Applying Chemical Ideas: 2025 Syllabus Study Guide

A water sample contains a metal ion at a concentration too low to see. A laboratory needs to identify the metal, measure how much is present, and decide whether the water is safe. Somewhere else, a mining company is working out whether an ore contains enough metal to justify extraction, while a battery manufacturer is using redox reactions to generate electricity.

These problems seem unrelated. They are connected by the same chemical ideas: identifying substances, analysing their composition, understanding how materials are produced, and tracking the transfer of electrons.

That is the central purpose of Applying Chemical Ideas in the new NSW Chemistry 11-12 (2025) syllabus.

There is one important detail to establish before you start studying: the new syllabus does not begin teaching in 2025. NESA has scheduled its implementation from 2028, with the first HSC examination under the new syllabus in 2029. The familiar Module 8 resources belong to the previous 2017 syllabus, even though the new course retains the Applying Chemical Ideas title.

01Which Chemistry syllabus applies to you?

The publication year of a syllabus and its implementation year are different. This matters when choosing textbooks, practice questions and revision resources.

As at 9 November 2026, NESA’s implementation schedule is:

The new syllabus is being prepared for during 2026 and 2027. New Year 11 teaching and assessment requirements begin in Term 1, 2028. New Year 12 teaching and assessment requirements begin in Term 4, 2028.

You can check the official dates through NESA’s Chemistry 11-12 syllabus implementation advice.

Where does Applying Chemical Ideas fit?

The new Year 12 Chemistry course contains four focus areas, each allocated 30 indicative hours.

All four focus areas are compulsory.

Applying Chemical Ideas therefore represents 30 of the 120 indicative teaching hours. This is a guide to teaching time, not a guarantee that the HSC examination will allocate exactly 25% of its marks to this focus area.

The focus area is associated with outcome CH-12-04:

Analyses the structures, production and uses of substances.

Notice the three parts of that outcome. You need to understand what substances are made of, how they are produced, and why their properties make them useful.

There is also an important course-wide requirement. Working scientifically is integrated throughout the four focus areas. It includes:

• Developing questions and hypotheses.
• Planning and conducting investigations.
• Processing qualitative and quantitative data.
• Analysing and evaluating evidence.
• Solving scientific problems.
• Communicating conclusions using appropriate scientific language.

Students must complete 10 hours of depth studies across the two-year course, including at least one depth study in Year 11 and one in Year 12.

You can find the complete structure and requirements in the official NESA Chemistry 11-12 syllabus.

02What does the new focus area actually cover?

A useful way to organise the content is into three connected strands:

1. Analysis of organic and inorganic substances.
2. The production and uses of materials.
3. Redox reactions, galvanic cells and battery technology.

These are useful study groupings rather than three separately named NESA focus areas. In particular, materials production is integrated into the organic and inorganic content.

The central idea is that chemistry must do more than describe a substance. It must also explain how we identify it, measure it, manufacture it and evaluate its effects.

Strand 1: Analysing organic substances

Imagine being given a small amount of an unknown liquid. You know it contains an organic compound, but you don’t know its structure.

How would you identify it?

You cannot simply look at the liquid and make a reliable identification. Instead, you collect different kinds of chemical evidence.

The new syllabus develops your ability to interpret several analytical techniques.

Mass spectrometry

Mass spectrometry provides information about the mass-to-charge ratio of ions.

A sample is ionised, and the resulting ions are separated according to their mass-to-charge ratios, commonly written as \(m/z\).

A mass spectrum contains peaks that provide information about the molecular mass and fragments of the compound.

The molecular ion peak can help establish the relative molecular mass. Fragment peaks provide additional clues about how the molecule is structured.

However, mass spectrometry alone does not necessarily identify a unique structure. Different compounds can produce similar fragments.

Infrared spectroscopy (IR)

Infrared spectroscopy helps identify functional groups.

Chemical bonds vibrate, and different bonds absorb infrared radiation at characteristic wavenumbers.

For example:

• O-H bonds produce characteristic absorption regions.
• C=O bonds produce a strong carbonyl absorption.
• C-H bonds produce absorptions in their own characteristic regions.

IR data can help distinguish compounds containing different functional groups.

Proton NMR (\(^{1}\mathrm{H}\) NMR)

Nuclear magnetic resonance spectroscopy provides information about the chemical environments of hydrogen atoms.

Three features are particularly important:

• Chemical shift: the position of a signal, which provides information about its chemical environment.
• Relative peak area: indicates the relative number of hydrogen atoms in each environment.
• Splitting pattern: provides information about neighbouring hydrogen atoms.

For simple spectra, the \(n+1\) rule helps predict splitting. Here, \(n\) represents the number of equivalent neighbouring hydrogen atoms.

Carbon-13 NMR (\(^{13}\mathrm{C}\) NMR)

Carbon-13 NMR provides information about the different carbon environments within a molecule.

The number of signals indicates the number of distinct carbon environments, while their chemical shifts help identify the types of environments present.

Combining the evidence

Think of each technique as a different witness describing the same molecule.

Mass spectrometry provides information about molecular mass and fragmentation. IR identifies functional groups. NMR provides information about the arrangement of atoms.

The analogy has a limit: these are physical measurements, not opinions. Each technique provides specific evidence that must agree with the proposed structure.

Consider ethanol.

Its molecular formula is \(\ce{C2H6O}\), giving a relative molecular mass of approximately 46.

