Question: Disclaimer: I'm going to be using the term "white dwarf" to refer to any spherical celestial body made of electron degenerate matter. If I had a better term, I would use it.

The Chandrasekhar limit for white dwarfs can be approximated with the equation $M\_{limit}=\frac{\omega\_3^0 \sqrt{3\pi}}{2} \left(\frac{\hbar c}{G}\right)^\frac32 \frac{1}{\left(\mu\_e m\_H\right)^2}$. The most relevant part of the equation here is $\frac{1}{\left(\mu\_e m\_H\right)^2}$. $m\_H$ is the mass of a hydrogen atom and $\mu\_e$ is the average molecular weight per electron, and it depends on what the white dwarf is made of. In real white dwarfs, $\mu\_e$ is around 2, since the elements in a white dwarf have mostly equal numbers of protons and neutrons, so the Chandrasekhar limit is around 1.4 $M\_\odot$.

But it seems like $\mu\_e$, and therefore $M\_{limit}$, could theoretically be very different if the white dwarf was made of different elements. If a white dwarf was (somehow) made of pure hydrogen, for example, $\mu\_e$ would be basically 1, and $M\_{limit}$ would evaluate to $5.7 M\_\odot$. If a white dwarf was made of mercury-204, which has the highest mass-to-electron ratio among stable nuclei, $\mu\_e$ would be around 2.55, and the Chandrasekhar for such a white dwarf would be around $0.88 M\_\odot$. Therefore, it looks like the Chandrasekhar limit for any stable white dwarf would have to be between $0.88 M\_\odot$ and $5.7 M\_\odot$. Do I have this right? Obviously we'd never have a white dwarf made of hydrogen or mercury.

Answer: 
Yes, you are correct in so far as what the Chandrasekhar limit is for a white dwarf supported by ideal electron degeneracy pressure and using a Newtonian equation of hydrostatic equilibrium.

Neither of these latter assumptions is true for real (or hypothetical) white dwarfs of any composition.

First, there is a negative correction to the ideal gas pressure caused by Coulomb interactions between the electrons and ions in the gas. Second, at high densities, inverse beta decay will start to remove free electrons and build nuclei with larger $\mu\_e$. This will destabilise the white dwarf at the critical density where the electron Fermi energy is high enough to start the inverse beta decay process. Third, General Relativity is required for high-mass white dwarfs. The GR equation of hydrostatic equilibrium has pressure on the right hand side, so that increasing pressure ultimately becomes self-defeating and causes instability at a finite density

All the above effects *lower* the Chandrasekhar mass from the ideal case you quote. By how much depends on which process may trigger instability first, which in turn depends on $\mu\_e$ and the inverse beta decay threshold for specific nuclei. For instance, it is thought to be about 1.38 solar masses for a carbon white dwarf, rather than the 1.44 solar masses that the ideal case would predict. I think the limit for a "hydrogen white dwarf" would be *much* lower than $5.7 M\_\odot$ because the threshold energy for inverse beta decay for protons is much lower than for carbon nuclei, so would occur at much lower densities.
The "vary wildly" point is not supported by the accepted stellar evolution models that always end up with some carbon-12/oxygen-16 mixture at the degenerate core stage near the Chandrasekhar limit.

Yes, the Chandrasekhar limit will be different for white dwarfs with different composition.

You are right for mercury and hydrogen "white dwarf" (i.e. a degeneracy-pressure supported object).

On the other hand, a white dwarf made of hydrogen will be VERY hard and slow to construct without failing either into a main-sequence star (if you just pile up hydrogen without cooling it) or an Ia-very-hydrogen-rich supernova (if you DO cool it carefully but fail at some point when there is already some amount of degenerate phase).

And whatever means to construct such a beast you invent, it will fail the "white dwarf" definition because it has to be cold in order to be stable. It will be a [black dwarf](https://en.wikipedia.org/wiki/Black_dwarf).

Or maybe an overweight rogue planet (since it has never been a main sequence star and is not a remnant of one)?

Mercury is easier - it won't fuse.