0.27 sample preparation - faculty.sdmiramar.edu

1 Sample Preparation 5/25/12

Dr. Fred Omega Garces Chemistry 251

Miramar College

0.27 Sample Preparation

Problems: 3, 11


Doping Catching the Dopers… coming soon this August


Chemical analysis Reliability Confidence in the analysis starts with meaningful samples


Chain of Custody and Sample Storage Besides choosing samples judiciously, sample must be stored such that cross contamination is eliminated. Samples can lose potency after as little as few hours of storage. Samples can also pick up contaminants if storage vials have not been properly cleaned.

The chain of custody must be also be observed and documented.


Statistics of Sampling For random errors, the overall variance so

2 is the sum of the variance of the analytical procedure sa

2 and the variance of the sampling operation ss2.

Uncertainty of sampling can best be understood by the following example: In an analysis of a barrel of powder, the standard deviation of sampling operation is + 4% and the standard deviation of the analytical procedure is +3%. What is the overall standard deviation? To what value must the sampling standard deviation be reduced so that the overall standard deviation is + 4%?

total analytical sampling = +

variance variance variance

!

"##

$##

so2 = s

a2 + s

s2

sa= 3% s

s = 4% : s

o2 = s

a2 + s

s2 : s

o2 = 3%2 + 4%2 = 9 + 16

so = 9 + 16 = 25 = 5

s0= 4% s

a= 3% s

s = ? : s

s2 = s

02 + s

a2 : s

s2 = 4%2 - 3%2 = 16 - 9

ss = 16 + 9 = 5 = 2.6%


Statistics of Sampling For random errors, the overall variance so

2 is the sum of the variance of the analytical procedure sa

2 and the variance of the sampling operation ss2.

Uncertainty of sampling can best be understood by the following example: In an analysis of a barrel of powder, the standard deviation of sampling operation is + 4% and the standard deviation of the analytical procedure is +3%. What is the overall standard deviation? To what value must the sampling standard deviation be reduced so that the overall standard deviation is + 4%?

overall analytical sampling = +

variance variance variance

!

"##

$##

so2 = s

a2 + s

s2

sa= 3% s

s = 4% : s

o2 = s

a2 + s

s2 : s

o2 = 3%2 + 4%2 = 9 + 16

so = 9 + 16 = 25 = 5

s0= 4% s

a= 3% s

s = ? : s

s2 = s

02 + s

a2 : s

s2 = 4%2 - 3%2 = 16 - 9

ss = 16 + 9 = 5 = 2.6%


Nature of Sampling The uncertainty when selecting a sample for analysis is express in terms of the probability of selecting A over B in a population of nT = nA +nB. If n particles are drawn, then the probability of selecting an A particle is npA. The standard deviation of many sampling is known as the binomial distribution An example of a mixture of 1-mm-diameter particles of KCl and KNO3 in a ratio 1:99. A sample containing 104 particles weights 11.0 g. i) What is the expected number in a sample weighing 11.0•102 g? and ii) relative standard deviation of KCl particles.

pA = possibility of drawing A =

nA

nA+ n

B

pB= probability of drawing B =

nB

nA+ n

B

= 1 - pA

standard deviationin sampling operation

!"#

$# s

n = n ⋅p

A⋅p

B

pKCl

= 1100

; 104 particles = 11.0 g; 106 particles = 1100 g: nKCl

= npA = 106 particles ⋅ 1

100 = 104

Relative standard deviation : s

KCl

nKCl

⋅100 = n ⋅ p

KCl⋅p

KNO3

nKCl

"

#

$$$

%

&

''' ⋅100 =

106 ⋅ 1100

⋅99100

nKCl

"

#

$$$$$

%

&

'''''

⋅100 = 0.99 ⋅%


Nature of Sampling The uncertainty when selecting a sample for analysis is express in terms of the probability of selecting A over B in a population of nT = nA +nB. If n particles are drawn, then the probability of selecting an A particle is npA. The standard deviation of many sampling is known as the binomial distribution An example of a mixture of 1-mm-diameter particles of KCl and KNO3 in a ratio 1:99. A sample containing 104 particles weights 11.0 g. i) What is the expected number in a sample weighing 11.0•102 g? and ii) relative standard deviation of KCl particles.