Its IR spectrum contains a broad O-H absorption. Its proton NMR spectrum contains three hydrogen environments:

• A three-hydrogen environment associated with the \(\ce{CH3}\) group.
• A two-hydrogen environment associated with the \(\ce{CH2}\) group.
• A hydrogen associated with the O-H group.

Its carbon-13 NMR spectrum contains two signals because ethanol has two distinct carbon environments.

Each technique supports the same structure: \(\ce{CH3CH2OH}\).

The new syllabus specifically limits its simple organic compounds to the homologous series studied in Organic chemistry:

• Alkanes.
• Alkenes.
• Alkynes.
• Alcohols.
• Haloalkanes.
• Carboxylic acids.
• Esters.

You should practise interpreting combined data rather than memorising isolated peak positions.

Petroleum and polymer materials

Organic analysis also connects to the production of useful materials.

Petroleum is a naturally occurring mixture of hydrocarbons formed from ancient organic material over geological time.

It is not a single compound. It contains hydrocarbons with different molecular sizes and boiling points.

Fractional distillation separates petroleum into fractions according to their boiling ranges.

A common misconception is that fractional distillation breaks large molecules into smaller ones. It does not. It is a physical separation process.

Cracking is different. It is a chemical process that breaks larger hydrocarbon molecules into smaller molecules, including useful alkenes such as ethene.

The new syllabus asks you to investigate the composition and origin of petroleum, fractional distillation, and the applications and environmental impacts of petroleum products.

It also examines how ethene can be obtained from petroleum and used to produce chloroethene.

Chloroethene, also called vinyl chloride, is the monomer used to manufacture polyvinyl chloride (PVC).

The syllabus specifically includes comparisons of four important polymers:

You need to understand how their structures relate to their properties and uses.

You also need to evaluate their environmental and health impacts. This means considering more than whether a material is useful. You should consider its production, lifespan, disposal, potential pollution and effects on human health.

Another interesting part of this content is the comparison of organic resins used by Aboriginal Peoples with modern polymer materials. This requires an appropriately researched secondary-source investigation that recognises the cultural context of the knowledge being discussed.

Strand 2: Analysing inorganic substances and producing metals

Now move from organic molecules to metal-containing materials.

A mining company discovers an ore containing copper. Before deciding whether to extract the copper, it needs to know how much copper the ore actually contains.

Why?

Because extracting metal requires energy, equipment, labour and chemical processing. An ore containing very little recoverable metal may not be economically worthwhile to process.

An ore generally contains valuable metal-bearing minerals mixed with unwanted material called gangue.

The proportion of the desired metal or mineral is therefore important.

The new syllabus expects you to:

• Describe the features of a base metal ore.
• Explain why the quantity of metal in an ore must be measured.
• Construct flow charts showing metal extraction and processing.
• Use chemical equations to represent the relevant reactions.
• Evaluate environmental, social and economic consequences.

For example, extracting copper from a copper sulfide ore involves several stages.

A simplified process may include:

1. Mining the ore.
2. Crushing and grinding it.
3. Concentrating the valuable mineral.
4. Extracting the copper through chemical processing.
5. Refining the recovered metal.

The exact process depends on the mineral being extracted.

You should be able to connect the chemical reactions to the physical processing stages rather than simply memorising a sequence of industrial steps.

Atomic absorption and emission spectroscopy

The new focus area also examines how scientists identify and quantify metal ions.

Three techniques are important to distinguish.

Why do different elements produce different wavelengths?

Electrons occupy particular energy levels within atoms. When electrons move between energy levels, they absorb or release specific amounts of energy.

These energy differences correspond to characteristic wavelengths of electromagnetic radiation.

This is why atomic emission and absorption techniques can be used for elemental analysis.

The important distinction is:

• AES measures emitted radiation.
• AAS measures absorbed radiation.

AAS is particularly useful when measuring low concentrations of metal ions in aqueous samples.

Understanding AAS calibration

Suppose an environmental scientist wants to determine the concentration of copper in a water sample.

The scientist prepares several solutions with known copper concentrations and measures their absorbance.

These measurements are used to produce a calibration curve.

The unknown sample is then analysed under the same conditions. Its absorbance is compared with the calibration curve to determine its concentration.

Within the linear calibration range, a greater absorbance indicates a greater concentration.

However, you cannot simply compare any two absorbance readings and assume their concentrations have the same relationship. The element, instrument conditions and calibration range must be appropriate.

Worked example: Calculate a metal concentration using AAS

An unknown water sample is diluted by transferring 5.00 mL into a 50.0 mL volumetric flask and making the solution up to the mark.

The diluted sample produces an absorbance of 0.180.

The calibration equation is:

\[ A=0.0200c \]

where \(A\) is absorbance and \(c\) is concentration in \(\mathrm{mg\,L^{-1}}\).

Calculate the concentration of the metal in the original sample.

Step 1: Calculate the concentration of the diluted sample.

Rearrange the calibration equation:

\[ c=\frac{A}{0.0200} \]

Substitute the measured absorbance:

\[ c=\frac{0.180}{0.0200} \]

\[ c=9.00\ \mathrm{mg\,L^{-1}} \]

Step 2: Calculate the dilution factor.

The original 5.00 mL sample was diluted to 50.0 mL.

\[ \text{Dilution factor} =\frac{50.0}{5.00}=10.0 \]

Step 3: Calculate the original concentration.

\[ c_{\text{original}} =9.00\times10.0 \]

\[ \boxed{c_{\text{original}}=90.0\ \mathrm{mg\,L^{-1}}} \]

The original sample contains 90.0 mg of the measured metal per litre.

