Question

1. The solubility of dulcin in water is 1.21 g/L at room temperature. Assuming that you achieved quantitative conversion (100% yield based on your measured amounts was present when you dissolved it for recrystallization) and used a total of 90.0 mL of water, what would be your maximum experimental yield after recrystallization if you had allowed the flask to cool to room temperature? Give your answer in mg.

2.One of the chemicals you added to the reaction contains potassium, but dulcin does not contain potassium. What happened to it? Be specific about exactly what action you performed and where the potassium went at that point. Examples of sufficiently detailed answers: "during the boiling to dissolve dulcin for recrystallization, it evaporated", "during the hot gravity filtration, it got caught on the filter paper as an insoluble impurity".

Answer 1

1. Solubility of Dulcin in water = 1.21g/L

Means

In 1000 ml of water, dissolves dulcin = 1.21 g

So, 90 ml of water will dissolve dulcin = (1.21 g x 90 ml)/1000 ml = 0.1089 g

1g = 1000 mg

0.1089 g = 1089 mg

Hence if yield of recrystallization is 100% so, dulcin formed = 1089 mg

2. See image 1 below

• In above process the dulcin is prepared from phenetidine. The pure dulcin is prepared from recrystallization of product of phenetidine and Isocyanic acid.
• The isocyanic acid is prepared by reaction of potassium cyanate and acetic acid forming isocyanic acid and potassium acetate.
• The Isocyanic acid reacts with phenetidine to form Dulcin.
• The dulcin I sthen dissolved in water by boiling and than filtered and dried to get pure crystals of dulcin.
• The potassium acetate formed inbetween the reaction dissolves in water during boiling the dulcin for recrystallization and filterated.