Chemistry AOS1 Unit 3 Notes

Thushan Hettige VCE Unit 3 Chemistry

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AREA OF STUDY 1: CHEMICAL ANALYSIS Principles: It’s about WHAT the stuff is, and HOW MUCH there is!

STUDY DESIGN: • volumetric analysis including determination of excess and limiting reagents and titration curves: simple and back titrations, acid-base and redox titrations • gravimetric analysis • calculations including amount of solids, liquids and gases; concentration; volume, pressure and temperature of gases • the writing of balanced chemical equations, including the use of oxidation numbers to write redox equations, and the application of chemical equations to volumetric and gravimetric analyses • principles and applications of chromatographic techniques (excluding features of instrumentation and operation), and interpretation of qualitative and quantitative data from: – thin layer chromatography (TLC), including calculation of Rf

– high performance liquid chromatography (HPLC) and gas chromatography (GC) including Rt

and the use of a calibration graph to determine amount of analyte • principles and applications of spectroscopic techniques (excluding features of instrumentation and operation), and interpretation of qualitative and quantitative data from: – atomic absorption spectroscopy (AAS) including electron transitions and use of calibration graph to determine amount of analyte – infrared spectroscopy (IR) including use of characteristic absorption bands to identify bonds – proton and carbon-13 nuclear magnetic resonance spectroscopy (NMR) including spin, the application of carbon-13 to determine number of equivalent carbon environments; and application of proton NMR to determine structure: chemical shift, areas under peak and peak splitting patterns (excluding coupling constants), and application of n+1 rule to simple compounds – visible and ultraviolet spectroscopy (visible-UV) including electron transitions and use of calibration graph to determine amount of analyte – mass spectroscopy including determination of molecular ion peak and relative molecular mass, and identification of simple fragments • matching analytical technique/s to a particular task: single and combined techniques. A summarised version of AOS 1: Wet-way analysis

1. Stoichiometry and Calculations 2. Gravimetric Analysis 3. Volumetric Analysis

Instrumental Analysis

4. Chromatography 5. Spectroscopy

*IR and NMR analysis will be in the AOS 2 notes, because they coincide with organic very well. In study design they are found in AOS 1. HIGHLY RECOMMEND this website: www.chemguide.co.uk


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Before we embark on our chemical analysis work, we need to be able to perform stoichiometric calculations. Hence, we will address this section first. 1. STOICHIOMETRY AND CALCULATIONS Formulas// Below is the list of formulas that you may be familiar with: n = m/M n = cV pV = nRT OR mol = g/(g mol-1) mol = mol L-1 x L You could memorise the formulas, or you could think about these conceptually, or you could play with units (as per the second line). However, because R = 8.31 and is a universal gas constant, p, V, n and T have to be in specific units: P is in kPa V is in L T is in Kelvins (NOT DEGREES) Excess reagent notes// Accuracy – “Sig Figs”//


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Here are the rules for quoting results in a nutshell: Adding/Subtracting: the result is only as accurate as the data with the least number

of decimal places eg. 2.463 (3 dp) + 1.2 (1 dp) = 3.7 (1 dp) Least accurate value is to 1 dp. Worked Example 1.1: It is known that 0.2402 mol of NaOH was added to a reaction

flask containing NH4+ ions, and that 0.2104 mol is remaining

in the flask after reaction with the ammonium ions. What is the amount of NaOH, in mol, that has reacted with the ammonium ions? (Express your answer to the correct accuracy)

Worked solution:


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Multiplying/Dividing: the result is only as accurate as the data with the least number of significant figures

eg. 1.265 (4 sf) x 1.31 (3 sf) = 1.66 (3 sf) Least accurate value is to 3 sf. Worked Example 1.2: 200 mL of a 0.1322 M HCl solution was added into a beaker. What is the amount, in mol, of HCl in the beaker? Worked solution:

Worked Example 1.3: A 1.294 g sample of Na2CO3 was dissolved into a 500.0 mL volumetric flask, and a 20.00 mL aliquot extracted and added into a conical flask. What is the amount, in mol, of Na2CO3 present in the conical flask? Express your answer to the correct number of significant figures. Worked solution:


