4. Consider the information below and answer the questions that follow (43 pts)

Question

4. Consider the information below and answer the questions that follow (43 pts) A student is informed that a liquid sample (9 mL total) is being studied to determine the contents. The student is told that the sample contains an unknown organic molecule of formula CsHsO dissolved in dichloromethane (boiling point 40°C). A 'H NMR spectrum of the sample shows two singlets at 2.05 and 5.30 ppm. The student first uses 1 mL total of the liquid sample in a gas chromatography (GC) analysis with helium as a carrier gas. Two peaks are observed with retention times of 0.4 (Peak A) and 0.6 (Peak B) minutes (relative area ratios: 3:1). The student then proceeds to separate the two components of the remaining sample via fractional distillation. In the process, the student observes condensation and collects a first fraction between 39°C and 40°C (4.0 mL total). The temperature increases quickly and a second fraction is collected between 56°C and 58°C (1.5 mL total). A GC analysis of the first fraction shows a single peak with a retention time of 0.4 while a GC analysis of the second fraction shows a single peak with a retention time of 0.6. A H NMR spectrum of the first fraction shows one peak (5.30 ppm, singlet) and a H NMR spectrum of the second fraction shows one peak (2.05 ppm).

Explanation / Answer

ANSWER:

The volume of compopund used for seperation is 8 mL ( initially it is 9 mL but from this 1 mL was used for GC-injection)

Amount of indiviual componet = (no. of parts of individual component in the ratio / totla no.of parts in the given ratio) * volume used for seperation

Here total no. of parts in the given ratio (3:1) is = 3+1= 4 and the volume used is 8 mL

Therefore, amount of compound observed at 0.4 mins in the mixture used for seperation is = (3/4) *8 = 06 mL

Similary, amount of compound observed at 0.6 mins  in the mixture used for seperation is = (1/4) *8 = 02 mL

Therefore % recovery of First fraction (at 0.4 mins, dichloromethane) = 4/6 * 100 = 66.6 %

% Recovery of second fraction (at 0.6 mins, acetone) = 1.5/2 * 100 = 75 %

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