Are we using an assumption in calculating the ratio of first state excited to unexcited chemical species using the Boltmann distribution?
The expression for the Boltzmann equation is:

$\eta_{i} = \dfrac{g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}{\sum_{i} g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}$,

wherein $\eta_{i}$ the probability is to encounter the ensemble in state $i$ is in the ensemble, $g_{i}$ the degeneracy of state $i$, and $\epsilon_{i}$ its energy (I reckon the other variables are familiar to you).

To calculate the ratio between two states, you can simply divide the above expression for two different states by which you obtain:

$\dfrac{\eta_{i}}{\eta_{j}} = \dfrac{g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}{\sum_{i} g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)} \cdot \dfrac{\sum_{i} g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}{g_{j} \exp \left( \dfrac{-\epsilon_{j}}{k_{b}T} \right)} = \dfrac{g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}{g_{j} \exp \left( \dfrac{-\epsilon_{j}}{k_{b}T} \right)}$

The expression above shows that because states $i$ and $j$ are of the same ensemble, you do not have to use (or even know) the energy levels of the other states in that ensemble if you are only interested in the probability ratio between the two states.