Study Notes

Stoichiometry is the study of the quantitative relationships between reactants and products in a chemical reaction.

A balanced chemical equation tells you the MOLAR RATIO of every substance involved.

Example: N₂ + 3H₂ → 2NH₃

• 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃

• Molar ratios: 1:3:2

• Conversion factor: (2 mol NH₃)/(1 mol N₂) OR (1 mol N₂)/(3 mol H₂), etc.

Using molar ratios as conversion factors:

To find moles of product from moles of reactant:

moles A × (moles B / moles A from balanced equation) = moles B

The molar ratio comes directly from the coefficients in the balanced equation.

Stoichiometry problems often require converting between grams, moles, and other units.

Steps for grams of A → grams of B:

1. Convert grams of A → moles of A (divide by molar mass of A)

2. Convert moles of A → moles of B (multiply by molar ratio B/A)

3. Convert moles of B → grams of B (multiply by molar mass of B)

Full chain as fractions:

grams A × (1 mol A / molar mass A) × (mol B / mol A) × (molar mass B / 1 mol B) = grams B

Example: Given N₂ + 3H₂ → 2NH₃, how many grams of NH₃ are produced from 28g of N₂?

Molar mass N₂ = 28 g/mol; Molar mass NH₃ = 17 g/mol

28g N₂ × (1 mol N₂ / 28g) × (2 mol NH₃ / 1 mol N₂) × (17g NH₃ / 1 mol NH₃)

= 1 × 2 × 17 = 34g NH₃

Identifying how many steps:

• Moles → Moles: 1 step (molar ratio only)

• Moles → Grams: 1 step (multiply by molar mass)

• Grams → Moles: 1 step (divide by molar mass)

• Grams → Grams: 3 steps (g→mol→mol→g)

In real reactions, one reactant is used up first — this is the LIMITING REAGENT. The other is in excess.

The limiting reagent DETERMINES how much product is formed.

The excess reagent is the one left over after the reaction stops.

Identifying the limiting reagent (given amounts in moles):

1. Calculate moles of product possible from each reactant using molar ratios

2. The reactant that produces LESS product is the limiting reagent

Example: N₂ + 3H₂ → 2NH₃

Given: 4 mol N₂ and 6 mol H₂

From N₂: 4 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 8 mol NH₃

From H₂: 6 mol H₂ × (2 mol NH₃ / 3 mol H₂) = 4 mol NH₃

H₂ produces less → H₂ is the LIMITING REAGENT

Maximum NH₃ produced = 4 mol (based on limiting reagent H₂)

Note: Exam questions on limiting reagents will give starting quantities in MOLES — only one conversion step needed.

In real-world chemistry, reactions rarely produce 100% of the theoretical (calculated) amount of product. The PERCENT YIELD measures efficiency.

Formula: % Yield = (Actual Yield / Theoretical Yield) × 100%

• Theoretical Yield: calculated maximum amount of product based on stoichiometry

• Actual Yield: the amount actually collected in the experiment

• Percent Yield: how efficient the reaction was (always ≤ 100% in real experiments)

Why is actual yield less than theoretical?

• Side reactions occur

• Some product is lost during collection

• Reaction doesn't go to completion

• Impure reactants

Example: A reaction should produce 50g of product (theoretical yield), but only 40g is collected (actual yield).

% Yield = (40g / 50g) × 100% = 80%

Note: Percent yield problems always have ONE starting quantity in excess — so you won't need to find the limiting reagent for these problems.