The Mole & Stoichiometry

Stoichiometry is the cornerstone of quantitative chemistry, allowing us to predict the amounts of reactants and products in a chemical reaction. The central unit for this is the mole.

### The Mole and Avogadro's Constant

A mole is the SI unit for the amount of a substance. It represents a specific number of particles (atoms, molecules, or ions). This number is known as Avogadro's constant (L), which has a value of 6.02 x 10²³ particles per mole. So, one mole of carbon contains 6.02 x 10²³ carbon atoms, and one mole of water contains 6.02 x 10²³ water molecules.

### Relative Masses and Molar Mass

To connect this count to a measurable mass, we use relative masses. The Relative Atomic Mass (Ar) is the weighted average mass of an atom of an element compared to 1/12th the mass of a carbon-12 atom. The Relative Molecular Mass (Mr) (for covalent molecules) or Relative Formula Mass (Mr) (for ionic compounds) is the sum of the Ar of all the atoms in a chemical formula.

For example, for water (H₂O):

Ar of H = 1, Ar of O = 16

Mr of H₂O = (2 × 1) + 16 = 18.

The Molar Mass (M) is the mass of one mole of a substance, expressed in grams per mole (g/mol). Conveniently, it is numerically equal to the Ar or Mr. Therefore, the molar mass of water is 18 g/mol.

### Core Calculations: The Mole Triangle

The fundamental relationship between mass, moles, and molar mass can be expressed with the formula:

Moles = Mass (g) / Molar Mass (g/mol)

This allows us to convert a known mass of a substance into moles, which is the crucial first step in most stoichiometric calculations.

### Stoichiometry: Calculations from Equations

A balanced chemical equation provides the molar ratio (or stoichiometric ratio) between reactants and products. For example, in the reaction 2H₂ + O₂ → 2H₂O, the ratio is 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.

A typical reacting mass calculation follows these steps:

### Calculations Involving Gases

At room temperature and pressure (r.t.p.), which is approximately 25°C and 1 atm, one mole of any gas occupies a volume of 24 dm³ (or 24,000 cm³). This is known as the molar volume of a gas.

The formula to use is:

Moles of Gas = Volume of Gas (dm³) / 24 dm³

This allows us to relate the volume of a gas directly to its number of moles, and vice versa, without needing to know its mass.

### Calculations Involving Solutions

The concentration of a solution tells us how much solute is dissolved in a certain volume of solvent. In chemistry, this is most often expressed as molarity, with units of moles per decimetre cubed (mol/dm³).

The key formula is:

Moles = Concentration (mol/dm³) x Volume (dm³)

*Note: Exam questions often give volume in cm³. Remember to convert to dm³ by dividing by 1000.* This relationship is fundamental for interpreting titration results, where a solution of known concentration is used to find the concentration of an unknown solution.

### Percentage Yield and Purity

In reality, chemical reactions rarely produce the maximum possible amount of product. The percentage yield compares the actual amount obtained in an experiment to the theoretical maximum calculated from stoichiometry.

Percentage Yield = (Actual Yield / Theoretical Yield) x 100%

Similarly, a reactant sample may not be 100% pure. Percentage purity helps determine the amount of the desired chemical in an impure sample.

Percentage Purity = (Mass of Pure Substance / Mass of Impure Sample) x 100%