You’ve been handed an unlabelled bottle. Is it lead nitrate or barium chloride? Is the liquid inside ethanol or propanone? Module 8 is where HSC Chemistry turns into detective work: every test, every instrument and every spectrum is a clue, and the answer comes from putting the clues together. This page is part of our HSC Chemistry notes, which summarise all four Year 12 modules in one place.

These HSC Chemistry Module 8 notes follow the three inquiry questions of Module 8: Applying Chemical Ideas in the NSW Chemistry Stage 6 Syllabus. They’re written from the lessons our tutors teach in class, and each topic opens with the exact syllabus dot point it covers. Module 8 pulls together everything before it: precipitation from our HSC Chemistry Module 5 notes, titration from Module 6 and functional groups (the reactive group of atoms that gives a family its reactions) from Module 7.

These are theory and revision notes. Reading notes doesn’t replace the practical investigations the syllabus asks you to do in class.

Key Takeaways

Ions are identified by a sequence of tests: precipitation with anions, flame colours, and coloured complexes. Order matters, because one test can hide another.
Gravimetric analysis finds how much of an ion is present by weighing a pure, dry precipitate; precipitation titrations do it by volume.
Colourimetry, UV-visible spectrophotometry and atomic absorption spectroscopy (AAS) measure concentration from absorbance using a calibration curve.
Functional groups are identified by simple tests: bromine water for C=C, oxidation for –OH, sodium carbonate for carboxylic acids.
Infrared spectroscopy shows which bonds are present, mass spectrometry gives the molar mass and fragments, and ¹³C and ¹H NMR (nuclear magnetic resonance) show the carbon and hydrogen environments.
Designing a synthesis means balancing reagents, conditions, yield, purity, uses and environmental, social and economic costs.
Working Scientifically skills (graphs, error, evaluating a method, command words) are marked in many Module 8 questions: see the skills section.

For how to study all this effectively, see how to actually understand HSC Chemistry.

What Does HSC Chemistry Module 8 Cover?

Module 8 is the last of four Year 12 modules in the NESA Chemistry Stage 6 Syllabus (2017). It’s organised around three inquiry questions, and these notes follow the same order:

Syllabus section	Inquiry question
Analysis of Inorganic Substances	How are the ions present in the environment identified and measured?
Analysis of Organic Substances	How is information about the reactivity and structure of organic compounds obtained?
Chemical Synthesis and Design	What are the implications for society of chemical synthesis and design?

How this module is examined

Module 8 in the 2019–2025 HSC papers

Marks: questions involving Module 8 were worth 23 to 33 of the 100 marks in each paper (12 to 28 from Module 8-only questions).

Every year: examined in both sections. Analysis of inorganic substances and analysis of organic substances appeared in Section II in all seven papers (including shared questions).

Often: chemical synthesis and design appeared in Section II in five papers (2020 and 2022 to 2025, including shared questions).

Full year-by-year table, official NESA paper links and marking feedback by module: HSC Chemistry past papers 2019–2025.

Module 8 Figure 1. Module 8 is examined heavily: questions involving it were worth 23 to 33 of the 100 marks in each HSC paper from 2019 to 2025, and the two analysis inquiry questions appeared in Section II every year. Tallied from the Mapping Grids in NESA's HSC Chemistry Marking Guidelines 2019–2025 (© NSW Education Standards Authority), which show what was examined, not what will be.

How Are the Ions Present in the Environment Identified and Measured?

Syllabus dot pointanalyse the need for monitoring the environment

Why do we monitor the environment?

Ions from mining, farming, industry and sewage end up in waterways and soil. Some are toxic even in tiny amounts: lead and mercury are heavy metals that build up in living things (bioaccumulation) and damage the nervous system. Others cause harm in larger amounts, like phosphate and nitrate from fertiliser, which cause algal blooms (eutrophication) that starve rivers of oxygen. Monitoring lets us check that drinking water is safe, that industries meet legal limits, and that problems are caught before they spread.

Three reasons to monitor ions in the environment — **Module 8 Figure 2.** Ions are monitored for three reasons: heavy metals such as Pb²⁺ and Hg²⁺ bioaccumulate and damage nerves, nutrients such as PO₄³⁻ and NO₃⁻ cause algal blooms, and drinking water must be checked against legal limits. Small concentrations can matter, because lead is harmful at parts per billion.

Limits for trace ions are usually given in mg L⁻¹, but most calculations give mol L⁻¹. To convert, multiply by the molar mass (giving g L⁻¹) and then by 1000. For example, water with 2.4 × 10⁻⁶ mol L⁻¹ of Pb²⁺ contains 2.4 × 10⁻⁶ × 207.2 × 1000 = 0.50 mg L⁻¹ of lead, 100 times the Australian drinking water guideline of 0.005 mg L⁻¹ (NHMRC, endorsed 2025). In an exam, use the limit you are given. If the sample was diluted before it was analysed, multiply the result by the dilution factor.

Three boxes joined by arrows: 2.4 × 10⁻⁶ mol L⁻¹ of lead, multiplied by the molar mass 207.2 g mol⁻¹, gives 4.97 × 10⁻⁴ g L⁻¹, and multiplied by 1000 gives 0.50 mg L⁻¹. A note says to multiply by the dilution factor if the sample was diluted. Below, bars on a log scale show the sample at 0.50 mg L⁻¹ against the NHMRC guideline of 0.005 mg L⁻¹, 100 times the limit. — **Module 8 Figure 3.** Converting a concentration from mol L⁻¹ to mg L⁻¹ takes two multiplications, by the molar mass and then by 1000, and only then can it be compared with a limit. The lead sample here, 0.50 mg L⁻¹, is 100 times the Australian drinking-water guideline of 0.005 mg L⁻¹.

Syllabus dot pointconduct qualitative investigations – using flame tests, precipitation and complexation reactions as appropriate – to test for the presence in aqueous solution of the following ions: cations: barium (Ba2+), calcium (Ca2+), magnesium (Mg2+), lead(II) (Pb2+), silver ion (Ag+), copper(II) (Cu2+), iron(II) (Fe2+), iron(III) (Fe3+); anions: chloride (Cl–), bromide (Br–), iodide (I–), hydroxide (OH–), acetate (CH3COO–), carbonate (CO32–), sulfate (SO42–), phosphate (PO43–)

How to test for cations

Qualitative analysis answers the question “which ion is there?” Three kinds of reaction are used:

Precipitation: adding an anion that forms an insoluble salt with the cation.
Flame tests: heating the metal ion excites its electrons, which emit a characteristic colour as they fall back.
Complexation: a molecule or ion bonds around a metal ion to form a coloured complex ion.

Three Bunsen burner flames coloured apple-green, brick-red and blue-green by barium, calcium and copper — **Module 8 Figure 4.** Heating a metal ion in a flame excites its electrons, which emit light of characteristic colours as they fall back: barium gives apple-green, calcium brick-red and copper blue-green. The colours are approximate and are easiest to see against the blue inner cone of the flame.

A common sequence for an unknown containing one cation:

Add dilute HCl. A white precipitate means Ag⁺ (AgCl) or Pb²⁺ (PbCl₂). Lead chloride dissolves again in hot water; silver chloride doesn’t.
Add dilute H₂SO₄. A white precipitate means Ba²⁺ (BaSO₄) or Ca²⁺ (CaSO₄, which is only slightly soluble, so it may form only faintly). A flame test separates them: barium is apple-green, calcium is brick-red.
Add NaOH solution. The hydroxide colour identifies what’s left: Mg²⁺ white, Cu²⁺ blue, Fe²⁺ green, Fe³⁺ red-brown.

Flow chart testing an unknown cation with hydrochloric acid, sulfuric acid then sodium hydroxide — **Module 8 Figure 5.** Testing an unknown cation with dilute HCl, then dilute H₂SO₄, then NaOH identifies it step by step, and the order matters because an ion picked out early, such as Pb²⁺, would also precipitate with the later reagents. The chloride test picks out Ag⁺ and Pb²⁺, the sulfate test Ba²⁺ and Ca²⁺, and the hydroxide colours identify Mg²⁺, Cu²⁺, Fe²⁺ and Fe³⁺. CaSO₄ is only slightly soluble, so dilute Ca²⁺ may give only a faint precipitate.

Complexation confirms the transition metals. Cu²⁺ with excess ammonia forms deep blue [Cu(NH₃)₄]²⁺. Fe³⁺ with thiocyanate forms blood-red FeSCN²⁺ (the same complex from Module 5), while Fe²⁺ gives no red colour. Copper also gives a blue-green flame.

A complex ion is a metal ion with molecules or ions bonded around it, and in exams complex ions usually come with an equilibrium. The classic one is cobalt(II) chloride:

[Co(H₂O)₆]²⁺(aq, pink) + 4Cl⁻(aq) ⇌ [CoCl₄]²⁻(aq, blue) + 6H₂O(l), ΔH > 0

Adding chloride (for example concentrated HCl) or heating shifts it right and the solution turns blue. Adding water or cooling shifts it left and it turns pink. Because complex ions are coloured, colourimetry can measure their concentration, which is how an equilibrium constant for complex formation can be found from a calibration curve.

