IB Chemistry: Calculation Skills Drill | IB化学：计算题专项训练

📚 IB Chemistry: Calculation Skills Drill | IB化学：计算题专项训练

Mastery of calculations is essential for success in IB Chemistry, from Paper 1 multiple‑choice questions to the structured problems in Paper 2 and the data‑based tasks in the Internal Assessment. This drill covers the core quantitative topics – mole concept, stoichiometry, gases, energetics, equilibrium, acids and bases, electrochemistry, kinetics, and error analysis – with step‑by‑step strategies, essential formulas, and exam‑style examples.

掌握计算是 IB 化学取得高分的关键，无论是 Paper 1 的选择题、Paper 2 的结构化题目还是内部评估中的数据处理任务，都离不开扎实的运算能力。这个专项训练覆盖了摩尔概念、化学计量、气体、能量学、平衡、酸碱、电化学、动力学和误差分析等核心计算主题，通过分步策略、必背公式和考试型例题帮助你系统提升。

1. Mole Concept and Avogadro’s Number | 摩尔概念与阿伏伽德罗常数

The mole is the SI unit for amount of substance, defined by Avogadro’s number (6.02 × 10²³ particles). The fundamental relationship is n = m / M, where n is the number of moles, m is mass in grams, and M is molar mass in g mol⁻¹. For gases at STP, 1 mol occupies 22.7 dm³ (IB uses 22.7 at 273 K, 100 kPa), and for solutions, n = c × V (volume in dm³). Always convert temperature to Kelvin (K = °C + 273) and pressure to kPa when using the ideal gas equation.

摩尔是物质的量的国际单位，由阿伏伽德罗常数（6.02 × 10²³ 个微粒）定义。基本关系式为 n = m / M，其中 n 是摩尔数，m 是质量（克），M 是摩尔质量（g mol⁻¹）。在标准状况下，1 mol 气体占 22.7 dm³（IB 采用 273 K、100 kPa 下的数值），而对于溶液则使用 n = c × V（体积以 dm³ 计）。使用理想气体方程时务必把温度换算成开尔文（K = °C + 273）、压强换算成 kPa。

n = m / M n = cV n = V(gas) / 22.7 dm³ (at STP)

Example: Calculate moles of NaCl in 5.85 g. n = 5.85 g / 58.44 g mol⁻¹ = 0.100 mol.
示例：计算 5.85 g NaCl 的摩尔数。n = 5.85 g / 58.44 g mol⁻¹ = 0.100 mol。

2. Empirical and Molecular Formulae | 实验式与分子式

Empirical formula is the simplest whole‑number ratio of atoms in a compound, determined from percentage composition or combustion analysis data. Molecular formula is a multiple of the empirical formula and requires the relative molecular mass (Mᵣ). To find the empirical formula: (1) convert % to grams, (2) convert grams to moles, (3) divide by the smallest mole value to obtain the ratio, (4) adjust to whole numbers if necessary. Molecular formula = (empirical formula)_n where n = Mᵣ / empirical formula mass.

实验式是化合物中原子的最简整数比，通过元素组成百分比或燃烧分析数据求得。分子式是实验式的整数倍，需要已知相对分子质量（Mᵣ）。求实验式的步骤为：(1) 把百分比视为质量（克），(2) 把质量换算成摩尔数，(3) 除以最小摩尔值得到比例，(4) 必要时调整为整数。分子式 = (实验式)_n，其中 n = Mᵣ / 实验式质量。

Example: A compound contains 40.0% C, 6.7% H, 53.3% O; Mᵣ = 60.0. Determine molecular formula. C:H:O mole ratio = (40.0/12.01):(6.7/1.01):(53.3/16.00) ≈ 3.33:6.63:3.33 → divide by 3.33 gives 1:2:1 → empirical formula CH₂O, mass = 30.03. n = 60.0/30.03 = 2 → molecular formula C₂H₄O₂.
示例：某化合物含 40.0% C, 6.7% H, 53.3% O，Mᵣ = 60.0。求分子式。C:H:O 摩尔比 = (40.0/12.01):(6.7/1.01):(53.3/16.00) ≈ 3.33:6.63:3.33，除以 3.33 得 1:2:1，实验式为 CH₂O，质量为 30.03。n = 60.0/30.03 = 2，分子式为 C₂H₄O₂。

