Analytical Chemistry
Stoichiometry
Mole: The mole is the amount of substance of a system which contains as many elementary entities as there are
carbon atoms in 12 grams of carbon-12.
Avogadro’s Number: Avogadro’s number, defined by NA is the number of carbon atoms in 12 grams of carbon 12.
This is equal to approximately 6.02×1023 atoms.
∴ There are 6.02×1023 particles in one mol.
Formulae:
 Mass:
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Particles:
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Solution:
Gas at STP/SLC:
o STP: 1 atm, 0OC, Vm=22.4 L/mol
o SLC: 1 atm, 25OC, Vm=24.5 L/mol
 General Gas Equation:
Calculations
Miscellaneous
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Determining Empirical Formulae
 Found by finding the simplest whole number mole ratios.
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Determining Molecular Formulae
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Acid-Base Reactions
Acids and Bases
 Acid: A substance that will react with a base to donate a proton (i.e. H+).
o Strong Acid: Acids in which every acidic particle will donate a proton to a base. Denoted by a full arrow
().
 E.g. Hydrochloric acid (HCl), Sulfuric Acid (H2SO4), Nitric Acid (HNO3).
o Weak Acid: Acids in which not every acidic particle will donate a proton to a base. Denoted by an
‘equilibrium arrow’ (⇌).
 E.g. Water (H2O), Ethanoic acid (CH3COOH), Ammonium ion (NH4+)
o Conjugate Base: Product a base forms when it has accepted a proton from an acid.
 E.g. OH- from H2O, NH3 from NH4+, CH3COO- from CH3COOH
o Polyprotic Acid: An acid which is able to donate more than one proton to a base.
 E.g. H2SO4 (Diprotic) and H3PO4 (Triprotic).
 i.e. H2SO4+H2OHSO4-+H3O+
 i.e. HSO4-+H2O ⇌ SO42-+H3O+
 Base:
o Strong Base: Acids in which every acidic particle will donate a proton to a base.
 E.g. Hydroxides (NaOH, KOH, LiOH).
o Weak Base: Acids in which not every acidic particle will donate a proton to a base.
 E.g. Water (H2O), Ammonia (NH3), Carbonate ion (CO32-)
o Conjugate Acid: Product a base forms when it has accepted a proton from an acid.
 E.g. H3O+ from H2O, NH4+ from NH3, HCO3- from CO32-
The Self-Ionisation of Water
 All aqueous solutions contain H+ and OH- ions.
 If no acids or bases have been added to the solution, then the hydronium ions (H+) and hydroxide ions (OH-) are
attributed to the self-ionisation of water.
o i.e. H2O+H2O ⇌H3O++OH As water is acting as an acid and a base in the above reaction, it is defined as amphiprotic.
 Amphiprotic: A substance that can donate or accept protons. Also known as amphoteric.
 This reaction occurs forwards and backwards, and at 25oC, will continue to occur so long as the concentration of
either ion is 10-7 mol L-1. This is referred to as the equilibrium state.
 When an acid or base is added, the H3O+ and OH- ions will no longer be equal, but however, their product will
be.
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, at 25oC, where Kw denotes the ionic product of water.
The pH Scale
 The pH scale is a logarithmic scale which measures the concentration of H+ ions, defined by;
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or
Forms of Analytical Chemistry
 Quantitative Analysis: A chemical analysis performed in order to determine the concentration of a particular
species within a mixture.
 Qualitative Analysis: A chemical analysis performed in order to determine which species are present in a
sample.
Gravimetric Analysis
Gravimetric analysis is the process of determining the composition of a sample by forming a precipitate and the
measurement of mass. The precipitate in a gravimetric analysis should have;
 a known chemical formula
 low solubility
 be stable when heated so it can be dried easily and not decompose into other substances
 be the product of only one reaction i.e. other ions in the sample do not form precipitates with the same
precipitating agent
Gravimetric Analysis by Precipitation
 When the component to be analysed is the only component soluble in a particular solvent that reacts with
the precipitating agent to form a precipitate
o The Procedure:
1. Weigh the sample to be analysed
2. Mix the sample with water to dissolve all the soluble ions
3. Filter the mixture and remove the insoluble ions in the filtrate
4. Add excess precipitating agent to the filtrate to form the precipitate
5. Filter precipitate and wash with water to ensure all soluble ions are dissolved
6. Dry precipitate in
 an oven or over a Bunsen burner with ashless filter paper if the component does not
decompose or melt on heating
 a dessicator with weighed dried filter paper if the component decomposes or melts on heating
 sunlight or room temperature heat if neither of the above two are feasible
7. Weigh the precipitate
8. Repeat steps 6 and 7 until constant mass
9. Perform calculations
o Examples:
 Determining the sulfate content of fertiliser
 Determining the salt content of potato chips
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Other Pathways of Gravimetric Analysis
When the component to be analysed is the only volatile component in the sample
o The sample is heated in an oven or with a Bunsen burner until constant mass to evaporate all of the
component, and the loss of mass is measured and identified as the mass of the component in the
sample
o Examples:
 Determining the degree of hydration of a salt
 Determining the moisture content in chocolate
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When the component to be analysed in the only component insoluble in a particular solvent
o The sample is dissolved in a solvent, and filtered to remove all soluble particles. The residue is rinsed
with small amounts of solvent, dried in the methods outlined above until constant mass and then
weighed
Sources of Error
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Increase in calculated mass
o If the component to be measured is not the only soluble or insoluble component in the sample
o If the precipitate still contains moisture
o Presence of contaminants in the precipitate
Decrease in calculated mass
o If splashes and inefficient rinsing out of containers occur in any stage in the procedure
o Because all substances have some solubility in a solvent, some of the precipitate will be lost
o Decomposition of samples during drying
Volumetric Analysis
Titrations
A titration is a quantitative analytical procedure where a solution of a substance with an unknown concentration is
reacted with a solution of a substance of a precisely known concentration.
