molecular mass determination by freezing point decrease
Weighing directly one molecule is impossible, so Molecular mass M must always be determined indirectly.
- Micro level: 1 molecule of any substance weigh M u(nit).
- Macro level: 1 mol of any substance weigh M gramme.
- 1 u(nit) is the average mass of one nuclear particle.
Weighing one mol is possible, but for that the molecular mass M must already be known.
The purpose is now to know how to determine the molecular mass of a substance.
The determination of M of an unknown substance often starts with carefully weighing of a certain amount of grammes; to be used during the determination.
For a good experiment you need a homogeneous substance, very regular divided inside a certain space.
The applied method depends on the phase of the substance:
- method I is applied when the substance is (s) or (l).
- method II is applied when the substance is (g).
If the substance is (s) or (l), the molecules must be divided homogenous.
In order to realise that, the substance...:
- must be dissolved in a propriate solvent; of that mixture (the solution) you must determine the freezing point (f.p.).
- must be evaporated.
Dissolved substances influence the f.p. of a solvent; a liquid will freeze more difficultly when strange particles are present in that liquid.
- When freezing, always a lattice is made (for example, an ionic lattice).
The formation of a lattice is more difficult when strange (not fitting) particles are present, come into that lattice.
So if a substance is dissolved in a liquid, it becomes more difficult to solidifacate that liquid. The realisation of a lattice has become more difficult.
In order to make, nevertheless, a lattice, you need a lower temperature: FREEZING POINT DECREASE (f.p.d.)
The fpd does not depend very much of the character of the dissolved particles (it is not very important what kind of particles were dissolved, like ions, atoms, big or small molecules), but depent in particular on the amount of dissolved particles (number of moles).
Note that at these determinations we do not use the usual unit mol/liter, but mol/100 g solvent.
Don't ask why! Has something to do with history.
The amount of unknown substance (for example p grammes) is dissolved in a weighed amount of known solvent (for example q grammes).
After measuring carefully the f.p. of the mixture (using for example a very accurate Beckmann thermometer), and knowing the f.p. of the pure solvent, than you can know the fpd.
The (unknown) molecular mass of the dissolved substance is M.
The number of dissolved particles is p/M. This number of mols determines the decrease.
The number of dissolved particles is p/M.
This number of mols of particles determines the increase of decrease.
The number of mols of solute is proportional with the fpd or bpi.
f.p.d. ≡ number of mol of solute
The real amount of solvent in practice rarely will be exactly 100 g; if you take less, then the effect will be stronger; if you take more solvent, then the effect will be less.
Suppose you take q grammes, then a correction factor is needed: 100/q
f.p.d. ≡ number of mols of solute x 100/q
To make the step to an =sign, a constant must be introduced: K*.
This constant K* is called: the molar f.p.d.
The value of K* is different per solvent and can be fount in tables.
If a substance is built up of ions, then this substance will dissociate in water in several particles (n particles)
f.p.d. = K* x number of mols of solute x 100/q · n
f.p.d = K* · p · 100 · n
M · q