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Exponentials: 10a = b If a is to p decimal places, b is to p significant figures eg. 10-4.76 = 1.7 x 10-5 (2 dp à 2 sf) Logarithms: log10 c = d If c is to p significant figures, d is to p decimal places

eg. log10 (1.74 x 10-4) = -3.759 (3 sf à 3 dp) Worked Example 1.4: Calculate [H+] for a solution of pH 1.34. Worked solution:

Worked Example 1.4: Calculate the pH of a solution that has a [H+] of 0.0342 M. Worked solution:


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Units The unit conversions are as follows: x 10-9 x 10-6 x 10-3 x 10-0 x 103 x 106

nano micro milli kilo mega n µ m k M The more ‘iffy’ unit conversions are the % (w/v), ppm and ppb. The three measurements are analogous to each other. The % is a part per 100, ppm a part per million and ppb a part per billion. Those three measurements are measuring a ratio between a mass and a volume, which is quite odd. The basis of these calculations is the assumption that: 1 g = 1 mL ßà 1000 g = 1 L as the density of water is about 1 g/mL, and (especially for dilute solutions) the mass of solute is considered negligible. This spawns the following equivalent measurements: 1% (w/v) = 0.01g/mL = 1 g/(100 mL) 1 ppm = 1 µg/mL = 1 mg/L 1 ppb = 1 ng/mL = 1 µg/L I’m being a little detailed here so that on the chance that you forget these equivalent measurements you could quickly derive them. So if we have 0.050 g of NaCl dissolved in 1 mL of water, we can say that we have a 0.050/1 x 100% = 5.0% (w/v) solution of NaCl. Worked Example 1.5: UV-visible spectroscopy was used to measure the concentration of iron in an iron sulphate tablet. It is found that the concentration of iron in a 20.00 mL solution with the iron tablet dissolved is 139 ppm. What is the amount, in mol, of iron present in the tablet? Worked solution:


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CHEMICAL ANALYSIS Now that we have the skills to perform calculations, it is now time to cover the analytical techniques. Before we do this though, we need to know the central concept involved in all of these techniques. The central concept is this: chemical analysis is about WHAT the stuff is, and HOW MUCH there is. To work out WHAT the stuff is, we need to use our senses – sight, hearing (NOT touch, taste or smell, for obvious reasons). The problem is that we cannot see individual atoms. So how do we work out what the chemical is? In some cases, such as determining the difference between silver chloride and silver iodide precipitates, we can do this by just looking at the solids. Silver chloride is white, silver iodide is light green. There’s a difference. Therefore we can tell which chemical is which. In other cases, this is impossible. Imagine trying to find a needle in a haystack or trying to tell the difference between sulphuric and hydrochloric acids and water – which are all clear. What we need to do is find a DIFFERENCE between all of these chemicals that we can test for and exploit. That is the basis of qualitative chemical analysis. To work out HOW MUCH there is, we need to work out a property of the chemical that increases or decreases proportionally according to the amount of chemical there is. For instance, the intensity of the blue colour of a CuSO4 solution will increase as the CuSO4 becomes more concentrated. We can exploit this and that yields our quantitative analysis. Hence, here are the definitions of qualitative and quantitative analysis: Qualitative: where one simply wants to determine the identity of a compound, with no

consideration as to its amount.

Quantitative: where one wants to determine the amount/concentration of a certain compound

– so how much compound there is.


