Determination of Molecular Mass - Continued

By silver salt method

Silver salt method is used for determining the molecular masses of carboxylic acids. This method is based on the fact that they form insoluble silver salts, which upon heating decompose to leave a residue of metallic silver.

Procedure

The unknown acid is dissolved in water and treated with a slight excess of ammonium hydroxide. The excess of ammonia is boiled off. To this sufficient quantity of silver nitrate is added when a white precipitate of silver salt is obtained. The precipitate is separated by filtration, washed successively with water, alcohol and ether and dried in the steam oven. About 0.2 g of the dry silver salt is weighed into a crucible and ignited until the decomposition is complete. Ignition is repeated until the crucible with the residue of silver has attained constant weight. The molecular mass of the acid is then calculated from the mass of the silver salt taken and the mass of the residue of metallic silver obtained from it.

For example, for a monocarboxylic acid, we can write

Calculations

Let,

The mass of silver salt taken for ignition be = W g

Mass of silver left behind after ignition = W₁ g

We have RCOOAg = Ag1 equivalent   1 equivalent

The equivalent mass of silver is 108. Hence, the mass of silver salt that would leave 108 g of residue (equivalent mass of the silver) is equal to the equivalent mass of the silver salt of the acid.

Then,

and,

Equivalent mass of acid = Equivalent mass of the silver salt - Equivalent mass of silver+ Equivalent mass of hydrogen

Or,

Then,

Molecular mass of the acid = Equivalent mass of the acid x Basicity of the acid.

The basicity of an acid is equal to the number of ionizable hydrogen atoms in a molecule of the acid.

Problem

6. About 0.759 g of silver salt of a dibasic acid was ignited when a residue of 0.463 g of metallic silver was left. Calculate the molecular mass of the acid.

Solution

Mass of silver salt of carboxylic acid, W₁= 0.759 g

Mass of silver left as residue, W = 0.463 g

= 177 - 108 + 1 = 70 amu

Molecular mass of acid = Equivalent mass x Basicity

= 70 x 2 amu = 140 amu

By chloro platinate method

This method is used to find the molecular masses of organic bases.

Principle

The given organic base is allowed to react with chloroplatinic acid (H₂PtCl₆) in concentrated HCl to form insoluble platinic chloride. This precipitate is separated, dried, weighed and is subsequently ignited till decomposition is complete. The residue left behind is platinum and is weighed again. The molecular mass is then calculated by knowing the mass of the platinic chloride salt and that of platinum left.

If B represents the molecule of monoacidic organic base, then the formula of platinum chloride salt is B₂H₂PtCl₆.

Calculations

Let the mass of platinic chloride salt be = W g

The mass of platinum residue left = x g

From the molar stoichiometry,

Number of mol of B₂H₂PtCl₆ = Number of mol of Pt

Or

If E is the equivalent mass of the organic base,

Then molecular mass of the salt (B₂H₂PtCl₆) = 2E + 2 + 195 + 213

= 2E + 410

Molecular mass of base = E x Acidity of the base

Problem

7. About 0.98 g of the chloroplatinate of some diacidic base when ignited left 0.39 g of platinum as residue. What is the molecular mass of the base? (Atomic mass of Pt = 198).

Solution

Mass of the chloroplatinate salt = 0.98 g

Mass of platinum = 0.39 g

Let E be the equivalent mass of the base, then

Molecular mass of salt (B₂H₂PtCl₆) = 2E + 410

Molecular mass of salt = 40 x 2 = 80.

Mass Spectroscopy

Mass spectroscopy is a technique that helps to measure the mass (molecular weight) of a molecule. In addition, one can also gain information on the structure of the organic compound.

In this method, a gaseous sample of the substance is bombarded with a high speed electron beam that has sufficient energy to do the following:

• Cause ionization of the molecules of the substance.

• Decompose the molecules into smaller positively charged fragments.

There are several different kinds of mass spectrometers, the most common one of which is the electron-impact, magnetic-sector instrument.

Fig:16.20 Schematic representation of an electronic impact, magnetic sector mass spectrometer

A small amount of the sample is introduced into the mass spectrometer and is bombarded by a stream of high-energy electrons (70 electron volts (eV) or 1600 kcal/mol). When the high-energy electron strikes an organic molecule, it removes a valence electron from the molecule, producing a cation radical (cation because the molecule has lost a negatively charged electron; radical because the molecule now has an odd number of electrons).

Electron bombardment transfers such a large amount of energy to the sample molecules that the cation radicals fragment. They fly apart into numerous smaller pieces, some of which retain a positive charge, and some of which are neutral. The fragments are then made to pass through a strong magnetic field applied perpendicularly, where they are deflected according to their mass-to-charge ratio (m/z).

Neutral fragments are not deflected by the magnetic field and are lost on the walls of the instrument. Positively charged fragments, however, are sorted by the mass spectrometer onto a detector, which records them as peaks. Since the number of charges 'z', is usually 1, the peaks represent the masses of the ions. The 'mass spectrum' of a compound is usually presented as a bar graph with 'unit' masses (m/z values) on the x-axis, and intensity (number of ions of a given m/z striking the detector) on the y-axis. The highest peak is called the 'base peak' and is arbitrarily assigned an intensity of 100%.

Fig:16.21 Mass spectra of methane, CH₄ (molecular weight = 16)

The mass spectrum of methane is relatively simple, since few fragmentations are possible. The base peak has m/z = 16, which corresponds to the unfragmented methane cation radical or the 'molecular ion' (M⁺). The mass spectrum also shows ions at m/z = 15 and 14, corresponding to cleavage of the molecular ion into CH and CH fragments.

For larger molecules, the mass spectral fragmentation patterns are usually complex, and the molecular ion is often not the highest (base) peak.

Interpreting mass spectra

The most obvious piece of information from mass spectra is the molecular weight of the compound. Measurement of an unknown's molecular weight allows us to narrow the possibilities for molecular formula down to only a few choices.

A second piece of information provided by the mass spectrum is a kind of molecular fingerprint. Each organic molecule fragments in a unique pattern depending on its structure, and the chance that two compounds will have identical mass spectra is small. Thus, we can often identify unknowns by matching their mass spectra to reference spectra.