The Procedure:
1. Rinse the burette with water, and then with the solution of precisely known concentration (i.e. the standard
solution) twice using the smallest funnel. After, fill the burette with the standard solution while avoiding
formation of air bubbles.
2. Remove the funnel and then take the initial reading of the burette to two decimal places. The initial reading
does not have to be zero.
3. Rinse the pipette with water, and then twice with the substance to be analysed. After, fill the pipette with the
substance up to the 20.00 mL mark.
4. Rinse a conical (titration) flask with deionised water twice and then deliver the aliquot into the flask. Add
indicator if necessary.
5. Deliver the solution from the burette slowly until the end point is reached and record the burette reading, and
hence calculate the titre values.
6. Repeat the above steps until 3 concordant results (i.e. all three titres are within 0.05mL (one drop) of one
another) are obtained.
Results of Titrations
Once the average titre is found, stoichiometry is used to standardise the unknown. However, the accuracy of this is
dependent on the accuracy of the end point.
 End Point: The experimentally determined point at which we deduce that the reaction is complete, as a result of
detecting a colour change, or achieving an almost vertical pH curve.
 Equivalence Point: The point in the titration at which the reaction is complete, as the reactants are in their
exact mole proportions.
As it is presumed the end point is equal to the equivalence point, ideally, the end point should be equal to the
equivalence point. However this is not the case and is a source of error as;
 Our eyes require a certain build up of colour before the colour change registers.
 The colour change from an indicator may not change at exactly the equivalence point, due to the choice of
indicator.
Conditions of analysis by titration:
The substance to be standardised must
 be soluble in water
 be present in the solution at a significant concentration
 have acid-base or redox properties
 and be the only constituent in the solution with such properties
Primary and Secondary Standards
The choice of standard is vital in a titration as the reaction must;
 be spontaneous, fast, and able to go to completion
 have a clearly defined end point
Primary Standards
A primary standard is a substance which can be used to directly standardise another substance. However, in order
to do so, they must be;
 readily obtainable in a very pure form
 has a known formula
 has a known degree of hydration
 does not absorb or react with any moisture or chemicals in the air
 does not decompose or lose water of hydration on storage or when exposed to the air
 has a reasonably high molar mass to minimise errors in weighing
 is completely soluble in water at room temperature
 is readily available and relatively inexpensive
Secondary Standards
Substances which do not qualify as primary standards are secondary standards. These are typically strong acids and
bases.
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Strong Acids: Strong acids are prepared in a manner in which their stated concentration is only an approximate.
As such, they must be standardised first before usage.
 Strong Bases: Strong bases such as NaOH and KOH are deliquescent, as it readily absorbs moisture from the air,
as well as acidic oxides and CO2.
Preparing a Standard
1. Accurately weigh out the required amount of solute
2. Using a funnel, transfer all the solute into a volumetric flask
3. Add a small amount of water and shake the bottle to dissolve all the solute
4. Add water to the volumetric flask up to the calibration mark
pH Curves
A pH curve is a graph obtained by titrating beyond the end point, and measuring the pH at regular intervals with a
pH meter. The end point can then be judged by finding the middle of the almost vertical section of the graph.
Strong acid with a strong base:
 E.g. HCl+NaOHNaCl+H2O
 The products do not affect pH, and as such, at the equivalence point, the pH will be 7.
 This produces a graph with a high initial pH, the centre of the graph at 7 and a low resultant pH.
Strong acid with a weak base:
 E.g. HCl+NH3NH4Cl+H2O
 As NH4+ is acidic, the pH will be below 7.
 This produces a graph with a fairly high initial pH, the centre of the graph at about 5 and a very low resultant
pH.
Weak acid with a strong base:
 E.g. CH3COOH+NaOHCH3COONa+H2O
 As CH3COO- is basic, the pH will be above 7.
 This produces a graph with a high initial pH, the centre of the graph at about 9 and a fairly low resultant pH.
Weak acid with a weak base:
 The reaction does not completely react because they are weak.
 Thus, this will produce a curved line with a very difficult to detect end point.
Dilution
If the substance to be standardised is too concentrated, this may;
 increase the margin of error per drop of solution
 may be a safety hazard
 may require large amounts of titre
Due to these consequences, substances which meet these conditions are diluted with water to increase the
accuracy of results, minimise hazards and minimise the amount of titre needed.
Calculating Dilutions
 As diluting with water does not change the mole of the substance in the solution, then;
o
o
 After diluting, a sample of the diluted solution must then be extracted. As the number of mole in the sample
differs from that in the diluted solution, but the concentration is the same, then;
o
o
Back Titrations
Soluble substances that have acid-base or redox properties are usually analysed by simple volumetric analysis.
However, substances which are;
o Toxic
o Volatile
o Gaseous and in a mixture of gases
o Fairly unreactive (undefined end point)
o Present at low concentrations
o A solid which is insoluble, but will react with an acid
can be analysed by subjecting the substance to preliminary treatment.
 In this preliminary treatment, excess amount of a reactant is added to the substance. We then obtain the
concentration of the substance by titrating either the;
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o remaining amount of excess reactant
o the amount of product produced in the reaction
In both cases, two reactions and thus two chemical equations should be made; one for the preliminary reaction,
and one for the titration
Sources of Error
Incorrect rinsing of apparatus
If the end point is too far from the equivalence point
If the end point is identified incorrectly
Glassware is contaminated with other substances
Inadvertent dilution of solutions
Random errors such as reading the burette wrong, measuring the mass of the primary standard incorrectly etc.
