Determination of Molecular Masses

1.0Role of Determination of Molecular Masses 

Understanding the determination of molecular masses is fundamental for JEE‑level chemistry. Whether solving stoichiometry problems, analyzing reaction mechanisms, or mastering physical chemistry, knowing how to calculate molecular mass underpins success. This guide covers both classical experimental methods and calculation strategies tailored to your exam needs.

2.0Molecular Mass vs. Molar Mass

• Molecular Mass (unit: atomic mass units, u): the mass of a single molecule, computed by summing atomic masses.
• Molar Mass (unit: g/mol): mass per mole of a substance. Numerically identical to molecular mass for molecules, but different units.
  Understanding this distinction is crucial for accurate conversions and chemical calculations.

3.0Experimental Methods to Determine Molecular Mass

Vapor Density Method

• Based on Avogadro’s law, vapor density (VD) is half the molar mass:
  [ \text{Molar Mass (M)} = 2 \times \text{Vapor Density} ]
• For example, if VD = 22.4 g/L at STP, then M = 44.8 g/mol.

Rast’s Method

• Uses a porous pot filled with the volatile liquid.
• RD: M = (mass of vapor × molar volume) / (mass of liquid evaporated)

Dumas Method

• Involves heating a weighed liquid in a flask, displacing the air, sealing, and measuring mass and volume.
• Calculation:
  [ M = \frac{mRT}{PV} ]
  where m = mass of vapor, R = gas constant, T = temperature in Kelvin, P = pressure, V = volume.

Cryoscopy (Freezing Point Depression)

• Colligative property: ΔTf = Kf × m (where m = molality of solute)
• Molar mass (M) from:
  [ M = \frac{Kf \times \text{mass of solute} \times 1000}{\Delta Tf \times \text{mass of solvent}} ]

Ebullioscopy (Boiling Point Elevation)

• ΔTb = Kb × m (where Kb is ebullioscopic constant)
• Molar mass (M) from:
  [ M = \frac{Kb \times \text{mass of solute} \times 1000}{\Delta Tb \times \text{mass of solvent}} ]

Solved Examples

Vapour Density: 

Example:

A gas has a vapour density of 11.2 g/L at STP. What is its molar mass?
Solution: M = 2 × 11.2 = 22.4 g/mol (under STP conditions).

Colligative Properties: Sample Numerical Problems

Freezing Point Depression

Example:

2 g of an unknown solute dissolved in 50 g of benzene causes ΔTf = 1.6°C. Kf of benzene = 5.12 °C·kg/mol.

Solution:

m = ΔTf / Kf = 1.6 / 5.12 = 0.3125 mol/kg

M = (mass solute × 1000) / (m × mass solvent) = (2 × 1000) / (0.3125 × 50) = 128 g/mol.