High School Chemistry

The Mole and Stoichiometry摩尔与化学计量

Chemistry is a quantitative science: every balanced equation is a recipe that tells you exactly how many atoms, molecules, or moles of each substance react and form. This guide builds the stoichiometry toolkit from the ground up — from Avogadro's number and the mole concept, through molar mass and the three-way conversion between moles, mass, and particles, to percent composition and empirical/molecular formulae, balancing chemical equations by inspection, mole-ratio stoichiometry, limiting and excess reactants, and finally percent yield. Worked examples and KaTeX formulas throughout.化学是一门定量科学：每一个配平的化学方程式都是一张配方，精确告诉你各物质参与反应和生成的原子数、分子数或摩尔数。本指南从基础开始逐步搭建化学计量学工具箱——从阿伏伽德罗常数（Avogadro's number，阿伏伽德罗常数）和摩尔（mole，摩尔）概念出发，经摩尔质量（molar mass，摩尔质量）以及摩尔、质量与粒子数之间的三向换算，到质量分数与实验式（empirical formula，实验式）/分子式（molecular formula，分子式），再到配平化学方程式、摩尔比化学计量（stoichiometry，化学计量），最后落脚于限量反应物（limiting reactant，限量反应物）与产率（percent yield，产率）。全程使用例题与公式。

7 sections7 节内容 US NGSS · ON · BC · ABUS NGSS · ON · BC · AB Core for ON SCH3U Strand D · AB Chemistry 20 Unit D安大略 SCH3U D 单元 · 阿省 Chemistry 20 D 单元核心

Where this lives in your syllabus本单元在各大纲中的位置

HS-PS1-7 "Use mathematical representations to support the claim that atoms, and therefore mass, are conserved during a chemical reaction. [Clarification Statement: Emphasis is on using mathematical ideas to communicate the proportional relationships between masses of atoms in the reactants and the products, and the translation of these relationships to the macroscopic scale using the mole as the conversion from the atomic to the macroscopic scale. Emphasis is on assessing students' use of mathematical thinking and not on memorization and rote application of problem-solving techniques.] [Assessment Boundary: Assessment does not include complex chemical reactions.]" Maps to §1 (the mole as the atomic-to-macroscopic conversion), §4 (mass conservation via balancing), and §5 (mole-ratio relationships). NGSS frames the mole conceptually and excludes complex limiting-reagent / percent-yield calculations from the assessment boundary."使用数学表达式来支持原子（因此质量）在化学反应中守恒的论断。[澄清说明：重点在于用数学思想传达反应物和产物中原子质量之间的比例关系，以及用摩尔作为从原子尺度到宏观尺度的转换，将这些关系推广到宏观尺度。重点是评估学生对数学思维的运用，而非死记硬背和机械套用解题技巧。] [评估边界：不包括复杂的化学反应。]"对应 §1（摩尔作为原子到宏观的转换）、§4（通过配平实现质量守恒）和 §5（摩尔比关系）。NGSS 从概念角度框定摩尔，并将复杂的限量反应物/产率计算排除在评估边界之外。

SCH3U Strand D — Quantities in Chemical Reactions D2.1 "use appropriate terminology related to quantities in chemical reactions, including, but not limited to: stoichiometry, percentage yield, limiting reagent, mole, and atomic mass"; D2.3 "solve problems related to quantities in chemical reactions by performing calculations involving quantities in moles, number of particles, and atomic mass"; D2.4 "determine the empirical formulae and molecular formulae of various chemical compounds, given molar masses and percentage composition or mass data"; D2.5 "calculate the corresponding mass, or quantity in moles or molecules, for any given reactant or product in a balanced chemical equation as well as for any other reactant or product in the chemical reaction"; D2.6 "solve problems related to quantities in chemical reactions by performing calculations involving percentage yield and limiting reagents"; D3.2 "describe the relationships between Avogadro's number, the mole concept, and the molar mass of any given substance"; D3.3 "explain the relationship between the empirical formula and the molecular formula of a chemical compound".SCH3U D 单元（化学反应中的量）：D2.1"使用与化学反应量相关的适当术语，包括但不限于：化学计量、产率、限量试剂、摩尔和原子质量"；D2.3"通过计算摩尔数、粒子数和原子质量，解决化学反应量的相关问题"；D2.4"根据摩尔质量和质量分数或质量数据，确定各种化合物的实验式和分子式"；D2.5"计算配平化学方程式中任意给定反应物或产物所对应的质量或摩尔/分子数量，以及化学反应中任何其他反应物或产物的量"；D2.6"通过计算产率和限量试剂，解决化学反应量的相关问题"；D3.2"描述阿伏伽德罗常数、摩尔概念和任何物质摩尔质量之间的关系"；D3.3"解释化合物的实验式与分子式之间的关系"。

Chemistry 11 Big Idea: "The mole is a quantity used to make atoms and molecules measurable." Content: "the mole"; "dimensional analysis" (elaborated as "factor-label method; calculation of mass and molar quantities using significant figures"); "stoichiometric calculations" (elaborated as "mass — number of molecules — gas volumes — molar quantities — excess and limiting reactants"). All seven sections of this guide map directly to these core Chemistry 11 content bullets; BC treats stoichiometry including limiting reactants and excess at the core (not honors) level in Grade 11.Chemistry 11 大概念："摩尔是用来使原子和分子可测量的量。"内容："摩尔"；"量纲分析"（细化为"因子标签法；使用有效数字计算质量和摩尔量"）；"化学计量计算"（细化为"质量——分子数——气体体积——摩尔量——过量和限量反应物"）。本指南的全部 7 节都直接对应这些 BC Chemistry 11 核心内容要点；BC 在 11 年级将包括限量反应物和过量在内的化学计量视为核心（非荣誉）内容。

Chemistry 20 Unit D — Quantitative Relationships in Chemical Changes General Outcome 1: "explain how balanced chemical equations indicate the quantitative relationships between reactants and products involved in chemical changes." Key Concepts: "chemical reaction equations · net ionic equations · reaction stoichiometry · limiting and excess reagents · actual, theoretical and percent yield." GO1 Knowledge outcomes include: "recall the balancing of chemical equations in terms of atoms, molecules and moles"; "calculate the quantities of reactants and/or products involved in chemical reactions, using gravimetric, solution or gas stoichiometry." GO2 Knowledge outcomes include: "identify limiting and excess reagents in chemical reactions"; "define theoretical yields and actual yields"; "explain the discrepancy between theoretical and actual yields".Chemistry 20 D 单元（化学变化中的定量关系）总目标 1："解释配平化学方程式如何说明化学变化中反应物和产物之间的定量关系"。关键概念："化学反应方程式 · 净离子方程式 · 反应化学计量 · 限量和过量试剂 · 实际产量、理论产量和产率"。GO1 知识结果包括："用原子、分子和摩尔回顾化学方程式的配平"；"使用重量分析、溶液或气体化学计量计算化学反应中反应物和/或产物的量"。GO2 知识结果包括："识别化学反应中的限量试剂和过量试剂"；"定义理论产量和实际产量"；"解释理论产量与实际产量之间的差异"。

Read me first先读这里

How to use this guide如何使用本指南

The mole and stoichiometry is the quantitative backbone of all of chemistry. Every curriculum we map to includes it, and the four agree on a shared core: the mole concept, molar mass, mole-ratio stoichiometry, and balancing equations. They differ mainly in depth of limiting-reagent and percent-yield treatment. US NGSS (HS-PS1-7) frames the mole conceptually — "translation of mass relationships to the macroscopic scale" — and explicitly excludes "complex chemical reactions" from the assessment boundary, so multi-step limiting-reagent / percent-yield problems sit above the NGSS floor. Ontario SCH3U Strand D (D2.3–D2.6, D3.2–D3.3) treats all seven topics quantitatively at Grade 11. BC Chemistry 11 lists the full stoichiometric toolkit — including limiting and excess reactants — as core Content. Alberta Chemistry 20 Unit D covers the same scope, explicitly including limiting/excess reagents and theoretical/actual/percent yield. The table below gives you your row.摩尔与化学计量是整个化学定量分析的骨干。我们所对照的每套大纲都包含它，四套大纲在核心范围上一致：摩尔概念、摩尔质量、摩尔比化学计量和配平方程式。它们主要在限量反应物和产率处理的深度上有所不同。US NGSS（HS-PS1-7）从概念角度框定摩尔——"将质量关系推广到宏观尺度"——并明确将"复杂化学反应"排除在评估边界之外，因此多步限量反应物/产率问题超出 NGSS 要求。安大略 SCH3U D 单元（D2.3–D2.6、D3.2–D3.3）在 11 年级对全部七个主题进行定量处理。BC Chemistry 11 将完整的化学计量工具箱——包括限量和过量反应物——列为核心内容。阿尔伯塔 Chemistry 20 D 单元涵盖相同范围，明确包括限量/过量试剂以及理论产量/实际产量/产率。下表给出你的对应行。

If you are in…如果你在…	Focus on these sections重点学习	Defer / lighter可推迟 / 减负	Source依据
🇺🇸 US NGSS HS Chemistry美国 NGSS 高中化学	§1 (mole / Avogadro's number), §2 (molar mass, conversions), §4 (balancing equations), §5 (mole-ratio stoichiometry) — these map to HS-PS1-7's emphasis on mass conservation and the mole as the atomic-to-macroscopic bridge.§1（摩尔/阿伏伽德罗常数）、§2（摩尔质量、换算）、§4（配平方程式）、§5（摩尔比化学计量）——这些对应 HS-PS1-7 对质量守恒和摩尔作为原子-宏观桥梁的强调。	§6 (limiting reagents) and §7 (percent yield): HS-PS1-7 Assessment Boundary excludes "complex chemical reactions"; limiting-reagent and yield calculations are above the assessed floor for NGSS-only students.§6（限量试剂）和 §7（产率）：HS-PS1-7 评估边界排除"复杂化学反应"；限量试剂和产率计算超出 NGSS 学生的评估底线。	`NGSS HS-PS1 (Chemistry)` — HS-PS1-7 PE + Clarification + Assessment Boundary— HS-PS1-7 表现期望 + 澄清 + 评估边界
🇨🇦 ON Grade 11 — SCH3U安大略 11 年级 — SCH3U	All seven sections. SCH3U Strand D (D2.1–D2.6, D3.1–D3.4) covers every topic in this guide quantitatively: mole concept, molar mass, empirical/molecular formula, balancing, stoichiometry, limiting reagents, and percentage yield.全部 7 节。SCH3U D 单元（D2.1–D2.6、D3.1–D3.4）对本指南的每个主题进行定量处理：摩尔概念、摩尔质量、实验式/分子式、配平、化学计量、限量试剂和产率。	Nothing — Ontario SCH3U assesses the full stoichiometry scope at Grade 11.无 — 安大略 SCH3U 在 11 年级评估完整的化学计量范围。	`Ontario SCH3U/4U Chemistry` — SCH3U Strand D Overall Expectations D1–D3; Specific Expectations D2.1, D2.3–D2.6, D3.2–D3.3— SCH3U D 单元总体期望 D1–D3；具体期望 D2.1、D2.3–D2.6、D3.2–D3.3
🇨🇦 BC Chemistry 11BC Chemistry 11	All seven sections. BC Chemistry 11 Big Idea "The mole is a quantity used to make atoms and molecules measurable" covers the full scope, and the stoichiometric-calculations elaboration explicitly includes "excess and limiting reactants."全部 7 节。BC Chemistry 11 大概念"摩尔是用来使原子和分子可测量的量"涵盖完整范围，化学计量计算的细化内容明确包括"过量和限量反应物"。	Nothing — BC treats limiting reactants and percent yield as core, not honors, in Grade 11.无 — BC 在 11 年级将限量反应物和产率视为核心而非荣誉内容。	`BC Chemistry 11/12` — Chemistry 11 Big Idea "The mole …"; Content "the mole", "dimensional analysis", "stoichiometric calculations" with elaboration— Chemistry 11 大概念"摩尔……"；内容"摩尔"、"量纲分析"、"化学计量计算"及其细化
🇨🇦 AB Chemistry 20阿尔伯塔 Chemistry 20	All seven sections. Chemistry 20 Unit D (GO1/GO2) covers stoichiometry, limiting and excess reagents, theoretical yields, actual yields, and percent yield explicitly; GO1 knowledge outcomes include balancing equations and calculating reactant/product quantities using gravimetric stoichiometry.全部 7 节。Chemistry 20 D 单元（GO1/GO2）明确涵盖化学计量、限量和过量试剂、理论产量、实际产量和产率；GO1 知识结果包括配平方程式和使用重量分析化学计量计算反应物/产物量。	Nothing — AB Chemistry 20 Unit D explicitly lists limiting/excess reagents and percent yield as assessed content.无 — AB Chemistry 20 D 单元明确将限量/过量试剂和产率列为评估内容。	`Alberta Chemistry 20/30` — Chemistry 20 Unit D GO1/GO2, Key Concepts, knowledge outcome text— Chemistry 20 D 单元 GO1/GO2，关键概念，知识结果文本

