Quinine in a 1.664-g antimalarial tablet was dissolved in sufficient 0.10 M to give 500 mL of solution. A 15.00-mL aliquot was then diluted to 100.0 mL with the acid. The fluorescence intensity for the diluted sample at 347.5 nm provided a reading of 327 on an arbitrary scale. A standard 100 ppm quinine solution registered 180 when measured under conditions identical to those for the diluted sample. Calculate the mass of quinine in milligrams in the tablet.

Answer:

605.6mg of quinine

Explanation:

Based on Lambert-Beer's law, the intensity of an optical measurement is directely proportional to its concentration.

Unknown concentration gives a reading of 327

100 ppm gives 180 of intensity

The concentration of the diluted quinine tablet is:

327 * (100ppm / 180) = 181.67 ppm. The final dilution

The concentration of the first diluted solution is:

181.67 ppm * (100.0mL / 15mL) = 1211ppm

ppm could be defined as mass of solute (In this case of quinine) in mg per liter of solution. That is:

1211mg / L

As the tablet was diluted to 500mL = 0.5L, the mass of the quinine is:

0.5L * (1211mg / L) =

An unknown compound was found to be soluble in concentrated sulfuric acid; however, the 2,4-DNP, chromic acid, and potassium permanganate gave negative results. The only positive result detected was the ferrous hydroxide test. What is the most likely functional group present in such compound