Chemical Formulae and Naming: 50-Mark Chemistry Practice Exam

This original tutorgum practice paper assesses the chemistry-2025 scope for chemical formulae and naming. It covers ionic and molecular nomenclature, valency, oxidation states, polyatomic ions, and the use of IUPAC naming conventions for common inorganic compounds.

The paper contains 50 marks of original questions, ranging from straightforward formula conversions to more challenging questions involving charge balance and oxidation-state reasoning. Complete worked solutions and a marking guide follow the question paper.

01Exam instructions

Paper details

This is an original tutorgum practice paper. It is not an official, endorsed, past or predicted NESA examination paper.

Assessed content

This paper assesses:

• Writing ionic compound formulae using valency and ionic charges.
• Naming ionic compounds using IUPAC conventions.
• Applying oxidation states and Roman numerals to metals with variable oxidation states.
• Writing and naming molecular compounds using prefixes.
• Recognising and naming common polyatomic ions.
• Distinguishing ionic compounds from molecular compounds.
• Applying nomenclature rules to common inorganic substances.

The paper does not assess chemical reactions, stoichiometry, organic nomenclature, bonding models, equilibrium or other Chemistry modules.

Conventions and reference information

1. Write ionic formulae using the lowest possible whole-number ratio of ions.
2. All ionic compounds must have an overall charge of zero.
3. Use Roman numerals in the names of metals with variable oxidation states.
4. Use molecular prefixes when naming binary molecular compounds.
5. Place brackets around a polyatomic ion when more than one of that ion is required.
6. Unless otherwise stated, assume oxygen has an oxidation state of −2 and hydrogen has an oxidation state of +1. The sum of oxidation states equals the overall charge.

The following common ions are supplied as reference information. Use their charges when constructing ionic formulae.

Common cations

Common anions

Molecular naming prefixes

For binary molecular compounds, omit mono- from the first element. The second element takes an -ide ending.

02Questions

Question 1 (6 marks)

Name each of the following ionic compounds using appropriate IUPAC nomenclature.

(a) \(\ce{MgCl2}\) (1 mark)

(b) \(\ce{Fe2O3}\) (1 mark)

(c) \(\ce{Cu(NO3)2}\) (1 mark)

(d) \(\ce{(NH4)2CO3}\) (1 mark)

(e) \(\ce{NaHCO3}\) (1 mark)

(f) \(\ce{Al2(SO4)3}\) (1 mark)

Question 2 (8 marks)

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

Show the ionic charges and charge balance used in each case.

(a) Calcium phosphate (2 marks)

(b) Iron(II) hydroxide (2 marks)

(c) Ammonium sulfate (2 marks)

(d) Copper(I) oxide (2 marks)

Question 3 (8 marks)

Apply molecular nomenclature conventions to answer the following questions.

(a) Name \(\ce{N2O4}\). (2 marks)

(b) Name \(\ce{PCl5}\). (2 marks)

(c) Write the chemical formula for sulfur trioxide. (2 marks)

(d) Write the chemical formula for dinitrogen monoxide. (2 marks)

Question 4 (8 marks)

Answer the following questions about oxidation states and ionic formulae.

(a) Determine the oxidation state of sulfur in the sulfite ion, \(\mathrm{SO_{3}}^{2-}\). Show your working. (2 marks)

(b) Determine the oxidation state of nitrogen in the ammonium ion, \(\ce{NH4+}\). Show your working. (2 marks)

(c) Write the simplest neutral ionic formula for a compound formed from aluminium ions, \(\mathrm{Al}^{3+}\), and oxide ions, \(\mathrm{O}^{2-}\). Show your charge balance. (2 marks)

(d) Write the simplest neutral ionic formula for a compound formed from copper(II) ions, \(\mathrm{Cu}^{2+}\), and phosphate ions, \(\mathrm{PO_{4}}^{3-}\). Show your charge balance. (2 marks)

Question 5 (6 marks)

A student makes the following two statements when naming and writing ionic compounds.

(a) The student identifies \(\ce{Cu2SO4}\) as copper(II) sulfate.