Notice that the dilution step is essential. Reporting 9.00 mg/L as the original concentration would underestimate the metal concentration by a factor of ten.

Gravimetric analysis

Another way to determine the concentration of an ion is through gravimetric analysis.

Instead of measuring light absorption, you measure the mass of a product formed in a chemical reaction.

For example, chloride ions can react with silver ions to form a white silver chloride precipitate:

\[ \ce{Ag+(aq) + Cl-(aq) -> AgCl(s)} \]

If you know the mass of the silver chloride produced, you can calculate the amount of chloride originally present.

The important relationship is the mole ratio between the precipitate and the ion being measured.

You should also understand the experimental limitations. Incomplete precipitation, contamination, loss of precipitate during filtration and insufficient drying can all affect the final result.

The new syllabus specifically includes calculations using AAS and gravimetric data to analyse pollutants in water.

Strand 3: Redox reactions and galvanic cells

The final strand connects chemical reactions to electrical energy.

Imagine placing a zinc strip into a copper sulfate solution.

You would observe copper forming on the zinc surface, while the zinc gradually dissolves.

The reaction is:

\[ \ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)} \]

What is happening to the electrons?

Zinc atoms lose electrons and become zinc ions. Copper ions gain those electrons and become solid copper.

This is a redox reaction.

Oxidation is the loss of electrons.

Reduction is the gain of electrons.

A useful way to remember the terminology is:

• The oxidising agent accepts electrons and is reduced.
• The reducing agent donates electrons and is oxidised.

In the zinc and copper reaction:

• Zinc is the reducing agent.
• Copper ions are the oxidising agent.

How does a galvanic cell produce electricity?

The zinc and copper reaction can be used to construct a galvanic cell.

The key change is that the oxidation and reduction reactions occur in separate half-cells.

This forces electrons to travel through an external wire rather than transferring directly between reactants in the same solution.

At the zinc electrode:

\[ \ce{Zn(s) -> Zn^2+(aq) + 2e-} \]

Zinc is oxidised. The zinc electrode is the anode.

At the copper electrode:

\[ \ce{Cu^2+(aq) + 2e- -> Cu(s)} \]

Copper ions are reduced. The copper electrode is the cathode.

The salt bridge allows ions to move between the half-cells and maintains electrical neutrality.

Electrons flow through the external circuit from the anode to the cathode.

The chemical reaction therefore produces an electrical potential difference.

Standard reduction potentials

How can you predict whether a galvanic cell will operate spontaneously?

You use a table of standard reduction potentials.

Each value describes the tendency of a half-reaction to undergo reduction under standard conditions.

A more positive reduction potential indicates a greater tendency for reduction.

The standard cell potential is calculated using:

\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Both values are taken from the standard reduction potential table.

A positive standard cell potential indicates that the reaction is spontaneous as written under standard conditions.

A negative value indicates that the reaction is non-spontaneous as written under those conditions.

Importantly, a positive cell potential does not tell you how quickly the reaction occurs. Spontaneity and reaction rate are different ideas.

Worked example: Calculate the standard potential of a galvanic cell

A galvanic cell contains zinc and copper electrodes.

The standard reduction potentials are:

Calculate the standard cell potential and identify the direction of electron flow.

Step 1: Identify the cathode and anode.

Copper has the more positive reduction potential, so copper ions undergo reduction.

Therefore:

• Copper is the cathode.
• Zinc is the anode.

Step 2: Write the half-equations.

Oxidation at the anode:

\[ \ce{Zn(s) -> Zn^2+(aq) + 2e-} \]

Reduction at the cathode:

\[ \ce{Cu^2+(aq) + 2e- -> Cu(s)} \]

Adding the half-equations gives:

\[ \ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)} \]

Step 3: Calculate the standard cell potential.

\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Substitute the values:

\[ E^\circ_{\text{cell}} =0.34-(-0.76) \]

\[ \boxed{E^\circ_{\text{cell}}=+1.10\ \mathrm{V}} \]

Step 4: Interpret the result.

The positive cell potential indicates that the reaction is spontaneous under standard conditions.

Electrons flow from the zinc electrode to the copper electrode.

The cell produces a standard potential difference of 1.10 V.

A common mistake is to reverse the sign of the zinc reduction potential before substituting it into the formula. Remember that the equation already subtracts the anode’s reduction potential.

You do not reverse the tabulated value before using this formula.

Connecting redox to battery technology

Galvanic cells are the basis of many battery technologies.

A battery converts chemical energy into electrical energy through spontaneous redox reactions.

Different battery designs use different electrode materials and electrolytes. These choices influence the voltage, energy capacity, operating life, cost and environmental impact of the battery.

The new syllabus expects you to investigate how understanding redox reactions has influenced battery technology.

You should be able to explain the chemical principles behind battery operation and evaluate how improvements in materials and electrochemical design have affected their applications.

03How to prepare for the practical and data-analysis requirements

Knowing the definitions is not enough.

The new focus area repeatedly asks you to interpret information, process measurements and justify conclusions. This means you need to understand how chemical evidence is collected and how reliable it is.

NESA recognises practical investigations and secondary-source investigations as part of the course. Practical investigations can involve laboratory experiments, simulations, modelling and processing primary data. Secondary-source investigations involve collecting and analysing existing information.

