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 FormulaeC3.1
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 Aluminium OxideC3.1
Problem Aluminium forms $\text{Al}^{3+}$ ions and oxygen forms $\text{O}^{2-}$ ions. What's the formula of aluminium oxide?
Step 1 Line up the charges: aluminium is 3+, oxygen is 2−. They don't cancel one-to-one.
Step 2 Swap the charge numbers down: the 3 goes under oxygen, the 2 goes under aluminium. You get two $\text{Al}^{3+}$ (6+) and three $\text{O}^{2-}$ (6−).
So the formula is $\text{Al}_2\text{O}_3$. Charges balanced, ratio already smallest. ✅
$$2\,\text{Al}^{3+} + 3\,\text{O}^{2-} \rightarrow \text{Al}_2\text{O}_3$$
Counting Atoms in a FormulaC3.1
When you count atoms, the only thing that catches people out is the brackets. A number outside a bracket multiplies everything inside it, not just the first atom. So in $\text{(NH}_4\text{)}_2\text{SO}_4$, that little 2 doubles the whole $\text{NH}_4$ group: both the nitrogen and all four hydrogens. Split the formula into "the bracket bit" and "the rest", count each, then add. Easy marks once you slow down.
> 🔑 Key idea: A subscript outside a bracket multiplies every atom inside it. Count the bracketed group first, then the rest, then add. 🧮
Worked Example: Atoms in Ammonium SulfateC3.1
Problem How many atoms in total are in one formula unit of ammonium sulfate, $\text{(NH}_4\text{)}_2\text{SO}_4$?
Step 1 Do the bracket first. $(\text{NH}_4)_2$ is two ammonium groups; each is 1 N + 4 H = 5 atoms, so two of them is 10 atoms (2 N, 8 H).
Step 2 Now the rest. $\text{SO}_4$ is 1 S + 4 O = 5 atoms. Add the two blocks.
15 atoms in total (but only 4 different elements). 🔥
$$10 + 5 = 15 \text{ atoms}$$
Balancing EquationsC3.1
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). And don't forget state symbols when they're asked for: $(s)$ solid, $(l)$ liquid, $(g)$ gas, $(aq)$ dissolved in water.
> 🔑 Key idea: Balance with the big front numbers, never the subscripts. Then add state symbols $(s)$, $(l)$, $(g)$, $(aq)$ where needed. ⚡
Worked Example: Potassium Plus OxygenC3.1
Problem Potassium reacts with oxygen to form potassium oxide, $\text{K}_2\text{O}$. Balance the equation.
Step 1 Write the formulae first (oxygen is $\text{O}_2$): $\text{K} + \text{O}_2 \rightarrow \text{K}_2\text{O}$. Oxygen is unbalanced (2 on the left, 1 on the right), so put a 2 in front of $\text{K}_2\text{O}$.
Step 2 Now there are 4 potassium on the right, so put a 4 in front of potassium on the left.
Both sides: 4 K, 2 O. Balanced. ✅
$$4\text{K} + \text{O}_2 \rightarrow 2\text{K}_2\text{O}$$
Relative Atomic MassC3.2
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 the average leans toward whichever isotope is more common, like a class average being pulled towards the bigger group. That's why chlorine's $A_r$ is 35.5 and not a tidy whole number.
> 🔑 Key idea: $A_r$ is the average isotope mass on the carbon-12 scale. It lies between the isotope masses, closer to the more abundant one. 🎯
Worked Example: The Mass of BoronC3.2
Problem Boron has two isotopes, of mass 10 and 11. The heavier one (mass 11) is more common. Which is the most likely $A_r$: 10.0, 10.2, 10.8 or 11.0?
Step 1 The average has to be between 10 and 11, so rule out 10.0 and 11.0 (those are the pure isotopes).
Step 2 The heavier isotope (mass 11) is more common, so the average gets pulled up above the half-way value of 10.5. Only 10.8 fits.
