Why is heptane more soluble in octane than in water? What intermolecular forces are responsible for this
The Correct Answer and Explanation is:
Heptane (C7H16) is more soluble in octane (C8H18) than in water due to the principle “like dissolves like.” This refers to the fact that solubility is typically highest when the solute and solvent have similar intermolecular forces.
Intermolecular forces at play:
- London Dispersion Forces (Van der Waals forces): Both heptane and octane are nonpolar molecules, meaning they have no significant dipole moment. The primary intermolecular force between these molecules is London dispersion forces, which arise from temporary fluctuations in electron density that induce instantaneous dipoles in neighboring molecules. These forces are relatively weak but are effective in nonpolar substances. Since heptane and octane are similar in structure and size, their intermolecular London dispersion forces are compatible, allowing them to mix readily.
- Hydrogen Bonding (in water): Water, on the other hand, is a highly polar molecule that exhibits strong hydrogen bonding due to the electronegativity of oxygen and the hydrogen atoms. Heptane, being nonpolar, cannot engage in hydrogen bonding with water molecules. This mismatch in intermolecular forces makes the interaction between water and heptane unfavorable, resulting in low solubility.
Thus, the lack of significant attractive forces between heptane and water molecules, coupled with the favorable dispersion interactions between heptane and octane molecules, explains why heptane is more soluble in octane than in water. This is a clear demonstration of the principle that nonpolar solvents dissolve nonpolar solutes more effectively than polar solvents.