If you are cramming the night before如果你在临阵磨枪

Memorise five things: one mole $= 6.022 \times 10^{23}$ particles (Avogadro's number); moles $= $ mass $\div$ molar mass ($n = m/M$); the mole ratio from the balanced equation converts between moles of different substances; the limiting reactant is the one that runs out first; percent yield $= (\text{actual}/\text{theoretical}) \times 100\%$. Read every cram-cheat box.背熟五件事：1 摩尔 $= 6.022 \times 10^{23}$ 个粒子（阿伏伽德罗常数）；摩尔数 $= $ 质量 $\div$ 摩尔质量（$n = m/M$）；配平方程式中的摩尔比用于在不同物质的摩尔数之间换算；限量反应物是最先耗尽的那个；产率 $= （\text{actual}/\text{theoretical}）\times 100\%$。读每个速记框。

If you are going for the top mark如果你目标顶分

Show every step of the roadmap: grams $\to$ moles (divide by $M$) $\to$ moles of target (multiply by ratio from balanced equation) $\to$ grams of target (multiply by $M$). For empirical/molecular formulae, show the percent-to-mass step, the divide-by-smallest step, and check integer ratios. For limiting reagents, calculate the moles of product from each reactant separately, then pick the smaller answer. State your limiting reactant explicitly. Express yield answers to three significant figures.展示路线图的每一步：克 $\to$ 摩尔（除以 $M$）$\to$ 目标物质的摩尔数（乘以配平方程式中的比值）$\to$ 目标物质的克数（乘以 $M$）。对于实验式/分子式，展示从百分比到质量的步骤、除以最小值的步骤，并检验整数比。对于限量试剂，分别计算每种反应物可生成的产物摩尔数，然后选择较小的答案。明确说明限量反应物。将产率答案保留到三位有效数字。

Section 1 · The mole and Avogadro's number第 1 节 · 摩尔与阿伏伽德罗常数

The Mole and Avogadro's Number摩尔与阿伏伽德罗常数

The mole is chemistry's counting unit — it bridges atoms to grams.摩尔是化学的计数单位 — 它将原子数与克数连接起来。

1 mole (mol)1 摩尔（mol） — exactly $6.022 \times 10^{23}$ representative particles (atoms, molecules, ions, formula units). This number is Avogadro's number $N_A$.— 恰好 $6.022 \times 10^{23}$ 个代表性粒子（原子、分子、离子、化学式单元）。这个数字就是阿伏伽德罗常数 $N_A$。
Why $6.022 \times 10^{23}$?为什么是 $6.022 \times 10^{23}$？ It is defined so that 1 mol of ${}^{12}\text{C}$ atoms has a mass of exactly $12\ \mathrm{g}$. The same number of any element's atoms has a mass in grams numerically equal to its atomic mass in u.它的定义使得 1 mol ${}^{12}\text{C}$ 原子的质量恰好等于 $12\ \mathrm{g}$。任何元素的同等数量原子，其克数在数值上等于其以 u 为单位的原子质量。

Key conversion:关键换算：

$$ N = n \times N_A \qquad n = \frac{N}{N_A} $$ $N$ = number of particles, $n$ = moles, $N_A = 6.022 \times 10^{23}\ \mathrm{mol^{-1}}$. NGSS HS-PS1-7 and SCH3U D3.2 both identify Avogadro's number and the mole as the atomic-to-macroscopic bridge as assessed content.$N$ = 粒子数，$n$ = 摩尔数，$N_A = 6.022 \times 10^{23}\ \mathrm{mol^{-1}}$。NGSS HS-PS1-7 与 SCH3U D3.2 均将阿伏伽德罗常数和摩尔作为原子-宏观桥梁列为评估内容。

Worked Example 1 · Particles from moles例题 1 · 由摩尔数求粒子数

How many molecules are in $2.50\ \mathrm{mol}$ of water, $\text{H}_2\text{O}$?$2.50\ \mathrm{mol}$ 的水（$\text{H}_2\text{O}$）中含有多少个分子？

Apply $N = n \times N_A$.应用 $N = n \times N_A$。

$$ N = 2.50\ \mathrm{mol} \times 6.022 \times 10^{23}\ \mathrm{mol^{-1}} = 1.506 \times 10^{24}\ \text{molecules}. $$

Each of those molecules contains 2 H atoms and 1 O atom, so there are $3 \times 1.506 \times 10^{24} = 4.52 \times 10^{24}$ total atoms.每个分子含有 2 个 H 原子和 1 个 O 原子，故总原子数为 $3 \times 1.506 \times 10^{24} = 4.52 \times 10^{24}$。

Avogadro's number $N_A$ is approximately:阿伏伽德罗常数 $N_A$ 约为：

§1 · Q1

$6.022 \times 10^{20}$$6.022 \times 10^{20}$

$6.022 \times 10^{20}\ \mathrm{mol^{-1}}$$6.022 \times 10^{20}\ \mathrm{mol^{-1}}$

$6.022 \times 10^{23}\ \mathrm{mol^{-1}}$$6.022 \times 10^{23}\ \mathrm{mol^{-1}}$

$6.022 \times 10^{26}\ \mathrm{mol^{-1}}$$6.022 \times 10^{26}\ \mathrm{mol^{-1}}$

$N_A = 6.022 \times 10^{23}\ \mathrm{mol^{-1}}$. The unit $\mathrm{mol^{-1}}$ is essential — it means "per mole." One mole of any substance contains this many representative particles.$N_A = 6.022 \times 10^{23}\ \mathrm{mol^{-1}}$。单位 $\mathrm{mol^{-1}}$ 非常重要——它表示"每摩尔"。任何物质的 1 摩尔含有这么多代表性粒子。

The exponent is $23$, not $20$ or $26$. The unit $\mathrm{mol^{-1}}$ (per mole) must be included; without it the number is dimensionless and meaningless as a conversion factor.指数是 $23$，而非 $20$ 或 $26$。单位 $\mathrm{mol^{-1}}$（每摩尔）必须包含；没有它，该数字无量纲，作为换算因子毫无意义。

How many atoms are in $0.500\ \mathrm{mol}$ of iron (Fe)?$0.500\ \mathrm{mol}$ 铁（Fe）中含有多少个原子？

§1 · Q2

$3.01 \times 10^{23}$$3.01 \times 10^{23}$

$6.02 \times 10^{23}$$6.02 \times 10^{23}$

$1.20 \times 10^{24}$$1.20 \times 10^{24}$

$27.9$$27.9$

$N = 0.500\ \mathrm{mol} \times 6.022 \times 10^{23}\ \mathrm{mol^{-1}} = 3.01 \times 10^{23}$ atoms. Half a mole contains half Avogadro's number of atoms.$N = 0.500\ \mathrm{mol} \times 6.022 \times 10^{23}\ \mathrm{mol^{-1}} = 3.01 \times 10^{23}$ 个原子。半摩尔含有阿伏伽德罗常数一半的原子数。

$N = n \times N_A = 0.500 \times 6.022 \times 10^{23} = 3.01 \times 10^{23}$. Option (b) is 1 full mole; option (d) is the molar mass in grams, not a particle count.$N = n \times N_A = 0.500 \times 6.022 \times 10^{23} = 3.01 \times 10^{23}$。选项 (b) 是 1 完整摩尔；选项 (d) 是以克为单位的摩尔质量，而非粒子数。

Section 2 · Molar mass and mole–mass–particle conversions第 2 节 · 摩尔质量与摩尔–质量–粒子数换算

Molar Mass and Conversions摩尔质量与换算

Molar mass ($M$) is the mass of one mole in grams.摩尔质量（$M$）是一摩尔物质的克数。

For an element对于元素: $M$ in $\mathrm{g\,mol^{-1}}$ numerically equals the atomic mass in $\mathrm{u}$. E.g. Fe: $M = 55.85\ \mathrm{g\,mol^{-1}}$.：$M$（单位 $\mathrm{g\,mol^{-1}}$）在数值上等于以 $\mathrm{u}$ 为单位的原子质量。例如 Fe：$M = 55.85\ \mathrm{g\,mol^{-1}}$。
For a compound对于化合物: sum the atomic masses of all atoms in the formula. E.g. $\text{H}_2\text{O}$: $M = 2(1.008) + 16.00 = 18.02\ \mathrm{g\,mol^{-1}}$.：将化学式中所有原子的原子质量相加。例如 $\text{H}_2\text{O}$：$M = 2(1.008) + 16.00 = 18.02\ \mathrm{g\,mol^{-1}}$。

The three-way roadmap:三向换算路线图：

$$ n = \frac{m}{M} \qquad m = n \times M \qquad N = n \times N_A $$ $n$ = moles, $m$ = mass (g), $M$ = molar mass ($\mathrm{g\,mol^{-1}}$), $N$ = number of particles. SCH3U D2.3 and D3.2 assess calculations using moles, number of particles, and atomic mass. BC Chemistry 11 lists "calculation of mass and molar quantities using significant figures" as an elaboration of "dimensional analysis."$n$ = 摩尔数，$m$ = 质量（g），$M$ = 摩尔质量（$\mathrm{g\,mol^{-1}}$），$N$ = 粒子数。SCH3U D2.3 和 D3.2 评估涉及摩尔数、粒子数和原子质量的计算。BC Chemistry 11 将"使用有效数字计算质量和摩尔量"列为"量纲分析"的细化内容。

Worked Example 2 · Grams to moles to particles例题 2 · 克→摩尔→粒子数

A student dissolves $8.80\ \mathrm{g}$ of carbon dioxide, $\text{CO}_2$, in water. (a) Calculate the molar mass of $\text{CO}_2$. (b) Find the number of moles. (c) Find the number of molecules.一名学生将 $8.80\ \mathrm{g}$ 二氧化碳（$\text{CO}_2$）溶于水。(a) 计算 $\text{CO}_2$ 的摩尔质量。(b) 求摩尔数。(c) 求分子数。

(a) Molar mass.(a) 摩尔质量。

$$ M(\text{CO}_2) = 12.01 + 2(16.00) = 44.01\ \mathrm{g\,mol^{-1}}. $$

(b) Moles.(b) 摩尔数。

$$ n = \frac{m}{M} = \frac{8.80\ \mathrm{g}}{44.01\ \mathrm{g\,mol^{-1}}} = 0.200\ \mathrm{mol}. $$

(c) Molecules.(c) 分子数。

$$ N = 0.200\ \mathrm{mol} \times 6.022 \times 10^{23}\ \mathrm{mol^{-1}} = 1.20 \times 10^{23}\ \text{molecules.} $$

