tl;dr:
For you, $\mu\simeq0.6$.
Mean molecular mass
The term you're looking for is the mean molecular mass $\mu$, more often called the mean molecular weight, despite not being a weight, despite usually not involving molecules, and despite being a unitless number, not a mass.
It is the average mass per particle, measured in terms of the proton mass (or sometimes the hydrogen mass which is almost the same, and sometimes the atomic mass unit $m_\mathrm{u} \equiv m_{^{12}\mathrm{C}}/12$), and it includes any free electrons.
That is, it is defined as
$$
\mu = \frac{\langle m \rangle}{m_p},
$$
Neutral gas
For a neutral gas (i.e. no free electrons) of mass fractions $X$, $Y$, and $Z$ of hydrogen, helium, and heavier elements (called "metals" in astronomy), you have
$$
\frac{1}{\mu_\mathrm{neut.}} = X + \frac{Y}{4} + \langle 1/A_n \rangle Z,
$$
because helium weighs four times as much as hydrogen, and where $\langle 1/A_n \rangle$ is the weighted average of all metals. For Solar abundances $\{X,Y,Z\}\simeq\{0.70,0.28,0.02\}$, and $\langle 1/A_n \rangle \simeq 1/15.5$ (e.g. Carroll & Ostlie 1996), so
$$
\mu_{\mathrm{neut.,\odot}} = \frac{1}{0.70+0.28/4+0.02/15.5} \simeq 1.30.
$$
For a primordial (i.e. metal-free) gas, $\{X,Y,Z\}\simeq\{0.75,0.25,0\}$ so
$$
\mu_{\mathrm{neut.,prim.}} = \frac{1}{0.75+0.25/4} \simeq 1.23.
$$
Ionized gas
For an ionized gas you have roughly twice the amount of particles, but the mass of half of them can be neglected, so $\mu$ is roughly half of the above:
$$
\frac{1}{\mu_{\mathrm{ion.}}} \simeq 2X + \frac{3Y}{4} + \frac{Z}{2}.
$$
Thus, a fully ionized primordial gas has
$$
\mu_{\mathrm{ion.,prim.}} = \frac{1}{2\times0.75 + 3\times0.25/4} \simeq 0.59.
$$
while a fully ionized metal-rich gas has
$$
\mu_{\mathrm{ion.,\odot}} = \frac{1}{2\times0.70 + 3\times0.28/4 + 0.02/2} \simeq 0.62.
$$
The answer for you
Since you're dealing with galaxy clusters, the bulk of the gas is ionized. As you can see above, the exact value of $\mu$ depends on the metallicity, but as an astronomer you will most likely not offend anyone by just using $\mu=0.6$.
Or even $\mu\sim1$. I mean if we can't even make a proper definition, why bother about anything more precise than an order-of-magnitude estimate…