📚 AS Chemistry: Mastering Calculation Questions | AS 化学：计算题专项训练
Welcome to the calculation drill designed specifically for AS Chemistry. Numeracy is at the heart of the subject, and whether you are determining an empirical formula, predicting the yield of a reaction or analysing a titration, a systematic approach to problem-solving will boost both your confidence and your exam performance. This article walks you through the core calculation types, complete with worked examples for each topic.
欢迎来到为 AS 化学量身定制的计算题专项训练。计算是化学的核心组成部分，无论你是在测定经验式、预测反应产率还是分析滴定数据，系统化的解题方法都能大幅提升你的自信心和考试表现。本文带你重温所有核心计算类型，并配有每个专题的完整范例。
1. Mole Concept and Molar Mass | 物质的量与摩尔质量
The mole is the chemist' counting unit. One mole of any substance contains exactly 6.022 × 10²³ particles (Avogadro' d in g mol⁻¹. The fundamental equation linking mass (m), molar mass (M) and amount (n) is:
摩尔是化学家的计数单位。1 摩尔任何物质都恰好包含 6.022 × 10²³ 个微粒（阿伏伽德罗常数）。1 摩尔该物质的质量即为其摩尔质量 (M)，单位为 g mol⁻¹。联系质量 (m)、摩尔质量 (M) 和物质的量 (n) 的基本公式为：
n = m / M
Worked Example: Calculate the mass of 0.50 mol of NaOH. (M = 40 g mol⁻¹)
示例： 计算 0.50 mol NaOH 的质量。(M = 40 g mol⁻¹)
Rearrange n = m/M → m = n × M = 0.50 mol × 40 g mol⁻¹ = 20 g.
变换公式：m = n × M = 0.50 mol × 40 g mol⁻¹ = 20 g。
Always write the units at each step – this reduces careless mistakes.
每一步都要写上单位——这能有效减少粗心错误。
2. Empirical Formula from Combustion Data | 由燃烧数据求经验式
An empirical formula shows the simplest whole-number ratio of atoms in a compound. When a compound containing carbon, hydrogen and perhaps oxygen is burned completely, all the carbon is converted to CO₂ and all the hydrogen to H₂O. By measuring the masses of CO₂ and H₂O produced, we can work out the mass of C and H originally present, and then deduce the mass of O by subtraction. Finally, we convert masses to moles and simplify the ratio.
经验式（最简式）表示化合物中各原子的最简整数比。当含有碳、氢、可能还有氧的化合物完全燃烧时，所有碳转化为 CO₂，所有氢转化为 H₂O。通过测定生成的 CO₂ 和 H₂O 的质量，我们可以计算出原化合物中碳和氢的质量，再用差减法求出氧的质量。最后将质量换算为物质的量并化简比例。
Example: 0.30 g of an organic compound gives 0.44 g CO₂ and 0.18 g H₂O on complete combustion. Determine its empirical formula.
例题： 0.30 g 某有机化合物完全燃烧生成 0.44 g CO₂ 和 0.18 g H₂O。求其经验式。
s of C = (12/44) × 0.44 g = 0.12 g; mass of H = (2/18) × 0.18 g = 0.02 g; mass of O = 0.30 – (0.12 + 0.02) = 0.16 g.
碳的质量 = (12/44) × 0.44 g = 0.12 g；氢的质量 = (2/18) × 0.18 g = 0.02 g；氧的质量 = 0.30 – (0.12+0.02) = 0.16 g。
= 0.16/16 = 0.010. Divide by smallest (0.010) → ratio 1 : 2 : 1. Empirical formula = CH₂O.
物质的量：C = 0.12/12 = 0.010 mol；H = 0.02/1 = 0.020 mol；O = 0.16/16 = 0.010 mol。除以最小值 0.010 → 比例 1:2:1。经验式为 CH₂O。
3. Molecular Formula Determination | 分子式的确定
The molecular formula is a whole-number multiple of the empirical formula. To find it, you need the relative molecular mass (Mᵣ) of the compound, usually obtained from mass spectrometry or given in the question.
分子式是经验式的整数倍。要确定分子式，需要知道化合物的相对分子质量 (Mᵣ)，通常通过质谱获得或由题目直接给出。
Example: A compound has the empirical formula CH₂O and a molar mass of 60 g mol⁻¹. What is its molecular formula?
例题： 某化合物的经验式为 CH₂O，摩尔质量为 60 g mol⁻¹。其分子式是什么？
ss = 12 + (2×1) + 16 = 30. Multiplier = 60 / 30 = 2. Molecular formula = (CH₂O)₂ = C₂H₄O₂.
经验式质量 = 12 + (2×1) + 16 = 30；倍数 = 60 / 30 = 2；分子式 = (CH₂O)₂ = C₂H₄O₂。
This simple check confirms whether the compound is, for example, ethanoic acid or methyl methanoate – both share the same molecular formula.
这个简单的检验可以帮助确认该化合物是乙酸还是甲酸甲酯——两者具有相同的分子式。
4. Reacting Masses | 反应质量计算
Stoichiometry uses the balanced equation to relate the moles of reactants and products. Once the number of moles of one substance is known, the masses of others can be calculated via the mole ratios from the equation.
