📚 IB Quantitative Chemistry: Mastering Moles, Stoichiometry & Calculations | IB定量化学：掌握摩尔、化学计量与计算
Quantitative chemistry forms the backbone of IB Chemistry, connecting macroscopic measurements with the particle nature of matter. From the mole concept to titrations and yield calculations, mastering these numeric tools enables you to predict reactant quantities, product masses, and the efficiency of reactions. This article systematically breaks down all essential topics covered in the IB Chemistry syllabus, ensuring a deep conceptual understanding paired with confident calculation skills.
定量化学是IB化学的支柱，它将宏观测量与物质的微粒本质联系起来。从摩尔概念到滴定和产率计算，掌握这些数字工具能让你预测反应物的量、生成物的质量以及反应的效率。本文系统梳理IB化学大纲涵盖的所有核心定量专题，确保你既能深入理解概念，又能熟练进行计算。
1. The Mole and Avogadro’s Constant | 摩尔与阿伏伽德罗常数
mentary entities (atoms, molecules, ions, electrons, or other particles). This number is Avogadro’s constant (NA).
摩尔是物质的量的国际单位。任何物质的一摩尔都恰好包含 6.02214076 × 10²³ 个基本单元（原子、分子、离子、电子或其他粒子）。这个数值就是阿伏伽德罗常数（NA）。
Number of particles = n × NA
Understanding the mole allows chemists to count particles by weighing. A mole of carbon‑12 atoms has a mass of exactly 12 g, which establishes the link between atomic mass units and grams.
理解摩尔让化学家可以通过称重来数粒子。一摩尔碳‑12 原子的质量恰好为 12 g，由此建立起原子质量单位与克之间的桥梁。
sample containing 3.01 × 10²³ water molecules corresponds to half a mole (0.5 mol) of H₂O.
含有 3.01 × 10²³ 个水分子的样品相当于 0.5 mol H₂O。
One mole of sodium chloride contains 6.02 × 10²³ Na⁺ ions and 6.02 × 10²³ Cl⁻ ions.
一摩尔氯化钠含有 6.02 × 10²³ 个 Na⁺ 离子和同样数量的 Cl⁻ 离子。
2. Molar Mass and Relative Masses | 摩尔质量与相对质量
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) but carries the unit g mol⁻¹.
摩尔质量（M）是一摩尔物质的质量，单位为 g mol⁻¹。它在数值上等于相对原子质量（Ar）或相对式量（Mr），但带有单位 g mol⁻¹。
n = m / M
e molar mass. For example, 36 g of water (M = 18.0 g mol⁻¹) gives n = 36 / 18.0 = 2.0 mol.
计算物质的量时，用质量（m）除以摩尔质量。例如，36 g 水（M = 18.0 g mol⁻¹）的物质的量 n = 36 / 18.0 = 2.0 mol。
Relative atomic mass is a weighted average of the masses of an element’s isotopes, taking into account their natural abundances. Relative formula mass is the sum of relative atomic masses of all atoms in the chemical formula.
相对原子质量是元素各种同位素质量的加权平均值，考虑了它们的天然丰度。相对式量则是化学式中所有原子相对原子质量的总和。
3. Empirical and Molecular Formulas | 经验式与分子式
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 a molecule; it is always a whole‑number multiple of the empirical formula.
经验式（最简式）表示化合物中各元素原子的最简单整数比。分子式表示一个分子中各元素原子的实际数目；它总是经验式的整数倍。
Molecular formula = (Empirical formula) × n
To determine the empirical formula from mass data, convert mass to moles, divide by the smallest number of moles, and adjust to the nearest whole‑number ratio. Combustion analysis or composition by mass often provides the initial data.
由质量数据确定经验式时，先将质量转换为物质的量，除以最小的摩尔数，再调整为最简整数比。通常通过燃烧分析或质量组成获得初始数据。
| Step | Action |
|---|---|
| 1 | Divide each element’s mass by its Ar to obtain moles. |
| 2 | Divide all mole values by the smallest mole value. |
| 3 | If necessary, multiply all ratios by a common factor to get integers. |
(Table translated) 步骤1：每元素质量除以Ar得到摩尔数；步骤2：各摩尔数除以最小摩尔数；步骤3：需要时乘以公倍数得到整数比。
4. Stoichiometry from Chemical Equations | 根据化学方程式进行化学计量
tive number of moles. For the reaction N₂ + 3H₂ → 2NH₃, the mole ratio N₂:H₂:NH₃ is 1:3:2.
