📚 Chemical Formulae, Equations and Stoichiometry | 化学式、方程式和化学计量学
In IGCSE Science, understanding chemical formulae and equations is crucial for explaining how atoms rearrange during reactions. This article covers the fundamental skills needed to write and balance equations, calculate reacting masses, use the mole concept, and apply stoichiometry to solve real-world chemical problems. Mastering these tools will not only help you in examinations but also deepen your appreciation of the quantitative nature of chemistry.
在IGCSE科学中，理解化学式和方程式对于解释原子在反应中如何重新排列至关重要。本文涵盖了书写和配平方程式、计算反应质量、使用摩尔概念以及应用化学计量学解决实际化学问题所需的基本技能。掌握这些工具不仅能帮助你在考试中取得好成绩，还能加深你对化学定量本质的理解。
1. Writing Chemical Formulae | 书写化学式
Every compound has a specific chemical formula that shows the types and numbers of atoms present. For ionic compounds, the formula is built by balancing the positive and negative charges so that the overall charge is zero. For example, sodium chloride consists of Na⁺ and Cl⁻ ions in a 1:1 ratio, so its formula is NaCl. Magnesium oxide contains Mg²⁺ and O²⁻ ions, giving MgO.
每种化合物都有特定的化学式，显示存在的原子种类和数量。对于离子化合物，化学式是通过平衡正负电荷使总电荷为零来构建的。例如，氯化钠由Na⁺和Cl⁻离子按1:1比例组成，因此其化学式为NaCl。氧化镁含有Mg²⁺和O²⁻离子，化学式为MgO。
When ions have different charges, criss-cross the charge numbers to form the correct ratio. For aluminium oxide, Al³⁺ and O²⁻ combine in a 2:3 ratio, resulting in the formula Al₂O₃. Brackets are used when a polyatomic ion appears more than once: magnesium nitrate is Mg(NO₃)₂ because the nitrate ion NO₃⁻ has a 1− charge and magnesium is 2+, so two nitrate ions are needed.
当离子带有不同电荷时，将电荷数交叉相乘形成正确的比例。对于氧化铝，Al³⁺和O²⁻以2:3的比例结合，化学式为Al₂O₃。当多原子离子出现不止一次时使用括号：硝酸镁的化学式是Mg(NO₃)₂，因为硝酸根离子NO₃⁻带1−电荷，而镁是2+，所以需要两个硝酸根离子。
2. Balancing Chemical Equations | 配平化学方程式
A balanced chemical equation has the same number of atoms of each element on both sides. This reflects the law of conservation of mass. Begin by writing the correct formulae for all reactants and products, then adjust coefficients (the numbers in front) – never change the subscript numbers inside the formulae.
一个配平的化学方程式在两边每种元素的原子数量相同。这反映了质量守恒定律。首先写出所有反应物和生成物的正确化学式，然后调整系数（前面的数字）——永远不要改动化学式内部的下标数字。
al 4 O on the right; place a 2 before O₂. The balanced equation is CH₄ + 2O₂ → CO₂ + 2H₂O.
对于甲烷的燃烧，骨架方程式是CH₄ + O₂ → CO₂ + H₂O。配平时从碳开始：两边各有一个C。接下来是氢：左边有4个H，所以在H₂O前面加上系数2使右边也有4个H。最后是氧：现在CO₂中有2个O，2H₂O中有2×1=2个O，右边共4个O；在O₂前加系数2。配平后的方程式为CH₄ + 2O₂ → CO₂ + 2H₂O。
3. Relative Atomic Mass and Relative Formula Mass | 相对原子质量和相对式量
The relative atomic mass (Aᵣ) of an element is the average mass of its atoms compared to 1/12th the mass of a carbon‑12 atom. It has no units. The relative formula mass (Mᵣ) is the sum of the Aᵣ values of all atoms in a formula unit. For simple molecules, Mᵣ is also called the relative molecular mass.
一种元素的相对原子质量（Aᵣ）是其原子的平均质量与一个碳‑12原子质量的1/12相比的比值，没有单位。相对式量（Mᵣ）是化学式中所有原子的Aᵣ之和。对于简单分子，Mᵣ也称为相对分子质量。
5 + 32 + (9 × 16) + (10 × 1) = 249.5. These Mᵣ values are then used with the mole concept.
