Molar mass from sedimentation and diffusion ($M=R_{\mathrm{gas}}Ts/[D(1-\bar{v}{\rho }_s)]$)

Let $M$ be molar mass in $\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$, $\bar{v}$ partial specific volume, ${\rho }_s$ solvent density, and $N_A$ the Avogadro constant. The buoyant molecular mass is $m_b=M(1-\bar{v}{\rho }_s)/N_A$. Combining sedimentation coefficient $s=m_b/f_h$ and the thermal diffusion-friction relation $D=k_{B}T/f_h$ eliminates the unknown friction and proves the displayed formula, with $R_{\mathrm{gas}}=N_A{k}_B$. Once friction is known, $R_h=f_h/(6\pi \eta )$ defines a hydrodynamic radius using Stokes's law. Mass and radius are different inferences: an extended molecule need not occupy the volume of that friction-equivalent sphere.