Low concentration replicate samples of Ni-EDTA gave the following counts in a ma
ID: 786523 • Letter: L
Question
Low concentration replicate samples of Ni-EDTA gave the following counts in a mass spectral measurement: 166.1, 124.7, 140.7, 155.7,124.7, 194.1, 145.7, 150.3, 139.7, 157.3. Ten measurements of a blank had a mean of 42.1 counts. A sample containing 1.00 ?M Ni-EDTA gave 1797 counts. Estimate the signal and concentration detection limits for Ni-EDTA.A) Calculate the signal detection limit.
B) Concentration detection limit.
For part A, I have used the equaiton ydI = yBlank + 3s
The equation for part B. is : concentration = 3s/m. where m = (ysample - yblank) counts/ (Ni-EDTA)
I calculated s which is the standard deviation= 6.48 and yBlank is given= 42.1. sub. all these value in the eqation gave me
42.1+(3*6.48) = 61.54 counts, but the system says my answer is incorrect
Can someone explain to me whereis my mistake, and how to do part B.
Low concentration replicate samples of Ni-EDTA gave the following counts in a mass spectral measurement: 166.1, 124.7, 140.7, 155.7,124.7, 194.1, 145.7, 150.3, 139.7, 157.3. Ten measurements of a blank had a mean of 42.1 counts. A sample containing 1.00 ?M Ni-EDTA gave 1797 counts. Estimate the signal and concentration detection limits for Ni-EDTA.
A) Calculate the signal detection limit.
B) Concentration detection limit.
For part A, I have used the equaiton ydI = yBlank + 3s
The equation for part B. is : concentration = 3s/m. where m = (ysample - yblank) counts/ (Ni-EDTA)
I calculated s which is the standard deviation= 6.48 and yBlank is given= 42.1. sub. all these value in the eqation gave me
42.1+(3*6.48) = 61.54 counts, but the system says my answer is incorrect
Can someone explain to me whereis my mistake, and how to do part B.
Explanation / Answer
The equations is correct.
m= (1797-42.1)/1=1754.9
concentration = 3(standard deviation)/m. The answer you got is way to high. It should be lower.
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