English
Stoichiometry 化学计量 is the study of the amounts of substances in a reaction — how much reacts and how much is made. This topic is mostly about counting atoms and doing calculations.
化学计量
IGCSE 化学 · 知识点 3
15:33
化学计量学
这是一个化学家的问题。单个原子太小看不见，而最小的尘埃里有数十亿个原子。那么如何计数呢？你…
英语旁白 · 英中内嵌字幕
3.1
化学式
教学大纲
| 核心 | 补充 |
|---|---|
| 1 写出学科内容中提及的元素和化合物的化学式 | |
| 2 将化合物的分子式定义为：一个分子中所含不同原子的种类和数量 | 5 将化合物的实验式（最简式）定义为：化合物中不同原子或离子之间的最简单整数比 |
| 3 根据模型或示意图中原子的相对数量，推导出简单化合物的化学式 | 6 根据模型或示意图中离子的相对数量，或根据离子的电荷，推导出离子化合物的化学式 |
| 4 构建文字方程式和符号方程式，以展示反应物如何转化为生成物，包括状态符号 | 7 构建带状态符号的符号方程式，包括离子方程式 |
| 8 根据相关信息，推导化学反应带状态符号的符号方程式 |
来源：Cambridge International 教学大纲
English
A formula 化学式 shows which atoms 原子 are in a substance, and how many of each. There are two kinds of formula you must know.
- The molecular formula 分子式 is the actual number of each atom in one molecule 分子. For glucose it is $\text{C}_6\text{H}_{12}\text{O}_6$.
- The empirical formula 实验式 is the simplest whole-number ratio 比例 of the atoms or ions 离子 in a compound 化合物. For glucose it is $\text{CH}_2\text{O}$.
Working out a formula
If you are given a model or diagram, just count the atoms of each element and write them as a formula.
For an ionic compound 离子化合物 you can work out the formula from the charges on the ions. The total positive charge must balance the total negative charge, because the compound has no overall charge. Some common ions:
| Positive ions | Negative ions |
|---|---|
| $\text{Na}^{+}$, $\text{K}^{+}$, $\text{H}^{+}$ | $\text{Cl}^{-}$, $\text{OH}^{-}$, $\text{NO}_3^{-}$ |
| $\text{Mg}^{2+}$, $\text{Ca}^{2+}$, $\text{Cu}^{2+}$ | $\text{O}^{2-}$, $\text{SO}_4^{2-}$, $\text{CO}_3^{2-}$ |
| $\text{Al}^{3+}$ | $\text{N}^{3-}$ |
For example, $\text{Na}^{+}$ and $\text{O}^{2-}$: you need two $\text{Na}^{+}$ to balance one $\text{O}^{2-}$, so the formula is $\text{Na}_2\text{O}$.
探索
书写化学式的步骤
遵循电荷或化合价得出中性化学式。
词汇
训练
| English | 中文 | 拼音 |
|---|---|---|
| stoichiometry/ˌstəʊɪkɪˈɒmətri/ | 化学计量 | huà xué jì liàng |
| formula/ˈfɔːmjʊlə/ | 化学式 | huà xué shì |
| atoms/ˈætəmz/ | 原子 | yuán zǐ |
| molecular formula/məˈlekjʊlə ˈfɔːmjʊlə/ | 分子式 | fēn zǐ shì |
| molecule/ˈmɒlɪkjuːl/ | 分子 | fēn zǐ |
| empirical formula/emˈpɪrɪkl ˈfɔːmjʊlə/ | 实验式 | shí yàn shì |
| ratio/ˈreɪʃɪəʊ/ | 比例 | bǐ lì |
| ions/ˈaɪɒnz/ | 离子 | lí zǐ |
| compound/ˈkɒmpaʊnd/ | 化合物 | huà hé wù |
| ionic compound/aɪˈɒnɪk ˈkɒmpaʊnd/ | 离子化合物 | lí zǐ huà hé wù |
3.1
书写方程式
English
An equation shows how reactants 反应物 (the starting substances) change into products 生成物 (the substances made).