3. Stoichiometry and Limiting Reactants | 化学计量与限量反应物

Stoichiometry uses the coefficients of a balanced equation to convert moles of one substance to moles of another. The limiting reactant is the one that is completely consumed first, determining the theoretical yield. To identify it, calculate the mole ratio of available reactants and compare with the required ratio from the equation. Theoretical yield is always calculated from the limiting reactant. Percentage yield = (actual yield / theoretical yield) × 100%. Always check that equation is balanced before starting any calculation.

化学计量利用配平方程式中的系数将一种物质的摩尔数换算为另一种物质的摩尔数。限量反应物是首先被完全消耗的反应物，它决定了理论产量。判断限量反应物的方法是：计算可用的反应物摩尔比，并与方程式所需比例进行比较。理论产量总是基于限量反应物计算。百分产率 = (实际产量 / 理论产量) × 100%。任何计算之前务必确认方程式已配平。

Example: 2Al + 3Cl₂ → 2AlCl₃. If 0.40 mol Al reacts with 0.50 mol Cl₂, which is limiting? Required Al:Cl₂ = 2:3 = 1:1.5. Available ratio 0.40:0.50 = 1:1.25, so Cl₂ is limiting. Theoretical moles of AlCl₃ = (0.50 mol Cl₂) × (2 mol AlCl₃ / 3 mol Cl₂) = 0.333 mol.
示例：2Al + 3Cl₂ → 2AlCl₃。若 0.40 mol Al 与 0.50 mol Cl₂ 反应，谁是限量反应物？所需 Al:Cl₂ = 2:3 = 1:1.5，实际比为 0.40:0.50 = 1:1.25，故 Cl₂ 限量。AlCl₃ 理论摩尔数 = (0.50 mol Cl₂) × (2 mol AlCl₃ / 3 mol Cl₂) = 0.333 mol。

4. Gas Law Calculations (Ideal Gas Equation) | 气体定律计算（理想气体方程）

The ideal gas equation, PV = nRT, links pressure (P in Pa or kPa), volume (V in m³ or dm³), amount (n), and temperature (T in K). The gas constant R has two common values: 8.31 J K⁻¹ mol⁻¹ when P is in Pa and V in m³, or 8.31 kPa dm³ K⁻¹ mol⁻¹ when P is in kPa and V in dm³. IB data booklet provides R = 8.31 J K⁻¹ mol⁻¹. Remember to convert units consistently: 1 atm = 101.3 kPa = 1.013 × 10⁵ Pa, 1 dm³ = 1 × 10⁻³ m³. Rearrange equation as needed to find unknown variable. For non‑STP conditions, use combined gas law P₁V₁/T₁ = P₂V₂/T₂ when amount of gas is constant.

理想气体方程 PV = nRT 建立了压强（P，单位 Pa 或 kPa）、体积（V，单位 m³ 或 dm³）、物质的量（n）和温度（T，单位 K）之间的关系。气体常数 R 有两个常用值：当 P 用 Pa、V 用 m³ 时 R = 8.31 J K⁻¹ mol⁻¹；或当 P 用 kPa、V 用 dm³ 时 R = 8.31 kPa dm³ K⁻¹ mol⁻¹。IB 数据手册给出 R = 8.31 J K⁻¹ mol⁻¹。记得单位换算：1 atm = 101.3 kPa = 1.013 × 10⁵ Pa，1 dm³ = 1 × 10⁻³ m³。根据需求变换方程求未知量。对于非标准状况且气体量不变的情况，使用联合气体定律 P₁V₁/T₁ = P₂V₂/T₂。

