Methods of Analysis and Detection Essay Sample

Methods of Analysis and Detection Pages
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* Separation techniques of analysis (PC / TLC / GLC / Electrophoresis)

* Mass Spectroscopy

* Atomic emission spectroscopy

* UV / Visible absorption spectroscopy

* Combined spectral techniques (NMR / IR / Mass Spectra)

By U6F Jacky Huang 2008/03 OR

A. Introduction to chromatography

All chromatography have the following characteristics:

1. They all have 2 phases, one is a stationary phase and the other one is mobile phase.

2. The dissolved compounds – solutes.

3. There are 2 possible mechanisms, one is partition and the other one is adsorption.

In partition:

The solutes move between the 2 phases. If it is in the mobile phase, the solutes will moves with it. Therefore, if we do spend more time in the mobile phase, it can move further.

In Adsorption:

The stationary phase is usually a polar solid and the solutes are polar molecules. The polar molecules will not enter the stationary phase, but it will hold on the surface of the polar stationary phase.

B. Paper Chromatography – The most basic chromatography

Stationary Phase: Cellulose fiber in the filter paper

Mobile Phase: Liquid Solvent

In paper chromatography, coloured compounds can be separated into many colours, but in some cases, colourless spots are involved. When this happen, chemicals like NINHYDRIN will be sprayed to those substances, and those substances will form coloured complexes with these coloured compounds.

The solutes can be identified in 2 ways:

1. Comparison to the reference compounds

2. Calculation of Retardation factor =

D moved by centre of solute spots

D moved by front of mobile phase

Sometimes a 2-way chromatography is carried out to ensure that all the solutions are fully separated.

C. Thin layer Chromatography – TLC- Simple, Cheap and Reliable

Stationary Phase: Thin layer of silica

Mobile Phase: Liquid (Silica Gel)

TLC is a very similar technique as paper chromatography, but it is 3 times faster than paper chromatography, works with small amount of sample, a wider range of mixtures separated.

In the stationary phase, the silica is heated to remove water, these acts as a polar solid and the solutes are transported by adsorption in mobile phase.

(When dry, both phases will be likely to occur)

However, silica will attract water and the presence of water becomes the stationary phase and the solutes are separated by partition.

Actual uses: Clinical diagnosis / Forensic testing / Quality control.

D. Gas/Liquid Chromatography – Use for very small sample

Stationary Phase: THE SAMPLE itself

Mobile Phase: Inert Gases or carrier gases

* GLC is worked by partition,

* Flows thru the column of stationary phase

* Separated due to different solubility,

* Or volatility,

* If the stationary phase is non polar –> Volatility.

* Stationary is polar –> polar molecules retain–> Take longer to appear.

The sample will be detected by the detector and appear on chromatogram.

Chromatogram tells us how much of a component is present, the area under the graph being related to the amount.

So how to Determine the % composition of a mixture by GLC:

The area of a peak = 1/2 (Base Height)

Therefore, a chromatogram must show all the components and the detector are respond equally to all components.

Green Tea Separation in GLC

? Retention time: Time between injection and appearance of a component.?

The retention time is specific to particular components for the same conditions:

*

* Same carrier gas

* Same flow rate

* Same stationary phase

* Same temperature

E. Mass Spectroscopy – A high resolution one (Double focusing)

In the detection of mass spectra, the following condition used in the experiment must be constant:

* The same Temperature

* The same ionizing voltage

* The same type of instrument

Analysis of an ion determine the Ar

The mass spectrum for zirconium

Number of isotope: 5 – 5 peaks can be observed.

Abundance of isotope:

Zirconium-9051.5%

Zirconium-9111.2 %

Zirconium-9217.1 %

Zirconium-9417.4 %

Zirconium-962.8 %

Relative atomic mass =

{900.515 + 910.112 + 920.171+ 940.174 + 960.028} / 100 = 91.3

Analysis of compounds – Determine Mr and formulae

In high resolution mass spectra, we can examine the relative isotopic mass up to 4 d.p.

1H

1.0078

14N

14.0031

12C

12.0000

16O

15.9949

This will be given in the exam

So we will be able to distinguish 2 compounds have similar Mr by using mass spectra analysis.

E.g. Mr= 44 – Propene, C3H8, & ethanal, CH3CHO

In a high resolution mass spectrometer, the molecular ion peaks for the two compounds give the following m/z values up to 4 d.p.