- A word equation 文字方程式 uses names: magnesium + oxygen → magnesium oxide.
- A symbol equation 化学方程式 uses formulae and must be balanced 配平 — the same number of each atom on both sides.
$$2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$$
You add state symbols 状态符号 in brackets to show the physical state: $(s)$ solid, $(l)$ liquid, $(g)$ gas, and $(aq)$ for aqueous 水溶液 (dissolved in water).
$$\text{Zn}(s) + 2\text{HCl}(aq) \rightarrow \text{ZnCl}_2(aq) + \text{H}_2(g)$$
Ionic equations
An ionic equation 离子方程式 shows only the ions that actually change. Ions that are the same on both sides are spectator ions 旁观离子 and are left out. For example, when an acid reacts with an alkali:
$$\text{H}^{+}(aq) + \text{OH}^{-}(aq) \rightarrow \text{H}_2\text{O}(l)$$
探索
配平方程式
调整系数直到每种元素在方程式两边具有相同数量的原子。
词汇
训练
| English | 中文 | 拼音 |
|---|---|---|
| reactants/rɪˈæktənts/ | 反应物 | fǎn yìng wù |
| products/ˈprɒdʌkts/ | 生成物 | shēng chéng wù |
| word equation/wɜːd ɪˈkweɪʒn/ | 文字方程式 | wén zì fāng chéng shì |
| symbol equation/ˈsɪmbl ɪˈkweɪʒn/ | 化学方程式 | huà xué fāng chéng shì |
| balanced/ˈbælənst/ | 配平 | pèi píng |
| state symbols/steɪt ˈsɪmblz/ | 状态符号 | zhuàng tài fú hào |
| aqueous/ˈeɪkwɪəs/ | 水溶液 | shuǐ róng yè |
| ionic equation/aɪˈɒnɪk ɪˈkweɪʒn/ | 离子方程式 | lí zi fāng chéng shì |
| spectator ions/spekˈteɪtə ˈaɪɒnz/ | 旁观离子 | páng guān lí zi |
3.2
原子与分子的相对质量
教学大纲
| 核心内容 | 补充内容 |
|---|---|
| 1 描述相对原子质量 $A_r$，即某元素各种同位素的平均质量与碳-12 ($^{12}\text{C}$) 原子质量的 1/12 相比的比值 | |
| 2 定义相对分子质量 $M_r$ 为各原子的相对原子质量之和。对于离子化合物，将使用相对式量 $M_r$ 来表示 | |
| 3 计算简单比例下的反应物质量。计算不涉及摩尔概念 |
来源：Cambridge International 教学大纲
English
The relative atomic mass 相对原子质量 ($A_r$) of an element 元素 is the average mass 质量 of its atoms compared to $\tfrac{1}{12}$ of the mass of one $^{12}\text{C}$ atom.
The relative molecular mass 相对分子质量 ($M_r$) is the sum of the relative atomic masses of all the atoms in the molecule. For ionic compounds we use the relative formula mass 相对式量, found the same way from the formula.
$$M_r(\text{H}_2\text{O}) = (2 \times 1) + 16 = 18$$
Reacting masses by simple proportion
You can sometimes find a reacting mass without the mole. If 24 g of magnesium makes 40 g of magnesium oxide, then 12 g of magnesium (half as much) makes 20 g of magnesium oxide.
