Stoichiometry and Chemical Calculations

Core Review

The Mole Concept

The mole (mol) is the chemist's counting unit. One mole of any substance contains exactly 6.02214076 × 10²³ particles (Avogadro's number, Nₐ). This number ties the microscopic world of atoms to the macroscopic world of grams. The molar mass (g/mol) of an element is numerically equal to its atomic mass from the periodic table — carbon's molar mass is 12.01 g/mol, meaning one mole of carbon atoms weighs 12.01 g.

Converting grams to moles:

moles = mass (g) / molar mass (g/mol)

For example, 24.02 g of carbon equals 24.02 / 12.01 = 2.00 mol C.

Balancing Chemical Equations

A balanced equation satisfies the law of conservation of mass — atoms are neither created nor destroyed. The number of atoms of each element must be identical on both sides. Balance by adjusting coefficients (never change subscripts). Start with elements appearing in the fewest compounds, save H and O for last.

Example: Combustion of methane

CH₄ + 2 O₂ → CO₂ + 2 H₂O

From this equation, the mole ratio CH₄ : O₂ : CO₂ : H₂O is 1 : 2 : 1 : 2.

Stoichiometric Calculations (Mass-to-Mass)

The standard pathway for mass-to-mass problems:

1. Convert grams of given substance to moles (÷ molar mass).
2. Use the mole ratio from the balanced equation to find moles of desired substance.
3. Convert moles of desired substance to grams (× molar mass).

Worked Example: How many grams of H₂O are produced when 16.0 g of CH₄ combusts completely?

• Step 1: moles CH₄ = 16.0 g / 16.04 g/mol = 0.9975 mol
• Step 2: moles H₂O = 0.9975 mol CH₄ × (2 mol H₂O / 1 mol CH₄) = 1.995 mol
• Step 3: grams H₂O = 1.995 mol × 18.02 g/mol = 35.9 g

Percent Composition, Empirical and Molecular Formulas

Percent composition is the mass percent of each element in a compound:

% element = (mass of element in 1 mol compound / molar mass of compound) × 100

Empirical formula is the simplest whole-number ratio of atoms. Determine it by converting percent to grams, grams to moles, then dividing by the smallest mole value to get integer subscripts.

Molecular formula is a whole-number multiple of the empirical formula:

n = molar mass / empirical formula mass

Worked Example: A compound is 40.0% C, 6.71% H, 53.3% O with a molar mass of 180.2 g/mol. Find both formulas.

• Assume 100 g sample: 40.0 g C, 6.71 g H, 53.3 g O
• moles C = 40.0 / 12.01 = 3.33; moles H = 6.71 / 1.008 = 6.66; moles O = 53.3 / 16.00 = 3.33
• Divide by 3.33: C₁H₂O₁ → empirical formula CH₂O (mass = 30.03 g/mol)
• n = 180.2 / 30.03 ≈ 6 → molecular formula = C₆H₁₂O₆

Limiting Reactants, Excess Reactants, and Yield

The limiting reactant is entirely consumed first and determines the maximum amount of product. The excess reactant is present in more than the stoichiometric amount; some remains unreacted.

To identify the limiting reactant: convert each reactant to moles of the same product using the balanced equation. Whichever produces fewer moles of product is limiting.

Theoretical yield = maximum product mass predictable from the limiting reactant (stoichiometric calculation).

Percent yield measures efficiency:

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

Worked Example: 10.0 g of N₂ reacts with 10.0 g of H₂ to produce NH₃. N₂ + 3 H₂ → 2 NH₃.

• moles N₂ = 10.0 / 28.02 = 0.357 mol → produces 0.357 × 2 = 0.714 mol NH₃
• moles H₂ = 10.0 / 2.016 = 4.96 mol → produces 4.96 × (2/3) = 3.31 mol NH₃
• N₂ is limiting. Theoretical yield = 0.714 mol × 17.03 g/mol = 12.2 g NH₃

Key Equations

Common Traps

• Using mass ratios instead of mole ratios for stoichiometric conversions.
• Forgetting to balance the equation before extracting mole ratios.
• Confusing the empirical formula with the molecular formula when molar mass is given.
• Identifying the limiting reactant by comparing grams of reactants directly (instead of moles of product formed).
• Rounding too aggressively in multi-step calculations — carry at least three significant figures through intermediate steps.