The $K_\mathrm{D}$ (distribution coefficient) of coffee water at $25\,°\mathrm{C}$ is $7.8$ in a methylene chloride/water system. If there is $2.0\,\mathrm{g}$ of caffeine in $200\,\mathrm{mL}$ of water, calculate the grams and the % of caffeine that can be extracted with a single $100\,\mathrm{mL}$ portion of methylene chloride.
My attempt:
$$\frac{x}{\frac{2\,\mathrm{g}}{200\,\mathrm{mL}}}=7.8$$
$$\frac{x}{0.01\,\mathrm{g\,mL^{-1}}}=7.8$$
$$x=0.078\,\mathrm{g\,mL^{-1}}$$
$$x=0.078\,\mathrm{g\,mL^{-1}}$$
$$0.078\,\mathrm{g\,mL^{-1}}\cdot 100\,\mathrm{mL}=7.8\,\mathrm{g}$$
But I think I'm wrong. Am I supposed to maybe use the molar mass of caffeine ($194\,\mathrm{g\,mol^{-1}}$)?