Transcript Principles of Analytical Chemistry (F13I11)

Principles of Analytical
Chemistry
(F13I11)
Recommended textbook:
“Fundamentals of Analytical Chemistry”
Skoog, West and Holler, 7th Ed., 1996
(Saunders College Publishing)
Applications of Analytical Chemistry
Industrial Processes: analysis for quality control, and “reverse engineering”
(i.e. finding out what your competitors are doing).
Environmental Analysis: familiar to those who attended the second year
“Environmental Chemistry” modules. A very wide range of problems and
types of analyte
Regulatory Agencies: dealing with many problems from first two.
Academic and Industrial Synthetic Chemistry: of great interest to many of my
colleagues. I will not be dealing with this type of problem.
The General Analytical Problem
Select sample
Extract analyte(s) from matrix
Separate analytes
Detect, identify and
quantify analytes
Determine reliability and
significance of results
Errors in Chemical Analysis
Impossible to eliminate errors.
How reliable are our data?
Data of unknown quality are useless!
•Carry out replicate measurements
•Analyse accurately known standards
•Perform statistical tests on data
Mean
N
 xi
Defined as follows:
x =
i=1
N
Where xi = individual values of x and N = number of replicate
measurements
Median
The middle result when data are arranged in order of size (for even
numbers the mean of middle two). Median can be preferred when
there is an “outlier” - one reading very different from rest. Median
less affected by outlier than is mean.
Illustration of “Mean” and “Median”
Results of 6 determinations of the Fe(III) content of a solution, known to
contain 20 ppm:
Note: The mean value is 19.78 ppm (i.e. 19.8ppm) - the median value is 19.7 ppm
Precision
Relates to reproducibility of results..
How similar are values obtained in exactly the same way?
Useful for measuring this:
Deviation from the mean:
di  xi  x
Accuracy
Measurement of agreement between experimental mean and
true value (which may not be known!).
Measures of accuracy:
Absolute error: E = xi - xt (where xt = true or accepted value)
Relative error:
x x
t 100%
E  i
r
x
t
(latter is more useful in practice)
Illustrating the difference between “accuracy” and “precision”
Low accuracy, low precision
Low accuracy, high precision
High accuracy, low precision
High accuracy, high precision
Some analytical data illustrating “accuracy” and “precision”
HN
H
S
NH3+ClH
Benzyl isothiourea
hydrochloride
O
OH
Analyst 4: imprecise, inaccurate
Analyst 3: precise, inaccurate
Analyst 2: imprecise, accurate
Analyst 1: precise, accurate
N
Nicotinic acid
Types of Error in Experimental
Data
Three types:
(1) Random (indeterminate) Error
Data scattered approx. symmetrically about a mean value.
Affects precision - dealt with statistically (see later).
(2) Systematic (determinate) Error
Several possible sources - later. Readings all too high
or too low. Affects accuracy.
(3) Gross Errors
Usually obvious - give “outlier” readings.
Detectable by carrying out sufficient replicate
measurements.
Sources of Systematic Error
1. Instrument Error
Need frequent calibration - both for apparatus such as
volumetric flasks, burettes etc., but also for electronic
devices such as spectrometers.
2. Method Error
Due to inadequacies in physical or chemical behaviour
of reagents or reactions (e.g. slow or incomplete reactions)
Example from earlier overhead - nicotinic acid does not
react completely under normal Kjeldahl conditions for
nitrogen determination.
3. Personal Error
e.g. insensitivity to colour changes; tendency to estimate
scale readings to improve precision; preconceived idea of
“true” value.
Systematic errors can be
constant (e.g. error in burette reading less important for larger values of reading) or
proportional (e.g. presence of given proportion of
interfering impurity in sample; equally significant
for all values of measurement)
Minimise instrument errors by careful recalibration and good
maintenance of equipment.
Minimise personal errors by care and self-discipline
Method errors - most difficult. “True” value may not be known.
Three approaches to minimise:
•analysis of certified standards
•use 2 or more independent methods
•analysis of blanks
Statistical Treatment of
Random Errors
There are always a large number of small, random errors
in making any measurement.
These can be small changes in temperature or pressure;
random responses of electronic detectors (“noise”) etc.
Suppose there are 4 small random errors possible.
Assume all are equally likely, and that each causes an error
of U in the reading.
Possible combinations of errors are shown on the next slide:
Combination of Random Errors
Total Error
No.
Relative Frequency
+U+U+U+U
+4U
1
1/16 = 0.0625
-U+U+U+U
+U-U+U+U
+U+U-U+U
+U+U+U-U
+2U
4
4/16 = 0.250
-U-U+U+U
-U+U-U+U
-U+U+U-U
+U-U-U+U
+U-U+U-U
+U+U-U-U
0
6
6/16 = 0.375
+U-U-U-U
-U+U-U-U
-U-U+U-U
-U-U-U+U
-2U
4
4/16 = 0.250
-U-U-U-U
-4U
1
1/16 = 0.01625
The next overhead shows this in graphical form
Frequency Distribution for
Measurements Containing Random Errors
4 random uncertainties
A very large number of
random uncertainties
10 random uncertainties
This is a
Gaussian or
normal error
curve.
Symmetrical about
the mean.
Replicate Data on the Calibration of a 10ml Pipette
No.
Vol, ml.
No.
Vol, ml.
No.
Vol, ml
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
9.988
9.973
9.986
9.980
9.975
9.982
9.986
9.982
9.981
9.990
9.980
9.989
9.978
9.971
9.982
9.983
9.988
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
9.975
9.980
9.994
9.992
9.984
9.981
9.987
9.978
9.983
9.982
9.991
9.981
9.969
9.985
9.977
9.976
9.983
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
9.976
9.990
9.988
9.971
9.986
9.978
9.986
9.982
9.977
9.977
9.986
9.978
9.983
9.980
9.983
9.979
Mean volume
Spread
9.982 ml
0.025 ml
Median volume
9.982 ml
Standard deviation 0.0056 ml
Calibration data in graphical form
A = histogram of experimental results
B = Gaussian curve with the same mean value, the same precision (see later)
and the same area under the curve as for the histogram.
SAMPLE = finite number of observations
POPULATION = total (infinite) number of observations
Properties of Gaussian curve defined in terms of population.
Then see where modifications needed for small samples of data
Main properties of Gaussian curve:
Population mean (m) : defined as earlier (N  ). In absence of systematic error,
m is the true value (maximum on Gaussian curve).
Remember, sample mean ( x ) defined for small values of N.
(Sample mean  population mean when N  20)
Population Standard Deviation (s) - defined on next overhead
s : measure of precision of a population of data,
given by:
N
s
2
(
x

