Density of saturated n-octane in air at 300 K?
Thermodynamic parameter values for many substances (not just water and steam) can be found using so-called steam tables. In the following, I use parameter values taken from the REFPROP – NIST Standard Reference Database 23, Version 9.0). For a simplified estimate, however, we only need the value of the vapour pressure of octane.

For simplicity’s sake, we may assume that the gas phase (containing air and octane) is an ideal mixture of non-reacting gases and that the liquid phase contains only pure octane. Thus, for the calculation of parameter values for octane, we may ignore the air and only consider the equilibrium in a system of pure liquid octane and pure gaseous octane.

Using tabulated data, we find that the saturation point for octane at a temperature of $T = 300\ \mathrm K$ corresponds to a pressure of $p_{\ce{C8H18(g)}} = 2060\ \mathrm{Pa}$. 

The corresponding density can be estimated using the ideal gas law:
$$\begin{align}
p\cdot V &amp;=n\cdot R\cdot T \\[6pt]
\frac nV &amp;= \frac p{R\cdot T} \\[6pt]
\frac m{M\cdot V} &amp;= \frac p{R\cdot T} \\[6pt]
\frac mV &amp;= \frac {M\cdot p}{R\cdot T} \\[6pt]
\rho &amp;= \frac {M\cdot p}{R\cdot T} \\[6pt]
&amp;= \frac {114.23\ \mathrm{g\ mol^{-1}}\times 2060\ \mathrm{Pa}}{8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}\times 300\ \mathrm K}} \\[6pt]
&amp;= \frac {0.11423\ \mathrm{kg\ mol^{-1}}\times 2060\ \mathrm{kg\ m^{-1}\ s^{-2}}}{8.314462618\ \mathrm{kg\ m^2\ s^{-2}\ mol^{-1}\ K^{-1}\times 300\ \mathrm K}} \\[6pt]
&amp;= 0.09434\ \mathrm{kg/m^3}
\end{align}$$

By way of comparison, the corresponding density given by REFPROP is $\rho_{\ce{C8H18(g)}} = 0.094649\ \mathrm{kg/m^3}$.

Assuming that the total pressure is equal to the sum of the partial pressures of the individual gases in accordance with Dalton’s law, we may calculate the partial pressure for the remaining air:
$$\begin{align}
p_\text{tot} &amp;= p_\text{air} + p_{\ce{C8H18(g)}} \\
p_\text{air} &amp;= p_\text{tot} - p_{\ce{C8H18(g)}} \\
&amp;= 101325\ \mathrm{Pa} - 2060\ \mathrm{Pa} \\
&amp;= 99265\ \mathrm{Pa}
\end{align}$$

The density of dry air at a temperature of $T = 300\ \mathrm K$ and a pressure of $p = 101325\ \mathrm{Pa}$ is $\rho_\text{air} = 1.1770\ \mathrm{kg/m^3}$. The corresponding density at a partial pressure of $p_\text{air} = 99265\ \mathrm{Pa}$ is $\rho_\text{air} = 1.1531\ \mathrm{kg/m^3}$.

Thus, the total density of the gas mixture is
$$\begin{align}
\rho_\mathrm g &amp;= \rho_{\ce{C8H18(g)}} + \rho_\text{air} \\
&amp;= 0.094649\ \mathrm{kg/m^3} + 1.1531\ \mathrm{kg/m^3} \\
&amp;= 1.2477\ \mathrm{kg/m^3}
\end{align}$$

By way of comparison, I made a quick simulation using REFPROP assuming that air is a mixture with the mole fractions $x_{\ce{N2}}=78.12\,\%$, $x_{\ce{O2}}=20.96\,\%$, and $x_{\ce{Ar}}=0.92\,\%$. I added an excess amount of octane $x_{\ce{C8H18}}=10\,\%$ (in order to obtain a liquid phase) and renormalized the composition to $100\,\%$.
The resulting density of the vapour phase at a temperature of $T = 300\ \mathrm K$ and a pressure of $p = 101325\ \mathrm{Pa}$ is $\rho_\text{g} = 1.2483\ \mathrm{kg/m^3}$.