Essential Chemistry for Cambridge IGCSE: Mastering Calculation Questions | 剑桥IGCSE化学基础：精通计算题型

📚 Essential Chemistry for Cambridge IGCSE: Mastering Calculation Questions | 剑桥IGCSE化学基础：精通计算题型

Calculation questions form the backbone of Cambridge IGCSE Chemistry, connecting theoretical concepts with quantitative problem-solving. They require you to convert between masses, moles, volumes and concentrations, and to apply mole ratios from balanced equations. Mastering these skills will not only boost your exam performance but also deepen your understanding of chemical quantities.

计算题是剑桥IGCSE化学的支柱，将理论概念与定量问题解决紧密结合。它们要求你在质量、摩尔、体积和浓度之间进行换算，并运用配平方程式的摩尔比。掌握这些技能不仅能提升考试成绩，还会加深你对化学量的理解。

1. Moles and Molar Mass | 摩尔与摩尔质量

The mole (mol) is the SI unit for amount of substance. One mole of any substance contains 6.02 × 10²³ formula units — this is Avogadro’s constant, Nₐ. It allows us to count atoms, ions or molecules by weighing.

摩尔（mol）是物质的量的SI单位。一摩尔任何物质含有6.02×10²³个基本单元——这就是阿伏伽德罗常数Nₐ。它使我们能够通过称量来计数原子、离子或分子。

Molar mass (M) is the mass of one mole of a substance, expressed in g/mol. For an element, the molar mass is numerically equal to its relative atomic mass (Aᵣ); for a compound, it equals its relative formula mass (Mᵣ) but with the unit g/mol.

摩尔质量（M）是一摩尔物质的质量，以g/mol表示。对于元素，摩尔质量在数值上等于其相对原子质量（Aᵣ）；对于化合物，则等于其相对式量（Mᵣ），但带有单位g/mol。

Amount (n) = mass (m) / molar mass (M)

Example: Find the number of moles in 10.6 g of sodium carbonate, Na₂CO₃ (M = 106 g/mol). n = 10.6 / 106 = 0.100 mol. This equation is the entry point for virtually all quantitative chemistry.

示例：计算10.6 g碳酸钠Na₂CO₃（M=106 g/mol）的摩尔数。n = 10.6 / 106 = 0.100 mol。这个方程式几乎是所有定量化学的入口。

2. Reacting Masses | 反应质量计算

Once you can calculate moles from a given mass, you can use the balanced equation to find masses of other substances. The coefficients in the equation give the mole ratio between reactants and products.

一旦你能从给定质量计算出摩尔数，就可以利用配平方程式求出其他物质的质量。方程式中的系数给出了反应物和产物之间的摩尔比。

Mass (g) = moles × molar mass (g/mol)

Worked example: What mass of zinc is produced when 8.10 g of zinc oxide is reduced by carbon? Equation: ZnO + C → Zn + CO. Molar masses: ZnO = 81.4 g/mol, Zn = 65.4 g/mol. Step 1: moles ZnO = 8.10 / 81.4 = 0.0995 mol. Step 2: mole ratio 1:1, so moles Zn = 0.0995 mol. Step 3: mass Zn = 0.0995 × 65.4 = 6.51 g (3 significant figures).

例题：用碳还原8.10 g氧化锌能生成多少克锌？方程式：ZnO + C → Zn + CO。摩尔质量：ZnO=81.4 g/mol，Zn=65.4 g/mol。步骤1：ZnO的摩尔数=8.10/81.4=0.0995 mol。步骤2：摩尔比1:1，因此Zn的摩尔数=0.0995 mol。步骤3：Zn的质量=0.0995×65.4=6.51 g（三位有效数字）。

3. Volumes of Gases | 气体体积计算

At room temperature and pressure (RTP, 20 °C and 1 atm), one mole of any gas occupies 24 dm³ (24,000 cm³). This molar gas volume allows you to link moles of a gas to its measured volume directly.

在室温和常压（RTP，20 °C，1 atm）下，一摩尔任何气体占据24 dm³（24000 cm³）的体积。这个摩尔气体体积使你能够直接将气体的摩尔数与其测量体积联系起来。

Volume (dm³) = moles × 24 dm³ mol⁻¹

Example: A reaction produces 0.500 mol of hydrogen gas. What volume will it occupy at RTP? Volume = 0.500 × 24 = 12.0 dm³ (12,000 cm³). Always convert to dm³ if the volume is given in cm³ (divide by 1000).

示例：某反应生成0.500 mol氢气。在RTP下它将占据多大体积？体积=0.500×24=12.0 dm³（12000 cm³）。如果给出的体积是cm³，请务必除以1000转换为dm³。Published by TutorHao | IGCSE Chemistry Revision Series | aleveler.com

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