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WET-WAY ANALYSIS// 2. GRAVIMETRIC ANALYSIS

Here, the difference between chemicals we are exploiting is their solubility properties (qualitative) and we know that we can easily measure the amount of chemical by working out their mass (qualitative) as we can isolate the precipitate. The theme behind gravimetric analysis is this: the formation and weighing of a precipitate. Hence, we need to know what common kinds of ionic compounds precipitate. Ion Solubility Exceptions* (except

trivial ones) Notes

Group 1, NH4+, NO3

- always soluble none (in scope of VCE)

Cl-, Br-, I- usually soluble AgX (s), PbX2(s) HgI2 is also insoluble SO4

2- usually soluble SrSO4 (s) BaSO4 (s) PbSO4 (s)

Ag2SO4 and CaSO4 are slightly soluble

OH- rarely soluble Ba(OH)2 Ca(OH)2 is slightly soluble

CO32-, PO4

3-. S2-, O2- rarely soluble In choosing which precipitate to form (if we want to precipitate for instance Ag+ ions, do we add CO3

2- ions or Cl- ions or NO3- ions) a desirable precipitate:

• has a known formula (so no variable water of hydration) • does not react with atmosphere • does not decompose under heating • should generally have a high molar mass General steps in gravimetric analysis: • Dissolve the sample in deionised/distilled water, or treat the solution such as to isolate

the ions of interest (for instance through filtration to get rid of insoluble impurities). • Add an excess of reagent that would form a precipitate. • Filter the solution, and wash the precipitate with cold distilled/deionised water. • Dry/heat the precipitate and weigh it – repeat until constant mass.


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Worked Example 2.1 (ATARnotes Chemistry Unit 3 Study Guide, AOS 1 Test 1, Question 1 modified) Dave wants to determine the percentage by mass of iron in a 0.320 g iron tablet. The iron in the tablet is present as iron (II) sulphate – FeSO4. Dave dissolves this iron tablet in a beaker of distilled water. To this solution she adds a calculated excess of BaCl2 (aq). A barium sulphate precipitate is formed. a. Write a balanced ionic equation for the precipitation reaction (including states).

2 marks

b. Why must the barium chloride be in excess?

1 mark

Dave then filters out the precipitate and washes it with cold de-ionised water. c. Why does Dave wash the precipitate with deionised water?

1 mark

d. Why is it optimal for the deionised water used for washing to be cold rather than hot?

1 mark

Dave then dries the precipitate and weighs it. She then proceeds to repeat this step multiple times until the mass obtained is constant. e. Why does Dave have to repeat the drying and weighing steps?

1 mark


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Dave finds the mass of precipitate to be 0.240 g.

f. Determine:

i. the amount, in moles, of sulphate ions that have been precipitated

ii. the mass, in grams, of iron in the iron tablet

iii. the percentage by mass of iron in the iron tablet

1 + 2 + 1 = 4 marks Another student, Pete, repeats the experiment but instead decides to intentionally add an excess of barium hydroxide - Ba(OH)2 – with the rest of the procedure being the same. g. Explain why adding Ba(OH)2 instead of BaCl2 could lead to a higher amount of

precipitate formed than when BaCl2 is used.

1 mark Total 11 marks


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3. VOLUMETRIC ANALYSIS In volumetric analysis we are exploiting acid-base or redox properties (in general) of the different chemicals. The basis of volumetric analysis is the titration, where we want to determine amounts or concentrations of stuff by using known amounts of stuff. However, how do we make up a known amount of stuff – a standard solution? We need a primary standard, something that we can weigh and dissolve. Characteristics of a primary standard:

• must have a known formula • must have a relatively high molar mass • must not react with the atmosphere • should be soluble in water

Examples of good primary standards are sodium carbonate and oxalic acid. In order to be able to tackle volumetric analysis, we need to have a sound understanding of acid-base chemistry. Acid-Base Chemistry • An acid is a proton donor. A base is a proton acceptor. • pH = -log10[H+] pOH = -log10[OH-] pH + pOH = 14

Strong acids are acids that completely donate its protons to water. Strong bases accept protons from water to completion. Weak acids partially donate, and weak bases partially accept.

