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.
Stoichiometry
IGCSE Chemistry · Topic 3
3.1
Chemical formulae
Syllabus
| Core | Supplement |
|---|---|
| 1 State the formulae of the elements and compounds named in the subject content | |
| 2 Define the molecular formula of a compound as the number and type of different atoms in one molecule | 5 Define the empirical formula of a compound as the simplest whole number ratio of the different atoms or ions in a compound |
| 3 Deduce the formula of a simple compound from the relative numbers of atoms present in a model or a diagrammatic representation | 6 Deduce the formula of an ionic compound from the relative numbers of the ions present in a model or a diagrammatic representation or from the charges on the ions |
| 4 Construct word equations and symbol equations to show how reactants form products, including state symbols | 7 Construct symbol equations with state symbols, including ionic equations |
| 8 Deduce the symbol equation with state symbols for a chemical reaction, given relevant information |
Source: Cambridge International syllabus
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 $\text{Na}_2\text{O}$.
For an ionic compound, the ion charges cross over to give the formula (here $\text{Al}_2\text{O}_3$)
Explore
Formula writing route
Follow charges or valencies to a neutral chemical formula.
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| stoichiometry | 化学计量 | huà xué jì liàng |
| formula | 化学式 | huà xué shì |
| atoms | 原子 | yuán zi |
| molecular formula | 分子式 | fēn zǐ shì |
| molecule | 分子 | fèn zǐ |
| empirical formula | 实验式 | shí yàn shì |
| ratio | 比例 | bǐ lì |
| ions | 离子 | lí zi |
| compound | 化合物 | huà hé wù |
| ionic compound | 离子化合物 | lí zi huà hé wù |
3.1
Writing equations
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.
$$2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$$
A balanced symbol equation has the same number of each atom on both sides
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:
$$\text{H}^{+}(aq) + \text{OH}^{-}(aq) \rightarrow \text{H}_2\text{O}(l)$$
Explore
Balance the equation
Step the coefficients until every element has the same number of atoms on both sides.
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| reactants | 反应物 | fǎn yìng wù |
| products | 生成物 | shēng chéng wù |
| word equation | 文字方程式 | wén zì fāng chéng shì |
| symbol equation | 化学方程式 | huà xué fāng chéng shì |
| balanced | 配平 | pèi píng |
| state symbols | 状态符号 | zhuàng tài fú hào |
| aqueous | 水溶液 | shuǐ róng yè |
| ionic equation | 离子方程式 | lí zi fāng chéng shì |
| spectator ions | 旁观离子 | páng guān lí zi |
3.2
Relative masses
Syllabus
| Core | Supplement |
|---|---|
| 1 Describe relative atomic mass, $A_r$, as the average mass of the isotopes of an element compared to 1/12th of the mass of an atom of $^{12}\text{C}$ | |
| 2 Define relative molecular mass, $M_r$, as the sum of the relative atomic masses. Relative formula mass, $M_r$, will be used for ionic compounds | |
| 3 Calculate reacting masses in simple proportions. Calculations will not involve the mole concept |
Source: Cambridge International syllabus
A laboratory balance measures mass — the basis of mole calculations.
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.
Explore
Formula mass lab
mass = moles x Mr
Change number of formula units and see total mass scale.
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| relative atomic mass | 相对原子质量 | xiāng duì yuán zi zhì liàng |
| element | 元素 | yuán sù |
| mass | 质量 | zhì liàng |
| relative molecular mass | 相对分子质量 | xiāng duì fèn zǐ zhì liàng |
| relative formula mass | 相对式量 | xiāng duì shì liàng |
3.3
The mole
Syllabus
| Core | Supplement |
|---|---|
| 2 State that the mole, mol, is the unit of amount of substance and that one mole contains $6.02 \times 10^{23}$ particles, e.g. atoms, ions, molecules; this number is the Avogadro constant | |
| 3 Use the relationship $\text{amount of substance (mol)} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$ to calculate: (a) amount of substance (b) mass (c) molar mass (d) relative atomic mass or relative molecular/formula mass (e) number of particles, using the value of the Avogadro constant | |
| 4 Use the molar gas volume, taken as $24\text{ dm}^3$ at room temperature and pressure, r.t.p., in calculations involving gases | |
| 1 State that concentration can be measured in $\text{g/dm}^3$ or $\text{mol/dm}^3$ | 5 Calculate stoichiometric reacting masses, limiting reactants, volumes of gases at r.t.p., volumes of solutions and concentrations of solutions expressed in $\text{g/dm}^3$ and $\text{mol/dm}^3$, including conversion between $\text{cm}^3$ and $\text{dm}^3$ |
| 6 Use experimental data from a titration to calculate the moles of solute, or the concentration or volume of a solution | |
| 7 Calculate empirical formulae and molecular formulae, given appropriate data | |
| 8 Calculate percentage yield, percentage composition by mass and percentage purity, given appropriate data |
Source: Cambridge International syllabus
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 $24 \text{ dm}^3$.
One mole of any gas fills 24 cubic decimetres at room temperature and pressure
$$\text{volume of gas (dm}^3) = \text{amount (mol)} \times 24$$
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.
Moles sit at the centre: multiply going outwards, divide coming back (concentration is moles ÷ volume)
Explore
Mole particle lab
particles = n x Avogadro constant
Change moles and see particle number scale with Avogadro constant.
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| mole | 摩尔 | mó ěr |
| amount of substance | 物质的量 | wù zhì dì liàng |
| particles | 粒子 | lì zi |
| Avogadro constant | 阿伏伽德罗常数 | ā fú gā dé luó cháng shù |
| molar mass | 摩尔质量 | mó ěr zhì liàng |
| volume | 体积 | tǐ jī |
| molar gas volume | 摩尔气体体积 | mó ěr qì tǐ tǐ jī |
| concentration | 浓度 | nóng dù |
| solution | 溶液 | róng yè |
3.3
Doing reaction calculations
A titration measures exactly how much acid reacts with a base.
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).
B runs out first, so it is the limiting reactant and sets how much product forms; the spare A is left over in excess
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 $25.0 \text{ cm}^3$:
$$\text{concentration} = \frac{0.0020}{25.0 / 1000} = 0.08 \text{ mol/dm}^3$$
The titration calculation: known solution to moles, ratio, then the unknown concentration
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.
$$\% \text{ purity} = \frac{45}{50} \times 100 = 90\%$$
Explore
Reaction calculation route
Follow a known mass through a balanced equation to the answer.
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| limiting reactant | 限量反应物 | xiàn liàng fǎn yìng wù |
| titration | 滴定 | dī dìng |
| solute | 溶质 | róng zhì |
| percentage yield | 产率 | chǎn lǜ |
| percentage composition by mass | 质量分数 | zhì liàng fēn shù |
| percentage purity | 纯度 | chún dù |
3.3
Exam tips
- 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.