C3H8

44.0624

CH3CHO

44.0261

M peak / M+1 peak / M+2 peak / M+4 peak

Peak

Original ions

Mr of isotope

Peak Height

M

Original Peak

M+1

Carbon -12

Carbon -13

Too small

M+2

Chlorine – 35

Chlorine – 37

3:1

M+4

Bromine – 79

Bromine – 81

1:1

In mass spectra, we will be also to be able to work out the number of carbon inside a compound. (Values will be given in the exam)

E.g. the M peak at m/e 112 has a height of 63.0mm and the M+1 peak at m/e 113 has a height of 4.1mm. (1.1% that a 13C will be presence)

Ans: Height of (M+1) Height of (M) peak 100

4.1mm / 63.0 mm 100 = 6.5 %

6.5 / 1.1 = 5 = 5 carbons in the compounds

E.g. Mass Spectra for CH3CH2CH2Br

m/z

Peak name

Responsible ions

122

M

M Peak

123

M+1

M Peak + C – 13

124

M+2

M Peak + Br – 81

Actual uses:

* Traces of toxins in food or water

* Testing drugs on racehorse

* Cancer Treatment – Urine analysis

F. Electrophoresis – Biological analysis

Electrophoresis is separation of charged particles by their movement in an electric field. The visual presentation is by using – Electropherogram.

Although the amino acid solution is colourless, its position after a time can be found by spraying it with a solution of ninhydrin. If the paper is allowed to dry and then heated gently, a coloured spot will be observed.

In zone Electrophoresis, the mixture is a solution or gel,

The supporting medium is a filter paper made by cellulose,

Separated by the charge and the size of compounds.

A highly charged ion will move faster in an electric field.

A larger ion moves slowly across the supporting medium.

The electrophoresis need to be worked in a stable environment. The temperature and pH need to be controlled by adding buffer solution.

Normal amino acids will be look like the one on the left. If there is an internal transfer of a hydrogen ion, a zwitterion will be released.

If the pH has increased, the zwitterions will then become,

If the pH has decreased, the zwitterions will then become,

One of the uses for electrophoresis will be used for genetic fingerprint:

1. Take a DNA strand

2. Cut – Restriction Enzyme

3. Separated by Electrophoresis

4. Transferred to Nylon membrane

5. P-32 DNA probes bind with DNA band

6. X – Ray exposed, Results.

There are also other uses for electrophoresis:

Forensic Science / GM Food / Medical Research / Establishing Relationships

G. Atomic Emission Spectroscopy – Interaction of EMR

This type of spectroscopy concerns the electromagnetic radiation with matter. The EMR have different frequency and different energy.

E.g. the radio waves change the orientation of spinning nuclei in a magnetic field and infrared cause’s changes in vibrational energy.

Calculation for E= h f and C= f ?

If a molecule absorbs EMR of 356nm, calculate the energy associated with EMR, directly to the energy gap between E1 and E2 or ?E

In the Wave equation we know: C = f ?

C = 2.998 x 108 // f = frequency //? = wavelength

2.998 x 108 = f x 3.567 –> f = 8.42 x1014 Hz.

This means the energy associated with the EMR –> E =h f

E = Energy // h = Plancks’ constant = 6.63 x10-34 // f = frequency

E = 6.63 x 10-34 x 8.42 x 1014 = 5.58 x 10-19J

If we want to calculate the energy for one mole we need to multiply the energy level by the Avogadro constant (6.023 x 1023).

E = 5.58 x 10-19 x 6.023 x1023 = 336223.6J = 336kJ mol -1

Explain the Hydrogen absorption/emission spectrum

The molecular H (g) at a low pressure is bombarded with electrons in a discharging tube. The bombardment dissociates the hydrogen molecules to give atom as well as excited atoms so that they are no longer in its ground state. The excited atoms relax and the EMR seen as a pink glow.

This radiation is then passing thru a spectrometer which splits the EMR according to frequency.

Hydrogen Absorption Spectrum

* Black lines on continuous spectrum – EMR absorbed.

Hydrogen Emission Spectrum

* Discrete lines – EMR missing from the absorption spectra.

When the energy of the photons exactly matches the energy difference between the ground state and an excited state, the radiation is absorbed by the atoms. An atom in an excited state will rapidly – relax – this means it will revert back to its ground energy level. So the EMR used this characteristic to let the atom to lose a photon.

Calculate the ionization energy

For hydrogen the convergence limit is 3.30 x 1015 Hz, so we can calculate the IE by E=h x f:

E = 6.63 x 10-34 x 3.30 x1015 x 6.02 x1023

E = 1318 kJ mol-1

Quantitative Analysis of flame emission spectroscopy

This is a method that we used for finding how much metal ion is present in clinical diagnosis. E.g. determination of Na+ in blood:

1. Moisten a nichrome or platinum wire with conc. HCl,

2. Dip into the sample and place the wire to a Bunsen flame,

3. If Na+ is present, a yellow / orange colour will be observed.

H. UV/Visible absorption spectroscopy – changes in electronic structure

An Electromagnetic spectrum see section G on p.9

Here are 2 definitions that you need to know:

* Chromophore – are unsaturated groups of atom in organic molecules that absorb radiation mainly in the UV/Vis regions.

* Conjugation – the system of alternate double bond and single bonds in a molecule. Organic molecules with extended conjugation absorb radiation in the visible part of the spectrum and are therefore colored.

The wavelength at which organic molecules absorb radiation depends on how tightly their electrons are bound.

