Chapter 8:
Sampling, Standardization, and
Calibration
TMHsiung@2014 1/59
Contents in Chapter 08
1.
2.
3.
4.
Analytical Samples and Methods
Sampling
The Least Squares and Calibration Curve
Calibration Methods
1) External Standard Calibration
- Matrix Effect
2) Standard Addition Method
3) Internal Standard
5. Figures of Merit for Analytical Methods
TMHsiung@2014 2/59
1. Analytical Samples and Methods
 Classification of analyses by sample size

Classification of constituent types by analyte level
TMHsiung@2014 3/59
Continued
 For example: Determining Hg in the ppb to ppm
range in a 1 μL (≈ 1 mg) sample of river water
would be a micro analysis of trace constituent.
* Interlaboratory error as a function of analyte
concentration:
TMHsiung@2014 4/59
2. Sampling
 Classification of sampling plan (Ref/Harvey)
- Random sampling
- Judgmental sampling
- Systematic sampling
- Systematic–judgmental sampling
- Stratified sampling
- Convenience sampling
 Random sample: A sample collected at random
from the target population. (provides and
unbiased estimate of the target population)
- using random number table
- generating from spreadsheet
TMHsiung@2014 5/59

Steps in obtaining a
laboratory sample
TMHsiung@2014 6/59
 Statistics of sampling
1) Sampling variance versus overall variance
so2 = ss2 + sm2
so2: overall variance
ss2 : variance of sampling process
sm 2: variance of analytical method
TMHsiung@2014 7/59
Example: A quantitative analysis for an analyte gives a mean
concentration of 12.6 mg/kg. The standard deviation for analytical
method (sm) is found to be 1.1 mg/kg, and that due to sampling
process (ss) is 2.1 mg/kg.
(a) What is the overall variance for the analysis?
(b) By how much does the overall variance change if sm is
improved 10% to 0.99 mg/kg?
(c) By how much does the overall variance change if ss is
improved 10% to 1.9 mg/kg ?
Solution:
(a) so2 = sa2 + ss2 = (1.1)2 + (2.1)2 = 5.6
(b) so2 = (0.99)2 + (2.1)2 = 5.4
(c) so2 = (1.1)2 + (1.9)2 = 4.8


In general, ss2 is larger than sm2
More significant improvement
in the so is focus on sampling
problems.
TMHsiung@2014 8/59
Example: The following data were particular drug concentration in
a animal-feed formulations. The data on the left were obtained
under conditions in which random errors in sampling and the
analytical procedure contribute to the overall variance. The data
on the right were obtained in a repeatedly analysing a single
sample. Determine the overall variance and the contributions from
sampling and the analytical procedure.
sm2
so 2
TMHsiung@2014 9/59
Solution:
From left data: so2 = 4.71x10–7
From right data: sm2 = 7.00x10–8
Sampling variance:
ss2 = so2 – sm2 = 4.71x10–7 – 7.00x10–8 = 4.01x10–7
s 2m
x100% 15%
2
so
s s2
x100% 85%
2
so
TMHsiung@2014 10/59
2) Drawing particles with two types particles
(binomial distribution)
Application example:
nA: analyte containing
nB: analyte free
N A  number of A particles
N B  number of B particles
Np  nA  n B
p  probability of drawing A 
nA
Np
Standard deviationof the thenumber of A particledraw ( A ) :
 A  N p (1 p)
Relativestandarddeviationof the thenumber of A particledraw ( r ) :
r 
A
Np

1 p
Np
T henumber of particles(N) needed to achievea given  r :
1 p
N
p r2
*
TMHsiung@2014 11/59
Example 1:
A mixture contains 1% KCl particles and 99% KNO3 particles. What
is the number of particle required for a sampling relative standard
deviation of 1% for KCl ?
Solution:
1 p
1  0.01
5
N


9.9x10
p r2 (0.01)(0.01) 2
TMHsiung@2014 12/59
Example 2:
A mixture contains 1% KCl particle and 99% KNO3 particle. The
average density of the particles is 2.108 g/cm3, and the average
diameter of the particle is 0.152 mm. What is the sample mass
equivalent to total 9.9x105 particles of the mixture?
Solution:
4
4
each particlevolume r 3   (0.076)3  1.84x103 mm3
3
3
totalparticlemass
m  Ndv
3
2
.
108
g
cm
3
3
 9.9 x105 