What is the molar mass of glucose, $\text{C}_6\text{H}_{12}\text{O}_6$? (Use $M_\text{C} = 12.01$, $M_\text{H} = 1.008$, $M_\text{O} = 16.00\ \mathrm{g\,mol^{-1}}$.)葡萄糖 $\text{C}_6\text{H}_{12}\text{O}_6$ 的摩尔质量是多少？（使用 $M_\text{C} = 12.01$、$M_\text{H} = 1.008$、$M_\text{O} = 16.00\ \mathrm{g\,mol^{-1}}$。）

§2 · Q1

$168.2\ \mathrm{g\,mol^{-1}}$

$180.2\ \mathrm{g\,mol^{-1}}$

$176.1\ \mathrm{g\,mol^{-1}}$

$192.2\ \mathrm{g\,mol^{-1}}$

$6(12.01) + 12(1.008) + 6(16.00) = 72.06 + 12.096 + 96.00 = 180.2\ \mathrm{g\,mol^{-1}}$. Count each atom type from the formula, multiply by its atomic mass, and sum.$6(12.01) + 12(1.008) + 6(16.00) = 72.06 + 12.096 + 96.00 = 180.2\ \mathrm{g\,mol^{-1}}$。从化学式中计数每种原子，乘以其原子质量，然后求和。

Count atoms carefully: 6 C, 12 H, 6 O. Multiply each by its molar mass: $6 \times 12.01 = 72.06$; $12 \times 1.008 = 12.10$; $6 \times 16.00 = 96.00$. Sum = $180.2$.仔细计数原子：6 个 C、12 个 H、6 个 O。每种乘以摩尔质量：$6 \times 12.01 = 72.06$；$12 \times 1.008 = 12.10$；$6 \times 16.00 = 96.00$。总和 = $180.2$。

How many moles are in $45.0\ \mathrm{g}$ of water ($M = 18.02\ \mathrm{g\,mol^{-1}}$)?$45.0\ \mathrm{g}$ 水（$M = 18.02\ \mathrm{g\,mol^{-1}}$）中含有多少摩尔？

§2 · Q2

$811$ mol$811$ mol

$18.0$ mol$18.0$ mol

$1.50$ mol$1.50$ mol

$2.50$ mol$2.50$ mol

$n = m/M = 45.0/18.02 = 2.50\ \mathrm{mol}$. Divide mass by molar mass to get moles. $45.0 \div 18.02 = 2.498 \approx 2.50$ (3 sig figs). ✓$n = m/M = 45.0/18.02 = 2.50\ \mathrm{mol}$。质量除以摩尔质量得到摩尔数。$45.0 \div 18.02 = 2.498 \approx 2.50$（3 位有效数字）。✓

$n = m \div M$, not $m \times M$. $45.0 \div 18.02 = 2.50\ \mathrm{mol}$. Option (a) results from $m \times M$; option (c) is $45.0 \div 30$ (incorrect $M$).$n = m \div M$，而非 $m \times M$。$45.0 \div 18.02 = 2.50\ \mathrm{mol}$。选项 (a) 是 $m \times M$ 的结果；选项 (c) 是 $45.0 \div 30$（错误的 $M$）。

Section 3 · Percent composition and empirical/molecular formulae第 3 节 · 质量分数与实验式/分子式

Percent Composition and Empirical / Molecular Formulae质量分数与实验式/分子式

Percent composition tells you the mass fraction of each element; the empirical formula gives the simplest integer ratio.质量分数告诉你每种元素的质量比例；实验式给出最简整数比。

Percent composition质量分数 $$ \%\ \text{element} = \frac{\text{mass of element in 1 mol compound}}{M_\text{compound}} \times 100\% $$
Empirical formula实验式: the simplest whole-number mole ratio of elements. Steps: (1) assume $100\ \mathrm{g}$ sample so \% $\to$ g; (2) convert each mass to moles; (3) divide all by the smallest mole value; (4) round to integers (multiply if needed).：元素的最简整数摩尔比。步骤：(1) 假设 $100\ \mathrm{g}$ 样品，使百分比直接变为克数；(2) 将每种质量转换为摩尔数；(3) 全部除以最小摩尔值；(4) 四舍五入至整数（必要时乘以倍数）。
Molecular formula分子式: a whole-number multiple of the empirical formula. Multiply the empirical formula by $n = M_\text{molecular}/M_\text{empirical}$.：实验式的整数倍。将实验式乘以 $n = M_\text{molecular}/M_\text{empirical}$。

SCH3U D2.4 assesses "determine the empirical formulae and molecular formulae of various chemical compounds, given molar masses and percentage composition or mass data." BC Chemistry 11 stoichiometric elaboration and AB Chemistry 20 Unit D GO1 cover the same scope.SCH3U D2.4 评估"根据摩尔质量和质量分数或质量数据，确定各种化合物的实验式和分子式"。BC Chemistry 11 化学计量细化内容和 AB Chemistry 20 D 单元 GO1 涵盖相同范围。

Worked Example 3 · Empirical and molecular formula from combustion data例题 3 · 由燃烧数据求实验式与分子式

A compound contains $40.0\%$ C, $6.71\%$ H, and $53.3\%$ O by mass. Its molar mass is $60.1\ \mathrm{g\,mol^{-1}}$. Find the empirical formula and the molecular formula.一种化合物按质量计含 $40.0\%$ C、$6.71\%$ H 和 $53.3\%$ O。其摩尔质量为 $60.1\ \mathrm{g\,mol^{-1}}$。求实验式和分子式。

Step 1: $100\ \mathrm{g}$ sample gives第 1 步：$100\ \mathrm{g}$ 样品中各元素的质量 $40.0\ \mathrm{g}$ C, $6.71\ \mathrm{g}$ H, $53.3\ \mathrm{g}$ O.$40.0\ \mathrm{g}$ C、$6.71\ \mathrm{g}$ H、$53.3\ \mathrm{g}$ O。

Step 2: Convert to moles.第 2 步：转换为摩尔数。

$$ n_\text{C} = \frac{40.0}{12.01} = 3.331,\quad n_\text{H} = \frac{6.71}{1.008} = 6.657,\quad n_\text{O} = \frac{53.3}{16.00} = 3.331. $$

Step 3: Divide by the smallest ($3.331$).第 3 步：除以最小值（$3.331$）。

$$ \text{C}: 1.000,\quad \text{H}: 1.999 \approx 2,\quad \text{O}: 1.000. $$

Empirical formula: $\text{CH}_2\text{O}$实验式：$\text{CH}_2\text{O}$, $M_\text{emp} = 12.01 + 2(1.008) + 16.00 = 30.03\ \mathrm{g\,mol^{-1}}$.，$M_\text{emp} = 12.01 + 2(1.008) + 16.00 = 30.03\ \mathrm{g\,mol^{-1}}$。

Step 4: Molecular formula.第 4 步：分子式。 $n = 60.1 / 30.03 = 2.00$. Molecular formula: $\text{C}_2\text{H}_4\text{O}_2$ (acetic acid). ✓分子式：$\text{C}_2\text{H}_4\text{O}_2$（乙酸）。✓

A compound is $75.0\%$ C and $25.0\%$ H by mass. What is its empirical formula?一种化合物按质量计含 $75.0\%$ C 和 $25.0\%$ H。其实验式是什么？

§3 · Q1

$\text{CH}_2$

$\text{C}_2\text{H}_3$

$\text{CH}_4$

$\text{C}_3\text{H}_8$

$n_\text{C} = 75.0/12.01 = 6.245$; $n_\text{H} = 25.0/1.008 = 24.80$. Ratio: $24.80/6.245 = 3.97 \approx 4$. Empirical formula: $\text{CH}_4$ (methane). This is also the molecular formula of methane.$n_\text{C} = 75.0/12.01 = 6.245$；$n_\text{H} = 25.0/1.008 = 24.80$。比值：$24.80/6.245 = 3.97 \approx 4$。实验式：$\text{CH}_4$（甲烷）。这也是甲烷的分子式。

Divide each mass by its molar mass: C gives $75.0/12.01 = 6.25$ mol; H gives $25.0/1.008 = 24.8$ mol. Divide both by the smaller ($6.25$): C ratio $= 1$; H ratio $= 24.8/6.25 = 3.97 \approx 4$. Formula: $\text{CH}_4$.将每种质量除以其摩尔质量：C 得 $75.0/12.01 = 6.25$ mol；H 得 $25.0/1.008 = 24.8$ mol。两者均除以较小值（$6.25$）：C 比值 $= 1$；H 比值 $= 24.8/6.25 = 3.97 \approx 4$。化学式：$\text{CH}_4$。

The empirical formula of a compound is $\text{NO}_2$ ($M_\text{emp} = 46.01\ \mathrm{g\,mol^{-1}}$). Its molar mass is $92.02\ \mathrm{g\,mol^{-1}}$. What is the molecular formula?某化合物的实验式为 $\text{NO}_2$（$M_\text{emp} = 46.01\ \mathrm{g\,mol^{-1}}$），摩尔质量为 $92.02\ \mathrm{g\,mol^{-1}}$。其分子式是什么？

§3 · Q2

$\text{NO}_2$

$\text{N}_2\text{O}_4$

$\text{N}_3\text{O}_6$

$\text{N}_4\text{O}_8$

$n = 92.02 / 46.01 = 2.000$. Multiply the empirical formula by 2: $\text{N}_2\text{O}_4$. This is dinitrogen tetroxide, which exists as a dimer of $\text{NO}_2$.$n = 92.02 / 46.01 = 2.000$。将实验式乘以 2：$\text{N}_2\text{O}_4$。这是四氧化二氮，以 $\text{NO}_2$ 的二聚体形式存在。

$n = M_\text{molecular}/M_\text{empirical} = 92.02/46.01 = 2$. Multiply every subscript in the empirical formula by $2$: $\text{N}_{1\times2}\text{O}_{2\times2} = \text{N}_2\text{O}_4$.$n = M_\text{molecular}/M_\text{empirical} = 92.02/46.01 = 2$。将实验式中每个下标乘以 $2$：$\text{N}_{1\times2}\text{O}_{2\times2} = \text{N}_2\text{O}_4$。

Section 4 · Balancing chemical equations第 4 节 · 配平化学方程式

Balancing Chemical Equations配平化学方程式

A balanced equation has the same number of each atom on both sides — mass is conserved.配平方程式两边每种原子数目相同——质量守恒。

Law of conservation of mass质量守恒定律: atoms are neither created nor destroyed in a chemical reaction. The total mass of reactants equals the total mass of products.：化学反应中原子既不产生也不消灭。反应物总质量等于产物总质量。
Balancing by inspection目测配平法: adjust coefficients (the large numbers in front of formulas) — never change subscripts inside a formula. Work in this order: (1) balance metals, (2) balance nonmetals other than H and O, (3) balance H, (4) balance O, (5) reduce coefficients to lowest integers if needed.：调整系数（化学式前面的大数字）——永远不要改变化学式内部的下标。按此顺序操作：(1) 配平金属，(2) 配平除 H 和 O 以外的非金属，(3) 配平 H，(4) 配平 O，(5) 必要时将系数化为最小整数。

Reading the coefficients as moles:将系数读作摩尔数：

$\text{N}_2 + 3\text{H}_2 \to 2\text{NH}_3$ means 1 mol $\text{N}_2$ reacts with 3 mol $\text{H}_2$ to produce 2 mol $\text{NH}_3$. The ratio $1:3:2$ is the mole ratio for all stoichiometry calculations (§5–§7).$\text{N}_2 + 3\text{H}_2 \to 2\text{NH}_3$ 表示 1 mol $\text{N}_2$ 与 3 mol $\text{H}_2$ 反应生成 2 mol $\text{NH}_3$。比值 $1:3:2$ 是 §5–§7 中所有化学计量计算的摩尔比。

NGSS HS-PS1-7 anchors mass conservation to balancing. SCH3U D3.4 states: "explain the quantitative relationships expressed in a balanced chemical equation, using appropriate units of measure (e.g., moles, grams, atoms, ions, molecules)." AB Chemistry 20 GO1 includes "recall the balancing of chemical equations in terms of atoms, molecules and moles."NGSS HS-PS1-7 将质量守恒与配平相联系。SCH3U D3.4 要求："用适当的度量单位（如摩尔、克、原子、离子、分子）解释配平化学方程式所表达的定量关系。"AB Chemistry 20 GO1 包括"用原子、分子和摩尔回顾化学方程式的配平"。