化学计量法利用配平的方程式将反应物和产物的物质的量联系起来。一旦知道了某种物质的物质的量，就可以通过方程式中的计量比计算其他物质的质量。
Example: cid: Zn + 2HCl → ZnCl₂ + H₂. What mass of zinc is needed to produce 0.50 g of hydrogen gas? (Aᵣ: Zn = 65)
例题： 锌与盐酸反应：Zn + 2HCl → ZnCl₂ + H₂。要制取 0.50 g 氢气需要多少克锌？(Aᵣ: Zn = 65)
ion, 1 mol Zn gives 1 mol H₂, so moles of Zn = 0.25 mol. Mass of Zn = 0.25 × 65 = 16.25 g.
H₂ 的物质的量 = 0.50/2.0 = 0.25 mol。根据方程式，1 mol Zn 生成 1 mol H₂，所以需要 Zn 0.25 mol。锌的质量 = 0.25 × 65 = 16.25 g。
Always check the balanced equation – a common error is using the wrong mole ratio.
务必核对配平后的方程式——常见的错误是用错了计量比。
5. Limiting Reactant and Excess | 限量反应物与过量
When two or more reactants are mixed, the one that is completely consumed first is the limiting reactant; it determines the maximum amount of product formed. The other reactant is in excess.
当多种反应物混合时，最先完全消耗的那种反应物称为限量反应物（限制试剂），它决定了产物的最大生成量。另一种反应物则处于过量状态。
Example: fy the limiting reactant and calculate the mass of MgO produced. (Aᵣ: Mg = 24.3, O = 16.0)
例题： 将 2.43 g 镁与 3.20 g 氧气加热反应：2Mg + O₂ → 2MgO。指出限量反应物并计算生成的 MgO 质量。(Aᵣ: Mg = 24.3, O = 16.0)
the equation, 2 mol Mg react with 1 mol O₂. So 0.100 mol Mg would need 0.050 mol O₂; but we have 0.100 mol O₂ – thus O₂ is in excess and Mg is the limiting reactant.
Mg 的物质的量 = 2.43/24.3 = 0.100 mol；O₂ 的物质的量 = 3.20/32.0 = 0.100 mol。根据方程式，2 mol Mg 与 1 mol O₂ 反应，所以 0.100 mol Mg 需要 0.050 mol O₂；而我们有 0.100 mol O₂ —— 因此 O₂ 过量，Mg 是限量反应物。
d = 0.100 mol. M of MgO = 24.3 + 16.0 = 40.3 g mol⁻¹. Mass of MgO = 0.100 × 40.3 = 4.03 g.
生成的 MgO 物质的量 = 消耗的 Mg 物质的量 = 0.100 mol。MgO 的摩尔质量 = 40.3 g mol⁻¹。MgO 质量 = 0.100 × 40.3 = 4.03 g。
6. Percentage Yield | 百分产率
The percentage yield compares the actual mass of product obtained in an experiment to the theoretical mass calculated from the limiting reactant. It is a measure of the efficiency of a reaction.
百分产率将实验中实际得到的产物质量与根据限量反应物计算出的理论质量进行比较，用于衡量反应的效率。
% Yield = (actual yield / theoretical yield) × 100
Example: In the MgO preparation above, if only 3.40 g of MgO is collected, calculate the percentage yield.
例题： 在上述制备 MgO 的实验中，若只收集到 3.40 g MgO，计算百分产率。
Theoretical yield = 4.03 g; actual yield = 3.40 g. % Yield = (3.40/4.03) × 100 = 84.4%.
理论产量 = 4.03 g；实际产量 = 3.40 g。百分产率 = (3.40/4.03) × 100 = 84.4%。
Yields are rarely 100% due to incomplete reactions, side reactions or losses during purification.
由于反应不完全、副反应或纯化过程中的损失，产率很少能达到 100%。
7. Atom Economy | 原子经济性
Atom economy evaluates how much of the reactants ends up in the desired product. It helps to assess the 'greenness' of a chemical process.
原子经济性评价了反应物中有多少原子最终进入了目标产物，有助于评估化学过程的“绿色程度”。
Atom economy = (Mᵣ of desired product / sum of Mᵣ of all reactants) × 100
Example: te: CuCO₃ → CuO + CO₂. Calculate the atom economy for CuO. (Mᵣ: CuO = 79.5, CuCO₃ = 123.5)
例题： 氧化铜可通过加热分解碳酸铜制得：CuCO₃ → CuO + CO₂。计算生成 CuO 的原子经济性。(Mᵣ: CuO = 79.5, CuCO₃ = 123.5)
Atom economy = (79.5 / 123.5) × 100 = 64.4%. Only about two-thirds of the reactant mass is incorporated into the useful product.
原子经济性 = (79.5 / 123.5) × 100 = 64.4%。只有约三分之二的反应物质量进入了有用产物。
A high atom economy is desirable for sustainable chemistry.