Examples Strong acid HCl, HBr, HI, H2SO4, HNO3 Strong base OH- Weak acid CH3COOH, HF, NH4

+, H3PO4, HSO4-

Weak base CO32-, SO4

2-, NH3, F-

An acid, when its proton is donated, becomes its conjugate base. (eg. HCl à Cl-) A base, when its proton is accepted, becomes its conjugate acid. (eg. Cl- à HCl) A conjugate acid-base pair differs by one proton: eg. HCl/Cl-

The conjugate of a strong acid/base is a really bad base/acid (eg. HCl is a strong acid, Cl- is a horrible base and CH3O- is a strong base, and CH3OH [methanol] is a horrible acid). The conjugate of a weak acid/base is a weak base/acid (eg. CH3COOH is a weak acid, CH3COO- is a weak base).


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Indicators are chemicals that allow you to see when the equivalence point has been reached. Acid-base indicators are themselves weak acids/bases themselves where the acid form (say HA) has a distinctive colour from its conjugate base (say A-). The relative proportions of the acid and conjugate base (and hence the colour) depend on the pH of the solution. So, indicator gives an indication (forgive the pun) of pH. Acid-Base Titration Curves Let’s say we measure the pH of an aliquot of a basic solution, where the equivalence point is 25 mL.

Strong acid – strong base eg. HCl + OH- à Cl- + H2O At equivalence point, only salt is in solution, and conjugate base is a horrible base (as the acid was strong), hence pH = 7.

Strong acid – weak base eg. HCl + NH3 à NH4

+ + Cl- At equivalence point, salt is in solution, and since we used a weak base, the conjugate acid is also a weak acid and this will be in solution, hence pH < 7.

Strong base – weak acid eg. OH- + CH3COOH à H2O + CH3COO- At equivalence point, salt is in solution, and since we used a weak acid, the conjugate base will also be a weak base, hence pH > 7.


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Weak base – weak acid CH3COOH + NH3 à NH4

+ + CH3COO- At equivalence point, we have a mixture of the conjugate base and acid (from weak acid and base), so pH ~ 7. This would NOT be a good titration to do.

Redox Titrations The principles are the same here. There may be indicators (eg. starch for iodine) or one of the reagents can act as an indicator (eg. MnO4

- which is intensely coloured). Worked Example 3.1 (ATARnotes Chemistry Unit 3 Study Guide, AOS 1 Test 1, Question 2 - modified) Michelle wants to standardise a solution of HCl, which has a concentration of approximately 0.1 M. a. What is a standard solution?

1 mark

She weighs a 1.305 g sample of sodium carbonate (Na2CO3), which she quantitatively transfers into a 250.0 mL volumetric flask, to which she makes up to the mark with distilled water. She pipettes a 20.00 mL aliquot of the sodium carbonate solution into a conical flask. She loads the HCl solution into a burette. b. Circle the substance the following pieces of volumetric glassware should be washed with

immediately prior to this experiment, and explain why washing with the other alternatives is not optimal:

i. burette

small volume of acid small volume of base deionised water

ii. pipette



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iii. conical flask


1 + 1 + 1 = 3 marks

c. Which of the following indicators is the most appropriate for the titration? Circle one

answer and explain in the space below.

methyl red phenolphthalein

3 marks

Michelle titrates the hydrochloric acid solution with the sodium carbonate solution, and obtains an average titre of 19.57 mL. d. Calculate

i. the amount, in moles, of sodium carbonate in the aliquot


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ii. the exact concentration of the HCl solution

2 + 2 = 4 marks

e. Explain how leaving the funnel on the burette during the titration may increase or decrease the calculated concentration of HCl:

2 marks

Total 13 marks


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Back Titrations This is a titration technique performed if (among other reasons): • direct titration of the analyte with the standard solution implies a ‘strong-weak’ acid-base

reaction and • the product of the first reaction (see next set of dot points) can be expelled (for instance

CO2 or NH3) The steps involved are: • addition of a known amount of excess reactant to react with all of the analyte and

expulsion of all reactive products (eg. NH3) • determination of the amount of reactant that has not reacted with the analyte through

titration To solve these kind of problems, it is essential to think through each step of the technique. This way, it is MUCH easier. Especially with back titrations, you MUST keep track of your accuracies (‘sig figs’) throughout the calculations, as you are performing subtractions (where you monitor decimal places) as well as multiplication/divisions (where you monitor significant figures)!