If the concentration of the standard solutions are misquoted
Chromatography
Chromatography analysis techniques involve the separation of components of a mixture by fluid flow. This relies
upon;
 Differing polarities of components in a mixture and hence differing solubilities
 Separation of the mixture by components adsorbing onto the stationary phase, and desorbing into the mobile
phase where it will be swept along the stationary phase.
 Greater desorption implies a greater distance moved along the stationary phase.
Definitions:
 Stationary Phase: An immobile surface that different components of a mixture can adhere to.
 Mobile Phase: A fluid or gas used to separate the components of a mixture along a stationary phase
Advantages:
 Much more rapid than volumetric and gravimetric analysis
 Can analyse substances present in low concentration
 Can analyse small samples of substances
Paper Chromatography
Features:
 Stationary Phase: Chromatography paper (Contains polar cellulose molecules)
 Mobile Phase: Liquid solvent such as water or ethanol
 Components: Must be water soluble and coloured, unless a developing agent is added or the chromatogram be
held under UV light
 Measures: Retention factor
(in decimal form)
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Type of Analysis:
o Qualitative analysis: through comparison of retention factors
Thin Layer Chromatography (TLC)
Features:
 Same as paper chromatography except;
 Stationary Phase: Glass/metal sheet covered in a thin layer of absorbent material e.g. Silica gel, Al2O3
 Advantages over paper chromatography:
o Faster
o Better chromatogram resolution
o More sensitive i.e. lower concentrations can be analysed
o Developing reagents that would destroy chromatography paper can be used
Preparative Thin Layer Chromatography (PTLC)
Features:
 Same as TLC except;
 Stationary Phase: Stationary phase is larger
 Type of Analysis:
o Qualitative analysis by comparison of retention factors
o Limited qualitative analysis as components are scraped off and weighed
Factors affecting Rf
 The molecular mass and polarity of the component and mobile phase
 Temperature around the chromatogram
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The moisture levels around the chromatogram
The type of stationary phase
Instrumental Analysis
Advantages over paper chromatography/TLC/PTLC
 Much faster
 Less sample preparation
Reasons why volumetric/gravimetric/paper chromatography are still used
 To calibrate analytical instruments
 To create standard solutions
 Is inexpensive and portable
Gas Chromatography (GC)
Features:
 Stationary Phase:
o Gas Liquid Chromatography (GLC):
 Liquid substance e.g. ester, high-boiling hydrocarbon
 Components dissolve into the liquid
 Also known as Partition Chromatography
o Gas Solid Chromatography (GSC):
 Solid material e.g. glass beads, brick
 Components adsorb onto the solid
 Also known as Adsorptive Chromatography
 Mobile Phase: Inert gas e.g. He, N2
o In GC, components do not bond to the mobile phase, but are just swept along by it
o Also known as the carrier gas
 Components:
o Gases
o Compounds which can be vaporised into gas i.e. weak intermolecular bonding, strong intramolecular
bonding
o Mr less than 300
 Measures: Retention time (Rt)– The time taken for a gas to pass through the column
 Type of Analysis:
o Qualitative analysis by comparison of retention times
o Quantitative analysis by comparison of area under each peak compared to the total area on the gas
chromatogram, which is standardised against a calibration curve
High Pressure Liquid Chromatography (HPLC)
Features:
 Same as GC except;
 Stationary Phase: Waxy, porous, polar solid with a high surface area
 Mobile Phase: Liquid substance (known as the eluent) e.g. water, methanol
 Components:
o Compounds which decompose if vaporised
o Mr greater than 300
Factors affecting Rt
In instrumental analysis:
 The molecular mass and polarity of the component and stationary phase
 The length of the column
 The temperature
In GC:
 The flow rate of the carrier gas (the type of carrier gas has no effect)
In HPLC
 The pressure at which the mobile phase is pumped
 The type and the density at which the stationary phase is packed
 The molecular mass and polarity of the eluent
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and vice versa.
The main regions we are concerned with in order of increasing wavelength are the;
o Ultraviolet (UV) region
o Visible light region
o Infrared (IR) region
Therefore infrared has the lowest frequency and energy, and ultraviolet has the highest.
Visible Light
The visible light region is the only wavelengths which are visible to the human eye.
The visible light region of the electromagnetic spectrum is composed primarily of these regions;
o Red
R
o OOrange
o YYellow
o GGreen
o BBlue
Io Indigo
o Violet
V
Substances
are perceived to be a certain colour as their particles absorb certain wavelengths of visible light,
while the remainder are transmitted. The transmitted light is the only wavelengths which we then perceive.
Ultraviolet-Visible Spectroscopy
Measurement under analysis:
o Absorbance of ultraviolet and visible light of a particular wavelength
Process:
o An absorbance spectrum is produced by passing light of different wavelengths and graphing absorbance
against wavelength for the component.
o The wavelength with the highest absorbance is used to measure the absorbance of a series of known
standards, as it provides the most accurate results.
o A blank (a pure sample of the solvent) is used first to calibrate the spectrometer against any reflecting,
scattering or absorbance of light by the cell and the solvent. Ideally, the blank should have zero
concentration of the component, and thus zero absorbance.
o A calibration curve is then produced to determine the concentration of the component from its
absorbance.
Principles:
o Absorption of ultraviolet and visible light corresponds to electronic changes throughout the molecule;
 Light is absorbed by electrons to provide enough energy for electrons to shift from a lower
energy level atomic orbital to one of higher energy level
 A particle will only absorb photons of light with the exact amount of energy required to perform
such a shift
 For different particles, this specific amount of energy differs, hence resulting in the range of
colours perceivable amongst different substances
o Absorbance is a measure of the intensity of light remaining after a substance has absorbed some,
relative to the intensity of light when none has been absorbed.
