Determination of Molecular Mass : The molecular mass of the organic compounds can be determined by various methods.  

         (i) Physical methods for volatile compounds

         (a) Victor Meyer's method  : Molecular mass of volatile liquids and solids can be easily determined from the application of Avogadro hypothesis according to which the mass of 22.4 litres or 22400ml of the vapour of any volatile substance at NTP is equal to the molecular mass of the substance.

         In Victor Meyer's method, a known mass of the volatile substance is vaporised in a Victor Meyer's tube. The vapours formed displace an equal volume of air into a graduated tube. The volume of air collected in graduated tube is measured under experimental conditions. This volume is converted to NTP conditions.    

         Calculations : Mass of the organic substance= Wg

         Let the volume of the air displaced be =V₁ml; 

         Temperature =T₁K

         Pressure (after deducting aqueous tension) =p₁mm

         Let the volume at NTP be=V₂ml

         Applying gas equation,    

          V₂ ml of vapours weight at NTP = Wg

          22400 ml of vapour weight at NTP    

         Alternate method: Vapour density of substance

or V. D. = (W/V₂)/0.00009     ( Mass of 1 ml of H₂ at NTP = 0.00009 g or 2/22400)

         or V. D. = W/(V₂ * 0.00009) ;

          Mol. Mass, 

         (b) Hofmann's method : The method is applied to those substances which are not stable at their boiling points, but which may be volatilised without decomposition under reduced pressure. A known mass of the substance is vaporised above a mercury column in a barometric tube and the volume of the vapour formed is recorded. It is then reduced to NTP conditions. The molecular mass of the organic substance can be calculated by the application of following relationship,

         Mol. Mass 

         (ii) Physical methods for Non-volatile substances : The molecular mass of a non-volatile organic compound can be determined by noting either the elevation in boiling point of the solvent (Ebullioscopic method) or the depression in freezing point of the solvent (Cryoscopic method) produced by dissolving a definite mass of the substance in a known mass of the solvent. The molecular mass of the compound can be calculated from the following mathematical relationships :

         (a) Elevation in boiling point :

Mol. Mass = 1000 Kb * w/ W *  ΔT

Where,  K_b = Molal elevation constant of the solvent, w = Mass of the compound, W= Mass of the solvent

         ΔT = Elevation in boiling point of the solvent (determined experimentally)   

         (b) Depression in freezing point :

Mol. Mass = 1000 Kf * w/ W *  ΔT

         Where, K_f = Molal depression constant of the solvent, w = Mass of the compound, W=Mass of the solvent

             ΔT = Depression in freezing point of the solvent (determined experimentally)

         (iii) Chemical methods

         (a) Silver salt method for acids : It is based on the fact that silver salt of an organic acid on heating gives residue of metallic silver.

         From the volume of the silver salt given and the mass of the silver residue obtained, the equivalent mass of the silver salt can be calculated.

         Knowing the equivalent mass of silver salt, the equivalent mass of the acid can be obtained. The molecular mass of an acid can be determined with the help of the following relationship,

         Mol. mass of the acid = Equivalent mass of the acid * basicity

         Calculations : (i) Mass of silver salt taken = wg (ii) Mass of metallic silver = xg

;   Eq. mass of silver salt = (w/x) * 108

         Let the equivalent mass of the acid be E. In the preparation of silver salt, a hydrogen atom of the carboxylic group is replaced by a silver atom.

         Thus, Equivalent mass of silver salt = E -1 + 108 = E + 107

         Thus, E + 107 = (w/x) * 108 or E = [(w/x) * 108-107]

         If n be the basicity of the acid, then  Mol. Mass of the acid = [(w/x) * 108-107] * n

         (b) Platinichloride method for bases : Organic bases combine with chloroplatinic acid, H2PtCl6 to form insoluble platinichlorides, which, on ignition, opt out a residue of metallic platinum. Knowing the mass of platinum salt and the mass of metallic platinum, the molecular mass of the platinum salt can be determined. Let B represents one molecule of the base. If the base is mono-acidic, the formula of the salt will be B₂H₂PtCl₆.

         Let E be the equivalent mass volume of the base.

         Molecular mass of the salt

= 2E +2 +195 +213 = 2E + 410

         So   ;

         Mol. mass of the base = Eq. mass * acidity = E * n

         where n is the acidity of the base.

         (c) Volumetric method for acids and bases : Molecular mass of an acid can be determined by dissolving a known mass of the acid in water and titrating the solution against a standard solution of an alkali using phenolphthalein as an indicator. Knowing the volume of alkali solution used, the mass of the acid, which will require 1000 ml of a normal alkali solution for complete neutralisation can be calculated. This mass of the acid will be its equivalent mass.

          One gram equivalent of the acid

         Calculations : Suppose w g of the organic acid requires V ml  N₁ alkali solution for complete neutralisation.

         V ml N₁ alkali solution = w gm acid

         So 1000 ml N₁ alkali solution = (w/V * N1) * 1000 g acid = one gram equivalent acid

           Equivalent mass of the acid = (w/V * N1) * 1000                   

         Thus, Molecular mass of the acid = Eq. mass * basicity

         In the case of organic bases, the known mass of the base is titrated against a standard solution of an acid. Knowing the volume of the acid solution used, the mass of the organic base which will require 1000 ml of a normal acid solution for complete neutralisation can be calculated. This mass will be the equivalent mass of the base.

                One gram equivalent of the base

         Molecular mass of the base = Eq. mass * acidity

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