Stoichiometry: The Math of Reactions

Stoichiometry allows chemists to predict the amounts of products that will form in a chemical reaction, based on the amounts of reactants provided.

At its core, stoichiometry relies on the Law of Conservation of Mass: matter is neither created nor destroyed. A balanced chemical equation tells you the molar ratio of reactants and products.

Balancing Equations

Before doing any math, the equation must be balanced. The number of atoms of each element on the reactant side must equal the number on the product side.

Unbalanced: CH₄ + O₂ → CO₂ + H₂O
Balanced:   CH₄ + 2O₂ → CO₂ + 2H₂O

The Limiting Reactant

In most real-world scenarios, reactants are not mixed in the exact stoichiometric ratio. One reactant will be entirely consumed first, stopping the reaction. This is the limiting reactant.

Worked Example

Problem: If 10.0g of H₂ and 50.0g of O₂ react to form water ($2H_2 + O_2 \rightarrow 2H_2O$), which is limiting, and what mass of water is formed?

1. Convert mass to moles:
   • Moles H₂ = $10.0\text{g} / 2.016 \text{g/mol} = 4.96 \text{ mol}$
   • Moles O₂ = $50.0\text{g} / 32.00 \text{g/mol} = 1.56 \text{ mol}$
2. Determine limiting reactant using the molar ratio (2 moles H₂ per 1 mole O₂):
   • 1.56 mol O₂ would require $1.56 \times 2 = 3.12$ mol H₂.
   • We have 4.96 mol H₂ (excess). Therefore, O₂ is the limiting reactant.
3. Calculate theoretical yield based on limiting reactant:
   • Ratio of O₂ to H₂O is 1:2. So, $1.56 \text{ mol O}_2 \rightarrow 3.12 \text{ mol H}_2\text{O}$.
   • Mass H₂O = $3.12 \text{ mol} \times 18.015 \text{ g/mol} = 56.2 \text{ grams of H}_2\text{O}$.

Percent Yield

Theoretical yield is what you calculated above. Actual yield is what you physically measure in the lab (which is almost always less due to side reactions or loss of product).

Percent Yield = (Actual Yield / Theoretical Yield) × 100%