Chemical Formulas and Naming: HSC Chemistry Practice Exam

This original tutorgum practice paper tests your ability to write and name ionic and covalent compounds using valency, oxidation states, and IUPAC nomenclature. It covers formula construction, variable oxidation states, polyatomic ions, and the differences between ionic and molecular naming conventions.

The paper is designed for NSW HSC Chemistry revision using the 2017 syllabus edition. It is a focused topic assessment, not a full-course examination. All questions are original tutorgum practice questions and are not official, endorsed, past, or predicted NESA questions.

01Exam instructions

Assessment conditions

Assessed syllabus content

This paper assesses:

• Writing ionic formulae using valency and ion charges.
• Naming ionic compounds using IUPAC conventions.
• Using oxidation states and Roman numerals for metals with variable oxidation states.
• Writing formulae and names for covalent molecular compounds using numerical prefixes.
• Distinguishing ionic compounds from molecular compounds.
• Using polyatomic ions when constructing and naming compounds.

Exclusions

This paper does not assess acid naming, organic nomenclature, chemical reaction equations, stoichiometry, or other Chemistry modules.

Reference data

The following polyatomic ions may be used throughout the paper.

Use the following assumptions:

• All compounds are electrically neutral unless a charge is explicitly shown.
• The sum of oxidation states in a neutral compound is zero.
• Oxygen has an oxidation state of -2 in the compounds used in this paper.
• Halide ions have a charge of -1.
• Group 1 metals have a charge of +1, Group 2 metals have a charge of +2, and aluminium has a charge of +3 in the compounds used here.
• Use the lowest whole-number ratio when writing ionic formulae.
• Use Roman numerals in ionic names when a metal can have more than one oxidation state.

Attempt all questions. The marks allocated to each part indicate the expected depth of your response.

02Questions

Question 1 (8 marks)

Write the chemical formula for each of the following ionic compounds.

Show your charge-balancing working for parts worth 2 marks.

(a) Calcium nitride. (1 mark)

(b) Aluminium sulfate. (2 marks)

(c) Iron(III) oxide. (2 marks)

(d) Ammonium carbonate. (2 marks)

(e) Potassium sulfide. (1 mark)

Question 2 (8 marks)

Write the full IUPAC name for each of the following ionic compounds.

Include Roman numerals where required.

(a) \(\ce{Na2S}\) (1 mark)

(b) \(\ce{CoBr2}\) (2 marks)

(c) \(\ce{Cr2(SO4)3}\) (2 marks)

(d) \(\ce{(NH4)2S}\) (2 marks)

(e) \(\ce{Al2O3}\) (1 mark)

Question 3 (10 marks)

This question assesses the naming and formula-writing conventions for covalent molecular compounds.

(a) Write the chemical formula for dinitrogen pentoxide. (2 marks)

(b) Write the chemical formula for sulfur hexafluoride. (2 marks)

(c) Write the chemical formula for carbon monoxide. (1 mark)

(d) Give the correct molecular name for \(\ce{PCl3}\). (2 marks)

(e) Give the correct molecular name for \(\ce{N2O4}\). (2 marks)

(f) Give the correct molecular name for \(\ce{CBr4}\). (1 mark)

Question 4 (8 marks)

This question assesses oxidation states and their use in ionic nomenclature.

(a) Calculate the oxidation state of iron in \(\ce{Fe2O3}\). Show your working. (2 marks)

(b) Determine the oxidation state of copper in \(\ce{CuCl}\), then write the full Stock name of the compound. (2 marks)

(c) Write the chemical formula for cobalt(III) phosphate. (2 marks)

(d) Calculate the oxidation state of vanadium in \(\ce{V2O5}\), then give the full Stock name of the compound. (2 marks)

Question 5 (8 marks)

A student is reviewing the differences between ionic and molecular compounds.

(a) Classify \(\ce{MgO}\) as either ionic or molecular. Give one structural clue that supports your classification. (1 mark)

(b) Classify \(\ce{CO2}\) as either ionic or molecular. Give one structural clue that supports your classification. (1 mark)

(c) Ammonium nitrate has the formula \(\ce{NH4NO3}\).

