[buffydvamp]

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

[buffydvamp]

Another way to do this is:

Conc=V/v (mass of front + mass of back - mass of blank)/Desorption efficiency x flow rate x time

Conc = 5 ml/800 ml (900 g + 300 g -0)/0.80 x 1.30 L/min x 24 min

V is volume of desorbing solvent, v is volume used for analysis.  Add 600 + 200 = 800 mL.

Density of front and back is given (1.5 g/cc), and volumes are 600 & 200 ml respectively, yielding the masses above

Conc = 0.30048 g/L or 3.0048 x 10 exp 5 mg/m3.

If you convert it to ppm, m(mg/m3)/50 x 24.45, 50 is Mol. wt.

Conc (ppm) = 1.47 x 10 exp 5 ppm

In this case, ppm is too large.

I really need your help guys.   Will wait for your responses.

Regards,
Buffy

[Dude]

Here's my take, and I didn't find enough information to get an answer.

Basically, it boils down to g contaminant / g air.  One assumption must be that the amount of charcoal used to adsorb the contaminant from the air is quantitative and in excess.  Therefore, it does not figure into the equation at all.

1.  Find the denominator.

MW of air ~ 29 g/mol for 80 % N2/20 % O2
1.3 L/min(24 min) = 31.2 L of air collected
ideal gas law @ 20 C V = 24 L for air
(31.2 L/24 L) = 1.3 mol(29 g/ 1 mol) = 37.7 g   This is your denominator.

2.  Find numerator.

A.  Run injection of known weight of known internal standard and known weight of known contaminant.  This enables you to find the response factor of the standard relative to the unknown.  For example, a known weight of hexane will produce a different response than a known weight of acetone on a flame ionization detector.
 Apply Ru = Rs (As / Au)(Wu / Ws)
A= area, R = response factor, W = weight, u = unknown s = standard, typically in a two component system Rs is set to 1

Sample area = (260 + 15) - blank = 272
standard area = (55)

(272 / 55) (Ru)(weight of internal standard after)(dilution factor) = weight contaminant

3.  Multiply weight contaminant by (1 / desorption efficiency), divide by 37.7 g and multiply by 1,000,000 for ppm answer