How to Solve Stoichiometry Problems: Mole-to-Mole, Mass-to-Mass, and Limiting Reagent

Every stoichiometry problem follows the same three-step framework: (1) convert what you are given to moles, (2) use the mole ratio from the balanced equation to convert to moles of what you need, and (3) convert from moles to the requested units (grams, liters, molecules). If the problem involves two reactants, compare the available moles to the required ratio to find the limiting reagent, then do your calculation from that reactant. This framework solves every stoichiometry problem you will see in general chemistry.

A balanced chemical equation is not just a description of what happens — it is a precise mathematical recipe. The coefficients tell you the exact mole ratio in which reactants combine and products form.

Take the combustion of methane: CH4 + 2O2 -> CO2 + 2H2O. The coefficients tell you that 1 mole of methane reacts with exactly 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water. These ratios are fixed by the chemistry. If you have 3 moles of CH4, you need 6 moles of O2 and you will produce 3 moles of CO2 and 6 moles of H2O.

The mole ratio is the bridge between any two substances in the equation. Want to go from moles of CH4 to moles of H2O? The ratio is 1:2, so multiply moles of CH4 by 2. Want to go from moles of O2 to moles of CO2? The ratio is 2:1, so divide moles of O2 by 2. Every stoichiometry conversion passes through moles and uses these ratios.

The single most common stoichiometry mistake is using grams directly with the mole ratio. Coefficients are mole ratios, not gram ratios. You must convert to moles first. Always. There are no exceptions.

Most textbook problems give you a mass of one substance and ask for a mass of another. Here is the complete workflow with a real example.

Problem: How many grams of CO2 are produced by burning 48.0 g of CH4? (CH4 + 2O2 -> CO2 + 2H2O)

Step 1: Convert given mass to moles. Molar mass of CH4 = 12.01 + 4(1.008) = 16.04 g/mol. Moles of CH4 = 48.0 g / 16.04 g/mol = 2.993 mol.

Step 2: Apply the mole ratio. From the balanced equation, CH4 to CO2 is 1:1. So moles of CO2 = 2.993 mol.

Step 3: Convert moles to grams. Molar mass of CO2 = 12.01 + 2(16.00) = 44.01 g/mol. Mass of CO2 = 2.993 mol x 44.01 g/mol = 131.7 g.

That is every mass-to-mass problem you will ever encounter. The numbers and substances change, but the three steps are identical every single time. If you internalize this workflow, stoichiometry becomes mechanical rather than mysterious.

When a problem gives you amounts of two or more reactants, one of them will run out first — that is the limiting reagent. It determines how much product you can make. The other reactant is in excess.

To find the limiting reagent, convert both reactant amounts to moles, then see which one would run out first based on the mole ratio. Here is an example.

Problem: You have 10.0 g of H2 and 80.0 g of O2. How many grams of water can be produced? (2H2 + O2 -> 2H2O)

Convert both to moles: H2 = 10.0 g / 2.016 g/mol = 4.96 mol. O2 = 80.0 g / 32.00 g/mol = 2.50 mol.

The equation says you need 2 mol H2 for every 1 mol O2. So 4.96 mol H2 would need 4.96/2 = 2.48 mol O2. You have 2.50 mol O2 — that is enough. Now check the other way: 2.50 mol O2 would need 2.50 x 2 = 5.00 mol H2. You only have 4.96 mol — not quite enough. So H2 is the limiting reagent.

Calculate product from the limiting reagent: 4.96 mol H2 x (2 mol H2O / 2 mol H2) = 4.96 mol H2O. Mass = 4.96 x 18.02 = 89.4 g H2O.

A shortcut some students use: divide each reactant's moles by its coefficient. The one with the smaller result is limiting. For H2: 4.96/2 = 2.48. For O2: 2.50/1 = 2.50. H2 gives the smaller number, so it is limiting. This shortcut works every time and is faster than the full comparison.

Theoretical yield is what stoichiometry predicts. Actual yield is what you measure in the lab. Percent yield tells you how efficient the reaction was: percent yield = (actual yield / theoretical yield) x 100.

If your calculation predicts 89.4 g of water but you only collect 76.0 g, your percent yield is (76.0 / 89.4) x 100 = 85.0%. Real reactions rarely give 100% yield because of side reactions, incomplete reactions, transfer losses, and purification steps.

Exam questions typically give you two of the three values (actual yield, theoretical yield, percent yield) and ask for the third. The algebra is straightforward — just rearrange the formula. The tricky part is calculating the theoretical yield correctly using the stoichiometry steps above, because errors in the theoretical yield propagate into the percent yield.

One thing that catches students: percent yield should never exceed 100% in a well-run experiment. If your calculation gives 105%, something is wrong — usually either the product was not fully dried (residual water adds mass) or there was a calculation error. ChemistryIQ practice problems include percent yield scenarios with common error-spotting challenges.