IB & OCR Chemistry: Stoichiometry Key Points | IB OCR 化学：化学计量考点精讲

📚 IB & OCR Chemistry: Stoichiometry Key Points | IB OCR 化学：化学计量考点精讲

Chemical stoichiometry is the quantitative backbone of chemistry, essential for both IB and OCR examinations. Understanding the mole concept, balancing equations, and performing calculations involving reacting masses, solutions, and gases is crucial. This article systematically breaks down the key principles, common pitfalls, and exam strategies to help you master stoichiometry.

化学计量是化学的定量基础，对于IB和OCR考试都至关重要。理解摩尔概念、配平化学方程式以及进行涉及反应质量、溶液和气体的计算是必须掌握的。本文系统梳理核心原理、常见误区和应试策略，助你攻克化学计量难关。

1. The Mole and Avogadro’s Constant | 摩尔与阿伏伽德罗常数

The mole is the SI unit for amount of substance. One mole contains exactly 6.022×10²³ elementary entities (Avogadro’s constant, Nₐ). This allows chemists to convert between the microscopic scale of atoms and the macroscopic scale of grams.

摩尔是物质的量的国际单位。1摩尔包含精确的6.022×10²³个基本单元（阿伏伽德罗常数，Nₐ）。化学家借此在原子微观尺度与宏观克数之间建立桥梁。

The relationship is: number of particles (N) = n × Nₐ, where n is the amount in moles. Always ensure your answer matches the context: atoms for elements like Fe, molecules for O₂, and ions for Ca²⁺.

关系式为：粒子数 (N) = n × Nₐ，其中n为摩尔数。务必根据语境明确粒子种类：元素如Fe用原子，O₂用分子，Ca²⁺用离子。

2. Molar Mass and Formula Mass | 摩尔质量与化学式量

Molar mass (M) is the mass of one mole of a substance, expressed in g mol⁻¹. It is numerically equal to the relative atomic mass (Aᵣ) or relative formula mass (Mᵣ) of the compound. For example, M(NaCl) = 23.0 + 35.5 = 58.5 g mol⁻¹.

摩尔质量 (M) 是一摩尔物质的质量，单位为g mol⁻¹。数值上等于该物质的相对原子质量（Aᵣ）或相对化学式量（Mᵣ）。例如，M(NaCl) = 23.0 + 35.5 = 58.5 g mol⁻¹。

To find the mass of a given amount: m = n × M. This formula is foundational for all stoichiometric calculations. Pay close attention to diatomic molecules: M(O₂) = 2 × 16.0 = 32.0 g mol⁻¹.

已知物质的量求质量：m = n × M。此公式是所有化学计量计算的基础。特别注意双原子分子：M(O₂) = 2 × 16.0 = 32.0 g mol⁻¹。

m = n × M

Substance	Formula mass (Mᵣ)	Molar mass (g mol⁻¹)
Water	18.0	18.0
Sulfuric acid	98.1	98.1
Calcium carbonate	100.1	100.1

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

A balanced equation obeys the law of conservation of mass: the number of atoms of each element is the same on both sides. Use whole number coefficients; never change the subscripts in a formula. Start with elements that appear in only one reactant and product.

配平的化学方程式遵循质量守恒定律：每种元素的原子数目在反应前后相等。使用最小整数系数，绝不可更改化学式中的下标。优先从只在一种反应物和生成物中出现的元素开始配平。

Example: combustion of propane: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. Check: C: 3→3; H: 8→8; O: 10→10. A common mistake with oxygen is forgetting it is O₂ when counting atoms. Always verify after balancing.

示例：丙烷燃烧：C₃H₈ + 5O₂ → 3CO₂ + 4H₂O。检查：C: 3→3; H: 8→8; O: 10→10。常见错误是计数氧原子时忘记它是O₂。配平后务必验证。

4. Stoichiometric Calculations (Mass-Mass, Mass-Volume) | 化学计量计算（质量-质量、质量-体积）

Mass-mass problems: Given the mass of one substance, calculate the mass of another using mole ratios from the balanced equation. Step 1: convert given mass to moles (n = m/M). Step 2: use mole ratio from coefficients. Step 3: convert moles of unknown back to mass.

