AP Chemistry 4.5: Stoichiometry

• Stoichiometry uses the numerical relationships in a balanced chemical equation to calculate how much reactant is required or how much product can form.

• Almost every stoichiometry problem is built around one central idea:

  • balanced equation coefficients create mole ratios.

• Khan Academy’s current Unit 4 stoichiometry material includes reactant/product calculations, limiting reactants, reaction yields, precipitation-based calculations, and connections between reaction stoichiometry and the ideal gas law.

Mole Ratios

• Consider:

  • 2H₂ + O₂ → 2H₂O
  • The equation tells us:
  • 2 mol H₂ : 1 mol O₂ : 2 mol H₂O

• These ratios can become conversion factors.

  • For example:
  • 2 mol H₂O / 1 mol O₂
  • or:
  • 1 mol O₂ / 2 mol H₂

• The correct ratio depends on what you are given and what you need to find.

• Suppose you begin with 3.00 mol O₂ and have excess H₂:

  • 3.00 mol O₂ × (2 mol H₂O / 1 mol O₂) = 6.00 mol H₂O

The Main Stoichiometry Path

• Most stoichiometry problems follow:

  • given quantity → convert to moles → use balanced-equation mole ratio → moles wanted → convert to requested unit

• The middle of nearly every problem is:

  • mol A → mole ratio → mol B

Mass-to-Mass Stoichiometry

• Suppose 4.00 g H₂ reacts with excess O₂. How many grams of H₂O can form?

  • Equation:
  • 2H₂ + O₂ → 2H₂O

• First convert H₂ mass into moles.

  • Molar mass H₂ ≈ 2.016 g/mol.
  • 4.00 g H₂ × (1 mol H₂ / 2.016 g H₂) ≈ 1.98 mol H₂

• Now use the coefficient ratio:

  • 1.98 mol H₂ × (2 mol H₂O / 2 mol H₂) ≈ 1.98 mol H₂O

• Convert to grams using the molar mass of water:

  • 1.98 mol H₂O × 18.02 g/mol ≈ 35.7 g H₂O

• The complete process was:

  • grams H₂ → mol H₂ → mol H₂O → grams H₂O

Solution Stoichiometry

• From Unit 3:

  • M = mol/L
  • Therefore:
  • mol = MV
  • where volume must be in liters.

• Suppose you have 50.0 mL of 0.200 M NaOH.

  • Convert:
  • 50.0 mL = 0.0500 L
  • Then:
  • n = MV
  • n = (0.200 mol/L)(0.0500 L)
  • n = 0.0100 mol NaOH

• Those moles can then be converted into another substance using the balanced equation.

  • Solution stoichiometry becomes extremely important in titration problems.

Gas Stoichiometry

• Stoichiometry can also connect to the ideal gas law:

  • PV = nRT

• If a problem gives pressure, volume, and temperature for a gas, use PV=nRT to find gas moles.

  • Then:
  • gas information → PV=nRT → mol gas → mole ratio → mol desired substance → requested unit
  • This is a direct connection between Units 3 and 4.

Stoichiometry From a Precipitate

• Some AP problems use the mass of a precipitate to determine the concentration or amount of a dissolved substance.

• The path is typically:

  • mass precipitate → mol precipitate → mole ratio → mol original ion/solute → concentration
  • For example, if CaCO₃ is collected from a reaction, its measured mass can be converted to moles using its molar mass. The balanced reaction then tells you how many moles of the original dissolved species were required to form that amount of CaCO₃.

Limiting Reactant

• The limiting reactant is the reactant that gets completely consumed first. Once it is gone, no additional product can form even if another reactant remains.

  • Because of this, the limiting reactant determines the maximum possible amount of product.

• Consider:

  • N₂ + 3H₂ → 2NH₃
  • Suppose you begin with:
  • 2.00 mol N₂ and 4.00 mol H₂

• Every 1 mol N₂ requires 3 mol H₂.

  • So 2.00 mol N₂ would require:
  • 2.00 × 3 = 6.00 mol H₂
  • But only 4.00 mol H₂ is available.

• Therefore:

  • H₂ is the limiting reactant
  • and:
  • N₂ is the excess reactant

The Most Reliable Limiting-Reactant Method

• Calculate the amount of the same product each reactant could make.

  • If reactant A could make 8.50 mol product but reactant B could make only 5.20 mol product, the reaction can only produce 5.20 mol.
  • Therefore B is limiting.

• This method avoids unreliable shortcuts such as assuming the reactant with the smaller mass or fewer moles must be limiting.

Excess Reactant

• The excess reactant is the substance left over after the limiting reactant has been completely consumed.

• To determine how much remains:

  • initial amount − amount consumed = amount remaining
  • You first use the limiting reactant to calculate how much of the excess reactant was required.

Theoretical Yield

• The theoretical yield is the maximum amount of product predicted by stoichiometry.

  • The theoretical yield must be calculated using the limiting reactant because that is the substance that determines when product formation stops.

Actual Yield

• The actual yield is the amount of product actually collected in an experiment.

  • Actual yield can be lower than theoretical yield because the reaction may not go completely to products, product may be lost during filtration or transfer, side reactions may occur, or measurement errors may happen.

Percent Yield

• Percent yield compares the amount actually obtained with the theoretical maximum:

  • Percent yield = (actual yield / theoretical yield) × 100%

• Suppose:

  • Theoretical yield = 20.0 g
  • Actual yield = 17.0 g
  • Then:
  • Percent yield = (17.0 / 20.0)(100%) = 85.0%

Common Mistakes

• Always balance the equation before doing stoichiometry. Mole ratios come from coefficients, not subscripts. Do not put grams directly into a mole ratio—convert grams to moles first.

• Do not assume the reactant with the smaller mass is limiting. Limiting reactant depends on both the number of moles and the coefficient ratio.

• Remember that theoretical yield comes from the limiting reactant, and percent yield uses:

  • actual / theoretical
  • not the reverse.

Remember This

• When stuck, return to:

  • given → moles → mole ratio → moles wanted → answer unit
  • That pathway solves most stoichiometry problems.