高原子经济性是可持续化学所追求的目标。
8. Gas Volume Calculations (Molar Volume) | 气体体积计算（摩尔体积）
At room temperature and pressure (r.t.p., 25 °C and 1 atm), one mole of any gas occupies 24 dm³ (or 24,000 cm³). This makes it straightforward to convert between moles and volume for gases.
在室温常压（r.t.p.，25 °C 和 1 atm）下，1 摩尔任何气体占据的体积为 24 dm³（或 24 000 cm³）。这使得气体的物质的量与体积之间的换算非常直接。
n = V / 24 (when V is in dm³)
Example: reacts with excess hydrochloric acid, what volume of carbon dioxide is produced at r.t.p.? (Mᵣ of CaCO₃ = 100)
例题： 2.0 g 碳酸钙与过量盐酸反应，在 r.t.p. 下能生成多大体积的二氧化碳？(Mᵣ of CaCO₃ = 100)
gives 1 mol CO₂, so moles of CO₂ = 0.020 mol. Volume = 0.020 × 24 = 0.48 dm³ (or 480 cm³).
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂。CaCO₃ 物质的量 = 2.0/100 = 0.020 mol。根据方程式 1 mol CaCO₃ 生成 1 mol CO₂，故 CO₂ 的物质的量 = 0.020 mol。体积 = 0.020 × 24 = 0.48 dm³（或 480 cm³）。
Remember to state the temperature and pressure if they differ from r.t.p., as molar volume changes accordingly.
如果温度和压强与 r.t.p. 不同，记得要说明，因为摩尔体积会随之改变。
9. Solution Concentration (mol dm⁻³) | 溶液浓度（mol dm⁻³）
The concentration of a solution is expressed in mol dm⁻³ and links the amount of solute to the volume of solution.
溶液的浓度以 mol dm⁻³ 表示，将溶质的物质的量与溶液体积联系起来。
c = n / V (V in dm³)
Example: What mass of sodium hydroxide is required to prepare 250 cm³ of a 0.100 mol dm⁻³ solution? (Mᵣ of NaOH = 40)
例题： 配制 250 cm³ 浓度为 0.100 mol dm⁻³ 的氢氧化钠溶液需要多少克 NaOH？(Mᵣ of NaOH = 40)
= 0.250 dm³. n = c × V = 0.100 × 0.250 = 0.0250 mol. Mass = n × M = 0.0250 × 40 = 1.00 g.
体积 (dm³) = 250/1000 = 0.250 dm³。n = c × V = 0.100 × 0.250 = 0.0250 mol。质量 = 0.0250 × 40 = 1.00 g。
The solute is dissolved in water and made up to the mark in a volumetric flask.
将溶质溶解于水并在容量瓶中定容至刻度线。
10. Acid-Base Titration Calculations | 酸碱滴定计算
Titration data allows you to determine the concentration of an unknown solution. At the equivalence point, the moles of H⁺ equal the moles of OH⁻ (for a strong monoprotic acid and strong base).
滴定数据可用于测定未知溶液的浓度。在等当点，H⁺ 的物质的量等于 OH⁻ 的物质的量（对于强一元酸和强碱而言）。
Example: A 25.0 cm³ sample of NaOH solution is titrated with 0.100 mol dm⁻³ HCl. The average titre is 23.50 cm³. Calculate the concentration of the NaOH solution.
例题： 用 0.100 mol dm⁻³ HCl 滴定 25.0 cm³ NaOH 溶液，平均滴定体积为 23.50 cm³。计算 NaOH 溶液的浓度。
aOH in 25.0 cm³ = 0.00235 mol. Concentration of NaOH = 0.00235 / 0.0250 = 0.0940 mol dm⁻³.
HCl + NaOH → NaCl + H₂O。HCl 物质的量 = 0.100 × (23.50/1000) = 0.00235 mol。根据 1:1 计量关系，25.0 cm³ 中的 NaOH 物质的量 = 0.00235 mol。NaOH 浓度 = 0.00235 / 0.0250 = 0.0940 mol dm⁻³。
Always average concordant titres (within 0.10 cm³) before calculation.
计算前务必取用重复性良好（偏差小于 0.10 cm³）的滴定体积的平均值。
11. Enthalpy Change from Calorimetry | 由量热法求焓变
e solution is calculated with q = mcΔT, and then converted to an enthalpy change per mole.
反应过程中以热形式传递的能量可用简易量热计测量。溶液吸收或放出的热量用 q = mcΔT 计算，再换算为每摩尔反应的焓变。
q = m c ΔT
Example: te the enthalpy change of neutralisation. (Assume density = 1 g cm⁻³, c = 4.18 J g⁻¹ °C⁻¹)
例题： 在聚苯乙烯杯中将 50 cm³ 1.0 mol dm⁻³ HCl 与 50 cm³ 1.0 mol dm⁻³ NaOH 混合，温度升高 6.5 °C。计算中和焓变。（假设密度 = 1 g cm⁻³，比热容 c = 4.18 J g⁻¹ °C⁻¹）
otal mass of solution = 100 g. q = 100 × 4.18 × 6.5 = 2717 J. Moles of HCl = 0.050 × 1.0 =
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