Classify this compound as ionic or molecular. Identify the positive and negative ions present, and explain why the fact that it contains only non-metal elements does not make it a molecular compound. (3 marks)

(d) Give the correct molecular name for \(\ce{N2O3}\).

Explain why the name “nitrogen oxide” is not sufficiently specific to identify this compound. (3 marks)

Question 6 (8 marks)

A hypothetical metal, X, forms two different positive ions:

• \(\mathrm{X}^{2+}\)
• \(\mathrm{X}^{3+}\)

The metal combines with sulfate ions, \(\mathrm{SO_{4}}^{2-}\), to form neutral ionic compounds.

(a) Write the chemical formula for X(II) sulfate. (2 marks)

(b) Write the chemical formula for X(III) sulfate. (2 marks)

(c) For the compound in part (b), state the simplest whole-number ratio of metal ions to sulfate ions. Demonstrate that the total charge of the formula unit is zero. (2 marks)

(d) If X represents iron, give the full IUPAC names of the compounds formed in parts (a) and (b). (2 marks)

03Worked solutions and marking guide

Solution 1

(a) Calcium nitride

Calcium forms \(\mathrm{Ca}^{2+}\) ions, while nitrogen forms \(\mathrm{N}^{3-}\) ions.

The total positive and negative charges must balance.

The lowest common multiple of 2 and 3 is 6, so we need three calcium ions and two nitride ions.

The formula is:

\[ \boxed{\ce{Ca3N2}} \]

Marking:

• 1 mark for the correct formula.

(b) Aluminium sulfate

Aluminium forms \(\mathrm{Al}^{3+}\) ions, while sulfate is \(\mathrm{SO_{4}}^{2-}\).

To balance the charges, we need two aluminium ions and three sulfate ions:

\[ 2(+3)+3(-2)=0 \]

This gives the formula:

\[ \boxed{\ce{Al2(SO4)3}} \]

Notice the brackets around sulfate. There are three complete sulfate ions, so the subscript 3 must apply to the entire polyatomic ion.

Marking:

• 1 mark for correctly balancing the aluminium and sulfate charges.
• 1 mark for the correct formula, including brackets.

(c) Iron(III) oxide

The Roman numeral III tells us that iron has an oxidation state of +3.

Therefore:

• Iron: \(\mathrm{Fe}^{3+}\)
• Oxide: \(\mathrm{O}^{2-}\)

The lowest common multiple of 3 and 2 is 6.

We need two iron ions and three oxide ions:

\[ 2(+3)+3(-2)=0 \]

The formula is:

\[ \boxed{\ce{Fe2O3}} \]

Marking:

• 1 mark for the correct charge ratio.
• 1 mark for the correct formula.

(d) Ammonium carbonate

Ammonium is a polyatomic ion with a charge of +1:

\[ \ce{NH4+} \]

Carbonate has a charge of -2:

\[ \mathrm{CO_{3}}^{2-} \]

Two ammonium ions are needed to balance one carbonate ion.

\[ 2(+1)+(-2)=0 \]

The formula is:

\[ \boxed{\ce{(NH4)2CO3}} \]

Brackets are essential because there are two complete ammonium ions.

Marking:

• 1 mark for the correct ion ratio.
• 1 mark for the correct formula, including brackets.

(e) Potassium sulfide

Potassium forms \(\mathrm{K}^{+}\) ions, while sulfide forms \(\mathrm{S}^{2-}\) ions.

Two potassium ions balance one sulfide ion.

The formula is:

\[ \boxed{\ce{K2S}} \]

Marking:

• 1 mark for the correct formula.

Question 1 total: 8 marks

Solution 2

(a) \(\ce{Na2S}\)

Sodium is a Group 1 metal and forms +1 ions. Sulfur forms sulfide ions with a charge of -2.

The correct name is:

\[ \boxed{\text{Sodium sulfide}} \]

Marking:

• 1 mark for the correct name.