PV = nRT R = 8.31 J K⁻¹ mol⁻¹

Example: What is the volume of 0.500 mol of an ideal gas at 25.0 °C and 100 kPa? T = 298 K, P = 100 kPa, use R = 8.31 kPa dm³ K⁻¹ mol⁻¹. V = nRT/P = (0.500 × 8.31 × 298) / 100 = 12.4 dm³.
示例：0.500 mol 理想气体在 25.0 °C、100 kPa 下的体积是多少？T = 298 K，P = 100 kPa，使用 R = 8.31 kPa dm³ K⁻¹ mol⁻¹。V = nRT/P = (0.500 × 8.31 × 298) / 100 = 12.4 dm³。

5. Molarity, Dilution, and Titration | 摩尔浓度、稀释与滴定

Concentration (c) is expressed as mol dm⁻³. The dilution formula is c₁V₁ = c₂V₂, where V must be in the same units. In titrations, the equivalence point is reached when the moles of acid equal the moles of base according to the stoichiometric ratio. Use a balanced equation and the formula nᴀcᴀVᴀ/coefficient = nᴀcᴀVᴀ/coefficient for the unknown. Standard titration procedure: (1) write balanced equation, (2) calculate moles of standard solution used, (3) use mole ratio to find moles of unknown, (4) calculate concentration or mass. Back titrations apply the same principles but involve an excess reactant.

浓度（c）以 mol dm⁻³ 表示。稀释公式为 c₁V₁ = c₂V₂，体积单位须一致。在滴定中，当酸和碱的摩尔数按照化学计量比相等时即达到等当点。使用配平方程式和关系式 nᴀcᴀVᴀ/系数 = nᴀcᴀVᴀ/系数来计算未知物浓度。标准滴定步骤：(1) 写出配平方程式，(2) 计算使用的标准溶液的摩尔数，(3) 通过摩尔比求出未知物的摩尔数，(4) 计算浓度或质量。返滴定应用相同原理，但涉及过量反应物。

Example: 25.0 cm³ of HCl was titrated with 0.100 mol dm⁻³ NaOH, requiring 20.0 cm³ for neutralization. HCl + NaOH → NaCl + H₂O. Moles NaOH = 0.100 × 0.0200 = 0.00200 mol. Mole ratio 1:1, so moles HCl = 0.00200 mol. [HCl] = 0.00200 / 0.0250 = 0.0800 mol dm⁻³.
示例：用 0.100 mol dm⁻³ NaOH 滴定 25.0 cm³ HCl，需 20.0 cm³ 中和。HCl + NaOH → NaCl + H₂O。NaOH 摩尔数 = 0.100 × 0.0200 = 0.00200 mol。摩尔比 1:1，因此 HCl 摩尔数 = 0.00200 mol。[HCl] = 0.00200 / 0.0250 = 0.0800 mol dm⁻³。

6. Energetics: Enthalpy Change Calculations | 能量学：焓变计算

Enthalpy change (ΔH) is calculated using q = mcΔT for solution or combustion experiments, where q is heat in J, m is mass of solution (g), c is specific heat capacity (4.18 J g⁻¹ K⁻¹ for water), and ΔT is temperature change. Then ΔH = –q / n (exothermic gives negative ΔH). For reactions, use Hess’s law or bond enthalpies: ΔH = Σ(bonds broken) – Σ(bonds formed). Standard enthalpy change of formation (ΔHf°) data allows calculation: ΔH° = ΣΔHf°(products) – ΣΔHf°(reactants). Always multiply by stoichiometric coefficients. Remember to give sign and units (kJ mol⁻¹).