Determine the correct name of this compound. Show how the oxidation state of copper can be determined from its formula. (3 marks)

(b) The student writes \(\ce{FeCO3}\) as the formula for iron(III) carbonate.

Write the correct formula for iron(III) carbonate. Explain how the charges of the ions determine the correct ratio. (3 marks)

Question 6 (6 marks)

Classify each of the following compounds as either ionic or molecular.

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

(b) \(\ce{CO2}\) (1 mark)

(c) \(\ce{Mg(NO3)2}\) (1 mark)

(d) \(\ce{PCl3}\) (1 mark)

(e) \(\ce{NH4Cl}\) (1 mark)

(f) \(\ce{SF6}\) (1 mark)

Question 7 (8 marks)

Consider an ionic compound containing a metal represented by \(M\) and sulfate ions.

(a) The metal forms ions with a charge of +3. Write the simplest neutral chemical formula for the compound. Show how you determine the ratio of ions. (2 marks)

(b) If the metal is chromium, provide the correct IUPAC name for the compound. (2 marks)

(c) Explain why the name iron oxide does not identify one unique ionic compound. (2 marks)

(d) Give the correct Stock names for the following compounds:

(i) \(\ce{FeO}\) (1 mark)

(ii) \(\ce{Fe2O3}\) (1 mark)

03Worked solutions and marking guide

Solution 1

The key to naming ionic compounds is identifying the cation and anion before writing the complete name.

For metals with a fixed charge, such as magnesium and aluminium, a Roman numeral is unnecessary. For metals with variable oxidation states, such as iron and copper, the Roman numeral identifies the metal’s oxidation state.

(a) \(\ce{MgCl2}\)

Answer: Magnesium chloride.

Magnesium forms \(\mathrm{Mg}^{2+}\) ions, while chlorine forms \(\mathrm{Cl}^{-}\) ions.

The compound contains one magnesium ion for every two chloride ions. Magnesium has a fixed charge, so no Roman numeral is needed.

Marking:

• 1 mark for the correct name.

(b) \(\ce{Fe2O3}\)

Answer: Iron(III) oxide.

Oxygen has an oxidation state of −2.

Three oxygen atoms contribute a total oxidation state of −6. Since the compound is neutral, the two iron atoms must contribute a total of +6.

Each iron atom therefore has an oxidation state of +3.

The correct name is iron(III) oxide.

Marking:

• 1 mark for the correct name, including the Roman numeral.

(c) \(\ce{Cu(NO3)2}\)

Answer: Copper(II) nitrate.

Nitrate has a charge of −1.

There are two nitrate ions, giving a total negative charge of −2. Copper must therefore have a charge of +2 to make the compound neutral.

The correct name is copper(II) nitrate.

Marking:

• 1 mark for the correct name, including the Roman numeral.

(d) \(\ce{(NH4)2CO3}\)

Answer: Ammonium carbonate.

The polyatomic ion \(\ce{NH4+}\) is ammonium, while \(\mathrm{CO_{3}}^{2-}\) is carbonate.

Two ammonium ions balance the charge of one carbonate ion.

Both ions have established names, so no Roman numeral or molecular prefix is required.

Marking:

• 1 mark for the correct name.

(e) \(\ce{NaHCO3}\)

Answer: Sodium hydrogen carbonate.

The cation is sodium, \(\mathrm{Na}^{+}\), and the anion is hydrogen carbonate, \(\ce{HCO3-}\).

Their charges balance in a 1:1 ratio.

The compound is therefore named sodium hydrogen carbonate. Sodium bicarbonate is also an accepted common name.

Marking:

• 1 mark for the correct name.

(f) \(\ce{Al2(SO4)3}\)

Answer: Aluminium sulfate.

The cation is aluminium, \(\mathrm{Al}^{3+}\), and the anion is sulfate, \(\mathrm{SO_{4}}^{2-}\).

Aluminium has a fixed +3 charge, so there is no need for a Roman numeral.

Marking:

• 1 mark for the correct name.

Total: 6 marks

Solution 2

When writing ionic formulae, the overall charge must equal zero.

A reliable method is to write the charges first, find the smallest whole-number ratio that balances them, and then write the formula.

