Stoichiometric Relationships: Fundamental Concepts & Calculations | 化学计量关系：基本概念与计算

📚 Stoichiometric Relationships: Fundamental Concepts & Calculations | 化学计量关系：基本概念与计算

Stoichiometry is the mathematical foundation of chemistry. It allows us to predict how much product can be formed from given reactants, to identify limiting reagents, and to determine the composition of substances through quantitative analysis. Mastering these relationships is essential for success in IB and CIE examinations.

化学计量学是化学的数学基础。它使我们能够预测在给定反应物下能生成多少产物、识别限量试剂，并通过定量分析确定物质的组成。掌握这些关系是通过 IB 和 CIE 考试的关键。

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1. The Mole and Avogadro’s Constant | 物质的量与阿伏伽德罗常数

The mole (mol) is the SI unit for the amount of substance. One mole of any entity contains exactly 6.02 × 10²³ particles — this number is Avogadro’s constant (L or Nₐ). These particles may be atoms, molecules, ions, or electrons.

摩尔（mol）是物质的量的国际单位。任何物质的 1 摩尔都恰好包含 6.02 × 10²³ 个微粒——这个数值就是阿伏伽德罗常数（L 或 Nₐ）。这些微粒可以是原子、分子、离子或电子。

n = N ÷ Nₐ

where n = amount in mol, N = number of particles, Nₐ = 6.02 × 10²³ mol⁻¹.

其中 n 为物质的量（单位 mol），N 为微粒数目，Nₐ = 6.02 × 10²³ mol⁻¹。

• 1 mol of carbon atoms contains 6.02 × 10²³ C atoms | 1 摩尔碳原子含有 6.02 × 10²³ 个碳原子
• 1 mol of water molecules contains 6.02 × 10²³ H₂O molecules | 1 摩尔水分子含有 6.02 × 10²³ 个水分子
• 1 mol of sodium ions contains 6.02 × 10²³ Na⁺ ions | 1 摩尔钠离子含有 6.02 × 10²³ 个 Na⁺ 离子

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2. Molar Mass and Relative Atomic Mass | 摩尔质量与相对原子质量

The molar mass (M) is the mass of one mole of a substance, expressed in g mol⁻¹. Numerically, it equals the relative atomic mass (Aᵣ) or relative molecular mass (Mᵣ) of the substance. For example, the molar mass of carbon-12 is exactly 12 g mol⁻¹.

摩尔质量（M）是 1 摩尔物质的质量，单位为 g mol⁻¹。其数值等于该物质的相对原子质量（Aᵣ）或相对分子质量（Mᵣ）。例如，碳-12 的摩尔质量正好是 12 g mol⁻¹。

n = m ÷ M

where m = mass in grams, M = molar mass in g mol⁻¹.

其中 m 为质量（单位 g），M 为摩尔质量（单位 g mol⁻¹）。

Substance | 物质	M / g mol⁻¹ | 摩尔质量
O₂ (oxygen gas) | 氧气	2 × 16.00 = 32.00
CO₂ (carbon dioxide) | 二氧化碳	12.01 + 2 × 16.00 = 44.01
H₂SO₄ (sulfuric acid) | 硫酸	2 × 1.01 + 32.07 + 4 × 16.00 = 98.09

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3. Empirical and Molecular Formulas | 实验式与分子式

The empirical formula gives the simplest whole-number ratio of atoms of each element in a compound. The molecular formula gives the actual number of atoms of each element in one molecule. They are related by the integer multiple n:

实验式表示化合物中各元素原子数的最简整数比。分子式表示一个分子中各元素原子的实际数目。两者通过整数倍 n 相关联：

molecular formula = (empirical formula)ₙ

分子式 = (实验式)ₙ

To find the empirical formula: convert percentage or mass data to moles, divide by the smallest mole value, and convert to whole numbers.

求实验式的方法：将百分比或质量数据换算成物质的量，除以最小的物质的量数值，再化简为整数比。

• Example: A compound contains 40.0% C, 6.7% H, and 53.3% O by mass. | 例：某化合物含 C 40.0%、H 6.7%、O 53.3%（质量分数）。
• C: 40.0 ÷ 12.01 = 3.33 mol; H: 6.7 ÷ 1.01 = 6.63 mol; O: 53.3 ÷ 16.00 = 3.33 mol | n(C) = 3.33 mol；n(H) = 6.63 mol；n(O) = 3.33 mol
• Divide by 3.33 → C₁H₂O₁, so empirical formula = CH₂O | 除以 3.33 → C₁H₂O₁，实验式为 CH₂O
• If Mᵣ = 180, then n = 180 ÷ 30 = 6, so molecular formula = C₆H₁₂O₆ | 若 Mᵣ = 180，则 n = 180 ÷ 30 = 6，分子式为 C₆H₁₂O₆

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

A balanced chemical equation shows the same number of atoms of each element on both sides. Balancing ensures that the law of conservation of mass is respected. Coefficients represent the relative amounts of substances in moles.

