IGCSE CIE Chemistry Stoichiometry Key Points | IGCSE CIE 化学：化学计量考点精讲

📚 IGCSE CIE Chemistry Stoichiometry Key Points | IGCSE CIE 化学：化学计量考点精讲

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It uses the mole concept to make predictions about mass, volume, and particle numbers, linking laboratory measurements to chemical equations. Mastering stoichiometry is essential for tackling calculations in the IGCSE CIE Chemistry exam, from balancing equations to determining limiting reactants and yields.

化学计量是化学中研究化学反应中反应物和生成物之间定量关系的一个分支。它利用摩尔概念，根据质量、体积和粒子数进行预测，把实验室测量值与化学方程式联系起来。掌握化学计量对于应对 IGCSE CIE 化学考试中的计算题至关重要，从配平方程式到确定限量试剂和产率，都离不开这一核心技能。

1. What is Stoichiometry? | 什么是化学计量？

Stoichiometry comes from the Greek words ‘stoicheion’ (element) and ‘metron’ (measure). In chemistry, it allows us to calculate the exact amounts of substances consumed and produced in a reaction. The heart of stoichiometry is the balanced chemical equation, which gives the mole ratio of reactants and products.

化学计量一词源于希腊语“元素”和“度量”。在化学中，它使我们能够准确计算反应中消耗和生成的物质的质量。化学计量的核心是配平的化学方程式，它给出反应物和生成物之间的摩尔比。

All stoichiometric calculations are based on the law of conservation of mass: matter is neither created nor destroyed in a chemical reaction. Therefore, the total mass of reactants equals the total mass of products, and the numbers of each type of atom are the same on both sides of the equation.

所有化学计量计算都基于质量守恒定律：化学反应中物质既不能被创造也不能被消灭。因此，反应物的总质量等于生成物的总质量，并且方程式两边各类原子的数目相同。

2. Relative Atomic Mass (Ar) and Relative Molecular Mass (Mr) | 相对原子质量 (Ar) 和相对分子质量 (Mr)

The relative atomic mass (Ar) of an element is the average mass of one atom of the element compared to 1/12 of the mass of one carbon‑12 atom. It has no units. For example, Ar of hydrogen is 1, carbon is 12, oxygen is 16, and chlorine is 35.5.

元素的相对原子质量 (Ar) 是该元素一个原子的平均质量与一个碳‑12 原子质量的 1/12 相比较得到的数值，没有单位。例如，氢的 Ar 为 1，碳为 12，氧为 16，氯为 35.5。

The relative molecular mass (Mr) is the sum of the relative atomic masses of all the atoms in a molecule. For ionic compounds, we use the term relative formula mass, but it is calculated in the same way. For example, Mr of H₂O = 2×1 + 16 = 18; Mr of CaCO₃ = 40 + 12 + 3×16 = 100.

相对分子质量 (Mr) 是一个分子中所有原子的相对原子质量之和。对于离子化合物，我们使用“相对式量”这个术语，但计算方法相同。例如，H₂O 的 Mr = 2×1 + 16 = 18；CaCO₃ 的 Mr = 40 + 12 + 3×16 = 100。

In the CIE exam, you must be able to calculate Mr from given Ar values and use these values to convert between mass and moles.

在 CIE 考试中，你必须能够根据给出的 Ar 值计算 Mr，并用这些值在质量和摩尔数之间进行转换。

3. The Mole and Avogadro’s Constant | 摩尔和阿伏伽德罗常数

One mole of any substance contains exactly 6.02 × 10²³ particles (atoms, molecules, ions, or electrons). This number is called Avogadro’s constant. The mole is the SI unit for amount of substance.

一摩尔任何物质都恰好含有 6.02 × 10²³ 个粒子（原子、分子、离子或电子）。这个数字被称为阿伏伽德罗常数。摩尔是物质的量的国际单位。

The mass of one mole of a substance is its molar mass, expressed in grams per mole (g mol⁻¹). Numerically, the molar mass is equal to the relative atomic mass (Ar) or relative molecular mass (Mr) of the substance.

一摩尔物质的质量就是它的摩尔质量，以克每摩尔（g mol⁻¹）表示。在数值上，摩尔质量等于该物质的相对原子质量 (Ar) 或相对分子质量 (Mr)。

Example: 1 mol of carbon atoms has a mass of 12 g; 1 mol of H₂O molecules has a mass of 18 g; 1 mol of NaCl formula units has a mass of 58.5 g.

