Need some help my job depends on it..calibration curve excel

Discussions about HPLC, CE, TLC, SFC, and other "liquid phase" separation techniques.

15 posts Page 1 of 1
Hello...ive been trying to determine for 3 days now a unknown. I have a calibration curve run on hplc with the following peak areas:
Conc (mg/L) Peak Area
0 0
0.001 43162
0.01 1027039
0.05 4623058
0.1 8742401

The unknown peak area is 1227456 so i should be able to put this in the equation of the line and be able to detemine the concentration!!...but i keep getting strange answers! 1227456-61347/9e+07=118.15. my equation of the line is
y= 9E+07x + 61347
i should be able to put the unknown into this equation and get a concentration but instead i get a strange answer....please help....i know its something simple im doing wrong
seamoro wrote:
Hello...ive been trying to determine for 3 days now a unknown. I have a calibration curve run on hplc with the following peak areas:
Conc (mg/L) Peak Area
0 0
0.001 43162
0.01 1027039
0.05 4623058
0.1 8742401

The unknown peak area is 1227456 so i should be able to put this in the equation of the line and be able to detemine the concentration!!...but i keep getting strange answers! 1227456-61347/9e+07=118.15. my equation of the line is
y= 9E+07x + 61347
i should be able to put the unknown into this equation and get a concentration but instead i get a strange answer....please help....i know its something simple im doing wrong


concentration is 0,0133

try to calculate:

(1227456-61347)/9e+07=
Which function are you using?

Code: Select all

. A      B
1 0      0
2 0.001  43,163
3 0.01   1,027,040
4 0.05   4,623,059
5 0.1    8,742,402
6 ?      1227456 (unknown)

Not constrained to zero: =FORECAST(B6,A1:A5,B1:B5)
Constrained to zero: =LINEST(A1:A5,B1:B5,FALSE,FALSE)*B6
in fact L.R implemented in HP 35s
Hope your problem is solved by now. If at all possible, you should consider using instrument software rather than Excel to do the calibration. Amongst other reasons, instrument software will almost always offer a multitude of other features, such as weighted least squares (rather than treating all points as equally important), or curved fits for calibrations that don't lie on a straight line. Many instrument software packages can also be set up to warn you if your values are way outside the range of the calibration curve, or other problems have happened that make the data unreliable.
Don't use the antilog function on your calculator to enter 9E+07, use the EXP key instead :mrgreen:
Good judgment comes from bad experience, and a lot of that comes from bad judgment.
incidentally, while we're on subjects like that, I sincerely hope you're not using 9E+07. I'd suggest working to a precision better than one significant figure...
JGK wrote:
Don't use the antilog function on your calculator to enter 9E+07, use the EXP key instead :mrgreen:


Well spoted.
I would never guess he calculated "e" to the power of 7, and really rounded slope value to one significant digit.
JGK wrote:
Don't use the antilog function on your calculator to enter 9E+07, use the EXP key instead :mrgreen:



Is it wrong that this made me laugh,

also your 0 = 0 point, did you get this number experimentally from a blank? Or are you just assuming 0 concentration will be a 0 area?
Well, if you haven't figured out what you did wrong, you had the axes incorrect. The equation you have X = Concentration and Y = Area

The correct equation is

Y = 1.13827E-08X - 0.00066

which gives the result of 0.013308371
damn right that your job is on the block if you can't solve a basic high school level linear equation...
michaelbarnes42 wrote:
Well, if you haven't figured out what you did wrong, you had the axes incorrect. The equation you have X = Concentration and Y = Area

The correct equation is

Y = 1.13827E-08X - 0.00066

which gives the result of 0.013308371


Are you kidding ?

Your equation is completely wrong (provided Y is peak area and X is concentration) - it gives peak area lower then zero for concentration 0.0133.
Gentelman, Ordinary Linear Regression is something which should be called "basics" for analytical chemist.

Time to summarize:

correct equation (point 0 , 0 included):

Y = 87 757 294 X + 61 347

correlation coefficient r = 0.9995 (BTW - residuals plot is acceptable)


concentration 0.0133 mg/L corresponds to peak area 1 227 456


I do hope it is not a concentration of API. :mrgreen:
lmh wrote:
Hope your problem is solved by now. If at all possible, you should consider using instrument software rather than Excel to do the calibration. ..........


New post strongly associated with this advise is waiting for the answer. :roll:

viewtopic.php?f=4&t=20370&start=0
dblux_ wrote:
michaelbarnes42 wrote:
Well, if you haven't figured out what you did wrong, you had the axes incorrect. The equation you have X = Concentration and Y = Area

The correct equation is

Y = 1.13827E-08X - 0.00066

which gives the result of 0.013308371


Are you kidding ?

Your equation is completely wrong (provided Y is peak area and X is concentration) - it gives peak area lower then zero for concentration 0.0133.



I am 100% not kidding, the equation he had has X = Concentration and Y = Area

He was unfortunatly putting Area in for X, which was giving him the incorrect answer.

The equation I gave was the one he should be using if he wanted to use X for area and Y for concentration.
michaelbarnes42 wrote:
I am 100% not kidding, the equation he had has X = Concentration and Y = Area


Correct. This form of equation is accepted by majority of analysts.
Y=bX + a where X is concentration

He was unfortunatly putting Area in for X, which was giving him the incorrect answer.



As JGK explained, he raised "e" (Euler's constant) to the power of 7 instead raising 10 to the power of 7.
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