Area of Study 1 – Chemical Analysis

Why do we need to analyze?

Identifying Chemicals

Using chemical reactions to isolate and

identify chemicals

Using chemical reactions to find the

quantities of chemicals that are present

in certain products

Instrumental Analysis of chemicals

Extended investigation on the analysis of

a commercial lawn fertilizer.

› 4 experimental activities that focus on

analyzing and determining the amounts of sulfates, nitrogen, iron and phosphates in the

fertilizer

› A summary and conclusion of the findings of

the investigation as a whole.

What are consumers? Anything that will use a product to ensure its

survival and comfort – Us!!

What are consumer items? Foods, cleaning products, clothes, fuel,

vehicles, buildings

All consumer products must be Quality

Control tested to ensure a consistent

quality is met with all products

When determining the composition of an

item we need to know two things:- What chemicals are present

How much of each chemical is present

Qualitative analysis finds out what

chemicals are present in the product

Eg. The list of ingredients on the side

of a food packet

Qualitative analysis is effective for

identifying toxins and impurities, food

allergies, reactivity with other products

Quantitative analysis finds out how much of each chemical is in a product

Eg. The nutritional information on the side of

a food packet

Quantitative analysis is effective for complying with health and Government requirements, correct dietary intake, alcohol consumption (.05)

Water is a component of most consumer

products whether it is added or has

occurred naturally.

Measuring the water content in a

product can be achieved by heating a

sample of the product to evapourate

the water and then comparing the

change in mass.

Weigh a sample

Heat the sample in an

oven at 110°

Allow the sample to cool in

a desiccator

Reweigh

The cycle must be

repeated until a

constant mass is

reached

As atoms and molecules vary in size and can be present in different forms (solid, liquid, gas) we must always convert our quantities to Number of Moles (n) or amount.

Each atom has its own molar mass. The molecular mass (M) of a molecule is the sum of the molar masses of each atom within the molecule.

Eg. The molecular mass of ammonia (NH4) is:

Nitrogen = 14 Hydrogen = 1

M(NH4) = 14 + 4x1 = 18

When given the mass of a particular

substance, the amount can be

calculated by the formula

M

mn

Where n= amount

m= mass

M= molecular mass

Calculate the amount

of H2O molecules in

9.0g of water.

n= ?

m= 9.0

M= 1x2 + 16 = 18

M

mn

18

9n

moln 5.0

The volume of one mole of gas will depend on the

temperature and pressure.

At Standard Temperature and Pressure (STP, 0°C and

1atm) the molar volume, Vm, is 22.4.

At Standard Laboratory Conditions (SLC, 25°C and 1atm)

the molar volume, Vm, is 24.5.

To calculate the amount of a gas at these conditions we

use the formula:

mV

Vn

Where n = amount

V = Volume (L)

Vm = Molar Vol

Calculate the amount of sulfur dioxide in 10.0L

of the gas measured at STP.

4.22

10n

moln 446.0

mV

Vn

n = ?

V = 10

Vm = 22.4

When the substance is in a gaseous state we

use the General Gas Equation to calculate the

amount of gas present.

nRTPV Where P = Pressure (kPa)

V = Volume (L)

R = 8.31

T = Temperature in Kelvin (K)

A steel cylinder with a volume of 30.0L is filled with

nitrogen gas to a pressure of 2atm at 25°C. What

mass of nitrogen does the cylinder contain?

n = ?

P = 2x101.3 = 202.6

V = 30

R = 8.31

T = 25+273 = 298

PV = nRT

202.6x30 = nx8.31x298

6078 = nx2476.38

6078 = n

2476.38

n = 2.45mol

The question asked for the mass of nitrogen

present!!

M

mn

2845.2

m

n = 2.45

m = ?

M = 14x2 = 28

(nitrogen gas is N2)

m = 2.45x28

m = 68.7g

Using chemical analysis we can find the

simplest whole number ratio of atoms in

a compound.

This is called the empirical formula.

Empirical formulas are determined

experimentally, usually by finding the

mass of each element in a given

compound.

Example 1

Chemical analysis of an oxide of sulfur present in

the gaseous emissions from a factory shows that it

contains 40% sulfur. Find its empirical formula.