pA = possibility of drawing A =

nA

nA+ n

B

pB= probability of drawing B =

nB

nA+ n

B

= 1 - pA

standard deviationin sampling operation

!"#

$# s

n = n ⋅p

A⋅p

B

pKCl

= 1100



100 = 104


KCl

nKCl

⋅100 = n ⋅ p

KCl⋅p

KNO3

nKCl

"

#

$$$

%

&

''' ⋅100 =

106 ⋅ 1100

⋅99100

nKCl

"

#

$$$$$

%

&

'''''

⋅100 = 0.99 ⋅%


Choosing a Sampling Size Often it is desired to have a specific sampling variance, the question is what sample size must be used? Consider the analysis of radioactive sodium in liver. The tissue is homogenized and contains an average of 237 cps / g with std dev = 13.1%. When the sample size is increase to 1.3 g, the std dev lowers to 5.5%. When the sample size increases to 5.8g, the std dev lowers to 2.4%.


Choosing the Number of Replicate Analysis

Often the overall uncertainty can be reduce by analyzing more samples How many 0.7 g samples must be analyzed to give 95% CI that the mean is known to within +/- 4 %

µ - x e

= e = t sn

or n = t se2

where,u is the true population meanx is the measured meann is the number of samplest is the student's tsx2 is the sampling variance

e is the sought -for uncertaintyµ - x = e = 4% = 0.04t∞ = 1.960 s

x2 = 7% or 0.07

First iteration

n = t se

!

"##

$

%&&

2

→ n = 1.960 0.070.04

!

"##

$

%&&

2

= 11.8 ≈ 12

Second iteration, t11

= 2.209

n = t se

!

"##

$

%&&

2

→ n = 2.209 0.070.04

!

"##

$

%&&

2

= 14.9 ≈ 15

Third iteration, t14

= 2.150

n = t se

!

"##

$

%&&

2

→ n = 2.150 0.070.04

!

"##

$

%&&

2

= 14.2 ≈ 14

Fourth iteration, t13

= 2.170

n = t se

!

"##

$

%&&

2

→ n = 2.170 0.070.04

!

"##

$

%&&

2

= 14.4 ≈ 14


The Dissolution Process or Digestion Grinding Dissolution in acid,

fusion, ashing


Sample Preparation Series of steps required to transform sample to state suitable for analysis

Liquid Extraction with extraction vessel in microwave

Supercritical extraction with vessel for high pressure extraction. GC of the product from technique.

Solid-Phase extraction: Shown are the steps for extraction process.


Summary of Sample processing in relation to the Analytical Strategy

Obtain the sample Representative of Bulk

Process the Sample Prep for analysis by weighing and

drying then dissolving...

Carry out the Analysis Method • Obtain weight or Vol Data

• Prep Referernce Std or Protecting group

• Std solution or Calib Equib • Obrain required Data

Work the Data • Calcuate the results work up

• Workup Statistic Analysis

Calculate and Report the

Results


Lecture Notes Exercise Answers Slide 5s

a= 3% s

s = 4% : s

o2 = s

a2 + s

s2 : s

o2 = 3%2 + 4%2 = 9 + 16

so = 9 + 16 = 25 = 5

s0= 4% s

a= 3% s

s = ? : s

s2 = s

02 + s

a2 : s

s2 = 4%2 - 3%2 = 16 - 9

ss = 16 + 9 = 5 = 2.6%

Slide 7

pKCl

= 1100



100 = 104


KCl

nKCl

⋅100 = n ⋅ p

KCl⋅p

KNO3

nKCl

"

#

$$$

%

&

''' ⋅100 =

106 ⋅ 1100

⋅99100

nKCl

"

#

$$$$$

%

&

'''''

⋅100 = 0.99 ⋅%

0.27 sample preparation - faculty.sdmiramar.edu

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