Pink hexaaquacobalt(II) and blue tetrachlorocobaltate(II) in equilibrium — **Module 8 Figure 6.** Pink [Co(H₂O)₆]²⁺ and blue [CoCl₄]²⁻ are in equilibrium, and the colour shows which way it has shifted. Because the forward reaction is endothermic, adding chloride or heating turns the solution blue, while adding water or cooling turns it pink, as Le Chatelier’s principle predicts.

A grid of eight cations with three rows. Key precipitate: Ag⁺ white AgCl and Pb²⁺ white PbCl₂ with dilute HCl; Ba²⁺ white BaSO₄ and Ca²⁺ faint white CaSO₄ with dilute H₂SO₄; with NaOH, Mg²⁺ white Mg(OH)₂, Cu²⁺ pale blue Cu(OH)₂, Fe²⁺ green Fe(OH)₂ and Fe³⁺ red-brown Fe(OH)₃. Flame colour: barium apple-green, calcium brick-red, copper blue-green, the rest none. Confirming test: AgCl does not dissolve in hot water; yellow PbI₂ with iodide; flame tests for barium and calcium; magnesium gives no sulfate precipitate and no flame colour; deep blue with excess ammonia for copper; no red with thiocyanate for Fe²⁺; blood-red with thiocyanate for Fe³⁺. — **Module 8 Figure 7.** Each of the eight cations has its own pattern across three tests: the precipitate it forms, its flame colour and a confirming reaction. Silver and lead are separated by hot water or iodide, barium and calcium by their flame colours, and copper and the two iron ions by complexes with ammonia or thiocyanate. Colours are as seen in a school lab; flame colours are easiest to see against the blue inner cone of a Bunsen flame.

How to test for anions

Carbonate (CO₃²⁻): add dilute nitric acid. Bubbles of CO₂ that turn limewater milky confirm carbonate. Test this first, because carbonate also precipitates with Ag⁺ and Ba²⁺.
Sulfate (SO₄²⁻): add acidified barium nitrate. A white BaSO₄ precipitate forms.
Phosphate (PO₄³⁻): add ammonium molybdate in nitric acid and warm. A yellow precipitate forms. (Phosphate also precipitates with Ba²⁺, but it dissolves in acid, unlike BaSO₄.)
Halides (Cl⁻, Br⁻, I⁻): add acidified silver nitrate. Cl⁻ gives white AgCl, Br⁻ cream AgBr, I⁻ yellow AgI. Their solubility in ammonia confirms which: AgCl dissolves in dilute NH₃, AgBr only in concentrated NH₃, and AgI not at all.
Hydroxide (OH⁻): once carbonate has been ruled out (it’s basic too), turns red litmus blue, and gives a brown Ag₂O precipitate with silver nitrate.
Acetate (the ethanoate ion, C₂H₃O₂⁻, drawn in Module 8 Figure 8): warm with dilute acid and the vinegar smell of ethanoic acid appears. Or warm with ethanol and concentrated sulfuric acid to make the fruity smell of an ester.

The acid is added first in most anion tests so that carbonate or hydroxide can’t form a misleading precipitate. Use a fresh sample for each anion test, and never acidify the portion you are testing for hydroxide, because the acid would use up the OH⁻ and hide it.

Tests for the eight anions, each on a fresh portion: 1 carbonate (test first) fizzes with dilute HNO₃ and the CO₂ turns limewater milky; 2 sulfate gives white BaSO₄ with acidified Ba(NO₃)₂; 3 phosphate gives a yellow precipitate with ammonium molybdate in HNO₃ when warmed; 4 halides with acidified AgNO₃ give white AgCl, cream AgBr and yellow AgI, and with ammonia AgCl dissolves in dilute NH₃, AgBr only in concentrated NH₃ and AgI not at all; 5 hydroxide, not acidified, turns red litmus blue and gives brown Ag₂O with AgNO₃; 6 acetate (the ethanoate ion, drawn) gives a vinegar smell when warmed with dilute acid, or a fruity ester smell with ethanol and concentrated H₂SO₄. — **Module 8 Figure 8.** Each anion is identified by its own test on a fresh portion of the sample: carbonate first, because it would also precipitate with Ag⁺ and Ba²⁺, then sulfate, phosphate and the halides in dilute nitric acid, with ammonia solubility telling the three silver halides apart. Hydroxide must not be acidified, because the acid would neutralise it, and acetate is detected by smell after warming with acid.

Syllabus dot pointconduct investigations and/or process data involving: gravimetric analysis; precipitation titrations

What is gravimetric analysis?

Gravimetric analysis finds the amount of an ion by converting it into a pure precipitate of known formula, then weighing it. For sulfate in fertiliser:

Weigh the sample and dissolve it in water.
Acidify with dilute HCl (so carbonate or phosphate can’t precipitate too), then add excess barium chloride solution so all the sulfate precipitates as BaSO₄.
Filter, and wash the precipitate with distilled water to remove soluble impurities.
Dry it in an oven and weigh it. Repeat drying and weighing until the mass stays constant.
Calculate: moles of BaSO₄ = mass ÷ 233.4 = moles of SO₄²⁻, then find the mass and percentage of sulfate.

Worked example. A 2.00 g fertiliser sample gives 1.62 g of dry BaSO₄. n(BaSO₄) = 1.62 ÷ 233.4 = 6.94 × 10⁻³ mol. Mass of SO₄²⁻ = 6.94 × 10⁻³ × 96.07 = 0.667 g, so the sample is 33.3% sulfate by mass.

Sources of error: an incompletely dried precipitate (too heavy, result too high), precipitate lost through the filter paper (too light, too low), or other ions co-precipitating (too high).

Filtration apparatus beside the five steps of a gravimetric analysis — **Module 8 Figure 9.** In gravimetric analysis, sulfate is converted into a pure BaSO₄ precipitate that is filtered, washed and dried, so its mass gives the moles of sulfate (mass ÷ 233.4). Drying to constant mass is what makes the result reliable, because leftover water would make the mass too high.

What is a precipitation titration?

A precipitation titration finds a concentration by titrating until an indicator shows all the analyte ion has precipitated (the analyte is the ion you are measuring). The standard example measures chloride by titrating with standard silver nitrate: Ag⁺ + Cl⁻ → AgCl(s). In the Mohr method, a little potassium chromate is added as the indicator. AgCl has a lower molar solubility than silver chromate, so AgCl forms first. The solution must be close to neutral pH for the indicator to work. Once all the chloride is used up, the next drop of Ag⁺ forms red-brown Ag₂CrO₄, marking the end point. The moles of Ag⁺ used equal the moles of Cl⁻ in the sample. In this case, adding silver nitrate to chloride, the end point is when the indicator’s precipitate first appears. As a shortcut for most indicator questions, the precipitate with the lower molar solubility tends to form first, but the real decider is the actual ion concentrations and whether the ion product exceeds K_sp. If the titration is run the other way, with chloride added to silver, the red Ag₂CrO₄ is there from the start and disappears at the end point (red to yellow), because the last silver ions are used up by chloride. Compare molar solubilities, not K_sp values, when the formulas have different ion ratios: Ag₂CrO₄ has the smaller K_sp but the higher molar solubility.

Bars on a log scale: Ksp of AgCl 1.77 × 10⁻¹⁰ and of Ag₂CrO₄ 1.12 × 10⁻¹², but molar solubility of AgCl 1.33 × 10⁻⁵ mol L⁻¹ (the square root of Ksp) and of Ag₂CrO₄ 6.5 × 10⁻⁵ mol L⁻¹ (the cube root of Ksp ÷ 4), so AgCl forms first. Below, flasks show the colour sequence: adding Ag⁺ to chloride, white AgCl forms in yellow solution and red-brown Ag₂CrO₄ appears at the end point; adding chloride to Ag⁺, the red Ag₂CrO₄ is present at first and disappears at the end point. — **Module 8 Figure 10.** In a Mohr titration silver chloride precipitates before silver chromate, because at the concentrations used its ion product exceeds its Ksp first, so the indicator reacts only once the chloride has run out. Silver chromate has the smaller Ksp but is the more soluble salt, because its formula contains two silver ions, so Ksp alone cannot predict the order, and very dilute chloride can reverse it. The direction of the titration decides the colour change: red-brown appears at the end point when silver is added to chloride, and disappears when chloride is added to silver.

Conical flask during a Mohr titration and at the end point — **Module 8 Figure 11.** In the Mohr titration, silver nitrate precipitates white AgCl first, and red-brown silver chromate forms only once the chloride has run out, which marks the end point. The moles of Ag⁺ added at the end point equal the moles of Cl⁻ in the sample.

