IB Chemistry: Common Question Types in Stoichiometric Calculations | IB化学：化学计算常见题型解析

📚 IB Chemistry: Common Question Types in Stoichiometric Calculations | IB化学：化学计算常见题型解析

Stoichiometry is the quantitative foundation of IB Chemistry. Almost every exam paper — whether Paper 1, Paper 2, or Paper 3 — contains at least several marks dedicated to calculations involving moles, concentrations, gas volumes, and limiting reagents. Mastering these recurring question types is one of the most efficient ways to secure a higher grade.

化学计量是 IB 化学的定量基础。几乎每次考试——无论是 Paper 1、Paper 2 还是 Paper 3——都会有至少几分考查摩尔、浓度、气体体积和限制反应物的计算。掌握这些高频出现的题型，是提高分数最有效的方法之一。

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1. Mole Calculations Using n = m / M | 摩尔质量计算

The most fundamental skill in IB stoichiometry is converting between mass and moles using the equation n = m / M, where n is the amount in moles, m is the mass in grams, and M is the molar mass in g mol⁻¹. You must be able to calculate molar mass from a formula — including hydrated salts, where water molecules are counted as part of the crystal structure.

IB 化学计量中最基础的技能是利用公式 n = m / M 在质量与摩尔之间进行换算，其中 n 为物质的量（摩尔），m 为质量（克），M 为摩尔质量（g mol⁻¹）。你必须能够根据化学式计算出摩尔质量——包括水合盐，其中结晶水也要计入总式量之中。

For example, for sodium carbonate decahydrate (Na₂CO₃·10H₂O), the molar mass is 2(22.99) + 12.01 + 3(16.00) + 10(18.02) = 286.15 g mol⁻¹. Many students forget to include the 10 water molecules and consequently obtain an incorrect answer.

例如，十水合碳酸钠（Na₂CO₃·10H₂O）的摩尔质量为 2(22.99) + 12.01 + 3(16.00) + 10(18.02) = 286.15 g mol⁻¹。许多学生忘记包含 10 个结晶水分子，从而得出错误答案。

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2. Empirical and Molecular Formula | 经验式与分子式

When given percentage composition by mass, the empirical formula is determined by dividing each percentage by its relative atomic mass, then dividing all values by the smallest ratio, and finally converting to whole numbers. The molecular formula is a whole-number multiple of the empirical formula, found using the molar mass.

当给出质量百分组成时，经验式的求法是：将每个百分数除以其相对原子质量，再全部除以最小比值，最后换算为最简整数比。分子式是经验式的整数倍，需利用摩尔质量来确定。

n(C) : n(H) : n(O) = (40.0 ÷ 12.01) : (6.7 ÷ 1.01) : (53.3 ÷ 16.00) = 3.33 : 6.63 : 3.33 = 1 : 2 : 1

If the empirical formula is CH₂O and the molar mass is 180 g mol⁻¹, then the molecular formula is C₆H₁₂O₆ because the formula mass of CH₂O is 30, and 180 ÷ 30 = 6.

如果经验式为 CH₂O，摩尔质量为 180 g mol⁻¹，则分子式为 C₆H₁₂O₆，因为 CH₂O 的式量为 30，而 180 ÷ 30 = 6。

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3. Gas Volume Calculations at STP and SLC | STP 与 SLC 条件下的气体体积计算

At standard temperature and pressure (STP: 0 °C, 1 atm), one mole of any ideal gas occupies 22.7 dm³. At standard laboratory conditions (SLC: 25 °C, 100 kPa), the molar volume is 24.0 dm³. Students commonly confuse these two values — memorise both clearly and check the conditions stated in the question.

