Hello guys! I need help on this Gas Chrom problem:

An air sample is collected by drawing air through a charcoal sampling tube. After a suitable sampling period, the charcoal tube is removed and analyzed using gas chromatography. Given the data listed below, compute the contaminant concentration in ppm. Make any assumptions you feel are necessary.

SAMPLING CONDITIONS

600 ml – charcoal volume in front section

200 ml – charcoal volume in rear section

1.30 liters per minute, sampling rate

24.00 minutes – sampling time

50 – molecular weight of contaminant

1.50 grams/cc, density of contaminant

5.00 ml (Volume of carbon disulfide used for desorption)

20 °C, temperature during collection

STANDARD

55.00 units, area under GC curve of standard

1.00 microliter injected for standard

80% desorption efficiency

10 micrograms/cc of contaminant in standard

RESULTS

260.00 units, area under sample GC curve (front section)

15.00 units, area under sample GC curve (rear section)

3.00 units, area under blank curve

1.00 microliter injected for each sample run.

This is what I did. I used the formula: Rf (c) = A (c) x w (IS) /A(IS) x w(c), where:

A(c) = area of component, w(IS) = weight of internal standard, A(IS)= area of internal standard and w(c)= weight of component. I added 260 + 15 = 275 to get the area of component. Mass of standard and sample is 0.010 ug, since density and volume of component and standard are given (density= 10 ug/cc, volume= 1uL).

Rf (c) = 275 x 0.010 ug/55 x 0.010 ug = 5

Then, % conc = A(c) /w(s) x w(IS)/A(IS) x 1/Rf(c) x 100% where w(s) is weight of sample and Rf is response factor. Since density of sample is 1.50 g/cc, volume of sample is 600 ml (front) + 200 ml (back) + 5 ml (CS2)= 805 ml, then mass of sample is 1207.5 g.

%conc = 275/1207.5 x 10 exp 6 ug x 0.010 ug/55 x 1/5 x 100% = 8.2 x 10-10%.

If you convert that to ppm, then conc is 8.2 x 10-6 ppm.

Is it correct? ppm I got is low. Will value your comments and suggestions.

Thanks a lot,

Buffy