Stoichiometry
A guided, interactive walkthrough of moles, formulas, and the math that connects mass to chemistry.
Chemistry is organized into ten major units. Each one builds on the last. This page covers Unit 3, Stoichiometry, which uses the periodic-table reasoning from Unit 2 and gives you the math toolkit you will lean on for every later unit.
By the end of this page you will be able to count atoms by mass using the mole, calculate the gram-formula mass of any compound, balance any chemical equation, classify it as one of the five reaction types, and use a balanced equation to find the mass of a missing reactant or product.
Atoms are tiny. Way too tiny to count one at a time. So chemistry uses a counting unit called the mole, the same way a baker uses a "dozen" to count eggs. The only difference is the size of the bundle.
A mole is just a unit of quantity, like a dozen (12) or a gross (144). It is big because the things we are counting are unimaginably small.
Why this exact number? Because chemists chose it so that the mass of one mole of an element, in grams, comes out to the atomic mass listed on the periodic table. That is the magic that makes stoichiometry possible.
1 mole of carbon = 12 g of carbon. 1 mole of oxygen = 16 g of oxygen.
A mole of marshmallows would cover the entire surface of the United States to a depth of about 1,000 km. A mole of seconds is roughly 19 quadrillion years, more than a million times the age of the universe. That is what makes atoms small enough to need such a huge counting unit.
For a single element, you read the molar mass off the periodic table. For a compound, you add up the masses of every atom in the formula. That total is called the gram-formula mass, or gfm. Some books call it molar mass or molecular weight. Same idea.
1. List every element in the formula.
2. Multiply each element's atomic mass by the number of atoms of that element in the formula (the subscript).
3. Add up all the results. The total is the gfm in grams per mole.
For compounds with parentheses, like Ca(OH)$_2$, the subscript outside applies to everything inside. Ca(OH)$_2$ has 1 Ca, 2 O, and 2 H. Be careful with these.
Atomic masses rounded to whole numbers (or the nearest tenth) for clarity. Real periodic tables often give one more decimal.
Once you can find a gram-formula mass, you can move freely between three different ways of measuring "how much" of a substance: moles, grams, and particles.
moles × gfm → grams
moles × $6.022 \times 10^{23}$ → particles
There is a famous diagram called the mole map with grams in one corner, particles in another, and moles in the middle. Whatever you are converting, you almost always go through moles first. Moles is the central currency.
Type in any one box and the other two update.
Most compounds can be written as a chemical formula in two different ways:
Molecular formulas show the exact number of each element in the compound. Sometimes the molecular and empirical formulas are the same (water is H$_2$O for both, since 2:1 cannot be simplified).
To get the empirical formula from the molecular formula, divide every subscript by the largest number that divides them all evenly (the greatest common factor).
Going the other way, you need extra information (typically the molar mass of the actual compound) to scale the empirical formula up to the molecular formula. The ratio between molar masses tells you what to multiply each subscript by.
The percent composition of a compound tells you what fraction of the compound's mass comes from each element. It is just a percent. Useful for figuring out how much of a target element you can extract from a given mass of compound.
Atoms cannot be created or destroyed in a chemical reaction. They just rearrange. Same atoms in, same atoms out, just bonded differently. Antoine Lavoisier discovered this in the 1770s by carefully measuring masses before and after burning, and it has been bedrock chemistry ever since.
What is also conserved in a chemical reaction: charge (total positive charge in equals total positive charge out) and energy (energy is not lost; it changes form between bonds, heat, and light).
If you start with 50 g of reactants in a closed container, you end with 50 g of products. Even if the products look totally different (a gas, a colored solid, whatever), the scale will read 50 g.
A chemical equation describes a reaction. The starting substances are on the left (reactants), an arrow points right, and the ending substances are on the right (products). To respect conservation of mass, the equation must be balanced: the same number of every kind of atom on both sides.
You may only change coefficients when balancing. Never change subscripts. Changing a subscript changes the substance itself, and you would no longer be talking about the same reaction.
2. Pick the most complex molecule. Balance its atoms first by putting coefficients elsewhere.
3. Balance any free elements (like O$_2$ or H$_2$) last. They are easy to adjust.
4. Use whole-number coefficients only. If you end up with a fraction, multiply every coefficient by the denominator.
5. Re-count every element on each side to confirm.
Almost every reaction you will meet falls into one of five categories. Recognizing the type lets you predict products and balance more confidently.
Two or more simple things combine into one bigger thing.
One bigger thing breaks apart into smaller pieces. Often needs heat or electricity.
A free element kicks another element out of a compound. Always one element on each side.
Two compounds swap partners. Both sides have only compounds, no free elements.
A hydrocarbon (or other fuel) reacts with oxygen, releasing energy. Products are almost always CO$_2$ and H$_2$O.
