📚 IGCSE CIE Chemistry: Calculation Drills | IGCSE CIE 化学：计算题专项训练
Mastering calculations is crucial for IGCSE CIE Chemistry. This article provides a comprehensive drill on mole concept, reacting masses, gas volumes, concentrations, titrations, and energy calculations. Practice is key to scoring high marks in Paper 4 (Extended).
掌握计算对IGCSE CIE化学至关重要。本文对摩尔概念、反应质量、气体体积、浓度、滴定和能量计算进行全面专项训练。练习是在Paper 4（拓展）中取得高分的关键。
1. Relative Atomic Mass and Relative Molecular Mass | 相对原子质量与相对分子质量
Relative atomic mass (Ar) is the weighted average mass of an atom of an element compared to 1/12 of the mass of a carbon-12 atom. Relative molecular mass (Mr) is the sum of the Ar values of all atoms in a molecule. For ionic compounds we use relative formula mass, but still call it Mr.
相对原子质量（Ar）是元素的原子平均质量与碳-12原子质量的1/12之比。相对分子质量（Mr）是分子中所有原子Ar值的总和。对于离子化合物我们使用相对式质量，但仍称之为Mr。
Example: Calculate the Mr of sulfuric acid, H₂SO₄. (Ar: H=1, S=32, O=16)
例题：计算硫酸 H₂SO₄ 的 Mr。（Ar: H=1, S=32, O=16）
Mr = 2×1 + 32 + 4×16 = 2 + 32 + 64 = 98.
Mr = 2×1 + 32 + 4×16 = 2 + 32 + 64 = 98。
2. The Mole and Avogadro Constant | 摩尔与阿伏伽德罗常数
The mole is the unit for amount of substance. One mole contains 6.02 × 10²³ particles (Avogadro constant). The mass of one mole of a substance is its molar mass (M) in g/mol, numerically equal to its Mr. The number of moles n = mass (g) ÷ molar mass (g/mol).
摩尔是物质的量的单位。1摩尔含有6.02 × 10²³个微粒（阿伏伽德罗常数）。1摩尔物质的质量是其摩尔质量（M），单位为g/mol，数值等于其Mr。摩尔数 n = 质量(g) ÷ 摩尔质量(g/mol)。
Example: How many moles are in 8.8 g of CO₂? (C=12, O=16)
例题：8.8 g CO₂ 是多少摩尔？
Mr(CO₂) = 12 + 2×16 = 44. n = 8.8 / 44 = 0.20 mol.
Mr(CO₂) = 12 + 2×16 = 44。n = 8.8 / 44 = 0.20 mol。
3. Reacting Masses (Stoichiometry) | 反应质量（化学计量学）
Using a balanced equation, we can calculate the mass of reactants or products. The mole ratio from the coefficients allows conversion between moles of different substances. Always start by calculating moles of the known substance.
利用配平的方程式，我们可以计算反应物或生成物的质量。系数给出的摩尔比允许在不同物质的摩尔之间转换。始终从已知物质的摩尔数开始计算。
Example: What mass of MgO is produced when 2.4 g of Mg burns completely? 2Mg + O₂ → 2MgO (Mg=24, O=16).
例题：2.4 g Mg 完全燃烧产生多少克 MgO？
Step 1: n(Mg) = 2.4 / 24 = 0.10 mol. Step 2: From equation, 2 mol Mg → 2 mol MgO, so ratio 1:1, n(MgO) = 0.10 mol. Step 3: Mr(MgO) = 24 + 16 = 40; mass = 0.10 × 40 = 4.0 g.
步骤1：n(Mg) = 2.4 / 24 = 0.10 mol。步骤2：由方程式，2 mol Mg → 2 mol MgO，摩尔比1:1，n(MgO) = 0.10 mol。步骤3：Mr(MgO) = 40；质量 = 0.10 × 40 = 4.0 g。
4. Volumes of Gases (Molar Volume) | 气体体积（摩尔体积）
At room temperature and pressure (r.t.p., 20 °C, 1 atm), one mole of any gas occupies 24 dm³ (24,000 cm³). This is the molar gas volume. Use this to convert between moles and volume of a gas.
