SME
Analytical Chemistry · Problem type
Detection and quantitation limits
Use this for “Ten blank readings have a standard deviation of 0.0035, and the calibration slope is 0.125 per ppm. Find the detection and quantitation limits.”
Chart
This is an example. Type your own values to start a new problem.
LOD (ppm)
=
3
×
m
LOQ (ppm)
=
10
×
m
Formulas
LOD
=
3
×
sb
m
0.1313
=
3
×
0.0021
0.048
LOQ
=
10
×
sb
m
0.4375
=
10
×
0.0021
0.048
Assumptions
•
sb is the standard deviation of repeated blank readings, in the same signal units as the slope.
•
The LOD is the concentration whose signal stands 3sb above the blank; the LOQ, at 10sb, is the lowest worth reporting as a number.
•
The calibration line stays straight down to these low concentrations.
Step-by-step
WE KNOW
sb = 0.0021, m = 0.048
FIND
limit of detection (LOD), limit of quantitation (LOQ)
Step 1 · Find limit of detection (LOD)
Use LOD = 3 × sb ÷ m
A signal 3sb above the blank, turned into a concentration by the slope.
LOD
=
3
×
0.0021
0.048
LOD
=
0.0063
0.048
LOD = 0.1313 ppm
Step 2 · Find limit of quantitation (LOQ)
Use LOQ = 10 × sb ÷ m
A signal 10sb above the blank, turned into a concentration by the slope.
LOQ
=
10
×
0.0021
0.048
LOQ
=
0.021
0.048
LOQ = 0.4375 ppm
Check
Put the numbers back in. Both sides should match.
0.1313
=
3
×
0.0021
0.048
✓
0.4375
=
10
×
0.0021
0.048
✓
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