在标准温度和压力下（STP：0 °C，1 atm），1 摩尔任何理想气体体积为 22.7 dm³。在标准实验室条件下（SLC：25 °C，100 kPa），摩尔体积为 24.0 dm³。学生经常混淆这两个数值——请清楚记忆两者，并仔细查看题目中给出的条件。

For the decomposition of hydrogen peroxide: 2H₂O₂(aq) → 2H₂O(l) + O₂(g). If 0.200 mol of H₂O₂ decomposes completely, the volume of O₂ produced at STP is:

对于过氧化氢的分解：2H₂O₂(aq) → 2H₂O(l) + O₂(g)。如果 0.200 mol H₂O₂ 完全分解，在 STP 条件下产生的 O₂ 体积为：

n(O₂) = 0.200 mol ÷ 2 = 0.100 mol; V(O₂) = 0.100 × 22.7 = 2.27 dm³

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4. Solution Concentration and Dilution | 溶液浓度与稀释

Concentration is expressed as mol dm⁻³ or g dm⁻³. The key equation is c = n / V, where V is in dm³. A crucial skill is converting cm³ to dm³ by dividing by 1000 — a recurring source of silly mistakes.

浓度以 mol dm⁻³ 或 g dm⁻³ 表示。关键公式为 c = n / V，其中 V 单位为 dm³。一个关键技能是将 cm³ 转换为 dm³（除以 1000）——这是反复出现低级错误的来源。

For dilution problems, the relationship c₁V₁ = c₂V₂ applies when the amount of solute remains unchanged. For example, diluting 25.0 cm³ of 0.400 mol dm⁻³ HCl to a final volume of 100.0 cm³ gives a new concentration of 0.100 mol dm⁻³.

对于稀释类问题，溶质物质的量不变时适用 c₁V₁ = c₂V₂。例如，将 25.0 cm³ 的 0.400 mol dm⁻³ HCl 稀释至终体积 100.0 cm³，新浓度为 0.100 mol dm⁻³。

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5. Limiting Reactant and Excess | 限制反应物与过量判断

Determining the limiting reactant is a classic calculation that appears in nearly every full-length IB paper. The method is: convert all reactant masses to moles, divide each by its stoichiometric coefficient, and the smallest value identifies the limiting reactant.

判断限制反应物是几乎每套 IB 完整试卷中都会出现的经典计算题。方法是：将所有反应物的质量转换为物质的量，再分别除以各自的化学计量系数，最小值对应的物质即为限制反应物。

For example, if 2.00 g of Mg reacts with 10.0 cm³ of 2.00 mol dm⁻³ HCl, for the reaction Mg + 2HCl → MgCl₂ + H₂:

例如，2.00 g Mg 与 10.0 cm³ 2.00 mol dm⁻³ HCl 反应，反应式为 Mg + 2HCl → MgCl₂ + H₂：

n(Mg) = 2.00 ÷ 24.31 = 0.0823 mol; n(HCl) = 2.00 × 0.0100 = 0.0200 mol

n(Mg) ÷ 1 = 0.0823; n(HCl) ÷ 2 = 0.0100 → HCl is the limiting reactant

Thus the theoretical yield of H₂ is 0.0200 ÷ 2 = 0.0100 mol, not the value based on magnesium. Students often make the mistake of choosing magnesium simply because it is a solid — always do the full division calculation.

因此 H₂ 的理论产量为 0.0200 ÷ 2 = 0.0100 mol，而非基于镁所计算出的值。学生常犯的错误是仅仅因为镁是固体就选择它——务必进行完整的除法计算。

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

Percentage yield compares actual yield to theoretical yield, calculated from the limiting reactant. Atom economy measures how many atoms in the reactants are incorporated into the desired product. Both are part of the IB Chemistry syllabus under “Stoichiometric Relationships”.

产率百分数比较实际产量与理论产量（理论产量由限制反应物计算得出）。原子经济性衡量反应物中有多少原子进入了目标产物。两者都属于 IB 化学课程大纲中“化学计量关系”部分的内容。

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

Atom economy = (molar mass of desired product sum ÷ total molar mass of all reactants) × 100%

For industrial processes such as the Haber process or the Contact process, the atom economy of the overall reaction path is often assessed conceptually. Be prepared to calculate both quantities from balanced equations and mass data.

对于哈伯法或接触法这样的工业过程，常会概念性地评估整条反应路径的原子经济性。要准备好从平衡方程式和质量数据出发，对这两个量进行实际计算。

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7. Titration Calculations | 滴定计算

Acid–base titration calculations are one of the most frequently examined types in IB Chemistry Paper 2 and the internal assessment. The core is the relationship n(acid) = n(base) at the equivalence point, adjusted by stoichiometry.