How to calculate a back titration for chloride

In a back titration, a known excess of silver nitrate is added to the chloride sample. All the chloride precipitates as AgCl, the AgCl is filtered off, and the Ag⁺ left over is titrated with standard potassium thiocyanate (KSCN), using Fe³⁺ as the indicator. Ag⁺ + SCN⁻ → AgSCN(s), and the first excess SCN⁻ forms red FeSCN²⁺. The chloride is whatever silver wasn’t left over:

n(Cl⁻) = n(Ag⁺ added) − n(SCN⁻ used)

Worked example. 20.00 mL of 0.1000 mol L⁻¹ AgNO₃ is added to a 25.00 mL water sample. After filtering, the leftover Ag⁺ needs 12.40 mL of 0.08000 mol L⁻¹ KSCN.

n(Ag⁺ added) = 0.1000 × 0.02000 = 2.000 × 10⁻³ mol
n(Ag⁺ left over) = n(SCN⁻) = 0.08000 × 0.01240 = 9.920 × 10⁻⁴ mol
n(Cl⁻) = 2.000 × 10⁻³ − 9.920 × 10⁻⁴ = 1.008 × 10⁻³ mol
c(Cl⁻) = 1.008 × 10⁻³ ÷ 0.02500 = 0.04032 mol L⁻¹
In mg L⁻¹: 0.04032 × 35.45 × 1000 = 1.429 × 10³ mg L⁻¹ of chloride

Two slips cost marks here: forgetting to subtract the leftover silver, and multiplying by the molar mass of AgCl (143.3) instead of Cl (35.45) when the question asks for chloride.

Bar showing the silver added split into the part that reacted with chloride and the part left over — **Module 8 Figure 12.** In a back titration, a known excess of silver is added and only the leftover is titrated, so the chloride is found by subtraction: 2.000 × 10⁻³ mol of Ag⁺ added minus 9.920 × 10⁻⁴ mol left over leaves 1.008 × 10⁻³ mol that reacted with Cl⁻. The chloride is the silver that was not left over.

Syllabus dot pointconduct investigations and/or process data to determine the concentration of coloured species and/or metal ions in aqueous solution, including but not limited to, the use of: colourimetry; ultraviolet-visible spectrophotometry; atomic absorption spectroscopy

How does colourimetry work?

A coloured solution absorbs some wavelengths of light and transmits the rest. A deep blue copper solution absorbs orange-red light, the complementary colour. The more concentrated the solution, the more light it absorbs. The absorbance (A) is proportional to concentration (the Beer–Lambert law, A = εcl, where ε is a constant for that substance at that wavelength, c is the concentration and l is the path length, the width of the cuvette), which makes a straight-line relationship.

$The light path of a UV-visible spectrophotometer: a lamp gives white light, a diffraction grating splits it into colours, a slit lets one wavelength through, the beam I₀ passes through a cuvette of blue sample of path length l, and a weaker beam I reaches a detector and an absorbance meter. A colour wheel shows complementary colours opposite each other, with an arrow from blue to orange.$

Module 8 Figure 13. A spectrophotometer measures how much of one chosen wavelength a sample absorbs: a grating and slit select the wavelength, the beam passes through a set path length l of solution, and the detector compares the transmitted light I with the incident light I₀. The wavelength chosen is the complementary colour of the solution, so a blue copper solution is measured with orange light; a colourimeter uses a coloured filter instead of the grating and slit.

A colourimeter shines light of one colour, chosen as the colour the sample absorbs most, through the sample and measures how much gets through. To find an unknown concentration:

Prepare a series of standard solutions of known concentration.
Measure the absorbance of each, and plot a calibration curve of absorbance against concentration.
Measure the unknown’s absorbance, then read its concentration off the curve.

The unknown must fall inside the range of the standards. If it’s too concentrated, dilute it and multiply back.

When you draw the calibration curve, follow the graphing conventions: label both axes with the quantity and its unit (absorbance has no unit), use an even scale, plot each point, and draw a straight line of best fit rather than joining the dots. If two substances share one graph, use a different symbol for each and add a key. The full method is in the Working Scientifically skills section at the end.

Calibration curve of absorbance against concentration with an unknown read off — **Module 8 Figure 14.** A calibration curve turns an absorbance into a concentration: the unknown’s absorbance of 0.46 is read across to the best-fit line and down to 0.25 mol L⁻¹. Only concentrations inside the range of the standards can be read reliably.

UV-visible spectrophotometry

A UV-visible spectrophotometer does the same job more precisely. It uses a prism or diffraction grating to select one exact wavelength, usually the wavelength of maximum absorbance (λ_max), where the measurement is most sensitive. It also reaches ultraviolet wavelengths, so it can measure colourless compounds that absorb UV light. A calibration curve is used in the same way.

Two choices make the measurement fair. Choose a solvent that absorbs almost nothing at the wavelength you are measuring, so it doesn’t add to the reading. In a mixture, pick a wavelength where your substance absorbs and the others absorb almost nothing, even if that is not the tallest peak of your substance.

Schematic absorbance spectra from 400 to 700 nm: substance X peaks near 520 nm with a shoulder near 455 nm; interfering substance Y peaks near 600 nm, still absorbs at 520 nm but absorbs almost nothing at 455 nm; the solvent is nearly flat at zero. A marker at 520 nm says not here, because Y absorbs too, and a marker at 455 nm says measure here, because Y absorbs almost nothing there. — **Module 8 Figure 15.** In a mixture the best wavelength is one where only your substance absorbs appreciably, even if it is not its tallest peak: at 520 nm the interfering substance would add to the absorbance, but at 455 nm it contributes less than 1% of the reading. The spectra are schematic.

Absorbance spectrum of deep blue tetraamminecopper(II) with its peak near 610 nanometres — **Module 8 Figure 16.** Measuring at λmax, the wavelength of maximum absorbance, gives the most sensitive reading. The absorbance spectrum shows how strongly deep blue [Cu(NH₃)₄]²⁺ absorbs at each wavelength, and its λmax is near 610 nm. Plain Cu²⁺(aq) absorbs further toward the red, with its peak near 800 nm in the near infrared.

How does atomic absorption spectroscopy (AAS) work?

AAS, developed in Australia by Alan Walsh at the CSIRO in the 1950s, measures the concentration of a specific metal, even at parts-per-million levels (ppm: parts per million, about mg per kg of sample).

A hollow cathode lamp made of the metal being tested emits exactly the wavelengths that metal’s atoms absorb.
The sample is sprayed into a flame, which breaks it into free, ground-state atoms.
Those atoms absorb the lamp’s light. Only the target metal absorbs these exact wavelengths, so other metals interfere very little.
A monochromator selects one wavelength and a detector measures how much light is left. Absorbance is compared against a calibration curve.

AAS is used to measure heavy metals like lead and mercury in water, soil and blood.

An atomic absorption spectrometer along one light beam: a hollow cathode lamp whose cathode is the metal being measured sends light of intensity I₀ through a flame; a nebuliser below draws sample solution up from a beaker and sprays it with fuel gas and air into the flame, where it becomes free atoms; a weaker beam I passes through a monochromator to a detector and a readout of absorbance. An inset shows a lead lamp emitting lines at 217 and 283 nm, those lines weakened after lead atoms in the flame absorb them, and copper absorbing only at 325 and 327 nm, where the lamp gives no light. — **Module 8 Figure 17.** AAS can measure one metal with very little interference from others, because the lamp’s cathode is made of the element being measured, so the lamp emits exactly the wavelengths that element’s free atoms absorb in the flame. Lead atoms absorb the lead lamp’s lines, while copper atoms absorb only at wavelengths the lamp does not emit, and the drop from I₀ to I gives the absorbance.

How Is Information About the Reactivity and Structure of Organic Compounds Obtained?

Syllabus dot pointconduct qualitative investigations to test for the presence in organic molecules of the following functional groups: carbon–carbon double bonds; hydroxyl groups; carboxylic acids (ACSCH130)

How to test for functional groups

C=C double bond: shake with bromine water. An alkene decolourises it immediately (orange-brown to colourless) through addition. An alkane doesn’t react with bromine water (without UV light), so the orange colour stays.
Hydroxyl group (–OH) in an alcohol: warm with acidified potassium dichromate. Primary and secondary alcohols turn it from orange to green; tertiary alcohols don’t. A second option is sodium metal, which fizzes with H₂, but it reacts with any –OH, including water and acids. To avoid a false positive, use a dry sample and first rule out a carboxylic acid with sodium carbonate.
Carboxylic acids (the carboxyl group): add sodium carbonate (or hydrogen carbonate). It fizzes with CO₂, which alcohols don’t do. Use a separate portion of the sample for this exclusion test. It also turns blue litmus red.
Oxidisable groups (alkenes, primary and secondary alcohols): acidified potassium permanganate goes from purple to colourless. It reacts with alkenes and with primary and secondary alcohols, but not with alkanes or typical saturated carboxylic acids such as propanoic acid (methanoic acid is an exception, because permanganate oxidises it). For a mixture, remember that small alcohols and acids mix with water but hydrocarbons do not.