焓变（ΔH）可通过溶液或燃烧实验中的 q = mcΔT 计算，其中 q 是热量（J），m 是溶液质量（g），c 是比热容（水为 4.18 J g⁻¹ K⁻¹），ΔT 是温度变化。然后 ΔH = –q / n（放热反应 ΔH 为负）。对于反应，可利用盖斯定律或键焓：ΔH = Σ(断裂键键焓) – Σ(形成键键焓)。利用标准生成焓（ΔHf°）数据：ΔH° = ΣΔHf°(生成物) – ΣΔHf°(反应物)。务必乘以化学计量系数。记得标注正负号和单位（kJ mol⁻¹）。

q = mcΔT ΔH = –q/n ΔH° = ΣΔHf°(products) – ΣΔHf°(reactants)

Example: 0.0500 mol of acid reacted in 50.0 g water, temperature rose by 6.5 °C. q = 50.0 × 4.18 × 6.5 = 1358.5 J. ΔH = –1358.5 / 0.0500 = –27170 J mol⁻¹ ≈ –27.2 kJ mol⁻¹.
示例：0.0500 mol 酸在 50.0 g 水中反应，温度升高 6.5 °C。q = 50.0 × 4.18 × 6.5 = 1358.5 J。ΔH = –1358.5 / 0.0500 = –27170 J mol⁻¹ ≈ –27.2 kJ mol⁻¹。

7. Equilibrium Constant (Kc) Calculations | 平衡常数 Kc 计算

For a general reaction aA + bB ⇌ cC + dD, the equilibrium constant expression is Kc = [C]c[D]d / [A]a[B]b, where square brackets denote equilibrium concentrations in mol dm⁻³. Kc has a fixed value at a given temperature. In calculations, an ICE table (Initial, Change, Equilibrium) is highly recommended to determine equilibrium concentrations. If Kc > 10³ the reaction goes to completion, if Kc < 10⁻³ the reaction hardly proceeds. Note that solids and pure liquids do not appear in Kc expression.

对于一般反应 aA + bB ⇌ cC + dD，平衡常数表达式为 Kc = [C]c[D]d / [A]a[B]b，方括号代表平衡浓度（mol dm⁻³）。Kc 在给定温度下为定值。强烈推荐使用 ICE 表格（初始浓度、变化量、平衡浓度）来确定平衡浓度。若 Kc > 10³，反应接近完全；若 Kc < 10⁻³，反应几乎不进行。注意固体和纯液体不出现在 Kc 表达式中。

Example: H₂ + I₂ ⇌ 2HI. Initially [H₂] = [I₂] = 0.100 M. At equilibrium [HI] = 0.150 M. Let x be change: 2x = 0.150 → x = 0.075. Equilibrium [H₂] = [I₂] = 0.100 – 0.075 = 0.025 M. Kc = (0.150)² / (0.025 × 0.025) = 36.
示例：H₂ + I₂ ⇌ 2HI。初始 [H₂] = [I₂] = 0.100 M，平衡时 [HI] = 0.150 M。设变化为 x：2x = 0.150 → x = 0.075。平衡 [H₂] = [I₂] = 0.100 – 0.075 = 0.025 M。Kc = (0.150)² / (0.025 × 0.025) = 36。

8. Acids and Bases: pH, pOH, Ka, Kb | 酸和碱：pH、pOH、Ka、Kb

pH = –log[H⁺] and pOH = –log[OH⁻], with pH + pOH = 14 at 298 K. For weak acids, Ka = [H⁺][A⁻]/[HA]; for weak bases, Kb = [BH⁺][OH⁻]/[B]. To calculate pH of a weak acid with known Ka and initial concentration c, assume [H⁺] ≈ √(Ka × c) when dissociation is small (c/Ka > 100). For buffer solutions, use the Henderson–Hasselbalch equation: pH = pKa + log([A⁻]/[HA]). Always check assumptions when approximating.

pH = –log[H⁺]，pOH = –log[OH⁻]，在 298 K 下有 pH + pOH = 14。对于弱酸，Ka = [H⁺][A⁻]/[HA]；对于弱碱，Kb = [BH⁺][OH⁻]/[B]。计算已知 Ka 和初始浓度 c 的弱酸 pH 时，当解离度很小（c/Ka > 100）可近似 [H⁺] ≈ √(Ka × c)。对于缓冲溶液，使用亨德森-哈塞尔巴尔赫方程：pH = pKa + log([A⁻]/[HA])。做近似时一定要检查前提条件。

pH = –log[H⁺] Ka = [H⁺][A⁻]/[HA] pH = pKa + log([A⁻]/[HA])