Applications:
o Quantitative: Determining the concentration of;
 Organic compounds e.g. food colourings and dyes
 Coloured cations and anions
 Cations and anions reacted with suitable reagents to produce a coloured compound
or
o Qualitative: Identifying unknown compounds by matching absorbance spectra
Increasing
energy
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Spectroscopy and Spectrometry
The Electromagnetic Spectrum
Electromagnetic radiation is the form of energy radiated out by the sun that is in the form of vibrating electric
and magnetic fields.
Electromagnetic radiation varies in wavelength, frequency and energy.
The relationship between the three is;
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Atomic Absorption Spectroscopy
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Measurement under analysis:
o Absorbance of ultraviolet and visible light of a particular wavelength
Process:
o The sample is vaporised in high temperatures to produce atomic vapour.
o The light is produced by a hollow cathode lamp which contains vaporised atoms of the metal to be
analysed. The light emitted by this lamp is pulsed.
o The metal atoms in the flame absorb some of light emitted by the lamp.
o The remaining light emitted is filtered by a monochromator and slit to select the wavelength which is
absorbed the most.
o The detector detects how much light is absorbed by measuring the intensity of the pulsed light before
and after absorption.
Principles:
o Absorption of ultraviolet and visible light corresponds to electronic changes particularly within the
incomplete 3d subshells of transition metals
o The lamp must contain the metal that is being measured as the wavelengths of light that are emitted by
an element are identical to the wavelengths that it absorbs.
o Only the pulsed light is measured because the thermal energy of the flame will cause the metal atoms
to continuously emit some light as well.
Applications:
o Quantitative: Determining the concentration of metals
Infrared Spectroscopy
Measurement under analysis:
o Percentage transmittance of a range of infrared radiation
Process:
o The sample is placed in a cell made of NaCl or KBr as they do not absorb IR radiation.
o IR radiation is emitted and split into two beams; one passed through the sample cell, and another
passed through the reference cell.
o The transmitted radiation is passed through a monochromator and slit which selects the wavelength of
light to be used for analysis, and then measured. The infrared spectrum is a graph measuring
percentage transmittance against wavenumber in cm-1
Principles:
o Absorption of infrared waves corresponds to vibrational and rotational changes throughout the
molecule;
 Molecules will absorb a discrete quantum of infrared radiation which corresponds to a
promotion in vibrational and rotational energy levels.
 Molecules are held together by covalent bonds which are not rigid. Therefore, the atoms in
molecules are able to contract, stretch, vibrate and bend.
o Features of an infrared spectrum are;
 Each ‘trough’ corresponds to absorption at that particular wavenumber
 The width of a trough indicates the number of vibrational changes which may have overlapping
energies
 The depth of a trough – often classified under strong, medium or weak – is indicative of the
concentration of a particular functional group
 Wavenumber is the reciprocal of the wavelength corresponding to a particular peak. As such,
increasing wavenumber corresponds to increasing frequency and decreasing wavelength.
 The region of wavenumbers above 1500 cm-1 tends to correspond to functional groups. This
section is useful for determining the structure of a compound.
 A broad and strong trough at around 3200 cm-1 will typically correspond to a hydroxyl
(OH) group, which could indicate an alcohol or carboxylic acid functional group
 A sharp and strong trough at around 1700 cm-1 will typically correspond to a carbonyl
(C=O) group, which could indicate a carboxylic acid or ester functional group
 The region of wavenumbers below 1500 cm-1 tends to correspond to the vibration and rotation
of the entire molecule and hence is unique to each molecule. This region is known as the
‘fingerprint region’, and is useful for identifying compounds by comparing against other spectra.
o The reference cell is used to account for interference of IR radiation by the;
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 Cell
 Solvent
 H2O, CO2 and temperature in the atmosphere
o The difference in transmittance between the reference cell and the sample cell can be attributed to
absorption by the sample
o The amount of transmittance decreases as the concentration of the component increases.
Applications:
o Qualitative: Identification of the types of bonds and functional groups in a molecule
 This allows identification of a molecule by matching infrared spectra – particularly the
‘fingerprint region’
 This also aids in determining the structure of a molecule
o Quantitative: Determining the concentration of molecules
Nuclear Magnetic Resonance Spectroscopy
Measurement under analysis:
o Resonance of radio waves and their corresponding chemical shifts
Principles:
o Absorption of radio waves corresponds to changes in the ‘spin’ of nuclei
 In the nucleus of atoms, nucleons (protons and neutrons) spin in a certain direction. In atoms
that have an even number of nucleons, the orientation of these spins is paired and thus cancels
out.
 However, in atoms such as 1H and 13C, the nucleus will have an overall spin which creates a
magnetic field around it.
 Without a magnetic field, the orientation of these spins is random, while in the presence of a
magnetic field, the majority of nuclei will spin parallel with the magnetic field and nuclei which
are in a higher energy state will be aligned anti-parallel with the magnetic field.
 Nuclei can be ‘spin-flipped’ into the higher energy state by applying the precise amount of
energy
 Nuclei do not maintain the higher energy state for long, and will ‘spin-flip’ back into the lower
energy state, thus emitting the precise radio frequency which was absorbed
o Chemical environments are formed by nuclear shielding
 However, in addition to the magnetic field generated by the nuclei, electrons also generate a
magnetic field which ‘shields’ the magnetic field of the nuclei, as these magnetic fields also
absorb some of the radio waves emitted
 If a hydrogen atom is bonded to a highly electronegative atom, then the shared electrons will
orbit closer to the electronegative element. Thus, the effect of the magnetic field of the
electrons is diminished, and the hydrogen atom is said to be ‘de-shielded’. This causes the
respective hydrogen atom to absorb and emit a stronger frequency.
 Deshielding causes hydrogen atoms to have increasing chemical shifts in NMR spectra, which is
referred to being ‘shifted downfield’.
