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Chapter 8: Sampling,

Standardization, and Calibration

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Contents in Chapter 08

1. Analytical Samples and Methods2. Sampling3. The Least Squares and Calibration

Curve4. Calibration Methods

1) External Standard Calibration - Matrix Effect2) Standard Addition Method3) Internal Standard

5. Figures of Merit for Analytical Methods

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1. Analytical Samples and Methods Classification of analyses by sample size

Classification of constituent types by analyte level

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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:

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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

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Steps in obtaining a laboratory sample

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so2 = ss

2 + sm2

so2: overall variance

ss2

: variance of sampling process

sm 2: variance of analytical method

Statistics of sampling1) Sampling variance versus overall variance

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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 = sa

2 + 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 sm

2

More significant improvement in the so is focus on sampling problems.

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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.

so2

sm2

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Solution:

From left data: so2 = 4.71x10–7

From right data: sm2 = 7.00x10–8

Sampling variance:

ss2 = so

2 – sm2 = 4.71x10–7 – 7.00x10–8 = 4.01x10–7

85%x100%s

s

15%x100%s

s

2o

2s

2o

2m

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2) Drawing particles with two types particles (binomial distribution)

2

r

pA

A

p

ABAp

B

A

1

:given a achieve toneeded (N) particles ofnumber The

1

:)( draw particleA ofnumber the theofdeviation standard Relative

p)(1N

:)( draw particleA ofnumber the theofdeviation Standard

N

nA drawing ofy probabilitp n nN

particles B ofnumber N

particlesA ofnumber N

r

r

Ar

p

pN

Np

p

Np

Application example:nA: analyte containingnB: analyte free

*

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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 ?

522

9.9x10)01.0)(01.0(

01.011

rp

pN

Solution:

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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?

g 84.3

mm 1084.1mm 1000

cm

cm

g 108.2109.9

vdNm

mass particle total

mm 1084.1)076.0(3

4

3

4 volumeparticleeach

333

3

35

3333

xx

xr Solution:

For solid sample:Finer particles require less mass of sample to reach same RSD of sampling operation

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ContinuedFor two types particles with different percentage of active ingredient and different density:

222

)())(1(P

PP

d

ddppN

r

BABA

dA: density of type A particledB: density of type B particled: the average density of the particlesPA: the percentage of the particle contain a

higher amount of active ingredientPB: the percentage of the particle contain a

less amount of active ingredientP: the overall average percent of active

ingredient

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Choosing a sample size (mount)

Ks = m (r 100)2

m: mass of the collected sampler : percent relative standard deviation of analyte

measurement due to samplingKs: Ingamells sampling constant

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Example: The following data were obtained for determination of the amount of inorganic ash in a breakfast cereal. (assume the Sm << Ss)

(a) Determine sampling constant Ks.(b) What is the amount of a single sample needed to give a

relative standard deviation for sampling of ±2.0%.(c) 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)

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Solution:(a)

Ks = m (r 100)2 = (1.0007)(2.46)2 = 6.06 g

Mean of m= 1.0007

Measurement mean = 1.298Ss = 0.03194

r = 2.46%

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(b) m = Ks/ (r 100)2 = 6.06 g/(2.0)2 = 1.5 g

0.0143100

98)(1.10)(1.2x

%10.1

1.10g 5.00

g 6.06

m

K100 (c) s

r

r

r

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Assuming Ss >> Sa, the number of defined mass samples (ns) to reach a desired overall error (e):

Number of replicate to be analyzed

deviation standard Sampling:s

samples mass defined ofnumber :N

s

22

22

r

s

x

stN

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3. The Least Squares and Calibration Curve1) 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:

yi: Experimental valueŷi: Calculated by y = mx + b

Regression line(Calibration curve)

Total squares of residual error = Σ(yi–ŷi)2

ŷi

yi

Residual error = yi–ŷi

Assumptions in method of least squares: error in y >> error in x Uncertainties (standard deviation) in all y values are same

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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 sampleCA : Analyte concentrationk: Slope between Smeas and CA

b: y-intercept of the linear equation

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3) Establishing Calibration Curve

D

x)y(xy)(xb :(b)intercept y

D

yx)y(xnm :(m) slope

)x()(xnD

iiii2i

iiii

2i

2i

Linear equation: y = mx + b

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Example:Establish a linear calibration curve by following data:

x (concentration) y (signal)

1 2

3 3

4 4

6 5

Solution:

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34615.1D

x)y(xy)(xb

61538.0D

yx)y(xnm

52)x()(xnD

iiii2i

iiii

2i

2i

Ans: Linear equation is y = 0.61538x + 1.34615

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4) Example of Applying Calibration CurveFollowing are results of spectrophotometric analysis of protein standard:

i) Reject datum of (0.392)(it is an outlier)

982_Ch04_Calibration_Curve_Application_Demo.xls

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ii) Reject data of 25.0 μg protein standard(out of the linear range)

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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 μg0.01628

iii)剩下 14 個數據建立 linear equationm = 0.01628

b = 0.10400

y = (0.01628)x + 0.10400

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4. Calibration Methods1) 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.