Your school does not necessarily need to own a mass spectrometer, NMR instrument or AAS machine for you to learn how to interpret the data they produce.

However, the syllabus explicitly includes a practical investigation involving the construction and comparison of galvanic cells.

Here are the skills worth developing.

A. Get comfortable with quantitative analysis

Refresh the following skills:

• Calculating moles from mass.
• Using molar masses.
• Applying mole ratios from balanced equations.
• Calculating concentrations in \(\mathrm{mol\,L^{-1}}\) and \(\mathrm{mg\,L^{-1}}\).
• Solving dilution problems.
• Converting between mass, moles and concentration.
• Interpreting calibration curves.

These are not isolated mathematical exercises. They allow you to turn experimental measurements into meaningful chemical conclusions.

B. Practise interpreting unfamiliar spectra

Don’t just memorise a list of chemical shifts.

For each spectrum, ask:

1. What is the signal measuring?
2. How many different environments are present?
3. What functional groups are supported by the evidence?
4. Does the proposed structure agree with the molecular formula?
5. Do the different analytical techniques support the same conclusion?

The fifth question is particularly important.

If an IR spectrum suggests a carboxylic acid but the NMR data are inconsistent with the proposed carbon skeleton, you need to reconsider your structure.

C. Understand the limitations of analytical methods

A method can produce precise measurements without necessarily producing an accurate result.

For example:

• An AAS instrument may produce consistent readings, but a poorly prepared calibration curve can produce inaccurate concentrations.
• Gravimetric analysis may underestimate an ion concentration if some precipitate is lost during filtration.
• An organic spectrum may provide useful information without being sufficient to identify a unique molecular structure.

Be prepared to evaluate reliability, accuracy, precision, validity and uncertainty.

D. Prepare properly for galvanic cell investigations

When comparing galvanic cells, you need to understand what is being changed and what is being measured.

For example, if you compare different electrode combinations, keep other experimental conditions as consistent as possible.

Consider:

• Electrode materials.
• Electrolyte identity and concentration.
• Temperature.
• Electrode surface condition.
• The measured potential difference.
• Repeated measurements and experimental variation.

You should also be able to compare experimental cell potentials with theoretical values calculated using standard reduction potentials.

Remember that standard potentials describe standard conditions. An actual experimental reading may differ because the cell is not operating under those conditions and because real cells have practical limitations.

E. Practise evaluating industrial processes

For ore extraction and polymer production, a strong response considers more than the chemistry of the reaction.

You should be able to discuss:

A good evaluation uses chemical evidence to support its conclusions.

04What changed from the old Module 8?

This is where students need to be particularly careful.

The 2017 Chemistry Stage 6 syllabus used eight modules. Module 8 was called Applying Chemical Ideas.

The new 2025 syllabus also uses the title Applying Chemical Ideas, but it is a focus area within a reorganised course.

The two are not interchangeable.

The old Module 8 contained three broad inquiry areas:

• Analysis of inorganic substances.
• Analysis of organic substances.
• Chemical synthesis and design.

The new focus area reorganises and extends this content.

Comparing the old and new content

The biggest changes are worth understanding individually.

1. Some analytical techniques remain, but their context changes.

Mass spectrometry, IR spectroscopy and NMR are still important.

AAS and gravimetric analysis also remain relevant.

However, you should not assume that every old Module 8 technique or investigation is automatically a required component of the new focus area.

For example, the old module included qualitative ion-testing procedures, precipitation titrations and colorimetry. These should not simply be transferred into your new study checklist without checking the 2025 syllabus.

This does not mean that all related chemistry has disappeared from the entire course. It means that you need to check where the content is actually located.

2. Materials production receives a different emphasis.

The new focus area explicitly includes petroleum, fractional distillation, polymers and base metal extraction.

You need to understand how chemistry connects the composition of raw materials to the properties and applications of manufactured products.

3. Redox and battery technology are explicit parts of the new focus area.

Galvanic cells, standard reduction potentials and battery technology are important new study areas within Applying Chemical Ideas.

These are not simply extensions of the old Module 8 spectroscopy questions.

Can you still use old Module 8 resources?

Yes, but you need to be selective.

Old resources are particularly useful when they cover concepts that remain relevant, such as:

• Mass spectrometry.
• Infrared spectroscopy.
• Proton and carbon-13 NMR.
• Atomic absorption spectroscopy.
• Gravimetric analysis.
• Scientific data interpretation.
• Experimental reliability and evaluation.

Other resources may cover content that is not an explicit requirement of the new focus area.

The NSW Department of Education has published a 2017 Chemistry Module 8 teaching guide and a Module 8 organic analysis resource page.

These are useful references for overlapping content, but they were developed for the 2017 syllabus.

For guidance on using past examination questions, see Can You Use Old HSC Chemistry Papers for the 2025 Syllabus?.

A sensible approach is to classify each old question into three groups:

1. Directly relevant to a listed 2025 syllabus dot point.
2. Useful for practising a transferable scientific or mathematical skill.
3. Specific to content that is not listed in the new focus area.

The first two groups can be useful. The third should not become a revision priority simply because it appears in an old HSC paper.

05Subject selection, school requirements and university entry

There are several different sets of rules that can affect your decision to study Chemistry. They should not be confused.

What NESA requires

NESA determines the syllabus content and formal course requirements.