探索
式量实验
质量 = 摩尔数 × Mr
改变式单位数量并观察总质量的变化。
词汇
训练
| English | 中文 | 拼音 |
|---|---|---|
| relative atomic mass/ˈrelətɪv əˈtɒmɪk mæs/ | 相对原子质量 | xiāng duì yuán zǐ zhì liàng |
| element/ˈelɪmənt/ | 元素 | yuán sù |
| mass/mæs/ | 质量 | zhì liàng |
| relative molecular mass/ˈrelətɪv məˈlekjʊlə mæs/ | 相对分子质量 | xiāng duì fēn zǐ zhì liàng |
| relative formula mass/ˈrelətɪv ˈfɔːmjʊlə mæs/ | 相对式量 | xiāng duì shì liàng |
3.3
摩尔与阿伏伽德罗常数
教学大纲
| 核心内容 | 补充内容 |
|---|---|
| 2 指出摩尔 (mol) 是物质的量的单位，且一摩尔包含 $6.02 \times 10^{23}$ 个粒子，例如原子、离子、分子；这个数字称为阿伏伽德罗常数 | |
| 3 使用关系式 $\text{amount of substance (mol)} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$ 计算：(a) 物质的量 (b) 质量 (c) 摩尔质量 (d) 相对原子质量或相对分子/式量 (e) 粒子数，使用阿伏伽德罗常数的值 | |
| 4 使用摩尔气体体积，在室温压力（r.t.p.）下取值为 $24\text{ dm}^3$，进行涉及气体的计算 | |
| 1 陈述浓度可以用 $\text{g/dm}^3$ 或 $\text{mol/dm}^3$ 来衡量 | 5 计算化学计量反应质量、限制反应物、室温压力下的气体体积、溶液体积以及以 $\text{g/dm}^3$ 和 $\text{mol/dm}^3$ 表示的溶液浓度，包括在 $\text{cm}^3$ 和 $\text{dm}^3$ 之间的转换 |
| 6 利用滴定实验数据计算溶质的摩尔数，或溶液的浓度或体积 | |
| 7 根据适当数据计算实验式式和分子式 | |
| 8 根据适当数据计算百分比产率、质量百分比组成和纯度百分比 |
来源：Cambridge International 教学大纲
English
The mole 摩尔 (symbol mol) is the unit for the amount of substance 物质的量. One mole of any substance contains $6.02 \times 10^{23}$ particles 粒子 (atoms, ions or molecules). This number is the Avogadro constant 阿伏伽德罗常数.
The molar mass 摩尔质量 is the mass of one mole, in grams per mole (g/mol). Its number is the same as the $A_r$ or $M_r$. The key relationship is:
$$\text{amount (mol)} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$$
Worked example. How many moles are in 36 g of water? Molar mass of water $= 18$ g/mol.
$$n = \frac{36}{18} = 2 \text{ mol}$$
To find the number of particles, multiply the moles by the Avogadro constant.
Volumes of gases
At room temperature and pressure (r.t.p.), one mole of any gas takes up the same volume 体积. This molar gas volume 摩尔气体体积 is $24 \text{ dm}^3$.
$$\text{volume of gas (dm}^3) = \text{amount (mol)} \times 24$$
Worked example. What volume does $0.25$ mol of carbon dioxide occupy at r.t.p.?
$$\text{volume} = 0.25 \times 24 = 6 \text{ dm}^3$$
Concentration of solutions
The concentration 浓度 of a solution 溶液 can be given in $\text{g/dm}^3$ or in $\text{mol/dm}^3$.
$$\text{concentration (mol/dm}^3) = \frac{\text{amount of solute (mol)}}{\text{volume (dm}^3)}$$
Remember to convert volume: $1 \text{ dm}^3 = 1000 \text{ cm}^3$, so divide a volume in $\text{cm}^3$ by 1000.
探索
摩尔粒子实验
粒子数 = n × 阿伏伽德罗常数
改变摩尔数并观察粒子数量随阿伏伽德罗常数的变化。
词汇
训练
| English | 中文 | 拼音 |
|---|---|---|
| mole/məʊl/ | 摩尔 | mó ěr |
| amount of substance/əˈmaʊnt ɒv ˈsʌbstəns/ | 物质的量 | wù zhì de liàng |
| particles/ˈpɑːtɪklz/ | 粒子 | lì zǐ |
| Avogadro constant/ˌævəˈɡædrəʊ ˈkɒnstənt/ | 阿伏伽德罗常数 | ā fú gā dé luó cháng shù |
| molar mass/ˈməʊlə mæs/ | 摩尔质量 | mó ěr zhì liàng |
| volume/ˈvɒljuːm/ | 体积 | tǐ jī |
| molar gas volume/ˈməʊlə ɡæs ˈvɒljuːm/ | 摩尔气体体积 | mó ěr qì tǐ tǐ jī |
| concentration/ˌkɒnsənˈtreɪʃn/ | 浓度 | nóng dù |
| solution/səˈluːʃn/ | 溶液 | róng yè |
3.3
进行反应计算
English
Most calculations follow the same steps: change the known amount into moles, use the balanced equation to find the moles of what you want, then change back into mass, volume or concentration.