· Two things on the left, one on the right? Synthesis.
· One thing on the left, two on the right? Decomposition.
· A free element plus a compound on each side? Single replacement.
· Two compounds on each side, no free elements? Double replacement.
· Hydrocarbon plus O$_2$ producing CO$_2$ and H$_2$O? Combustion.
Green = correct, red = try again. The right answer locks in once you find it.
Here is a beautiful fact about gases that simplifies a huge amount of chemistry.
1 L of helium has the same number of molecules as 1 L of carbon dioxide at the same conditions. The molecules of CO$_2$ are heavier, but the count is the same.
Why? Because gas molecules are so far apart that the molecule's size barely matters. What matters is the empty space they bounce around in, and that depends on temperature and pressure, not on identity.
A useful consequence: at standard temperature and pressure (STP), one mole of any gas occupies 22.4 liters. This is sometimes called the molar volume of a gas.
Each box is the same size and contains the same number of molecules, even though the gases are different. The masses are different (heavier molecules → more grams) but the count is the same.
Here is the payoff. A balanced equation is a recipe. The coefficients tell you the mole ratio, the proportion in which the substances combine. Once you know that, you can predict how much product you will get from a given amount of reactant, or how much of one reactant you need to consume another.
1. Make sure the equation is balanced.
2. Convert the given mass to moles (divide by gfm).
3. Use the coefficient ratio to convert to moles of the target substance.
4. Convert moles back to mass (multiply by the new gfm).
Grams → Moles → Moles (different substance) → Grams. Through moles, every time.
Sometimes you do not need a full mole calculation. If you know all but one mass and the equation is balanced, conservation of mass alone can give you the answer. Total mass in = total mass out.
Every chemical reaction either releases energy or absorbs energy. There is no third option, because rearranging atoms always changes the total stored energy in their bonds. The flavor of energy can be heat, light, or even electricity, but on a thermometer the result is the same: things either get warmer or colder.
Endothermic (Greek endo-, "in"): the reaction absorbs energy from the surroundings. Things get colder, or you have to keep adding energy (light, heat) to keep the reaction going. Bonds in the products are weaker than bonds in the reactants. $\Delta H$ is positive.
Even an exothermic reaction needs a small upfront energy input to get started. That small input is called the activation energy. Think of a match: rubbing it on the box gives the chemicals enough push to start, then the burning supplies its own energy and keeps going. Endothermic reactions need that initial push plus a continuous supply of energy.
Sometimes the energy is written right in the equation. If you see + heat or + energy on the product side, the reaction is exothermic (energy is coming out). If it appears on the reactant side, the reaction is endothermic (energy must be supplied for it to happen).
Endo: 2870 kJ + 6CO$_2$ + 6H$_2$O → C$_6$H$_{12}$O$_6$ + 6O$_2$
Stoichiometry is not a chemistry-only topic. Two of the most important reactions in your biology textbook are perfect examples of balanced equations, and they happen to be exact opposites of each other.
Plants use light energy from the sun to combine carbon dioxide and water into glucose (food) and oxygen. Notice how the equation is balanced: 6 C, 18 H, 18 O on each side. Energy is on the left because it has to be supplied for the reaction to happen.
Every cell in your body, plant or animal, runs this in reverse. Glucose plus oxygen breaks down into carbon dioxide and water, releasing the stored energy as ATP, the molecule your cells use to do work. Same atoms, same balanced count.
Plants do photosynthesis. Every living thing, including plants, does respiration. The two reactions form a continuous cycle on a planetary scale: the carbon and oxygen atoms cycle endlessly between living things and the air, while energy flows in from the sun and eventually radiates away as heat. The atoms are conserved (Unit 3 stuff). The energy is not destroyed, just changes form.
Photosynthesis and respiration are written with single arrows, like most reactions you have seen. The arrow says: this happens, in this direction. But not every reaction is a one-way street.
When the forward and reverse rates are equal, the reaction is at chemical equilibrium. The amounts of reactants and products stop changing, even though both directions are still happening underneath. This is the focus of Unit 6.
A familiar everyday example: dissolved CO$_2$ in soda. CO$_2$(g) $\rightleftharpoons$ CO$_2$(aq). When the bottle is sealed, equilibrium is reached and the bubbles stop forming. When you open it, you remove CO$_2$ from the system, the equilibrium shifts, and bubbles burst out until a new balance is reached.
For now, in stoichiometry, we mostly assume reactions go fully to completion (single arrow, all the way). Just remember that for some reactions, especially in biology and aqueous chemistry, "going to completion" is an idealization.
When ionic compounds dissolve in water, they break apart into individual ions. So in a reaction between dissolved ionic compounds, what is "really" reacting is not the formulas you see on paper, but the loose ions floating around. Net ionic equations help us see only the part of the reaction that actually changes.
