Stoichiometry · ⁨สโตइโอเมทรี⁩

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:

For example, $\text{Na}^{+}$ and $\text{O}^{2-}$: you need two $\text{Na}^{+}$ to balance one $\text{O}^{2-}$, so the formula is $\text{Na}_2\text{O}$.

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.

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).

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:

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.

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.

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:

Worked example. How many moles are in 36 g of water? Molar mass of water $= 18$ g/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 $24 \text{ dm}^3$.

Worked example. What volume does $0.25$ mol of carbon dioxide occupy at r.t.p.?

Concentration of solutions

The concentration 浓度 of a solution 溶液 can be given in $\text{g/dm}^3$ or in $\text{mol/dm}^3$.

Remember to convert volume: $1 \text{ dm}^3 = 1000 \text{ cm}^3$, so divide a volume in $\text{cm}^3$ by 1000.

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 $25.0 \text{ cm}^3$:

Empirical and molecular formulae from data

To find an empirical formula from masses or percentages:

1. Divide each element's mass (or %) by its $A_r$.
2. 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

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.

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.

• 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.