Worked example. What mass of magnesium oxide forms when 12 g of magnesium burns completely? $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$. ($A_r$: Mg $= 24$, O $= 16$.)
Moles of Mg $= 12 / 24 = 0.5$ mol. The ratio Mg : MgO is $1 : 1$, so $0.5$ mol of MgO forms. Its molar mass is $24 + 16 = 40$ g/mol, so the mass is $0.5 \times 40 = 20$ g — the same answer as the simple-proportion method above.
Limiting reactant
When two reactants are mixed, often one runs out first. This is the limiting reactant 限量反应物. It decides how much product can form; any other reactant is in excess (left over).
Worked example. 8 g of hydrogen reacts with 8 g of oxygen: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$. Which is the limiting reactant, and what mass of water forms? ($A_r$: H $= 1$, O $= 16$.)
Moles of $\text{H}_2 = 8 / 2 = 4$ mol; moles of $\text{O}_2 = 8 / 32 = 0.25$ mol. The equation needs $2\text{H}_2$ for every one $\text{O}_2$, so $0.25$ mol of $\text{O}_2$ reacts with only $0.5$ mol of $\text{H}_2$. There is far more hydrogen than that, so oxygen is the limiting reactant and the hydrogen is in excess. The oxygen makes $2 \times 0.25 = 0.5$ mol of water, so the mass is $0.5 \times 18 = 9$ g. Always work in moles, not masses — the reactant with the smaller mass is not always the one that runs out.
Titration calculation
In a titration 滴定 you measure the volume of one solution that reacts with another. From the volume and concentration you find the moles of one solute 溶质, then use the equation ratio to find the moles, concentration or volume of the other.
Worked example. $25.0 \text{ cm}^3$ of sodium hydroxide solution is exactly neutralised by $20.0 \text{ cm}^3$ of $0.10 \text{ mol/dm}^3$ hydrochloric acid. $\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$. Find the concentration of the sodium hydroxide.
Moles of HCl $= 0.10 \times \dfrac{20.0}{1000} = 0.0020$ mol. The ratio is $1 : 1$, so there are $0.0020$ mol of NaOH in $25.0 \text{ cm}^3$:
$$\text{concentration} = \frac{0.0020}{25.0 / 1000} = 0.08 \text{ mol/dm}^3$$
Empirical and molecular formulae from data
To find an empirical formula from masses or percentages:
- Divide each element's mass (or %) by its $A_r$.
- Divide all the answers by the smallest one to get the simplest ratio.
To find the molecular formula, compare the empirical formula mass with the real $M_r$ and multiply up.
Worked example. A compound contains $40\%$ calcium, $12\%$ carbon and $48\%$ oxygen by mass. Find its empirical formula. ($A_r$: Ca $= 40$, C $= 12$, O $= 16$.)
Divide each percentage by its $A_r$: Ca $= 40/40 = 1$, C $= 12/12 = 1$, O $= 48/16 = 3$. The ratio is $1 : 1 : 3$, so the empirical formula is $\text{CaCO}_3$.
Percentages
- Percentage yield 产率 compares how much product you actually got with the most you could get: $\dfrac{\text{actual}}{\text{theoretical}} \times 100$.
- Percentage composition by mass 质量分数 of an element $= \dfrac{\text{mass of the element}}{M_r} \times 100$.