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Worked Example 3.2 (VCAA 2008, Question 8)


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4. REDOX CHEMISTRY

• Oxidation is loss of electrons (or increase in oxidation number) • Reduction is gain of electrons (or decrease in oxidation number)

Writing redox half equations (we are going to assume that it’s in acidic solution):

1. Balance all heteroatoms (not H or O). 2. Balance O by adding H2O. 3. Balance H by adding H+. 4. Balance charge by adding electrons.

Combining half equations:

• Multiply each equation by the appropriate factor, so when you add the two equations the electrons cancel out, and you have your full equation. Don’t forget to cancel out any water or H+ on both sides!

Worked Example 4.1

Derive an overall equation for the oxidation of Fe2+ to Fe3+ by MnO4- ions, which

themselves get reduced to Mn2+ ions.

Worked Solution


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Oxidation Number This is a tool, an invention. But it works. The oxidation number system bases itself on the assumption that in a polar bond P—N (where N is more electronegative so the electron pair is closer to the N), both electrons ‘belong’ to the N. Rules for oxidation numbers: • Oxidation number of an element is 0. • The sum of the oxidation numbers of each atom is equal to charge of species (eg. NO3

- [N = +5, O = -2])

• H, being very electropositive, generally has +1, has -1 in hydrides when bonded with even more electropositive metals

• O, being second most electronegative atom, will be -2 , except peroxides (-1) and superoxides (-0.5), OF2 (F even more electronegative, O = +2)

• F = -1 unless its F2

Worked Example 4.2

Work out the oxidation numbers of the atom in brackets in the following species:

a. FeCl3 (Fe) 3 b. NaOH (Na) 1 c. NO (N) 2 d. Cl2O (Cl) +1 e. K2O2 (O) -1


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5. CHROMATOGRAPHY The main principle here is that we separate substances according to their interactions with the stationary phase and the mobile phase. The stationary phase is the substance that remains ‘still’ throughout the analysis and the mobile phase is the substance that passes through/over the stationary phase. The different types of chromatography we study are: • Thin-layer Chromatography (TLC) • Gas Chromatography (GLC or GSC) • High Performance Liquid Chromatography General Principles// • What determines the speed of the component through the stationary phase?

Intermolecular bonding. Consider hydrogen bonding, dipole-dipole interactions, dispersion forces.

• The less soluble/more adsorbed to the stationary phase a substance is, the slower it will

travel. • The more soluble/less adsorbed to the stationary phase a substance is, the faster it will

travel. These principles apply to all of the following chromatographic methods. Let us now have a look at each of the chromatographic methods in turn. Thin-layer chromatography// Of the chromatographic techniques you learn in Unit 3 Chemistry, this is the simplest and cheapest method. It is quite versatile – it can be used for many compounds, from drugs to amino acids. Thin-layer chromatography is primarily a qualitative technique; using this technique we can identify compounds, but we cannot accurately determine their amount.


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The setup is as follows (apologise for the lack of artistry here!): Setup

card coated with alumina or silica (stationary phase)

spots of sample to be analysed

To prepare a TLC, we use a capillary tube to ‘spot’ tiny samples of each compound to be analysed on a pencilled line on the card. The card is then placed in a cylindrical container containing eluent – the solvent that acts as the mobile phase. The eluent starts to move up the plate, carrying the sample with it. The spotted samples on the pencilled lines MUST be above the line of eluent initially, otherwise the samples will dissolve into the eluent instead of moving up the plate. After the eluent has moved up the plate a sufficient distance, the plate is removed from solution, a second pencilled line drawn where the eluent line (the solvent front) was, and the spots on the TLC plate observed under visible or UV light. Now, how do we differentiate between compounds using TLC? We differentiate them by how fast they travel up the column. Remember, the greater the affinity of the compound for the mobile phase and the less the adsorption of the compound to the stationary phase, the faster the compound will travel through the column. The measurement for how fast the compound travels through the column is the retardation factor (Rf) value. It is calculated as follows: Rf = component/solvent; ratio of distances travelled by the solvent and the component. Note that Rf value depends on the compound, the eluent, the stationary phase and temperature. How do we use this Rf value? Well, suppose that we perform a TLC on an unknown compound, and after the development of the chromatogram we have the solvent front travelling a distance of 5.0 cm, and the substance travelling 3.0 cm. What is the Rf value? It is found that, in previous experiments, paracetamol has an Rf value the same as that calculated in this experiment. This means that it is possible that this compound is paracetamol. Why cannot we definitively say that this compound is paracetamol?