Four candidate liquids drawn skeletally: propanoic acid, hex-1-ene, propan-1-ol and hexane. A test order using a fresh portion each time: 1, sodium carbonate solution fizzes with a carboxylic acid; if not, 2, bromine water turns from orange to colourless with an alkene; if not, 3, warm acidified potassium dichromate turns from orange to green with a primary or secondary alcohol; no change in any test means an alkane. Below, skeletal equations: propene plus Br₂ gives 1,2-dibromopropane; propan-1-ol is oxidised by Cr₂O₇²⁻/H⁺ to propanoic acid; two propanoic acid plus Na₂CO₃ give two sodium propanoate plus CO₂ plus H₂O. — **Module 8 Figure 18.** Three functional-group tests, done in a sensible order on fresh portions, identify an unknown as one of the four liquids shown: sodium carbonate fizzes only with a carboxylic acid, bromine water is decolourised by the addition reaction of an alkene, and acidified dichromate turns green when it oxidises a primary or secondary alcohol. An alkane gives no change in any of them. A negative result leaves the reagent unchanged.

Syllabus dot pointinvestigate the processes used to analyse the structure of simple organic compounds addressed in the course, including but not limited to: proton and carbon-13 NMR; mass spectrometry; infrared spectroscopy (ACSCH130)

How does mass spectrometry work?

A mass spectrometer measures the mass of molecules and of the pieces they break into.

The sample is vaporised and bombarded with high-energy electrons, which knock an electron off each molecule to form a positive molecular ion (M⁺). Many molecular ions break apart into smaller fragments.
The ions are accelerated and deflected by a magnetic field. Lighter ions are deflected more, so ions are sorted by their mass-to-charge ratio (m/z).
A detector records how many ions arrive at each m/z value.

A magnetic-sector mass spectrometer from left to right: a heated sample inlet; an ionisation chamber where an electron beam from a hot filament ionises the molecules (stage 1); charged plates that accelerate the positive ions into a beam (stage 2); a curved flight tube in a magnetic field, where light ions curve most and hit the inner wall, heavy ions curve least and hit the outer wall, and ions of one m/z reach the exit slit (stage 3); a vacuum pump on the tube; and a detector linked to a computer that builds the spectrum (stage 4). A box gives M + e⁻ → M⁺ + 2e⁻. — **Module 8 Figure 19.** A mass spectrometer sorts ions by mass-to-charge ratio (m/z): it turns molecules into positive ions, accelerates them, and deflects them in a magnetic field, where lighter ions are deflected more, so only ions of one m/z reach the detector at a given field strength. Scanning the field brings each m/z to the detector in turn, and the ion counts become the mass spectrum. The tube is kept under vacuum so the ions do not collide with air molecules.

Reading a mass spectrum:

The molecular ion peak (the peak at the highest m/z, ignoring the small M+1 peak and halogen isotope peaks) gives the molar mass of the compound.
The tallest peak is the base peak, the most abundant ion (usually a stable fragment), set to 100%.
The gaps between peaks show which pieces broke off: a loss of 15 is a CH₃, a loss of 17 an OH, a loss of 29 a C₂H₅ or CHO.
Chlorine and bromine show up as pairs of peaks two units apart, because each has two common isotopes. Chlorine is about 75% ³⁵Cl and 25% ³⁷Cl, so an ion with one Cl gives peaks at M and M+2 in a 3 : 1 ratio. Bromine is about half ⁷⁹Br and half ⁸¹Br, so one Br gives two equal peaks at M and M+2. With two Cl atoms you see three peaks, M, M+2 and M+4, in a ratio of about 9 : 6 : 1. The tallest is then the M peak (both atoms ³⁵Cl), not the peak furthest to the right, so don’t pick the furthest-right peak automatically. The 2023 HSC question on a dichlorinated compound used m/z 114 and also accepted 112, so state which peak you used.

For ethanol, the molecular ion is at m/z 46, and the base peak at m/z 31 is the [CH₃O]⁺ fragment, the carbon bonded to oxygen with its hydrogens, formed when the methyl group (15) breaks off.

Ethanol drawn as a full structural formula with all six hydrogens, with brackets marking the m/z 29 piece (the two carbons and five hydrogens) and the m/z 31 piece (the carbon bonded to oxygen, the oxygen and the O–H hydrogen), above its mass spectrum: molecular ion at 46; 45 from loss of the H on the carbon bonded to oxygen, drawn as an ion with a C=O⁺ double bond and the O–H hydrogen kept; base peak at 31 from loss of CH₃; and peaks at 29 and 27 ([C₂H₃]⁺). — **Module 8 Figure 20.** The mass spectrum of ethanol shows its molecular ion at m/z 46, and the base peak at m/z 31 comes from losing the CH₃ group (46 − 15), leaving the piece that contains the oxygen. The brackets on the drawn structure show which atoms each main fragment contains. Peak heights from the NIST Chemistry WebBook, simplified.

Molecular ion peaks of chloroethane (m/z 64 and 66, heights 3 : 1) and bromoethane (m/z 108 and 110, heights 1 : 1), each drawn skeletally, and below them the molecular-ion cluster of 1,2-dichloropropane at m/z 112, 114 and 116 in the ratio about 9 : 6 : 1, labelled with the chlorine isotopes each peak contains. — **Module 8 Figure 21.** Chlorine and bromine each have two common isotopes, so ions containing them show peaks two mass units apart: one Cl gives M and M + 2 in a 3 : 1 ratio, one Br gives two nearly equal peaks, and two Cl give M, M + 2 and M + 4 in about 9 : 6 : 1. With two chlorines the tallest peak of the cluster is M, not the right-most peak, so the pattern tells you how many halogen atoms the ion contains.

How to read an infrared spectrum

Covalent bonds vibrate, and each type of bond absorbs infrared light at a characteristic frequency, given as a wavenumber in cm⁻¹. An IR spectrum plots transmittance against wavenumber, so each absorption shows as a dip.

The HSC data sheet lists the key absorptions, and Module 8 Figure 22 places each one on the wavenumber axis. The ones to recognise on sight are the broad, strong alcohol O–H (3230–3550 cm⁻¹), the very broad acid O–H (2500–3000 cm⁻¹), N–H in amines and amides (3300–3500 cm⁻¹), C–H (2850–3300 cm⁻¹ on the data sheet, 2850–3000 for alkane C–H, present in almost every organic molecule), the sharp, strong C=O of aldehydes, ketones, acids, esters and amides (1680–1750 cm⁻¹) and C=C (1620–1680 cm⁻¹).

Chart of the HSC data sheet infrared absorptions on a wavenumber axis from 4000 to 500 per centimetre, each row with a small structure and the vibrating bond coloured: N–H 3300 to 3500, alcohol O–H 3230 to 3550 broad and strong, C–H 2850 to 3300, acid O–H 2500 to 3000 very broad, C≡N 2220 to 2260, C=O 1680 to 1750 strong, C=C 1620 to 1680, C–O 1000 to 1300 and C–C 750 to 1100 in the fingerprint region. Three boxes: C=O with a very broad O–H is a carboxylic acid; C=O without O–H is an aldehyde, ketone, ester or amide; a broad O–H without C=O is an alcohol. — **Module 8 Figure 22.** Each bond absorbs in a fixed band of wavenumbers, so the position and shape of a dip identify the bond: alcohol O–H is broad and strong (very broad in acids), C=O is sharp and strong near 1700 cm⁻¹, and everything below 1500 cm⁻¹ is the fingerprint region. A functional group is identified from the combination of bands, such as C=O plus a very broad O–H for a carboxylic acid. Ranges are from the NESA HSC data sheet.

The region below about 1500 cm⁻¹ is the fingerprint region, with a pattern unique to each molecule. It’s used to match an unknown against a reference spectrum, not to identify groups.

So IR tells you which functional groups are present: a C=O dip plus a broad O–H means a carboxylic acid; a C=O without O–H could indicate an aldehyde, ketone, ester or amide. Combine the molecular formula, other absorptions and NMR evidence to distinguish them.

Butan-1-ol above its simplified infrared spectrum — **Module 8 Figure 23.** A broad O–H dip with no C=O dip identifies an alcohol rather than an acid or aldehyde. The infrared spectrum of butan-1-ol shows a broad O–H dip near 3300 cm⁻¹, C–H near 2950 cm⁻¹ and a C–O dip in the fingerprint region, but no C=O near 1700 cm⁻¹.

How does carbon-13 NMR work?

Nuclear magnetic resonance (NMR) uses the fact that some nuclei, like ¹H and ¹³C, behave like tiny magnets. In a strong magnetic field they absorb radio waves at frequencies that depend on their chemical environment, meaning the atoms around them. The position of each signal is its chemical shift (δ) in ppm, measured relative to a reference compound, TMS (tetramethylsilane), set at 0 ppm.

In a ¹³C NMR spectrum:

The number of peaks equals the number of different carbon environments. Carbons in identical positions by symmetry give one peak.
The chemical shift shows what each carbon is bonded to. Carbons near electronegative atoms appear further left (higher δ): about 5–40 ppm for C–C, 50–90 ppm for C–O, and 160–220 ppm for C=O. The data sheet gives the full table, charted in Module 8 Figure 24.

