DOSY-NMR allows for the determination of the diffusion coefficient of the analyte molecule. How do I calculate the volume/radius of the molecule from the diffusion coefficient?
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asked Oct 29, 2016 at 11:27
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You can try and use the Stokes-Einstein equation for the diffusion coefficient $$D=\frac{k_\mathrm{B}T}{\zeta}$$ where the friction term is $$\zeta = 6\pi\eta r$$ and $k_\mathrm{B}$ is the Boltzmann constant, $\eta$ the solvent viscosity $(1\ \mathrm{cP} = 10^{-3}\ \mathrm{kg~m^{-1}~s^{-1}} = 10^{-3}\ \mathrm{Pa~s})$ and r the molecule's radius . You only have one number in diffusion constant so you can only get radius or volume assuming the molecule is a sphere.
The model assumes that the molecules are larger than the solvent molecules, i.e. solvent is effectively continuous not molecular. There are slightly different expressions for the friction, 4 instead of 6 for example depending on whether 'slip' or 'stick' limits are assumed.
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$\begingroup$ Are there other ways to calculate the volume without assuming the molecules to be a sphere. Can I calculate the diffusion constants using e.g. MM or DFT? $\endgroup$ Oct 29, 2016 at 12:54
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$\begingroup$ @ketbra What exactly are you after? Don't you know what substance you are measureing? $\endgroup$– KarlOct 29, 2016 at 21:37
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$\begingroup$ @Karl I know the substance but whant to show wheather its a monomer/dimer/tetramer and want to argue with the size of the molecule. $\endgroup$ Oct 30, 2016 at 8:01
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$\begingroup$ If you only have one measurement then you can only assume a sphere, you do not have enough information to assume a prolate or oblate spheroid. You should be able to distinguish monomer etc just from the diffusion coef as the molecules probably coil up or cluster in some way . You can certainly get size & shape from calculation. If you could measure rotational diffusion, say by fluorescence anisotropy, then this gives a better idea of shape. $\endgroup$ Oct 30, 2016 at 8:56
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$\begingroup$ @ketbra You can surely differentiate betwen monomers, dimers and trimers spectroscopically. After that the diffusion cofficients are distinctive. $\endgroup$– KarlOct 30, 2016 at 20:42