So the most likely relative atomic mass is 10.8. ✅
$$A_r \approx 10.8$$
Relative Formula MassC3.2
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 HydroxideC3.2
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 MassC3.2
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: Burning Magnesium by ProportionC3.2
Problem 48 g of magnesium burns to form 80 g of magnesium oxide. What mass of magnesium oxide forms from 24 g of magnesium?
Step 1 Compare the magnesium: 24 g is a half of 48 g.
Step 2 So you make a half of the product.
40 g of magnesium oxide. No moles needed, just proportion. ✅
$$\frac{1}{2} \times 80 = 40 \text{ g}$$
The Mole and Molar MassC3.3
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 of WaterC3.3
Problem The relative molecular mass of water, $\text{H}_2\text{O}$, is 18. How many moles are in 36 g of water?
Step 1 Use amount = mass ÷ molar mass. The molar mass of water is 18 g/mol. $$n = \frac{36}{18} = 2.0 \text{ mol}$$
Step 2 Sanity check: 36 g is exactly two lots of 18 g, so it's 2 moles. ✅
Particles and the Avogadro ConstantC3.3
Once you've got moles, the number of actual particles is one more step: multiply by the Avogadro constant, $6.02 \times 10^{23}$. "Particles" can mean atoms, molecules, or ions, so read the substance carefully, one mole of $\text{O}_2$ is $6.02 \times 10^{23}$ molecules but twice as many atoms. The neatest case is the definition itself: exactly one mole of any substance contains exactly one Avogadro constant of particles.
> 🔑 Key formula: number of particles $=$ amount (mol) $\times 6.02 \times 10^{23}$. Watch whether the question wants atoms, molecules, or ions. 🔬
Worked Example: Atoms in CarbonC3.3
Problem The relative atomic mass of carbon is 12. How many atoms are in 12 g of carbon? ($N_A = 6.02 \times 10^{23}$.)
Step 1 First get to moles: amount = mass ÷ molar mass. $$n = \frac{12}{12} = 1.0 \text{ mol}$$
Step 2 Number of atoms = moles × Avogadro constant.
12 g of carbon is exactly one mole, so it's exactly one Avogadro constant of atoms. 💎
$$N = 1.0 \times (6.02 \times 10^{23}) = 6.02 \times 10^{23} \text{ atoms}$$
Reacting Masses from EquationsC3.3
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. The only "chemistry" step is the middle one, where you use the big balancing numbers as the ratio; everything else is just unit conversion with the molar mass.
> 🔑 Key formula: reacting-mass route: mass → ÷ $M$ → moles → × ratio → moles → × $M$ → mass. The ratio comes from the big balancing numbers. 🧮
Worked Example: Magnesium to Magnesium OxideC3.3
Problem $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$. The $A_r$ of magnesium is 24 and the $M_r$ of magnesium oxide is 40. What mass of magnesium oxide forms from 9.6 g of magnesium?
Step 1 Get to moles of Mg, then cross the equation ratio. $$n(\text{Mg}) = \frac{9.6}{24} = 0.4 \text{ mol}, \qquad \text{ratio } 2:2 = 1:1 \Rightarrow n(\text{MgO}) = 0.4 \text{ mol}$$
Step 2 Leave moles: turn moles of MgO into mass.
16.0 g of magnesium oxide. Three moves, done. 🎯
$$\text{mass} = 0.4 \times 40 = 16.0 \text{ g}$$
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; bracket polyatomic groups used more than once.
- A molecular formula shows the number and type of atoms; two differently drawn molecules can share one.
- Balance equations with the big front numbers only, then add state symbols $(s)$, $(l)$, $(g)$, $(aq)$.
- $A_r$ is the average isotope mass on the carbon-12 scale, closer to the more abundant isotope; $M_r$ is the sum of $A_r$ values.
- Percentage by mass of an element = (total $A_r$ of the element ÷ $M_r$) × 100.
- Mass is always conserved; apparent changes are gas going in or out.
- The mole is the hub: $n = m/M$, particles $= n \times 6.02 \times 10^{23}$, concentration $= n/V$.
- Every reacting-mass question is: get to moles → cross the ratio → leave moles.