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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?

LggLC

Sslope

s

ds 11

tan 2709.075.1

474.0

11

33.12709.0

361.0

gLLgslope

SC

sampA

Solution:

Ans: 1.33 μgL–1

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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:

111

33.1296.0

003.0397.0

296.0

003.0

gL

LgLg

SC sampA

Ans: 1.33 μgL-1

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– 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

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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.

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3) Standard Addition Methodi) Single-point standard addition:

Vo of [X]i

Initial sample Spiked sample

Add Vs of [S]i

Vo of [X]i

XS

X

So

Si

So

oi

i

So

Si

So

oiXS

iX

I

I

VV

V[S]

VV

V[X]

[X]

VV

V[S]

VV

V[X]kI

k[X]I

[X]i: Analyte conc. of sampleVo: Vol. of sampleIX : Signal for [X]i

[S]i: Analyte conc. in standardVS: Vol. of spiked standardIS+X: Signal of spiked sample

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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:

Ans: 1.33 μgL–1

1i

i1

i

3

31

3i

i

XS

X

So

Si

So

oi

i

μgL 1.33[X]

1.546[X]μgL 1.1090.7113[X]

1.546

0.712

mL105.00mL 5.00

mL105.00μgL 1560

mL105.00mL 5.00

mL 5.00[X]

[X]

I

I

VV

V[S]

VV

V[X]

[X]

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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)

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o

Si

i

XX

oXS

Si

i

XoXXS

Si

oiXS

Si

oiXS

iX

V

V[S]

[X]

II

V

VI

V

V[S]

[X]

I

V

VII

V

V[S]

V

Vk[X]I

V

V[S]

V

V[X]kI

k[X]I

k

[X]i: Analyte conc. of sampleIX : Signal for [X]i

Vo: Vol. of sample (constant)[S]i: Analyte conc. in standard (constant)VS: Vol. of added standard (variable)IS+X: Signal of spiked sample (variable)V: Final vol. after diluting

y = b + a x

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o

Si

i

XX

oXS V

V[S]

[X]

II

V

VI

y = b + a x

io

Si [X]

V

V[S]intercept -x

0yWhen

2n dof :value-For t

y ofmean :y

xofmean :x

xof valuesindividual :x

curvelonear theof slpoe :m

points data ofnumber :n

y of SD :s

)x(xm

y

n

1

m

s

intercept- xof SD

i

i

i

y

2i

2

2y

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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.

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Solution:

Ans: 2.89 mM

o

Si

i

XX

oXS V

V[S]

[X]

II

V

VI

y = b + a x

mM 89.2[X]

[X]nM 89.26463.0

8687.1intercept -x

0yWhen

x0.6463 1.8687 y

i

i

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4) Internal Standardi) 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.

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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。

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iii) Internal standard interpretation以一個製備樣品其溶劑為揮發性者為例：。

Sig

nal

Absorption wavelength (nm)200 300

AnalyteStandard

InternalStandard

Sig

nal

Absorption wavelength (nm)200 300

AnalyteStandard

InternalStandard

a) 溶劑未揮發

b) 部分溶劑揮發

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

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iv) Establishing response factor (F)

[S]

AF

[X]

A Sx

F: Response factorAX: Signal of analyte standard[X]:Conc. of analyte standardAS: Signal of internal standard[S]: Conc. of internal standard

v) Calculate analyte in unknown

[S]

AF

[X]

A Sx

F: Response factorAX: Signal of analyte in unknown[X]:Conc. of analyte in unknownAS: Signal of internal standard[S]: Conc. of internal standard

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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.

1Sx

Sx

μgL 1.33[X] 2.25

13.05

[X]

1.80

[S]

AF

[X]

A

3.05F 2.25

1F

1.75

2.37

[S]

AF

[X]

A

Solution:

Ans: 1.33 μgL–1

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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.

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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.

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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 +b5) LOD (concentration) = 3s/m

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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:

μM 7.nA/μA 0.229

nA) (3)(0.5

m

3sLOD

nA 0.5s

36

6

Ans: 7.3 μM

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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

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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.

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Duplicate analysis: (for checking precision)

(Relative Percent Difference, RPD)

重複分析相對差異值百分比

100%)x1/2(x

x-x=RPD

21

21

Xl 、 X2: 樣品重覆分析二次之個別分析值

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Recovery of SRM analysis: (for checking accuracy)

R ： 回收率 (recovery)Ct ： 已知參考值或確認值Cm ： 實際分析值

100%C

C=R

t

m

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Recovery of Spike analysis:(for checking matrix effect/accuracy)

R ： 回收率 (recovery)T ： 添加於樣品中之標準品的之濃度 ( 已

知值 )S ： 添加樣品之測定濃度X ： 未添樣品之測定濃度

100%

TX-S

=R

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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

NLCL

NUCL

3

3

Example: A control chart for a modern analytical balance

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 Problems8-48-58-98-138-22

End of Chapter 08