Question

After you done organic chemistry lab, the product was analyzed by GC. Your product produced a chromatogram containing two peaks with retention times of 3.4 min and 5.6 min, respectively. Your instructor said that your product contains two product compounds, A and B. (total 50 pts) GC signal analysis, C (concentration) = RF (Yesponse factor)XA (area) From the test above, you find that peak areas of compounds A and B are 2500 and 3500. Assuming that the response factor of these two compounds are equal, what are the % compositions of A and B in your product? (20 pts) However, the composition analysis done question 3b above is not accurate. Therefore, organic lab instructor conducted additional GC analysis using a standard sample containing 5 mg of A and 5 mg of B. From this standard sample, he found areas of peaks of A and B in a chromatogram was 2000 and 3000, respectively. From this data, recalculate the compositions (%) of A and B of your product. (30 pts)

Answer 1

Formula given:

Concentration = Retention factor * Area

C = RF * A

1. Given that: Area of peak for component A, AA = 2500

Area of peak for component B, AB = 3500

assuming and RFA = RFB

Ratio of the concnetrations of A and B in the sample is:

CA/CB = RFA*AA/(RFB*AB)

or CA/CB = 2500/3500

Add 1 both side:

CA/CB +1 = 2500/3500 +1

(CA + CB)/CB = (2500+3500)/3500

or, CB/(CA + CB) = 3500/6000 = 58.33

% composition of B = CB/(CA + CB) *100= 58.33%

thus % composition of A = 100 - 58.33 = 41.67%

2. GC for standard samples:

Note* -Assuming sample volumes same for A and B

Given that: Concentration of A, CA,std = 5 mg and Peak area, AA,std = 2000

Concentration of B, CB,std = 5 mg and Peak area, AB,std = 3000

For sample A:

CA(unknown) = RFA* AA(unknown)

and CA,std = RFA* AA,std

Dividing the equations we get:

CA(unknown) / CA,std =  AA(unknown) / AA,std

CA(unknown) = CA,std * AA(unknown) / AA,std

$\therefore$ CA(unknown) = 5 * 2500 / 2000

​​​​CA(unknown) = 6.25 mg

Similarly for sample B:

CB(unknown) / CB,std =  AB(unknown) / AB,std

CB(unknown) = CB,std * AB(unknown) / AB,std

$\therefore$ CB(unknown) = 5 * 3500 / 3000

CB(unknown) = 5.83 mg

thus % composition of A = CA(unknown)*100/(CA(unknown) + CB(unknown))

= 6.25*100/(6.25+5.83)

= 51.74%

% composition of B = 100 - %A= 100- 51.74 = 48.26%