Introduction
Alright, let's talk stoichiometry. Sounds like a final-boss word, but it's basically chemistry's version of a recipe: how much of this do I mix to get that much of the other thing? Bakers do it with cups, chemists do it with the mole.
Here's the cheat code for the whole chapter: your balance only reads grams, but reactions actually happen in ratios of particles. The mole is the translator between the two. Lock that in and every calculation becomes the same three little steps. This is your fast refresh, not the full textbook: one key formula per idea, one quick worked example, and the exact move the marker wants. Let's count like a chemist! 🧪
Writing Chemical FormulaeE3.1
A formula just says how many atoms of each element are in there. The little number sticks to the symbol right before it, so $\text{H}_2\text{O}$ is two H and one O. For ionic compounds (a metal + a non-metal, or anything with a group like sulfate), the trick is that the whole thing has to be electrically neutral. So you balance the positive and negative charges until they cancel, then write the smallest whole-number ratio.
> 🔑 Key idea: An ionic formula is the smallest ratio of ions whose charges cancel. If a group like sulfate appears more than once, wrap it in brackets: $\text{Fe}_2(\text{SO}_4)_3$. 🎯
Worked Example: Building Magnesium ChlorideE3.1
Problem Magnesium forms $\text{Mg}^{2+}$ ions and chlorine forms $\text{Cl}^{-}$ ions. What's the formula of magnesium chloride?
Step 1 Line up the charges: magnesium is 2+, chlorine is 1−. They don't cancel one-to-one.
Step 2 You need two 1− chlorides to balance one 2+ magnesium, so the ratio is 1 : 2.
So the formula is $\text{MgCl}_2$. One magnesium, two chlorines, charges balanced. ✅
$$\text{Mg}^{2+} + 2\,\text{Cl}^{-} \rightarrow \text{MgCl}_2$$
Empirical and Molecular FormulaeE3.1
Two flavours of formula, and people mix them up constantly. The molecular formula is the actual atom count in one molecule (ethane = $\text{C}_2\text{H}_6$). The empirical formula is the simplest ratio (ethane = $\text{CH}_3$). Think of a pizza cut into slices: the empirical formula is the "1 cheese : 2 pepperoni per slice" ratio, the molecular formula is the whole pie. To find an empirical formula from masses or percentages, you turn each into moles (divide by $A_r$), then divide by the smallest.
> 🔑 Key idea: Empirical = simplest ratio, molecular = real count. To get empirical from % or mass: divide each by $A_r$, then divide by the smallest number. Never compare the percentages directly. 🧮
Worked Example: A Carbon-Hydrogen CompoundE3.1
Problem A compound is 85.7% carbon and 14.3% hydrogen by mass. Find its empirical formula. ($A_r$: C = 12, H = 1.)
Step 1 Treat the percentages as masses out of 100 g and divide each by its $A_r$ to get moles. $$\text{C} = \frac{85.7}{12} = 7.14, \qquad \text{H} = \frac{14.3}{1} = 14.3$$
Step 2 Divide both by the smaller number (7.14).
So the empirical formula is $\text{CH}_2$. Clean and simple. 🔥
$$\text{C} : \text{H} = 1 : 2$$
Balancing and Ionic EquationsE3.1
A balanced equation is just a promise that no atoms vanish: the same number of each element on both sides. You make it balance by changing the big numbers in front only, never the little subscripts (that would change the actual substance). An ionic equation is the stripped-down version: you keep only the ions that actually do something and ditch the spectators who just watch.
> 🔑 Key idea: Balance with the big front numbers, never the subscripts. Ionic equations keep only what changes and must balance both atoms and charge. ⚡
Worked Example: The Thermite ReactionE3.1
Problem In the thermite reaction, aluminium reacts with iron(III) oxide to make aluminium oxide and iron. Balance the equation.
Step 1 Write the correct formulae first, no balancing yet. $$\text{Al} + \text{Fe}_2\text{O}_3 \rightarrow \text{Al}_2\text{O}_3 + \text{Fe}$$
Step 2 Count up: the right side has 2 Al, the left has only 1; the left has 2 Fe, the right has only 1. Put a 2 in front of aluminium and a 2 in front of iron.