配平的化学方程式表示反应前后各元素原子数目相等。配平保证了质量守恒定律成立。系数代表各物质之间相对的物质的量比例。

CH₄ + 2O₂ → CO₂ + 2H₂O

CH₄ + 2O₂ → CO₂ + 2H₂O

• Left side: 1 C, 4 H, 4 O | 左侧：1 C、4 H、4 O
• Right side: 1 C, 4 H, 4 O | 右侧：1 C、4 H、4 O
• Mole ratio: CH₄ : O₂ : CO₂ : H₂O = 1 : 2 : 1 : 2 | 物质的量之比：CH₄ : O₂ : CO₂ : H₂O = 1 : 2 : 1 : 2

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5. Mass Calculations from Balanced Equations | 由化学方程式进行质量计算

The mole ratio from a balanced equation is used as a conversion factor. The general workflow is: mass → moles → molar ratio → target moles → target mass.

化学方程式中的物质的量之比可作为换算因子。一般流程为：质量 → 物质的量 → 摩尔比 → 目标物质的量 → 目标质量。

2H₂ + O₂ → 2H₂O

2H₂ + O₂ → 2H₂O

How many grams of water are produced from 4.0 g of hydrogen? n(H₂) = 4.0 ÷ 2.02 = 1.98 mol. Mole ratio H₂ : H₂O = 1 : 1, so n(H₂O) = 1.98 mol. Mass of H₂O = 1.98 × 18.02 = 35.7 g.

4.0 g 氢气能生成多少克水？n(H₂) = 4.0 ÷ 2.02 = 1.98 mol。H₂ : H₂O 的摩尔比为 1 : 1，因此 n(H₂O) = 1.98 mol。水的质量 = 1.98 × 18.02 = 35.7 g。

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6. Limiting Reactant and Excess Reactant | 限量反应物与过量反应物

The limiting reactant is the substance that is completely consumed first in a chemical reaction. It determines the maximum amount of product that can be formed. All other reactants are present in excess.

限量反应物是在化学反应中最先被完全消耗的物质。它决定了能生成产物的最大量。其他反应物则为过量。

N₂ + 3H₂ → 2NH₃

N₂ + 3H₂ → 2NH₃

If 1.0 mol N₂ is mixed with 3.0 mol H₂, the ratio exactly matches 1 : 3, so neither is limiting. If 1.0 mol N₂ is mixed with 2.0 mol H₂, H₂ is limiting because 3 mol H₂ would be required for all N₂ to react. The maximum NH₃ is 2 × 2.0 ÷ 3 = 1.33 mol.

若将 1.0 mol N₂ 与 3.0 mol H₂ 混合，物质的量之比恰为 1 : 3，两者均不是限量反应物。若将 1.0 mol N₂ 与 2.0 mol H₂ 混合，则 H₂ 为限量反应物，因为要使全部 N₂ 反应需要 3 mol H₂。NH₃ 的最大生成量为 2 × 2.0 ÷ 3 = 1.33 mol。

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7. Theoretical, Actual, and Percent Yield | 理论产量、实际产量与产率

Theoretical yield is the maximum mass of product calculated from the balanced equation, assuming complete reaction. Actual yield is the mass obtained experimentally. Percent yield compares the two.

理论产量是基于配平方程式、假设反应完全进行时计算出的最大产物质量。实际产量是实验中实际得到的质量。百分产率比较两者。

percent yield = (actual yield ÷ theoretical yield) × 100%

百分产率 = (实际产量 ÷ 理论产量) × 100%

• Theoretical yield is determined by the limiting reactant | 理论产量由限量反应物决定
• Actual yield is always less than theoretical yield due to incomplete reactions, side reactions, or loss during separation | 由于反应不完全、副反应或分离过程中的损失，实际产量总是小于理论产量
• Percent yield can be greater than 100% only if the product is impure or contaminated | 只有在产物不纯或被污染时，百分产率才会大于 100%

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8. Concentration and Solution Stoichiometry | 浓度与溶液化学计量

Concentration is the amount of solute dissolved in a given volume of solution. The most common unit in IB and CIE is mol dm⁻³, often written as M.

浓度是单位体积溶液中所含溶质的量。IB 和 CIE 中最常用的单位是 mol dm⁻³，通常写作 M。

c = n ÷ V

where c = concentration in mol dm⁻³, n = amount in mol, V = volume in dm³.

其中 c 为浓度（单位 mol dm⁻³），n 为物质的量（单位 mol），V 为体积（单位 dm³）。

For titration calculations, use the relationship:

对于滴定计算，使用以下关系：

cₐVₐ ÷ c_bV_b = a ÷ b

where a and b are the stoichiometric coefficients in the balanced equation.

其中 a 和 b 是配平方程式中的化学计量系数。

Example: 25.0 cm³ of 0.100 mol dm⁻³ HCl neutralises 20.0 cm³ of NaOH. Find the concentration of NaOH. HCl: n = 0.100 × 0.0250 = 0.00250 mol. Since HCl : NaOH = 1 : 1, n(NaOH) = 0.00250 mol. c(NaOH) = 0.00250 ÷ 0.0200 = 0.125 mol dm⁻³.