Example 2

Find the empirical formula of a hydrocarbon

extracted from natural gas if, upon complete

combustion, a sample of the hydrocarbon

produces 7.75L of carbon dioxide, measured at

SLC, and 7.59g of water.

Example 3

After the heating of 2.95g of hydrated

magnesium sulfate (MgSO4.xH2O), the

mass of the residue is found to be 1.44g.

Find the value of x in the formula of the

compound.

The molecular formula of a compound

gives the actual number of atoms

present in a molecule of a compound.

For example, the empirical formula of

glucose is CH2O. Its molecular formula is

6 times larger – C6H12O6.

A molecular formula can be deduced

from the empirical formula of a

compound if the molar mass is known.

Using a balanced equation we can use ratios to determine the mass of product produced or the mass of reactants consumed.

The following steps outline this process: Write balanced equation.

Calculate the number of mol from the known quantities

Find the ratio of amounts.

Calculate n of unknown.

Calculate the unknown mass.

Calculate the mass of lead iodide that

can be made from a reaction using

30.0g of potassium iodide and lead

nitrate.

Step 1: Write a balanced equation

Pb(NO3)(aq) + KI(aq) PbI2(s) +

KNO3(aq)

2 2

)(

)()(

KIM

KImKIn

n= ?

m= 30.0

M= 39 + 127 = 166

166

30)( KIn

molKIn 181.0)(

Pb(NO3)(aq) + 2KI(aq) PbI2(s) + 2 KNO3(aq)

2

1

)(

)( 2 KIn

PbIn

)(2

1)( 2 KInPbIn

mol0905.0181.02

1

n= 0.0905

m= ?

M= 207 + 2x127 = 461

)(

)()(

2

22

PbIM

PbImPbIn

)()()( 222 PbIMPbInPbIm

4610905.0)( 2 PbIm

g7.41

In most reactions, one of the reactants is usually in

excess.

When doing calculations on these reactions, the

calculations must be based around the reactant

that is totally consumed.

This reactant is called the limiting reactant.

Example

Calculate the mass of silver bromide that can be

formed if a solution containing 15.0g of silver nitrate

is allowed to react with a solution containing 10.0g

of calcium bromide.

When trying to determine the amount of

a particular ingredient in a substance we

use a process called Gravimetric

Analysis.

The substance in question can be

isolated by dissolving it water and

filtering away all excess solids.

The substance can then be isolated by

reacting it with a chemical that will form

a precipitate.

The precipitate can then be filtered and washed thoroughly to remove any excess ions

The precipitate can then be dried and weighed to determine the mass of the product formed from the reaction.

Figure 2.7 Steps in the analysis

of the chloride content in peanut

butter.

When doing a Gravimetric Analysis it is

important that you make the correct

choices when attempting to form a

precipitate. The precipitate should:

•Have a known formula

•Have low solubility

•Be stable when heated

•Not form other precipitates with other ions that may be

present

A 7.802g sample of baby cereal was blended with water and filtered. The solution obtained was mixed with excess silver nitrate solution, causing silver chloride solution to precipitate. The precipitate was collected by filtration, dried and weighed. A mass of 0.112g was obtained. What is the percentage of salt in the baby food?

Full equation for the precipitation reaction is:

3)(3 )()( NaNOsAgClaqNaClAgNO aq

)(

)()(

AgClM

AgClmAgCln

9.130

112.0)( AgCln

n(AgCl) = ?

m(AgCl) = 0.112g

M(AgCl) = 107.9 + 23 = 130.9

molAgCln 000855.0)(

From the equation:

3)(3 )()( NaNOsAgClaqNaClAgNO aq

1

1

)(

)(

AgCln

NaCln

)(1)( AgClnNaCln

molNaCln 000855.0)(

)(

)()(

NaClM

NaClmNaCln

)()()( NaClMNaClnNaClm

5.58000855.0)( NaCln

gNaClm 050.0)(

n = 0.000855

m = ?

M = 23 + 35.5 = 58.5

So, the 7.802g sample contains 0.05g of NaCl.

Therefore:

100802.7

05.0% NaCl

%64.0

Concentration is how much solute

dissolved in a solvent.

Molar concentration or molarity is how

many mole of solute in one L of solvent

given in the units mol L-1(moles per litre)

or M.

