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Chemistry · January 2026

Stoichiometry and Mole Calculations

Stoichiometry is the branch of chemistry that uses balanced equations to calculate the amounts of reactants and products involved in a chemical reaction. It is the mathematical heart of chemistry — and the mole is its central unit.

The Mole Concept

Atoms and molecules are far too small to count individually in a laboratory. Chemists instead measure amounts in moles. One mole of any substance contains exactly 6.022 × 1023 particles (atoms, molecules, or ions). This number — Avogadro's number — was chosen so that one mole of carbon-12 atoms has a mass of exactly 12 grams. The same logic extends to everything else: one mole of water (H2O) has a mass of 18.015 g; one mole of sodium chloride (NaCl) has a mass of 58.44 g.

The molar mass of any element equals its atomic mass in grams per mole (found on the periodic table). For compounds, add the molar masses of all atoms in one formula unit: H2O = (2 × 1.008) + 16.00 = 18.016 g/mol.

Converting Between Mass, Moles, and Particles

Three quantities are interconvertible using molar mass and Avogadro's number:

  • Grams to moles: moles = mass (g) ÷ molar mass (g/mol)
  • Moles to grams: mass = moles × molar mass
  • Moles to particles: particles = moles × 6.022 × 1023
  • Particles to moles: moles = particles ÷ 6.022 × 1023

Worked example: How many moles are in 50.0 g of calcium carbonate (CaCO3)?
Molar mass of CaCO3 = 40.08 + 12.01 + (3 × 16.00) = 100.09 g/mol
Moles = 50.0 ÷ 100.09 = 0.500 mol

Reading a Balanced Equation

The coefficients in a balanced chemical equation give the mole ratios in which reactants combine and products form. For the combustion of methane:

CH4 + 2 O2 → CO2 + 2 H2O

This means 1 mol of methane reacts with 2 mol of oxygen to produce 1 mol of carbon dioxide and 2 mol of water. These ratios are fixed by the law of conservation of mass and can be used as conversion factors in stoichiometric calculations.

Mole-to-Mole and Mass-to-Mass Calculations

The standard stoichiometry calculation follows a three-step process:

  1. Convert the given quantity to moles (if it is in grams, divide by molar mass).
  2. Use the mole ratio from the balanced equation to convert to moles of the desired substance.
  3. Convert the answer to the required unit (if grams are needed, multiply by molar mass).

Worked example: How many grams of CO2 are produced when 8.00 g of CH4 burns completely?

Step 1: Moles of CH4 = 8.00 ÷ 16.05 = 0.499 mol
Step 2: Mole ratio CH4 : CO2 = 1 : 1, so moles of CO2 = 0.499 mol
Step 3: Mass of CO2 = 0.499 × 44.01 = 21.96 g

Limiting Reagent

In most real reactions, one reactant is used up before the others. The limiting reagent is the one that runs out first and determines the maximum amount of product that can be formed. All other reactants are in excess.

To find the limiting reagent: convert each reactant to moles, then calculate how many moles of product each would yield (using the mole ratio). The reactant that produces the smaller amount of product is the limiting reagent.

Worked example: 5.00 g of H2 and 40.0 g of O2 react: 2 H2 + O2 → 2 H2O. Which is limiting?
Moles of H2 = 5.00 ÷ 2.016 = 2.48 mol. From mole ratio (2 H2 : 2 H2O), this gives 2.48 mol H2O.
Moles of O2 = 40.0 ÷ 32.00 = 1.25 mol. From mole ratio (1 O2 : 2 H2O), this gives 2.50 mol H2O.
H2 produces fewer moles of product (2.48 vs. 2.50), so H2 is the limiting reagent.

Excess Reagent

Once the limiting reagent is identified, the amount of excess reagent remaining can be calculated. In the example above, O2 is in excess. The H2 uses 2.48/2 = 1.24 mol O2; the amount of O2 remaining = 1.25 − 1.24 = 0.01 mol, which equals 0.32 g.

Theoretical Yield and Percentage Yield

The theoretical yield is the maximum amount of product calculated from stoichiometry, assuming the limiting reagent is completely consumed and the reaction goes to completion with no losses. In practice, reactions are rarely 100% efficient — side reactions occur, product is lost in transfer, or the reaction reaches equilibrium before going to completion. The amount actually obtained is the actual yield.

Percentage yield = (actual yield ÷ theoretical yield) × 100%

If the theoretical yield is 22.0 g but you obtained 18.7 g: percentage yield = (18.7 ÷ 22.0) × 100 = 85.0%. A yield above 100% is impossible — if you calculate one, check that you have not mixed up actual and theoretical yields, or that the product is not contaminated with water or unreacted reagent.

Summary

Stoichiometry uses the mole as a bridge between the atomic scale and the laboratory scale. Molar mass converts grams to moles; balanced equation coefficients provide mole ratios; these ratios convert moles of one substance to moles of another. The limiting reagent determines the maximum product obtainable. Percentage yield measures reaction efficiency by comparing actual to theoretical output. Together, these tools allow chemists to plan reactions, scale up industrial processes, and account for every gram of reactant and product.