m
)
 i
i 1
N
Where m = population mean; N is very large.
The equation for a Gaussian curve is defined in terms of m and s, as follows:
y
e
 ( x  m ) 2 / 2s 2
s 2
Two Gaussian curves with two different
standard deviations, sA and sB (=2sA)
General Gaussian curve plotted in
units of z, where
z = (x - m)/s
i.e. deviation from the mean of a
datum in units of standard
deviation. Plot can be used for
data with given value of mean,
and any standard deviation.
Area under a Gaussian Curve
From equation above, and illustrated by the previous curves,
68.3% of the data lie within s of the mean (m), i.e. 68.3% of
the area under the curve lies between s of m.
Similarly, 95.5% of the area lies between s, and 99.7%
between s.
There are 68.3 chances in 100 that for a single datum the
random error in the measurement will not exceed s.
The chances are 95.5 in 100 that the error will not exceed s.
Sample Standard Deviation, s
The equation for s must be modified for small samples of data, i.e. small N
N
s
2
(
x

x
)
 i
i 1
N 1
Two differences cf. to equation for s:
1.
Use sample mean instead of population mean.
2.
Use degrees of freedom, N - 1, instead of N.
Reason is that in working out the mean, the sum of the
differences from the mean must be zero. If N - 1 values are
known, the last value is defined. Thus only N - 1 degrees
of freedom. For large values of N, used in calculating
s, N and N - 1 are effectively equal.
Alternative Expression for s
(suitable for calculators)
N
N
s
(  xi ) 
2
(  xi ) 2
i 1
i 1
N
N 1
Note: NEVER round off figures before the end of the calculation
Reproducibility of a method for determining
the % of selenium in foods. 9 measurements
were made on a single batch of brown rice.
Standard Deviation of a Sample
Sample
1
2
3
4
5
6
7
8
9
xi2
0.0049
0.0049
0.0064
0.0049
0.0049
0.0064
0.0064
0.0081
0.0064
Selenium content (mg/g) (xI)
0.07
0.07
0.08
0.07
0.07
0.08
0.08
0.09
0.08
Sxi
=
0.69
Mean = Sxi/N= 0.077mg/g
Standard deviation:
Sxi2=
0.0533
(Sxi)2/N = 0.4761/9 = 0.0529
s
0.0533  0.0529
 0.00707106  0.007
9 1
Coefficient of variance = 9.2% Concentration = 0.077 ± 0.007 mg/g
Standard Error of a Mean
The standard deviation relates to the probable error in a single measurement.
If we take a series of N measurements, the probable error of the mean is less than
the probable error of any one measurement.
The standard error of the mean, is defined as follows:
sm  s
N
Pooled Data
To achieve a value of s which is a good approximation to s, i.e. N  20,
it is sometimes necessary to pool data from a number of sets of measurements
(all taken in the same way).
Suppose that there are t small sets of data, comprising N1, N2,….Nt measurements.
The equation for the resultant sample standard deviation is:
s pooled 
N1
N2
N3
i 1
i 1
i 1
 ( xi  x1 ) 2   ( xi  x2 ) 2   ( xi  x3 ) 2 ....
N 1  N 2  N 3 ...... t
(Note: one degree of freedom is lost for each set of data)
Pooled Standard Deviation
Analysis of 6 bottles of wine
for residual sugar.
Bottle Sugar % (w/v) No. of obs.
Deviations from mean
1
0.94
3
0.05, 0.10, 0.08
2
1.08
4
0.06, 0.05, 0.09, 0.06
3
1.20
5
0.05, 0.12, 0.07, 0.00, 0.08
4
0.67
4
0.05, 0.10, 0.06, 0.09
5
0.83
3
0.07, 0.09, 0.10
6
0.76
4
0.06, 0.12, 0.04, 0.03