(a) Calcium phosphate

Answer: \(\ce{Ca3(PO4)2}\)

Calcium forms \(\mathrm{Ca}^{2+}\), while phosphate is \(\mathrm{PO_{4}}^{3-}\).

We need the total positive and negative charges to balance.

The lowest common multiple of 2 and 3 is 6.

Therefore, we need:

• Three calcium ions, contributing +6.
• Two phosphate ions, contributing −6.

The resulting formula is:

\[ \boxed{\ce{Ca3(PO4)2}} \]

Notice the brackets around phosphate. There are two complete phosphate ions in the formula.

Marking:

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

(b) Iron(II) hydroxide

Answer: \(\ce{Fe(OH)2}\)

Iron(II) forms \(\mathrm{Fe}^{2+}\).

Hydroxide is \(\ce{OH-}\).

One iron ion has a +2 charge, so two hydroxide ions are required to provide a total charge of −2.

\[ \boxed{\ce{Fe(OH)2}} \]

The brackets are necessary because the hydroxide ion contains both oxygen and hydrogen, and two complete hydroxide ions are present.

Marking:

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

(c) Ammonium sulfate

Answer: \(\ce{(NH4)2SO4}\)

Ammonium has a charge of +1:

\[ \ce{NH4+} \]

Sulfate has a charge of −2:

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

Two ammonium ions are required to balance one sulfate ion.

\[ \boxed{\ce{(NH4)2SO4}} \]

The brackets indicate that the entire ammonium ion is repeated twice.

Marking:

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

(d) Copper(I) oxide

Answer: \(\ce{Cu2O}\)

Copper(I) has a charge of +1:

\[ \mathrm{Cu}^{+} \]

Oxide has a charge of −2:

\[ \mathrm{O}^{2-} \]

Two copper(I) ions are needed to balance one oxide ion.

\[ \boxed{\ce{Cu2O}} \]

Marking:

• 1 mark for identifying the correct 2:1 ion ratio.
• 1 mark for the correct formula.

Total: 8 marks

Solution 3

Molecular compounds use prefixes to indicate the number of atoms of each element.

This differs from ionic nomenclature. Molecular compounds do not use Roman numerals to indicate the charge of a metal ion.

(a) Name \(\ce{N2O4}\).

Answer: Dinitrogen tetroxide.

There are two nitrogen atoms, so the prefix di- is used.

There are four oxygen atoms, so the prefix tetra- is used. When naming this oxide, the ending combines with oxide to give tetroxide.

The complete name is dinitrogen tetroxide.

Marking:

• 1 mark for dinitrogen.
• 1 mark for tetroxide.

(b) Name \(\ce{PCl5}\).

Answer: Phosphorus pentachloride.

There is one phosphorus atom. We omit mono- from the first element.

There are five chlorine atoms, so we use penta-. Chlorine changes to chloride in the compound name.

The complete name is phosphorus pentachloride.

Marking:

• 1 mark for phosphorus.
• 1 mark for pentachloride.

(c) Write the formula for sulfur trioxide.

Answer: \(\ce{SO3}\)

Sulfur has no prefix, meaning there is one sulfur atom.

Trioxide indicates three oxygen atoms.

Therefore:

\[ \boxed{\ce{SO3}} \]

Marking:

• 1 mark for the correct element symbols and order.
• 1 mark for the correct subscripts.

(d) Write the formula for dinitrogen monoxide.

Answer: \(\ce{N2O}\)

Dinitrogen indicates two nitrogen atoms.

Monoxide indicates one oxygen atom.

Therefore:

\[ \boxed{\ce{N2O}} \]

Marking:

• 1 mark for the correct element symbols and order.
• 1 mark for the correct subscripts.

Total: 8 marks

Solution 4

Oxidation states help us track how the charges within a compound or polyatomic ion balance.

For oxidation-state calculations, remember:

• Oxygen is usually −2.
• Hydrogen is usually +1.
• The sum of oxidation states equals the overall charge.

(a) Oxidation state of sulfur in \(\mathrm{SO_{3}}^{2-}\)

Answer: +4.

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

There are three oxygen atoms, each with an oxidation state of −2.

