How does the chemical potential allow for open systems to be considered? Gibbs free energy can be defined as: $$dG=VdP-SdT+\sum_i\mu_idn_i$$ where $\mu=(\frac{\partial G}{\partial n})_{P,T}$. Last term, $\sum_i\mu_idn_i$ allows for open systems to be considered (where $dn$ does not equal zero). However, this can also be used for state functions like enthalpy too:
$$dH=TdS+VdP+\sum_i\mu_idn_i$$
But my question is why can this be used for enthalpy, considering that $\mu=(\frac{\partial G}{\partial n})_{P,T}$ (i.e it's a partial derivative of $G$ not $H$). Perhaps it's clearer for me to write the total derivative out like this: $$dH=(\frac{\partial H}{\partial S})_{P,n}dS+(\frac{\partial H}{\partial P})_{S,n}dP+(\frac{\partial G}{\partial n})_{P,T}dn$$ where I have considered a single-component open system (the component may enter or leave the system)
Actually, there is a completely valid definition of chemical potential in terms of enthalpy (I realize your confusion on this point may be my fault, since I initially told you the opposite on another post.)

The correct form for chemical potential as a partial molar enthalpy is (as I think you already suspected):

$$
\mu_i=(\frac {\partial H}{\partial n_i})_{p,S,n_{j\neq i}}
$$

So the total exact differential is expressible as you had originally expected:

$$
dH=TdS+VdP+\sum_i\mu_idn_i=(\frac {\partial H}{\partial S})_{p,n_i}dS+(\frac {\partial H}{\partial p})_{S,n_i}dS+\sum_i(\frac {\partial H}{\partial n_i})_{p,S,n_{j\neq i}}dn_i
$$
  The definition of chemical potential as a partial molar Gibbs free energy is only valid at constant temperature and pressure, so it wouldn't make any sense to write it that way in the expression for the exact differential of enthalpy.

Sorry for any confusion my initial errors may have caused you.  Hope this clears it up.