Mole Concepts

The mole is the chemical engineer's counting unit. Learn molar mass, moles and kmol, mass/mole fractions and concentrations, ideal-gas molar volume, stoichiometry, limiting and excess reactants, conversion, selectivity and yield.

Chemical reactions happen molecule by molecule, so engineers count in moles: 1 mol = 6.022 × 10²³ particles; in plants we use kmol. Converting between mass (what we weigh and ship) and moles (what reacts) is the first skill of chemical engineering.

Stoichiometry then tells us how much of each reactant is needed and how much product forms.

An analogy

A recipe says '2 eggs per cup of flour' — stoichiometry is the recipe of a reaction, and moles are how you count the eggs.

Moles, mass and molar mass

Molar mass is the mass of one mole: H₂O 18.015, CO₂ 44.01, CH₄ 16.04 g/mol. Divide mass by molar mass to get moles. Air is ≈ 21 % O₂ / 79 % N₂ by moles, average M ≈ 28.97.

Composition measures

Mass fraction, mole fraction, molarity (mol/L), molality (mol/kg solvent), ppm, and partial pressure for gases (p_i = y_i P). Liquids are often specified by mass, gases by moles/volume.

Stoichiometry

Balanced equations give mole ratios. The limiting reactant runs out first; others are in excess. Combustion uses excess air (10–30 %) to ensure complete burning. Conversion, selectivity (desired/undesired product) and yield (product formed/maximum possible) measure reactor performance.

Extent of reaction and atom balances

For multiple reactions, the extent ξ tracks progress: n_i = n_i0 + Σν_ij ξ_j. Element (atom) balances are always conserved and help check complex reacting systems.

Formulas

• Moles: n = m / M
• Mass → mole fraction: x_i = (w_i/M_i) / Σ(w_j/M_j)
• Ideal-gas molar volume: 22.414 m³/kmol at 0 °C, 1 atm (STP)
• Conversion: X = (n_in − n_out) / n_in (of limiting reactant)
• Percent excess: % excess = (fed − stoichiometric)/stoichiometric × 100

Worked example: Combustion of methane

Given: 100 kg CH₄; CH₄ + 2O₂ → CO₂ + 2H₂O; 20 % excess air

Find: O₂ required, air fed, CO₂ produced.

• n = 100/16.04 = 6.23 kmol CH₄
• O₂ theoretical = 12.47 kmol (399 kg) → air = 12.47/0.21 = 59.4 kmol
• With 20 % excess: 71.3 kmol air
• CO₂ = 6.23 × 44.01 = 274 kg

Answer: 399 kg O₂, 71.3 kmol air, 274 kg CO₂

Worked example: Mass to mole fraction

Given: 40 wt % ethanol (M 46.07) in water (M 18.015)

Find: Mole fraction ethanol.

• Per 100 g: ethanol 0.868 mol, water 3.331 mol
• x = 0.868/4.199 = 0.207

Answer: x ≈ 0.21

Common mistakes

• Wrong: Balancing on mass ratios from the equation. — Right: Equations give mole ratios — convert masses to moles first.
• Wrong: Forgetting air is only 21 % oxygen. — Right: Air = O₂ required / 0.21 (plus excess).

Applications

• Reactor feed design: Feed ratios set from stoichiometry plus excess of cheap reactant.
• Combustion & emissions: Air requirements and CO₂ reporting for furnaces and boilers.
• Laboratory to plant: Recipes scaled from grams to tonnes via moles.

Key takeaways

• n = m/M; reactions balance in moles.
• Identify limiting and excess reactants.
• Conversion, selectivity, yield rate reactors.
• Air = O₂/0.21 (+ excess).

Going further

• Extent of reaction: Single variable per reaction tracks all species changes.
• Non-ideal gases: Compressibility factor Z corrects molar volumes at high pressure.
• Atom economy: Green chemistry metric: mass of desired product / mass of all reactants.