Take this classic double-replacement: silver nitrate plus sodium chloride forms silver chloride (a white solid) and sodium nitrate.
A precipitate is an insoluble solid that forms when two solutions are mixed. We write it with the (s) state symbol. The precipitate is what you actually see happening: the cloudy stuff falling out of solution. AgCl(s) is the precipitate in our example.
2. Split every (aq) ionic compound into its component ions (with charges). Leave (s), (l), (g) compounds and molecular substances (like H$_2$O, gases) intact.
3. Cancel ions that appear identically on both sides. Those are your spectators.
4. Write what remains. That is the net ionic equation.
A precipitate forms only if at least one of the products is insoluble in water. Use this short rule sheet for common ionic compounds:
Run the test: of the two products in your double-replacement, is either one on the "insoluble" list? If yes, that is the precipitate, and you have a real reaction. If both products are soluble, no precipitate forms and there is technically no net reaction (everything stays as floating ions).
Net ionic equations are not just for precipitates. When a strong acid neutralizes a strong base, the molecular equation is HCl(aq) + NaOH(aq) → NaCl(aq) + H$_2$O(l). Splitting the soluble ionic compounds, you get H$^+$(aq) + Cl$^-$(aq) + Na$^+$(aq) + OH$^-$(aq) → Na$^+$(aq) + Cl$^-$(aq) + H$_2$O(l). The Na$^+$ and Cl$^-$ are spectators, leaving the famous net ionic for any strong acid + strong base reaction:
Whether you start with HCl + NaOH or HNO$_3$ + KOH, the net ionic is identical. That is the value of net ionic equations: they reveal the underlying chemistry.
Balancing is a skill, and the only way to get fast at it is reps. The interactive balancer in §7 already has the basic patterns. Here are five more, ranging from "easy warm-up" to "needs a big least-common-multiple."
2. Polyatomic ions stay together. If SO$_4^{2-}$ shows up on both sides, treat it as one unit and balance it as a whole.
3. If you get stuck on fractions, double everything. Coefficients have to be whole numbers, but you can balance with fractions then multiply through.
4. Combustion shortcut: for C$_x$H$_y$ + O$_2$ → CO$_2$ + H$_2$O, balance C first (set coefficient on CO$_2$ to $x$), then H (set coefficient on H$_2$O to $y/2$), then O last by counting and dividing.
Click the card to flip it. Use the arrows to move through the deck. Shuffle whenever you want a different order.
Try each one before opening the answer.
(a) 2KClO$_3$ → 2KCl + 3O$_2$
(b) Mg + 2HCl → MgCl$_2$ + H$_2$
(c) C$_3$H$_8$ + 5O$_2$ → 3CO$_2$ + 4H$_2$O
(d) Pb(NO$_3$)$_2$ + 2KI → PbI$_2$ + 2KNO$_3$
(e) N$_2$ + 3H$_2$ → 2NH$_3$
(b) Single replacement. Free Mg displaces H from HCl.
(c) Combustion. Hydrocarbon + O$_2$ → CO$_2$ + H$_2$O.
(d) Double replacement. Two compounds swap partners.
(e) Synthesis. Two simpler substances combine into one.
Step 2: Mole ratio: 2 mol H$_2$ → 2 mol H$_2$O. So 2 mol of H$_2$O are produced.
Step 3: Mass of H$_2$O = 2 × 18 = 36 g.
2 mol × $6.022 \times 10^{23}$ = $1.2 \times 10^{24}$ molecules.
Molecules: $0.5 \times 6.022 \times 10^{23} = $ $3.011 \times 10^{23}$.
Step 1: Balance C first → 2 CO$_2$ for each C$_2$H$_6$.
Step 2: Balance H next → 3 H$_2$O for each C$_2$H$_6$.
Step 3: Count O on right (4 + 3 = 7), so 7/2 O$_2$. Multiply everything by 2 to clear the fraction.
Cancel Na$^+$ and NO$_3^-$ (appear on both sides): they are the spectator ions.
Net ionic: Ag$^+$(aq) + Cl$^-$(aq) → AgCl(s).
Precipitate: AgCl(s) (silver chloride is insoluble).
C: 6 on the left (6 from 6CO$_2$), 6 on the right (in C$_6$H$_{12}$O$_6$). ✓
H: 12 on the left (6 × 2 in 6H$_2$O), 12 on the right. ✓
O: 18 on the left (12 from 6CO$_2$ + 6 from 6H$_2$O), 18 on the right (6 in glucose + 12 in 6O$_2$). ✓
Endothermic because the products (glucose and oxygen) store more energy than the reactants. The plant absorbs light energy from the sun to drive this uphill reaction. The energy doesn't come from nowhere — it comes from the sun.