When given the concentration of a

particular substance, the amount can

be calculated by the formula:

Vcn Where:

n = number of moles

c = concentration

V = volume in Litres

Example:

Calculate the molarity (concentration) of a 250mL solution containing 0.5mol of HCl.

Vcn

250.05.0 c

250.0

5.0c

Mc 2

n = 0.5

c = ?

V = 250mL

= 0.250L

When converting molarity to gL-1 you

simply multiply by the molar mass

Eg. A 2.0M solution of HCl (molar mass of HCl is

1+ 35.5 = 36.5)

2.0 x 36.5 = 73gL-1

When converting gL-1 to molarity you

simply divide by the molar mass

Eg. A solution containing 18.25gL-1 of HCl

18.25/36.5 = 0.5M

Solutions called Primary Standards can

be to test the contents of unknown

substances.

For Primary standards to be useful they

should:

› Have a known formula

› Be readily obtainable in pure form

› Be easy to store

Figure 3.4 Steps in preparing a standard solution. A known mass of solute is used to make a known volume of solution.

Volumetric analysis is used to find the

concentration of an unknown solution by reacting

a measured volume of a standard solution with a

measured volume of the unknown solution until it

reaches equivalence point (just reacted

completely or neutralised).

An acid-base indicator is used to show when the

equivalence point is reached.

Figure 3.6 A pipette must

always be filled using a

safety pipette filler. When

the pipette is filled correctly,

the meniscus sits on the

graduation line.

Figure 3.7 The volume of

solution in a burette is read

going down from the top of

the burette.

Acids and bases can be defined in the

following ways:

Acids are proton donors

Bases are proton acceptors

Acid-Base reactions involve the transfer

of a proton from an acid to a base.

Consider the following Acid-Base reaction:

)()(3)(2)( aqaqlaq ClOHOHHCl

HCl is a strong acid as it ionises completely in water

to for H3O+ and Cl-.

•During the reaction there is a proton (H+) transfer

between the HCl and the H2O.

•HCl acts as the acid and donates the proton to form

Cl-.

•H2O acts as a base and accepts the proton to form

H3O+.

The choice of an appropriate acid-base

indicator is very important when

conducting a volumetric analysis.

An indicator will change colour at a

particular pH. This is called the end

point.

The end point of the indicator must

closely match the equivalence point of

the acid-base reaction.

Figure 4.4 pH curves showing change of pH during a titration of a a

strong base with a strong acid, and b a weak base with a strong acid.

Phenolphthalein, which changes colour in the pH range 8.2–10, gives

a sharp end point in a but a broad end point in b. Methyl orange, which

changes colour between pH 3.1 and 4.5, would be a more suitable

indicator for the second titration.

a b

Figure 4.7 pH Change in pH during a titration of:

a a strong acid with a strong base; b a strong acid with a weak base

c weak acid with a strong base; d weak acid with a weak base.

a b

c d

A commercial concrete cleaner contains

concentrated hydrochloric acid. A 25.0mL volume

of cleaner was diluted to 250.0mL in a volumetric

flask.

A 20.0mL aliquot of 0.4480M sodium carbonate

solution was placed in a conical flask. Methyl

Orange indicator was added and the solution was

titrated with the diluted cleaner.

The indicator changed permanently from yellow to

pink when 19.84mL of the cleaner was added.

Calculate the concentration of hydrochloric acid

in the concrete cleaner.

First, start with a balanced equation:

)(2)(2)()(32)( 22 glaqaqaq COOHNaClCONaHCl

)()()( 323232 CONaVCONacCONan

Calculate the number of moles in the known solution:

0200.04480.0)( 32 CONan

mol008960.0n(Na2CO3)= ?

c(Na2CO3)= 0.4440

V(Na2CO3)= 20.0mL

= 0.0200L

Using the equation find the number of moles of the unknown:

)(2)(2)()(32)( 22 glaqaqaq COOHNaClCONaHCl

1

2

)(

)(

32

CONan

HCln

)(2)( 32CONanHCln

008960.02)( HCln

mol01792.0

Calculate the concentration of the unknown:

MHClc

HClc

HClc

HClVHClcHCln

9032.0)(

01984.0

01792.0)(

01984.0)(01792.0

)()()(

n(HCl)=0.01792

c(HCl)=?