计算碳酸钙CaCO₃的Mᵣ：Ca的Aᵣ = 40，C = 12，O = 16。所以Mᵣ = 40 + 12 + (3 × 16) = 100。对于五水合硫酸铜CuSO₄·5H₂O：Cu = 63.5，S = 32，O = 16（CuSO₄中4个 + 水中5个 = 9个O原子），H = 1（×10）。Mᵣ = 63.5 + 32 + (9 × 16) + (10 × 1) = 249.5。这些Mᵣ值随后会与摩尔概念一起使用。
4. The Mole and Avogadro's Constant | 摩尔与阿伏伽德罗常数
× 10²³ particles (atoms, molecules, ions or formula units). This number is called Avogadro' 1 mol of carbon atoms has a mass of 12 g, and 1 mol of water molecules has a mass of 18 g.
一摩尔任何物质恰好含有6.02 × 10²³个粒子（原子、分子、离子或式单元）。这个数字称为阿伏伽德罗常数。一摩尔物质的质量（以克为单位）在数值上等于其Mᵣ。例如，1摩尔碳原子的质量是12克，1摩尔水分子的质量是18克。
tanding this triangle (mass, moles, Mᵣ) allows you to plan experiments and predict yields.
关系式 摩尔数 = 质量 ÷ Mᵣ 是定量化学的基石。如果你有4.0克氢氧化钠（NaOH，Mᵣ = 40），物质的量为4.0 ÷ 40 = 0.10 mol。反过来，0.25摩尔硫酸（H₂SO₄，Mᵣ = 98）的质量为0.25 × 98 = 24.5克。理解这个三角关系（质量、摩尔数、Mᵣ）可以帮助你设计实验和预测产率。
5. Reacting Mass Calculations | 反应质量计算
Using a balanced equation and moles, you can calculate the mass of one reactant or product from the mass of another. Steps: convert the given mass to moles, use the molar ratio from the equation to find moles of the target substance, then convert moles to mass.
利用配平的方程式和摩尔，你可以从一种物质的质量计算出另一种反应物或生成物的质量。步骤：将已知质量转换为摩尔数，利用方程式中的摩尔比求出目标物质的摩尔数，再将摩尔数转换为质量。
What mass of magnesium oxide is produced when 6.0 g of magnesium burns in excess oxygen? so mole ratio is 1:1. Therefore moles of MgO = 0.25 mol. Mass of MgO = 0.25 × 40 = 10.0 g.
当6.0克镁在过量氧气中燃烧，会产生多少克氧化镁？方程式：2Mg + O₂ → 2MgO。Mg的Mᵣ = 24，MgO = 40。Mg的摩尔数 = 6.0 ÷ 24 = 0.25 mol。从方程式看，2 mol Mg生成2 mol MgO，摩尔比为1:1。因此MgO的摩尔数 = 0.25 mol。MgO的质量 = 0.25 × 40 = 10.0 g。
For problems involving a limiting reactant, identify which reactant runs out first by comparing the mole ratio given in the equation with the actual mole ratio available. The amount of product is determined entirely by the limiting reactant.
对于涉及限量反应物的问题，通过比较方程式给出的摩尔比和实际可用的摩尔比来确定哪一种反应物首先耗尽。产物的量完全由限量反应物决定。
6. Concentration in mol/dm³ | 摩尔浓度的计算
mol) ÷ volume (dm³). You can also use the triangle to find moles = concentration × volume.
浓度是衡量一定体积溶液中溶解了多少溶质的量。连接浓度、摩尔数和体积的公式是：浓度（mol/dm³）= 摩尔数（mol） ÷ 体积（dm³）。你也可以利用三角关系求出 摩尔数 = 浓度 × 体积。
If 0.50 mol of sodium chloride is dissolved in 250 cm³ of water, what is the concentration? 50 ÷ 0.250 = 2.0 mol/dm³. This type of calculation is essential for titration experiments.
如果将0.50摩尔氯化钠溶解在250 cm³水中，浓度是多少？体积（dm³）= 250 ÷ 1000 = 0.250 dm³。浓度 = 0.50 ÷ 0.250 = 2.0 mol/dm³。这类计算对于滴定实验至关重要。
on containing 9.8 g/dm³ of H₂SO₄ (Mᵣ = 98) has a concentration of 9.8 ÷ 98 = 0.10 mol/dm³.
你经常还需要将g/dm³为单位的浓度转换为mol/dm³。用质量浓度除以溶质的Mᵣ。例如，含有9.8 g/dm³ H₂SO₄（Mᵣ = 98）的溶液的浓度为9.8 ÷ 98 = 0.10 mol/dm³。
7. Volumes of Gases | 气体的体积
4. This molar volume is remarkably constant for all gases because the sizes of the molecules are negligible compared to the distances between them.