Carbon-13 chemical shift ranges from the HSC data sheet on an axis from 220 to 0 ppm, each with a drawn fragment and the measured carbon coloured: alkyl C 5 to 40, C–Cl or C–Br 10 to 70, carbon next to C=O 20 to 50, C–N 25 to 60, C–O in alcohols, ethers or esters 50 to 90, C=C 90 to 150, C≡N 110 to 125, benzene ring 110 to 160, acid or ester C=O 160 to 185, aldehyde or ketone C=O 190 to 220. — **Module 8 Figure 24.** Each carbon environment has its own chemical-shift range, and carbons bonded to oxygen or another electronegative atom sit further left: alkyl carbons at 5–40 ppm, carbons single-bonded to oxygen at 50–90 ppm, and C=O carbons at 160–220 ppm, with aldehyde and ketone C=O furthest left. C=C and benzene carbons (90–160 ppm) also sit well left without an electronegative atom. Ranges are from the NESA HSC data sheet.

Propan-1-ol has 3 carbon environments, but propan-2-ol has only 2, because its two CH₃ groups are identical. Propanone has 2 as well, with one peak above 200 ppm for the C=O.

Propan-1-ol and propan-2-ol drawn as full structural formulas with their carbons lettered, above their carbon-13 NMR spectra: propan-1-ol has three peaks near 64 (C–O), 26 and 10 ppm; propan-2-ol has two peaks near 64 (C–O) and 25 ppm, because its two end carbons are identical. — **Module 8 Figure 25.** The number of ¹³C peaks equals the number of carbon environments, so symmetry reduces the count: propan-1-ol’s three carbons are all different and give three peaks, but propan-2-ol’s two end carbons are identical and its three carbons give only two. In both, the carbon bonded to oxygen appears furthest left, near 64 ppm.

The same counting lets you follow a reaction. When propanone is reduced to propan-2-ol, the C=O peak near 207 ppm shrinks while a C–O peak near 64 ppm grows, and a sample taken partway through shows all four peaks.

Three carbon-13 spectra on one axis from 220 to 0 ppm for the reduction of propanone to propan-2-ol: at the start, propanone gives peaks near 207 (C=O) and 31 ppm; partway, the mixture gives four peaks at 207, 64, 31 and 25; at the end, propan-2-ol gives peaks near 64 (C–O) and 25 ppm. — **Module 8 Figure 26.** Carbon-13 NMR can follow a reaction because reactant and product have different carbon environments: as propanone is reduced to propan-2-ol, the C=O peak near 207 ppm disappears and a C–O peak near 64 ppm appears, while the methyl peak near 31 ppm is replaced by one near 25 ppm; partway, the mixture shows all four peaks. Both molecules are symmetrical, so each gives only two peaks.

How to read a proton (¹H) NMR spectrum

Proton NMR gives four pieces of information about the hydrogens:

Number of signals (a split group counts as one signal) = number of different hydrogen environments, unless two environments have such similar shifts that they overlap.
Chemical shift (δ): hydrogens near electronegative atoms appear further left. Roughly 0.7–2.1 ppm for C–H in a hydrocarbon chain (most CH₃/CH₂ at 0.9–1.5), 2–2.7 ppm on a carbon next to C=O, 3.3–4.5 ppm on a carbon bonded to O, 4.5–6.5 ppm on a C=C, 9–10 ppm for an aldehyde H and 10–12 ppm for a carboxylic acid H (charted in Module 8 Figure 27). An HSC question supplies its own table, so use the one you are given.
Relative peak area (integration): the area of each peak is proportional to the number of hydrogens in that environment. (Integration measures area, so spectra show it as a stepped line whose step heights are in the ratio of the numbers of hydrogens, as in the ethanol spectrum (Module 8 Figure 29).) If two environments overlap into one signal, their areas add: a 2H environment and a 4H environment show as one 6H signal.
Splitting pattern (the n + 1 rule): a peak is split by the hydrogens on the neighbouring carbons. If there are n neighbouring H, the peak splits into n + 1 lines: a singlet (0 neighbours), doublet (1), triplet (2), quartet (3).

Typical proton NMR chemical shift ranges on an axis from 13 to 0 ppm, each with a drawn fragment and the hydrogen coloured: C–H in a hydrocarbon chain 0.7 to 2.1, H on a carbon next to C=O 2.0 to 2.7, H on a carbon bonded to O 3.3 to 4.5, H on a C=C 4.5 to 6.5, aldehyde H 9.0 to 10.0, carboxylic acid H 10.0 to 12.0, and an alcohol O–H that varies from about 1 to 5. — **Module 8 Figure 27.** Each hydrogen environment has its own chemical-shift range, and hydrogens nearer oxygen or a C=O group sit further left: chain C–H at 0.7–2.1 ppm, H on a carbon bonded to oxygen at 3.3–4.5 ppm, and aldehyde and acid hydrogens far to the left at 9–12 ppm. H on a C=C (4.5–6.5 ppm) also sits left without oxygen, and an alcohol O–H varies, from about 1 to 5 ppm. These are typical ranges; an HSC question supplies its own table.

Five drawn molecules with the observed hydrogens in red and neighbouring hydrogens in blue, beside their line patterns: in ethanoyl chloride the methyl has no neighbouring H and gives a singlet; in 1,1-dibromo-2,2-dichloroethane each H has one neighbour and gives a doublet 1 : 1; in chloroethane the methyl has two neighbours and gives a triplet 1 : 2 : 1, and the CH₂ has three and gives a quartet 1 : 3 : 3 : 1; in 2-chloropropane the CH has six equivalent neighbours and gives a septet 1 : 6 : 15 : 20 : 15 : 6 : 1. — **Module 8 Figure 28.** A proton signal is split into n + 1 lines by the n equivalent hydrogens on neighbouring carbons, with line heights from Pascal’s triangle: no neighbours give a singlet, three give a quartet, and the six hydrogens of two methyl groups split a CH into a septet. Count the hydrogens next door, not the hydrogens in the group itself: equivalent hydrogens do not split one another’s signal.

For ethanol (Module 8 Figure 29):

the CH₃ (3H) is a triplet near 1.2 ppm, split by the 2 H on the CH₂
the CH₂ (2H) is a quartet near 3.7 ppm, split by the 3 H on the CH₃ and shifted left because it’s next to oxygen
the OH (1H) is a singlet, because its hydrogen exchanges too quickly to split

No single instrument solves the structure. Mass spectrometry gives the molar mass, IR gives the functional groups, and NMR gives the carbon skeleton.

Ethanol drawn as a full structural formula with its hydrogens coloured by environment, above its proton NMR spectrum: a quartet near 3.7 ppm (2 H next to oxygen), a singlet near 2.6 ppm (the O–H hydrogen, whose position varies) and a triplet near 1.2 ppm (3 H), with a stepped integration line in the ratio 2 : 1 : 3. — **Module 8 Figure 29.** Ethanol has three hydrogen environments, so its proton NMR spectrum has three signals with areas in the ratio 3 : 2 : 1. The CH₂ hydrogens next to oxygen appear furthest left as a quartet, split by the three neighbouring CH₃ hydrogens; the CH₃ is a triplet, split by the two CH₂ hydrogens; and the O–H is a singlet whose position varies.

Combining spectra to identify a compound

Exam questions give several spectra of one unknown. A reliable order:

Molar mass and formula: the molecular ion peak gives the molar mass; combine it with percentage composition, if given, to get the molecular formula.
Functional groups: from the key IR absorptions (O–H, C=O, N–H).
Carbon skeleton: count carbon environments in the ¹³C NMR, then use the proton NMR shifts, integrations and splitting to connect the pieces.
Check: does your proposed structure predict every peak you were given?

Worked example. An unknown has its molecular ion at m/z 74. Its IR spectrum shows a very broad dip from 2500 to 3000 cm⁻¹ and a strong dip at 1710 cm⁻¹. Its ¹³C NMR has 3 peaks, near 180, 27 and 9 ppm. Its ¹H NMR has a singlet at 11.7 ppm (1H), a quartet at 2.4 ppm (2H) and a triplet at 1.2 ppm (3H). All four spectra are drawn in Module 8 Figure 30, numbered in the order of the steps.

Propanoic acid drawn as a full structural formula with carbons C1 (the C=O carbon), C2 and C3 and three colour-coded hydrogen environments, above four spectra. 1, IR: a very broad O–H dip across 2500 to 3000 per centimetre and a strong C=O dip at 1710. 2, mass spectrum: molecular ion and base peak at m/z 74, with peaks at 57 (loss of OH, 17), 45 (the carboxyl group, loss of 29) and 29 (the ethyl piece). 3, carbon-13 NMR: three peaks near 180 (C1), 27 (C2) and 9 ppm (C3). 4, proton NMR: a broad singlet near 11.7 ppm (1 H), a quartet near 2.4 (2 H) and a triplet near 1.2 (3 H), integration 1 : 2 : 3. — **Module 8 Figure 30.** Each spectrum of propanoic acid contributes one piece of the structure: the IR shows a carboxylic acid (very broad O–H with C=O), the molecular ion gives M = 74, the three ¹³C peaks show three carbon environments with C=O near 180 ppm, and the ¹H quartet–triplet pair shows an ethyl group. Colours link each carbon and hydrogen environment in the drawn structure to its peak. Mass spectrum from the NIST Chemistry WebBook; NMR and IR values are standard reference values, simplified.