A B C D E


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Polarity of Stationary and Mobile Phases Often, the stationary phase is a polar substance such as silica or alumina and the mobile phase (eluent) is a non-polar organic solvent such as ethyl ethanoate. Now, remember that like dissolves like. This means that polar substances tend to adsorb onto the polar stationary phase to a greater degree and non-polar substance tend to dissolve into the mobile phase better. Therefore, using these general principles, which substance will travel further up the plate – the polar or non-polar substances? Worked Example (VCAA 2008, Question 3b)


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Column Chromatographic Methods Before we embark on the other two chromatographic methods – HPLC and GC – let us explore the simple setup of column chromatography:

This column is like a burette without markings. The sample is pipetted onto the packed silica column, eluent added on top of the sample, and the tap opened. The eluent is allowed to go through the column and drip out of the burette, carrying the sample with it. Just like in TLC, the sample will adsorb onto and desorb from the silica. Again, the stronger the intermolecular forces between the silica and the sample (and the weaker the intermolecular bonds between the eluent and the sample), the slower the sample goes through the column. In TLC, we measured how fast the sample goes up the plate by measuring the retardation factor, a ratio of distances travelled by the eluent and the sample. In column chromatographic methods though, we measure the time taken for the sample to travel through the column. This is called the retention time. Question: the stronger the intermolecular bonds between the sample and the stationary phase, the higher or lower the retention time?


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The difference between this and TLC is that this method allows for quantitative analysis of the samples. We can actually collect the fractions of sample as it comes out of the column and perform analyses that will test for their amount. Now, one thing. You’d rather use finer pieces of silica than coarse – because there is a higher surface area of silica with which the sample can potentially adsorb. This means that the more adsorptive components of the sample will separate better from the less adsorptive; using finer pieces of silica has the same effect as using a longer column. There’s a downside to this though; imagine trying to pass water through a column packed with big gravel pieces, then passing water through a column packed with sand. The water will pass much more slowly through the sand, wouldn’t it? It is also said that time is money – so time is valuable, therefore using finer pieces of silica has a downside to it – it takes longer. However, there is a way to get around this. How about using a hand-pump to pump the eluent under pressure through the column? This will make the eluent pass through the column faster, saving time. Even then though, it would make sense if we could use extremely fine pieces of silica to increase separation – but hand pumps would not be able to pressurise the eluent to 14000 kPa (about 140 atmospheres of pressure). How about if we use a really powerful machine to pump the eluent at 14,000 kPa through a really robust column able to withstand such pressures – and use an electronic detector to measure how much of each compound in the sample is passed through the column? Well, we are doing this already: it’s called high performance liquid chromatography (HPLC). High Performance Liquid Chromatography High performance liquid chromatography is effectively identical to the basic form of column chromatography, except that it is mechanised. The stationary phase is generally alumina or silica, and the mobile phase generally a non-polar organic solvent. As a method of column chromatography, the Rt value is used to identify compounds. How to interpret HPLC chromatograms will be discussed later. Gas Chromatography Gas chromatography is another form of column chromatography; it is the most accurate and sensitive of chromatographic techniques – and the most expensive. Gas chromatography is somewhat unusual in that: • the stationary phase is either a solid (eg. alumina in gas-solid chromatography) or a liquid

(high boiling point hydrocarbon or ester in gas-liquid chromatography) coated on a porous solid

• the mobile phase (eluent) is an inert gas (eg. H2, N2) • the sample is in the gaseous phase


