Stoichiometry

The balanced equation is the starting point

For N₂ + 3H₂ → 2NH₃, one N₂ molecule reacts with three H₂ molecules to form two NH₃ molecules. The same ratio applies to moles:

The ratio does not say 1 g N₂ reacts with 3 g H₂ because different substances have different molar masses.

Worked mass-to-mass calculation

Suppose 10.0 g H₂ reacts with excess N₂. Using M(H₂) ≈ 2.016 g mol⁻¹:

Writing units at every step makes the dimensional logic visible and catches many setup errors.

Limiting reactant and theoretical yield

When more than one reactant amount is given, calculate the product each could make. The smaller predicted amount comes from the limiting reactant.

The theoretical yield is the maximum product allowed by stoichiometry from the limiting reactant. Excess reagent is left over after the limiting reagent is consumed.

Actual yield and percent yield

Real experiments can lose product, stop before completion or form side products. Percent yield compares recovered product with the theoretical maximum:

A yield above 100% usually signals wet/impure product, measurement error or an incorrect assumed composition—not a reaction that created extra matter.

Mole-ratio calculator

The calculator handles only the mole-ratio step. A complete problem still requires a balanced equation and any necessary mass, volume or concentration conversions.

Solutions, gases and stoichiometry

For a solution, moles can come from n = cV when concentration c is in mol L⁻¹ and volume V is in liters. Gas amounts can be obtained from an appropriate equation of state, then related through the balanced reaction.

The central rule stays the same: convert the measured quantity to moles, apply the coefficient ratio, then convert to the requested quantity.

Excess reactant and material left over

After identifying the limiting reagent, the amount of excess reagent consumed can be calculated from the same mole ratio. Subtract that from the initial amount to find what remains.

For N₂ + 3H₂ → 2NH₃ with 2.00 mol N₂ and 4.00 mol H₂, hydrogen is limiting. Consuming 4.00 mol H₂ requires 4.00/3 = 1.33 mol N₂, leaving 0.67 mol N₂.

This leftover calculation is important in reactor design and laboratory planning because excess reagent can affect purification, cost and safety.

Solution stoichiometry and titration

If concentration c and volume V are known, the amount in solution is n = cV. Once moles are found, the balanced equation supplies the stoichiometric ratio exactly as in a mass problem.

Titration measures the volume required to reach a stoichiometric reaction point. Indicator color or an instrumental signal is used to locate an experimental endpoint close to that equivalence point.

Gas stoichiometry

Gas amounts can be converted to moles through an equation of state such as PV = nRT when ideal-gas behavior is adequate. At the same temperature and pressure, gas volumes are proportional to mole amounts, so balanced coefficients can also become volume ratios.

For 2H₂ + O₂ → 2H₂O(g), two volumes of H₂ react with one volume of O₂ to produce two volumes of water vapor at the same T and P, provided the gases behave ideally and water remains gaseous.

Purity, hydrates and the chemical formula actually present

A weighed sample may not be pure reactant. If a solid is 92.0% active compound by mass, only 0.920 times the sample mass should enter the mole calculation for that compound.

Hydrated salts require the molar mass of the actual hydrate. CuSO₄·5H₂O and anhydrous CuSO₄ contain the same amount of CuSO₄ per mole of formula units but very different mass fractions because five waters are included in the hydrate mass.

Many apparent stoichiometry mistakes are actually formula or sample-composition mistakes made before the coefficient ratio is ever used.

Reaction extent unifies all stoichiometric changes

For advanced bookkeeping, a reaction extent ξ measures how far a balanced reaction has proceeded. If νᵢ is the signed stoichiometric coefficient of species i, then its amount changes according to Δnᵢ = νᵢξ.

For N₂ + 3H₂ → 2NH₃, an extent of 1.00 mol consumes 1.00 mol N₂ and 3.00 mol H₂ while producing 2.00 mol NH₃. The familiar mole ratios are the component form of this single relation.

Significant figures and exact coefficients

Stoichiometric coefficients in a correctly balanced equation are exact integers for the stated reaction and do not limit significant figures. Measurement precision comes from masses, volumes, concentrations and molar masses.

Carry extra digits through intermediate steps and round at the end. Premature rounding can noticeably distort a limiting-reactant comparison when two predicted product amounts are close.

Exercises

Mole ratio

For 2H₂ + O₂ → 2H₂O, how many moles H₂O can 3.0 mol O₂ form with excess H₂?

Limiting reagent

For N₂ + 3H₂ → 2NH₃, which is limiting: 1.0 mol N₂ or 2.0 mol H₂?

Percent yield

The theoretical yield is 25.0 g and actual yield 20.0 g. Find percent yield.