For the new 2025 Chemistry course:

• All four Year 12 focus areas must be studied.
• Working scientifically outcomes must be integrated throughout the course.
• Students must complete the required depth studies.
• The Year 12 Chemistry course lists Year 11 Chemistry (2 units) as a prerequisite.

That last point is a NESA course-enrolment requirement. It means you need to complete Year 11 Chemistry to progress into the new Year 12 Chemistry course.

It is not a requirement that applies to every university degree.

For official course completion and assessment requirements, refer to NESA’s ACE Rules.

What your school decides

Schools develop their own teaching and assessment programs within NESA’s requirements.

This means your school determines matters such as:

• The sequence in which topics are taught.
• The timing of assessment tasks.
• The particular practical activities used.
• The equipment and secondary-source data available.
• The topic and organisation of depth studies.

Your school may also have local subject-selection recommendations or placement requirements.

These are school-level decisions, not automatically statewide NESA rules.

If you are unsure whether you can enrol in Chemistry at your school, ask your science faculty directly.

Do you need HSC Chemistry for university?

Not necessarily.

University admission requirements are set by individual institutions and courses. Some courses have subject prerequisites, while others list assumed knowledge or recommended studies.

These terms mean different things.

• A prerequisite is a requirement that must be met for admission.
• Assumed knowledge is knowledge the university expects you to have when commencing the course.
• Recommended studies are subjects that may help you prepare but are not necessarily required.

For example, UNSW’s assumed knowledge guidance states that it does not have formal subject prerequisites for its undergraduate degrees, but some degrees have assumed knowledge.

The University of Sydney’s explanation of prerequisites and assumed knowledge makes the same important distinction.

A more specific example appears in the University’s 2026 Chemistry handbook. HSC Chemistry is listed as assumed knowledge for its CHEM1111 Chemistry 1A unit. Students without HSC Chemistry are strongly advised to undertake appropriate bridging preparation for CHEM1011.

These are university-specific examples, not statewide HSC requirements. The 2026 handbook is also not a guarantee of the requirements for a future 2029 university intake.

If you are considering university, check the exact entry requirements for your intended course and admission year.

UAC’s Steps to Uni for Year 10 Students: 2026 edition is a useful starting point for understanding course selection and admission terminology. However, its institution course listings relate to courses planned for 2026, so later cohorts should check updated university information.

Does Chemistry scale well?

You should not choose Chemistry simply because you expect it to receive a scaling bonus.

In NSW, your ATAR is calculated using the aggregate of scaled marks from 10 units of HSC courses:

• Your best 2 units of English.
• Your best 8 units from your remaining courses.

UAC uses a statistical scaling process to make results from different courses comparable.

The process considers the academic performance and distribution of marks among students studying each course. It changes from year to year.

There is no fixed bonus for studying Chemistry, and no guarantee that choosing a particular subject will increase your ATAR.

A high HSC mark is still important. Scaling does not automatically turn a weaker performance into a stronger result.

You can read the details in UAC’s explanation of ATAR calculations and its ATAR frequently asked questions.

UAC also recommends selecting subjects based on your interests, abilities and future study plans rather than trying to predict which subjects will scale best.

Is Chemistry the right choice for you?

Beyond formal requirements, consider what you actually enjoy studying.

Chemistry may suit you if you:

• Enjoy understanding how substances are structured and how they react.
• Like interpreting experimental data.
• Are interested in pharmaceuticals, materials, environmental science or chemical manufacturing.
• Are comfortable developing mathematical problem-solving skills.
• Want to understand electrochemical processes and energy technologies.

It is also worth considering how Chemistry fits with your other subjects and your intended university course.

Your personal preference matters. A subject that interests you and aligns with your strengths is generally a more sensible choice than one selected solely because of assumptions about scaling.

06Questions and solutions

The following questions progress from straightforward analytical calculations to more challenging problems that combine several parts of the new focus area.

Question 1

A water sample with a volume of 100.0 mL is analysed for chloride ions using gravimetric analysis.

An excess of silver nitrate is added, producing a white silver chloride precipitate. After filtration, washing and drying, the precipitate has a mass of 0.287 g.

Assume the precipitation is complete and the precipitate is pure.

Given:

• \(M(\ce{AgCl})=143.32\ \mathrm{g\,mol^{-1}}\)
• \(M(\ce{Cl})=35.45\ \mathrm{g\,mol^{-1}}\)

Calculate the chloride ion concentration in:

a) \(\mathrm{mol\,L^{-1}}\)

b) \(\mathrm{mg\,L^{-1}}\)

Solution 1

The chloride concentration is \(0.0200\ \mathrm{mol\,L^{-1}}\), equivalent to approximately \(710\ \mathrm{mg\,L^{-1}}\).

The key is to use the mass of the precipitate to calculate the moles of chloride ions.

Step 1: Write the precipitation equation.

\[ \ce{Ag+(aq) + Cl-(aq) -> AgCl(s)} \]

The mole ratio between silver chloride and chloride ions is 1:1.

Step 2: Calculate the moles of silver chloride.

\[ n=\frac{m}{M} \]

\[ n(\ce{AgCl})=\frac{0.287}{143.32} \]

\[ n(\ce{AgCl})=0.00200\ \mathrm{mol} \]

Therefore:

\[ n(\mathrm{Cl}^{-})=0.00200\ \mathrm{mol} \]

Step 3: Calculate the chloride concentration.

