In: Chemistry
Low concentrations of Ni-EDTA near the detection limit gave the following counts in a mass spectral measurement: 184, 148, 148, 148, 136, 170., 196, 152, 156, 175. Ten measurements of a blank had a mean of 45 counts. A sample containing 1.00 µM Ni-EDTA gave 1797 counts.
(a) Find the mean.
___ counts
(b) Find the standard deviation.
___ counts
(c) Estimate the detection limit for Ni-EDTA.
___ counts
___ M
(a) The mean (average) of given ten measurements = (184+148 +148 +148 +136 +170 +196 +152 +156+175)/10
Therefore, the mean = 161.3 counts
(b) The standard deviation of given counts (s)for the given measurements is
s = [{(184-161.3)2 + 3*(148-161.3)2 + (136-161.3)2 + (170-161.3)2 + (196-161.3)2 + (152-161.3)2 + (156-161.3)2 + (175-161.3)2}/(10-1)]1/2
s = {(515.29 + 530.67 + 640.09 + 75.69 + 1204.09 + 86.49 + 28.09 + 187.69)/9}1/2.
s = (3268.1/9)1/2
s = (363.12)1/2.
Therefore, the standard deviation = 19.0557 counts
(c) The detection limit for Ni-EDTA (ydl) = yblank + 3*s
i.e. ydl = yblank + 3*s
= 45 + 3*19.0557
Therefore, the detection limit = 102.1671 counts
Conversion of counts to molarity (M)
m = (ysample - yblank) / sample concentration
= (1797 - 45) / 10-6. (since sample concentration = 1 µM, i.e. 10-6 M)
= 1.752 * 109 counts/M
The minimum detectable concentration, c = 3*s/m
c = 3*19.0557/(1.752 * 109), i.e. 3.26296 * 10-8 M