质量-质量计算：已知一种物质的质量，利用配平方程式中的摩尔比求另一物质的质量。步骤1：将已知质量转换为摩尔 (n = m/M)；步骤2：利用系数比确定摩尔数；步骤3：将所求摩尔数转换为质量。

Mass-volume at RTP (room temperature and pressure) or STP: 1 mol of gas occupies 24 dm³ at RTP (298 K, 1 atm) or 22.7 dm³ at STP (273 K, 100 kPa) for IB. OCR uses 24 dm³ at RTP. Use V = n × 24 or V = n × 22.7 after moles are found.

室温常压或标准状况下的质量-体积计算：IB中，1 mol气体在RTP (298 K, 1 atm) 下体积为24 dm³，或在STP (273 K, 100 kPa) 下为22.7 dm³；OCR使用RTP下24 dm³。求出摩尔数后利用V = n × 24或V = n × 22.7计算体积。

n = mass / M and V(gas) = n × 24 dm³ (RTP)

5. Limiting and Excess Reagents | 限量试剂与过量试剂

In many reactions, one reactant is used up before others, limiting the amount of product formed. The limiting reagent is the one that produces the least moles of product, as determined by dividing initial moles by stoichiometric coefficient. The other reactants are in excess.

许多反应中，一种反应物会先被耗尽，从而限制产物生成量。将各反应物初始摩尔除以其化学计量系数，商值最小者即为限量试剂，其余则为过量试剂。

Example: 2H₂ + O₂ → 2H₂O. If 5 mol H₂ and 3 mol O₂ are present, H₂: 5/2 = 2.5; O₂: 3/1 = 3. Thus H₂ is limiting. All calculations must be based on the limiting reagent; do not use the excess amount.

例：2H₂ + O₂ → 2H₂O。若现有5 mol H₂和3 mol O₂，H₂: 5/2 = 2.5；O₂: 3/1 = 3。因此H₂为限量试剂。所有后续计算必须基于限量试剂，切勿使用过量反应物的量。

6. Percentage Yield and Atom Economy | 产率百分数与原子经济性

Percentage yield compares the actual mass of product to the theoretical maximum, reflecting losses during practical work. Atom economy measures how much of the total mass of reactants ends up in the desired product, assessing the ‘greenness’ of a reaction.

产率百分数对比实际产物质量与理论最大产量，体现实验过程中的损失。原子经济性衡量所有反应物总质量中有多少进入了目标产物，用于评估反应的 “绿色性”。

% Yield = (actual mass / theoretical mass) × 100

% Atom Economy = (mass of desired product / total mass of reactants) × 100

IB and OCR both emphasize these concepts for sustainable chemistry. High atom economy processes minimise waste. In exam questions, identify the limiting reagent first to find theoretical yield.

IB与OCR均强调这些概念在可持续化学中的意义。高原子经济性过程可减少废物。考题中应先确定限量试剂以求得理论产量。

7. Empirical and Molecular Formulae | 经验式与分子式

The empirical formula gives the simplest whole number ratio of atoms in a compound. The molecular formula is a multiple of the empirical formula. Combustion data or percentage composition can be used to deduce the empirical formula: assume 100 g, convert mass to moles, then divide by smallest mole value.

经验式表示化合物中各原子的最简整数比。分子式则是经验式的整数倍。通过燃烧数据或元素百分组成可推导经验式：假设样品100 g，将各组分质量转为摩尔数，然后除以最小摩尔数。

Molecular formula is found by comparing the empirical formula mass with the molar mass: n = Mᵣ (given) / empirical formula mass. Then multiply the empirical formula subscripts by n. Example: empirical formula CH₂, molar mass 56 g mol⁻¹ → n = 56/14 = 4, so molecular formula C₄H₈.

分子式的确定：经验式量除以摩尔质量得到倍数n = Mᵣ（已知）/ 经验式量，再将经验式下标乘以n。例如经验式CH₂，摩尔质量56 g mol⁻¹ → n = 56/14 = 4，分子式为C₄H₈。

8. Solution Stoichiometry and Concentration | 溶液计量与浓度

Concentration is commonly expressed as mol dm⁻³ (molarity). The key formula is: c = n / V, where V is in dm³. For titrations, at the equivalence point the reacting moles are related by the balanced equation. Always remember to convert cm³ to dm³ by dividing by 1000.