 A combination of influences of chemical shifting caused not only by the atom which the
hydrogen is bonded to, but also what atoms are bonded to that atom create the chemical
environment in which a hydrogen exists.
 Each different chemical environment will be represented by a peak on the spectrum, with its
relative peak height corresponding to the relative number of hydrogen atoms in that
environment.
o These individual peaks may then split on high-resolution 1H NMR spectra due to what is known as ‘spinspin coupling’. The number of peaks visible within a single peak set is equal to the number of adjacent
hydrogen atoms located in non-equivalent environments to the environment of concern, plus one. This
is known as the n+1 rule
o Tetramethylsilane (TMS) is the standard against which 1H NMR is measured. TMS is used because;
 It has 12 hydrogen atoms in the same environment, thus producing a very strong signal
 It is chemically inert and will not react with the sample
 Its hydrogen atoms produce a signal well away from the signals generally emitted by organic
compounds
o NMR spectra measure chemical environments in terms of chemical shift to standardise results among
different spectrometers. Chemical shift is a measure of the difference of resonance between the
hydrogen atom in the sample, and the resonance of TMS, over the operating frequency of the
spectrometer.
o
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13
C NMR is also used to analyse the chemical environments of carbons in a molecule. 13C NMR works
quite similarly, except it does not produce peak splitting because the 13C isotope is of low abundance,
and thus, it is uncommon to find two adjacent 13C within a molecule which exert ‘spin-spin coupling’.
Consequently, the 13C spectrometer even suppresses the effects of 13C coupling to avoid unnecessarily
complicated spectra.
Applications:
o Qualitative: Identification of the structural arrangement of a molecule
 This allows the precise qualitative analysis of a molecule as it allows identification of isomers
 This is through matching NMR spectra or by analysis of the spectra
Mass Spectroscopy
Measurement under analysis:
o Mass/charge ratio
Process:
o Vaporisation
 The sample is vaporised into a gas before entering the ionisation chamber
o Ionisation
 A beam of electrons is fired which causes collisions between the molecules and the electrons,
causing electrons to be dislodged from the molecules.
 The molecules break up into a variety of fragments, which some more probable than others
 In most cases, a positively charged fragment and an uncharged fragment is created.
 The positively charged fragment is pushed out of the ionisation chamber by the ion repeller; an
electric plate with a slightly positive charge
o Acceleration
 The charged ions are focused into narrow beams and accelerated to high speeds through an
electric field.
o Deflection
 The ions enter a magnetic field which causes them to separate into streams of ions that have
the same mass/charge ratio
 This is caused by the curvature of this part of the spectrometer, so depending on the strength of
the magnetic field, only a particular mass/charge ratio will reach the detector
 Others are either deflected too much or not enough, and hit the sides of the chamber where
they collect electrons and lose their charge
 The magnetic field is progressively increased, or the voltage of the electric field in acceleration
is progressively increased to allow all the streams to reach the collector
o Detection
 The detector measures the current created due to the ions reaching the detector, and produces
a mass spectrum
Principles:
o The greatest deflection is caused by lowest mass and the highest positive charge. i.e. lowest
mass/charge ratio
o The parent/molecular peak refers to the peak which corresponds to the ion which consists of the
unbroken entire molecule.
o The base peak refers to the peak which has the greatest relative abundance, and hence corresponds to
the ion which is of the greatest stability
o Slight variations from expected results may occur due to isotopes. For example, the parent peak may
not be the largest due to 13C or other isotopes.
Applications:
o Qualitative:
 Identification of the molecular mass of the compound
 Identification of the structural formula of the compound by deducing fragments which can be
formed, or by matching mass spectra
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Organic Families
Hydrocarbons are organic compounds that contain only hydrogen and carbon atoms
o Hydrocarbons cannot be synthesised, and are extracted by fractional distillation from crude oil
o Carbon and hydrogen have similar electronegativity, and due to the symmetrical structure of
hydrocarbons, hydrocarbons are classified as non-polar compounds
o Thus, the only intermolecular forces between hydrocarbons are dispersion forces
Instantaneous Dipoles and Dispersion Forces
o An instantaneous dipole is created from the momentary non-symmetrical electron distribution around
an atom or molecule that results in a temporary dipole arrangement of charge.
o This momentary dipole arrangement will form very weak, temporary, non-directional bonds with other
instantaneous dipoles
o Dispersion forces are stronger with;
 Larger atoms as there are more regions of charge to attract
 Molecules which can pack closely, which depends on the shape of the molecule
 Low temperatures
General properties of hydrocarbons
o As the carbon chain length increases:
 The strength of dispersion forces increase
 The boiling point increases
 The melting point increases
 The viscosity (resistance to flowing) of the liquid form increases
o Hydrocarbons are insoluble in water
Hydrocarbons are highly flammable, and thus can be combusted.
o In excess oxygen, the chemical equation for the complete combustion is;
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Note: If a non-integer number is obtained, multiply all coefficients by two
Homologous Series
Alkanes:
 General Formula: CnH2n+2
 Suffix: -ane
 States: C1 to C4 are gases at room temperature, with C5 to C15 being liquids, and the rest solids
 Alkanes are saturated because they only have carbon-carbon single bonds, and thus have the maximum number
of hydrogen atoms possible bonded to its carbons.
Alkenes
 General Formula: CnH2n
 Suffix: -ene
 States: C1 to C4 are gases at room temperature, with C5 to C15 being liquids, and the rest solids
 Alkenes are unsaturated because they have a single carbon-carbon double bond, and thus do not have the
maximum number of hydrogen atoms possible bonded to its carbons.
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The Bromine Test
Bromine – Br2(aq) – is a brown liquid
In the presence of an unsaturated hydrocarbon:
o Bromine will undergo an addition reaction with the unsaturated hydrocarbon and hence, the bromine
solution will fade from a brown colour to colourless.
o NOTE: If this occurs, this only indicates the presence of an unsaturated hydrocarbon, not a carboncarbon double or triple bond specifically.