1
.
84
x
10
mm
cm3
1000mm3
 3.84 g
For solid sample:
Finer particles require less mass of sample to
reach same RSD of sampling operation
TMHsiung@2014 13/59
Continued
For two types particles with different percentage of
active ingredient and different density:
d A d B 2 PA  PB 2
N  p(1  p)( 2 ) (
)
rP
d
dA: density of type A particle
dB: density of type B particle
d: the average density of the particles
PA: the percentage of the particle contain a
higher amount of active ingredient
PB: the percentage of the particle contain a
less amount of active ingredient
P: the overall average percent of active
ingredient
TMHsiung@2014 14/59
TMHsiung@2014 15/59
 Choosing a sample size (mount)
Ks = m  (r  100)2
m: mass of the collected sample
r : percent relative standard deviation of analyte
measurement due to sampling
Ks: Ingamells sampling constant
TMHsiung@2014 16/59
Example: The following data were obtained for determination of
the amount of inorganic ash in a breakfast cereal. (assume the Sm
<< Ss)
(a)
(b)
(c)
Determine sampling constant Ks.
What is the amount of a single sample needed to give a
relative standard deviation for sampling of ±2.0%.
Predict the relative standard deviation and the absolute
standard deviation if a single sample of 5 g are collected.
(assume same average of percentage ash)
TMHsiung@2014 17/59
Solution:
(a)
Mean of m
= 1.0007
Measurement mean = 1.298
Ss = 0.03194
r = 2.46%
Ks = m (r 100)2 = (1.0007)(2.46)2 = 6.06 g
TMHsiung@2014 18/59
(b)
m = Ks/ (r 100)2 = 6.06 g/(2.0)2 = 1.5 g
(c)
Ks
6.06g
 r 100 

 1.10
m
5.00g
 r  1.10%
(1.10)(1.298)
 r x 
 0.0143
100
TMHsiung@2014 19/59
 Number of replicate to be analyzed
Assuming Ss >> Sa, the number of defined mass samples (ns) to
reach a desired overall error (e):
N
t 2 ss2
2
x  r2
N : number of defined mass samples
s s : Samplingstandarddeviation
TMHsiung@2014 20/59
TMHsiung@2014 21/59
3. The Least Squares and Calibration Curve
1) Method of Least Squares
A mathematical optimization technique that attempts to find a
"best fit" to a set of data by attempting to minimize the total
squares of residual error, between the fitted function and the
data. A linear least squares for example:
Regression line
(Calibration curve)
ŷi
Residual error = yi–ŷi
yi
yi: Experimental value
ŷi: Calculated by y = mx + b
Total squares of residual error
= Σ(yi–ŷi)2
Assumptions in method of least squares:
 error in y >> error in x
 Uncertainties (standard deviation) in all y values areTMHsiung@2014
same
22/59
2) Linear Calibration Curve
The linear relationship between the amount of
analyte and a method’s measuring signal:
Smeas = kCA + b (i.e., y = mx + b)
(established by the method of least squares)
Smeas: Signal response by sample
CA : Analyte concentration
k:
Slope between Smeas and CA
b:
y-intercept of the linear equation
TMHsiung@2014 23/59
3) Establishing Calibration Curve
D  n  (x i2 )  ( x i ) 2
slope (m) :
y  intercept(b) :
n  (x i y i )   x i  y i
m
D
b
2
 (x i ) y i   (x i y i ) x i
D
Linear equation: y = mx + b
TMHsiung@2014 24/59
Example:
Establish a linear calibration curve by following data:
x (concentration)
y (signal)
1
2
3
3
4
4
6
5
Solution:
TMHsiung@2014 25/59
D  n  (x i2 )  ( x i ) 2  52
n  (x i y i )   x i  y i
m
 0.61538
D
b
2
 (x i ) y i   (x i y i ) x i
D
 1.34615
Ans: Linear equation is y = 0.61538x + 1.34615
TMHsiung@2014 26/59
4) Example of Applying Calibration Curve
Following are results of spectrophotometric analysis of protein
standard:
982_Ch04_Calibration_C
urve_Application_Demo.
xls
i)
Reject datum of (0.392)
(it is an outlier)
TMHsiung@2014 27/59
ii) Reject data of 25.0 μg protein standard
(out of the linear range)
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iii) 剩下 14 個數據建立 linear equation
m = 0.01628
b = 0.10400
y = (0.01628)x + 0.10400
iv) 計算 unknown sample 中分析物含量
Example:
已知 calibration curve: y = (0.01628)x + 0.10400
樣品分析結果 吸收值為 0.406，樣品中 protein 含量為多少？
Solution:
0.406  0.104
μg of protein 
 18.55 μg
0.01628
TMHsiung@2014 29/59
4. Calibration Methods
1) External Standard Calibration
 A calibration curve, y = ax + b, established by
standard containing known amount of analyte.
 External standard calibration is feasible if the
standards and the sample’s matrix has no effect on the
value of the slope of the calibration curve.
TMHsiung@2014 30/59
Example:
A spectrophotometric method for the quantitative
determination of Pb2+ levels in blood yields an absorbance
of 0.474 for a standard whose concentration of Pb2+ is 1.75
μg/L. How many μg/L of Pb2+ occur in a sample of blood if
its absorbance is 0.361?
Solution:
S s tan d
0.474
slope 