The total oxidation state must equal the overall charge of −2.

\[ \begin{aligned} x+3(-2)&=-2\\ x-6&=-2\\ x&=+4 \end{aligned} \]

Therefore, sulfur has an oxidation state of +4.

Marking:

• 1 mark for correctly setting up the oxidation-state equation.
• 1 mark for the correct answer of +4.

(b) Oxidation state of nitrogen in \(\ce{NH4+}\)

Answer: −3.

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

There are four hydrogen atoms, each with an oxidation state of +1.

The total oxidation state must equal the overall charge of +1.

\[ \begin{aligned} x+4(+1)&=+1\\ x+4&=+1\\ x&=-3 \end{aligned} \]

Therefore, nitrogen has an oxidation state of −3.

Marking:

• 1 mark for correctly setting up the oxidation-state equation.
• 1 mark for the correct answer of −3.

(c) Formula from aluminium and oxide ions

Answer: \(\ce{Al2O3}\).

The ions are:

• Aluminium: \(\mathrm{Al}^{3+}\)
• Oxide: \(\mathrm{O}^{2-}\)

The lowest common multiple of 3 and 2 is 6.

Two aluminium ions provide a total charge of +6, while three oxide ions provide a total charge of −6.

The charges cancel:

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

Therefore, the simplest neutral formula is:

\[ \boxed{\ce{Al2O3}} \]

Marking:

• 1 mark for the correct 2:3 ion ratio.
• 1 mark for the correct formula.

(d) Formula from copper(II) and phosphate ions

Answer: \(\ce{Cu3(PO4)2}\).

The ions are:

• Copper(II): \(\mathrm{Cu}^{2+}\)
• Phosphate: \(\mathrm{PO_{4}}^{3-}\)

The lowest common multiple of 2 and 3 is 6.

Three copper(II) ions contribute +6, while two phosphate ions contribute −6.

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

Therefore, the simplest neutral formula is:

\[ \boxed{\ce{Cu3(PO4)2}} \]

Marking:

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

Total: 8 marks

Solution 5

This question tests two common mistakes:

1. Assuming a metal always has the same oxidation state.
2. Writing an ionic formula without checking whether the charges balance.

(a) Correct name for \(\ce{Cu2SO4}\)

Answer: Copper(I) sulfate.

The sulfate ion has a charge of −2:

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

The compound is neutral, so the two copper atoms must contribute a total charge of +2.

Each copper atom must therefore have an oxidation state of +1.

We can show this using:

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

\[ 2x=+2 \]

\[ x=+1 \]

Copper has an oxidation state of +1, so the correct name is copper(I) sulfate.

The student’s answer of copper(II) sulfate is incorrect because the Roman numeral does not match the oxidation state of copper in the formula.

Marking:

• 1 mark for identifying the sulfate charge and determining the total positive charge required.
• 1 mark for calculating the copper oxidation state as +1.
• 1 mark for the correct name, copper(I) sulfate.

(b) Correct formula for iron(III) carbonate

Answer: \(\ce{Fe2(CO3)3}\).

Iron(III) has a charge of +3:

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

Carbonate has a charge of −2:

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

The lowest common multiple of 3 and 2 is 6.

Therefore:

• Two iron(III) ions contribute +6.
• Three carbonate ions contribute −6.

The charges balance:

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

The correct formula is:

\[ \boxed{\ce{Fe2(CO3)3}} \]

The student’s formula, \(\ce{FeCO3}\), is incorrect because one iron(III) ion and one carbonate ion would produce an overall charge of +1.

Marking:

• 1 mark for identifying the correct ions and their charges.
• 1 mark for establishing the correct 2:3 ratio.
• 1 mark for the correct formula, including brackets.

Total: 6 marks

Solution 6

Answer:

Why?

Ionic compounds consist of positively charged cations and negatively charged anions.

A common example is a metal combined with a non-metal. Ionic compounds can also contain polyatomic ions, such as ammonium and nitrate.

Molecular compounds consist of non-metal atoms joined through covalent bonds.

Let’s examine each compound.

(a) \(\ce{KBr}\): Ionic

Potassium is a metal and bromine is a non-metal. The compound consists of potassium cations and bromide anions.