In the presence of a saturated hydrocarbon:
o Bromine cannot react with a saturated hydrocarbon and hence, the bromine solution will remain a
brown colour
Functional Groups
A functional group is an atom or a group of atoms that gives a characteristic set of chemical properties to a molecule
containing that group
 Alkyl group
o Formula: CnH2n+1
o Prefix: -yl
o Effects:
 Decreases the density, and thus boiling and melting points of the resultant hydrocarbon, as the
molecules will be separated further apart which disrupts dispersion forces
 Halogen group
o Formula: -X (where X is a halogen i.e. Cl)
o Prefix:
 Chlorine – Chloro
 Bromine - Bromo
o Effects:
 Halogens are very polar, allowing solubility in water
 Halogens are very electronegative, which means they are very reactive
 Amine group
o Formula: H2No Prefix: aminoo Suffix: -amine
o Effects:
 The amine group is very polar and allows solubility in water when the hydrocarbon chain is
short
 The amine group is weakly basic, and in acidic solutions will turn into H3N+
 Carboxylic acids
o Formula: COOH i.e. carbonyl group (C=O) and hydroxyl group (O-H) bonded to the same carbon
o Suffix:
 -oic acid
 Note: the carbon which the carboxyl group is attached to is always denoted the 1st carbon and
does not need to be stated
o Effects:
 The carboxyl group is very polar and allows solubility in water when the hydrocarbon chain is
short
 The carboxyl group is weakly acidic, and in basic solutions will turn into COO Alcohol
o Formula: -OH
o Suffix: -ol
o Effects:
 The hydroxyl group is very polar and allows solubility in water when the hydrocarbon chain is
short
 The hydroxyl group allows hydrogen bonding, and causes all to be liquids at room temperature
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Fractional Distillation
Crude oil is a mixture of many different hydrocarbons of varying carbon-chain lengths
The mixture of hydrocarbons is separated by the process called fractional distillation
The process of fractional distillation for crude oil involves;
o Step 1: Pre-treatment
 Water, salt and sulfur is removed from the crude oil
o Step 2: Atmospheric Distillation
 The crude oil is heated to high temperatures. As the vapours rise up the fractionating column,
the temperatures become progressively lower, and they condense into liquid hydrocarbons at
their respective boiling points
This produces fractions of hydrocarbons of similar boiling points and hence carbon chain length
Production Pathways
Production of Alkanes Alkenes
CH2CH2(g)+H2(g)
CH3CH3(g)
CH3CH3(g)
CH2CH2(g)
ethene+H2O ethane
ethane
ethene
*Addition *Hydrogenation
*Catalytic cracking
Production of Halogenated Hydrocarbons
CH3CH3(g)+Cl2(g)
CH3CH2Cl(g)+HCl(g)
CH2CH2(g)+HCl(g)
ethane+Cl2
1-chloroethane+HCl
*Halogenation *Substitution *Chlorination
CH3CH2OH(aq)
CH3CH2Cl(g)
ethene+HCl
1-chloroethane
*Halogenation *Addition *Chlorination
CH3CH2Cl
CH2CH2(g)+Cl2(g)
CH2ClCH2Cl(g)
ethanol
1-chloroethane
ethene+Cl2 1,2-dichloroethane
*Halogenation *Substitution *Chlorination
*Halogenation *Addition *Chlorination
Production of Alcohols
CH3CH2Cl(g)
CH3CH2OH(aq)+NaCl(g)
CH2CH2(g)+H2O(g)
CH3CH2OH(aq)
1-chloroethane
ethanol+NaCl
ethene+H2O
Ethanol
*Substitution
*Addition *Hydrolysis
Production of Carboxylic Acids
CH3CH2OH(aq)
CH3COOH(aq)
ethanol
ethanoic acid
*Oxidation
Production of Esters
CH3CH2OH(aq)+CH3COOH(aq)
H3CCOOCH2CH3(g)+H2O(l)
ethanol+ethanoic acid
ethylethanoate+H2O
*Esterification *Condensation
Production of Amines
CH3CH2Cl(g)
1-chloroethane
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CH3CH2NH2(aq)+HCl(g)
ethan-1-amine+HCl(g)
*Substitution
Catalysts
When substances react, often an intermediate product known as an activation complex is formed before the
final product is obtained
The steps involved is called the reaction mechanism
A catalyst is a substance which increases the rate of a chemical reaction by lowering the required activation
energy, without being consumed or undergoing any permanent chemical change itself
Cracking
Cracking is the process of making smaller hydrocarbons from larger ones through;
o Thermal cracking: Application of heat to break the covalent bonds in the larger hydrocarbon
o Catalytic cracking: Application of a catalyst the break the covalent bonds in the larger hydrocarbon
Biochemical Fuels
Fossil fuels are a non-renewable source of energy as their synthesis takes millions of years. As such, biofuels are now
of interest as they can be synthesised from renewable, sustainable and carbon-neutral sources.
 Renewable refers to naturally occurring sources of energy which can be replenished by ecological cycles.
 Sustainable refers to sources of energy which can be consumed indefinitely without depletion.
 Carbon-neutral refers a substance whose amount of carbon dioxide released from combustion is offset by the
carbon dioxide absorbed from the atmosphere in its synthesis.
Types of Biofuels
Biomass:
 Composition: Living and recently dead biological material such as plants
 Synthesis:
o Biomass is usually just collected and not refined in any way
Bioethanol:
 Composition: Ethanol (99.7%)
 Synthesis:
o The most common method is fermentation (anaerobic respiration) of sugar-based crops by yeast,
shown by
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C6H12O6(aq)
2CH3CH2OH(aq)+2CO2(g)
 glucose
ethanol + carbon dioxide
As ethanol is poisonous, this process occurs until the ethanol concentration in the solution reaches
about 10% at which the yeast organisms die.
o This solution is then distilled to collect the ethanol at a concentration of about 95%.