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Given that the states of the stationary and mobile phases are different, we need to consider our intermolecular bonding a little differently. Recall from Unit 2 Chemistry that the ideal gas law (pV = nRT) is based on the assumption that there are no intermolecular bonds between gas molecules. This means that there are no intermolecular bonds between the sample and the mobile phase. This means that we can only consider intermolecular forces between the stationary phase and the sample. So, suppose we are trying to separate a mixture of straight-chain alkanes using gas-liquid chromatography where the stationary phase is a high boiling point hydrocarbon. Which alkane will have the greatest retention time? The fact that the sample has to be in the gaseous phase in the column raises two points: • compounds that are not easily vaporised – they don’t have a low boiling point – cannot

be detected using gas chromatography; the general rule is if the molar mass is > 300 g mol-1 the compound cannot be detected

• compounds that decompose at high temperatures (eg. monosaccharides and amino acids) also cannot be analysed as attempting to vaporise them in the oven will make them decompose

Analysing chromatograms for GC and HPLC The chromatograms for GC and HPLC have identical features. Consider the following examples of a chromatogram (on page 27): Remember that the retention time identifies the compound; therefore, the location of the peak identifies the compound. The area under the peak that the compound gives rise to gives information about the amount of that compound. It is found that, at low amounts, the area under the peak is proportional to the amount of compound. However, how do we know what amount corresponds to what area? If I get an area of 12,000 square units, how do I know whether that corresponds to 1 mol, or 2 mol, or 0.1 mol? This is where calibration curves come in. We can plot a calibration curve by taking samples of the compound of known amount and determining the area under the peak at its retention time. With this calibration curve, we now know a relationship between the area under the peak and the amount of compound – and therefore when we measure the area under the peak of an unknown amount of compound to be 12,000 square units, we can tell what amount of compound that corresponds to. Another note: calibration graphs cannot be extrapolated accurately; if the area under the peak of an unknown amount is outside the range of the calibration curve, it is not valid to use that curve to determine the amount of compound in that sample.


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Worked Example (VCAA 2011, Question 3)


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6. SPECTROSCOPY (AAS + UV-VIS) Just as chromatography differentiated between compounds by their relative attractions to the stationary and mobile phases, spectroscopy does it by their different patterns of absorption of electromagnetic radiation. Note that this section will cover atomic absorption spectroscopy and UV-visible spectroscopy only. Infra-red and nuclear magnetic resonance spectroscopies – and mass spectrometry – will be covered in the AOS 2 section (although they are AOS 1 material), as it is better to cover these in conjunction with organic chemistry. Electromagnetic radiation One way of looking at light is that light travels as particles of a particular frequency, measured in Hz. Below is a table of different types of light that is used in spectroscopy, from low to high frequency: Frequency Wavelength Type of Radiation Spectroscopy in

which it is used Low High Radio NMR . . Infra-red IR . . Visible AAS/UV-visible High Low Ultraviolet AAS/UV-visible Here are a few points to notes – these need not necessarily be memorised, but they should be recognised: • the frequency of the light ‘particle’ is proportional to the energy of the photon • the wavelength of the light ‘particle’ is inversely proportional to the frequency of the

particle – hence the wavelength is inversely proportional to the energy Now that we have an understanding of electromagnetic radiation, we can turn our attention to AAS and UV-visible spectroscopy. Atomic Absorption Spectroscopy An electron in an atom can be excited to a higher shell – a higher energy level. This electron will be at an unstable position, so will immediately drop down to its stable low energy level. In doing so, this electron emits a light particle (a photon) of a specific energy, hence a specific frequency/wavelength. It follows that if an electron in an atom can emit a photon of light of a particular energy (hence frequency/wavelength) in going down from its unstable energy level to its stable energy level, that same electron can emit a photon of the same energy (hence frequency/wavelength) in going from its stable energy level to its unstable energy level.


