2
SPECTROSCOPIC ANALYSIS
2.1 Introduction
Chemical analysis falls into two basic categories:
 qualitative – what is present
 quantitative – how much is present
Spectroscopy is capable of both types of analysis, though some forms are better than others at one or
the other of the types of analysis. Table 2.1 summarises the relative strengths of the forms of
spectroscopy examined in this subject.
TABLE 2.1 Analytical capabilities of spectroscopic techniques
Technique
Absorption vs
emission
Absorption
Qualitative
Quantitative
UV-visible
Atomic vs
molecular
Molecular
Poor
Excellent
Infrared
Molecular
Absorption
Excellent
Poor
Atomic absorption
Atomic
Absorption
Useless
Excellent
Atomic emission
Atomic
Emission
Good
Excellent
As you look at each of these techniques in some detail, you will see why each is good or bad in the
two forms of analysis.
2.2 Qualitative Analysis
Because each species – atom or molecule – has an individual set of energy states, which are different
in energy and number to all others, the spectrum (absorption or emission) produced will be different
for each species. This is known as a spectral fingerprint for that particular species, as shown in
Figure 2.1.
Li
Na
K
300
350
400
450
500
550
Wavelength (nm)
600
650
700
FIGURE 2.1 Atomic emission spectra of sodium, potassium and lithium (the vertical lines indicate an emission
line)
Species can then be identified by comparing the spectrum of the sample with standard spectra, since
the position of the peaks in the spectrum of a chemical species is the same, regardless whether it is
pure or in a mixture.
2. Spectroscopic Analysis
EXAMPLE 2.1
Below is the atomic emission spectrum of a sample. Using Figure 2.1, identify any elements
present.
Na
300
?
350
400
?
450
NaNa Li
500
550
Wavelength (nm)
600
Li
650
700
Comparing the sample spectrum to those of known substances, you must be sure that all
peaks for a particular species are found in the sample to be sure that that species is present
in the sample.
Each peak in the spectrum of lithium and of sodium is present in the sample spectrum, so
both those elements are present. However, the peaks at 400 and 480 nm are not
identifiable. While potassium has a peak at 400 nm, it also has peaks and 350 and 690 nm,
neither of which is present in the sample. Therefore, at least one other species must be
present in the sample to cause the other peaks.
CLASS EXERCISE 2.1
What can you tell about the sample giving the following spectrum?
300
350
400
450
500
550
Wavelength (nm)
600
650
700
2.3 Quantitative Analysis
While qualitative analysis is important – you can’t determine how much of a species is in a sample if
you don’t know if it’s there - the major use of spectroscopy is quantitative: the determination of
concentrations of analytes (normally in solution). Spectrophotometric methods are extremely good at
accurately and precisely determining extremely low levels of concentration, down to microgram/litre
levels (ppb), depending on the technique and analyte.
Steps in a quantitative spectrophotometric analysis
There are a series of steps common to all spectrophotometric analyses, and they are briefly outlined
below. More detail will be provided in the sections devoted to the four spectroscopic techniques
examined in this subject.
1. SAMPLE PREPARATION
The sample has to be in a physical form that is useable in the instrument. Most often, this will mean
that it needs to be dissolved, and possibly diluted.
Sci Inst Analysis (Spectro/Chrom)
2.2
2. Spectroscopic Analysis
2. SELECTION OF THE WAVELENGTH TO MAKE MEASUREMENTS
If a spectrum of the analyte is not available, then it should be recorded and a decision made from its
appearance. The general rule is to choose the wavelength of maximum absorption/emission for the
analyte, since this limits the errors in the analysis. If a wavelength was chosen where the analyte
doesn’t absorb/emit strongly, then there will be minimal difference in response between low and high
concentrations.
The spectrum of the solvent and any other reagents used in the preparation of the sample should
also be recorded, since in certain circumstances, these non-analyte species may absorb radiation at the
wavelength of maximum analyte absorption. With absorption measurement, the general rule is that if
the solvent/reagents has an absorbance of greater than 0.2 at a given wavelength, then a different
analyte peak should be chosen.