(b) \(\ce{CoBr2}\)

Bromide ions have a charge of -1.

There are two bromide ions, giving a total negative charge of -2.

Since the compound is neutral, cobalt must have an oxidation state of +2.

The correct name is:

\[ \boxed{\text{Cobalt(II) bromide}} \]

The Roman numeral II identifies the oxidation state of cobalt.

Marking:

• 1 mark for identifying cobalt’s oxidation state as +2.
• 1 mark for the correct full name.

(c) \(\ce{Cr2(SO4)3}\)

Each sulfate ion has a charge of -2.

Three sulfate ions contribute a total charge of:

\[ 3(-2)=-6 \]

There are two chromium atoms, so their combined oxidation state must be +6.

Each chromium therefore has an oxidation state of +3.

The correct name is:

\[ \boxed{\text{Chromium(III) sulfate}} \]

Marking:

• 1 mark for calculating the chromium oxidation state as +3.
• 1 mark for the correct full name.

(d) \(\ce{(NH4)2S}\)

This compound contains ammonium ions and sulfide ions.

The positive ion is ammonium, and the negative ion is sulfide.

The correct name is:

\[ \boxed{\text{Ammonium sulfide}} \]

There is no need for a Roman numeral because ammonium has a fixed charge.

Marking:

• 1 mark for correctly naming ammonium.
• 1 mark for correctly naming sulfide and giving the complete compound name.

(e) \(\ce{Al2O3}\)

Aluminium has a fixed oxidation state of +3 in this compound, and oxygen forms oxide ions.

The correct name is:

\[ \boxed{\text{Aluminium oxide}} \]

No Roman numeral is needed because aluminium has a fixed charge in the compounds considered here.

Marking:

• 1 mark for the correct name.

Question 2 total: 8 marks

Solution 3

Covalent molecular compounds are named using numerical prefixes to indicate the number of atoms of each element.

The prefixes relevant to this question include:

Remember that the first element generally does not receive the prefix mono- when only one atom is present. The second element takes an -ide ending.

(a) Dinitrogen pentoxide

The prefix di- indicates two nitrogen atoms.

The prefix pent- indicates five oxygen atoms.

Therefore:

\[ \boxed{\ce{N2O5}} \]

Marking:

• 1 mark for identifying the correct atom counts.
• 1 mark for the correct formula.

(b) Sulfur hexafluoride

There is no prefix before sulfur, so there is one sulfur atom.

The prefix hexa- indicates six fluorine atoms.

Therefore:

\[ \boxed{\ce{SF6}} \]

Marking:

• 1 mark for identifying the correct atom counts.
• 1 mark for the correct formula.

(c) Carbon monoxide

Carbon has no prefix, so there is one carbon atom.

Monoxide indicates one oxygen atom.

Therefore:

\[ \boxed{\ce{CO}} \]

Notice that mono- is retained in monoxide, even though it is omitted from the first element.

Marking:

• 1 mark for the correct formula.

(d) \(\ce{PCl3}\)

The first element is phosphorus.

There are three chlorine atoms, so the second element receives the prefix tri- and changes its ending to -ide.

The correct name is:

\[ \boxed{\text{Phosphorus trichloride}} \]

Marking:

• 1 mark for correctly identifying phosphorus.
• 1 mark for correctly naming the three chlorine atoms as trichloride.

(e) \(\ce{N2O4}\)

There are two nitrogen atoms, giving the prefix di-.

There are four oxygen atoms, giving the prefix tetra-.

The correct name is:

\[ \boxed{\text{Dinitrogen tetroxide}} \]

Notice the spelling tetroxide rather than tetraoxide. The final vowel of the prefix is dropped for easier pronunciation.

Marking:

• 1 mark for the correct first element and prefix.
• 1 mark for the correct second element and prefix.

(f) \(\ce{CBr4}\)

There is one carbon atom and four bromine atoms.

The correct name is:

\[ \boxed{\text{Carbon tetrabromide}} \]

Marking:

• 1 mark for the correct name.

