A 25.0 mL aqueous solution contains 2.50 g of compound Λ. Calculate the mass of compound Λ in three 5.0 mL extractions with ethyl acetate if the partition coefficient (CEtOAc/Cwater) for the extraction is 1.5 (Λ being more soluble in ethyl acetate)
To calculate the mass of compound Λ in three 5.0 mL extractions with ethyl acetate, we need to determine the amount of compound Λ extracted in each extraction and then sum them up.
First, let's calculate the amount (in grams) of compound Λ in the aqueous solution. We are given that the 25.0 mL aqueous solution contains 2.50 g of compound Λ. This means that the concentration of compound Λ in the solution is:
c(Λ) = 2.50 g / 25.0 mL
= 0.100 g/mL
Since each extraction is performed with 5.0 mL of ethyl acetate, the volume of the aqueous solution remaining after each extraction is:
25.0 mL - 5.0 mL = 20.0 mL
Next, we need to determine the fraction of compound Λ that partitions into the ethyl acetate in each extraction. The partition coefficient (CEtOAc/Cwater) is given as 1.5, indicating that the ethyl acetate is 1.5 times more effective in extracting compound Λ than water.
Let's denote the fraction of compound Λ that partitions into the ethyl acetate as x. This means that the fraction remaining in the aqueous solution is (1 - x).
From the partition coefficient, we can set up the equation:
x / (1 - x) = 1.5
Solving this equation, we find:
x = 1.5 - (1.5 * x)
2.5 * x = 1.5
x = 0.60
This means that in each extraction, 60% of compound Λ partitions into the ethyl acetate, and 40% remains in the aqueous solution.
Now let's calculate the amount of compound Λ extracted in each extraction:
Extraction 1:
Amount extracted = 0.60 * 0.100 g/mL * 5.0 mL = 0.030 g
After Extraction 1, the remaining amount of compound Λ in the aqueous solution is:
(1 - 0.60) * 0.100 g/mL * 20.0 mL = 0.040 g
Extraction 2:
Amount extracted = 0.60 * 0.040 g/mL * 5.0 mL = 0.012 g
After Extraction 2, the remaining amount of compound Λ in the aqueous solution is:
(1 - 0.60) * 0.040 g/mL * 15.0 mL = 0.024 g
Extraction 3:
Amount extracted = 0.60 * 0.024 g/mL * 5.0 mL = 0.0072 g
The total amount of compound Λ extracted in three 5.0 mL extractions is:
0.030 g + 0.012 g + 0.0072 g = 0.0492 g
Therefore, the mass of compound Λ in three 5.0 mL extractions with ethyl acetate is 0.0492 g.