在室温和常压（r.t.p., 20 °C, 1 atm）下，1摩尔任何气体占据24 dm³（24,000 cm³）。这就是气体摩尔体积。用它来转换气体的摩尔数和体积。
Example: What volume of CO₂ (at r.t.p.) is produced when 5.0 g of CaCO₃ decomposes? CaCO₃ → CaO + CO₂ (Ca=40, C=12, O=16).
例题：5.0 g CaCO₃ 分解产生多少体积 CO₂（r.t.p.）？
Mr(CaCO₃) = 40 + 12 + 48 = 100. n(CaCO₃) = 5.0 / 100 = 0.050 mol. From equation, 1 mol CaCO₃ → 1 mol CO₂, so n(CO₂) = 0.050 mol. Volume = 0.050 × 24 = 1.2 dm³ (or 1200 cm³).
Mr(CaCO₃) = 40 + 12 + 48 = 100。n(CaCO₃) = 5.0 / 100 = 0.050 mol。由方程式，1 mol CaCO₃ → 1 mol CO₂，所以n(CO₂) = 0.050 mol。体积 = 0.050 × 24 = 1.2 dm³（即1200 cm³）。
5. Concentrations of Solutions | 溶液浓度
Concentration (c) is the amount of solute per unit volume of solution, c = n / V, where n is in mol and V is in dm³. If the volume is given in cm³, divide by 1000 to convert to dm³.
浓度（c）是单位体积溶液中溶质的物质的量，c = n / V，n单位为mol，V单位为dm³。如果体积以cm³给出，除以1000转换为dm³。
Example: Calculate the concentration of a solution containing 4.0 g of NaOH in 250 cm³ of water. (Na=23, O=16, H=1)
例题：计算含4.0 g NaOH的250 cm³溶液的浓度。
Mr(NaOH) = 40. n = 4.0 / 40 = 0.10 mol. V = 250 / 1000 = 0.25 dm³. c = 0.10 / 0.25 = 0.40 mol/dm³.
Mr(NaOH) = 40。n = 4.0 / 40 = 0.10 mol。V = 250 / 1000 = 0.25 dm³。c = 0.10 / 0.25 = 0.40 mol/dm³。
6. Empirical and Molecular Formulae | 实验式与分子式
Empirical formula gives the simplest whole-number ratio of atoms in a compound. Molecular formula is a multiple of the empirical formula. To find the empirical formula from percentage composition, divide the mass percentage of each element by its Ar, then find the simplest ratio.
实验式给出化合物中原子的最简整数比。分子式是实验式的整数倍。由质量百分比求实验式，将各元素的质量百分数除以各自的Ar，然后求最简比。
Example: A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Find its empirical formula. (C=12, H=1, O=16)
例题：一种化合物含碳40.0%、氢6.7%和氧53.3%。求其实验式。
C: 40.0 / 12 = 3.33; H: 6.7 / 1 = 6.7; O: 53.3 / 16 = 3.33. Divide by smallest (3.33): C = 1, H = 2, O = 1. Empirical formula is CH₂O.
C: 40.0 / 12 = 3.33；H: 6.7 / 1 = 6.7；O: 53.3 / 16 = 3.33。除以最小值3.33：C = 1，H = 2，O = 1。实验式为CH₂O。
7. Percentage Yield and Purity | 百分产率与纯度
Percentage yield = (actual mass obtained / theoretical mass) × 100%. It measures the efficiency of a reaction. Percentage purity = (mass of pure substance / mass of impure sample) × 100%.
百分产率 = (实际得到的质量 / 理论质量) × 100%。它衡量反应的效率。百分纯度 = (纯物质质量 / 不纯样品质量) × 100%。
Example: In a preparation, theoretical mass of product is 5.0 g, but only 3.5 g is obtained. Calculate percentage yield.
例题：
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