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Topic: hplc calculation .plz help me  (Read 5454 times)

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profmsg

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hplc calculation .plz help me
« on: May 23, 2006, 05:46:56 PM »
procedure:
Using a stock solution of ethanol (100%), a series of standards of ethanol in water at concentration of 0,2,4,6,8,10 were prepared and 1ml of standard 1-butanol was added to each of the sample.
% ethanol   EtOH
ml   BuOH
ml   Water
ml   Total volume
ml
0   0   1   24.0
2   0.5   1   23.5   25
4   1   1   23   25
6   1.5   1   22.5   25
8   2   1   22   25
10   2.5   1   21.5   25

•   1ul of each of the standards were injected and recorded the retention time and the corresponding peak areas for ethanol and 1-butanol.
•   1ul of supplied gripe water was injected onto the GC and recorded a chromatogram for 4min approx. and recorded the peak areas.
•   A calibration curve of peak area ratio (PAR) and ethanol concentration (%) was prepared using the data obtained from the ethanol standards.
•    The PAR for the gripe water sample was calculated and the % ethanol in the sample was determined by interpolation.
RESULT:-

= 1.44%
% ethanol   Peak area
EtOH
Peak area
BuOH
PAR   Total volume
ml
0   0   285506   0   25
2   76691   175598   0.436   25
4   139683   183237   0.762   25
6   275208   188446   1.0460   25
8   329509   178345   1.0847   25
10   375101   156907   2.390   25
Gripe water    417610   240070   1.739   1 ul

Calculation
from graph
y = 0.2026x - 0.0597
1.739 = 0.2026x – 0.0597
X = 8.87
calculation

Error calculation:
The actual alcohol % = 90%
how can i find the percentage error ..is my calculation correct or not??? i got huge error ...what i do.....plz help me

Borek

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Re: hplc calculation .plz help me
« Reply #1 on: May 23, 2006, 06:16:36 PM »
Try to reformat message using tables, now it doesn't make much sense.

Code: [Select]
[table][tr][td]MeOH[/td][td]HCl[/td][/tr][tr][td]0.1[/td][td]0.2[/td][/tr][tr][td]0.3[/td][td]0.4[/td][/tr][/table]
gives

 MeOH HCl 0.1 0.2 0.3 0.4

Much easier to follow.
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