Molarity of Pure Substances

The molarity (aka molar concentration) of a pure substance is the ratio of the density of the pure substance to its molar mass. It is calculated by the following formula.

Here, c is the molarity, ρ is the density, and M is the molar mass of pure compound.

Derivation

Molarity is the number of moles of a substance (usually a solute) present in a litre of the solution.

Here, c_i is the molarity of a solute, m_i is moles of the solute, and V is the volume of the solution.

In a pure substance, there is no solute or solvent, only one chemical. Thus, c_i = c and n_i = n.

Volume is mass divided by density; .

Here, is molar mass (mass divided by the number of moles).

Note: The term the molarity of a pure substance is inappropriate because molarity is a property of solutions. A solution is a homogeneous mixture of two or substances. And pure substances are not solutions. Molar density is the appropriate term to use here, not molarity or molar concentration. The ratio of density of a substance to the molar mass of the substance gives its molar density.

In short, molar density and molarity are the same for pure substances ignorantly.

Examples

The molar concentration (or molar density correctly) of some common chemicals is calculated below.

Example 1: Water

ρ_H2O = 1000 g L⁻¹ and M_H2O = 18.0 g mol⁻¹.

Example 2: Ethanol

ρ_C2H6O = 789 g L⁻¹ and M_C2H6O = 46.1 g mol⁻¹.

Example 3: Methanol

ρ_CH4O = 792 g L⁻¹ and M_CH4O = 32.0 g mol⁻¹.

Example 4: Benzene

ρ_C6H6 = 876 g L⁻¹ and M_C6H6 = 78.1 g mol⁻¹.

Example 5: Acetic acid

ρ_C2H4O2 = 1050 g L⁻¹ and M_C2H4O2 = 60.1 g mol⁻¹.

Example 6: Glycerol

ρ_C3H8O3 = 1260 g L⁻¹ and M_C3H8O3 = 92.1 g mol⁻¹.

Associated articles

• Molarity