5.3 Stoichiometry & the Mole Concept

Stoichiometry is the arithmetic of chemistry: it converts a balanced chemical equation into real-world quantities — grams, moles, and molecules — that you can measure on a scale or read off a gauge. Whether you're calculating how much fuel a reaction consumes or how much product a shipboard water-treatment process generates, stoichiometry is the tool that connects the equation to the numbers. This section covers Avogadro's number, molar mass calculation, mole-to-mole and mass-to-mass conversions, and a full worked example converting grams of reactant to grams of product.

The Mole and Avogadro's Number

Chemists need a counting unit for atoms and molecules because individual particles are far too small and numerous to count directly. That unit is the mole (mol) — the amount of substance that contains exactly 6.022 × 10²³ particles, a quantity known as Avogadro's number (NA), after the Italian scientist Amedeo Avogadro. One mole of anything — atoms, molecules, ions — always contains this same number of particles, even though different substances have very different masses per particle.

• 1 mole of carbon atoms = 6.022 × 10²³ atoms = 12.01 grams (carbon's molar mass).
• 1 mole of water molecules = 6.022 × 10²³ molecules = 18.02 grams (water's molar mass).

The conversion works both directions:

• Number of particles = moles × 6.022 × 10²³
• Moles = number of particles ÷ 6.022 × 10²³

For example, 3.00 moles of water contain 3.00 × 6.022 × 10²³ = 1.81 × 10²⁴ molecules — over a thousand times more than a single mole, which is why Avogadro's number is so large: it has to bridge the gap between everyday masses (grams) and the truly enormous number of atoms it takes to make up that mass.

Molar Mass Calculation

Molar mass is the mass, in grams, of exactly one mole of a substance (units: g/mol). For an element, molar mass is numerically equal to the atomic mass shown on the periodic table. For a compound, add up the atomic mass of every atom in the formula, multiplying by its subscript:

A common mistake is forgetting to multiply by the subscript — for Ca(OH)2, that means counting only one hydroxide (OH) group instead of two, which gives an incorrect 57.09 g/mol instead of the correct 74.10 g/mol. Always check that every atom inside (and outside) a set of parentheses gets multiplied by its subscript.

Mole-to-Mole and Mass-to-Mass Stoichiometry

A balanced chemical equation's coefficients don't just balance atoms — they also give the mole ratio between any two substances in the reaction. That ratio is the bridge for every stoichiometry calculation.

Mole-to-mole conversion: moles of B = moles of A × (coefficient of B ÷ coefficient of A).

Worked example: for N2 + 3H2 → 2NH3, the mole ratio of N2 to NH3 is 1:2. Starting with 5.00 mol of N2 (and excess H2 available), you can produce 5.00 mol N2 × (2 mol NH3 ÷ 1 mol N2) = 10.0 mol NH3.

Mass-to-mass conversion adds a molar-mass step on each end, following a four-step roadmap: convert grams of A to moles of A (divide by molar mass of A), convert moles of A to moles of B (multiply by the mole ratio from the balanced equation), then convert moles of B to grams of B (multiply by molar mass of B).

Worked Example: Grams of Reactant to Grams of Product

Problem: Propane burns completely according to C3H8 + 5O2 → 3CO2 + 4H2O. How many grams of carbon dioxide (CO2) are produced when 44.0 g of propane (C3H8) is burned in excess oxygen?

Step 1 — Molar mass of the reactant. C3H8 = 3 × C (12.01) + 8 × H (1.008) = 36.03 + 8.064 = 44.09 ≈ 44.1 g/mol.

Step 2 — Grams to moles. 44.0 g ÷ 44.1 g/mol ≈ 1.00 mol C3H8.

Step 3 — Mole ratio to the product. The balanced equation gives a mole ratio of 1 mol C3H8 to 3 mol CO2, so: 1.00 mol C3H8 × (3 mol CO2 ÷ 1 mol C3H8) = 3.00 mol CO2.

Step 4 — Moles to grams of product. Molar mass of CO2 = 12.01 + 2(16.00) = 44.01 g/mol. So: 3.00 mol CO2 × 44.01 g/mol ≈ 132 g of CO2.

Answer: Burning 44.0 g of propane completely produces about 132 g of CO2. Notice that the same four-step roadmap — grams → moles → moles → grams — solves every mass-to-mass stoichiometry problem, no matter which reaction or which substances are involved.

Key Takeaways

• One mole of any substance contains Avogadro's number of particles: 6.022 × 10²³.
• Molar mass (g/mol) comes from summing each element's atomic mass, multiplied by its subscript, e.g., Ca(OH)2 = 74.10 g/mol.
• A balanced equation's coefficients supply the mole ratio used in both mole-to-mole and mass-to-mass conversions.
• Mass-to-mass stoichiometry always follows the same roadmap: grams of A → moles of A → moles of B → grams of B.
• Burning 44.0 g of propane (C3H8 + 5O2 → 3CO2 + 4H2O) produces about 132 g of CO2 — a complete worked example of the grams-to-grams roadmap.