It is also advisable to choose the top of the absorption peak, rather than either side where the
steep slope of the peak means that slight changes in wavelength (caused by instrument error) can lead
to significant changes in readings, as shown in Figure 2.2.
a change in wavelength causes no
difference in response
response
a change in wavelength causes a
large difference in response
wavelength
FIGURE 2.2 Choosing the top of the peak
3. PREPARATION OF THE CALIBRATION GRAPH
Response
Spectroscopic instruments must be calibrated each time you use them, because they cannot be
guaranteed to give exactly the same reading today as they did yesterday. Calibration means running a
number of standards – normally 3 or 4 – to produce a measure of how the instrument responds to the
analyte across a range of concentrations.
In principle, the response is related to the concentration – the more atoms or molecules of the
analyte there are, the more radiation will be absorbed/emitted. However, there are a number of
restrictions in practice, which limit the range of concentrations that can be used.
The most important requirement for a calibration graph is that it should be linear – the response
is directly proportional to the concentration. If this is not the case, then errors in terms of drawing the
graph are greater. This immediately eliminates transmittance as a possible measure because even in
theory it is not linear with concentration. However, absorbance (for absorption measurements) and
intensity (for emission measurements) are linear within limits.
Figure 2.3 shows a typical response curve for a spectroscopic measurement.
linear region
Concentration
FIGURE 2.3 Response of analyte at different concentrations
Sci Inst Analysis (Spectro/Chrom)
2.3
2. Spectroscopic Analysis
Different analytes and different techniques have different concentration ranges that lie in the linear
region. If you don’t know what concentration range is appropriate for a given analysis, then you have
to do some trial-and-error checking to find it (see below for further details regarding this). For some
techniques, the linear region has a range of concentrations of 10 times, other it may be more than 100
times.
Having determined the linear region, the standards are prepared and measured, and the
calibration graph drawn up as you have done in your Laboratory Calculations/Mathematics subject.
EXAMPLE 2.2
Draw a calibration graph, given the following information.
Standard conc. (mg/L)
0
5
10
15
20
Reading
0.01
0.12
0.25
0.34
0.49
0.6
0.5
Reading
0.4
0.3
0.2
0.1
0
0
5
10
15
20
25
Conc. (mg/L)
CLASS EXERCISE 2.2
Draw a calibration graph, given the following information. Standards were prepared by
pipetting aliquots of 0, 5, 10, 15 and 20 mL of 500 mg/L iron into 200 mL volumetric flasks.
each flask was made up to the mark, and the solutions measured.
Vol. of 500 mg/L std (mL)
0
5
10
15
20
Sci Inst Analysis (Spectro/Chrom)
Conc mg/L
Reading
2
61
125
189
248
2.4
2. Spectroscopic Analysis
4. SAMPLE MEASUREMENT
The sample is measured in the same way that the standard were, and the calibration graph to determine
the concentration of analyte in the sample.
EXAMPLE 2.3
Determine the analyte concentration from the calibration graph in Example 2.2, if the sample
reading was 0.33.
0.6
0.5
Reading
0.4
0.3
0.2
13.7 mg/L
0.1
0
0
5
10
15
20
25
Conc. (mg/L)
CLASS EXERCISE 2.3
Determine the concentration of analyte from your calibration graph in Exercise 2.2 is the
sample has a reading of 181.
Matrix interference
It is essential that the response of analyte in a sample is exactly the same as the response of the same
concentration of analyte in a standard. If this was not the case, then the answer obtained for the
sample from the calibration graph would be incorrect.
For example, a 100 mg/L standard gives an absorbance of 0.4. Thus, a sample of absorbance
0.4 should have the same concentration. However, matrix elements reduce the absorbance of the
sample, meaning that a higher concentration of analyte gives a similar absorbance to the standard. In
fact, the sample has a concentration of 150 mg/L, which would have been expected to produce an
absorbance of about 0.6.
Matrix interference in the absorption of radiation by the analyte species can be a considerable
problem, less so in molecular spectroscopy than atomic spectroscopy where it is very common. We
will leave discussion about correction of matrix-induced errors until a later chapter.
Determining the linear range for absorbance measurements
Beer's Law indicates that absorbance is directly proportional to concentration without limitation.
However, it has been found that maximum accuracy and linearity occurs for solutions of absorbance
between 0.2 and 0.8 (with outer limits of 0.1 and 1.0).