酸碱滴定计算是 IB 化学 Paper 2 和内部评估中考查频率最高的题型之一。核心是在等当点时满足 n(酸) = n(碱) 的关系，再根据化学计量进行调整。

A typical question: 25.0 cm³ of 0.100 mol dm⁻³ NaOH requires 20.0 cm³ of H₂SO₄ for complete neutralisation. What is the concentration of H₂SO₄?

典型题目：25.0 cm³ 0.100 mol dm⁻³ NaOH 恰好需要 20.0 cm³ H₂SO₄ 完全中和。求 H₂SO₄ 的浓度。

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

n(NaOH) = 0.100 × 0.0250 = 0.00250 mol; n(H₂SO₄) = 0.00250 ÷ 2 = 0.00125 mol; c(H₂SO₄) = 0.00125 ÷ 0.0200 = 0.0625 mol dm⁻³

Notice the 1:2 ratio between H₂SO₄ and NaOH — failing to divide by 2 is the most common error in this problem type.

注意 H₂SO₄ 与 NaOH 的 1:2 比例——忘记除以 2 是此类题目最常见的错误。

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8. Ideal Gas Equation PV = nRT | 理想气体方程 PV = nRT

When gases are involved under non-standard conditions, the ideal gas equation becomes essential. You must use R = 8.314 J K⁻¹ mol⁻¹ when P is in Pa and V is in m³. Practise converting units: 1 cm³ = 10⁻⁶ m³, 1 dm³ = 10⁻³ m³, 1 kPa = 1000 Pa.

当气体处于非标准条件时，理想气体方程就必不可少。当 P 以 Pa、V 以 m³ 为单位时，必须使用 R = 8.314 J K⁻¹ mol⁻¹。请练习单位换算：1 cm³ = 10⁻⁶ m³、1 dm³ = 10⁻³ m³、1 kPa = 1000 Pa。

A common question asks: a 200 cm³ flask contains 0.500 g of N₂ at 27 °C. Calculate the pressure in kPa.

常见问题：一个 200 cm³ 的烧瓶在 27 °C 下装有 0.500 g N₂。计算以 kPa 为单位的压力。

n = 0.500 ÷ 28.02 = 0.0178 mol; T = 300 K; V = 2.00 × 10⁻⁴ m³

P = nRT ÷ V = 0.0178 × 8.314 × 300 ÷ (2.00 × 10⁻⁴) = 222,000 Pa = 222 kPa

Students lose marks when they forget to convert °C to K or forget to cube the volume conversion. Always write out units in each step to verify consistency.

学生忘记将 °C 转换为 K 或忘记体积立方换算时就会丢分。务必在每一步写出单位以验证一致性。

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9. Back Titration | 返滴定

Back titration appears when the substance being analysed is insoluble, volatile, or reacts slowly. Typically, an excess of a known reagent is added, the reaction is allowed to complete, and the remaining unreacted excess is titrated with a second standard solution.

当被测物质不溶、易挥发或反应缓慢时，常采用返滴定法。通常先加入过量且已知浓度的试剂，待反应完成后，用另一种标准溶液滴定剩余的未反应部分。

A standard IB example: 1.00 g of impure CaCO₃ is treated with 50.0 cm³ of 1.00 mol dm⁻³ HCl. The excess HCl requires 25.0 cm³ of 0.500 mol dm⁻³ NaOH. Calculate the purity of the CaCO₃.

一个标准的 IB 例题：将 1.00 g 不纯的 CaCO₃ 与 50.0 cm³ 1.00 mol dm⁻³ HCl 反应，剩余的 HCl 需要 25.0 cm³ 0.500 mol dm⁻³ NaOH 中和。计算 CaCO₃ 的纯度。

n(HCl initial) = 0.0500 mol; n(NaOH) = 0.0125 mol = n(HCl excess)

n(HCl reacted) = 0.0500 − 0.0125 = 0.0375 mol; n(CaCO₃) = 0.0375 ÷ 2 = 0.01875 mol

mass(CaCO₃) = 0.01875 × 100.09 = 1.877 g; purity = 87.5%

The sign of a good answer is clarity: labelling initial, excess, and reacted amounts explicitly prevents sign errors and confusion in later steps.