Question 3 total: 10 marks

Solution 4

(a) Oxidation state of iron in \(\ce{Fe2O3}\)

We know that:

• Oxygen has an oxidation state of -2.
• The compound is neutral.
• There are two iron atoms and three oxygen atoms.

Let the oxidation state of iron be \(x\).

The sum of oxidation states must equal zero:

\[ 2x+3(-2)=0 \]

Expanding:

\[ 2x-6=0 \]

\[ 2x=6 \]

\[ \boxed{x=+3} \]

Therefore, iron has an oxidation state of +3.

Marking:

• 1 mark for correctly setting up the oxidation-state equation.
• 1 mark for calculating +3.

(b) Oxidation state and name of \(\ce{CuCl}\)

Chloride has an oxidation state of -1.

Since the compound is neutral, copper must have an oxidation state of +1.

\[ x+(-1)=0 \]

\[ x=+1 \]

The correct Stock name is:

\[ \boxed{\text{Copper(I) chloride}} \]

The Roman numeral I indicates that copper has an oxidation state of +1.

Marking:

• 1 mark for identifying copper as +1.
• 1 mark for the correct Stock name.

(c) Formula for cobalt(III) phosphate

Cobalt(III) means the cobalt ion has a charge of +3:

\[ \mathrm{Co}^{3+} \]

Phosphate has a charge of -3:

\[ \mathrm{PO_{4}}^{3-} \]

The charges cancel in a 1:1 ratio.

\[ (+3)+(-3)=0 \]

Therefore, the formula is:

\[ \boxed{\ce{CoPO4}} \]

Marking:

• 1 mark for identifying the correct ion charges and ratio.
• 1 mark for the correct formula.

(d) Oxidation state and name of \(\ce{V2O5}\)

Oxygen has an oxidation state of -2.

Let the oxidation state of vanadium be \(x\).

The compound is neutral:

\[ 2x+5(-2)=0 \]

\[ 2x-10=0 \]

\[ 2x=10 \]

\[ \boxed{x=+5} \]

Vanadium therefore has an oxidation state of +5.

The correct Stock name is:

\[ \boxed{\text{Vanadium(V) oxide}} \]

Marking:

• 1 mark for correctly calculating the oxidation state of +5.
• 1 mark for the correct Stock name.

Question 4 total: 8 marks

Solution 5

(a) Classification of \(\ce{MgO}\)

\(\ce{MgO}\) is an ionic compound.

Magnesium is a metal, while oxygen is a non-metal. The compound consists of positively and negatively charged ions.

Marking:

• 1 mark for identifying the compound as ionic and giving an appropriate supporting clue.

(b) Classification of \(\ce{CO2}\)

\(\ce{CO2}\) is a molecular (covalent) compound.

Carbon and oxygen are both non-metals. They form a compound using covalent bonding rather than a combination of metal and non-metal ions.

Marking:

• 1 mark for identifying the compound as molecular and giving an appropriate supporting clue.

(c) Classification of \(\ce{NH4NO3}\)

Ammonium nitrate is an ionic compound.

It contains two polyatomic ions:

Positive ion:

\[ \ce{NH4+} \]

This is the ammonium ion.

Negative ion:

\[ \ce{NO3-} \]

This is the nitrate ion.

Although both ions contain only non-metal elements, they carry electrical charges. The compound is formed from the electrostatic attraction between the positively charged ammonium ions and negatively charged nitrate ions.

Therefore, it is classified as ionic rather than molecular.

This is an important exception to the simple metal-plus-non-metal rule. The presence of polyatomic ions means that a compound can be ionic even when it contains no metals.

Marking:

• 1 mark for identifying ammonium as the positive ion.
• 1 mark for identifying nitrate as the negative ion.
• 1 mark for correctly explaining why the compound is ionic despite containing only non-metal elements.

(d) Naming \(\ce{N2O3}\)

There are two nitrogen atoms and three oxygen atoms.

The prefix di- indicates two nitrogen atoms.

The prefix tri- indicates three oxygen atoms.