- Percentage purity 纯度 $= \dfrac{\text{mass of pure substance}}{\text{mass of impure sample}} \times 100$.
Worked example. Find the percentage by mass of nitrogen in ammonium nitrate, $\text{NH}_4\text{NO}_3$. ($A_r$: N $= 14$, H $= 1$, O $= 16$.)
$M_r = (2 \times 14) + (4 \times 1) + (3 \times 16) = 80$. The mass of nitrogen is $2 \times 14 = 28$, so
$$\% \text{ nitrogen} = \frac{28}{80} \times 100 = 35\%$$
Worked example. A reaction could make at most 5.0 g of product, but only 4.0 g is actually collected. Find the percentage yield.
$$\% \text{ yield} = \frac{4.0}{5.0} \times 100 = 80\%$$
Some product is always lost — left behind in the apparatus, or used up in side reactions — so the yield is almost never $100\%$.
Worked example. A 50 g sample of impure calcium carbonate contains 45 g of pure calcium carbonate. Find its percentage purity.
$$\% \text{ purity} = \frac{45}{50} \times 100 = 90\%$$
Common mistakes.
- Writing a hydroxide with a missing bracket: $\text{CaOH}_2$. When an ion made of several atoms appears more than once, put it in brackets: $\text{Ca(OH)}_2$ means one $\text{Ca}^{2+}$ and two $\text{OH}^{-}$.
- Using the mass of $\text{O}$ (16) for oxygen gas. Oxygen gas is $\text{O}_2$, so its $M_r$ is 32; the same goes for $\text{H}_2$, $\text{N}_2$ and $\text{Cl}_2$.
- Forgetting the equation ratio. In $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$, 1 mol of $\text{O}_2$ makes 2 mol of water, not 1.
- Using the volume in $\text{cm}^3$ in a concentration formula. Divide by 1000 first: $25.0\ \text{cm}^3 = 0.0250\ \text{dm}^3$.
- Using 24 dm³ for a solid or a solution. The molar gas volume applies only to gases, at room temperature and pressure.
- Rounding the empirical-formula ratio too early. A ratio such as $1 : 1.5$ is not $1 : 2$; multiply by 2 to get $2 : 3$.
- Turning the percentage-yield fraction upside down. It is actual ÷ theoretical, so the answer can never be above 100%.
探索
反应计算路径
根据已知质量通过配平的方程式推导出答案。
词汇
训练
| English | 中文 | 拼音 |
|---|---|---|
| limiting reactant/ˈlɪmɪtɪŋ rɪˈæktənt/ | 限量反应物 | xiàn liàng fǎn yìng wù |
| titration/taɪˈtreɪʃn/ | 滴定 | dī dìng |
| solute/ˈsɒljuːt/ | 溶质 | róng zhì |
| percentage yield/pəˈsentɪdʒ jiːld/ | 产率 | chǎn lǜ |
| percentage composition by mass/pəˈsentɪdʒ ˌkɒmpəˈzɪʃn baɪ mæs/ | 质量分数 | zhì liàng fēn shù |
| percentage purity/pəˈsentɪdʒ ˈpjʊərɪti/ | 纯度 | chún dù |
3.3
考试技巧
English
- Always turn masses, gas volumes and concentrations into moles first, use the balanced equation's ratio, then convert back. Never compare masses directly across an equation.
- Key formulas: $n = \dfrac{\text{mass}}{M_r}$, gas volume $= n \times 24\ \text{dm}^3$ at r.t.p., and concentration $= \dfrac{n}{\text{volume in dm}^3}$. Change cm³ to dm³ by dividing by 1000.
- To balance a symbol equation you may only change the big numbers in front of a formula, never the small subscripts inside it.
- The limiting reactant is the one that runs out (fewest moles once you allow for the ratio); it sets how much product forms. The other reactant is in excess.
- Percentage yield = actual ÷ theoretical × 100; percentage purity = mass of pure substance ÷ mass of sample × 100. Both are always 100% or less.
该知识点的互动课程
逐步完成，配合即时检查练习。