Question

Calculate the resolution factors of the first 2 peaks for the solvent ratios 30/70 and 50/50 methanol/water. Using the resolution factors calculated, discuss the separation of the compounds with respect to the different solvent ratios

지 30% methanol-UV.pdf-Adobe Reader ㄧ一 2.s 지 50% methanol-UV.pdf-Adobe Reader *File Edit View Window Help File Edit ViewWindow Help de , 2目台 1 / 1 7% L ; Tools Sign Comment at KB M 1 / 1 7% ▼ ; Tools Sign Comrment | | / 1 | | 66.796 | 1 | / | | | 66.7% | ▼ 04 March 2016 12:31:35 FZ1029 FZ1029 Analysis : 50% methanol Analysis : 30% methanol 7 50% methanol-UV 30% methanol 7-UV 500000 750000 400000 5000001 E 30000 200000 250000 100000 12 Minutes 6 18 2 2 4 6 8 Minutes Integration results Integration results # Peak name Rt. 1 C+T+T 2 Toluene % Area 5.39 41970751.56 90.38 4.45 3 Ethylbenzene 21.35 2399776.12 5.17 4643781241 100.00 Area Peak name Rt. Area %Area 1 Theobromine 3.26 8093851.43 34.18 2 Theophyline 3.86 8896868.67 37.57 4.49 6691577.35 28.26 15.80 2067284.73 3 Caffeine SUM 8.26 x11.69 in 8.26 x11.69 in 14:02 23/03/2016

Answer 1

1) for 30/70 solvent ratio,

Resolution=R=t_R2-t_R1/[1/2(W1+W2)]

where t_R1=retention time for peak 1=3.26 min

t_R2=retention time for peak 2=3.86 min

W1=peak width for peak 1=0.59 min

W2=peak width for peak 2=0.25 min

R=3.86-3.26/1/2(0.59+0.25)=0.6/0.42=1.43

R=1.43

2) for solvent ratio 50/50

t_R1=retention time for peak 1=5.39 min

t_R2=retention time for peak 2=15.80 min

W1=peak width for peak 1=2.05 min

W2=peak width for peak 2=1.4 min

R=15.80-5.39/[1/2(2.05+1.4)]=10.41/1.72=6.05

R=6.05

Resolution is more in case of solvent ratio 50/50 is higher ,this indicates better peak separation in case of solvent ratio 50/50 (in other words the mutual overlap of the peak is less in case of higher resolution)