V(HCl)=19.84mL

=0.01984L

Because the original cleaner had been

diluted by a factor of 250.0/25.0 (10

times) the concentration of the original

cleaner must be:

0.9032 x 10 = 9.032M

The concrete cleaner has a HCl

concentration of 9.032M

Back titrations are used to find the end point of

very weak acids and bases that do not give

definite colour changes.

The weak acid is first reacted with excess base

A titration can then be carried out to determine

the amount of base that didn’t react.

The difference between the amount that reacted

and the original amount of excess base added will

be the amount that reacted with the weak acid.

The amount of the weak acid can then be

determined.

A 1.50g sample of fertiliser was boiled

with 25.00mL of 0.9987M sodium

hydroxide solution. Once no further

ammonia gas was evolved from the

mixture, it was cooled and titrated with

0.2132M hydrochloric acid, using

phenolphthalein as an indicator. A titre

of 19.78mL was required.

Calculate the percentage of ammonium

ions in the fertiliser.

Write a balanced equation for the titration:

)(2)()()( laqaqaq OHNaClNaOHHCl

molHCln

HCln

HClVHClcHCln

004217.0)(

01978.02132.0)(

)()()(

Find the amount of the known substance:

n(HCl)=?

c(HCl)=0.2132

V(HCl)=19.78ml

=0.01978L

Use the equation to calculate the amount of the base

used:

)(2)()()( laqaqaq OHNaClNaOHHCl

molNaOHn

HClnNaOHn

HCln

NaOHn

004217.0)(

)()(

1

1

)(

)(

There were 0.004217mol reacted in the second

reaction, therefore there was 0.004217mol unreacted

in the initial reaction.

Use this value to calculate the amount of base

that reacted with the weak acid in the original

reaction:

The amount of base added in the original reaction:

molNaOHn

NaOHn

NaOHVNaOHcNaOHn

02497.0)(

025.09987.0)(

)()()(

The difference between the amount added and the

amount unreacted will be the amount of base that

reacted with the weak acid.

n(NaOH) used originally = 0.02497 – 0.004217

= 0.02075mol

Use a balanced equation for the original

reaction to calculate the amount of the weak

acid that reacted:

)(2)(3)()(4 laqaqaqOHNHOHNH

molNHn

OHnNHn

OHn

NHn

02074.0)(

)()(

1

1

)(

)(

4

4

4

To calculate the percentage we must first

calculate the mass:

gNHm

NHm

NHm

NHM

NHmNHn

3737.0)(

1802075.0)(

18

)(02075.0

)(

)()(

4

4

4

4

44

n(NH+4)=0.02075

m(NH+4)= ?

M(NH+4)= 14+4 = 18

The percentage mass can then be calculated:

%91.2410050.1

3737.0

Oxidation/Reduction reactions (REDOX)

are two reactions that occur

simultaneously.

The oxidation part of a redox reaction is

where a substance loses electrons (OIL –

Oxidation Is Loss)

The reduction part of a redox reaction is

where a substance gains electrons (RIG –

Reduction Is Gain)

Consider the Redox reaction:

)()(2)( 22 sgs MgOOMg

eMgMg 22

The magnesium will undergo oxidation by losing

electrons to form magnesium ions:

2

2 24 OeO

The oxygen gas will undergo reduction by gaining

electrons to form oxide ions:

Redox reactions can be identified by

observing a change in Oxidation

Numbers of the substances involved.

Oxidation numbers are determined using the following rules:

1. Free elements have an oxidation number equal to 0. Eg Na(s), C(s), Cl2(g).

2. In ionic compounds the oxidation number s equal to the charge on the ion. Eg CaCl2: Ca2+=+2 and Cl- = -1.

3. Oxygen usually has an oxidation number of

-2 and hydrogen has +1 but there are a few exceptions.

4. The sum of oxidation numbers in a

neutral compound is 0 and in a

polyatomic ion is equal to the charge of

the ion.

By defining oxidation numbers for the

atoms involved in a reaction we can

look for increases and decreases in

oxidation numbers.

An increase in oxidation number means

the element has undergone oxidation.

A decrease in oxidation number means

the element has undergone reduction.