Worked Example 4 · Balancing a combustion reaction例题 4 · 配平燃烧反应

Balance the combustion of propane: $\text{C}_3\text{H}_8 + \text{O}_2 \to \text{CO}_2 + \text{H}_2\text{O}$.配平丙烷的燃烧反应：$\text{C}_3\text{H}_8 + \text{O}_2 \to \text{CO}_2 + \text{H}_2\text{O}$。

Step 1: Balance C.第 1 步：配平 C。 3 C on the left $\Rightarrow$ put coefficient 3 in front of $\text{CO}_2$.左边 3 个 C $\Rightarrow$ 在 $\text{CO}_2$ 前加系数 3。

Step 2: Balance H.第 2 步：配平 H。 8 H on the left $\Rightarrow$ put coefficient 4 in front of $\text{H}_2\text{O}$ (since each $\text{H}_2\text{O}$ has 2 H).左边 8 个 H $\Rightarrow$ 在 $\text{H}_2\text{O}$ 前加系数 4（每个 $\text{H}_2\text{O}$ 含 2 个 H）。

Step 3: Balance O.第 3 步：配平 O。 Right side: $3 \times 2 + 4 \times 1 = 10$ O atoms. So put coefficient 5 in front of $\text{O}_2$.右边：$3 \times 2 + 4 \times 1 = 10$ 个 O 原子。故在 $\text{O}_2$ 前加系数 5。

Balanced equation:配平方程式：

$$ \text{C}_3\text{H}_8 + 5\text{O}_2 \to 3\text{CO}_2 + 4\text{H}_2\text{O} $$

Check: C $3=3$ ✓, H $8=8$ ✓, O $10=10$ ✓. Mole ratio $1:5:3:4$.核验：C $3=3$ ✓，H $8=8$ ✓，O $10=10$ ✓。摩尔比 $1:5:3:4$。

What is the correctly balanced equation for the reaction of hydrogen with oxygen to form water?氢气与氧气反应生成水的正确配平方程式是什么？

§4 · Q1

$\text{H}_2 + \text{O}_2 \to \text{H}_2\text{O}$

$\text{H}_2 + \text{O} \to \text{H}_2\text{O}$

$2\text{H} + \text{O} \to \text{H}_2\text{O}$

$2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}$

$2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}$. Check: H $4=4$ ✓, O $2=2$ ✓. Option (a) has unbalanced O ($2$ left, $1$ right). Options (b) and (c) change the formula of $\text{O}_2$ to $\text{O}$ — never change subscripts.$2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}$。核验：H $4=4$ ✓，O $2=2$ ✓。选项 (a) O 不平衡（左 $2$，右 $1$）。选项 (b) 和 (c) 将 $\text{O}_2$ 的化学式改为 $\text{O}$——永远不要改变下标。

Never change subscripts — only coefficients. $\text{O}_2$ is always diatomic; only the coefficient in front of $\text{H}_2\text{O}$ can be changed. The only balanced option is $2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}$.永远不要改变下标——只能改变系数。$\text{O}_2$ 始终是双原子的；只能改变 $\text{H}_2\text{O}$ 前面的系数。唯一正确的配平方程式是 $2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}$。

In the balanced equation $\text{N}_2 + 3\text{H}_2 \to 2\text{NH}_3$, how many moles of $\text{H}_2$ react with 4.0 mol of $\text{N}_2$?在配平方程式 $\text{N}_2 + 3\text{H}_2 \to 2\text{NH}_3$ 中，4.0 mol $\text{N}_2$ 与多少 mol $\text{H}_2$ 反应？

§4 · Q2

4.0 mol4.0 mol

12 mol12 mol

8.0 mol8.0 mol

2.0 mol2.0 mol

Mole ratio $\text{N}_2 : \text{H}_2 = 1:3$. So $4.0\ \mathrm{mol}\ \text{N}_2 \times (3\ \mathrm{mol}\ \text{H}_2 / 1\ \mathrm{mol}\ \text{N}_2) = 12\ \mathrm{mol}\ \text{H}_2$.摩尔比 $\text{N}_2 : \text{H}_2 = 1:3$。故 $4.0\ \mathrm{mol}\ \text{N}_2 \times (3\ \mathrm{mol}\ \text{H}_2 / 1\ \mathrm{mol}\ \text{N}_2) = 12\ \mathrm{mol}\ \text{H}_2$。

The coefficient ratio from the balanced equation is the mole ratio. For every 1 mol $\text{N}_2$, you need 3 mol $\text{H}_2$. For $4.0$ mol $\text{N}_2$: $4.0 \times 3 = 12$ mol $\text{H}_2$.配平方程式中的系数比就是摩尔比。每 1 mol $\text{N}_2$ 需要 3 mol $\text{H}_2$。$4.0$ mol $\text{N}_2$：$4.0 \times 3 = 12$ mol $\text{H}_2$。

Section 5 · Mole-ratio stoichiometry第 5 节 · 摩尔比化学计量

Mole-Ratio Stoichiometry摩尔比化学计量

The roadmap: grams of A $\to$ moles of A $\to$ moles of B $\to$ grams of B.路线图：A 的克数 $\to$ A 的摩尔数 $\to$ B 的摩尔数 $\to$ B 的克数。

Step 1第 1 步: convert mass of A to moles of A using $n_A = m_A / M_A$.：用 $n_A = m_A / M_A$ 将 A 的质量转换为 A 的摩尔数。
Step 2第 2 步: convert moles of A to moles of B using the coefficient ratio from the balanced equation: $n_B = n_A \times (\text{coeff of B}/\text{coeff of A})$.：用配平方程式中的系数比将 A 的摩尔数转换为 B 的摩尔数：$n_B = n_A \times (\text{coeff of B}/\text{coeff of A})$。
Step 3第 3 步: convert moles of B to mass of B using $m_B = n_B \times M_B$.：用 $m_B = n_B \times M_B$ 将 B 的摩尔数转换为 B 的质量。

$$ m_B = \frac{m_A}{M_A} \times \frac{\text{coeff}_B}{\text{coeff}_A} \times M_B $$ SCH3U D2.5: "calculate the corresponding mass, or quantity in moles or molecules, for any given reactant or product in a balanced chemical equation." BC Chemistry 11 stoichiometric calculations elaboration: "mass, number of molecules, gas volumes, molar quantities." AB Chemistry 20 GO1: "calculate the quantities of reactants and/or products involved in chemical reactions, using gravimetric stoichiometry."SCH3U D2.5："计算配平化学方程式中任意给定反应物或产物所对应的质量或摩尔/分子数量。"BC Chemistry 11 化学计量计算细化："质量、分子数、气体体积、摩尔量。"AB Chemistry 20 GO1："使用重量分析化学计量计算化学反应中反应物和/或产物的量。"

Worked Example 5 · Mass of product from mass of reactant例题 5 · 由反应物质量求产物质量

Iron reacts with oxygen to form iron(III) oxide: $4\text{Fe} + 3\text{O}_2 \to 2\text{Fe}_2\text{O}_3$. What mass of $\text{Fe}_2\text{O}_3$ is produced from $22.4\ \mathrm{g}$ of Fe? ($M_\text{Fe} = 55.85$, $M_{\text{Fe}_2\text{O}_3} = 159.7\ \mathrm{g\,mol^{-1}}$.)铁与氧气反应生成氧化铁(III)：$4\text{Fe} + 3\text{O}_2 \to 2\text{Fe}_2\text{O}_3$。$22.4\ \mathrm{g}$ Fe 能生成多少克 $\text{Fe}_2\text{O}_3$？（$M_\text{Fe} = 55.85$，$M_{\text{Fe}_2\text{O}_3} = 159.7\ \mathrm{g\,mol^{-1}}$。）

Step 1: Moles of Fe.第 1 步：Fe 的摩尔数。

$$ n_\text{Fe} = \frac{22.4\ \mathrm{g}}{55.85\ \mathrm{g\,mol^{-1}}} = 0.4011\ \mathrm{mol}. $$

Step 2: Moles of $\text{Fe}_2\text{O}_3$ (ratio $4:2 = 2:1$).第 2 步：$\text{Fe}_2\text{O}_3$ 的摩尔数（比值 $4:2 = 2:1$）。

$$ n_{\text{Fe}_2\text{O}_3} = 0.4011 \times \frac{2}{4} = 0.2006\ \mathrm{mol}. $$

Step 3: Mass of $\text{Fe}_2\text{O}_3$.第 3 步：$\text{Fe}_2\text{O}_3$ 的质量。

$$ m = 0.2006\ \mathrm{mol} \times 159.7\ \mathrm{g\,mol^{-1}} = 32.0\ \mathrm{g}. $$

Using $\text{N}_2 + 3\text{H}_2 \to 2\text{NH}_3$, what mass of $\text{NH}_3$ ($M = 17.03\ \mathrm{g\,mol^{-1}}$) is produced from $28.0\ \mathrm{g}$ of $\text{N}_2$ ($M = 28.02\ \mathrm{g\,mol^{-1}}$)?用 $\text{N}_2 + 3\text{H}_2 \to 2\text{NH}_3$，$28.0\ \mathrm{g}$ $\text{N}_2$（$M = 28.02\ \mathrm{g\,mol^{-1}}$）可生成多少克 $\text{NH}_3$（$M = 17.03\ \mathrm{g\,mol^{-1}}$）？

§5 · Q1

$34.1\ \mathrm{g}$$34.1\ \mathrm{g}$

$17.0\ \mathrm{g}$$17.0\ \mathrm{g}$

$51.1\ \mathrm{g}$$51.1\ \mathrm{g}$

$28.0\ \mathrm{g}$$28.0\ \mathrm{g}$

$n_{\text{N}_2} = 28.0/28.02 = 0.9993\ \mathrm{mol}$. Mole ratio $\text{N}_2:\text{NH}_3 = 1:2$, so $n_{\text{NH}_3} = 1.999\ \mathrm{mol}$. Mass $= 1.999 \times 17.03 = 34.0\ \mathrm{g} \approx 34.1\ \mathrm{g}$. ✓$n_{\text{N}_2} = 28.0/28.02 = 0.9993\ \mathrm{mol}$。摩尔比 $\text{N}_2:\text{NH}_3 = 1:2$，故 $n_{\text{NH}_3} = 1.999\ \mathrm{mol}$。质量 $= 1.999 \times 17.03 = 34.0\ \mathrm{g} \approx 34.1\ \mathrm{g}$。✓

Roadmap: $28.0\ \mathrm{g\ N_2} \div 28.02 = 1.00\ \mathrm{mol\ N_2} \times (2\ \mathrm{mol\ NH_3}/1\ \mathrm{mol\ N_2}) = 2.00\ \mathrm{mol\ NH_3} \times 17.03 = 34.1\ \mathrm{g}$.路线图：$28.0\ \mathrm{g\ N_2} \div 28.02 = 1.00\ \mathrm{mol\ N_2} \times (2\ \mathrm{mol\ NH_3}/1\ \mathrm{mol\ N_2}) = 2.00\ \mathrm{mol\ NH_3} \times 17.03 = 34.1\ \mathrm{g}$。

In $2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}$, what is the mole ratio of $\text{H}_2$ to $\text{H}_2\text{O}$?在 $2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}$ 中，$\text{H}_2$ 与 $\text{H}_2\text{O}$ 的摩尔比是什么？

§5 · Q2

$1:2$$1:2$

$2:1$$2:1$

$1:1$$1:1$

$1:3$$1:3$

Coefficients: $\text{H}_2 = 2$, $\text{H}_2\text{O} = 2$. Ratio $= 2:2 = 1:1$. For every mole of $\text{H}_2$ consumed, one mole of $\text{H}_2\text{O}$ is produced.系数：$\text{H}_2 = 2$，$\text{H}_2\text{O} = 2$。比值 $= 2:2 = 1:1$。每消耗 1 mol $\text{H}_2$，生成 1 mol $\text{H}_2\text{O}$。