(0.05) 2  (010
. ) 2  (0.08) 2
0.0189
s1 

 0.0972  0.097
2
2
and similarly for all sn .
Set n  ( x  x )
1
0.0189
2
0.0178
3
0.0282
4
0.0242
5
0.0230
6
0.0205
Total
0.1326
2
i
sn
0.097
0.077
0.084
0.090
0.107
0.083
spooled
01326
.

 0.088%
23  6
Two alternative methods for measuring the precision of a set of results:
VARIANCE:
This is the square of the standard deviation:
N
s2 
2
2
(
x

x
)
 i
i 1
N 1
COEFFICIENT OF VARIANCE (CV)
(or RELATIVE STANDARD DEVIATION):
Divide the standard deviation by the mean value and express as a percentage:
s
CV  ( )  100%
x
Use of Statistics in Data
Evaluation
How can we relate the observed mean value (
x ) to the true mean (m)?
The latter can never be known exactly.
The range of uncertainty depends how closely s corresponds to s.
We can calculate the limits (above and below) around
with a given degree of probability.
x that m must lie,
Define some terms:
CONFIDENCE LIMITS
interval around the mean that probably contains m.
CONFIDENCE INTERVAL
the magnitude of the confidence limits
CONFIDENCE LEVEL
fixes the level of probability that the mean is within the confidence limits
Examples later.
First assume that the known s is a good
approximation to s.
Percentages of area under Gaussian curves between certain limits of z (= x - m/s)
50%
80%
90%
95%
99%
of area lies between
“
“
“
“
0.67s
1.29s
1.64s
1.96s
2.58s
What this means, for example, is that 80 times out of 100 the true mean will lie
between 1.29s of any measurement we make.
Thus, at a confidence level of 80%, the confidence limits are 1.29s.
For a single measurement: CL for m = x  zs (values of z on next overhead)
For the sample mean of N measurements (
x
), the equivalent expression is:
CL for m  x  zs
N
Values of z for determining Confidence
Limits
Confidence level, %
50
68
80
90
95
96
99
99.7
99.9
Note:
z
0.67
1.0
1.29
1.64
1.96
2.00
2.58
3.00
3.29
these figures assume that an excellent approximation
to the real standard deviation is known.
Confidence Limits when s is known
Atomic absorption analysis for copper concentration in aircraft engine oil gave a value
of 8.53 mg Cu/ml. Pooled results of many analyses showed s  s = 0.32 mg Cu/ml.
Calculate 90% and 99% confidence limits if the above result were based on (a) 1, (b) 4,
(c) 16 measurements.
(b)
(a)
(164
. )(0.32)
 8.53  0.52 mg / ml
1
i.e. 8.5  0.5mg / ml
(164
. )(0.32)
 8.53  0.26mg / ml
4
i.e. 8.5  0.3mg / ml
90% CL  8.53 
(2.58)(0.32)
99% CL  8.53 
 8.53  0.83mg / ml
1
i.e. 8.5  0.8mg / ml
90% CL  8.53 
(2.58)(0.32)
 8.53  0.41mg / ml
4
i.e. 8.5  0.4 mg / ml
99% CL  8.53 
90% CL  8.53 
(c)
(164
. )(0.32)
16
 8.53  013
. mg / ml
i.e. 8.5  01
. mg / ml
(2.58)(0.32)
 8.53  0.21mg / ml
16
i.e. 8.5  0.2 mg / ml
99% CL  8.53 
If we have no information on s, and only have a value for s the confidence interval is larger,
i.e. there is a greater uncertainty.
Instead of z, it is necessary to use the parameter t, defined as follows:
t = (x - m)/s
i.e. just like z, but using s instead of s.
By analogy we have:
CL for m  x  ts
N
(where x = sample mean for N measurements)
The calculated values of t are given on the next overhead
Values of t for various levels of probability
Degrees of freedom
(N-1)
1
2
3
4
5
6
7
8
9
19
59