Example: Calculate pH of 0.100 M CH₃COOH (Ka = 1.8 × 10⁻⁵). [H⁺] = √(1.8 × 10⁻⁵ × 0.100) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ M. pH = –log(1.34 × 10⁻³) = 2.87.
示例：计算 0.100 M CH₃COOH（Ka = 1.8 × 10⁻⁵）的 pH。[H⁺] = √(1.8 × 10⁻⁵ × 0.100) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ M。pH = –log(1.34 × 10⁻³) = 2.87。

9. Electrochemistry: Cell Potential and Faraday’s Laws | 电化学：电池电位与法拉第定律

Standard cell potential E°cell = E°cathode – E°anode (reduction potentials). A spontaneous reaction has positive E°cell. The relationship with Gibbs free energy is ΔG° = –nFE°cell, where n is number of electrons transferred and F is Faraday’s constant (96 500 C mol⁻¹). In electrolytic cells, Faraday’s laws allow calculation of mass deposited: Q = It, and n(e⁻) = Q / F. Mass of substance = (n(e⁻) × M) / (electrons per ion). Always draw half‑equations to identify number of electrons.

标准电池电位 E°cell = E°阴极 – E°阳极（采用还原电位）。自发反应的 E°cell 为正值。与吉布斯自由能的关系为 ΔG° = –nFE°cell，其中 n 为转移电子数，F 为法拉第常数（96 500 C mol⁻¹）。在电解池中，用法拉第定律计算沉积质量：Q = It，n(e⁻) = Q / F。物质质量 = (n(e⁻) × M) / (每离子电子数)。务必先写半反应式以确定电子数。

Example: Calculate mass of copper deposited by a current of 2.00 A for 30 minutes. Q = 2.00 × (30 × 60) = 3600 C. n(e⁻) = 3600 / 96500 = 0.0373 mol. Cu²⁺ + 2e⁻ → Cu, so 2 mol e⁻ deposit 1 mol Cu. n(Cu) = 0.0373 / 2 = 0.01865 mol. Mass = 0.01865 × 63.55 = 1.19 g.
示例：计算电流 2.00 A 通电 30 分钟所沉积的铜的质量。Q = 2.00 × (30 × 60) = 3600 C。n(e⁻) = 3600 / 96500 = 0.0373 mol。Cu²⁺ + 2e⁻ → Cu，故 2 mol e⁻ 沉积 1 mol Cu。n(Cu) = 0.0373 / 2 = 0.01865 mol。质量 = 0.01865 × 63.55 = 1.19 g。

10. Rate of Reaction and Orders | 反应速率与级数

The rate equation rate = k[A]m[B]n expresses how rate depends on concentrations, with m and n being the partial orders determined experimentally (not from stoichiometry). Units of k vary with overall order: zero order mol dm⁻³ s⁻¹, first order s⁻¹, second order dm³ mol⁻¹ s⁻¹. Use initial rates method or integrated rate laws to find order. For a first‑order reaction, ln[A] = ln[A]₀ – kt and half‑life t½ = ln2 / k. For data, compare experiments where only one concentration changes to deduce order.

速率方程 rate = k[A]m[B]n 表示速率与浓度的关系，m 和 n 是由实验测定的分反应级数（不是从化学计量数得出）。速率常数 k 的单位随总级数变化：零级 mol dm⁻³ s⁻¹，一级 s⁻¹，二级 dm³ mol⁻¹ s⁻¹。使用初始速率法或积分速率方程求级数。对于一级反应，ln[A] = ln[A]₀ – kt，半衰期 t½ = ln2 / k。处理数据时，比较只有一种浓度变化的实验来推断级数。