Therefore, the correct molecular name is:

\[ \boxed{\text{Dinitrogen trioxide}} \]

The name “nitrogen oxide” is insufficiently specific because nitrogen and oxygen can form several different molecular compounds.

For example, \(\ce{NO}\), \(\ce{NO2}\), and \(\ce{N2O3}\) have different numbers of nitrogen and oxygen atoms.

The numerical prefixes identify the exact composition of the compound.

Marking:

• 1 mark for the correct name, dinitrogen trioxide.
• 1 mark for correctly explaining the numerical prefixes.
• 1 mark for explaining why “nitrogen oxide” is ambiguous.

Question 5 total: 8 marks

Solution 6

This question brings together valency, charge balancing, and ionic nomenclature.

The metal X can form two different ions:

\[ \mathrm{X}^{2+} \qquad \mathrm{X}^{3+} \]

Sulfate has a charge of -2:

\[ \mathrm{SO_{4}}^{2-} \]

(a) Formula for X(II) sulfate

X(II) has a charge of +2.

Sulfate has a charge of -2.

One ion of each type is sufficient to produce a neutral compound:

\[ (+2)+(-2)=0 \]

Therefore:

\[ \boxed{\ce{XSO4}} \]

Marking:

• 1 mark for identifying the 1:1 ion ratio.
• 1 mark for the correct formula.

(b) Formula for X(III) sulfate

X(III) has a charge of +3.

Sulfate has a charge of -2.

The lowest common multiple of 3 and 2 is 6.

We therefore need:

• Two X ions, contributing a total charge of +6.
• Three sulfate ions, contributing a total charge of -6.

The formula is:

\[ \boxed{\ce{X2(SO4)3}} \]

The brackets are required because there are three complete sulfate ions.

Marking:

• 1 mark for identifying the correct 2:3 ion ratio.
• 1 mark for the correct formula, including brackets.

(c) Ion ratio and charge verification

From the formula:

\[ \ce{X2(SO4)3} \]

There are two metal ions for every three sulfate ions.

The simplest whole-number ratio is:

\[ \boxed{2:3} \]

We can verify that the compound is neutral by calculating its total charge:

\[ 2(+3)+3(-2) \]

\[ =+6-6 \]

\[ \boxed{=0} \]

The formula is therefore electrically neutral.

Marking:

• 1 mark for the correct 2:3 ratio.
• 1 mark for correctly demonstrating that the total charge is zero.

(d) Naming both iron compounds

If X represents iron, the two metal ions are:

• \(\mathrm{Fe}^{2+}\), iron(II).
• \(\mathrm{Fe}^{3+}\), iron(III).

Sulfate retains the same name in both compounds.

Therefore:

For part (a):

\[ \ce{FeSO4} \]

\[ \boxed{\text{Iron(II) sulfate}} \]

For part (b):

\[ \ce{Fe2(SO4)3} \]

\[ \boxed{\text{Iron(III) sulfate}} \]

The Roman numerals distinguish the oxidation states of iron. Without them, the names would not clearly identify which iron compound is being described.

Marking:

• 1 mark for iron(II) sulfate.
• 1 mark for iron(III) sulfate.

Question 6 total: 8 marks

04Final mark summary

05Checking your understanding

After marking your paper, pay particular attention to the type of mistake you made. Different errors point to different gaps in your understanding.

A final checklist

Before moving on from this topic, make sure you can:

• [ ] Convert an ionic compound name into its correct chemical formula.
• [ ] Balance ion charges using the lowest whole-number ratio.
• [ ] Use brackets correctly when a polyatomic ion appears more than once.
• [ ] Calculate the oxidation state of a metal in a neutral compound.
• [ ] Use Roman numerals correctly in ionic compound names.
• [ ] Convert covalent molecular names into chemical formulae using prefixes.
• [ ] Name covalent compounds using the correct numerical prefixes.
• [ ] Explain why some compounds containing only non-metals are still ionic.

Being able to move confidently in both directions – from name to formula and from formula to name – is the main skill this topic requires. Practise both directions rather than only memorising a list of compound names.