Now both sides have 2 Al, 2 Fe, 3 O. Balanced. ✅
$$2\text{Al} + \text{Fe}_2\text{O}_3 \rightarrow \text{Al}_2\text{O}_3 + 2\text{Fe}$$
Relative Atomic MassE3.2
Atoms are way too tiny to weigh one at a time, so chemists compare everything to a carbon-12 atom (set at exactly 12). The relative atomic mass ($A_r$) is the average mass of an element's atoms. It's an average because most elements come as a mix of isotopes (same element, different masses). And it's a weighted average, like working out your grade when the final exam counts for more than homework. That's why chlorine's $A_r$ is 35.5 and not a tidy whole number.
> 🔑 Key formula: $A_r = \dfrac{\sum(\text{isotope mass} \times \text{percentage})}{100}$. Only use a plain average if the isotopes are split 50/50. 🎯
Worked Example: The Mass of ChlorineE3.2
Problem Chlorine is 75% mass-35 atoms and 25% mass-37 atoms. Find its relative atomic mass.
Step 1 Multiply each isotope mass by its percentage, then add. $$(75 \times 35) + (25 \times 37) = 2625 + 925 = 3550$$
Step 2 Divide by 100.
It lands closer to 35 because there's more of the lighter isotope. Makes sense. ✅
$$A_r = \frac{3550}{100} = 35.5$$
Relative Formula MassE3.2
Once you know the $A_r$ of each element, the relative molecular (or formula) mass $M_r$ is just the total mass of all the atoms in the formula. Add them up. The one thing that trips everyone is brackets: a number outside a bracket multiplies everything inside it, so $\text{Ca(OH)}_2$ has two oxygens and two hydrogens, not just two oxygens.
> 🔑 Key idea: $M_r$ = sum of all the $A_r$ values in the formula. A bracket number multiplies every atom inside the bracket. 🧮
Worked Example: Calcium HydroxideE3.2
Problem Find the relative formula mass of calcium hydroxide, $\text{Ca(OH)}_2$. ($A_r$: Ca = 40, O = 16, H = 1.)
Step 1 The 2 outside the bracket doubles both the O and the H. So you've got 1 Ca, 2 O, 2 H.
Step 2 Add the masses.
So $M_r = 74$. Forget the bracket and you'd wrongly get 57 or 58. 👀
$$M_r = 40 + (2 \times 16) + (2 \times 1) = 40 + 32 + 2 = 74$$
Conservation of MassE3.2
In any reaction, atoms just get rearranged, never created or destroyed. So the total mass of products equals the total mass of reactants. If mass looks like it changed, a gas snuck in or escaped: burning magnesium gets heavier (grabs oxygen from the air), a heated carbonate gets lighter (lets carbon dioxide escape). And because a reaction is fixed, if you scale the amounts up or down, every mass scales by the same factor, so you can answer loads of questions with simple proportion.
> 🔑 Key idea: Mass of products = mass of reactants, always. Scale a fixed reaction up or down and every mass scales together. ⚖️
Worked Example: Iron Plus SulfurE3.2
Problem 28 g of iron reacts to form 44 g of iron sulfide. What mass of iron sulfide forms from 7 g of iron?
Step 1 Compare the iron: 7 g is a quarter of 28 g.
Step 2 So you make a quarter of the product.
11 g of iron sulfide. No moles needed, just proportion. ✅
$$\frac{1}{4} \times 44 = 11 \text{ g}$$
The Mole and Molar MassE3.3
The mole is just a "dozen" but enormous: one mole is $6.02 \times 10^{23}$ particles (the Avogadro constant), like how "a dozen" always means 12 whether it's eggs or doughnuts. The genius bit: one mole of a substance weighs its relative mass in grams, called the molar mass. So 1 mol of water (18) is 18 g, 1 mol of carbon (12) is 12 g. That's the link between grams on your balance and particles in the reaction.