IR: a very broad O–H across 2500–3000 plus C=O at 1710 means a carboxylic acid.
Mass: the carboxyl group (the C=O carbon with its two oxygens and the acid H, red in the drawn structure) accounts for 45 of the 74, leaving 29, which is C₂H₅, an ethyl group. So the molecular formula is C₃H₆O₂. The peaks at 45 and 29 are those two pieces, and 57 is the molecular ion minus OH (17).
¹³C NMR: 3 carbon environments, with the one near 180 ppm being the C=O.
¹H NMR: the 11.7 ppm singlet is the acid H. A quartet (2H) next to a triplet (3H) is the signature of an ethyl group (a CH₃ bonded to a CH₂), with the CH₂ shifted to 2.4 ppm because it’s next to the C=O.

The compound is propanoic acid, C₃H₆O₂ (M = 74.08), drawn in full in Module 8 Figure 30. Checking it: 3 carbons in different environments, 3 hydrogen environments in a 3 : 2 : 1 ratio, and the acid’s O–H and C=O are both present.

Second worked example: a ketone. Many questions turn on a C=O with no O–H, on symmetry and on unusual splitting. An unknown has its molecular ion at m/z 86 and its base peak at m/z 43. Its IR spectrum shows a strong dip at 1715 cm⁻¹ and no broad O–H. Its ¹³C NMR has 4 peaks, near 212, 42, 27 and 18 ppm. Its ¹H NMR has a septet at 2.6 ppm (1H), a singlet at 2.1 ppm (3H) and a doublet at 1.1 ppm (6H).

IR: C=O but no O–H, so it is not an acid or an alcohol. There is no aldehyde H near 9–10 ppm, so it is a ketone (or an ester).
Mass and formula: M = 86, but mass alone cannot fix the formula, because other formulas such as C₄H₆O₂ also have a mass of 86. Combine the evidence. The ¹H integrations add to 1 + 3 + 6 = 10 H. The ¹³C spectrum shows the carbon skeleton: a ketone C=O near 212 ppm and, since symmetry makes two carbons share a peak, 5 carbons in 4 environments. Five carbons and ten hydrogens make 5 × 12 + 10 = 70, leaving 16 of the 86 for one oxygen, the ketone oxygen, so the formula is C₅H₁₀O. The peak at 71 is a loss of 15 (a CH₃), and the base peak at 43 is the C₂H₃O⁺ ion, a CH₃ still attached to the C=O.
¹³C NMR: 5 carbons but only 4 peaks, so two carbons are identical. The peak near 212 ppm is a ketone C=O.
¹H NMR: a doublet (6H) with a septet (1H) is the signature of two identical CH₃ groups on one CH (n + 1: the 6 H have 1 neighbour, the 1 H has 6). The 3H singlet is a CH₃ with no neighbouring H, shifted to 2.1 ppm because it is bonded to the C=O.

The compound is 3-methylbutan-2-one, C₅H₁₀O (M = 86.13), the compound in the 2022 HSC spectra question.

3-methylbutan-2-one drawn as a full structural formula with its environments colour-coded, above four spectra. 1, IR: a strong C=O dip at 1715 per centimetre, C–H dips near 2900 and no O–H band. 2, mass spectrum: molecular ion at m/z 86, a small peak at 71 (loss of 15) and the base peak at 43. 3, carbon-13 NMR: four peaks near 212, 42, 27 and 18 ppm for five carbons. 4, proton NMR: a septet near 2.6 ppm (1 H), a singlet near 2.1 (3 H) and a doublet near 1.1 (6 H), integration 1 : 3 : 6. — **Module 8 Figure 31.** The four spectra together fit the ketone 3-methylbutan-2-one, and its two identical CH₃ groups explain the peak counts. They give one ¹³C peak (four peaks for five carbons) and one 6H doublet, split by the single H beside them, which is itself a septet. The IR shows a C=O with no O–H band, which an aldehyde or ester would also give, so the other spectra decide. The base peak at m/z 43 comes from breaking the bond beside the C=O. Mass spectrum from the NIST Chemistry WebBook; NMR and IR values are standard reference values, simplified.

What Are the Implications for Society of Chemical Synthesis and Design?

Syllabus dot pointevaluate the factors that need to be considered when designing a chemical synthesis process, including but not limited to: availability of reagents; reaction conditions (ACSCH133); yield and purity (ACSCH134); industrial uses (eg pharmaceutical, cosmetics, cleaning products, fuels) (ACSCH131); environmental, social and economic issues

What factors go into designing a chemical synthesis?

Industrial chemists have to balance many factors at once. The main ones:

Availability of reagents: raw materials should be cheap, plentiful and safe to transport. Ammonia production relies on N₂ from air and H₂ from natural gas.
Reaction conditions: temperature, pressure and concentration affect both the rate and the equilibrium yield, while a catalyst affects the rate without changing the equilibrium yield (Module 5). Higher temperature speeds a reaction up but lowers the yield of an exothermic one, and high pressure costs money and adds safety risk, so industry picks a compromise.
Yield and purity: percentage yield = actual ÷ theoretical × 100. Recycling unreacted reagents raises it. Purity matters most for medicines, where impurities can be harmful, so extra purification steps are worth the cost.
Industrial uses: what the product is for sets the quality needed and so the process. Pharmaceuticals must be extremely pure, so they justify slow, expensive purification and small batches. Cosmetics need consistent purity and must be safe on skin. Cleaning products such as soap are made in bulk from cheap fats and NaOH, where low cost matters more than high purity. Fuels are made on an enormous scale, so energy use, raw material cost and yield dominate.
Environmental, social and economic issues: waste and by-products, energy use and emissions, safety of workers and nearby communities, jobs created, and whether the product is affordable. Green chemistry aims to reduce waste, use renewable feedstocks, and design safer chemicals and processes.

Five factors in designing a chemical synthesis — **Module 8 Figure 32.** Designing a chemical synthesis means balancing five factors at once: reagents that are cheap and safe, conditions that trade rate against yield, yield and purity, the purity the use demands, and the effects on society such as waste and jobs. Green chemistry adds less waste, renewable feedstocks and safer processes.

How to calculate atom economy and percentage yield

Atom economy measures how much of the reactants’ mass ends up in the product you want, if the reaction worked perfectly:

atom economy = molar mass of desired product ÷ total molar mass of all reactants × 100

Use the balanced equation, including the coefficients. Making ethanol by hydrating ethene, C₂H₄ + H₂O → C₂H₆O, has an atom economy of 100%, because every atom ends up in the ethanol. Making it by fermentation, C₆H₁₂O₆ → 2C₂H₆O + 2CO₂ (both equations are drawn with full structures in Module 8 Figure 34), has an atom economy of 2 × 46.07 ÷ 180.16 × 100 = 51.1%, because the CO₂ is a by-product. Green chemistry favours routes with high atom economy, since there’s less waste to deal with.

Percentage yield is different: it measures how much product you actually got, compared with the most you could get from the limiting reagent.

Worked example. 6.00 g of ethanoic acid is refluxed (heated with a condenser so vapour drips back into the flask) with 5.00 g of ethanol, giving 5.28 g of ethyl ethanoate.

n(ethanoic acid) = 6.00 ÷ 60.05 = 0.0999 mol; n(ethanol) = 5.00 ÷ 46.07 = 0.109 mol.
They react 1 : 1, so ethanoic acid is the limiting reagent.
Theoretical mass of ethyl ethanoate = 0.0999 × 88.11 = 8.80 g.
Percentage yield = 5.28 ÷ 8.80 × 100 = 60.0%.

Ethanoic acid plus ethanol, with concentrated sulfuric acid under reflux, in equilibrium with ethyl ethanoate plus water, all drawn as full structural formulas. Under the reactants: ethanoic acid 6.00 g ÷ 60.05 = 0.0999 mol, the limiting reagent; ethanol 5.00 g ÷ 46.07 = 0.109 mol, in excess. Under the ester: theoretical mass 0.0999 × 88.11 = 8.80 g, actual 5.28 g. A bar shows the percentage yield, 60.0%. — **Module 8 Figure 33.** The limiting reagent sets the most product you could make: the acid and alcohol react 1 : 1, and there is less ethanoic acid (0.0999 mol) than ethanol (0.109 mol), so the acid is limiting. The theoretical mass of ethyl ethanoate is 8.80 g, and the 5.28 g actually collected is a 60.0% yield. The drawn structures show the ester joining the acid and alcohol, with water as the second product.

A yield below 100% comes from the equilibrium not going to completion, side reactions, and product lost while it is being separated and purified.