Reasons for non-adherence to Beer's law outside the range of 0.1-1.0 are numerous, and tend to
vary between techniques. However, there are some general observations that can be made for
molecular absorption in solution:
Sci Inst Analysis (Spectro/Chrom)
2.5
2. Spectroscopic Analysis


at low absorbances, and hence low concentrations, errors arise from the detection of very low
levels of absorption, and through preparation of low concentration standards, where either small
masses are weighed (with the attendant relative errors that this produces) or through a sequence
of dilutions (which accumulate errors); the direction of failure of Beer's Law at these levels is
unpredictable;
at high absorbances, there are errors caused by detection problems of low levels of radiation, and
also the interaction of analyte molecules with each other at the higher concentrations.
We can use this 0.2-0.8 absorbance range to allow us to work out the appropriate concentration range
for a series of standards.
Step 1 - Determining the approximate relationship between A and c
Calculate the constant (k = ab) from the absorbance and concentration of a solution of
approximately known concentration, which must have an absorbance of less than 2.0.
k
A
c
EXAMPLE 2.4A
A 1000 mg/L standard has an absorbance of 3.45. It is diluted roughly 10 to 100, and this
solution has an absorbance of 1.23.
We are not trying to calculate k exactly, so our approximately 100 mg/L solution is good
enough.
k
1.23
 0.0123
100
Step 2 - Determining the concentration of the 0.2 absorbance standard
Calculate the concentration of a solution of the analyte with an absorbance of 0.2, using the
value of k calculated in Step 1. Round the concentration from step 2 to a manageable value
(nearest 5 or 10).
c
0.2
k
EXAMPLE 2.4B
The concentration of a solution with an absorbance of approximately 0.2 is:
c
0.2
 16.3 mg / L  15 mg / L
0.0123
Step 3 - Determining a suitable concentration range for standards
The other standards are 2, 3 and 4 times the concentration of the 0.2 standard. These will
absorbances of approximately 0.4, 0.6 and 0.8,
EXAMPLE 2.4C
The concentration of the other standards are 30, 45 and 60 mg/L.
Sci Inst Analysis (Spectro/Chrom)
2.6
2. Spectroscopic Analysis
Step 4 - Preparation of the standards
The calibration standards can be most conveniently and accurately prepared by dilution of 5,
10, 15 and 20 mL aliquots, respectively, of a more concentrated stock solution (X mg/L).
To calculate the concentration of X, assume that 100 mL of the final standards are
being prepared. Therefore, 5 mL of X is diluted to produce 100 mL of 0.2 Abs std.
X = 20 x conc. of 0.2 std
(if 100 mL vol. flasks used)
EXAMPLE 2.4D
The concentration of the stock solution is 20 x 15 = 300 mg/L.
Step 5 - Preparation of X
If X is greater than 500 mg/L, than it can be prepared directly. Otherwise, it should be
prepared by diluting a more concentrated standard. Using pipettes, the most convenient and
accurate dilutions are 2 (50 => 100), 4 (25 => 100), 5 (20 => 100), 10 (10 => 100) or 20 (5 =>
100). A burette could, however, be used. You need at least 100 mL of stock standard X
(5+10+15+20 = 50).
EXAMPLE 2.4E
To make a 300 mg/L solution requires dilution from a more concentrated solution. 30 mL of
1000 mg/L made up to 100 mL would work, but there are plenty of other options.
CLASS EXERCISE 2.4
A 1000 mg/L standard solution has an absorbance of 3.811. Diluted to 100 mg/L, it gives an
absorbance of 1.128. Determine the appropriate standard concentrations and method of
preparation.
Sci Inst Analysis (Spectro/Chrom)
2.7
2. Spectroscopic Analysis
What You Need To Be Able To Do
 describe the basic idea behind qualitative analysis by spectroscopy
 describe the basic ideas behind quantitative analysis by spectroscopy
 carry out analysis calculations using simple calibration graphs
 describe the steps in a quantitative spectroscopic analysis
 indicate the absorbance range where Beer's law is obeyed
 explain the reasons for non-adherence to Beer's Law outside this region
 determine an appropriate series of standards for a calibration graph
Questions
1.
How can be spectroscopy be used as a means for identifying a particular chemical species?
Which techniques studied in this subject are best for qualitative analysis?
2.
Why are limits placed on the concentration of solutions that can be prepared for use in
quantitative spectrophotometric analysis?
3.