Convert the sample volume into litres:

\[ 100.0\ \mathrm{mL}=0.1000\ \mathrm{L} \]

Use:

\[ c=\frac{n}{V} \]

\[ c=\frac{0.00200}{0.1000} \]

\[ \boxed{c=0.0200\ \mathrm{mol\,L^{-1}}} \]

Step 4: Convert to mg/L.

First calculate the mass of chloride per litre:

\[ m=c\times M \]

\[ m=0.0200\times35.45 \]

\[ m=0.709\ \mathrm{g\,L^{-1}} \]

Convert grams into milligrams:

\[ 0.709\times1000=709\ \mathrm{mg\,L^{-1}} \]

Therefore:

\[ \boxed{c=710\ \mathrm{mg\,L^{-1}}} \]

The result represents the chloride concentration in the original water sample.

Notice that the measured mass was the mass of silver chloride, not chloride alone. The mole ratio allowed us to calculate the amount of chloride originally present.

Question 2

An unknown organic compound has the molecular formula \(\ce{C3H6O2}\).

The following analytical information is obtained:

Mass spectrometry

• A molecular ion peak occurs at \(m/z=74\).

Infrared spectroscopy

• A broad absorption occurs between approximately \(2500\) and \(3300\ \mathrm{cm^{-1}}\).
• A strong absorption occurs near \(1710\ \mathrm{cm^{-1}}\).

Proton NMR

Carbon-13 NMR

Three signals occur at approximately:

• 9 ppm.
• 27 ppm.
• 180 ppm.

Identify the compound and explain how each analytical technique supports your conclusion.

Why would an ester structure be a less suitable interpretation of this evidence?

Solution 2

The compound is propanoic acid, \(\ce{CH3CH2COOH}\).

The strongest evidence is the combination of the carboxylic acid functional group, the two-carbon alkyl arrangement and the three distinct carbon environments.

Step 1: Use mass spectrometry to check the molecular mass.

The molecular formula is:

\[ \ce{C3H6O2} \]

Its relative molecular mass is:

\[ M_r=3(12)+6(1)+2(16) \]

\[ M_r=74 \]

This agrees with the molecular ion peak at \(m/z=74\), supporting the proposed formula.

Step 2: Identify the functional group using IR spectroscopy.

The broad absorption between approximately 2500 and 3300 cm\(^{-1}\) is consistent with the O-H bond of a carboxylic acid.

The strong absorption near 1710 cm\(^{-1}\) indicates a carbonyl group.

Together, these signals strongly support a carboxylic acid.

Step 3: Interpret the proton NMR spectrum.

The triplet integrating to 3H indicates a \(\ce{CH3}\) group next to a carbon containing two neighbouring hydrogens.

The quartet integrating to 2H indicates a \(\ce{CH2}\) group next to three equivalent hydrogen atoms.

These signals are consistent with:

\[ \ce{CH3-CH2-} \]

The broad signal near 11 ppm is consistent with the hydrogen attached to the carboxylic acid group.

Step 4: Check the carbon-13 NMR spectrum.

Three signals indicate three distinct carbon environments.

• 9 ppm corresponds to the terminal \(\ce{CH3}\) carbon.
• 27 ppm corresponds to the \(\ce{CH2}\) carbon.
• 180 ppm corresponds to the carboxylic acid carbonyl carbon.

All three signals agree with propanoic acid.

Step 5: Explain why an ester is less suitable.

An ester also contains a carbonyl group, so the IR carbonyl absorption alone would not distinguish it from a carboxylic acid.

However, the broad O-H absorption and the proton NMR signal near 11 ppm support a carboxylic acid.

An ester does not contain the carboxylic acid O-H group.

The combined evidence therefore makes propanoic acid the most suitable structure.

The important lesson is that no single spectrum should determine your answer when several analytical techniques are available.

Question 3

A galvanic cell is constructed using zinc and silver electrodes.

The half-cells contain 1.0 mol/L solutions of their respective ions at 25°C.

The standard reduction potentials are:

a) Identify the anode and cathode.

b) Write the balanced overall cell reaction.

c) Calculate the standard cell potential.

d) Identify the oxidising agent and reducing agent.

e) A student doubles the mass of the silver electrode and predicts that the cell voltage will double. Explain why this prediction is incorrect.

Solution 3

Zinc is the anode, silver is the cathode, and the standard cell potential is \(+1.56\ \mathrm{V}\).

Step 1: Identify the electrode reactions.

Silver has the more positive standard reduction potential, so silver ions undergo reduction.

Therefore:

• Silver is the cathode.
• Zinc is the anode.

Step 2: Write the half-equations.

Oxidation:

\[ \ce{Zn(s) -> Zn^2+(aq) + 2e-} \]

Reduction:

\[ \ce{Ag+(aq) + e- -> Ag(s)} \]

The silver half-equation requires two electrons to balance with the zinc half-equation:

\[ \ce{2Ag+(aq) + 2e- -> 2Ag(s)} \]

Step 3: Construct the overall equation.

\[ \ce{Zn(s) + 2Ag+(aq) -> Zn^2+(aq) + 2Ag(s)} \]

Step 4: Calculate the standard cell potential.

\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

\[ =0.80-(-0.76) \]

\[ \boxed{E^\circ_{\text{cell}}=+1.56\ \mathrm{V}} \]

The positive result confirms that the reaction is spontaneous under standard conditions.

Step 5: Identify the oxidising and reducing agents.

