5.3 Stoichiometry & the Mole Concept
Key Takeaways
- The Avogadro constant is exactly 6.02214076 × 10²³ mol⁻¹; calculations often round it to about 6.022 × 10²³.
- Molar mass (g/mol) is calculated by summing each element's atomic mass multiplied by its subscript in the chemical formula, e.g., Ca(OH)2 = 74.10 g/mol.
- A balanced equation's coefficients supply the mole ratio needed for both mole-to-mole and mass-to-mass stoichiometry conversions.
- Mass-to-mass stoichiometry always follows a four-step roadmap: grams of A → moles of A (÷ molar mass) → moles of B (× mole ratio) → grams of B (× molar mass).
- 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 stoichiometry roadmap.
Stoichiometry connects balanced equations to moles and mass in standard introductory chemistry. This section is part of the site's editorial STEM framework; its worked examples are not claims about NAPT content or Navy operations.
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 containing 6.02214076 × 10²³ specified entities, the exact SI-defined Avogadro constant (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 and has a molar mass of about 12.01 g/mol.
- 1 mole of water molecules ≈ 6.022 × 10²³ molecules and has a molar mass of about 18.02 g/mol.
The conversion works both directions:
- Number of particles ≈ moles × 6.022 × 10²³
- Moles ≈ number of particles ÷ 6.022 × 10²³
These calculation forms use a rounded value; the exact constant is 6.02214076 × 10²³ mol⁻¹.
For example, using the rounded constant, 3.00 moles of water contain about 3.00 × 6.022 × 10²³ = 1.81 × 10²⁴ molecules—exactly three times the number of molecules in 1.00 mole. The constant is large because it bridges macroscopic amounts and particle counts.
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:
| Compound | Calculation | Molar Mass |
|---|---|---|
| NaCl | Na (22.99) + Cl (35.45) | 58.44 g/mol |
| CO2 | C (12.01) + 2 × O (16.00) | 44.01 g/mol |
| Ca(OH)2 | Ca (40.08) + 2 × O (16.00) + 2 × H (1.008) | 74.10 g/mol |
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
- The exact Avogadro constant is 6.02214076 × 10²³ mol⁻¹; this section rounds it for arithmetic.
- 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.
How many individual water molecules are present in 3.00 moles of H2O?
What is the molar mass of calcium hydroxide, Ca(OH)2? (Ca = 40.08 g/mol, O = 16.00 g/mol, H = 1.008 g/mol)
Using 2H2(g) + O2(g) → 2H2O(l), how many grams of water form when 10.0 g of H2 reacts completely with excess oxygen? (Molar mass H2 = 2.016 g/mol; H2O = 18.02 g/mol)
For the reaction N2(g) + 3H2(g) → 2NH3(g), how many moles of NH3 can be produced from 5.00 moles of N2 (with excess H2 available)?