Question

13. A Chemistry major obtains a high paying summer job in an analytical laboratory and is given the task of determining the ethanol content (in vol %) of a commercial bourbon sample. He/she decides to do this by gas chromatography because the GC experiment in Chem 209 was so darn enjoyable. Accordingly, the student takes exactly 10.0 mL of the bourbon sample and dilutes this to exactly 50 mL with deionized water. He/she then injects 2.0 yL of the diluted sample into the Gc and obtains a chromatogram containing two peaks. The first peak, which has the same 56-secend retention time as an ethanol standard, has an area of 666 while the second peak, whose T17-second retention time matches that of pure water, has an area of 333. The student then makes up several standard solutions ranging in concentration from 0 to 30% ethanol (vol %), injects 2.0 L of each into the GC, and uses the areas of the ethanol peaks to create a calibration curve whose linear regression equation is y71.2 x 33 ares what is the ethanol concentration (vol %) of the initial mouthwash sample? D) 23.4% E) 46.8% A) 1.87% B) C) 4.67% 9.35% 46,8

Answer 1

Calibration curve has linear regression equation is y=71.2 x.

y= area of the etanol peaks.

x= concentration of ethanol sample (vol %)

for 50 ml. diluted sample two peaks were observed.(666 and 333)

for 666 peak concentration will be = (666/71.2) = 9.35%(calculaed from equation given).

from 333 peak concentration will be=(333/71.2)=4.67%(calculated from equation given).

so if we take the initial 10ml bourbon sample the concentration will be 5 times of the diluted sample as 50ml solution changes to 10ml. with no dilution.

so the changed concentrations will be

1) 9.35*5 = 46.75% $\simeq$ 46.8%(for 666 peak)

2) 4.67*5=23.35% $\simeq$ 23.4%(for the 333 peak)

If we take 666 data as our peak area of ethanol then the ethanol concentration of initial mouth wash sample will be 46.8%(vol %).