Silver ions accept electrons and are reduced.

Therefore, \(\mathrm{Ag}^{+}\) is the oxidising agent.

Zinc loses electrons and is oxidised.

Therefore, zinc is the reducing agent.

Step 6: Evaluate the student’s prediction.

Doubling the mass of the silver electrode does not double the standard cell potential.

The standard reduction potential is a property of the half-reaction under specified conditions. It does not depend on the mass of solid silver present.

Increasing the electrode mass may affect how much material is available, but it does not change the standard potential.

The tempting mistake is to assume that more electrode material must produce a greater voltage. Voltage depends on the electrochemical potential difference between the two half-cells, not simply the amount of metal used.

Question 4

A mining company analyses a 250.0 g sample of copper ore.

The sample contains 18.0% copper(I) sulfide, \(\ce{Cu2S}\), by mass. Assume that this is the only copper-containing mineral in the ore and that copper extraction is 100% efficient.

Use the following molar masses:

• \(M(\ce{Cu})=63.55\ \mathrm{g\,mol^{-1}}\)
• \(M(\ce{S})=32.06\ \mathrm{g\,mol^{-1}}\)

a) Calculate the theoretical mass of copper that can be extracted.

b) A simplified extraction process involves oxidation of copper(I) sulfide to copper(I) oxide, followed by a reaction between the two compounds to produce copper. Write balanced equations for these reactions.

c) Explain two environmental or social consequences that should be considered when evaluating the extraction process.

Solution 4

The theoretical mass of copper is 35.9 g.

The extraction process also produces sulfur dioxide, making environmental management an important consideration.

Step 1: Calculate the mass of copper(I) sulfide in the ore.

\[ m(\ce{Cu2S})=250.0\times0.180 \]

\[ m(\ce{Cu2S})=45.0\ \mathrm{g} \]

Step 2: Calculate the molar mass of copper(I) sulfide.

\[ M(\ce{Cu2S})=2(63.55)+32.06 \]

\[ M(\ce{Cu2S})=159.16\ \mathrm{g\,mol^{-1}} \]

Step 3: Calculate the moles of copper(I) sulfide.

\[ n=\frac{45.0}{159.16} \]

\[ n(\ce{Cu2S})=0.2827\ \mathrm{mol} \]

Step 4: Use the mole ratio to calculate the moles of copper.

Each mole of \(\ce{Cu2S}\) contains two moles of copper atoms.

\[ n(\ce{Cu})=2(0.2827) \]

\[ n(\ce{Cu})=0.5654\ \mathrm{mol} \]

Step 5: Calculate the theoretical copper mass.

\[ m=nM \]

\[ m=0.5654\times63.55 \]

\[ \boxed{m(\ce{Cu})=35.9\ \mathrm{g}} \]

This is the maximum theoretical mass under the stated assumption of complete recovery.

Step 6: Write the balanced extraction equations.

The first reaction involves oxidation of copper(I) sulfide:

\[ \ce{2Cu2S(s) + 3O2(g) -> 2Cu2O(s) + 2SO2(g)} \]

The second reaction produces copper:

\[ \ce{Cu2S(s) + 2Cu2O(s) -> 6Cu(s) + SO2(g)} \]

Both equations are balanced for atoms.

Step 7: Evaluate the environmental and social consequences.

One important environmental consequence is the production of sulfur dioxide.

If released into the atmosphere, sulfur dioxide contributes to air pollution and can contribute to acid deposition. Industrial facilities therefore need appropriate gas treatment and emission controls.

Another consideration is the environmental disturbance caused by mining. Extracting and processing ore can require substantial land, water and energy, while producing waste rock and tailings.

There can also be social and economic benefits. Mining may create employment, support local infrastructure and provide valuable metals for manufacturing and technology.

A strong evaluation recognises both benefits and costs, then considers how environmental management and improved processing technologies can reduce negative impacts.

Question 5

This question combines analytical chemistry, gravimetric analysis and redox reactions.

A mining company is investigating a leachate solution produced during metal extraction.

A 250.0 mL sample contains copper(II) ions and silver ions.

Atomic absorption spectroscopy determines that the initial copper ion concentration is:

\[ [\mathrm{Cu}^{2+}]=0.0200\ \mathrm{mol\,L^{-1}} \]

The company adds an excess of solid iron to the solution.

Assume that:

• All copper(II) and silver ions are completely reduced.
• No other metal ions are present.
• The solid deposit contains only copper and silver.
• No deposited metal is lost during filtration or washing.
• The final deposit is completely dry.

The total mass of the recovered solid deposit is 0.895 g.

The standard reduction potentials are:

Use the following molar masses:

• \(M(\ce{Cu})=63.55\ \mathrm{g\,mol^{-1}}\)
• \(M(\ce{Ag})=107.87\ \mathrm{g\,mol^{-1}}\)

a) Determine whether iron can spontaneously reduce copper(II) ions and silver ions under standard conditions. Support your answer with cell potential calculations and balanced equations.

b) Calculate the mass of copper present in the recovered deposit.

c) Calculate the initial concentration of silver ions in the leachate.

d) A student argues that because silver has a more positive standard reduction potential than copper, only silver ions should be reduced by iron. The student also claims that the total mass of the recovered deposit can be reported as the mass of copper extracted.

Evaluate both claims.

Solution 5

The initial silver ion concentration is \(0.0214\ \mathrm{mol\,L^{-1}}\). Both copper(II) ions and silver ions can be reduced by iron under standard conditions.

