= Solution
In a spherical self-gravitating equilibrium with positive <mass density>, $d\psi/dr=GM(r)/r^2>0$ away from the center. Thus $r$ can locally be replaced by $\psi$, so both $p(r)$ and $\rho(r)$ can be expressed as functions of $\psi$. <Hydrostatic equilibrium> further gives
$$
\frac{dp}{d\psi}=-\rho.
$$
Since $p$ decreases outward wherever $\rho>0$, eliminating $\psi$ gives a <barotropic equation of state> along this equilibrium. This is a relation describing its structure; it does not require all thermodynamic perturbations of an arbitrary material to be barotropic.
For $p=K\rho^{1+1/n}$, integrate the <polytropic enthalpy> balance $dp/\rho=-d\psi$:
$$
(n+1)K\rho^{1/n}+\psi=\psi_T,\qquad
\boxed{\rho=\left[\frac{\psi_T-\psi}{(n+1)K}\right]^n.}
$$
For a finite-radius zero-pressure boundary with $K>0$ and $n>0$, $\rho(R)=0$ and therefore $\psi_T=\psi(R)$. With the gravitational potential zero at infinity, an isolated sphere has $\psi_T=-GM/R$. It is an integration constant tied to that boundary, not necessarily a zero of the <gravitational potential>.
Put $\theta=(\rho/\rho_c)^{1/n}$ and $\xi=\alpha r$, with $\alpha^2=4\pi G\rho_c^{1-1/n}/[(n+1)K]$. The preceding relation gives $\psi=\psi_T-(n+1)K\rho_c^{1/n}\theta$. Substitution into the spherical gravitational <Poisson equation> yields the <Lane-Emden equation>
$$
\boxed{\frac1{\xi^2}\frac d{d\xi}\left(\xi^2\frac{d\theta}{d\xi}\right)=-\theta^n,\qquad
\theta(0)=1,\quad\theta'(0)=0.}
$$
Let $\xi_1$ be the first zero of the regular <Lane-Emden function> and $Y_n=-\xi_1^2\theta'(\xi_1)>0$. Then the <Lane-Emden mass formula> and radius are
$$
R=\frac{\xi_1}{\alpha},\qquad
M=\frac{4\pi\rho_c}{\alpha^3}Y_n.
$$
For a common fixed <polytropic index> with a finite first zero, their scalings are
$$
\boxed{M\propto K^{3/2}\rho_c^{3/(2n)-1/2},\qquad
R\propto K^{1/2}\rho_c^{1/(2n)-1/2}.}
$$
The usual regular zero-pressure models have a finite first zero for $0<n<5$; the $n=5$ model has infinite radius, so a finite-radius formula cannot simply be applied to it.
For an <ideal gas> of fixed composition, $T_c\propto K\rho_c^{1/n}$. Consequently <homologous polytropes at fixed central temperature> have $K\propto\rho_c^{-1/n}$ and both $M,R\propto\rho_c^{-1/2}$, recovering $M\propto R$. The common central <temperature>, composition and dimensionless structure are the relevant assumptions. Keeping $K$ fixed instead gives the different <polytropic mass-radius relation>, and no finite index makes its generic fixed-$K$ exponent exactly one.
For adiabatic monatomic <hydrogen>, $\gamma=5/3$ and $n=3/2$. At a common fixed <specific entropy>, hence common $K$, the same relations give
$$
\boxed{R\propto KM^{-1/3},\qquad M\propto R^{-3},\qquad\bar\rho\propto M^2\quad(K\text{ fixed}).}
$$
Here $\bar\rho=3M/(4\pi R^3)$, and the omitted constant in the last relation depends on $K$. A population with different cloud entropies need not follow that <mass> sequence.
Efficient <radiative cooling> in <molecular clouds> prevents compression from forcing the <temperature> upward as steeply as in an adiabatic monatomic gas. Rotational transitions of molecules and radiation by dust provide low-temperature cooling channels; which dominates depends on composition, opacity and <temperature>. It is therefore possible to assemble dense massive cold clouds while disposing of compressional <heat>. This explains the direction of the idealized comparison, rather than proving a universal cloud-mass cutoff.
For a common <isothermal sound speed> $c_T$, a spherical isothermal equilibrium has $GM/R$ equal to $c_T^2$ times a dimensionless structural factor. Holding that factor fixed gives the <homologous isothermal mass-radius relation>
$$
\boxed{M\propto R\quad(T\text{ and dimensionless structure fixed}).}
$$
For example, the <singular isothermal sphere> has $M(R)=2c_T^2R/G$. A regular finite cloud requires a <pressure-confined isothermal sphere> or another outer boundary: an unconfined <isothermal sphere> does not end at a finite vacuum radius. Equal <temperature> alone, while allowing arbitrary truncation or density contrast, does not make the coefficient universal.
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