 0.2709g 1L
Cs
1.75gL1
S samp
0.361
CA 

 1.33 gL1
slope
0.2709 g 1 L
Ans: 1.33 μgL–1
TMHsiung@2014 31/59
Example:
A spectrophotometric method for the quantitative
determination of Pb2+ levels in blood gives a linear normal
calibration curve for which Sstand = (0.296 μg –1L)CS +
0.003. What is the Pb2+ level (in ppb) in a sample of blood
if Ssamp is 0.397?
Solution:
S samp  0.003 0.397 0.003
1
CA 


1
.
33

gL
0.296g 1L
0.296g 1L
Ans: 1.33 μgL-1
TMHsiung@2014 32/59
–
Matrix Effect
 Matrix: Everything in the unknown sample other than
analyte
 Matrix effect: The change in the analytical signal
caused by matrix.
 Value of slope versus
matrix effect:
Analyzing groundwater
ClO4– for example
Signal of sample
Incorrect
Actual
TMHsiung@2014 33/59
 Solutions for matrix effect:
- Matrix matching:
Adjusting the matrix of an external standard so that it
is the same as the matrix of the samples to be analyzed.
- Standard addition:
By adding aliquots of standard solution into sample,
and resolving analyte concentration by adequate
algebra.
Cautionary for Standard Addition Method:
 Amount of spiked analyte (standard) should close to the
amount of analyte in the sample
 Volume of the spiked standard should be designed to avoid the
matrix difference.
 Drawback of standard addition: It can not avoid the signal from
blank.
TMHsiung@2014 34/59
3) Standard Addition Method
i) Single-point standard addition:
 Initial sample  Spiked sample
Add Vs of [S]i
Vo of [X]i
I X  k[X]i
Vo of [X]i
[X]i:
Vo:
IX :
[S]i:
V S:
IS+X:

 Vo 
 VS 
  [S] i 

I S  X  k [X]i 

 Vo  VS 
 Vo  VS 
[X]i
I
 X
 Vo 
 VS  I S  X
  [S] i 

[X]i 
 Vo  VS 
 Vo  VS 
Analyte conc. of sample
Vol. of sample
Signal for [X]i
Analyte conc. in standard
Vol. of spiked standard
Signal of spiked sample
TMHsiung@2014 35/59
Example:
A spectrophotometric method for measurong Pb2+ in blood sample
yields an absorbance f 0.712. A 5.00 mL blood sample after spiking
with 5.00 μL of a 1560 μgL–1 Pb2+ standard, an absorbance of 1.546
is measured. Determine the concentration of Pb2+ in the original
sample of blood.
Solution:
[X]i
IX

 Vo 
 VS  I S  X
  [S] i 

[X]i 
 Vo  VS 
 Vo  VS 
[X]i




5.00mL
5.00 10 3 mL
1 

  1560μgL
[X]i 
 5.00mL  5.00 10 3 mL 
 5.00mL  5.00 10 3 mL 



0.712
1.546
0.7113[X]i  1.109μgL1  1.546[X]i
[X]i  1.33μgL1
Ans: 1.33 μgL–1
TMHsiung@2014 36/59
ii) Multiple-point standard addition (Linear Graphic
application):
Experimental example:
Place constant volume
(Vo) of sample with conc. of [X]i
Add different volume
(VS) of standard with conc. of [S]i
Dilute to a particular constant
volume (Vf)
TMHsiung@2014 37/59
[X]i:
IX :
Vo:
[S]i:
VS:
IS+X:
V:
Analyte conc. of sample
Signal for [X]i
Vol. of sample (constant)
Analyte conc. in standard (constant)
Vol. of added standard (variable)
Signal of spiked sample (variable)
Final vol. after diluting
I X  k[X]i