浓度通常以 mol dm⁻³（体积摩尔浓度）表示。核心公式为 c = n / V，V 单位为 dm³。滴定分析中，等量点处反应物的摩尔数经由配平方程式关联。切记将 cm³ 转换为 dm³（除以1000）。

n = c × V (volume in dm³)

In IB, standard solution preparation and back titration are common. OCR often includes acid-base and redox titrations. Both require careful volume readings and use of the mole ratio.

IB中常见标准溶液配制和返滴定。OCR常涉及酸碱滴定和氧化还原滴定。两者均需精确读数并正确使用摩尔比。

9. Gas Stoichiometry and Molar Volume | 气体计量与摩尔体积

Reactions involving gases can be solved using reacting volumes directly if conditions are constant, because volume ratio equals mole ratio (Avogadro’s law). For example, N₂ + 3H₂ → 2NH₃: 1 dm³ N₂ reacts with 3 dm³ H₂ to give 2 dm³ NH₃.

若反应条件恒定，涉及气体的反应可直接通过体积比求解，因为体积比等于摩尔比（阿伏伽德罗定律）。例如 N₂ + 3H₂ → 2NH₃：1 dm³ N₂ 与 3 dm³ H₂ 反应生成 2 dm³ NH₃。

When conditions differ, use the ideal gas equation PV = nRT, where R = 8.31 J K⁻¹ mol⁻¹, T in Kelvin, P in Pa, and V in m³. For OCR, this is often optional, but IB requires use of the ideal gas equation in various scenarios.

条件变化时使用理想气体状态方程 PV = nRT，其中 R = 8.31 J K⁻¹ mol⁻¹，T 为开尔文温度，P 单位为 Pa，V 单位为 m³。OCR 对此通常不强制要求，但 IB 需要灵活应用此方程。

10. Back Titrations and Advanced Calculations | 返滴定与进阶计算

Back titration is used when direct titration is not feasible, e.g., with volatile substances or insoluble solids. A known excess of reagent is added, reaction allowed to proceed, and the remaining excess is titrated. Subtract to find moles reacted with the sample.

当直接滴定不可行（如分析挥发性物质或不溶性固体）时使用返滴定。加入已知过量的一种试剂，反应完全后滴定剩余过量部分，差值即为与样品反应的摩尔数。

Example: determination of calcium carbonate in limestone. Add excess HCl, then back-titrate with NaOH. Moles of CaCO₃ = (initial moles HCl – moles NaOH used) / 2, based on 2HCl + CaCO₃ → CaCl₂ + CO₂ + H₂O.

示例：测定石灰石中碳酸钙含量。加入过量 HCl，然后用 NaOH 返滴定。根据反应 2HCl + CaCO₃ → CaCl₂ + CO₂ + H₂O，CaCO₃ 摩尔数 = (初始 HCl 摩尔数 – NaOH 用去的摩尔数) / 2。

11. Common Exam Pitfalls and How to Avoid Them | 常见考试误区与应对

Pitfall 1: Forgetting to convert units (cm³ to dm³, °C to K). Always write units in every step and check consistency.

误区1：忘记单位换算（cm³→dm³，°C→K）。每一步都标注单位并检查一致性。

Pitfall 2: Using the wrong mole ratio by misreading the balanced equation. Highlight the coefficients or draw brackets. Double-check which substance is the unknown.

误区2：因误读配平方程式而用错摩尔比。用高亮或括号标出系数，并再次确认未知物是哪个。

Pitfall 3: Applying the mass conservation in open systems. Some mass might appear to decrease if gas escapes. Consider the system thoroughly in percentage yield questions.

误区3：在开放体系中简单套用质量守恒。若气体逸出，表观质量会减少。在产率计算题中需全面分析体系。

Pitfall 4: Rounding prematurely. Keep intermediate values in calculator memory and round only the final answer to the correct number of significant figures, matching the data given.

误区4：过早取舍。保留计算中间值，仅在最终答案根据给定数据的有效数字进行合理修约。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

IB & OCR Chemistry: Stoichiometry Key Points | IB OCR 化学：化学计量 考点精讲