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How can we exploit this property to differentiate between different atoms? To do so, we need to work out how this pattern of light absorption is different between atoms and why this pattern is different. It turns out that (put simply) different atoms will have different differences in energies between electron shells, and therefore will absorb photons of specific different energies hence frequencies/wavelengths. This is because different elements have a different electronic configuration and nuclear charge. Note that the closer an electron is to the nucleus, the lower its energy. A higher nuclear charge brings all the electron shells closer to the nucleus (and therefore perhaps closer to each other as well), but too many electrons in the same shell means there is more repulsion between electrons in the same shell, and the energies of those electrons will be higher (electrons become more distant from nucleus). The setup is as follows:

The components of the atomic absorption spectrometer are: • cathode ray lamp – this is a lamp containing a small amount of the metal of interest

(there is a Na lamp, a K lamp, and so on). The electrons in the metal in the lamp are excited by a voltage and when they go back down to their stable energy levels, they emit light – which is shot at the sample to be analysed. The awesome thing about this is that all of the light emitted will be absorbed by the metal of interest and only the metal of interest. Why?

• atomiser – this is composed of a flame (fyi its due to combustion of C2H2) which turns the metal ion sample into the gaseous state and reduces the metal ion to the element. So, for instance, if we sprayed a sample of NaCl (aq) into this flame, the resulting species of interest will be Na (g) [not Na+ (g)]. The gaseous metal atoms floating in the flame will absorb the light shot at them from the lamp

• monochromator/detector – select wavelengths of light and detect light. Not too important to know the mechanisms of their function here.

Now, if there are only a few gaseous metal atoms floating in the flame, only a few of the emitted photons from the lamp will happen to hit the atoms and be absorbed – most will go through. If there are a lot of gaseous atoms in the flame, more of the emitted photons will happen to hit the metal atoms and be absorbed – less light will go through. Hence, we can exploit this and find the relationship between the amount of light absorbed (absorbance) and the concentration of the metal ion (in the sample that is sprayed onto the flame).


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It turns out that for low concentrations, absorbance and concentration of metal ion is proportional. UV-Visible Spectroscopy Put simply, UV-visible spectroscopy exploits bonding electrons in molecules being excited to higher energy levels, just as AAS exploits electrons in atoms being excited to higher energy levels. The exact nature of the energy levels in bonding electrons in molecules is well beyond the scope of the course. The principles of UV-visible spectroscopy are effectively the same – and so is the setup. The main difference is that you choose the wavelength to be absorbed (unlike in AAS). Therefore, you need to determine the most appropriate wavelength of light to be absorbed. This wavelength must satisfy the following conditions:

• the compound to be analysed must absorb quite strongly at that wavelength • no other compound that could be in the sample can absorb well at that wavelength,

otherwise it will interfere with the measurement of the compound to be analysed In order to select the right wavelength, we need a UV-visible spectrum, like the one below: Consider the following question – VCAA 2010 Question 9


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The mechanics are effectively the same. Light of a specific wavelength is shone through the sample (which is stored in a cuvette, a small clear cuboidal container), and the light remaining is detected, and the amount of light absorbed measured (again, measured in absorbance). Again, if there is a high concentration of compound, more light will be absorbed as more photons will hit the molecules and be absorbed. It also turns out that absorbance and concentration of the compound are proportional at low concentrations of compound. Determining concentrations of compound using AAS and UV-Vis Again, if we are given the absorbance of a solution at a particular wavelength, how does this translate into a concentration of compound in the sample? Well, remember that absorbance and concentration of compound are proportional – just like in GC and HPLC. How did we combat those? We used calibration curves. We can do it again here. We can get samples of known concentration and run them through the AAS or UV-Visible spectrometer and get their absorbances at the specific wavelength, and then draw a calibration curve plotting these points. Then we can run our actual sample through the AAS or UV-visible spectrometer and get its absorbance at the wavelength, and use the calibration line to work out the concentration of the compound. Like in the calibration curves in HPLC and GC, one cannot extrapolate outside the range of the calibration curve accurately.


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Worked Example (VCAA 2008, Question 2)


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END OF AOS 1

Chemistry AOS1 Unit 3 Notes

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