For the reaction:

)(2)(2)( 22 ggg COOCO +2 -2 0 +4 -2

Carbon in the carbon monoxide has gone from +2

to +4 which means that oxidation has occurred.

Oxygen in the oxygen gas has gone from 0 to -2

which means that reduction has occurred.

As both oxidation and reduction have occurred the

reaction is a redox reaction.

Although most half equations are quite easy to

write, some involving polyatomic ions can be more

difficult.

The following steps will make balancing these half

equations easier:

1. Balance all elements except O and H in the half

equation.

2. Balance the O atoms by adding water.

3. Balance the H atoms by adding H+ ions.

4. Balance the charge by adding electrons (e) and

then add states.

A green solution containing Fe2+ ions is mixed

with a purple solution containing MnO4- ions.

Fe3+ and Mn2+ ions are formed.

Write a balanced equation for this reaction.

The half equation involving the iron ions is quite

simple:

Fe2+(aq) Fe3+

+ e-

The half equation involving the manganese is a

little more difficult:

Step 1: Balance all elements except for O and H

MnO4- Mn2+

Step 2: Balance O by adding water

MnO4- Mn2+ + 4H2O

Step 3: Balance H atom by adding H+

MnO4- + 8H+ Mn2+ + 4H2O

Step 4: Balance the charge with electrons

MnO4- + 8H+ + 5e- Mn2+ + 4H2O

To complete the full balanced equation we must balance the electrons in each half equation:

Fe2+ Fe3+ + e-

MnO4- + 8H+ + 5e- Mn2+ + 4H2O

5Fe2+ 5Fe3+ + 5e-

Combine the two equations: MnO4

- + 8H+ + 5e- + 5Fe2+ Mn2+ + 4H2O + 5Fe3+ + 5e-

Cancel out the electrons:

MnO4- + 8H+ + 5Fe2+ Mn2+ + 4H2O + 5Fe3+

(X5)

Volumetric analysis to determine unknown concentration or mass can be performed in the same way that it is with acid-base reactions.

Equivalence point of most redox reactions will be indicated by a colour change in the reacting solutions.

Volumetric analysis involving redox reactions are used to determine the composition of substances such as Fruit Juice, Wine, Bleach and Hair Bleach.

Chromatography is used to separate the substances present in a mixture.

A chromatogram consists of a stationary phase (filter paper or powdered alumina) and a mobile phase or solvent (usually water).

The solvent breaks the bonds between the sample and the stationary phase as it absorbs.

Each different substance wants to ‘re-bond’ or desorb back on to the stationary phase.

Each different substance desorbs at a different rate.

Each different substance can be determined by

how far it has traveled along the stationary

phase in comparison to how far the mobile

phase has traveled.

This is called the Rf value. This can be

calculated by:

Rf = distance moved from origin by

component

distance moved from origin by solvent

Every component has its own individual Rf

value.

HPLC is similar to paper or thin-layer chromatography as it passes a liquid mobile phase (sample) through a tightly packed solid stationary phase under pressure.

The constituents separate and pass through a detector at the bottom of the column.

Each constituent has a unique retention time (Rt value).

This method is used to detect certain drugs present in blood.

GLC is the most sensitive of the chromatographic techniques.

It can detect as little as 10-2g of a compound.

Sample is injected into carrier gas (usually N) and heated to vaporize. It then sent through the column which runs through an oven in a series of tubes filled with a liquid stationary phase of hydrocarbon.

Don’t just copy things out without understanding them!

Retention times are taken and a Flame Ionisation Detector is used to detect which compounds are present.

Mainly used to detect the presence of illegal substances in the blood.

The chromatogram shows the response of the

detector against time.

The solvent gives a flat baseline and as a

component goes through the detector a peak will

form.

The time taken to pass through the column is called

the retention time (Rt).

Each component has its own unique retention time

when done at the same conditions.

A comparison is conducted against

chromatograms of known substances to determine

the composition of samples.

Figure 6.11 Gas

chromatogram of a petrol

sample.

Figure 6.12 Chromatogram

of a reference sample

containing a mixture

of butane, 2-methylbutane,

hexane, benzene and 2-

methylhexane.

Spiking a sample with a known

substance will also help define individual

peaks as the area under the peak will

increase with increased concentration.

Figure 6.13 Petrol

sample spiked with

benzene.