Case study: making ethanol industrially

Ethanol shows how the same product can be made by two very different processes:

Hydration of ethene: ethene from crude oil reacts with steam at about 300 °C and 60–70 atm over a phosphoric acid catalyst. It’s fast, continuous and gives very pure ethanol, and recycling the unreacted ethene raises the overall conversion to about 95%. But it uses a non-renewable feedstock and a lot of energy.
Fermentation: sugars from crops such as sugarcane are fermented by yeast at 30–40 °C. It uses a renewable feedstock and low temperatures, but it is slow, works in batches, stops at about 15% ethanol so it needs energy-hungry distillation, and has a lower atom economy.

Which route wins depends on the price of oil versus crops, and on how much the environmental cost of each is valued.

Two routes to ethanol with equations drawn as full structural formulas. Hydration of ethene: ethene plus water, with a phosphoric acid catalyst, gives ethanol; ethene from crude oil and steam at 300 °C and 60–70 atm; fast, continuous and very pure, but non-renewable; atom economy 100%. Fermentation: glucose, C₆H₁₂O₆, with yeast gives two ethanol plus two CO₂; renewable sugars at 30–40 °C; slow, batch, stops near 15% ethanol so it needs distilling; atom economy 51.1%. — **Module 8 Figure 34.** Of the two routes to ethanol, hydrating ethene has the higher atom economy, 100%, because every atom of the reactants ends up in the product, while fermenting glucose loses two CO₂ for every two ethanol, giving 51.1%. Atom economy is worked out from the balanced equation alone, so it ignores the energy each route uses and whether its feedstock is renewable, and here the route with the lower atom economy is the renewable one.

Two other processes appear in exam questions. The Ostwald process for nitric acid oxidises ammonia over a platinum–rhodium catalyst at about 850 °C, then converts the nitrogen monoxide to NO₂ and absorbs it in water. Methanol is made from CO and H₂ (CO + 2H₂ ⇌ CH₄O, ΔH < 0) at about 250 °C and 50–100 atm over a copper-based catalyst, the same rate-versus-yield compromise as the Haber process.

Case study: the Haber process

The Haber process, N₂ + 3H₂ ⇌ 2NH₃ (ΔH < 0), shows these trade-offs in action:

Temperature about 400–450 °C: a lower temperature would give more ammonia at equilibrium, but it would be too slow, so this is a compromise between rate and yield.
Pressure about 200 atm: high pressure shifts the equilibrium towards the side with fewer gas moles, but higher pressures need expensive, thick-walled equipment.
Iron catalyst: speeds up the reaction without changing the equilibrium position.
Recycling: ammonia is liquefied and removed, and unreacted N₂ and H₂ are recycled, so the overall conversion approaches 98% even though each pass gives only about 15%.

A block flow diagram of an ammonia plant: nitrogen from air and hydrogen from natural gas, a compressor at about 200 atm, a reactor with an iron catalyst at 400–450 °C converting about 15% per pass, and a condenser where ammonia liquefies and is removed, with unreacted nitrogen and hydrogen recycled to the reactor. Notes explain the temperature compromise, the effect of pressure, the catalyst and recycling to about 98% overall conversion. — **Module 8 Figure 35.** A Haber process plant runs each pass under compromise conditions that convert only about 15% of the gases, then recovers the yield by removing the ammonia as a liquid and recycling the unreacted nitrogen and hydrogen. Each block addresses a different factor: the temperature trades rate against equilibrium yield; the high pressure raises both the rate and the yield, limited by the cost and safety of a strong plant; the catalyst raises only the rate; and recycling raises the overall conversion to about 98%.

The ammonia goes on to make fertilisers (feeding billions of people), explosives and nylon, so one synthesis has enormous social, economic and environmental effects.

Equilibrium yield falling and rate rising with temperature in the Haber process — **Module 8 Figure 36.** In the Haber process the equilibrium yield of ammonia falls as temperature rises, because the reaction is exothermic, while the rate rises, so a compromise temperature of 400–450 °C gives an acceptable yield fast enough. The curves are schematic: they show the trade-off, not exact values.

Working Scientifically skills for Module 8

Module 8 is the most practical module in the course, so a lot of its marks reward Working Scientifically skills (the investigating and communicating skills in the syllabus: questioning, planning, collecting data, analysing and communicating). You are asked to read a graph, judge a method, or decide whether a result is believable. This section teaches those skills using Module 8 examples. Numbers marked illustrative are made up to show the method, not real results.

Hypotheses: a testable prediction

A hypothesis is a testable statement that predicts how one thing affects another. A useful hypothesis identifies measurable variables and predicts a relationship. Giving its direction makes the prediction more specific. “Absorbance changes with concentration” can be tested, but it is a weak prediction because almost any result fits it. “Absorbance increases as the concentration of Cu²⁺ increases from 0.05 to 0.30 mol L⁻¹” can be shown to be right or wrong, because the prediction covers the range you test. Starting with “because…” (for example, because more ions absorb more light) is a helpful scaffold, but a mechanism isn’t required. If your data disagree, you modify the hypothesis using the evidence, instead of ignoring the result. For instance, if absorbance stops rising at high concentration, you would limit the hypothesis to the range where the line is straight.

Forming an investigable question

A good inquiry question names a measurable quantity you can change and a measurable quantity you can observe. “How does pH affect the colour of the complex?” is vague. “How does the absorbance of a Cu²⁺ solution at a fixed wavelength change as its concentration increases from 0.05 to 0.30 mol L⁻¹?” is investigable, because both variables can be measured with the equipment you have. Check that the question can be answered safely, in the time and with the equipment available.

Variables and controls

The independent variable (IV) is the one thing you change. The dependent variable (DV) is what you measure. Controlled variables are everything else you keep the same.
For a colourimetry investigation: IV = concentration of the standard; DV = absorbance; controlled = wavelength, the same cuvette (so the path length l stays the same), the same instrument and temperature.
An experimental control is a comparison run that shows what happens without the IV. In colourimetry, the blank (solvent only, 0.0 concentration) is the control: it should read about zero, and it zeroes the instrument.
Replicates are separately prepared samples (you make the solution again and measure it again). A repeat is measuring the same sample again. Replicates show whether your preparation is consistent; repeats show whether your measuring is.

Module 8 Figure 37. In a colourimetry investigation only the concentration of the standards is changed and only the absorbance is measured, while the wavelength, cuvette, instrument and temperature are held constant. The blank is the control that zeroes the instrument, and the unknown is read only if it falls inside the range of the standards. Readings are illustrative.

Planning: risk, ethics and materials

For every hazard, state the hazard, the possible harm, and a specific precaution. “Be careful” earns nothing.

Silver nitrate stains skin and is toxic to aquatic life, so wear gloves and goggles and collect the silver waste instead of washing it down the sink.
Bromine water is corrosive and its vapour is harmful to breathe, so use small amounts in a well-ventilated area or fume cupboard, and wear gloves and goggles.
Lead salts and barium salts are toxic, so students should use a teacher demonstration or secondary data, and waste is collected for disposal.

Ethics in this module is mostly environmental: collect water samples only with permission, don’t release heavy-metal waste, and report results honestly. Materials and method need a sensible order. In the 2023 HSC anion question, the marks rewarded a testing sequence with a reason for each step (carbonate first, because it would give a false positive with silver) plus the expected observations. The ordering logic is in How to test for anions.

Accuracy, precision, validity and reliability

Accuracy: how close a result is to the true value. Precision: how close repeated measurements are to each other. A dirty cuvette gives precise but inaccurate readings.
Validity: the investigation really answers the question it claims to. A colourimetry result is only valid if the wavelength suits the substance and only the concentration changes.
Reliability: repeating the investigation gives consistent results.
Calibration checks an instrument against known values. A calibration curve is calibration made visible: the standards show how your instrument responds, so the unknown is read against it. The instrument is zeroed with a blank first.

Four targets whose centre is the true value: accurate and precise, a tight cluster on the centre; precise but not accurate, a tight cluster off centre, caused by a systematic error such as undried BaSO₄ making every mass high; accurate but not precise, a wide scatter around the centre, caused by random error such as a balance flickering by ±0.002 g; and neither, a wide scatter off centre. — **Module 8 Figure 38.** Accuracy is how close results are to the true value and precision is how close they are to each other, so the two can vary independently. Random error scatters results and averaging repeats reduces it, while systematic error shifts every result the same way and only a better method or calibration removes it.

Uncertainty and error

Random error makes results scatter either side of the true value (for example, reading a burette meniscus slightly differently each time). Averaging independent repeats reduces it.
Systematic error pushes every result the same way: a constant offset (a balance that has not been zeroed and reads 0.05 g too high every time) or a proportional one (a standard solution made 5% too strong). Residual water left in a pipette is a different case: it dilutes the solution, so it is a method error that changes the concentration. Averaging cannot remove it. You fix it with calibration or a corrected method.
Example (gravimetric analysis): if the precipitate is not dried to constant mass, leftover water makes the mass too high every time, which is a systematic error. A balance that flickers by ±0.002 g is random.
Uncertainty is the size of doubt about a measurement. As an illustrative example, a 25 mL pipette with an uncertainty of ±0.05 mL has a percentage uncertainty of 0.05 ÷ 25 × 100 = 0.2%. The same uncertainty is a bigger percentage of a small volume, so measure larger quantities where you can.
In titrations, concordant titres (usually within 0.10 mL) show good precision. Average only those.

