GCSE OCR Chemistry: Calculation Practice | GCSE OCR 化学：计算题专项训练

📚 GCSE OCR Chemistry: Calculation Practice | GCSE OCR 化学：计算题专项训练

Quantitative chemistry calculations form a core component of the OCR GCSE Chemistry specification. Mastery of these skills is essential for success in all three exam papers and the practical endorsement. This article provides a structured walkthrough of the most common calculation types, from formula masses to bond energy problems, with step-by-step methods, worked examples, and bilingual explanations.

定量化学计算是OCR GCSE化学考试的核心组成部分。掌握这些技能对于在三份试卷和实践考核中取得成功至关重要。本文系统地梳理了最常见的计算题型，从式量到键能问题，提供分步骤的方法、例题和双语讲解。

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

Relative atomic mass (Ar) is the average mass of an atom of an element compared to 1/12th of the mass of a carbon-12 atom. The relative formula mass (Mr) of a substance is the sum of the Ar values of all atoms in its formula unit. For molecules like water, Mr = 2×Ar(H) + Ar(O) = 2×1 + 16 = 18.

相对原子质量（Ar）是一个元素原子的平均质量与碳-12原子质量的1/12之比。一种物质的相对式量（Mr）是其化学式单元中所有原子的Ar值之和。对于水分子，Mr = 2×Ar(H) + Ar(O) = 2×1 + 16 = 18。

You will need to use the Ar values from the Periodic Table provided in the exam. Common values include:

你需要使用考卷提供的周期表中的Ar数值。常见值包括：

Element	Symbol	Ar	元素
Hydrogen	H	1	氢
Carbon	C	12	碳
Oxygen	O	16	氧
Sodium	Na	23	钠
Chlorine	Cl	35.5	氯
Calcium	Ca	40	钙

Example: Calculate the Mr of calcium carbonate, CaCO₃.

例题：计算碳酸钙CaCO₃的Mr。

Mr(CaCO₃) = Ar(Ca) + Ar(C) + 3×Ar(O) = 40 + 12 + (3×16) = 100

The same principle applies to compounds with brackets or water of crystallisation. For Mg(OH)₂, the subscript outside the bracket multiplies everything inside: Mr = 24.3 + 2×(16 + 1) = 58.3. For hydrated salts like CuSO₄·5H₂O, add the mass of the water molecules.

同样的原则适用于带括号或结晶水的化合物。对于Mg(OH)₂，括号外的下标乘以括号内所有原子：Mr = 24.3 + 2×(16 + 1) = 58.3。对于结晶水合物如CuSO₄·5H₂O，需要加上水分子的质量。

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). This number of atoms, molecules or ions has a mass in grams equal to its relative formula mass (Mr), which is called the molar mass (M) in g/mol. For instance, the molar mass of O₂ is 32 g/mol.

一摩尔的任何物质都精确包含6.02×10²³个粒子（阿伏伽德罗常数）。这个数量的原子、分子或离子的质量（以克为单位）等于其相对式量（Mr），称为摩尔质量（M），单位g/mol。例如，O₂的摩尔质量为32 g/mol。

The key relationship for converting between mass and moles is:

质量和摩尔数之间转换的关键关系是：

number of moles (n) = mass (m) ÷ molar mass (M) (n = m/M)

Worked example: How many moles are in 4.0 g of sodium hydroxide, NaOH? (Ar values: Na=23, O=16, H=1)

例题：4.0 g 氢氧化钠（NaOH）中有多少摩尔？（Ar: Na=23, O=16, H=1）

M(NaOH) = 23 + 16 + 1 = 40 g/mol. Therefore, n = 4.0 g / 40 g/mol = 0.10 mol.

M(NaOH) = 23 + 16 + 1 = 40 g/mol。因此，n = 4.0 g ÷ 40 g/mol = 0.10 mol。

3. Mass, Moles and Particles | 质量、摩尔与粒子数

To find the number of formula units, atoms or molecules, use Avogadro’s constant:

要计算结构单元、原子或分子的数量，使用阿伏伽德罗常数：

number of particles = moles × 6.02 × 10²³

If you are given the mass first, convert to moles, then multiply. Example: Determine the number of water molecules in 9.0 g of H₂O.

