Question

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

Answer 1

(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 (y_dl) = y_blank + 3*s

i.e. y_dl = y_blank + 3*s

= 45 + 3*19.0557

Therefore, the detection limit = 102.1671 counts

Conversion of counts to molarity (M)

m = (y_sample - y_blank) / 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