Question

A 36.0 mg sample of an organic compound (molar mass = 84.0 g/mol) is dissolved in 10.0 mL of water. This aqueous solution is extracted with 5.0 mL of hexane. Separation and analysis of the aqueous phase shows that it now contains 12.0 mg of the organic compound. Calculate the partition coefficient for the compound between hexane and water (Kphexane/water). Your answer should reflect the proper number of significant figures.

Answer 1

Answer:

K = 4.00

Explanation:

The partition is equal to the concentration of the solute in phase 2 (hexane) divided by the concentration of the solute in phase 1 (water)

K = [S]₂ / [S]₁

The concentration is in moles per liter, so first we must find the number of moles of the solute that remained in the aqueous phase:

(12.0 mg)(mol/84.0 g) = 0.142857 mmol in phase 1

Additional significant figures are kept to avoid round-off errors.

We calculate the concentration in phase 1 by dividing by the volume:

[S]₁ = n/V = (0.142857 mmol) / (10.0 mL) = 0.0142857 mol/L

Next, we calculate the number of moles that were extracted into the organic phase:

(36.0 mg - 12.0 mg)(mol/84.0g) = 0.285714 mmol in phase 2

Similary, we divide by the volume to get the concentration:

[S]₂ = n/V = (0.285714 mmol) / (5.0 mL) = 0.0571428 mol/L

Finally, we can calculate the partition coefficient:

K = [S]₂ / [S]₁ = (0.0571428 mol/L) / (0.0142857 mol/L)

K = 4.00

Answer 2

Answer:

$K=4.0$

Explanation:

Hello,

In this case, considering the involved substances, after the extraction we will have two phases, an organic which is rich in hexane and an aqueous which is rich in water. In this manner, the given definition of the partition coefficient involves the calculation of the sample's concentration in both solvents, it means water and hexane as shown below:

$[solute]_{water}=\frac{12mg*\frac{1mmol}{84.0mg}}{10.0mL}*\frac{1000mL}{1L}=14.3mM$

$[solute]_{hexane}=\frac{(36.0-12.0)mg*\frac{1mmol}{84.0mg}}{5mL}*\frac{1000mL}{1L}=57mM$

Now, the partition coefficient turns out being:

$K=\frac{[solute]_{hexane}}{[solute]_{water}} = \frac{57mM}{14.3mM} \\K=4.0$

Best regards.