配平的化学方程式显示了反应物与生成物之间的摩尔比。化学计量数代表了相对摩尔数。对于反应 N₂ + 3H₂ → 2NH₃，摩尔比 N₂:H₂:NH₃ 为 1:3:2。
Using the mole ratio, you can calculate the amount of any substance consumed or produced from a known amount of another substance. Always convert masses to moles first, apply the ratio, then convert back to mass if required.
利用摩尔比，可以从已知的某一物质的量计算出任一其他物质消耗或生成的量。务必先将质量转换为摩尔，再乘以化学计量比，最后根据需要转换回质量。
What mass of NH₃ is produced from 14.0 g of N₂? 14.0/28.0=0.500 mol; moles of NH₃ = 2 × 0.500 = 1.00 mol; mass NH₃ = 1.00 × 17.0 = 17.0 g.
由 14.0 g N₂ 能产生多少克 NH₃？n(N₂)=14.0/28.0=0.500 mol；NH₃ 的摩尔数=2×0.500=1.00 mol；质量 NH₃ = 1.00 × 17.0 = 17.0 g。
5. Limiting Reactant | 限量反应物
The limiting reactant is the substance that is completely consumed when a reaction goes to completion. It determines the maximum amount of product that can form. The other reactants are present in excess.
限量反应物是在反应完全进行时被完全消耗的物质。它决定了所能生成的最大产物量。其他反应物以过量形式存在。
To identify the limiting reactant, calculate the moles of each reactant. Divide each mole value by its stoichiometric coefficient; the reactant with the smallest resulting ratio is limiting.
为了确定限量反应物，先计算每种反应物的摩尔数。将每种反应物的摩尔数除以其化学计量数；得到比值最小的反应物即为限量反应物。
Limiting ratio = nreactant / coefficient
limiting because 1.5 < 2.0. The amount of H₂O formed is determined by H₂: 3.0 mol H₂O.
对 2H₂ + O₂ → 2H₂O，若混合 3.0 mol H₂ 和 2.0 mol O₂：H₂ 比值 = 3.0/2 = 1.5，O₂ 比值 = 2.0/1 = 2.0。由于 1.5 小于 2.0，H₂ 是限量反应物。生成 H₂O 的量由 H₂ 决定：3.0 mol H₂O。
6. Yield Calculations: Theoretical, Actual and Percent | 产率计算：理论、实际和百分比
The theoretical yield is the maximum mass of product calculated from stoichiometry, assuming complete conversion of the limiting reactant. The actual yield is the amount of product experimentally obtained. Percent yield compares the two.
理论产率是根据化学计量计算得出的最大产物质量，假设限量反应物完全转化。实际产率是实验实际获得的产物量。百分比产率用于比较两者。
Percent yield = (actual yield / theoretical yield) × 100%
Yields are rarely 100% due to incomplete reactions, side reactions, or mechanical losses during purification. IB questions often ask you to evaluate the efficiency of a synthesis or to determine the purity of a product.
由于反应不完全、副反应或提纯过程中的机械损失，产率很少达到 100%。IB 试题经常要求评估合成效率或确定产品纯度。
ield of aspirin = 5.00 g; actual mass obtained = 4.20 g. Percent yield = (4.20/5.00) × 100% = 84.0%.
阿司匹林的理论产率为 5.00 g，实际获得 4.20 g，百分比产率 = (4.20/5.00) × 100% = 84.0%。
7. Molar Volume of Gases at STP and Ideal Gas Law | 标准状况下气体摩尔体积与理想气体定律
At standard temperature and pressure (STP: 0 °C, 100 kPa according to IB current syllabus), one mole of any ideal gas occupies a volume of 22.7 dm³. This value, called the molar gas volume (Vm), links amount and volume directly.
在标准状况（STP：0 °C、100 kPa，按照IB现行大纲）下，一摩尔任何理想气体的体积为 22.7 dm³。这个数值称为气体摩尔体积（Vm），直接将物质的量与体积联系起来。
n = V / Vm (at STP)
For conditions that are not STP, the ideal gas equation is used: PV = nRT les, T is temperature in kelvin (K), and R = 8.31 J K⁻¹ mol⁻¹ (or 8.31 kPa dm³ K⁻¹ mol⁻¹).