Explain how Beer's Law is used in quantitative spectroscopic analysis. What techniques studied
in this subject are best suited to quantitative analysis?
4.
If a compound has a high absorption coefficient (a in Beer’s Law), what levels of concentrations
(high or low) can be used to analyse it quantitatively?
5.
A 50.0 mL sample of well water is treated with excess thiocyanate to yield the red colour, and
diluted to 100 mL. Standard solutions of the iron-SCN compound are made and their
absorbances recorded as below. Determine the concentration of iron in the sample of well water
if the diluted solution made from the sample exhibited an absorbance of 0.54.
Std conc. (mg/L)
0
5
10
15
6.
In preparing a calibration graph for a series of standard permanganate solutions and unknown
samples, the technician mistakenly recorded percent transmittance. Determine the concentration
of the unknown.
Std conc. (mg/L)
0
1
2
3
4
Unknown
7.
Absorbance
0.00
0.24
0.48
0.72
% Transmittance
100
66
44
29
19
40
Nickel levels in contaminated soil were analysed. A calibration graph was produced from the
data below. A steel sample weighing 5.437 g was dissolved, and the solution diluted to 250
mL. A 10.0 mL aliquot of this was further diluted to 100 mL, and this registered an intensity of
264. What was the percentage of nickel in the steel?
Std conc. (mg/L)
0
4
8
12
Sci Inst Analysis (Spectro/Chrom)
Intensity
1
171
353
516
2.8
2. Spectroscopic Analysis
8.
Manganese (II) ions can be analysed by spectrophotometry if converted by oxidation to
permanganate. Aliquots of a 100 mg/L stock solution of permanganate were diluted to 100.0
mL, their absorbances measured and a calibration graph prepared. 1.0381 g of steel was
dissolved, treated with an oxidant and diluted to 100 mL. A 10.0 mL aliquot of this solution
was further diluted to 100 mL, and its absorbance was determined to be 0.324. Calculate the %
Mn in the steel.
Vol. of 100 mg/L std (mL)
0
5
10
20
9.
The iron content of meat was analysed by atomic absorption spectroscopy. 63.8539 g of meat
was decomposed, and the iron-containing residue dissolved in dilute acid and the solution made
up to 100.0 mL. Iron standards were prepared, and the absorbances of each solution measured.
Calculate the level of iron in the meat sample (mg/kg). The results are given below:
Std conc. (mg/L)
0
10
20
30
40
Sample
10.
Absorbance
0.000
0.182
0.355
0.719
Absorbance
0.000
0.136
0.278
0.404
0.551
0.359
The green colouring in plants is due to the compound, chlorophyll. Samples of spinach were
analysed for its chlorophyll content by absorbance measurements. Determine the level (in
mg/kg) of chlorophyll in uncooked spinach.
Mass of uncooked spinach sample: 15.728 g
Sample volume: 50.0 mL
Std conc. (mg/L)
0
4
8
12
16
Spinach
11.
Absorbance
0.000
0.149
0.301
0.447
0.609
0.462
The aspirin content of a headache tablet was analysed by ultraviolet spectroscopy. Standards
were prepared by diluting aliquots of a 500 mg/L stock solution of aspirin to 250 mL. Ground
tablets weighing 0.7362 g was dissolved in 200 mL of solution, and a 5 mL aliquot of this
diluted to 250 mL. Given the following data, calculate the %w/w of aspirin in the tablet
mixture.
Vol. of 500 mg/L std (mL)
0
5
10
15
20
Sample
Sci Inst Analysis (Spectro/Chrom)
Absorbance
0.000
0.143
0.269
0.419
0.658
0.442
2.9
2. Spectroscopic Analysis
12.
Lead in industrial effluent was analysed by emission spectroscopy. Determine the concentration
of lead given the information below.
Std conc. (ug/L)
0
40
80
120
160
Sample
13.
Intensity
1
658
1340
2019
2754
1569
Given the following data, determine an appropriate series of standards for a Beer's Law
calibration graph, and the concentrations of any stock and intermediate standards. Also
determine a suitable dilution of the sample for analysis.
(a)
(b)
(c)
(d)
Std conc. (mg/L)
100
50
250
1200
Sci Inst Analysis (Spectro/Chrom)
Std Absorbance
1.583
2.014
0.792
1.825
Sample Absorbance
1.291
0.932
1.339
1.617
2.10