Question

chemist attempted to recrystallize 2.00 g of fordhamic acid (FA) by dissolving it in 60.0 mL of boiling ethanol (p = 78'C Upon cooling the resulting clear solution in an ice water bath (0 °C), a relatively small amount of crystalline solid appeared from solution. She then consulted a standard reference source that had a table (see below) containing some useful solubility data on some carboxylic acids: solubility (B/15.0 ml) solubility te/25.0 ml) MW mpwater ethanol hexane water ethanol hexane acid name (abbrev.) | t/mole) (°C) 100 °C 78°C 68°C 0°C 0°C 0°C notredamic acid (NA) 189.23 120 0.500 3.00 0 .800 | 0.0100 0.400 0.0100 loyolaic acid (LA) 159.82 | 1250.800 1.00 0.500 0.0500 0.250 0.0100 fordhamic acid (FA) 184.10 1291.00 5.00 0 .500 0.100 0.500 0.0100 Xavieric acid (XA) 190.25 130 2.50 8.50 100 0.500 0.600 0.0200

a.) based on the information in the above table, how many mL of ethanol must be evaporated from 60.0 mL solution to achieve minimum volume of ethanol needed to dissolve the entire 2.00 g sample of the acid (FA)?

b.) after the evaporation of the 60.0 mL of ethanol down to rhe minimum volume needed to dissolve the 2.00 g of fordhamic acid (FA), how many grams of the acid will recrystallize out of the resulting solution when it is cooled to 0 degrees C ?

c.) when the chemist did NOT bothed to evaporate any ethanol and the volume of the ethanol present still remainsd at 60.0mL, how many gramd of the acid (FA) recrystallized out of solution when it cooled to 0 degrees C?

d.) the label went missing from a bottle sitting on a lab shelf used exclusively for storage bottles of pure fordhamic acid(FA) and ohre xavieric acid(XA); no other chemicals were stored on that shelf. explain how someonw could positively identify the contents od the unlabeled bottle as either fordhamic or xavieric acid. assume the person has only completed a simple/fractional distillation, recrystallization, and liquid extraction experiments.

Answer 1

a) At 78°C, 5.00 g of FA can be dissolved in 15 mL of ethanol. For 2.00 g, the volume required would be:

V = 15.0mL - . 2.00g = 6.00 mL 5.009

Hence, 54 mL of ethanol would have to be evaporated.

b) At 0°C, the solubility is 0.500 g per each 25 mL. If we have 6.00 mL, the mass dissolved would be:

m = 750mL 0.500gmL -. 6.00mL = 0.12 g

If we originally had 2.00 g, the precipitated mass would be 1.88 g.

c) Similar to item b, but with 60.0 mL:

0.500gm L. 60.0mL = 1.20 g m = 750mL

It means that 0.80 g would crystallize in this case.

d) It can be determined by a determination of the melting point with an addition. Mix a little unknown with a little FA. If the mixture melts in a short temperature range around 129°C, the you had FA, since this corresponds to the melting point of a pure substance (you've just added more FA to the original FA). If, after the addition of FA, the melting happens at a temperature lower than 129°C and in a broad temperature range, then the unkwon was XA, since these conditions correspond to the melting of a mixture (the original XA + FA).