The concentrations of unknown samples

can also be found using gas

chromatography by comparing the

areas under the peaks of accurately

known samples.

Example

The concentration of benzene in a sample of petrol was

determined by gas chromatography. A series of standards

with accurately known concentrations of benzene were run

under the same conditions as the sample.

Figure 6.14

Chromatogram of

standard samples of

benzene and an unknown

sample.

A calibration graph can be plotted using the peak

areas of the standards. Using the height of the

sample it can then be found that the

concentration of the sample is 0.6%.

All forms of spectroscopy use a part of

the electromagnetic spectrum to give

us information about the materials

around us.

Different types of electromagnetic

radiation will react with atoms and

molecules in specific ways.

The various spectroscopic techniques provide us with information about:

› The type of atom or molecule present.

› How much of an atom or molecule is present.

› The structure and bonding of the molecule.

Spectroscopic techniques utilise the fact that:

› Particles absorb and emit electromagnetic radiation of specific energies.

› Particles undergo a change when they absorb electromagnetic radiation.

› Different parts of the electromagnetic spectrum affect different parts of the particle.

Flame tests identify different metals in

chemicals by the colour they produce

Electrons in inner shells absorb heat

energy and jump to an outer shell. The

atom becomes unstable and jumps

back emitting the absorbed energy as

light

Different metals will produce different

colours

Atomic Emission Spectroscopy also identifies metals in samples but is much more accurate than the flame tests

AES uses a prism to break the light into wavelengths. This produces an emission spectrum that is very much like a DNA fingerprint

AAS is one of the most widely used modern instrumental techniques.

Atoms will absorb light that is emitted at a certain wavelength that has the energy to promote an electron to a higher energy level.

Light of a particular wavelength is passed through a sample that is sprayed into a flame and then the absorption is measured and compared to calibration graph.

Used for analysis of deficiency of metals in blood and urine, toxic metals in foods, testing for pollution, finding the amounts and types of metals in minerals, soil and oils.

A colorimeter is used to measure the concentration of a metal ions and some simple anions in solutions..

It passes a light beam through the sample and measures the amount of light that passes through.

Concentration of a solution is directly related the amount of light it will absorb.

The samples with higher concentrations will absorb more light.

The absorption is compared to a

Calibration Graph – a graph of the

different absorptions of standard

solutions with known concentrations – to

determine the concentration of the

solution.

This method is similar to colorimetry in that it uses a light beam to measure light absorption.

UV – Visible is more effective in that it uses a monochromator to select specific wavelengths to best detect the presence of specific ions.

A light detector identifies which wavelength the sample absorbs and this is compared with the wavelength absorbed by reference samples.

UVVS is particularly useful for detecting metal ions in uncoloured solutions.

Some uses include detecting substances in blood and urine, and determining the amount of coloured dye in plastics.

Although UV-Spec is mainly used for

finding the concentration of certain

elements within a solution, it can also be

helpful in qualitative analysis.

When a substance absorbs light it

appears coloured.

The colour observed is the complement

of the absorbed colour because this is

what remains to reach our eyes.

Figure 7.16 Visible

spectrum of chlorophyll.

Chlorophyll absorbs

strongly in the violet (420

nm) and red (660 nm)

regions of the spectrum,

so it appears green in

colour.

The energy from Infrared radiation is not enough to promote an electron into a higher energy level but it is enough to cause changes to the bonds in molecules.

Just as electrons can occupy discrete energy levels, molecules occupy discrete vibrational energy levels.

All molecules absorb infrared radiation and will absorb different levels of radiation depending on which bonds have been distorted.

Figure 7.22 Bending (scissoring) in water molecules

Figure 7.23 Stretching and bending motions in diatomic and

triatomic molecules.

Figure 7.24 As the molecule moves to a higher vibrational energy

level the frequency of the stretching vibration increases.

The range of energies absorbed

depends on the strength of the bonds

involved.

The mass of the atoms attached to

the bond also affects the energy

absorbed.

Infrared absorption is measured in

wavenumbers (cm-1).

Wavenumbers are the reciprocal of

wavelengths – 1/λ.

The spectrum produced will show us

areas of absorption particular bonds

and functional groups.

