How to Approach Stoichiometry
Almost every stoichiometry problem follows the same path: get to moles, use the balanced equation to cross over to the substance you want, then convert back to the units the question asks for. Master that one road map and the whole topic stops feeling like a collection of unrelated problem types.
The four-step road map
- Balance the equation. The coefficients are the mole ratios you will use, so the equation has to be balanced before anything else.
- Convert the known amount to moles. From a mass, use n = m ÷ M. From a gas, use PV = nRT. From a solution, use n = M × V.
- Apply the mole ratio. Multiply by the ratio of coefficients from the balanced equation to cross over to the substance you want.
- Convert to the requested units. Moles back to grams (m = n × M), to a gas volume, to a solution volume, or to particles (N = n × NA).
Why moles are the hub
A balanced equation does not tell you anything about grams or millilitres directly. Its coefficients only ever count particles — and the mole is just a convenient package of particles. That is why grams, gas volumes and concentrations all have to be turned into moles before the equation can be used, and converted back afterwards. Think of moles as the central station every quantity passes through.
The conversions you reach for
| Starting information | To get moles, use |
|---|---|
| Mass of a pure substance | n = m ÷ M (mass ÷ molar mass) |
| Volume + concentration of a solution | n = M × V (with V in litres) |
| A gas at known P, V, T | n = PV ÷ RT |
| A gas at STP | n = V ÷ 22.4 L·mol⁻¹ |
| A number of particles | n = N ÷ NA |
Worked example: mass to mass
How many grams of FeCl₃ form when 38.5 g of Cl₂ reacts with excess iron, given 2 Fe + 3 Cl₂ → 2 FeCl₃?
The mole ratio Cl₂ → FeCl₃ is 3 : 2, so n(FeCl₃) = 0.543 × (2 ÷ 3) = 0.362 mol. Then m = 0.362 × 162.20 = 58.7 g of FeCl₃. Notice "excess iron" was a signal that iron cannot be the limiting reactant, so the chlorine controls the yield.
Worked example: a gas and a solution
What volume of 0.250 M HCl reacts completely with 0.448 L of CO₂ gas at STP, given CaCO₃ + 2 HCl → CaCl₂ + H₂O + CO₂?
n(HCl) = 0.0200 × (2 ÷ 1) = 0.0400 mol
V(HCl) = 0.0400 ÷ 0.250 = 0.160 L = 160 mL
The same chain handled a gas on the way in and a solution on the way out — only the first and last conversions changed.
Common pitfalls
- Not balancing first, so the mole ratio is wrong from the start.
- Using a mass ratio instead of the mole ratio. Coefficients count particles, never grams.
- Skipping unit conversions, especially mL → L and °C → K.
- Inverting the ratio. Write it as (moles wanted ÷ moles known) so the known cancels.
- Rounding moles too early. Carry extra digits and round only the final answer.
Start by finding molar masses with the Molar Mass Calculator, then carry an amount through with the Stoichiometry Calculator. When two reactants are given, see Limiting Reagents Explained, and make sure the equation is right first with Balancing Chemical Equations.
The General Chemistry Workbook has a full stoichiometry chapter with worked examples and a complete answer key.
Frequently Asked Questions
Balance the equation, convert the known quantity to moles, apply the mole ratio from the coefficients, then convert the moles of the unknown into the units the question asks for. Every problem is a variation on this chain.
A balanced equation only counts particles, and the mole is a package of particles. Grams, gas volumes and concentrations carry no ratio information on their own, so they must be turned into moles before the equation can be used and converted back afterwards.
Write it as moles of the substance you want over moles of the substance you know, so the known unit cancels. For 2 Fe + 3 Cl₂ → 2 FeCl₃, going from Cl₂ to FeCl₃ uses (2 mol FeCl₃ ÷ 3 mol Cl₂).
Forgetting to balance the equation first, which makes every mole ratio wrong. A close second is using a mass ratio instead of a mole ratio, since coefficients never count grams directly.
Yes. Only the first and last conversions change. A gas uses PV = nRT or 22.4 L/mol at STP, and a solution uses n = M × V. The mole-ratio step in the middle is identical.