Chemistry

The mole and stoichiometry

Moles, molar mass, empirical formulas, and yield.

Basics

Mole

One mole is Avogadro’s number of particles (about 6.022×10²³). Mass (g) = moles × molar mass (g/mol). For gases, molar volume is about 22.4 L/mol near STP — check the STP definition in use.

Limiting reactant

Convert to moles, apply the coefficient ratio, and the reactant that runs out first sets the product amount. The rest is in excess.

Formulas

Amount in moles

n = m / M

Mass divided by molar mass.

Symbols

  • n amount (mol)
  • m mass (g)
  • M molar mass (g/mol)

Mass percent

%X = (m_X / m_total) × 100%

Mass fraction of element X in a compound.

Symbols

  • m_X mass of element X
  • m_total total mass

Percent yield

%yield = (actual / theoretical) × 100%

Actual product versus the stoichiometric maximum.

Symbols

  • actual actual yield
  • theoretical theoretical yield

Molarity

c = n / V

Moles of solute per litre of solution.

Symbols

  • c molarity (mol/L, M)
  • n moles of solute
  • V solution volume (L)

Dilution

c₁ V₁ = c₂ V₂

Same moles of solute; add solvent only.

Symbols

  • c₁, V₁ initial concentration and volume
  • c₂, V₂ final concentration and volume

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