The spectrum will also give us an idea of

other bonds present in the molecule that

will determine the structure.

a

b

No two molecular compounds are identical in bonds and bonding environment and so infrared spectroscopy can give us a characteristic fingerprint for each compound.

The infrared spectrum above 1000cm-1 is used to identify functional groups.

The spectrum at frequencies less than 1000cm-1 is characteristic of a particular compound.

This area is known as the fingerprint region.

Figure 7.32 Typical regions of absorption in the infrared spectrum.

NMR uses energy in the Radio Frequency range of the electromagnetic spectrum.

NMR is one of the most powerful techniques for identifying the structure of complex biochemical molecules.

NMR uses a magnetic field to change the spin state of the nucleus of particular atoms (usually 1H or 13C) within a molecule and measuring the change in radio frequency emitted as it relaxes back to its original spin state.

Because different NMR machines can differ in

results under different lab conditions, the

machines are standardised against a reference

sample, usually tetramethylsilane (TMS).

TMS has a symmetrical, tetrahedral structure

and therefore all H atoms are in an identical

environment.

The difference in energy needed to change

spin state in the sample compared to TMS is

called the chemical shift.

The chemical shift of TMS is defined as zero.

Table 7.9 Some

characteristic

chemical shifts in

proton NMR relative

to TMS (protons

highlighted). (R is

any alkyl group; see

Chapter 9.)

Proton NMR will indicate the bonding

nature of the hydrogen atoms in the

molecule.

In Low Resolution NMR, the number of

peaks is equal to the number of different

bonding environments of the H atoms.

The relative areas under the peaks will

give the ratio of H atoms present in each

environment.

High Resolution NMR shows the NMR in more detail and will give the nature of the neighbouring atoms.

Each peak will be split into finer peaks.

The number of peaks caused by splitting equals n+1 where n is the number of H atoms on the neighbouring atom.

› CH splits the signal from Hs attached to neighbouring atoms into 2 peaks.

› CH2 splits the signal from Hs attached to neighbouring atoms into 3 peaks.

› CH3 splits the signal from Hs attached to neighbouring atoms into 4 peaks.

a

b

Figure 7.46 The likely

structure for the compound

C4H8.

13C NMR is used to identify the

different carbon atom environments in

a molecule.

Mass Spectroscopy is one of the most

common and useful analytical tools used

in combination of other techniques.

MS can be used for quantitative and

qualitative analysis.

The key principle is that a charged particle passing

through a magnetic field is deflected along a

circular path of radius proportional to the mass:

charge ratio (m/e)

The sample enters the tube.

Positive ions are formed in the ionisation chamber

when electrons are dislodged from the sample by

an electron beam.

The ions enter a magnetic field that cause the ions

to move in a curved path with a particular radius

that depends on the m/e.

Only charged particles moving with a

specific radius will reach the detector.

The magnetic field and accelerating

voltage can be adjusted so that other

particles can then reach the detector.

The collector measures the current due

to the ions reaching the detector and

the data is recorded as a mass

spectrum.

A molecular substance will give a range

of peaks in the spectrum.

The peaks can be caused due to the

molecule fragmenting into a large

number of different positive ions.

They are also caused due to the

occurrence of different isotopes of

atoms that make up the molecules.

The high energy electron beam can

knock just one electron from a molecule

to form a positive ion.

This ion is called the parent molecular

ion.

The parent molecular ion is a radical and

is chemically unstable. These can split

into smaller fragments consisting of

smaller ions and uncharged free

radicals.

In the same spectrum additional

peaks can be formed due to the

occurrence of different isotopes of an

element.

The height of the peak will give the relative concentration of the ions present.

The highest peak is known as the base peak and is assigned an intensity of 100%.

The intensities of the other peaks are measured relative to the base peak.

The mass spectrum is unique to the compound and can be used to identify a substance by comparing it to a data bank of known spectrums.

The peak with the greatest mass will

generally give the relative atomic

mass of the molecule.

Many instruments combine two techniques to

provide more detailed and rapid information

about the sample.

The most commonly used combination is that of

mass spectroscopy and chromatography.

The advantage of gas chromatography-mass

spectroscopy (GC-MS) and high pressure liquid

chromatography-mass spectroscopy (HPLC-MS) is

that the chromatography can separate the

sample into its individual components and then the

components can be identified using MS.