* Bonds – the shared electrons in single bonds do not easily give absorption spectra. But the electrons in double and triple bonds are more loosely held and easier to get excited.

* Lone pair – Lone pair that is localized around atoms (O2 / N) is loosely bound and are absorb in the same region of the spectrum.

Ultraviolet and visible radiation are absorbed by the outmost electrons in organic molecules, which then move from lower to higher energy levels.

The outmost electrons can be:

* Electrons directly involved in bond formation,

* Unshared outer electrons.

?It is the movement of electrons between molecular orbital give rises to UV/Vis spectra.?

Some of the organic molecules have colours, because, they absorb certain frequencies of radiation within the visible region. (Wavelength 400 – 750 nm)

The frequencies of radiation absorbed by various chromophores can be used for the identification.

The wavelengths and the intensities of the absorption bands due to chromophores are altered by the conjugation in molecules and delocalized electrons.

In benzene, the electrons are evenly distributed

around the rings.

The conjugation of chromophores in a molecule causing ? orbital interact with each other then become delocalized –> Formation of new orbital –> Energy gap reduced –> absorption bands shift to longer wavelength –> intensity increase.

Effects of additional Chromophores will shift to an organic molecule more from the UV region to the visible region. See table below for more details.

Number of CH=CH

Region of Electronic transition

Colour

1 / 2

UV

Colourless

3

Visible

Yellow

15

Visible

Dark Green

Dyes and colour change in acid – base indicators

1. Phenolphthalein

Acidic = Colourless // Alkaline = Red

2. Methyl Orange

Acidic = Red // Alkaline = Yellow

3. Azo Dyes – coloured due to presence of chromophores

Benzene Diazonium + Napthalene-2-ol –> Red azo dyes + HCl

I. Combined spectral Techniques – Analysis of mixture and compound

Now, we will be able to use evidence from N.M.R. / I.R. / Mass Spectra to suggest a probable structure for an unknown compound.

The following questions are from the OCR text book – methods of analysis and detection.

See page 50 -53 in OCR Chemistry – Methods of Analysis and Detection.

J. Infrared Spectroscopy (IR) – using wave numbers

Here, IR used the characteristics that different bonds in compounds absorb different frequencies in the infrared region of the spectrum. In infrared analysis the wave numbers are used instead of frequencies or wavelengths.

In IR, note the following:

Compound

Things to note

Alcohol

Peak at 3400 (OH)

Carboxylic acid

Peak at 2500-3500 (Broad for OH)

Peak at 1700 (C=O)

Peak at 1000 -1250 (C-O)

Carbonyl Compounds

Peak at 1700 (C=O)

Ester

Peak at 1700 (C=O)

Example of an IR spectrum – Ethanoic Acid

K. Nuclear Magnetic Resonance Spectroscopy (NMR) -Splitting Pattern

The peak in the NMR gives many information of compound; this is a summary in a table of what it tells you:

What it tells you

How this is shown in the spectrum

How many types of proton present

The number of distant peak = number of proton

How many of each type of proton

The relative peak area under the peak = relative number of proton

What type of proton

The value of ? and ten use the table provided in the exam

The number of chemically different protons adjacent to a particular type of proton

By the splitting of the distant peak. If there is n chemically different type of protons next to a particular type of proton then the peak for that proton is split to n+1 peaks

Use D2O with -OH group

Deuterium is an isotope of hydrogen that does not produce an NMR peak.

If an alcohol or carboxylic acid is shaken with D2O, the H of the OH group is replaced by D and the peak disappears from the spectrum.

Example of a NMR spectrum – C4H6O2

1. 4 peaks and 4 Cs, 2 of them are in the same environment.

2. The peak just over 50 must be C-O

3. 2 peaks around 130 must be 2 carbons at either end of a C=C.

4. The peak at just less then 170 is the C in a C=O.

5. By using IR and Mass Spectroscopy, we can work out the structural formula.

L. Additional Notes and Materials

Colour wheel can be used to determine a compounds colour:

Colours

nm

Red

625-740

Yellow

565-590

Green

520-565

Cyan

500-520

Blue

435-500

Magneta

380-435

Colours directly opposite each other on the colour wheel are said to be complementary colours. Blue and yellow are complementary colours; red and cyan are complementary; and so are green and magenta.

E.g. Beta-carotene absorbs throughout the ultra-violet region into the violet – but particularly strongly in the visible region between about 400 and 500 nm with a peak about 470 nm and if we use the table above, we know it will absorb blue colour so the colour we will see will between red and yellow.

M. Bibliography and Reference

1. http://www.chemguide .co.uk

2. A-Z Chemistry Hand book

3. Revise A2 Chemistry by Helen Eccles and Mike Wooster

4. OCR Text Book – Methods of analysis and detection

5. OCR Past Examination Paper – 2002 ~ 2007

6. Sherborne School resources

7. Mr. N.C. Scorer – Teachers in chemistry of Sherborne School

8. Yahoo! Image Search Engine

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