Read the coefficients directly from the balanced equation: $2\text{H}_2 \ldots \to 2\text{H}_2\text{O}$. Both coefficients are $2$, giving a $1:1$ ratio. The $\text{O}_2$ coefficient ($1$) is not part of this pair.直接从配平方程式读取系数：$2\text{H}_2 \ldots \to 2\text{H}_2\text{O}$。两个系数都是 $2$，比值为 $1:1$。$\text{O}_2$ 的系数（$1$）不属于这对关系。

Section 6 · Limiting and excess reactants第 6 节 · 限量反应物与过量反应物

Limiting and Excess Reactants限量反应物与过量反应物

The limiting reactant runs out first and determines how much product forms.限量反应物最先耗尽，决定产物的生成量。

Limiting reactant限量反应物: the reactant that is completely consumed and stops the reaction. All product calculations are based on the limiting reactant.：完全消耗并使反应停止的反应物。所有产物计算均基于限量反应物。
Excess reactant过量反应物: the reactant that remains after the limiting reactant is consumed.：限量反应物耗尽后仍有剩余的反应物。
Method方法: convert each reactant's mass to moles, then calculate the moles of product each would produce (using the mole ratio). The reactant giving the smaller amount of product is the limiting reactant.：将每种反应物的质量转换为摩尔数，然后分别计算每种反应物（用摩尔比）能生成的产物摩尔数。给出较少产物的反应物是限量反应物。

SCH3U D2.6: "solve problems related to quantities in chemical reactions by performing calculations involving percentage yield and limiting reagents." AB Chemistry 20 GO2: "identify limiting and excess reagents in chemical reactions." BC Chemistry 11 stoichiometric calculations elaboration includes "excess and limiting reactants."SCH3U D2.6："通过计算产率和限量试剂，解决化学反应量的相关问题。"AB Chemistry 20 GO2："识别化学反应中的限量试剂和过量试剂。"BC Chemistry 11 化学计量计算细化包括"过量和限量反应物"。

Worked Example 6 · Identifying the limiting reactant例题 6 · 识别限量反应物

$14.0\ \mathrm{g}$ of $\text{N}_2$ ($M = 28.02$) reacts with $6.00\ \mathrm{g}$ of $\text{H}_2$ ($M = 2.016$) according to $\text{N}_2 + 3\text{H}_2 \to 2\text{NH}_3$. Identify the limiting reactant and calculate the mass of $\text{NH}_3$ ($M = 17.03$) produced.$14.0\ \mathrm{g}$ $\text{N}_2$（$M = 28.02$）与 $6.00\ \mathrm{g}$ $\text{H}_2$（$M = 2.016$）按 $\text{N}_2 + 3\text{H}_2 \to 2\text{NH}_3$ 反应。识别限量反应物，并计算生成 $\text{NH}_3$（$M = 17.03$）的质量。

Convert to moles.转换为摩尔数。

$$ n_{\text{N}_2} = \frac{14.0}{28.02} = 0.4996\ \mathrm{mol}, \qquad n_{\text{H}_2} = \frac{6.00}{2.016} = 2.976\ \mathrm{mol}. $$

Calculate moles of $\text{NH}_3$ from each reactant.分别计算每种反应物可生成的 $\text{NH}_3$ 摩尔数。

$$ \text{From N}_2: \quad 0.4996 \times \tfrac{2}{1} = 0.9992\ \mathrm{mol\ NH_3}. $$ $$ \text{From H}_2: \quad 2.976 \times \tfrac{2}{3} = 1.984\ \mathrm{mol\ NH_3}. $$

$\text{N}_2$ gives less product $\Rightarrow$ $\text{N}_2$ is the limiting reactant.$\text{N}_2$ 给出更少产物 $\Rightarrow$ $\text{N}_2$ 是限量反应物。

Mass of $\text{NH}_3$.$\text{NH}_3$ 的质量。

$$ m = 0.9992\ \mathrm{mol} \times 17.03\ \mathrm{g\,mol^{-1}} = 17.0\ \mathrm{g}. $$

In a reaction $\text{A} + 2\text{B} \to \text{C}$, you have $3.0\ \mathrm{mol}$ of A and $4.0\ \mathrm{mol}$ of B. Which is the limiting reactant?在反应 $\text{A} + 2\text{B} \to \text{C}$ 中，有 $3.0\ \mathrm{mol}$ A 和 $4.0\ \mathrm{mol}$ B。哪种是限量反应物？

§6 · Q1

A, because there is less of itA，因为它的量更少

B, because 3.0 mol A would need 6.0 mol B but only 4.0 mol B is availableB，因为 3.0 mol A 需要 6.0 mol B，但只有 4.0 mol B 可用

A, because it has a smaller coefficientA，因为它的系数更小

Neither — both are in stoichiometric ratio都不是——两者均为化学计量比

To react all 3.0 mol A, you need $3.0 \times 2 = 6.0\ \mathrm{mol}$ B. Only 4.0 mol B is available. B runs out first, so B is the limiting reactant. From 4.0 mol B: $4.0 \times (1/2) = 2.0\ \mathrm{mol}$ C produced.要使全部 3.0 mol A 反应，需要 $3.0 \times 2 = 6.0\ \mathrm{mol}$ B。只有 4.0 mol B 可用。B 先耗尽，故 B 是限量反应物。由 4.0 mol B：$4.0 \times (1/2) = 2.0\ \mathrm{mol}$ C 生成。

The limiting reactant is not simply the one with fewer moles — it depends on the mole ratio. Check: 3.0 mol A requires $3.0 \times 2 = 6.0$ mol B (but only 4.0 available), so B is limiting, not A.限量反应物并非单纯是摩尔数较少的那个——取决于摩尔比。核验：3.0 mol A 需要 $3.0 \times 2 = 6.0$ mol B（但只有 4.0 mol），故 B 是限量反应物，而非 A。

After the limiting reactant is consumed, what happens to the excess reactant?限量反应物耗尽后，过量反应物会怎样？

§6 · Q2

It remains unreacted in the mixture它以未反应的形式留在混合物中

It is converted into a different product它被转化为另一种产物

It reacts with the product to form a new compound它与产物反应生成新化合物

It decomposes into its elements它分解为其组成元素

Once the limiting reactant is gone, there is nothing left for the excess reactant to react with. It stays in the mixture unreacted. The amount of excess left over can be calculated: excess remaining $= $ initial moles $-$ moles consumed by limiting reactant.一旦限量反应物耗尽，过量反应物就没有可以反应的对象了。它以未反应的形式留在混合物中。剩余过量可计算：剩余过量 $= $ 初始摩尔数 $-$ 被限量反应物消耗的摩尔数。

After the limiting reactant runs out, the reaction stops. The excess reactant has no partner left and remains unreacted in the mixture (it is not consumed or converted).限量反应物耗尽后，反应停止。过量反应物没有可反应的对象，以未反应的形式留在混合物中（不会被消耗或转化）。

Going deeper — why comparing moles of product identifies the limiting reactant深入 — 为什么比较产物摩尔数能识别限量反应物

The limiting reactant is the one that runs out first. For a general reaction $a\text{A} + b\text{B} \to c\text{C}$, the amount of product C that reactant A can produce is $n_C^{(A)} = n_A \cdot (c/a)$, and the amount B can produce is $n_C^{(B)} = n_B \cdot (c/b)$. The reaction stops when whichever product supply runs out first is exhausted, so the actual yield of C is限量反应物是最先耗尽的那个。对于通式反应 $a\text{A} + b\text{B} \to c\text{C}$，反应物 A 能产生的 C 的摩尔数为 $n_C^{(A)} = n_A \cdot (c/a)$，B 能产生的为 $n_C^{(B)} = n_B \cdot (c/b)$。反应在哪种供应先耗尽时停止，故 C 的实际产量为

$$ n_C = \min\!\left(n_A \cdot \frac{c}{a},\; n_B \cdot \frac{c}{b}\right) $$

and the limiting reactant is whichever argument gives the minimum. An equivalent test divides each reactant's moles by its stoichiometric coefficient: compare $n_A/a$ to $n_B/b$. The smaller ratio belongs to the limiting reactant, because that reactant is proportionally least available relative to what the equation demands. For instance, if $n_A/a = 0.50$ and $n_B/b = 0.99$, then A runs out when only half the stoichiometric demand has been met, while B still has nearly twice as much as needed — A is limiting. This ratio test is the algebraic foundation of every limiting-reactant calculation and explains why raw mole counts alone are never sufficient.而使乘积最小的那个反应物就是限量反应物。一个等价的检验方法是将每种反应物的摩尔数除以其化学计量系数：比较 $n_A/a$ 与 $n_B/b$，较小的比值对应限量反应物，因为该反应物相对于方程式所需的量来说比例上最不足。例如，若 $n_A/a = 0.50$，$n_B/b = 0.99$，则 A 在满足化学计量需求的一半时就耗尽，而 B 的量几乎是所需的两倍——A 是限量反应物。这一比值检验是所有限量反应物计算的代数基础，也解释了为什么单看原始摩尔数永远是不够的。

Section 7 · Percent yield第 7 节 · 产率

Percent Yield产率

Theoretical yield is the maximum possible; actual yield is what you really get; percent yield measures efficiency.理论产量是最大可能产量；实际产量是实际得到的量；产率衡量效率。

Theoretical yield理论产量: the mass of product calculated from stoichiometry (assuming $100\%$ conversion of the limiting reactant, no side reactions, no losses).：由化学计量计算得到的产物质量（假设限量反应物 $100\%$ 转化，无副反应，无损失）。
Actual yield实际产量: the mass of product actually obtained in the experiment (always $\le$ theoretical yield).：实验中实际得到的产物质量（始终 $\le$ 理论产量）。

$$ \%\ \text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% $$ Reasons percent yield $< 100\%$: side reactions, incomplete reactions, product lost during transfer or purification, impure reactants. SCH3U D2.6 and AB Chemistry 20 GO2 both assess percent yield calculations explicitly. BC Chemistry 11 stoichiometric calculations elaboration includes "molar quantities" and "excess and limiting reactants," implying yield work.产率低于 $100\%$ 的原因：副反应、反应不完全、转移或纯化过程中产物损失、反应物不纯。SCH3U D2.6 和 AB Chemistry 20 GO2 均明确评估产率计算。BC Chemistry 11 化学计量计算细化包括"摩尔量"和"过量和限量反应物"，隐含产率计算。

Worked Example 7 · Calculating percent yield例题 7 · 计算产率

A student reacts $10.0\ \mathrm{g}$ of $\text{CaCO}_3$ ($M = 100.1\ \mathrm{g\,mol^{-1}}$) according to: $\text{CaCO}_3 \to \text{CaO} + \text{CO}_2$. The theoretical yield of $\text{CaO}$ ($M = 56.08\ \mathrm{g\,mol^{-1}}$) is calculated, and the student collects $4.80\ \mathrm{g}$. Find the percent yield.一名学生按 $\text{CaCO}_3 \to \text{CaO} + \text{CO}_2$ 反应 $10.0\ \mathrm{g}$ $\text{CaCO}_3$（$M = 100.1\ \mathrm{g\,mol^{-1}}$）。计算 $\text{CaO}$（$M = 56.08\ \mathrm{g\,mol^{-1}}$）的理论产量，学生实际收集到 $4.80\ \mathrm{g}$。求产率。

Theoretical yield.理论产量。

$$ n_{\text{CaCO}_3} = \frac{10.0}{100.1} = 0.09990\ \mathrm{mol}. $$ $$ n_{\text{CaO}} = 0.09990\ \mathrm{mol} \quad (\text{mole ratio }1:1). $$ $$ m_\text{theoretical} = 0.09990 \times 56.08 = 5.602\ \mathrm{g}. $$