Evaluating and improving an investigation

The 2021 HSC asked students to judge a sequence of tests for identifying an unknown salt. The marking guideline rewards listing positive and negative aspects and then making an informed judgement. The sample answer pointed out that one test was not needed, because sodium never precipitates with the possible anions and chloride never precipitates with the possible cations, while the silver ion test was a good choice. The method that works:

For each step, say what it detects and whether the result changes the decision.
Name strengths (selective, simple, clear observation) and weaknesses (unnecessary, a false positive, a missed ion).
Finish with a clear judgement, then suggest a change (remove, reorder or replace a test) that would improve validity, reliability or safety.

Presenting data: choosing and drawing graphs

A column graph for categories beside a line graph with a best-fit line for continuous data — **Module 8 Figure 39.** The type of data decides the graph: separate categories such as sampling sites get a column graph, while measured numbers such as concentration get a line graph with a best-fit line, straight if the trend is linear and curved if it is not. The data are illustrative.

Data are categorical when they sort into groups (sampling sites, metal ions) and continuous when they are measured numbers that can take any value (concentration, absorbance). Use a column graph for categories and a line graph for continuous data. Continuous does not mean straight: draw a straight best-fit line when the relationship is linear, and an appropriate smooth curve when it is not.

The 2019 HSC gave absorbance data for two metals and asked for calibration curves (7 marks). The marking guideline rewarded these graph features:

Label both axes with the quantity and unit. Put the independent variable (concentration) on the x-axis and the dependent variable (absorbance) on the y-axis. Absorbance has no unit.
Use an even scale that fills most of the grid, with equal steps (0, 1, 2, 3…), not squashed into one corner.
Plot every point for both data sets.
Draw a line of best fit (straight here, because calibration data are linear) through the trend, with points scattered either side. Don’t join the dots when showing a relationship or making predictions. (For spectra, time series, or when the question gives an instruction, follow the question.)
Use a different symbol for each data set and add a key.

Two data sets plotted with different symbols, each with a straight best-fit line and a key — **Module 8 Figure 40.** A graph earns full marks when it has both axes labelled with quantity and unit, an even scale, every point plotted, a best-fit line rather than dot-to-dot, and a different symbol for each data set with a key. The data are illustrative.

The graph was only part of the question. The rest rewarded reading each unknown from its curve, converting mol L⁻¹ to mg L⁻¹ (see Why do we monitor the environment?), comparing to the discharge limit and judging how effective the treatment was. A graph is a tool for a decision, so always finish by answering the question that was asked.

Interpolation (reading a value between your standards) is reliable. Extrapolation (reading beyond them) is not, because the line might bend. That’s why an unknown must sit inside the calibration range, as in How does colourimetry work?.

Quantitative methods

Keep units in every line and convert mg L⁻¹, g L⁻¹ and mol L⁻¹ carefully.
Keep significant figures consistent with the data given (the 2019 and 2022 guidelines award a mark for this), and round only at the end.
Show main steps so marks can be awarded even if the last number is wrong. Back titrations and gravimetric percentages follow the same layout: moles first, then the ratio, then the answer.

Models and secondary sources

A model is a simplified picture used to explain or predict. The n + 1 rule for NMR splitting and the 9 : 6 : 1 pattern for two chlorine atoms are models: they work for simple cases, so state when a model might fail (an OH proton, or overlapping signals).

Secondary sources are information someone else collected, such as a spectrum database or a guideline value. Check who published it, how recent it is and whether it matches your conditions. The drinking-water limit for lead was once 0.01 mg L⁻¹ and is now 0.005 mg L⁻¹, so an old table can mislead. In an exam, use the limit you are given.

When you use a secondary source, reference it so that someone else can find it: author or organisation, title, year, link and the date you accessed it, set out the same way every time. For example, in Harvard style: National Health and Medical Research Council (2025) Australian Drinking Water Guidelines: lead, NHMRC, accessed 17 October 2026, https://guidelines.nhmrc.gov.au/australian-drinking-water-guidelines/part-5/physical-chemical-characteristics/lead.

A quick peer check also helps before you finish a report. Swap with a classmate, who checks that axes are labelled with units, that the conclusion matches the data, and that someone could repeat the method from your steps alone. Then fix what they find.

Evidence-based arguments

A claim is backed by evidence and a reason linking them. “The unknown contains chloride” is a claim; “a white precipitate formed with silver nitrate and dissolved in dilute ammonia” is the evidence; “AgCl is white and dissolves in dilute ammonia, while AgBr and AgI do not” is the reason. For “to what extent” questions, give your judgement with evidence for and against, such as a result that is below the limit but close to it.

Command words: what the question wants

These are guides, not formulas: read the whole question and its marks.

Identify / state: give the name or fact, with no explanation.
Outline: give the main points briefly.
Describe: give characteristics and features, such as the test and the expected observation.
Explain: give the reason, using “because” or “so”.
Analyse: break the information into parts and show the relationships and trends, using evidence from the data or spectrum.
Discuss: give considerations for and against, with evidence and implications, and reach a conclusion.
Assess / evaluate: make a judgement supported by evidence, usually weighing strengths and limitations.
Justify: support a choice or conclusion with reasons and evidence.

The full list is in the NESA glossary of key words. Match the effort to the marks: 1 mark is usually one idea. Use the command word and the marking demands: seven marks may reward calculations, graphs, structural deductions or an extended explanation.

HSC Chemistry Module 8 Revision Summary

Module 8

Revision summary

Analysis of inorganic substances

Identify ions by precipitation, flame colour and complexes. Measure them by gravimetric analysis, precipitation titration, colourimetry, UV-vis and AAS, using calibration curves.

Analysis of organic substances

Bromine water for C=C, dichromate for –OH, Na₂CO₃ for carboxylic acids. Mass spectrometry gives molar mass, IR gives functional groups, ¹³C and ¹H NMR give environments, integration and n + 1 splitting.

Chemical synthesis and design

Balance reagents, conditions, yield, purity and uses against environmental, social and economic costs. The Haber process is the classic rate-versus-yield compromise.

Module 8 Figure 41. Module 8 has three parts: identifying and measuring ions, identifying organic compounds by tests and spectroscopy, and designing a chemical synthesis. If you can explain each row without looking, you have covered the module.

How Does Module 8 Fit the Rest of HSC Chemistry?

Module 8 is the most cumulative module in the course, so it’s worth revising alongside the others:

The precipitation, Ksp and Haber process equilibrium ideas come from our HSC Chemistry Module 5 notes.
Titration technique and standard solutions come from our Module 6 notes.
The functional groups being tested and analysed come from our Module 7 notes.

For the bigger picture, our ultimate guide to scoring Band 6 in Chemistry covers how to study the HSC course, and how I hacked HSC Chemistry covers how to use past papers.

HSC Chemistry Tutoring in Blacktown

Module 8 questions often combine three or four modules in one long response, which is where students lose the most marks. In our HSC Chemistry tutoring in Blacktown, theory is broken down by our tutors into the smallest possible parts and practised in classes of 1 to 5 students, a few minutes from Blacktown Station. Check our pricing page, or book a free trial with no obligations.

Frequently Asked Questions

What is Module 8 in HSC Chemistry?

Module 8 is Applying Chemical Ideas, the last of four Year 12 modules in the NSW Chemistry Stage 6 Syllabus. It covers identifying and measuring ions, identifying organic compounds with chemical tests and spectroscopy, and the factors involved in designing a chemical synthesis.

What are the flame test colours in HSC Chemistry?

Barium gives an apple-green flame, calcium a brick-red flame and copper a blue-green flame. Magnesium, lead, silver and iron don't give useful flame colours, so they are identified with precipitation and complexation tests instead.

What is the difference between AAS and colourimetry?

Both measure concentration from absorbance using a calibration curve. Colourimetry measures a coloured solution with light of one colour. AAS atomises the sample in a flame and uses a lamp made of the target metal, so it measures one specific metal, even at parts-per-million levels, with very little interference from other metals.

What does each spectroscopy technique tell you?

Mass spectrometry gives the molar mass from the molecular ion peak and clues from fragments. Infrared spectroscopy shows which bonds and functional groups are present, such as O–H and C=O. Carbon-13 NMR counts the carbon environments, and proton NMR shows hydrogen environments, how many hydrogens are in each, and their neighbours through splitting.

What is the n + 1 rule in proton NMR?

A proton NMR signal is split into n + 1 lines, where n is the number of hydrogens on the neighbouring carbon atoms. A CH₃ next to a CH₂ appears as a triplet, and the CH₂ next to that CH₃ appears as a quartet. Hydrogens on OH groups usually appear as singlets.

What factors affect the design of a chemical synthesis?

The syllabus lists availability of reagents, reaction conditions, yield and purity, industrial uses, and environmental, social and economic issues. Real processes like the Haber process use compromise conditions that balance reaction rate, equilibrium yield, cost and safety.

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HSC Chemistry Module 8 Full Syllabus Notes: Applying Chemical Ideas