Question

You want to decaffeinate your coffee by extracting the caffeine out with dichloromethane (they don't do it that way anymore, but they used to). It's too late to extract the caffeine from the coffee beans, because you've already brewed yourself a 200 mL cup of coffee. Your particular brand of coffee contains 100 mg of caffeine in that 200 mL cup. The partition coefficient of caffeine in dichloromethane/water is 9.0 (it is more soluble in the dichloromethane).

How much caffeine would still be in your 200 mL if you did:_____.
A. One extraction using 200 mL ofdichloromethane
B. Two extractions using 100 mL ofdichloromethan each.

Answer 1

Partition coefficient Kd

`$=\frac{\frac{\text{mass of caffeine in } CH_2Cl_2}{\text{volume of } CH_2Cl_2}}{\frac{\text{mass of caffeine in water}}{\text{volume of water}}}$`

= 9.0

A). 1 x 200 mL extraction :

Let m be the mass of caffeine in the water

Mass of caffeine in
`$CH_2Cl_2$` = 100 - m

`$((100-m)/(200))/((m)/(200)) = 9.0$`

`$=(100-m)/(m) = 9.0$`

`$= 10 m = 100$`

m = 10

Mass remaining in coffee = m = 10 mg

B). 2 x 100 mL extraction :

First extraction :

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

Mass of caffeine in
`$CH_2Cl_2$` =
`$100-m_1$`

`$((100-m_1)/(100))/((m_1)/(200)) = 9.0$`

`$=(100-m_1)/(m_1) = 4.5$`

`$ 5.5 \ m_1 = 100$`

`$m_1$` = 18.18

Mass remaining in coffee =
`$m_1$` = 18.18 mg

Second Extraction :

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

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

`$((18.18-m_2)/(100))/((m_2)/(200)) = 9.0$`

`$=(18.18-m_2)/(m_2) = 4.5$`

`$ 5.5 \ m_2 = 18.18$`

`$m_2$` = 3.3

Mass remaining in coffee =
`$m_2$` = 3.3 mg