The key is to recognise that the standard reduction potential predicts thermodynamic favourability, not exclusive reaction selection or reaction rate.

Step 1: Determine whether iron can reduce copper(II) ions.

The relevant reduction potential is:

\[ E^\circ_{\ce{Cu^2+/Cu}}=+0.34\ \mathrm{V} \]

Iron is oxidised:

\[ \ce{Fe(s) -> Fe^2+(aq) + 2e-} \]

The standard cell potential is:

\[ E^\circ_{\text{cell}} =0.34-(-0.44) \]

\[ \boxed{E^\circ_{\text{cell}}=+0.78\ \mathrm{V}} \]

The positive potential means the reaction is spontaneous under standard conditions.

The balanced reaction is:

\[ \ce{Fe(s) + Cu^2+(aq) -> Fe^2+(aq) + Cu(s)} \]

Step 2: Determine whether iron can reduce silver ions.

The silver reduction potential is:

\[ E^\circ_{\ce{Ag+/Ag}}=+0.80\ \mathrm{V} \]

Using the same iron oxidation reaction:

\[ E^\circ_{\text{cell}} =0.80-(-0.44) \]

\[ \boxed{E^\circ_{\text{cell}}=+1.24\ \mathrm{V}} \]

This reaction is also spontaneous under standard conditions.

The balanced equation is:

\[ \ce{Fe(s) + 2Ag+(aq) -> Fe^2+(aq) + 2Ag(s)} \]

Therefore, both copper and silver can be deposited.

Step 3: Calculate the moles of copper initially present.

The original solution volume is:

\[ 250.0\ \mathrm{mL}=0.2500\ \mathrm{L} \]

Using:

\[ n=cV \]

\[ n(\mathrm{Cu}^{2+}) =0.0200\times0.2500 \]

\[ n(\mathrm{Cu}^{2+})=0.00500\ \mathrm{mol} \]

The mole ratio between copper ions and copper metal is 1:1.

Therefore:

\[ n(\ce{Cu})=0.00500\ \mathrm{mol} \]

Step 4: Calculate the mass of copper deposited.

\[ m=nM \]

\[ m(\ce{Cu})=0.00500\times63.55 \]

\[ \boxed{m(\ce{Cu})=0.318\ \mathrm{g}} \]

Step 5: Calculate the mass of silver deposited.

The total deposit has a mass of 0.895 g.

Since it contains only copper and silver:

\[ m(\ce{Ag})=0.895-0.318 \]

\[ \boxed{m(\ce{Ag})=0.577\ \mathrm{g}} \]

Step 6: Calculate the moles of silver deposited.

\[ n=\frac{m}{M} \]

\[ n(\ce{Ag})=\frac{0.57725}{107.87} \]

\[ n(\ce{Ag})=0.00535\ \mathrm{mol} \]

Each silver ion requires one electron to form one silver atom.

Therefore, the mole ratio between silver ions and silver metal is 1:1.

\[ n(\mathrm{Ag}^{+})=0.00535\ \mathrm{mol} \]

Step 7: Calculate the initial silver ion concentration.

\[ c=\frac{n}{V} \]

\[ c(\mathrm{Ag}^{+})=\frac{0.00535}{0.2500} \]

\[ \boxed{c(\mathrm{Ag}^{+})=0.0214\ \mathrm{mol\,L^{-1}}} \]

Step 8: Evaluate the student’s first claim.

The claim that only silver ions are reduced is incorrect.

Silver has a more positive standard reduction potential than copper, meaning its reduction is more thermodynamically favourable under standard conditions.

However, this does not mean copper ions cannot also be reduced.

Both calculated cell potentials are positive:

• Copper reduction by iron: \(+0.78\ \mathrm{V}\).
• Silver reduction by iron: \(+1.24\ \mathrm{V}\).

Both reactions are spontaneous under standard conditions.

The standard reduction potential alone does not establish the order or rate at which the two reactions occur.

Step 9: Evaluate the student’s second claim.

The total deposit mass cannot be reported as the mass of copper.

The deposit contains both copper and silver.

Gravimetric analysis measures the total mass of the recovered solid. It does not automatically identify the individual substances within that solid.

AAS provides element-specific information about the copper concentration, allowing the copper mass to be calculated separately. The remaining deposit mass can then be attributed to silver under the stated assumptions.

This is the important distinction between identifying a substance and measuring the total amount of material collected.

The problem combines three different types of evidence:

• AAS determines the initial concentration of a particular metal ion.
• Standard reduction potentials predict which redox reactions are spontaneous.
• Gravimetric analysis measures the total mass of the recovered solid.

None of these measurements should be interpreted without considering what it actually measures.

06What should you study next?

Applying Chemical Ideas brings together several concepts that you have developed throughout Chemistry.

The next step is to connect these applications back to the other Year 12 focus areas.

Equilibrium helps explain how reaction conditions affect the production and concentration of substances. Acid-base chemistry helps explain the behaviour of ions in aqueous solutions. Organic chemistry provides the structural and reaction knowledge needed to understand petroleum processing and polymer production.

The strongest understanding comes from connecting these ideas rather than studying each instrument, reaction or industrial process as an isolated fact.

When you can move from an experimental signal to a chemical conclusion, explain how a material is produced, and justify the behaviour of a redox system, you are applying the central outcome of the new focus area.