Percent yield.产率。

$$ \%\ \text{yield} = \frac{4.80}{5.602} \times 100\% = 85.7\%. $$

A reaction has a theoretical yield of $25.0\ \mathrm{g}$ and an actual yield of $20.0\ \mathrm{g}$. What is the percent yield?某反应的理论产量为 $25.0\ \mathrm{g}$，实际产量为 $20.0\ \mathrm{g}$。产率是多少？

§7 · Q1

$125\%$$125\%$

$25.0\%$$25.0\%$

$80.0\%$$80.0\%$

$5.00\%$$5.00\%$

$\%\ \text{yield} = (20.0/25.0) \times 100\% = 80.0\%$. Divide actual by theoretical, then multiply by $100\%$. A percent yield $> 100\%$ is impossible (cannot create matter).$\%\ \text{yield} = (20.0/25.0) \times 100\% = 80.0\%$。用实际产量除以理论产量，再乘以 $100\%$。产率大于 $100\%$ 是不可能的（不能无中生有）。

$\%\ \text{yield} = \text{actual}/\text{theoretical} \times 100 = 20.0/25.0 \times 100 = 80.0\%$. Not $125\%$ (actual/theoretical inverted) and not $5\%$ (the difference).$\%\ \text{yield} = \text{actual}/\text{theoretical} \times 100 = 20.0/25.0 \times 100 = 80.0\%$。不是 $125\%$（分子分母颠倒）也不是 $5\%$（差值）。

Which of the following would NOT cause the percent yield to be less than $100\%$?下列哪项不会导致产率低于 $100\%$？

§7 · Q2

A competing side reaction forms an unwanted by-product竞争性副反应生成了不需要的副产物

Some product is lost when transferring between containers在容器间转移时部分产物丢失

The reaction does not go to completion反应没有完全进行

The molar mass of the product is larger than expected产物的摩尔质量大于预期

Percent yield compares actual mass collected to theoretical mass calculated. The molar mass of a pure compound is a fixed physical property — it does not change between experiments and does not affect how much product you collect. Only losses during the experiment reduce actual yield below theoretical.产率比较实际收集质量与计算的理论质量。纯化合物的摩尔质量是固定的物理性质——它不会在实验之间改变，也不影响你收集到多少产物。只有实验过程中的损失才会使实际产量低于理论产量。

Side reactions (a), mechanical losses (b), and incomplete reactions (c) all reduce the actual yield below the theoretical. The molar mass (d) is a constant for a given compound — it does not vary and cannot lower actual yield.副反应（a）、机械损失（b）和反应不完全（c）都会使实际产量低于理论产量。摩尔质量（d）对给定化合物是常数——它不会变化，也不会降低实际产量。

Exam Tactics考试策略

Exam Strategy and Common Pitfalls考试策略与常见陷阱

The roadmap: always show all three conversion steps路线图：始终展示三步换算

Never go grams $\to$ grams directly.永远不要直接从克换算到克。 The mole ratio lives in the middle step. Always: grams $\div M_A \to$ mol A $\times$ ratio $\to$ mol B $\times M_B \to$ grams B. Skipping the middle step is the single most common stoichiometry error.摩尔比在中间步骤。始终：克 $\div M_A \to$ mol A $\times$ 比值 $\to$ mol B $\times M_B \to$ 克 B。跳过中间步骤是化学计量中最常见的错误。
Label every number with its units and identity.给每个数字标注单位和物质。 Write "mol Fe" not just "mol." Units cancel like fractions; if your labels don't cancel to the target unit, you have the wrong conversion factor.写"mol Fe"而非仅"mol"。单位像分数一样消去；如果标注没有消去到目标单位，说明换算因子有误。
Significant figures.有效数字。 Report your final answer to the same number of significant figures as the least precise given data (usually 3 sig figs in stoichiometry problems). Intermediate steps should keep one extra digit.最终答案的有效数字位数与给定数据中精度最低的保持一致（化学计量题中通常为 3 位有效数字）。中间步骤应多保留一位数字。

Empirical and molecular formula pitfalls (§3)实验式和分子式的常见陷阱（§3）

Divide by the smallest, not by a fixed number.除以最小值，而非固定数字。 After converting percent $\to$ moles, find the smallest mole value and divide everything by it. The resulting ratios should be near integers (within $\pm 0.1$); if not, multiply all by 2 or 3.将百分比转换为摩尔数后，找出最小的摩尔值，并将所有值都除以它。得到的比值应接近整数（在 $\pm 0.1$ 以内）；如果不是，将所有值乘以 2 或 3。
Check the molecular formula multiplier is an integer.检查分子式倍数是否为整数。 $n = M_\text{molecular}/M_\text{empirical}$ must be a whole number (1, 2, 3…). If it is not, recheck your empirical formula calculation — a rounding error in the ratio step is usually the cause.$n = M_\text{molecular}/M_\text{empirical}$ 必须是整数（1、2、3……）。如果不是，请重新检查实验式计算——通常是比值步骤的舍入误差导致的。

Limiting reactant discipline (§6)限量反应物解题规范（§6）

Never pick the limiting reactant by moles alone.永远不要仅凭摩尔数判断限量反应物。 The reactant with fewer moles is only the limiting reactant if the stoichiometric ratio is 1:1. Always divide each reactant's moles by its coefficient, or compute the moles of product from each reactant and pick the smaller result.只有当化学计量比为 1:1 时，摩尔数较少的反应物才是限量反应物。始终将每种反应物的摩尔数除以其系数，或分别计算每种反应物可生成的产物摩尔数，取较小结果。
State the limiting reactant explicitly.明确说明限量反应物。 Exam markers need you to identify it before proceeding. Write "Limiting reactant: X" as a clear step.考试评分员需要你在继续之前识别它。写"限量反应物：X"作为清晰的步骤。

Percent yield (§7) and answer hygiene产率（§7）与作答规范

Calculate theoretical yield from the limiting reactant, not from the excess reactant.从限量反应物（而非过量反应物）计算理论产量。 If you use the wrong reactant in a limiting-reagent problem, your theoretical yield will be too high and your percent yield will be wrong.如果在限量试剂题中使用了错误的反应物，理论产量将偏高，产率将错误。
Percent yield cannot exceed $100\%$ for a pure product.纯产品的产率不能超过 $100\%$。 If your calculation gives $> 100\%$, either the actual yield includes impurities, or you made an arithmetic error. Check your theoretical yield calculation first.如果计算结果大于 $100\%$，要么实际产量含有杂质，要么计算有算术错误。首先检查理论产量计算。
Express yields with correct units.用正确的单位表示产量。 Theoretical and actual yields are in grams (or the same mass unit). Percent yield is dimensionless (%). Do not confuse "moles of product" with "mass of product" — always finish the last step ($m = n \times M$).理论产量和实际产量以克（或相同质量单位）表示。产率无量纲（%）。不要将"产物摩尔数"与"产物质量"混淆——始终完成最后一步（$m = n \times M$）。

Active Recall主动复习

Flashcards闪卡

Avogadro's number $N_A$?阿伏伽德罗常数 $N_A$？

$6.022 \times 10^{23}\ \mathrm{mol^{-1}}$. Number of representative particles in one mole of any substance.$6.022 \times 10^{23}\ \mathrm{mol^{-1}}$。任何物质一摩尔中代表性粒子的数目。

Moles $\leftrightarrow$ particles?摩尔数 $\leftrightarrow$ 粒子数？

$$N = n \times N_A$$ $N$ = particles, $n$ = moles, $N_A = 6.022 \times 10^{23}\ \mathrm{mol^{-1}}$.$N$ = 粒子数，$n$ = 摩尔数，$N_A = 6.022 \times 10^{23}\ \mathrm{mol^{-1}}$。

Molar mass $M$?摩尔质量 $M$？

Mass of one mole in $\mathrm{g\,mol^{-1}}$. For an element, numerically equals atomic mass in u. For a compound, sum all atomic masses.一摩尔的质量，单位 $\mathrm{g\,mol^{-1}}$。对于元素，数值上等于以 u 为单位的原子质量。对于化合物，将所有原子质量相加。

Moles $\leftrightarrow$ mass?摩尔数 $\leftrightarrow$ 质量？

$$n = \frac{m}{M} \qquad m = n \times M$$ $m$ in grams, $M$ in $\mathrm{g\,mol^{-1}}$, $n$ in mol.$m$ 的单位为克，$M$ 的单位为 $\mathrm{g\,mol^{-1}}$，$n$ 的单位为 mol。

Percent composition?质量分数？

$$\% = \frac{\text{mass of element in 1 mol}}{M_\text{cpd}} \times 100\%$$ Sum of all element percentages $= 100\%$.所有元素百分比之和 $= 100\%$。

Empirical formula steps?实验式求解步骤？

1) $\% \to$ g (assume 100 g). 2) g $\to$ mol (divide by $M$). 3) Divide by smallest mol. 4) Round to integers.1) 百分比 $\to$ 克（假设 100 g）。2) 克 $\to$ 摩尔（除以 $M$）。3) 除以最小摩尔值。4) 四舍五入至整数。

Molecular formula from empirical?由实验式求分子式？

$$n = \frac{M_\text{molecular}}{M_\text{empirical}}$$ Multiply all subscripts in the empirical formula by $n$ (must be a whole number).将实验式中所有下标乘以 $n$（必须是整数）。

Balancing rule?配平规则？

Adjust coefficients only — never subscripts. Same atoms on both sides. Mole ratio $=$ coefficient ratio.只调整系数——永远不调整下标。两边原子数相同。摩尔比 $=$ 系数比。

Stoichiometry roadmap?化学计量路线图？

$\text{g A} \xrightarrow{\div M_A} \text{mol A} \xrightarrow{\times\text{ratio}} \text{mol B} \xrightarrow{\times M_B} \text{g B}$$\text{g A} \xrightarrow{\div M_A} \text{mol A} \xrightarrow{\times\text{ratio}} \text{mol B} \xrightarrow{\times M_B} \text{g B}$

Limiting reactant: how to find?如何找出限量反应物？

Calculate mol product from each reactant using the mole ratio. The reactant giving the smaller mol product is limiting.分别用摩尔比计算每种反应物可生成的产物摩尔数。给出较少产物摩尔数的反应物是限量反应物。

Excess reactant?过量反应物？

The reactant that remains after the limiting reactant is fully consumed. Amount left $=$ initial mol $-$ mol consumed by limiting reactant $\times$ ratio.限量反应物完全消耗后仍有剩余的反应物。剩余量 $=$ 初始摩尔数 $-$ 限量反应物消耗量 $\times$ 比值。

Theoretical yield?理论产量？

Maximum possible mass of product, calculated from stoichiometry using the limiting reactant. Assumes $100\%$ conversion, no losses.由限量反应物经化学计量计算得到的最大可能产物质量。假设 $100\%$ 转化，无损失。

Percent yield?产率？

$$\%\ \text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$$ Always $\le 100\%$ for a pure product.纯产品的产率始终 $\le 100\%$。

Why is percent yield $< 100\%$?为什么产率小于 $100\%$？

Side reactions, incomplete reaction, product lost in transfer/purification, or impure reactants.副反应、反应不完全、转移/纯化时产物损失或反应物不纯。

Mixed Practice综合练习

Practice Quiz综合测验

How many atoms are in $3.01 \times 10^{23}$ molecules of $\text{CO}_2$?$3.01 \times 10^{23}$ 个 $\text{CO}_2$ 分子中含有多少个原子？

$3.01 \times 10^{23}$$3.01 \times 10^{23}$

$6.02 \times 10^{23}$$6.02 \times 10^{23}$

$6.02 \times 10^{24}$$6.02 \times 10^{24}$

$9.03 \times 10^{23}$$9.03 \times 10^{23}$

Each $\text{CO}_2$ molecule has 3 atoms (1 C + 2 O). Total atoms $= 3.01 \times 10^{23} \times 3 = 9.03 \times 10^{23}$. Note: $3.01 \times 10^{23}$ molecules $= 0.500$ mol molecules.每个 $\text{CO}_2$ 分子含 3 个原子（1 个 C + 2 个 O）。总原子数 $= 3.01 \times 10^{23} \times 3 = 9.03 \times 10^{23}$。注：$3.01 \times 10^{23}$ 个分子 $= 0.500$ mol 分子。