如果先给出质量，先转换为摩尔数，再乘以常数。例题：求9.0 g H₂O中的水分子数。

M(H₂O) = 18 g/mol. Moles of H₂O = 9.0/18 = 0.50 mol. Number of molecules = 0.50 × 6.02×10²³ = 3.01×10²³.

M(H₂O) = 18 g/mol。H₂O的摩尔数 = 9.0 / 18 = 0.50 mol。分子数 = 0.50 × 6.02×10²³ = 3.01×10²³。

This calculation is also useful for determining the number of ions in ionic compounds. For 0.10 mol of NaCl, there are 0.10 mol of Na⁺ ions and 0.10 mol of Cl⁻ ions, so the total number of ions is 0.20 × 6.02×10²³.

此计算也可用于确定离子化合物中的离子数。对于0.10 mol NaCl，有0.10 mol Na⁺离子和0.10 mol Cl⁻离子，因此离子总数为0.20 × 6.02×10²³。

4. 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. To find it from percentage composition or experimental mass data:

经验式给出化合物中原子的最简整数比。若要从百分组成或实验质量数据求出经验式：

Divide the mass (or percentage) of each element by its Ar to obtain the number of moles.

将每种元素的质量（或百分比）除以其Ar，得到摩尔数。
Divide all mole values by the smallest mole value to get the simplest ratio.

将所有摩尔数除以最小的摩尔数，得到最简比。
If the ratio is close to a fraction (e.g., 1.5), multiply all numbers by 2 to obtain whole numbers.

如果比值接近分数（如1.5），将所有数值乘以2得到整数。

Example: A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Deduce its empirical formula.

例题：某化合物含碳40.0%、氢6.7%和氧53.3%（质量分数）。推导其实验式。

Assume 100 g: moles C = 40.0/12 = 3.33; H = 6.7/1 = 6.7; O = 53.3/16 = 3.33. Divide by 3.33: C=1, H≈2.01, O=1. Empirical formula: CH₂O.

设100 g：C的摩尔数 = 40.0/12 = 3.33；H = 6.7/1 = 6.7；O = 53.3/16 = 3.33。各除以3.33：C=1，H≈2.01，O=1。经验式：CH₂O。

The molecular formula is a multiple of the empirical formula. You need the relative molecular mass (Mr) to find the multiplier. If the Mr of the compound above is 60, then (12+2+16) = 30 is the empirical mass; multiplier = 60/30 = 2, giving molecular formula C₂H₄O₂.

分子式是经验式的倍数。需要相对分子质量（Mr）来求倍数。若上述化合物的Mr为60，经验式质量为30；倍数 = 60/30 = 2，分子式为C₂H₄O₂。

5. Reacting Mass Calculations | 反应质量计算

In a balanced equation, the coefficients give the mole ratio of reactants and products. To calculate reacting masses:

在配平的方程式中，系数给出反应物和产物的摩尔比。计算反应质量的方法：

Write the balanced symbol equation.

写出配平的化学方程式。
Convert the known mass of a substance to moles (n=m/M).

将已知物质的质量转换为摩尔（n=m/M）。
Use the mole ratio to find moles of the target substance.

使用摩尔比求出目标物质的摩尔数。
Convert the target moles to mass (m=n×M).

将目标摩尔数转换为质量（m=n×M）。

Worked example: What mass of calcium oxide (CaO) is produced when 10.0 g of calcium carbonate decomposes on heating? Equation: CaCO₃ → CaO + CO₂.

例题：加热分解10.0 g碳酸钙，会生成多少质量的氧化钙（CaO）？方程式：CaCO₃ → CaO + CO₂。

Mr(CaCO₃)=100, Mr(CaO)=56. Moles CaCO₃ = 10.0/100 = 0.100 mol. From the 1:1 mole ratio, moles CaO = 0.100 mol. Mass CaO = 0.100 × 56 = 5.6 g.

Mr(CaCO₃)=100，Mr(CaO)=56。CaCO₃摩尔数 = 10.0/100 = 0.100 mol。根据1:1摩尔比，CaO的摩尔数 = 0.100 mol。CaO质量 = 0.100 × 56 = 5.6 g。

Always check the equation is balanced. In questions where masses of both reactants are given, you may need to identify the limiting reactant first (see next section).