例：25.0 cm³ 的 0.100 mol dm⁻³ HCl 恰好中和 20.0 cm³ 的 NaOH。求 NaOH 的浓度。HCl：n = 0.100 × 0.0250 = 0.00250 mol。由于 HCl : NaOH = 1 : 1，n(NaOH) = 0.00250 mol。c(NaOH) = 0.00250 ÷ 0.0200 = 0.125 mol dm⁻³。

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9. Gas Stoichiometry and Molar Volume | 气体化学计量与摩尔体积

At standard temperature and pressure (STP: 0 °C, 1 atm), one mole of any ideal gas occupies 22.4 dm³. At room temperature and pressure (RTP: 25 °C, 1 atm), the molar volume is approximately 24.0 dm³ mol⁻¹.

在标准状况（STP：0 °C，1 atm）下，1 摩尔任何理想气体占 22.4 dm³。在室温常压（RTP：25 °C，1 atm）下，摩尔体积约为 24.0 dm³ mol⁻¹。

n = V ÷ Vₘ

where Vₘ = 22.4 dm³ mol⁻¹ at STP or 24.0 dm³ mol⁻¹ at RTP.

其中 Vₘ 在 STP 下为 22.4 dm³ mol⁻¹，在 RTP 下为 24.0 dm³ mol⁻¹。

Example: What volume of CO₂ at RTP is produced when 10.0 g of CaCO₃ decomposes completely? CaCO₃ → CaO + CO₂. n(CaCO₃) = 10.0 ÷ 100.09 = 0.0999 mol. Mole ratio 1 : 1, so n(CO₂) = 0.0999 mol. Volume = 0.0999 × 24.0 = 2.40 dm³.

例：10.0 g CaCO₃ 完全分解时，在 RTP 下产生多少体积的 CO₂？CaCO₃ → CaO + CO₂。n(CaCO₃) = 10.0 ÷ 100.09 = 0.0999 mol。摩尔比 1 : 1，因此 n(CO₂) = 0.0999 mol。体积 = 0.0999 × 24.0 = 2.40 dm³。

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10. Water of Crystallisation | 结晶水

Hydrated salts contain water molecules within their crystal structure. Heating drives off this water, leaving the anhydrous salt. The formula of a hydrate is written as salt·xH₂O, where x is the number of moles of water per mole of salt.

水合盐的晶体结构中含有水分子。加热可除去结晶水，留下无水盐。水合物的化学式写作 盐·xH₂O，其中 x 是每摩尔盐对应的水的物质的量。

CuSO₄·5H₂O → CuSO₄ + 5H₂O

CuSO₄·5H₂O → CuSO₄ + 5H₂O

To find x: measure the mass of hydrate, heat to constant mass, measure the mass of anhydrous salt. The difference in mass is water. Convert both masses to moles and find the simplest ratio.

求 x 的方法：称量水合物的质量，加热至恒重，再称量无水盐的质量。质量差即为水的质量。将两者质量换算为物质的量，并求最简整数比。

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11. Application of Stoichiometry in Analysis | 化学计量在分析中的应用

Stoichiometry is applied in gravimetric analysis (measuring mass of a precipitate), titrimetric analysis (measuring volume of a standard solution), and volumetric analysis of gases. These methods rely on exact mole ratios and precise measurements.

化学计量广泛应用于重量分析（测定沉淀质量）、滴定分析（测定标准溶液体积）和气体体积分析。这些方法依赖于准确的摩尔比和精确的测量。

Method | 方法	Measurement | 测量量	Key Relationship | 关键关系
Gravimetric | 重量分析	Mass of precipitate | 沉淀质量	n = m ÷ M
Titration | 滴定	Volume of standard solution | 标准溶液体积	cV ratio from balanced equation | 配平方程中的 cV 比
Gas analysis | 气体分析	Gas volume | 气体体积	n = V ÷ Vₘ

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12. Common Pitfalls and Exam Tips | 常见错误与考试提示

Students often forget to balance equations, use the wrong molar mass, or confuse cm³ with dm³ when applying c = n ÷ V. Always convert volume to dm³ by dividing cm³ by 1000.

学生常犯的错误包括：忘记配平方程式、使用错误摩尔质量、或在使用 c = n ÷ V 时混淆 cm³ 与 dm³。务必进行单位换算：将 cm³ 除以 1000 得到 dm³。

• Always state the unit in every step | 每一步都要标明单位
• Check significant figures according to the data given | 按照题给数据保留有效数字
• Identify the limiting reactant before calculating yield | 计算产量前先确定限量反应物
• Use the full balanced equation, not the net ionic equation, for mole ratios | 使用完整的配平方程式而非净离子方程式来确定摩尔比

By treating every calculation as a chain of simple mole conversions, you can systematically solve even the most complex stoichiometric problems. Practice each type until the units cancel naturally and the ratios become second nature.

把每一个计算都视为一系列简单的摩尔换算，你就能系统地解决最复杂的化学计量问题。反复练习每种题型，直到单位能够自然抵消、比例关系烂熟于心。

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