Combustion: CH3CH2OH(l)+3O2(g)2CO2(g)+3H2O(l)
Note: Ethanol which is synthesised directly from ethene or ethane is not considered bioethanol because ethene
and ethane are not renewable sources, as hydrocarbons take millions of years to synthesise
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Biogas:
 Composition: Methane (CH4) and carbon dioxide (CO2)
 Synthesis:
o Methane and carbon dioxide is released from the decomposition of organic matter by micro-organisms
under anaerobic conditions
 Combustion: CH4(g)+2O2(g)CO2(g)+2H2O(g)
Biodiesel:
 Composition: Esters (usually methyl esters)
 Synthesis:
o Vegetable oil or animal fats (which consist of triglycerides) are hydrolysed by heating them in methanol
(CH3OH) and potassium hydroxide (KOH) solution, as shown by
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C3H5(OOCR)3(l)+CH3OH(l)
H3COOCR(l)+C3H5(OH)3(l)
 triglyceride +methanol
methyl ester+ glycerol
Combustion: e.g. H3COOC(C16H29)(l)+26O2(g)18CO2(g)+16H2O(g)
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Biochemistry
Condensation and Hydrolysis Reactions
Polymerisation, the process of forming a polymer from monomer subunits among organic molecules is created by
condensation reactions.
Condensation reactions occur when two monomers, when supplied with energy, bond together to form a larger
molecule and release one water molecule.
Hydrolysis reactions occur when a larger molecule receives one water molecule and breaks its bonds to release
energy.
Biomacromolecules
Carbohydrates
Elements: Carbon, hydrogen, oxygen
Type of Bonding:
o Ether link (–C–O–C–)
o Formed between two hydroxyl (OH-) groups
Monomer: Monosaccharide
o Structure: Monosaccharides are either;
 Hexose sugars: A monosaccharide which has a six-carbon ring e.g. glucose
 Pentose sugars: A monosaccharide which has a five-carbon ring e.g. fructose
o Monosaccharides have many hydroxyl groups and thus are very polar and soluble in water
Dimer: Disaccharide
o A disaccharide consists of two monosaccharides bonded by an ether link e.g. maltose, sucrose, lactose
Polymer: Polysaccharide
o A polysaccharide consists of over ten monosaccharides bonded together in a chain by ether links e.g.
starch, glycogen, cellulose
Lipids
Elements: Carbon, Hydrogen, Oxygen
Type of Bonding:
o Ester link (–O–CO–)
o Formed between a hydroxyl group and a carboxylic group
Fatty Acids:
o A hydrocarbon chain with a carboxyl group at one end.
o Saturated Fatty Acids:
 In a saturated fatty acid, there is the maximum number of hydrogen atoms possible and
therefore there are no carbon-carbon double or triple bonds present.
 A saturated fatty acid is represented by CnH2n+1COOH or CnH2nO2
o Unsaturated Fatty Acids:
 In an unsaturated fatty acid, there is less than the maximum number of hydrogen atoms
possible and therefore there are carbon-carbon double or triple bonds present.
 In a monounsaturated fatty acid, there is one double or triple bond present. In a
polyunsaturated, there are two or more double or triple bonds present.
Glycerol:
o A three carbon alcohol molecule with the sub-structural formula C3H5(OH)3
o Is non-polar and hydrophobic.
Triglycerides
o Structure: Glycerol ester bonded to three fatty acid chains.
o Properties:
 Liquid triglycerides with short fatty acid chains or unsaturated fatty acids are called oils
 Unsaturated fatty acids have double or triple bonds which disrupt the linearity of the
straight-chain and therefore disrupt dispersion forces between other fatty acid
molecules; thus making them liquid at room temperature
 Solid triglycerides with long fatty acid chains or saturated fatty acids are called fats
 Saturated fatty acids have straight carbon chains which have relatively strong dispersion
forces and thus are solid at room temperature
o Fatty acid chains are variable on triglycerides, and provide varied properties.
Proteins
 Elements: Carbon, hydrogen, oxygen, nitrogen and sometimes sulfur.
 Type of Bonding:
o Amide link (–CO–NH–)
o Also known as a peptide link in proteins
o Formed between a carboxyl group and an amine group
 Monomer: α-amino acids
o Structure: Carbon atom bonded to a hydrogen atom, a carboxyl group (COOH), an amine group (NH2)
and a variable R group
o The R group of an amino acid determines its properties, such as whether it is polar or non polar,
whether it is acidic or basic, and whether it is charged.
o α-amino acids change structure depending on the pH of their surroundings
 At their isoelectric point, α-amino acids have both a positive and negative charge and are known
as zwitterions
 In a basic solution that has a pH greater than that of the isoelectric point, the carboxyl group
will donate a hydrogen ion and become –COO In an acidic solution that has a pH less than that of the isoelectric point, the amine group will
accept a hydrogen ion and become –NH3+
 Dimer: Dipeptide
o Consists of two amino acids bonded by a peptide link
 Polymer: Polypeptide
o Primary Structure:
 Type of Bonding:
 Peptide bonds between the carboxylic and amine groups of adjacent amino acids
 The primary structure refers to the specific number, type and linear sequence of amino acids in
a polypeptide chain
o Secondary Structure:
 Type of Bonding:
 Hydrogen bonds between the –C=O and –N–H or –O–H groups of non-adjacent amino
acids
 The secondary structure refers to the type of shape created by the coiling of the polypeptide
due to hydrogen bonds at regular intervals
 Possible shapes in α-helices and β-pleated sheets, as well as random coils
o Tertiary Structure:
 Type of Bonding:
 Hydrogen bonds between the –C=O and –N–H or –O–H groups of non-adjacent amino
acids
 Di-sulfide link (–S–S–) between non-adjacent cysteine amino acids
 Ionic bonds between –NH3+ and –COO Hydrophobic interactions between polar R groups to other non-adjacent polar or nonpolar groups
 The tertiary structure refers to the overall three-dimensional shape created as a result of
interaction between atoms in the variable groups of the amino acids.