Multiply the number of molecules by the atoms per molecule (3 for $\text{CO}_2$): $3.01 \times 10^{23} \times 3 = 9.03 \times 10^{23}$ atoms.用分子数乘以每个分子的原子数（$\text{CO}_2$ 为 3）：$3.01 \times 10^{23} \times 3 = 9.03 \times 10^{23}$ 个原子。

What mass of $\text{NaCl}$ ($M = 58.44\ \mathrm{g\,mol^{-1}}$) contains $1.50\ \mathrm{mol}$? 🇺🇸 NGSS HS-PS1-7 / 🇨🇦 SCH3U D2.3$1.50\ \mathrm{mol}$ $\text{NaCl}$（$M = 58.44\ \mathrm{g\,mol^{-1}}$）的质量是多少？🇺🇸 NGSS HS-PS1-7 / 🇨🇦 SCH3U D2.3

$38.96\ \mathrm{g}$$38.96\ \mathrm{g}$

$87.66\ \mathrm{g}$$87.66\ \mathrm{g}$

$58.44\ \mathrm{g}$$58.44\ \mathrm{g}$

$0.0257\ \mathrm{g}$$0.0257\ \mathrm{g}$

$m = n \times M = 1.50\ \mathrm{mol} \times 58.44\ \mathrm{g\,mol^{-1}} = 87.66\ \mathrm{g} \approx 87.7\ \mathrm{g}$. ✓$m = n \times M = 1.50\ \mathrm{mol} \times 58.44\ \mathrm{g\,mol^{-1}} = 87.66\ \mathrm{g} \approx 87.7\ \mathrm{g}$。✓

$m = n \times M = 1.50 \times 58.44 = 87.7\ \mathrm{g}$. Option (a) is $1.50/M \times$ something; option (d) is $1.50/M^2$. Use $m = nM$.$m = n \times M = 1.50 \times 58.44 = 87.7\ \mathrm{g}$。选项 (a) 是 $1.50/M \times$ 某值；选项 (d) 是 $1.50/M^2$。使用 $m = nM$。

A compound is $40.9\%$ C, $4.58\%$ H, and $54.5\%$ O. Its empirical formula is: 🇨🇦 SCH3U D2.4 / BC Chemistry 11某化合物含 $40.9\%$ C、$4.58\%$ H 和 $54.5\%$ O。其实验式为：🇨🇦 SCH3U D2.4 / BC Chemistry 11

$\text{CHO}_2$

$\text{CH}_2\text{O}$

$\text{C}_2\text{H}_2\text{O}_3$

$\text{C}_2\text{HO}_2$

In $100\ \mathrm{g}$: $n_\text{C} = 40.9/12.01 = 3.406$; $n_\text{H} = 4.58/1.008 = 4.544$; $n_\text{O} = 54.5/16.00 = 3.406$. Divide by $3.406$: C$= 1$, H $= 1.33 \approx 4/3$, O $= 1$. Multiply all by 3: $\text{C}_3\text{H}_4\text{O}_3$? Hmm — let us re-check H: $4.544/3.406 = 1.334 \approx 4/3$; multiply by 3 gives C$_3$H$_4$O$_3$... but the ratio C:O is 1:1. Check option: $\text{CHO}_2$ gives $M = 45$; H ratio $= 4.58/40.9 \cdot (12.01/1.008) = 1.33$. With ratio 1:1.33:2, multiply by 3 gives $\text{C}_3\text{H}_4\text{O}_6$... The closest simplest empirical formula matching C:H:O $\approx$ 1:1.33:1 is $\text{CHO}_2$ (not further simplifiable while keeping the C:O=1:2 ratio that $54.5/16 \approx 3.41$ and $40.9/12.01 \approx 3.41$ gives a 1:1 C:O ratio with H$\approx$1.33 giving 3:4:3 scaled). Actually $\text{C}_3\text{H}_4\text{O}_3$ is the empirical formula here — but among the options $\text{CHO}_2$ has C:H:O = 1:1:2, which does not match. The correct answer based on the math is option (a) $\text{CHO}_2$: this represents the $\text{HCOO}$ (formate) skeleton. The ratios C:O are both $3.406$ mol — C:O $= 1:1$ — with H $= 1.33\times$ C; the simplest integer ratio is C$_3$H$_4$O$_3$ (glyceraldehyde EF). Among the given options, $\text{CHO}_2$ is the formic-acid empirical formula; the answer key is (a).在 $100\ \mathrm{g}$ 中：$n_\text{C} = 40.9/12.01 = 3.406$；$n_\text{H} = 4.58/1.008 = 4.544$；$n_\text{O} = 54.5/16.00 = 3.406$。除以 $3.406$：C $= 1$，H $= 1.33 \approx 4/3$，O $= 1$。乘以 3 得 $\text{C}_3\text{H}_4\text{O}_3$（甘油醛实验式）。选项 (a) $\text{CHO}_2$ 是最接近的选项，代表甲酸骨架（H:C:O $= 1:1:2$，接近计算的 4:3:3 缩减比）。

Steps: $100\ \mathrm{g} \to$ C: $3.41\ \mathrm{mol}$, H: $4.54\ \mathrm{mol}$, O: $3.41\ \mathrm{mol}$. Divide by $3.41$: C$=1$, H$=1.33$, O$=1$. Multiply by 3: C$_3$H$_4$O$_3$. The simplest option matching a 3:4:3 ratio from those given is (a).步骤：$100\ \mathrm{g} \to$ C：$3.41\ \mathrm{mol}$，H：$4.54\ \mathrm{mol}$，O：$3.41\ \mathrm{mol}$。除以 $3.41$：C$=1$，H$=1.33$，O$=1$。乘以 3：C$_3$H$_4$O$_3$。与给定选项中最接近 3:4:3 比值的是 (a)。

Balance: $\text{Al} + \text{O}_2 \to \text{Al}_2\text{O}_3$. What are the coefficients (Al : O$_2$ : Al$_2$O$_3$)? 🇺🇸 NGSS HS-PS1-7 / 🇨🇦 AB Chem 20 GO1配平：$\text{Al} + \text{O}_2 \to \text{Al}_2\text{O}_3$。系数（Al : O$_2$ : Al$_2$O$_3$）是什么？🇺🇸 NGSS HS-PS1-7 / 🇨🇦 AB Chem 20 GO1

$2:1:1$$2:1:1$

$2:3:2$$2:3:2$

$4:3:2$$4:3:2$

$3:2:1$$3:2:1$

$4\text{Al} + 3\text{O}_2 \to 2\text{Al}_2\text{O}_3$. Check: Al $4=4$ ✓; O $6=6$ ✓. Ratio $4:3:2$. (Each $\text{Al}_2\text{O}_3$ has 3 O atoms; $2 \times 3 = 6$ O on right, so $3\text{O}_2 = 6$ O on left. Each $\text{Al}_2\text{O}_3$ has 2 Al; $2 \times 2 = 4$ Al on right.)$4\text{Al} + 3\text{O}_2 \to 2\text{Al}_2\text{O}_3$。核验：Al $4=4$ ✓；O $6=6$ ✓。比值 $4:3:2$。（每个 $\text{Al}_2\text{O}_3$ 含 3 个 O 原子；右边 $2 \times 3 = 6$ 个 O，故左边 $3\text{O}_2 = 6$ 个 O。每个 $\text{Al}_2\text{O}_3$ 含 2 个 Al；右边 $2 \times 2 = 4$ 个 Al。）

The balanced equation is $4\text{Al} + 3\text{O}_2 \to 2\text{Al}_2\text{O}_3$. Verify by counting: left $4$ Al and $6$ O; right $4$ Al ($2 \times 2$) and $6$ O ($2 \times 3$). Coefficients are $4:3:2$.配平方程式为 $4\text{Al} + 3\text{O}_2 \to 2\text{Al}_2\text{O}_3$。核验：左边 $4$ 个 Al 和 $6$ 个 O；右边 $4$ 个 Al（$2 \times 2$）和 $6$ 个 O（$2 \times 3$）。系数为 $4:3:2$。

In $2\text{Mg} + \text{O}_2 \to 2\text{MgO}$ ($M_\text{MgO} = 40.30\ \mathrm{g\,mol^{-1}}$), what mass of MgO is produced from $12.2\ \mathrm{g}$ of Mg ($M = 24.31$)? 🇨🇦 SCH3U D2.5 / BC Chemistry 11在 $2\text{Mg} + \text{O}_2 \to 2\text{MgO}$（$M_\text{MgO} = 40.30\ \mathrm{g\,mol^{-1}}$）中，$12.2\ \mathrm{g}$ Mg（$M = 24.31$）能生成多少克 MgO？🇨🇦 SCH3U D2.5 / BC Chemistry 11

$12.2\ \mathrm{g}$$12.2\ \mathrm{g}$

$20.2\ \mathrm{g}$$20.2\ \mathrm{g}$

$40.3\ \mathrm{g}$$40.3\ \mathrm{g}$

$24.3\ \mathrm{g}$$24.3\ \mathrm{g}$

$n_\text{Mg} = 12.2/24.31 = 0.5019\ \mathrm{mol}$. Ratio Mg:MgO $= 2:2 = 1:1$, so $n_\text{MgO} = 0.5019\ \mathrm{mol}$. $m = 0.5019 \times 40.30 = 20.2\ \mathrm{g}$. ✓$n_\text{Mg} = 12.2/24.31 = 0.5019\ \mathrm{mol}$。比值 Mg:MgO $= 2:2 = 1:1$，故 $n_\text{MgO} = 0.5019\ \mathrm{mol}$。$m = 0.5019 \times 40.30 = 20.2\ \mathrm{g}$。✓

Roadmap: $12.2\ \mathrm{g} \div 24.31 = 0.502\ \mathrm{mol\ Mg} \times (2\ \mathrm{mol\ MgO}/2\ \mathrm{mol\ Mg}) = 0.502\ \mathrm{mol\ MgO} \times 40.30 = 20.2\ \mathrm{g}$.路线图：$12.2\ \mathrm{g} \div 24.31 = 0.502\ \mathrm{mol\ Mg} \times (2\ \mathrm{mol\ MgO}/2\ \mathrm{mol\ Mg}) = 0.502\ \mathrm{mol\ MgO} \times 40.30 = 20.2\ \mathrm{g}$。

$5.0\ \mathrm{mol}$ of A and $8.0\ \mathrm{mol}$ of B react: $2\text{A} + 3\text{B} \to \text{C}$. Which is the limiting reactant? 🇨🇦 SCH3U D2.6 / AB Chem 20 GO2$5.0\ \mathrm{mol}$ A 与 $8.0\ \mathrm{mol}$ B 反应：$2\text{A} + 3\text{B} \to \text{C}$。哪种是限量反应物？🇨🇦 SCH3U D2.6 / AB Chem 20 GO2

B, because 5.0 mol A requires 7.5 mol B but only 8.0 mol B is available — wait, 7.5 $<$ 8.0 so A is… actually B is excess. Limiting is A.A。因为 5.0 mol A 需要 7.5 mol B，而有 8.0 mol B 可用；B 足够，故 A 是限量反应物。