Example: Experiment 1: [A]=0.10, [B]=0.10, rate=0.0020. Experiment 2: [A]=0.20, [B]=0.10, rate=0.0080. Doubling [A] quadruples rate → second order in A.
示例：实验 1：[A]=0.10, [B]=0.10，速率=0.0020。实验 2：[A]=0.20, [B]=0.10，速率=0.0080。[A] 加倍导致速率变四倍，故对 A 为二级。

11. Spectrophotometry and Beer–Lambert Law | 分光光度法与比尔-朗伯定律

Absorbance (A) is related to concentration by A = ε c l, where ε is the molar absorptivity (dm³ mol⁻¹ cm⁻¹), c is concentration (mol dm⁻³), and l is path length (cm). This equation is linear, enabling construction of a calibration curve. In the IA and exam questions, often you are asked to determine concentration from absorbance readings or to calculate ε from a graph slope. Always use the wavelength of maximum absorbance (λ_max) for better sensitivity. Remember that absorbance has no units.

吸光度（A）与浓度的关系为 A = ε c l，其中 ε 为摩尔吸光系数（dm³ mol⁻¹ cm⁻¹），c 为浓度（mol dm⁻³），l 为光程长度（cm）。该方程为线性关系，可用于制作标准曲线。在 IA 和考题中，常需通过吸光度读数求浓度，或通过曲线斜率计算 ε。为提高灵敏度，应使用最大吸收波长（λ_max）。注意吸光度没有单位。

A = ε c l

Example: A solution with ε = 150 dm³ mol⁻¹ cm⁻¹ and l = 1.0 cm gives A = 0.450. c = A/(εl) = 0.450/(150×1.0) = 0.00300 mol dm⁻³.
示例：某溶液 ε = 150 dm³ mol⁻¹ cm⁻¹，l = 1.0 cm，测得 A = 0.450。则 c = A/(εl) = 0.450/(150×1.0) = 0.00300 mol dm⁻³。

12. Error Analysis and Significant Figures | 误差分析与有效数字

Calculated results must reflect the precision of the given data. Use the correct number of significant figures (sf): in multiplication/division, answer should have the same number of sf as the least precise measurement; in addition/subtraction, same number of decimal places as the least precise measurement. For example, 5.11 g ÷ 0.220 mol = 23.2 g mol⁻¹ (3 sf). In IA, include propagation of uncertainties. Percentage uncertainty = (absolute uncertainty / measurement) × 100%. Combined uncertainties are added for sums/differences; for products/quotients percentage uncertainties are added. Always show units and final rounding.

计算结果必须反映原始数据的精度。使用正确的有效数字（sf）位数：乘除运算结果的有效数字位数与测量值中最少的有效数字位数相同；加减运算结果的小数位数与最不精确的测量值相同。例如，5.11 g ÷ 0.220 mol = 23.2 g mol⁻¹（3 位有效数字）。在 IA 中需要包含不确定度传递。百分不确定度 = (绝对不确定度 / 测量值) × 100%。加减运算中绝对不确定度相叠；乘除运算中百分不确定度相叠。始终标明单位并在最后进行修约。

Example: Mass = 2.50 ± 0.05 g, volume = 10.0 ± 0.1 cm³. Density = 2.50/10.0 = 0.250 g cm⁻³. % uncertainty mass = (0.05/2.50)×100 = 2.0%, volume = (0.1/10.0)×100 = 1.0%. Total % uncertainty = 3.0%. Absolute uncertainty = 0.250 × 3.0% = 0.0075 → 0.008 g cm⁻³. Density = 0.250 ± 0.008 g cm⁻³.
示例：质量 = 2.50 ± 0.05 g，体积 = 10.0 ± 0.1 cm³。密度 = 2.50/10.0 = 0.250 g cm⁻³。质量的百分不确定度 = (0.05/2.50)×100 = 2.0%，体积 = (0.1/10.0)×100 = 1.0%。总百分不确定度 = 3.0%。绝对不确定度 = 0.250 × 3.0% = 0.0075 → 0.008 g cm⁻³。密度 = 0.250 ± 0.008 g cm⁻³。

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IB Chemistry: Calculation Skills Drill | IB化学：计算题专项训练