例子：1 mol 碳原子的质量为 12 g；1 mol H₂O 分子的质量为 18 g；1 mol NaCl 化学式单元的质量为 58.5 g。

4. Molar Mass and Molar Gas Volume | 摩尔质量和摩尔气体体积

Molar mass (M) links mass and moles through the formula:

number of moles = mass (g) ÷ molar mass (g mol⁻¹)

摩尔质量 (M) 通过以下公式将质量和摩尔数联系起来：

摩尔数 = 质量 (g) ÷ 摩尔质量 (g mol⁻¹)

For gases, CIE specifies that at room temperature and pressure (r.t.p., 20 °C and 1 atm), one mole of any gas occupies a volume of 24 dm³ (24 000 cm³). This is the molar gas volume. The formula is:

number of moles of gas = volume (dm³) ÷ 24 dm³ mol⁻¹

对于气体，CIE 规定在常温常压下（r.t.p.，20 °C 和 1 atm），一摩尔任何气体所占的体积为 24 dm³（24 000 cm³）。这就是摩尔气体体积。其公式为：

气体的摩尔数 = 体积 (dm³) ÷ 24 dm³ mol⁻¹

Always check whether the volume given is in cm³ or dm³. If the volume is in cm³, divide by 24 000 cm³ mol⁻¹.

务必检查给出的体积单位是 cm³ 还是 dm³。如果体积单位是 cm³，则除以 24 000 cm³ mol⁻¹。

5. Calculating Moles from Mass and Volume | 从质量和体积计算摩尔数

These two relationships are the backbone of stoichiometry. Learn to apply them interchangeably:

这两个关系式是化学计量的支柱。要学会灵活运用它们：

Known quantity	Formula to find moles	已知量	求摩尔数的公式
Mass (g)	moles = mass / molar mass	质量 (g)	摩尔数 = 质量 / 摩尔质量
Gas volume (dm³ at r.t.p.)	moles = volume / 24	气体体积 (dm³, 在 r.t.p.)	摩尔数 = 体积 / 24
Number of particles	moles = number of particles / (6.02×10²³)	粒子数	摩尔数 = 粒子数 / (6.02×10²³)
Concentration & volume (aq)	moles = concentration (mol dm⁻³) × volume (dm³)	浓度和体积 (溶液)	摩尔数 = 浓度 (mol dm⁻³) × 体积 (dm³)

Always use the balanced equation to relate the moles of one substance to another. The mole ratio comes from the coefficients in front of each formula.

始终利用配平的方程式将一种物质的摩尔数与另一种物质联系起来。摩尔比来自每种化学式前面的系数。

6. Empirical and Molecular Formulae | 经验式和分子式

The empirical formula is the simplest whole‑number ratio of atoms of each element in a compound. The molecular formula shows the actual number of atoms of each element in a molecule.

经验式是化合物中各元素原子的最简整数比。分子式则表示一个分子中各元素原子的实际数目。

To find the empirical formula from masses or percentages:

Divide the mass (or percentage) of each element by its Ar to get moles.
Divide each mole value by the smallest number of moles to obtain the simplest ratio.
If necessary, multiply to get whole numbers.

从质量或百分含量求经验式的步骤：

用每种元素的质量（或百分比）除以它的 Ar，得到摩尔数。
将每个摩尔数值除以最小的摩尔数，得到最简比。
如果需要，乘以整数得到最简整数比。

Example: A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Moles: C = 40/12 = 3.33, H = 6.7/1 = 6.7, O = 53.3/16 = 3.33. Divide by 3.33 gives ratio C:H:O = 1:2:1. Empirical formula = CH₂O.

例子：某化合物含有 40.0% 碳、6.7% 氢和 53.3% 氧。摩尔数：C = 40/12 = 3.33，H = 6.7/1 = 6.7，O = 53.3/16 = 3.33。除以 3.33 得到比例 C:H:O = 1:2:1。经验式 = CH₂O。

To determine the molecular formula, you need the relative molecular mass (Mr). Divide the Mr by the empirical formula mass to find the multiplier, then multiply the empirical formula by this number.

要确定分子式，需要知道相对分子质量 (Mr)。将 Mr 除以经验式的式量得到倍数，然后将经验式乘以该数值即可。

7. Balancing Chemical Equations | 配平化学方程式

A balanced chemical equation has the same number of each type of atom on both sides. Balancing is done by adjusting coefficients, never by changing subscripts in chemical formulas.

配平的化学方程式两边每种原子的数目相同。配平时只能调整系数，绝对不能更改化学式中的下标。

Steps to balance:

Write the unbalanced equation with correct formulas.
Count the atoms of each element on both sides.
Add coefficients to balance one element at a time, starting with elements that appear in only one reactant and one product.
Balance hydrogen and oxygen last if present.
Check that all coefficients are in the smallest whole‑number ratio.

配平步骤：

写出正确化学式的未配平方程式。
统计两边每种元素的原子数。
添加系数，一次配平一种元素，从只出现在一种反应物和一种生成物中的元素开始。
如果有氢和氧，最后再配平它们。
检查所有系数是否处于最简整数比。

Example: Fe₂O₃ + CO → Fe + CO₂. Unbalanced. Balance Fe: 1 Fe₂O₃ gives 2 Fe. Balance C and O: Fe₂O₃ + 3CO → 2Fe + 3CO₂. Now Fe: 2=2, C: 3=3, O: 3+3=6 on left, 3×2=6 on right. Balanced.