在室温和常压下（r.t.p.，通常为20 °C和1个大气压），一摩尔任何气体占据的体积为24 dm³。这使得气体计算变得简单：体积（dm³）= 摩尔数 × 24。所有气体的摩尔体积惊人地恒定，因为分子本身的尺寸与它们之间的距离相比可以忽略不计。
lume of carbon dioxide is produced when 5.0 g of calcium carbonate is thermally decomposed? mol CaCO₃ gives 1 mol CO₂. So moles CO₂ = 0.050 mol. Volume of CO₂ = 0.050 × 24 = 1.2 dm³.
将5.0克碳酸钙热分解，会产生多少体积的二氧化碳？CaCO₃ → CaO + CO₂。CaCO₃的Mᵣ = 100。CaCO₃的摩尔数 = 5.0 ÷ 100 = 0.050 mol。从方程式看，1 mol CaCO₃产生1 mol CO₂。所以CO₂的摩尔数 = 0.050 mol。CO₂的体积 = 0.050 × 24 = 1.2 dm³。
, 2 volumes of hydrogen react with 1 volume of oxygen to produce 2 volumes of water vapour.
对于只涉及气体的反应，体积比与摩尔比相同。例如，在2H₂ + O₂ → 2H₂O（气态）中，2体积氢气与1体积氧气反应生成2体积水蒸气。
8. Empirical and Molecular Formulae | 实验式和分子式
The empirical formula shows the simplest whole‑number ratio of atoms in a compound. The molecular formula shows the actual number of atoms of each element in a molecule. For many ionic compounds, only the empirical formula exists. To determine the empirical formula, divide the mass or percentage of each element by its Aᵣ, then simplify the ratio to whole numbers.
实验式表示化合物中原子的最简整数比。分子式表示分子中每种原子的实际数量。对于许多离子化合物，只有实验式存在。要确定实验式，将每种元素的质量或百分比除以它的Aᵣ，然后将比例简化为整数。
A compound contains 40.0 % carbon, 6.7 % hydrogen and 53.3 % rical formula mass. If the Mᵣ is 180, n = 180/30 = 6, so the molecular formula is C₆H₁₂O₆.
某化合物按质量含40.0%碳、6.7%氢和53.3%氧。假设100克：质量为40.0 g C、6.7 g H、53.3 g O。摩尔数：C = 40.0/12 = 3.33；H = 6.7/1 = 6.7；O = 53.3/16 = 3.33。除以最小的3.33，得到比例C:H:O = 1:2:1。实验式为CH₂O。要找到分子式，用相对分子质量除以实验式质量。如果Mᵣ是180，n = 180/30 = 6，所以分子式为C₆H₁₂O₆。
9. Water of Crystallisation | 结晶水
Many hydrated salts contain water molecules as part of their crystal structure. The number of water molecules per formula unit is shown after a dot, e.g. CuSO₄·5H₂O. You can determine the value of n in hydrated salt · nH₂O by heating the salt to drive off the water and measuring the mass change.
许多水合盐在晶体结构中含有水分子。每个式单元的水分子数用点后数字表示，例如CuSO₄·5H₂O。你可以通过加热盐驱逐水分并测量质量变化来确定水合盐·nH₂O中的n值。
/18 = 0.111 mol. Divide both by 0.0115 gives ratio 1 : 9.7, which rounds to 10, so n = 10.
将3.22克水合碳酸钠（Na₂CO₃·nH₂O）加热，剩余1.22克无水Na₂CO₃。失去水的质量 = 3.22 − 1.22 = 2.00 g。Na₂CO₃的摩尔数 = 1.22/106 = 0.0115 mol。H₂O的摩尔数 = 2.00/18 = 0.111 mol。两者除以0.0115得到比例1 : 9.7，约等于10，所以n = 10。
10. Percentage Yield and Atom Economy | 产率百分比和原子经济性
eld ÷ theoretical yield) × 100. A high percentage yield indicates an efficient conversion.
在实际化学过程中，由于反应不完全、副反应或产物损失，实际产量往往低于理论产量。产率百分比 =（实际产量 ÷ 理论产量）× 100。高百分比产率表示转化效率高。
ble. For example, the hydration of ethene (C₂H₄ + H₂O → C₂H₅OH) has an atom economy of 100% because all reactant atoms become part of the single product.
原子经济性衡量反应物原子有多少最终进入了目标产物。原子经济性 =（目标产物的Mᵣ ÷ 所有反应物Mᵣ之和）× 100。高原子经济性的反应产生的废物更少，更具可持续性。例如，乙烯的水合反应（C₂H₄ + H₂O → C₂H₅OH）的原子经济性为100%，因为所有反应物原子都成为唯一产物的一部分。
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