Given 500 mL of an aqueous solution containing 8 g of benzoic acid. (a) How many grams of benzoic acid could be removed in a single?

(a)
Let y g be the mass of benzoic acid removed in the single extraction.
[benzoic acid in diethyl ether] : [benzoic acid in water] = 10
(y/150) : [(8 - y)/500] = 10 : 1
(y/150) × 1 = [(8 - y)/500] × 10
50y = 1200 - 150y
200y = 1200
y = 6
Mass of benzoic acid removed = 6 g


(b)
For the 1st extraction:
Let z g be the mass of benzoic acid removed in the first extraction.
[benzoic acid in diethyl ether] : [benzoic acid in water] = 10
(z/50) : [(8 - z)/500] = 10 : 1
(z/50) × 1 = [(8 - z)/500] × 10
50z = 400 - 50z
100z = 400
z = 4
Mass of benzoic acid removed in the 1st extraction = 4 g
Mass of benzoic acid left in the aqueous solution in the 1st extraction = (8 - 4) g = 4 g
This means that (4 g)/(8 g) = 1/2 of benzoic acid is removed for each attraction with 50 mL of diethyl ether.

For the 2nd extraction:
Mass of benzoic acid removed in the 2nd extraction = (4 g) × (1/2) = 2 g
Mass of benzoic acid left in the aqueous solution in the 2nd extraction = (4 - 2) g = 2 g

For the 3rd extraction:
Mass of benzoic acid removed in the 3rd extraction = (2 g) × (1/2) = 1 g

Total mass of benzoic acid removed = (4 + 2 + 1) g = 7 g