 Vo 
 VS 
 Vo 
 VS 
I S  X  k [X]i 
  [S] i 
  I S  X  k[X]i 
  k[S] i 

 V 
 V 
 V 
 V 

V  I
V 
I S  X  I X  o   X [S] i  S 
 V  [X]i
 V 
 V
I S  X 
 Vo
y=

V 
I
  I X  X [S] i  S 
[X]i

 Vo 
b + a x
TMHsiung@2014 38/59
 V 
 VS 
IX



IS  X 
 IX 
[S] i 

[X]i
 Vo 
 Vo 
y=
b + a
x
SD of x - intercept

sy
m
2
1
y

n m 2  (x i  x ) 2
s y : SD of y
n : number of data points
m : slpoe of thelonearcurve
x i : individual valuesof x
x : mean of x i
y : mean of y i
When y  0
 VS 
  [X]i
x - int ercept [S] i 
 Vo 
For t - value : dof  n  2
TMHsiung@2014 39/59
Example:
Ascorbic acid in orange juice was determined by an electrochemical
method. Following data and a linear graph were constructed by eight
standard addition. Calculate the concentration of ascorbic acid in the
sample.
TMHsiung@2014 40/59
Solution:
 V 
V 
I
  I X  X [S] i  S 
IS  X 
[X]i
 Vo 
 Vo 
y=
b + a
x
y  1.8687 0.6463x
When y  0
1.8687
x - intercept 
 2.89 nM  [X]i
0.6463
[X]i  2.89 mM
Ans: 2.89 mM
TMHsiung@2014 41/59
4) Internal Standard
i) Definition:
 Adding known amount of a reference substance
(internal standard) other than analyte is added to all
blank, analyte standard and samples.
 Establish a response factor based on the ratio of
standard analyte signal/internal standard signal versus
standard analyte conc. /internal standard conc.
 Analyte concentration in unknown is then calculated
according to the established response factor.
-
Standard addition: the “standard” is the same substance as
the analyte.
Internal standard: The internal “standard” is a different
substance from the analyte.
TMHsiung@2014 42/59
ii) Applying internal standard:
 常用於具有 multichannel dtecting 功能 (e.g., mass)
的分析方法。
 常用於 Chromatography (層析) 法。
 克服配製溶液其溶劑為揮發性的問題。
 克服上機時樣品體積不易控制的問題。
 克服訊號容易受到操作環境干擾的問題。
TMHsiung@2014 43/59
iii) Internal standard interpretation
以一個製備樣品其溶劑為揮發性者為例：。
Signal
a) 溶劑未揮發
Analyte
Standard
200
Internal
Standard
300
Absorption wavelength (nm)
b) 部分溶劑揮發
Internal
Standard
Signal
Analyte
Standard
 Analyte b) 之訊
號值高於 a)
 a) 和 b) Analyte
standard 訊號值
/Internal
standard 訊號值
之比例不變
200
300
Absorption wavelength (nm)
TMHsiung@2014 44/59
iv) Establishing response factor (F)
Ax
 AS 

 F
[X]
 [S] 
F: Response factor
AX: Signal of analyte standard
[X]:Conc. of analyte standard
AS: Signal of internal standard
[S]: Conc. of internal standard
v) Calculate analyte in unknown
Ax
 AS 

 F
[X]
 [S] 
F: Response factor
AX: Signal of analyte in unknown
[X]: Conc. of analyte in unknown
AS: Signal of internal standard
[S]: Conc. of internal standard
TMHsiung@2014 45/59
Example:
A spectrophotometric method for the quantitative determination of
Pb2+ levels in blood uses Cu2+ as an internal standard. A standard
containing 1.75 μgL–1 Pb2+ and 2.25 μgL–1 Cu2+ yields a ratio of
AX/AS of 2.37. A sample of blood is spiked with the same
concentration of Cu2+, giving a signal ratio of 1.80. Determine the
concentration of Pb2+ in the sample of blood.
Solution:
Ax
 AS 