例子：Fe₂O₃ + CO → Fe + CO₂。未配平。先配 Fe：1 Fe₂O₃ 生成 2 Fe。再配 C 和 O：Fe₂O₃ + 3CO → 2Fe + 3CO₂。此时 Fe: 2=2，C: 3=3，O: 左边 3+3=6，右边 3×2=6。已配平。

State symbols are often required in CIE: (s) solid, (l) liquid, (g) gas, (aq) aqueous solution.

CIE 经常要求注明状态符号：(s) 固体，(l) 液体，(g) 气体，(aq) 水溶液。

8. Reacting Mass Calculations | 反应质量计算

Once the equation is balanced, you can calculate the mass of a reactant needed or the mass of a product formed.

一旦方程式配平，你就可以计算所需反应物的质量或生成的产物的质量。

Standard method:

Write the balanced equation.
Convert the given mass to moles (using mass / molar mass).
Use the mole ratio from the equation to find moles of the unknown substance.
Convert moles of the unknown back to mass (mass = moles × molar mass).

标准方法：

写出配平的方程式。
将已知质量转换为摩尔数（质量 / 摩尔质量）。
利用方程式中的摩尔比求出未知物质的摩尔数。
将未知物质的摩尔数换算回质量（质量 = 摩尔数 × 摩尔质量）。

Example: What mass of magnesium oxide (MgO) is formed when 6.0 g of magnesium burns completely in oxygen? 2Mg + O₂ → 2MgO. Moles of Mg = 6.0 / 24 = 0.25 mol. Mole ratio Mg:MgO = 2:2 = 1:1, so moles of MgO = 0.25. Mass of MgO = 0.25 × (24+16) = 0.25 × 40 = 10.0 g.

例子：6.0 g 镁在氧气中完全燃烧，生成多少克氧化镁 (MgO)？2Mg + O₂ → 2MgO。Mg 的摩尔数 = 6.0 / 24 = 0.25 mol。摩尔比 Mg:MgO = 2:2 = 1:1，因此 MgO 的摩尔数 = 0.25。MgO 的质量 = 0.25 × (24+16) = 0.25 × 40 = 10.0 g。

9. Limiting Reactant | 限量试剂

In many reactions, one reactant is used up completely before the others. This is the limiting reactant; it determines the maximum amount of product that can be formed. The other reactants are in excess.

在许多反应中，一种反应物会先于其他反应物完全消耗。这就是限量试剂；它决定了能生成的产物的最大量。其他反应物则处于过量状态。

To identify the limiting reactant, calculate the moles of each reactant. Then use the balanced equation to see which one gives the smallest amount of product. That reactant is limiting.

要识别限量试剂，计算每种反应物的摩尔数。然后利用配平方程式，看哪一种反应物生成的产物量最少。该反应物即为限量试剂。

Example: 4.8 g of magnesium is added to 7.3 g of hydrochloric acid: Mg + 2HCl → MgCl₂ + H₂. Moles of Mg = 4.8/24 = 0.20 mol. Moles of HCl = 7.3/36.5 = 0.20 mol. The equation requires 2 mol HCl per 1 mol Mg. 0.20 mol Mg needs 0.40 mol HCl, but only 0.20 mol HCl is available. Therefore HCl is the limiting reactant.

例子：将 4.8 g 镁加入 7.3 g 盐酸中：Mg + 2HCl → MgCl₂ + H₂。Mg 的摩尔数 = 4.8/24 = 0.20 mol。HCl 的摩尔数 = 7.3/36.5 = 0.20 mol。方程式要求每 1 mol Mg 需 2 mol HCl。0.20 mol Mg 需要 0.40 mol HCl，但可用的 HCl 只有 0.20 mol。因此 HCl 是限量试剂。

All further calculations, such as mass of product, must be based on the moles of the limiting reactant.

所有后续计算，如产物的质量，都必须基于限量试剂的摩尔数。

10. Percentage Yield and Percentage Purity | 产率百分比和纯度百分比

The percentage yield compares the actual mass of product obtained in an experiment (actual yield) to the maximum theoretical mass (theoretical yield) calculated from the limiting reactant.

percentage yield = (actual yield / theoretical yield) × 100%

产率百分比将实验获得的实际产物质量（实际产率）与由限量试剂算出的最大理论质量（理论产率）进行比较。

产率百分比 = (实际产率 / 理论产率) × 100%

Yields are often below 100% due to incomplete reactions, side reactions, loss during purification, or reversible reactions. In CIE calculations, always use the balanced equation to find theoretical yield.

产率常低于 100%，原因包括反应不完全、副反应、提纯过程中的损失或可逆反应。在 CIE 计算中，一定要用配平方程式求理论产率。

Percentage purity is used when the reactant is not pure:

percentage purity = (mass of pure substance / mass of impure sample) × 100%

当反应物不纯时，使用纯度百分比：

纯度百分比 = (纯物质质量 / 不纯样品质量) × 100%

First calculate the mass of pure reactant using moles, then find its percentage in the original sample.

先用摩尔数计算纯反应物的质量，再求出它在原始样品中的百分比。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

IGCSE CIE Chemistry Stoichiometry Key Points | IGCSE CIE 化学：化学计量 考点精讲