> 🔑 Key formula: amount (mol) $= \dfrac{\text{mass}}{\text{molar mass}}$, and number of particles $=$ amount $\times 6.02 \times 10^{23}$. 🎯
Worked Example: Moles and Atoms of ArgonE3.3
Problem How many moles, and then how many atoms, are in 2.00 g of argon? ($A_r = 40$, $N_A = 6.02 \times 10^{23}$.)
Step 1 Get to moles: amount = mass ÷ molar mass. $$n = \frac{2.00}{40} = 0.0500 \text{ mol}$$
Step 2 Argon is single atoms, so number of atoms = moles $\times N_A$.
Less than a mole, so fewer than $6.02 \times 10^{23}$ atoms. Checks out. 🔬
$$N = 0.0500 \times (6.02 \times 10^{23}) = 3.01 \times 10^{22} \text{ atoms}$$
Gas Volumes and ConcentrationE3.3
Two more arrows off the mole map. For gases: at room temperature and pressure (r.t.p.), one mole of any gas takes up the same 24 dm³, so amount = volume ÷ 24. For solutions: concentration = moles ÷ volume (in dm³). The classic trap is units, so always change cm³ to dm³ first (divide by 1000). A reaction calc is still just the three moves: get to moles, cross the equation ratio, leave moles.
> 🔑 Key formula: gas volume (dm³) $=$ moles $\times 24$ at r.t.p.; concentration (mol/dm³) $= \dfrac{\text{moles}}{\text{volume in dm}^3}$. And $1 \text{ dm}^3 = 1000 \text{ cm}^3$. 🧪
Worked Example: Carbon Dioxide from a CarbonateE3.3
Problem $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$. What volume of carbon dioxide at r.t.p. comes from 0.5 mol of calcium carbonate? (24 dm³/mol.)
Step 1 Get to moles: you already have 0.5 mol of $\text{CaCO}_3$. Cross the equation: the ratio is 1 : 1, so you make 0.5 mol of $\text{CO}_2$.
Step 2 Leave moles: turn moles of gas into a volume.
12 dm³ of carbon dioxide. Same three moves every time. ✅
$$V = 0.5 \times 24 = 12 \text{ dm}^3$$
Reacting Masses and YieldE3.3
This is the payoff. A balanced equation plus the three moves lets you predict exactly how much product you should get: mass of reactant → moles → (cross the ratio) → moles of product → mass of product. Real life never quite hits the predicted maximum, so chemists report a percentage yield (what you actually got ÷ the theoretical max × 100). It can never beat 100%, so if you ever get more, you flipped the fraction.
> 🔑 Key formula: percentage yield $= \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100$. Reacting-mass route: mass → ÷ $M$ → moles → × ratio → moles → × $M$ → mass. 🧮
Worked Example: Burning MagnesiumE3.3
Problem $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$. What mass of magnesium oxide forms from 4.8 g of magnesium? ($A_r$: Mg = 24, O = 16.)
Step 1 Get to moles of Mg, then cross the equation ratio. $$n(\text{Mg}) = \frac{4.8}{24} = 0.2 \text{ mol}, \qquad \text{ratio } 2:2 = 1:1 \Rightarrow n(\text{MgO}) = 0.2 \text{ mol}$$
Step 2 Leave moles: turn moles of MgO into mass ($M_r = 24 + 16 = 40$).
8.0 g of magnesium oxide. Three moves, done. 🎯
$$\text{mass} = 0.2 \times 40 = 8.0 \text{ g}$$
Key Points to Remember
- A formula says how many atoms; a bracket number multiplies everything inside it.
- Ionic formulae are the smallest ratio of ions whose charges cancel.
- Empirical = simplest ratio; molecular = actual count. For empirical from %/mass: ÷ $A_r$, then ÷ smallest.
- Balance equations with the big front numbers only; ionic equations keep only what changes.
- $A_r$ is the weighted average of isotope masses; $M_r$ is the sum of $A_r$ values.
- Mass is always conserved; apparent changes are gas going in or out.
- The mole is the hub: $n = m/M$, particles $= n \times 6.02 \times 10^{23}$, gas volume $= n \times 24$ dm³, concentration $= n/V$.
- Every reacting-amount question is: get to moles → cross the ratio → leave moles. Yield never beats 100%.