Note:
(1)
(2)
80%
90%
95%
99%
3.08
1.89
1.64
1.53
1.48
1.44
1.42
1.40
1.38
1.33
1.30
1.29
6.31
2.92
2.35
2.13
2.02
1.94
1.90
1.86
1.83
1.73
1.67
1.64
12.7
4.30
3.18
2.78
2.57
2.45
2.36
2.31
2.26
2.10
2.00
1.96
63.7
9.92
5.84
4.60
4.03
3.71
3.50
3.36
3.25
2.88
2.66
2.58
As (N-1)  , so t  z
For all values of (N-1) < , t > z, I.e. greater uncertainty
Confidence Limits where s is not known
Analysis of an insecticide gave the following values for % of the chemical lindane:
7.47, 6.98, 7.27. Calculate the CL for the mean value at the 90% confidence level.
xi%
7.47
6.98
7.27
Sxi = 21.72
xi2
55.8009
48.7204
52.8529
Sxi2 = 157.3742
(  xi ) 2
(2172
. )2
x  N
157.3742 
3
s

N 1
2
 0.246  0.25%
2
i
If repeated analyses showed that s s = 0.28%:
x

x
2172
.

 7.24
N
3
i
(2.92)(0.25)
 7.24 
N
3
 7.24  0.42%
90% CL  x  ts
90% CL  x  zs
 7.24 
N
 7.24  0.27%
(164
. )(0.28)
3
Testing a Hypothesis
Carry out measurements on an accurately known standard.
Experimental value is different from the true value.
Is the difference due to a systematic error (bias) in the method - or simply to random error?
Assume that there is no bias
(NULL HYPOTHESIS),
and calculate the probability
that the experimental error
is due to random errors.
Figure shows (A) the curve for
the true value (mA = mt) and
(B) the experimental curve (mB)
Bias = mB- mA = mB - xt.
Test for bias by comparing x  xt with the
difference caused by random error
Remember confidence limit for m (assumed to be xt, i.e. assume no bias)
is given by:
CL for m  x 
ts
N
 at desired confidence level, random
errors can lead to:
ts
x  xt  
N
 if x  xt 
ts
, then at the desired
N
confidence level bias (systematic error)
is likely (and vice versa).
Detection of Systematic Error (Bias)
A standard material known to contain
38.9% Hg was analysed by
atomic absorption spectroscopy.
The results were 38.9%, 37.4%
and 37.1%. At the 95% confidence level,
is there any evidence for
a systematic error in the method?
x  37.8%
 x  xt  11%
.
 xi  113.4
s 
 xi2  4208.30
4208.30  (113.4) 2 3
 0.943%
2
Assume null hypothesis (no bias). Only reject this if
x  xt   ts
N
But t (from Table) = 4.30, s (calc. above) = 0.943% and N = 3
ts
N  4.30  0.943
 x  xt   ts
3  2.342%
N
Therefore the null hypothesis is maintained, and there is no
evidence for systematic error at the 95% confidence level.
Are two sets of measurements significantly different?
Suppose two samples are analysed under identical conditions.
Sample 1  x1 from N 1 replicate analyses
Sample 2  x2 from N 2 replicate analyses
Are these significantly different?
Using definition of pooled standard deviation, the equation on the last
overhead can be re-arranged:
x1  x2  ts pooled
N1  N 2
N1 N 2
Only if the difference between the two samples is greater than the term on
the right-hand side can we assume a real difference between the samples.
Test for significant difference between two sets of data
Two different methods for the analysis of boron in plant samples
gave the following results (mg/g):
(spectrophotometry)
(fluorimetry)
Each based on 5 replicate measurements.
At the 99% confidence level, are the mean values significantly
different?
Calculate spooled = 0.267. There are 8 degrees of freedom,
therefore (Table) t = 3.36 (99% level).
Level for rejecting null hypothesis is
 ts N1  N2 N1 N2 - i.e.  (336
. )(0.267) 10 25
i.e. ± 0.5674, or ±0.57 mg/g.
But x1  x2  28.0  26.25  1.75mg / g
i.e. x1  x2  ts pooled
N1  N 2
N1 N 2
Therefore, at this confidence level, there is a significant
difference, and there must be a systematic error in at least
one of the methods of analysis.
Detection of Gross Errors
A set of results may contain an outlying result
- out of line with the others.
Should it be retained or rejected?
There is no universal criterion for deciding this.
One rule that can give guidance is the Q test.
Consider a set of results
The parameter Qexp is defined as follows:
Qexp  x q  x n /w
where xq = questionable result
xn = nearest neighbour
w = spread of entire set

