Question

You want to decaffeinate your coffee by extracting the caffeine outwith dichloromethane. It's too late to extract the caffeinefrom the coffee beans because you've already brewed yourself a 200mL cup of coffee. Your particular brand of coffee contains100 mg of caffeine in that 200mL cup. The partitioncoefficient of caffeine in dichloromethane/water is 9.0.
How much caffeine would still be in your 200 mL if you did:
A. One extraction using 200 mL o fdichloro methane
B. Two extractions using 100 mL of dichloro methane each.

Answer 1

Solution :

The partition coefficient

$k_d= \frac{\text{(mass of caffeine in }CH_2Cl_2 / \text{volume of }CH_2Cl_2)}{\text{(mass of caffeine in water/ volume of water)}}$

   = 9.0

A). 1 x 200 mL extraction

Let m be the mass of caffeine in water

Mass of caffeine in $CH_2Cl_2$ = 100 - m

∴ $\frac{(100-m)/200}{m/200}=9$

  $\frac{100-m}{m}=9$

  $10 \ m = 100$

$m=\frac{100}{10}$

     = 10

Therefore, the mass remaining in the coffee is m = 10 mg

B). 2 x 100 mL extraction

First extraction :

Let $m_1$ be the mass of the caffeine in water.

Mass of caffeine in $CH_2Cl_2$ = 100 - m

∴ $\frac{(100-m_1)/100}{m_1/200}=9$

  $\frac{100-m_1}{m_1}=9$

  $5.5 \ m_1 = 100$

$m_1=\frac{100}{5.5}$

     = 18.18

Mass remaining in the coffee after the 1st extraction $m_1$ = 18.18 mg

Second extraction:

Let $m_2$ be the mass of the caffeine in water.

Mass of caffeine in $CH_2Cl_2$ = 18.18 - $m_2$

∴ $\frac{(18.18-m_2)/100}{m_2/200}=9$

  $\frac{18.18-m_2}{m_2}=9$

  $5.5 \ m_2 = 18.18$

$m_1=\frac{18.18}{5.5}$

     = 3.3

Mass remaining in the coffee after the 1st extraction $m_2$ = 3.3 mg