 → 0.75
O₂ 的摩尔 = 16.0 / 32 = 0.50 mol；除以系数 1 → 0.50
0.50 < 0.75，因此 O₂ 是限制反应物。

Once you know the limiting reactant, use its moles and the mole ratio to find the moles and mass of any product. The excess reactant is left over; you can calculate how much remains by finding the moles that actually react and subtracting.

一旦知道限制反应物，就可以用它的摩尔数和摩尔比求出任何产物的摩尔数和质量。过量反应物会有剩余；你可以通过计算实际反应的摩尔数，然后减去初始摩尔数，求出剩余量。

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6. Concentration of Solutions | 溶液的浓度计算

Concentration can be expressed in two common ways: grams per cubic decimetre (g dm⁻³) or moles per cubic decimetre (mol dm⁻³). Both are linked by the M_r of the solute.

浓度常用两种方式表示：克每立方分米 (g dm⁻³) 或摩尔每立方分米 (mol dm⁻³)。两者通过溶质的 M_r 联系起来。

The key formulas are:

concentration (g dm⁻³) = mass of solute (g) / volume of solution (dm³)

concentration (mol dm⁻³) = number of moles / volume (dm³)

关键公式：

浓度 (g dm⁻³) = 溶质质量 (g) / 溶液体积 (dm³)

浓度 (mol dm⁻³) = 摩尔数 / 体积 (dm³)

Remember to convert cm³ to dm³ by dividing by 1000 (1 dm³ = 1000 cm³). Often you will need to calculate the M_r first to switch between the two concentration units.

记住要将 cm³ 转换为 dm³，除以 1000 (1 dm³ = 1000 cm³)。通常你需要先计算 M_r，才能在两种浓度单位之间转换。

Example: 5.85 g of sodium chloride (NaCl) is dissolved in water to make 250 cm³ of solution. Calculate the concentration in both g dm⁻³ and mol dm⁻³. (A_r: Na = 23, Cl = 35.5)
Volume in dm³ = 250 / 1000 = 0.250 dm³
Concentration (g dm⁻³) = 5.85 / 0.250 = 23.4 g dm⁻³
M_r of NaCl = 23 + 35.5 = 58.5
Moles = 5.85 / 58.5 = 0.100 mol
Concentration (mol dm⁻³) = 0.100 / 0.250 = 0.400 mol dm⁻³

例题：将 5.85 g 氯化钠 (NaCl) 溶于水中，配成 250 cm³ 溶液。计算其以 g dm⁻³ 和 mol dm⁻³ 表示的浓度。(A_r: Na = 23, Cl = 35.5)
体积 (dm³) = 250 / 1000 = 0.250 dm³
浓度 (g dm⁻³) = 5.85 / 0.250 = 23.4 g dm⁻³
NaCl 的 M_r = 23 + 35.5 = 58.5
摩尔数 = 5.85 / 58.5 = 0.100 mol
浓度 (mol dm⁻³) = 0.100 / 0.250 = 0.400 mol dm⁻³

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7. Titration Calculations | 滴定计算

Titration problems require you to find the unknown concentration of an acid or alkali using a neutralisation reaction. The general method is to record the volumes used, calculate the moles of the known solution, use the mole ratio from the balanced equation, and then find the unknown concentration.

滴定计算要求你利用中和反应求算未知酸或碱的浓度。一般方法是记录所用体积，计算已知溶液的摩尔数，利用配平方程式中的摩尔比，再求出未知浓度。

Follow these steps:
1. Write the balanced equation.
2. Calculate moles of the solution with known concentration and volume (moles = concentration × volume in dm³).
3. Use the mole ratio to determine moles of the other reactant.
4. Find its concentration using concentration = moles / volume (in dm³).