Qexp is then compared to a set of values Qcrit:
Qcrit (reject if Qexpt > Qcrit)
No. of observations
90%
95%
99% confidencelevel
3
0.941
0.970
0.994
4
0.765
0.829
0.926
5
0.642
0.710
0.821
6
0.560
0.625
0.740
7
0.507
0.568
0.680
8
0.468
0.526
0.634
9
0.437
0.493
0.598
10
0.412
0.466
0.568
Rejection of outlier recommended if Qexp > Qcrit for the desired confidence level.
Note:1.
The higher the confidence level, the less likely is
rejection to be recommended.
2. Rejection of outliers can have a marked effect on mean
and standard deviation, esp. when there are only a few
data points. Always try to obtain more data.
3. If outliers are to be retained, it is often better to report
the median value rather than the mean.
Q Test for Rejection
of Outliers
The following values were obtained for
the concentration of nitrite ions in a sample
of river water: 0.403, 0.410, 0.401, 0.380 mg/l.
Should the last reading be rejected?
Qexp  0.380  0.401 (0.410  0.380)  0.7
But Qcrit = 0.829 (at 95% level) for 4 values
Therefore, Qexp < Qcrit, and we cannot reject the suspect value.
Suppose 3 further measurements taken, giving total values of:
0.403, 0.410, 0.401, 0.380, 0.400, 0.413, 0.411 mg/l. Should
0.380 still be retained?
Qexp  0.380  0.400 (0.413  0.380)  0.606
But Qcrit = 0.568 (at 95% level) for 7 values
Therefore, Qexp > Qcrit, and rejection of 0.380 is recommended.
But note that 5 times in 100 it will be wrong to reject this suspect value!
Also note that if 0.380 is retained, s = 0.011 mg/l, but if it is rejected,
s = 0.0056 mg/l, i.e. precision appears to be twice as good, just by
rejecting one value.
Obtaining a representative sample
Homogeneous gaseous or liquid sample
No problem – any sample representative.
Solid sample - no gross heterogeneity
Take a number of small samples at random from throughout the bulk - this will
give a suitable representative sample.
Solid sample - obvious heterogeneity
Take small samples from each homogeneous region and
mix these in the same proportions as between each
region and the whole.
If it is suspected, but not certain, that a bulk material is heterogeneous, then
it is necessary to grind the sample to a fine powder, and mix this very
thoroughly before taking random samples from the bulk.
For a very large sample - a train-load of metal ore, or soil in a field - it is always
necessary to take a large number of random samples from throughout the whole.
Sample Preparation
and Extraction
May be many analytes present - separation - see later.
May be small amounts of analyte(s) in bulk material.
Need to concentrate these before analysis.e.g. heavy metals in
animal tissue, additives in polymers, herbicide residues in flour etc. etc.
May be helpful to concentrate complex mixtures selectively.
Most general type of pre-treatment: EXTRACTION.
Classical extraction method is:
(named after developer).
Apparatus
Sample in porous
thimble.
Exhaustive reflux for
up to 1 - 2 days.
Solution of analyte(s)
in volatile solvent
(e.g. CH2Cl2, CHCl3 etc.)
Evaporate to dryness or
suitable concentration,
for separation/analysis.
SOXHLET EXTRACTION