# Distribution coeffecient without concentrations

From what I understand about distribution coefficient is straight from my book — which does not give any practice examples — is that

$$D = \frac{C_\mathrm{A}(\text{ext})}{C_\mathrm{A}(\text{orig})}$$

where $$C_\mathrm{A}(\text{ext})$$ and $$C_\mathrm{A}(\text{orig})$$ represent the total concentration of all analyte species present in the two phases regardless of chemical state. Below is a homework problem that I'm try to solve, but having no luck, since I'm not given any concentrations, only weight and volume.

In an extraction experiment, it is found that $$\pu{0.0376 g}$$ of an analyte are extracted into $$\pu{50 mL}$$ of solvent from $$\pu{150 mL}$$ of a water sample. If there was originally $$\pu{0.192 g}$$ of analyte in this volume of the water sample, what is the distribution coefﬁcient?

The answer for this problem is $$0.731\%$$, but no matter which way I plug in the numbers that are given do I reach this answer. Any ideas on what I'm missing?

Molar concentration can be expressed via mass $$m$$ and volume $$V$$ all right:

$$C_i = \frac{n_i}{V_i} = \frac{m_i}{M_iV_i}$$

Analyte doesn't change during the extraction and its molar mass $$M$$ remains the same $$(M_i = \text{const})$$ so the distribution coefficient $$D$$ can be rewritten as such:

$$D = \frac{C_\mathrm{s}}{C_\mathrm{w}} = \frac{m_\mathrm{s}V_\mathrm{w}}{m_\mathrm{w}V_\mathrm{s}}$$

where "s" refers to the solvent phase and "w" to aqueous phase. At equilibrium

$$m_\mathrm{w} = m_0 - m_\mathrm{s}$$

where $$m_0$$ is the initial mass of the analyte. Finally, the distribution coefficient is

$$D = \frac{m_\mathrm{s}V_\mathrm{w}}{(m_0 - m_\mathrm{s})V_\mathrm{s}} = \frac{\pu{0.0376 g}\cdot\pu{150 mL}}{(\pu{0.192 g} - \pu{0.0376 g})\cdot\pu{50 mL}} = 0.731$$