按照以下步骤：
1. 写出配平方程式。
2. 计算已知浓度和体积的溶液的摩尔数 (摩尔 = 浓度 × 体积 dm³)。
3. 利用摩尔比确定另一反应物的摩尔数。
4. 用浓度 = 摩尔 / 体积 (dm³) 求出其浓度。

Example: 25.0 cm³ of sulfuric acid (H₂SO₄) required 30.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide (NaOH) for complete neutralisation. Find the concentration of the acid.
Equation: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
Moles NaOH = 0.100 × (30.0 / 1000) = 0.00300 mol
Mole ratio NaOH : H₂SO₄ = 2 : 1, so moles H₂SO₄ = 0.00300 / 2 = 0.00150 mol
Volume of acid in dm³ = 25.0 / 1000 = 0.0250 dm³
Concentration H₂SO₄ = 0.00150 / 0.0250 = 0.0600 mol dm⁻³

例题：25.0 cm³ 硫酸 (H₂SO₄) 需要 30.0 cm³ 的 0.100 mol dm⁻³ 氢氧化钠 (NaOH) 才能完全中和。求硫酸的浓度。
方程式：H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
NaOH 的摩尔数 = 0.100 × (30.0 / 1000) = 0.00300 mol
摩尔比 NaOH : H₂SO₄ = 2 : 1，因此 H₂SO₄ 的摩尔数 = 0.00300 / 2 = 0.00150 mol
酸的体积 (dm³) = 25.0 / 1000 = 0.0250 dm³
H₂SO₄ 浓度 = 0.00150 / 0.0250 = 0.0600 mol dm⁻³

Always pay attention to units and significant figures. If a titration is repeated, you may need to calculate the mean titre, discarding any anomalous results.

始终注意单位和有效数字。如果滴定重复进行，你可能需要计算平均滴定体积，并舍弃任何异常值。

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8. Gas Volume Calculations | 气体体积计算

At room temperature and pressure (RTP), one mole of any gas occupies 24 dm³ (or 24 000 cm³). This simplifies calculations when you know the moles of a gaseous reactant or product. Use:

volume of gas (dm³) = moles of gas × 24

在室温常压 (RTP) 下，一摩尔任何气体占据 24 dm³（或 24 000 cm³）。当你知道气态反应物或产物的摩尔数时，这能简化计算。使用：

气体体积 (dm³) = 气体摩尔 × 24

If the question involves a reacting mass and a gas, first find the moles of the known substance, apply the mole ratio, then convert to gas volume.

如果题目涉及反应质量和气体，首先求出已知物质的摩尔数，应用摩尔比，再转换为气体体积。

Example: What volume of carbon dioxide (at RTP) is produced when 10.0 g of calcium carbonate decomposes fully?
Equation: CaCO₃ → CaO + CO₂
M_r of CaCO₃ = 100
Moles CaCO₃ = 10.0 / 100 = 0.100 mol
Mole ratio 1:1, so moles CO₂ = 0.100 mol
Volume CO₂ = 0.100 × 24 = 2.40 dm³

例题：10.0 g 碳酸钙完全分解时，在 RTP 下生成多少体积的二氧化碳？
方程式：CaCO₃ → CaO + CO₂
CaCO₃ 的 M_r = 100
CaCO₃ 的摩尔数 = 10.0 / 100 = 0.100 mol
摩尔比 1:1，因此 CO₂ 的摩尔数 = 0.100 mol
CO₂ 的体积 = 0.100 × 24 = 2.40 dm³

Sometimes you may need to calculate the mass of a gas given its volume: first convert volume to moles by dividing by 24, then multiply by M_r to find the mass.

有时候你需要根据气体体积求其质量：首先将体积除以 24 得到摩尔数，再乘以 M_r 得到质量。

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9. Percentage Yield | 产率

The percentage yield compares the actual mass of product obtained in an experiment to the theoretical mass calculated from the balanced equation. It indicates the efficiency of a reaction.

产率将实验中实际获得的产物质量与根据配平方程式计算的理论质量进行比较。它反映了反应的效率。

percentage yield = (actual yield / theoretical yield) × 100

产率 = (实际产量 / 理论产量) × 100

A yield of 100% means all the reactants converted perfectly into products. In practice, yields are often less than 100% due to incomplete reactions, side reactions, or product lost during purification.

产率 100% 意味着所有反应物都完美转化为产物。实际上，由于反应不完全、副反应或提纯过程中的损失，产率往往低于 100%。

Example: In a reaction, the theoretical yield of copper sulfate crystals is 2.50 g. After carrying out the experiment, a student collects 1.95 g. Calculate the percentage yield.
Percentage yield = (1.95 / 2.50) × 100 = 78.0%

例题：某反应中

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