当条件不是标准状况时，使用理想气体方程：PV = nRT。其中 P 为压力（kPa），V 为体积（dm³），n 为摩尔，T 为开氏温度（K），R = 8.31 J K⁻¹ mol⁻¹（或 8.31 kPa dm³ K⁻¹ mol⁻¹）。
t 25 °C and 101.3 kPa. T = 298 K. n = PV / RT = (101.3 × 2.50) / (8.31 × 298) ≈ 0.102 mol.
计算 25 °C、101.3 kPa 下收集到 2.50 dm³ 气体时产生的 CO₂ 的物质的量。T = 298 K，n = PV / RT = (101.3 × 2.50) / (8.31 × 298) ≈ 0.102 mol。
8. Concentration and Dilution | 浓度与稀释
Concentration (c) is the amount of solute dissolved in a unit volume of solution, typically expressed in mol dm⁻³. It is calculated as:
浓度（c）是溶解在单位体积溶液中的溶质的量，通常以 mol dm⁻³ 表示。计算公式为：
c = n / V
Dilution reduces the concentration by adding more solvent. The number of moles of solute remains constant before and after dilution, which gives the useful relationship:
稀释通过加入更多溶剂来降低浓度。溶质的摩尔数在稀释前后保持不变，由此得到一个实用的关系式：
c₁V₁ = c₂V₂
000 cm³. When a solid is dissolved to make a standard solution, weigh the solid, dissolve, and make up to a known volume in a volumetric flask.
务必把体积转换为相同单位（dm³ 或 cm³）。记住 1 dm³ = 1000 cm³。当用固体配制标准溶液时，先称量固体，溶解后在容量瓶中定容到已知体积。
9. Titration and Back Titration | 滴定与反滴定
An acid‑base titration determines the unknown concentration of a solution by reacting it with a standard solution of known concentration. The end point is usually detected by a colour change of an indicator. From the titre volume and known concentration, the moles of the known solution are calculated, and the stoichiometric ratio yields moles of the analyte.
酸碱滴定通过让未知浓度溶液与已知浓度的标准溶液反应来测定其浓度。滴定终点通常靠指示剂的颜色变化来判断。由滴定体积和已知浓度算出已知溶液的摩尔数，再根据化学计量比得到分析物的摩尔数。
nacid = (cbase × Vbase) × (mole ratio acid/base)
A back titration is used when the reaction is slow or the sample is impure. An excess of a standard reagent is added, allowed to react, and the leftover amount is titrated with a second standard solution. The difference gives the amount that reacted with the sample.
当反应较慢或样品不纯时，使用反滴定。先加入过量的一种标准试剂，让其充分反应，然后用第二标准溶液滴定剩余的量。差值即为与样品反应的量。
Back titration example: Determining the mass of calcium carbonate in an indigestion tablet by reacting with excess HCl and titrating the unreacted HCl with NaOH.
反滴定实例：用过量 HCl 与一片胃药反应，未反应的 HCl 用 NaOH 滴定，从而求出碳酸钙的质量。
10. Relative Atomic Mass and Isotopic Abundance | 相对原子质量与同位素丰度
The relative atomic mass (Ar) is a weighted mean of the masses of an element’s naturally occurring isotopes. It is calculated from the mass spectrum or from given percentage abundances. The formula is:
相对原子质量（Ar）是该元素天然同位素质量的加权平均值。可从质谱图或给定的丰度百分比计算得出。公式为：
Ar = Σ (isotopic mass × % abundance) / 100
For example, chlorine has two main isotopes: ³⁵Cl (75.77% abundance, mass = 34.97 u) and ³⁷Cl (24.23%, mass = 36.97 u). Ar .97×75.77 + 36.97×24.23)/100 ≈ 35.45. This explains why the periodic table often shows non‑integer masses.
例如，氯有两种主要同位素：³⁵Cl（丰度 75.77%，质量 34.97 u）和 ³⁷Cl（丰度 24.23%，质量 36.97 u）。Ar(Cl) = (34.97×75.77 + 36.97×24.23)/100 ≈ 35.45。这就解释了为何周期表中的相对原子质量常不是整数。
IB questions may also present mass spectrum data where you identify the ion peaks, their relative intensities, and calculate the relative atomic mass. The concept extends to determining molecular mass from molecular ion peaks.
IB 试题也可能给出质谱图数据，要求识别离子峰及其相对丰度，并计算相对原子质量。这一概念可推广到通过分子离子峰确定分子量。
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