务必检查方程式已配平。在给出两种反应物质量的题目中，可能需要先找出限量反应物（见下一节）。

6. Limiting Reactants | 限量反应物

The limiting reactant is the substance that is completely used up in a reaction, determining the amount of product formed. The other reactant is in excess. To find the limiting reactant, compare the mole ratio from the balanced equation to the actual moles available.

限量反应物是在反应中完全消耗掉的物质，它决定了产物的生成量。另一种反应物是过量的。要找出限量反应物，需将配平方程式的摩尔比与实际可用的摩尔数进行比较。

Example: 4.0 g of hydrogen gas reacts with 32.0 g of oxygen gas to produce water: 2H₂ + O₂ → 2H₂O. Which is the limiting reactant?

例题：4.0 g氢气与32.0 g氧气反应生成水：2H₂ + O₂ → 2H₂O。哪种是限量反应物？

M(H₂) = 2 g/mol, so moles H₂ = 4.0/2 = 2.0 mol. M(O₂) = 32 g/mol, so moles O₂ = 32.0/32 = 1.0 mol. The equation requires 2 mol H₂ per 1 mol O₂. Here we have exactly 2.0 mol H₂ and 1.0 mol O₂ — they match perfectly, so neither is limiting (they react completely). If we had 3.0 mol H₂ and 1.0 mol O₂, O₂ would be limiting because only 2.0 mol H₂ can react.

M(H₂) = 2 g/mol，因此H₂的摩尔数 = 4.0/2 = 2.0 mol。M(O₂) = 32 g/mol，O₂的摩尔数 = 32.0/32 = 1.0 mol。方程式要求每1 mol O₂对应2 mol H₂。这里恰好有2.0 mol H₂和1.0 mol O₂——反应物比例完全匹配，都没有过量（完全反应）。如果有3.0 mol H₂和1.0 mol O₂，则O₂是限量反应物，因为仅有2.0 mol H₂可参与反应。

To determine the mass of water produced, use the limiting moles. In the exact-match case above, either reactant gives 2.0 mol H₂O (mass = 2.0 × 18 = 36 g).

要确定生成水的质量，使用限量反应物的摩尔数。在上述恰好匹配的例子中，任一反应物均生成2.0 mol H₂O（质量 = 2.0 × 18 = 36 g）。

7. Percentage Yield | 产率

Percentage yield compares the actual mass of product obtained from an experiment to the theoretical mass calculated from the chemical equation.

产率将实验中实际获得的产品质量与根据化学方程式计算的理论质量进行比较。

% yield = (actual yield ÷ theoretical yield) × 100

Example: In the CaCO₃ decomposition example earlier, the theoretical yield of CaO was 5.6 g. If only 4.2 g of CaO was collected, the percentage yield is (4.2/5.6) × 100 = 75%.

例题：在前述CaCO₃分解的例子中，CaO的理论产量是5.6 g。如果只收集到4.2 g CaO，产率为（4.2/5.6）× 100 = 75%。

Yields are often less than 100% due to incomplete reactions, side reactions, or product lost during purification (e.g., filtration, evaporation). Reasoning about these practical losses is frequently examined.

由于反应不完全、副反应或产品在提纯过程中（如过滤、蒸发）的损失，产率通常低于100%。对这些实际损失的原因进行解释是常见考点。

8. Atom Economy | 原子经济

Atom economy measures the efficiency of a reaction in incorporating atoms from the reactants into the desired product. A higher atom economy means less waste and a more sustainable process.

原子经济衡量的是反应将反应物中的原子整合到目标产物中的效率。原子经济越高，意味着废物越少，过程更可持续。

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

Example: Calculate the atom economy for producing ethanol (C₂H₅OH) by the fermentation of glucose: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂. (Ar: C=12, H=1, O=16)

例题：计算葡萄糖发酵生产乙醇（C₂H₅OH）的原子经济：C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂。（Ar：C=12，H=1，O=16）

Mr(C₆H₁₂O₆) = (6×12)+(12×1)+(6×16) = 180. Mr(desired C₂H₅OH) = (2×12)+(6×1)+(1×16) = 46, but the equation produces 2 moles, so mass of desired product = 2×46 = 92. Atom economy = (92/180) × 100 = 51.1%.