 F
[X]
 [S] 
2.37
 1 
 F

1.75
 2.25
Ax
 AS 

 F
[X]
 [S] 
1.80
 1 
 3.05

[X]
 2.25
F  3.05
[X]  1.33μgL1
Ans: 1.33 μgL–1
TMHsiung@2014 46/59
5. Figures of merit for analytical methods
 Linearity:
A measure of how well data in a graph follow a
straight line, e.g., square of the correlation
coefficient R2 required close to 1.
 Linear dynamic range: The concentration interval
over which linearity, accuracy, and precision are
acceptable.
 (Calibration) Sensitivity: The change in the
response signal per unit change in analytical
concentration, thus, the slope of the calibration
curve.
TMHsiung@2014 47/59
 Limit of Detection (LOD):
The smallest quantity of analyte that is “significant”
different from blank, e.g., 99% chance of being
greater than the blank.


Only 1% of measurement
for a blank are expected to
exceed the detection limit.
But, 50% of measurement
for a sample containing
analyte at LOD will
below LOD.
TMHsiung@2014 48/59
 LOD estimation procedure:
1) Prepare sample whose analyte concentration is 1~5
times the LOD.
2) Measure 7 replicate samples.
3) Compute the signal standard deviation (s) of the 7
measurements.
4) Establish a calibration curve: y = mx +b
5) LOD (concentration) = 3s/m
TMHsiung@2014 49/59
Example:
Signal from 7 replicate sample with a concentration 3 times the
detection limit were 5.0, 5.0, 5.2, 4.2, 4.6, 6.0, and 4.9 nA. The
slope of the calibration curve is m=0.229 nA/μM. What is the
concentration of the limit of detection?
Solution:
s  0.56 nA
3s (3)(0.56 nA)
LOD 

 7.3 μM
m 0.229nA/μA
Ans: 7.3 μM
TMHsiung@2014 50/59
 Limit of quantitation (LOQ):
The minimum quantity of analyte that can be
measured “accurately”, often calculated by 10s/m.
s: signal standard deviation (s) of replicate
measurements.
m: slope in the calibration curve
 Instrument detection limit (IDL):
Obtained by 7 replicate measurements of one sample.
 Method detection limit (MDL):
Obtained preparing 7 individual samples and
analyzing each one once.
Not detected (ND):
Reported “ND” if the analytical result below the reporting
limit of detection
TMHsiung@2014 51/59
 Quality Assurance (QA)
 The protocol designed to demonstrate whether data is
meeting criteria that has been established to ensure
adequate data quality
 Process includes quality control and documentation of
procedures and results.
 Quality control (QC):
The measures taken to ensure that an analysis meets
the required accuracy and precision.
TMHsiung@2014 52/59
 Duplicate analysis: (for checking precision)
(Relative Percent Difference, RPD)
重複分析相對差異值百分比
x1 - x 2
RPD =
 100%
1/2(x1  x 2 )
Xl、X2:樣品重覆分析二次之個別分析值
TMHsiung@2014 53/59
 Recovery of SRM analysis: (for checking accuracy)
Cm
R=
 100%
Ct
R： 回收率 (recovery)
Ct： 已知參考值或確認值
Cm： 實際分析值
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 Recovery of Spike analysis:
(for checking matrix effect/accuracy)
R=
R：
T：
S：
X：
S - X 
 100%
T
回收率 (recovery)
添加於樣品中之標準品的之濃度(已知值)
添加樣品之測定濃度
未添樣品之測定濃度
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 Calibration check (calibration verification):
例：如每十個樣品，加入一個已知濃度的樣品，確認所建
立的檢量線是否仍然適用。
 Performance test samples (QC samples, blind
sample:
由外部提供之樣品，分析者視同一般樣品進行分析，數據
彙整後據以評估該分析人員或該實驗室分析數據的品質。
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 Control chart
Example: A control chart for a modern analytical
balance
UCL   
LCL   
3
N
3
N
失控時，需停止分析工作進行故障排除。
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Example of QC application:
BearLab 石墨爐原子
吸光譜分析試算表
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Homework (Due 2014/3/20)
Skoog 9th edition, Chapter 08 Questions and Problems
8-4
8-5
8-9
8-13
8-22
End of Chapter 08
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