 Common shapes are globular (e.g. all enzymes) and fibrous (e.g. collagen).
Nucleic Acids
 Elements: Carbon, hydrogen, oxygen, nitrogen, phosphorus.
 Type of Bonding: Covalent bonding and hydrogen bonding.
 Monomer: DNA Nucleotide
o Structure: Deoxyribose sugar bonded to a phosphorus group at 5’, with a nitrogenous base bonded to
the sugar at 1’.
 Nitrogenous bases:
 Purines: 2 ringed molecules
o Adenine: Pairs with T and U by 2 hydrogen bonds
o Guanine: Pairs with C by 3 hydrogen bonds
 Pyramidines: 3 ringed molecules
o Thymine: Pairs with A by 2 hydrogen bonds
o Cytosine: Pairs with G by 3 hydrogen bonds
o Uracil: Pairs with A by 2 hydrogen bonds
o Complementary base pairing rules state that A can only hydrogen bond with T, G with C and vice versa
 Polymer: Deoxyribonucleic Acid (DNA)
o Primary Structure:
 Type of bonding:
 Covalent bonds from the phosphate group on 5’ of one nucleotide, to the OH group on
3’ of an adjacent nucleotide to form the ‘sugar-phosphate backbone’.
 This covalent bond –O–P–O– is called a phosphodiester bond.
 The primary structure of DNA refers to the sequence of nucleotides and their respective bases
in a polynucleotide chain.
o Secondary Structure:
 Type of bonding:
 Hydrogen bonds between complementary bases in two separate polynucleotide chains
 The secondary structure of DNA refers to the winding of a double-stranded polymer into a
double helix.
 Polymer: Ribonucleic Acid (RNA)
o Primary Structure:
 RNA has the same primary structure as DNA except it uses RNA nucleotides instead, which has;
 Ribose sugars instead of a deoxyribose sugars
o A ribose sugar has a hydroxyl group at 2’ instead of a hydrogen atom
 The nitrogenous base uracil instead of thymine
o Thymine has a methyl group at the 5’ carbon instead of a hydrogen atom
o Secondary structure:
 RNA does not have complementary base pairing with other RNA strands, and therefore does
not form double-stranded polymers that wind into a double helix.
Applications of Biochemistry
Enzymes
 Enzymes are protein molecules with the ability to catalyse metabolic reactions by lowering the required
activation energy, thus increasing the rate of reaction, without being consumed or undergoing any permanent
chemical change itself.
Enzyme Action
 Enzymes are globular proteins which have a cleft in them called the active site, into which the substrate
molecule are attracted to, and bond to
 When the substrate molecule bonds to the enzyme, it forms an enzyme-substrate complex
 Lock and Key Model: The lock and key model states the active site matches the shape of the substrate
molecules.
Denaturation of Proteins
 Denaturation is the disruption of the overall functional shape of a protein which results in a loss of biological
activity
 Denaturation only involves the change of the protein’s secondary and tertiary structure, and may be caused by;
o Temperature:
 High temperatures will permanently denature enzymes as the bonding holding the functional
shape together will break apart
 Low temperatures will cause the collisions with substrate molecules to occur too infrequently
o pH Levels:
 Extremes of pH outside of an enzyme’s optimum pH will cause enzymes to be denatured as it
changes the structure of the amino acids’ variable groups, thus changing the tertiary bonding
that occurs
Disease Markers
A protein or molecule which can be reliably used for diagnosis of a disease is known as a disease marker. Certain
conditions relate to abnormal concentrations of particular proteins and molecules in the blood. For example;
 Phenylketonuria results in abnormally high concentrations of the molecule phenylpyruvate in the blood
 Following a heart attack, abnormally high concentrations of cardiac enzymes exist in the blood
 Prostate cancer in men results in abnormally high concentrations of prostate specific antigen (PSA), which is
normally found in low concentrations in the blood
Forensic Science
 Polymerase chain reaction:
o PCR is used to amplify the amount of DNA from a small sample by replication
 Restriction enzymes:
o Restriction enzymes cut DNA into specific fragments
 Gel electrophoresis:
o Gel electrophoresis is a technique which separates DNA based on its size.
o The technique relies on DNA’s overall negative charge given by the phosphate groups, and causes DNA
to migrate across a gel matrix when a charge is applied.
o Larger DNA fragments move through the gel at a slower rate while smaller DNA fragments move
through the gel at a faster rate
o This separates the DNA, and produces a unique ‘fingerprint’ for each sample’s DNA, as each sample of
DNA will have been cut at different locations by the restriction enzyme
o By comparing these fingerprints, identification of DNA can be made
o Note: Samples having the same DNA fingerprint does not absolutely mean that they come from the
same person, or from the same family
Drug Synthesis
Aspirin is a molecule which fits into the active site of the protein responsible for inflammation and deactivates the
physiological response. It is synthesised by;
C6H4(OH)COOH(s)+CH3COOCOCH3(aq)C6H4(COOH)OCOCH3(s)+CH3COOH(aq)
salicylic acid+acetic anhydrideacetylsalicylic acid+ethanoic acid
*esterification
or
C6H4(OH)COOH(s)+CH3COOH(aq)C6H4(COOH)OCOCH3(s)+H2O(l)
salicylic acid+ethanoic acid acetylsalicylic acid+water
*esterification *condensation