Mr(C₆H₁₂O₆) = 180。目标产物C₂H₅OH的Mr = 46，但方程式生成2摩尔，所以目标产物总质量 = 2×46 = 92。原子经济 = （92/180）× 100 = 51.1%。

Note: Sum of Mr of all reactants means the total mass of all reactants that appear in the equation. If one reactant is in excess, its mass is still included. This is a common examiner trick.

注意：所有反应物的Mr总和指的是方程式中出现的所有反应物的总质量。即使某种反应物过量，其质量仍要计入。这是考官的常见陷阱。

9. Concentration of Solutions | 溶液浓度

Concentration can be expressed in g/dm³ or mol/dm³. The two are related by the molar mass:

浓度可以用g/dm³或mol/dm³表示。二者通过摩尔质量联系：

concentration (mol/dm³) = concentration (g/dm³) ÷ M or c = n/V

where n is moles of solute and V is solution volume in dm³. Remember: 1 dm³ = 1000 cm³. To convert cm³ to dm³, divide by 1000.

其中n是溶质的摩尔数，V是溶液体积，单位dm³。记住：1 dm³ = 1000 cm³。要将cm³转换为dm³，除以1000。

Titration calculations rely on concentration. For a reaction with a 1:1 mole ratio (e.g., HCl + NaOH → NaCl + H₂O), the formula c₁V₁ = c₂V₂ can be used, where c and V are the concentration and volume of the acid and base respectively. Make sure both volumes are in the same unit.

滴定计算依赖浓度。对于摩尔比为1:1的反应（如HCl + NaOH → NaCl + H₂O），可用公式c₁V₁ = c₂V₂，其中c和V分别是酸和碱的浓度和体积。确保两者体积单位一致。

Worked example: 25.0 cm³ of NaOH solution required 30.0 cm³ of 0.100 mol/dm³ HCl for neutralisation. Find the concentration of NaOH in mol/dm³ and g/dm³. (Mr NaOH=40)

例题：25.0 cm³ NaOH溶液需30.0 cm³ 0.100 mol/dm³ HCl中和。求NaOH的浓度，以mol/dm³和g/dm³表示。（Mr NaOH=40）

Moles HCl = (30.0/1000) × 0.100 = 0.00300 mol. Since NaOH:HCl is 1:1, moles NaOH = 0.00300 mol. Concentration NaOH (mol/dm³) = 0.00300 / (25.0/1000) = 0.120 mol/dm³. In g/dm³: 0.120 × 40 = 4.80 g/dm³.

HCl的摩尔数 = (30.0/1000) × 0.100 = 0.00300 mol。NaOH与HCl为1:1反应，故NaOH摩尔数 = 0.00300 mol。NaOH浓度（mol/dm³）= 0.00300 / (25.0/1000) = 0.120 mol/dm³。以g/dm³计：0.120 × 40 = 4.80 g/dm³。

10. 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 is known as the molar gas volume. You can relate gas volume to moles and vice versa:

在常温常压（RTP）下，一摩尔任何气体的体积为24 dm³（或24 000 cm³）。这称为气体摩尔体积。你可以将气体体积与摩尔数联系起来：

volume (dm³) = moles × 24 or moles = volume (dm³) ÷ 24

If you are working with cm³, use moles = volume (cm³) ÷ 24 000.

如果使用cm³，则用摩尔数 = 体积（cm³）÷ 24 000。

Example: A student reacts 0.50 g of magnesium with excess hydrochloric acid: Mg + 2HCl → MgCl₂ + H₂. Calculate the volume of hydrogen gas produced at RTP. (Ar Mg=24)

例题：一名学生将0.50 g镁与过量盐酸反应：Mg + 2HCl → MgCl₂ + H₂。计算在RTP下生成氢气的体积。（Ar Mg = 24）

Moles Mg = 0.50/24 = 0.02083 mol. From the equation, 1 mol Mg gives 1 mol H₂, so

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GCSE OCR Chemistry: Calculation Practice | GCSE OCR 化学：计算题专项训练