好答案的标志是清晰：明确标注初始量、剩余量和反应量，可以防止符号错误及后续步骤的混乱。

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10. Combustion Analysis | 燃烧分析

Combustion analysis in IB involves burning an organic compound to produce CO₂ and H₂O, then using the masses of these products to determine the empirical formula. This is the classic “Find the formula” problem.

IB 中的燃烧分析涉及将有机物燃烧生成 CO₂ 和 H₂O，再根据产物的质量来确定经验式。这是经典的“求化学式”问题。

For example, 0.100 g of a hydrocarbon produces 0.314 g of CO₂ and 0.128 g of H₂O. Calculate the empirical formula.

例如，0.100 g 某烃完全燃烧生成 0.314 g CO₂ 和 0.128 g H₂O。求经验式。

n(CO₂) = 0.314 ÷ 44.01 = 0.00714 mol; n(C) = 0.00714 mol

n(H₂O) = 0.128 ÷ 18.02 = 0.00710 mol; n(H) = 0.0142 mol

C:H = 0.00714 : 0.0142 = 1 : 1.99 ≈ 1:2 → CH₂

The most common pitfall is forgetting to multiply n(H₂O) by 2 to obtain n(H). Always keep track of where each atom comes from.

最常见的陷阱是忘记将 n(H₂O) 乘以 2 来求 n(H)。始终追踪每个原子的来源。

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11. Redox Titration — KMnO₄ and I₂/Na₂S₂O₃ | 氧化还原滴定——KMnO₄ 与 I₂/Na₂S₂O₃

In both SL and HL, redox titrations are assessed conceptually, and at HL level often numerically. The most common systems: MnO₄⁻ / C₂O₄²⁻ in acidic solution, and I₂ / S₂O₃²⁻. Write down the half-equations before doing any arithmetic.

在 SL 和 HL 中，氧化还原滴定都会被从概念上考查，而 HL 通常会进行定量计算。最常见的体系：酸性溶液中 MnO₄⁻ / C₂O₄²⁻，以及 I₂ / S₂O₃²⁻。进行任何计算之前，先写出半反应方程式。

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

C₂O₄²⁻ → 2CO₂ + 2e⁻

This yields a MnO₄⁻ : C₂O₄²⁻ ratio of 2:5. For I₂ / S₂O₃²⁻, the ratio is 1:2 because 4S₂O₃²⁻ → 2S₄O₆²⁻ + 4e⁻ and I₂ + 2e⁻ → 2I⁻. Having the half-equations on paper before calculating is your best strategy.

这得出 MnO₄⁻ 与 C₂O₄²⁻ 的化学计量比为 2:5。对于 I₂ / S₂O₃²⁻ 体系，比例为 1:2，因为 4S₂O₃²⁻ → 2S₄O₆²⁻ + 4e⁻，而 I₂ + 2e⁻ → 2I⁻。在计算前先在纸上写出半反应方程式是最佳策略。

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12. Common Mistakes and Exam Strategy | 常见错误与考试策略

The most frequent sources of lost marks in IB stoichiometry questions are: forgetting to convert cm³ to dm³, using STP instead of SLC, not dividing by stoichiometric coefficients when identifying limiting reactants, confusing mass and moles, and omitting units. All of these are easily avoidable with a disciplined checklist approach.

在 IB 化学计量题目中，最常见的失分原因包括：忘记将 cm³ 转换为 dm³、使用 STP 而非 SLC、判断限制反应物时未除以化学计量系数、混淆质量与摩尔量，以及漏写单位。所有这些都可以通过有条理的检查清单方法来轻松避免。

Quantity	Unit	Symbol
Amount of substance	mol	n
Mass	g, kg	m
Molar mass	g mol⁻¹	M
Concentration	mol dm⁻³	c
Volume	dm³, cm³, m³	V
Pressure	Pa, kPa, atm	P

Before attempting questions under time pressure, set a routine: write the balanced equation, identify the known, label the unknown, convert to moles, apply the stoichiometric ratio, and convert to the final required units. This sequence works for every type of calculation discussed here.

在限时答题前，养成固定流程：写出平衡方程式、找出已知量、标明未知量、换算成摩尔、应用化学计量比、转换到最终所需单位。这套流程适用于以上讨论的所有计算题型。

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