Every stoichiometry problem is the same three moves: convert what you are given into moles, cross over to the wanted substance with the mole ratio from the balanced equation, then convert those moles into the units the question asks for. Grams, particle counts, and gas volumes never convert into each other directly; each one converts to and from moles by a single operation, so the mole is the hub of every calculation. Learn that road map once and mass-to-mass, gas volume, limiting reagent, and percent yield problems all become one calculation with at most one extra step.
One pattern, not five problem types
Stoichiometry gets taught as a parade of problem types, each with what looks like its own recipe. There is only one recipe. Moles sit at the hub of every calculation, every other quantity is one conversion away, and the only place two different substances meet is the mole ratio read from the balanced equation. Mass-to-mass, gas volume, particle count, limiting reagent, and percent yield problems differ only in which spoke you start on and which you end on.
The mole itself is the bridge between scales: one mole is 6.022 × 10²³ particles, chosen so that one mole of a substance has a mass in grams numerically equal to its relative atomic or molecular mass. Carbon's relative atomic mass is 12.0, so one mole of carbon weighs 12.0 g. That link is what lets a balance, which measures grams, effectively count particles.
The mole road map
Conversion
Operation
grams to moles
divide by molar mass
moles to grams
multiply by molar mass
moles to particles
multiply by 6.022 × 10²³
particles to moles
divide by 6.022 × 10²³
moles of gas to liters at STP
multiply by 22.4 L/mol
liters of gas at STP to moles
divide by 22.4 L/mol
moles of one substance to moles of another
multiply by the mole ratio from the balanced equation
There is never a direct route from grams to particles, or grams to liters, that skips the mole. A 9.0 g sample of water is 9.0 ÷ 18.0 = 0.50 mol, which is about 3.0 × 10²³ molecules; the count only appears at the last step. The 22.4 L/mol spoke belongs only to gases at STP (0 °C and 1 atm), and never to liquids, solids, or solutions.
The three moves, fully worked
How many grams of water form when 8.0 g of hydrogen burns in excess oxygen? The equation is 2H₂ + O₂ → 2H₂O, already balanced.
Move one, grams to moles: hydrogen gas is H₂ with molar mass 2.0 g/mol, so 8.0 g ÷ 2.0 g/mol = 4.0 mol H₂.
Move two, mole ratio: H₂ to H₂O is 2:2, so 4.0 mol of H₂ produces 4.0 mol of H₂O.
Move three, moles to grams: 4.0 mol × 18.0 g/mol = 72 g of water.
Every mass-to-mass problem is exactly this. The traps live in the details: use each substance's own molar mass on its own grams, remember H₂ is 2.0 and O₂ is 32.0 because they are diatomic, and carry molar masses to one decimal place, rounding only at the end.
Limiting reagent and percent yield: the same moves plus one
Limiting reagent problems add a comparison before the three moves. Convert every reactant to moles, then check which falls short of the coefficient ratio: in N₂ + 3H₂ → 2NH₃ with 28.0 g of nitrogen (1.00 mol) and 3.0 g of hydrogen (1.5 mol), nitrogen demands 3.0 mol of hydrogen and only 1.5 mol exists, so hydrogen limits even though its mole amount is larger. Every product is then computed from the limiting reagent alone: 1.5 mol H₂ × (2 mol NH₃ ÷ 3 mol H₂) = 1.0 mol NH₃, which is 17 g. Mass conservation gives a built-in check: 31.0 g of reactants in, 17.0 g of ammonia plus 14.0 g of leftover nitrogen out.
Percent yield adds a division after the three moves. The calculation gives the theoretical yield, the balance gives the actual yield, and percent yield = actual ÷ theoretical × 100. A yield above 100% signals an error, usually wet or impure product. Learn these as extensions of the same route, not as new problem types, and drill them from a blank page: rereading worked examples feels productive, but the retention gap between self-testing and passive review is roughly 80% versus 36% after a week.
The 25-minute drill plan
5 minutes
Write the mole road map from memory: every spoke in both directions with its operation. Check and fix. The map is the entire subject.
10 minutes
Two mass-to-mass problems from a blank page, three moves labeled explicitly. Start with: how many grams of water form when 8.0 g of hydrogen burns in excess oxygen? Target: 72 g.
7 minutes
One limiting reagent problem: N₂ + 3H₂ → 2NH₃ from 28.0 g of nitrogen and 3.0 g of hydrogen. Find the limiting reagent in moles, compute the ammonia, verify 31.0 g in equals 31.0 g out.
3 minutes
Rework the conversion you missed, from scratch, without notes. Whatever still fails is tomorrow's first drill.
Frequently asked questions
Three moves, always in the same order. First, convert the given quantity to moles: divide grams by molar mass, particles by 6.022 × 10²³, or liters of gas at STP by 22.4. Second, cross over to the wanted substance with the mole ratio from the balanced equation, written as coefficient of wanted over coefficient of given. Third, convert the resulting moles into the units the question asks for. Before any of it, confirm the equation is balanced, because the ratio in move two comes from the coefficients.
Convert every reactant to moles, then compare against the coefficient ratio; the limiting reagent is the one that falls short of what the ratio demands, not the one with the smaller mass. In N₂ + 3H₂ → 2NH₃, starting from 28.0 g of nitrogen (1.00 mol) and 3.0 g of hydrogen (1.5 mol), the equation demands 3 mol of H₂ per mol of N₂, so 1.00 mol of N₂ needs 3.0 mol of H₂ and only 1.5 mol is available. Hydrogen limits despite the larger mole amount. Then compute every product from the limiting reagent alone.
Only for gases, and only at STP, defined in most school courses as 0 °C and 1 atm. At those conditions one mole of any ideal gas occupies 22.4 liters regardless of identity: hydrogen, oxygen, and carbon dioxide each fill 22.4 L per mole. So 11.2 L of any gas at STP is 0.500 mol. Never apply it to liquids, solids, solutions, or gases at other temperatures and pressures; that misuse is one of the most common ways to get a confident wrong answer.
Because the coefficients of the balanced equation are the mole ratios, and the mole ratio is the only step where the given substance and the wanted substance meet. An unbalanced equation gives ratios that are simply wrong, and every calculation built on them inherits the error no matter how careful the arithmetic is. The check costs seconds: count each element on both sides before starting. For gas reactions the coefficients do double duty as volume ratios too, which only makes balancing more load-bearing.
They are numerically equal but conceptually different. The relative molecular mass is a unitless ratio measured against 1/12 of a carbon-12 atom; the molar mass is the same number carrying the units g/mol, and it always refers to one mole of the complete formula. Water's relative molecular mass is 18.0 and its molar mass is 18.0 g/mol. The practical traps are elsewhere: diatomic elements (O₂ is 32.0 g/mol, not 16.0), parentheses that multiply everything inside, and hydrated salts whose water of crystallization counts.