1 mark for ionic.

(b) \(\ce{CO2}\): Molecular

Carbon and oxygen are both non-metals. They form a molecular compound.

1 mark for molecular.

(c) \(\ce{Mg(NO3)2}\): Ionic

Magnesium is a metal, and nitrate is a polyatomic anion. The compound consists of magnesium cations and nitrate anions.

1 mark for ionic.

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

Phosphorus and chlorine are both non-metals. They form a molecular compound.

1 mark for molecular.

(e) \(\ce{NH4Cl}\): Ionic

Ammonium is a positively charged polyatomic ion, while chloride is a negatively charged ion.

Although ammonium contains non-metal elements, the compound is ionic because it consists of oppositely charged ions.

1 mark for ionic.

(f) \(\ce{SF6}\): Molecular

Sulfur and fluorine are both non-metals. They form a molecular compound.

1 mark for molecular.

Total: 6 marks

Solution 7

This question combines charge balancing, oxidation-state naming and the distinction between metals with fixed and variable oxidation states.

(a) Formula of the compound containing \(M^{3+}\) and sulfate

Answer: \(\ce{M2(SO4)3}\).

The metal ion has a charge of +3:

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

Sulfate has a charge of −2:

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

The lowest common multiple of 3 and 2 is 6.

We therefore require:

• Two metal ions, contributing +6.
• Three sulfate ions, contributing −6.

The resulting formula is:

\[ \boxed{\ce{M2(SO4)3}} \]

Marking:

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

(b) Name of the chromium compound

Answer: Chromium(III) sulfate.

The metal is chromium, and its charge is +3.

The Roman numeral III identifies the oxidation state of chromium.

The anion is sulfate.

Combining these names gives chromium(III) sulfate.

Marking:

• 1 mark for identifying chromium as the metal.
• 1 mark for the correct oxidation-state numeral and sulfate name.

(c) Why is iron oxide an insufficient name?

Iron can form more than one stable positive oxidation state.

In particular, iron commonly forms:

• Iron(II), with a charge of +2.
• Iron(III), with a charge of +3.

These different charges produce different compounds when iron combines with oxide ions.

The name iron oxide does not specify which oxidation state of iron is present.

Stock nomenclature solves this problem by including a Roman numeral in the compound name. The numeral identifies the oxidation state of the metal.

For example, iron(II) oxide and iron(III) oxide are two distinct compounds.

Marking:

• 1 mark for identifying that iron can form ions with different oxidation states.
• 1 mark for explaining that the Roman numeral distinguishes the oxidation state and therefore the compound.

(d) Stock names for iron oxides

(i) \(\ce{FeO}\)

Answer: Iron(II) oxide.

Oxygen has an oxidation state of −2.

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

Therefore, the correct name is iron(II) oxide.

Marking:

• 1 mark for the correct name, iron(II) oxide.

(ii) \(\ce{Fe2O3}\)

Answer: Iron(III) oxide.

Three oxygen atoms contribute a total oxidation state of −6.

The two iron atoms must contribute a total of +6. Each iron atom therefore has an oxidation state of +3.

The correct name is iron(III) oxide.

Marking:

• 1 mark for the correct name, iron(III) oxide.

Total: 8 marks

04Final mark summary

Checking your performance

When reviewing your answers, pay particular attention to these distinctions:

• If you lost marks in Questions 1 or 7, revisit the use of Roman numerals for metals with variable oxidation states.
• If you struggled with Question 2, practise balancing ionic charges before writing the final formula.
• If you lost marks in Question 3, revise molecular prefixes and remember that mono- is omitted from the first element.
• If Question 4 was difficult, practise setting up oxidation-state equations using the overall charge of the compound or ion.
• If Question 5 was challenging, check that every ionic formula you write has an overall charge of zero.
• If Question 6 caused problems, remember that compounds containing polyatomic ions can still be ionic.

A useful final check for any ionic formula is to multiply each ion’s charge by the number of ions present. The positive and negative totals must cancel.

For molecular compounds, don’t try to balance ionic charges. The prefixes tell you how many atoms of each element are present.

These are different naming systems, and recognising which one applies is the foundation of this topic.