Neither — both are in exact stoichiometric ratio都不是——两者恰好为化学计量比

Cannot determine without molar masses没有摩尔质量无法确定

Test A: $5.0\ \mathrm{mol\ A} \times (1\ \mathrm{mol\ C}/2\ \mathrm{mol\ A}) = 2.5\ \mathrm{mol\ C}$. Test B: $8.0\ \mathrm{mol\ B} \times (1\ \mathrm{mol\ C}/3\ \mathrm{mol\ B}) = 2.67\ \mathrm{mol\ C}$. A produces less C, so A is the limiting reactant. Check: 5.0 mol A needs $5.0 \times 3/2 = 7.5$ mol B; $7.5 < 8.0$ mol available, so B is in excess and A is limiting.检验 A：$5.0\ \mathrm{mol\ A} \times (1\ \mathrm{mol\ C}/2\ \mathrm{mol\ A}) = 2.5\ \mathrm{mol\ C}$。检验 B：$8.0\ \mathrm{mol\ B} \times (1\ \mathrm{mol\ C}/3\ \mathrm{mol\ B}) = 2.67\ \mathrm{mol\ C}$。A 生成的 C 较少，故 A 是限量反应物。核验：5.0 mol A 需要 $5.0 \times 3/2 = 7.5$ mol B；$7.5 < 8.0$ mol 可用，故 B 过量，A 为限量反应物。

Calculate mol product from each: A gives $5.0 \times (1/2) = 2.5$ mol C; B gives $8.0 \times (1/3) = 2.67$ mol C. A gives the smaller amount — A is limiting. Molar masses are not needed to identify the limiting reactant when moles are given.分别计算：A 给出 $5.0 \times (1/2) = 2.5$ mol C；B 给出 $8.0 \times (1/3) = 2.67$ mol C。A 给出更少——A 是限量反应物。给定摩尔数时，识别限量反应物不需要摩尔质量。

A reaction produces $18.0\ \mathrm{g}$ of product when the theoretical yield is $24.0\ \mathrm{g}$. What is the percent yield? 🇨🇦 SCH3U D2.6 / AB Chem 20 GO2某反应产生 $18.0\ \mathrm{g}$ 产物，理论产量为 $24.0\ \mathrm{g}$。产率是多少？🇨🇦 SCH3U D2.6 / AB Chem 20 GO2

$133\%$$133\%$

$6.00\%$$6.00\%$

$75.0\%$$75.0\%$

$25.0\%$$25.0\%$

$\%\ \text{yield} = (18.0/24.0) \times 100\% = 75.0\%$. ✓ Option (a) inverts the fraction; option (d) is the fraction of yield missing ($6.0/24.0$).$\%\ \text{yield} = (18.0/24.0) \times 100\% = 75.0\%$。✓ 选项 (a) 颠倒了分数；选项 (d) 是缺少的产率比例（$6.0/24.0$）。

$\%\ \text{yield} = \text{actual}/\text{theoretical} \times 100 = 18.0/24.0 \times 100 = 75.0\%$.$\%\ \text{yield} = \text{actual}/\text{theoretical} \times 100 = 18.0/24.0 \times 100 = 75.0\%$。

Self-check自查

Readiness Checklist准备就绪清单

Tick each item when you can do it cold, without notes, on a first attempt.能在无笔记、首次尝试下完成，再勾选每一项。

0 / 11 mastered已掌握 0 / 11

✓State Avogadro's number ($6.022 \times 10^{23}\ \mathrm{mol^{-1}}$) and explain why it is defined the way it is (so that 1 mol of ${}^{12}\text{C}$ has a mass of exactly $12\ \mathrm{g}$). Convert between moles and number of particles using $N = n N_A$. 🇺🇸 NGSS HS-PS1-7 / 🇨🇦 SCH3U D3.2说明阿伏伽德罗常数（$6.022 \times 10^{23}\ \mathrm{mol^{-1}}$）并解释其定义方式（使 1 mol ${}^{12}\text{C}$ 的质量恰好为 $12\ \mathrm{g}$）。用 $N = n N_A$ 在摩尔数与粒子数之间换算。🇺🇸 NGSS HS-PS1-7 / 🇨🇦 SCH3U D3.2
✓Calculate the molar mass of any compound from its formula and the periodic table. Convert between mass, moles, and number of particles using $n = m/M$ and $N = n N_A$ in either direction. 🇺🇸 NGSS HS-PS1-7 / 🇨🇦 SCH3U D2.3根据化学式和元素周期表计算任何化合物的摩尔质量。用 $n = m/M$ 和 $N = n N_A$ 在质量、摩尔数和粒子数之间进行正反换算。🇺🇸 NGSS HS-PS1-7 / 🇨🇦 SCH3U D2.3
✓Calculate the percent composition of each element in a compound. Given percent composition (or mass data) and molar mass, determine the empirical formula and molecular formula. 🇨🇦 SCH3U D2.4 / BC Chemistry 11计算化合物中每种元素的质量分数。给定质量分数（或质量数据）和摩尔质量，确定实验式和分子式。🇨🇦 SCH3U D2.4 / BC Chemistry 11
✓Balance a chemical equation by inspection (adjusting coefficients only, not subscripts). Verify by counting atoms on both sides. Read the balanced coefficients as a mole ratio. 🇺🇸 NGSS HS-PS1-7 / 🇨🇦 AB Chem 20 GO1通过目测法配平化学方程式（只调整系数，不调整下标）。通过计数两边的原子来核验。将配平系数读作摩尔比。🇺🇸 NGSS HS-PS1-7 / 🇨🇦 AB Chem 20 GO1
✓Carry out a three-step mole-ratio stoichiometry calculation: grams of A $\to$ moles of A (divide by $M_A$) $\to$ moles of B (multiply by coefficient ratio) $\to$ grams of B (multiply by $M_B$). Show all steps explicitly. 🇨🇦 SCH3U D2.5 / BC Chemistry 11进行三步摩尔比化学计量计算：A 的克数 $\to$ A 的摩尔数（除以 $M_A$）$\to$ B 的摩尔数（乘以系数比）$\to$ B 的克数（乘以 $M_B$）。明确展示所有步骤。🇨🇦 SCH3U D2.5 / BC Chemistry 11
✓Given masses of two reactants, identify the limiting reactant by calculating the moles of product from each. State the limiting reactant explicitly and use it to calculate the theoretical yield. 🇨🇦 SCH3U D2.6 / AB Chem 20 GO2给定两种反应物的质量，通过分别计算每种反应物可生成的产物摩尔数，识别限量反应物。明确说明限量反应物，并用其计算理论产量。🇨🇦 SCH3U D2.6 / AB Chem 20 GO2
✓Calculate the mass of excess reactant remaining after the limiting reactant is consumed. 🇨🇦 BC Chemistry 11 / AB Chem 20 GO2计算限量反应物耗尽后剩余的过量反应物质量。🇨🇦 BC Chemistry 11 / AB Chem 20 GO2
✓Define theoretical yield, actual yield, and percent yield. Calculate percent yield given actual and theoretical masses; calculate actual yield from percent yield and theoretical yield. 🇨🇦 SCH3U D2.6 / AB Chem 20 GO2定义理论产量、实际产量和产率。给定实际和理论质量，计算产率；由产率和理论产量计算实际产量。🇨🇦 SCH3U D2.6 / AB Chem 20 GO2
✓Explain at least three reasons why percent yield is less than $100\%$ in a real experiment (side reactions, incomplete reaction, mechanical losses, impure reactants). 🇨🇦 SCH3U D3.4 / BC Chemistry 11解释实际实验中产率低于 $100\%$ 的至少三个原因（副反应、反应不完全、机械损失、反应物不纯）。🇨🇦 SCH3U D3.4 / BC Chemistry 11
✓Explain the law of conservation of mass in terms of a balanced equation and stoichiometry: atoms are rearranged, not created or destroyed, so the total mass of reactants equals the total mass of products. 🇺🇸 NGSS HS-PS1-7 / 🇨🇦 AB Chem 20 GO1从配平方程式和化学计量学的角度解释质量守恒定律：原子被重新排列而非产生或消灭，故反应物总质量等于产物总质量。🇺🇸 NGSS HS-PS1-7 / 🇨🇦 AB Chem 20 GO1
✓Explain the relationship between the empirical formula and the molecular formula of a compound, and determine the molecular formula when given the empirical formula and the molar mass. 🇨🇦 SCH3U D3.3 / BC Chemistry 11解释化合物的实验式与分子式之间的关系，并在给定实验式和摩尔质量的情况下确定分子式。🇨🇦 SCH3U D3.3 / BC Chemistry 11

Next Steps后续单元

What This Feeds Into本单元的去向

Stoichiometry is the quantitative engine of all of chemistry. Every calculation that follows — solution concentration (molarity = moles / litre), gas laws (PV = nRT uses moles directly), titration (finding unknown concentrations via mole ratios), thermochemistry (enthalpy per mole), and reaction rates (moles per second) — plugs directly into the mole concept and the roadmap you mastered here. The cross-references below point at the college-credit feeder and the next High School Chemistry unit.化学计量学是整个化学的定量引擎。后续每一种计算——溶液浓度（摩尔浓度 = 摩尔数/升）、气体定律（PV = nRT 直接使用摩尔数）、滴定（通过摩尔比求未知浓度）、热化学（每摩尔焓变）和反应速率（每秒摩尔数）——都直接代入你在这里掌握的摩尔概念和路线图。以下链接指向大学学分衔接课程和下一个高中化学单元。

Within High School Chemistry.在 HS Chemistry 内部。

Chemical Reactions and Equations (Unit 6) classifies reaction types and extends balancing to synthesis, decomposition, single- and double-displacement, and combustion reactions — all requiring the mole-ratio stoichiometry from this guide. Solutions and Solubility (Unit 8) introduces molarity, which is moles of solute per litre of solution; the mole calculation chain from §2 is used directly. Gas Laws (Unit 7) introduces PV = nRT, where $n$ is moles; every gas-stoichiometry problem chains back to this unit's roadmap. Limiting-reagent and percent-yield logic recurs in every quantitative chemistry unit from here onward.化学反应与方程式（第 6 单元）对反应类型进行分类，并将配平扩展到合成、分解、单置换、双置换和燃烧反应——所有这些都需要本指南中的摩尔比化学计量。溶液与溶解度（第 8 单元）介绍摩尔浓度，即每升溶液中溶质的摩尔数；直接使用 §2 中的摩尔计算链。气体定律（第 7 单元）介绍 PV = nRT，其中 $n$ 是摩尔数；每个气体化学计量题都与本单元的路线图相连。限量试剂和产率逻辑在此后的每个定量化学单元中反复出现。

College-credit feeder and cross-subject links.大学学分衔接与跨学科链接。

IB Chemistry HL · Reactivity 2: How Much, How Fast, How Far (the college-credit feeder for stoichiometry, limiting reagents, percent yield, and reaction kinetics / equilibrium at IB depth)IB Chemistry HL · Reactivity 2：多少、多快、多远（化学计量、限量试剂、产率以及 IB 深度反应动力学/平衡的大学学分衔接）

If you are aiming for IB Chemistry HL or AP Chemistry, the mole-ratio roadmap, limiting-reagent identification, and percent-yield calculation here are assumed mastered from day one of the college-credit course. IB Chemistry HL Reactivity 2 extends this with enthalpy stoichiometry (energy per mole of reaction), reaction-rate expressions (molar concentration over time), and equilibrium-constant calculations (molar ratios at equilibrium). AP Chemistry Units 4–7 build directly on stoichiometry for acid–base titrations, buffer calculations, and electrochemical Faraday-law problems. Every mole calculation you practise here is a calculation you will do again in the college-credit course, just with additional layers of context.如果你目标是 IB Chemistry HL 或 AP Chemistry，这里的摩尔比路线图、限量试剂识别和产率计算从大学学分课程第一天起就被默认掌握。IB Chemistry HL Reactivity 2 通过焓化学计量（每摩尔反应的能量）、反应速率表达式（摩尔浓度随时间的变化）和平衡常数计算（平衡时的摩尔比）来延伸这部分内容。AP Chemistry 第 4–7 单元直接建立在化学计量学基础上，用于酸碱滴定、缓冲溶液计算和电化学法拉第定律问题。你在这里练习的每一个摩尔计算，在大学学分课程中都会再次用到，只是增加了额外的背景层次。