Worksheet

GCE Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry

(Nuffield) (9086) Issue 3 October 2004

Workbook for GCE students

GCE

Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080)

Edexcel Advanced GCE in Chemistry (Nuffield) (9086)

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Introduction

This workbook has been developed from an earlier version offering support to students in transition from GCSE Science (Double Award) and the GCE Advanced Subsidiary.

The aim of the booklet is to help students to practise their skills in the areas of formulae, equations and simple mole equations. The booklet gives examples for students to work through to help build their confidence. There are some sections involving multi-step calculations.

Edexcel acknowledges the help and support received from teachers in updating this latest edition. It replaces previous versions issued in January 1998 and August 2000.

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Section 1 Atoms

Exercise 1 Calculation of Molar Mass of compounds

Section 2 Chemical formulae

Exercise 2 Writing formulae from names

Section 3 Naming of compounds

Exercise 3 Names from formulae

Section 4 The mole

Exercise 4a Calculation of the number of moles of material in a given mass of that material

Exercise 4b Calculation of the mass of material in a given number of moles of that material

Exercise 4c Calculation of the volume of a given number of moles of a gas

Exercise 4d Calculation of the number of moles of gas in a given volume of that gas

Exercise 4e Calculation of the mass of a given volume of gas

Exercise 4f Calculation of the volume of a given mass of gas

Exercise 4g Calculation of the Relative Molecular Mass of a gas from mass and volume data for the gas

Section 5 Using the idea of moles to find formulae

Exercise 5 Calculation of formulae from experimental data

Section 6 Chemical equations; equations in words; writing formulae; balancing the equation

Exercise 6a Balancing equations

Exercise 6b What’s wrong here?

Exercise 6c Writing equations in symbols from equations in words

Section 7 How equations are found by experiment

Exercise 7 Writing equations from experimental data

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Section 8 Amounts of Substances

Exercise 8 Calculations of amounts of products/reactants based on equations

Section 9 Reactions involving gases

Exercise 9 Calculations based on equations involving only gases

Section 10 Ions and ionic equations; structure of ionic compounds

Exercise 10 Ionic equations

Section 11 Calculations involving chemicals in solution

Exercise 11a Calculations based on concentrations in solution

Exercise 11b Simple volumetric calculations

Section 12 Data — The periodic table Answers

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

Section 1

Atoms

All matter is made of particles. At one time, it was thought that the tiniest particle was the atom; the word comes from the Greek word meaning ‘indivisible’.

We now know that atoms can be split and that there are smaller particles than atoms, the so- called sub-atomic particles, electrons, protons and neutrons. You will need to know something about these particles which make up the different kinds of atoms.

However, you must understand that chemistry is all about rearrangements of atoms that do not themselves change.

Atoms are very small. The hydrogen atom, the smallest and lightest of all atoms, has a diameter of about 108 mm. 1 g of hydrogen atoms contains about 6 x 1023 atoms. It is very difficult to ‘see’ an individual atom and to find its mass.

An atom is the smallest, electrically neutral, particle of an element that can take part in a chemical change.

A molecule is the smallest, electrically neutral, particle of an element or compound that can exist on its own.

An ion is an atom, or group of atoms, which carries an electric charge.

You need to know these definitions by heart, but you also need to be able to recognise the formulae of atoms and molecules when you see them. Li, O, Cl, C are all formulae which represent atoms. Some of these can exist on their own, but not all of them. Oxygen, for example, always exists as oxygen molecules, O2, which contain two atoms, unless it is combined with something else. Water contains only one atom of oxygen but here it is combined with two hydrogen atoms.

Make sure that you really understand these ideas:

• a single oxygen atom, O, cannot exist on its own
• a single oxygen atom can exist when it is combined with something else, but then it is part of a molecule
• an oxygen molecule has two oxygen atoms, O2
• a few elements exist as single atoms: for these elements, an atom is the same as a molecule.

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

Structure of the atom

The atom is composed of electrons, neutrons and protons. You have to remember the relative mass of, and the electric charge on, each.

Particle Relative mass (Carbon -12 scale) Relative charge (on scale electron charge = -1 unit)

Proton +1

Electron 1/1840 –1

Neutron

The atom is mostly empty space. It has a solid core or nucleus, the centre that contains the protons and neutrons. The electrons circulate round the nucleus in specific orbits or shells.

We can picture the hydrogen atom - the simplest of all atoms with one electron, and one proton in the nucleus - by considering a pea placed in the centre of a football pitch, to represent the nucleus with its proton. On this scale the electron will revolve in a circular orbit round the goal posts. Between the electron and the nucleus is empty space.

Atoms are the particles whose symbols are found in the periodic table given in all your examination papers and also on page 113 of this book. You can see there are only about 100 of them. The middle part of the atom, the nucleus, contains one or more protons. It is the number of protons that make the atom what it is. An atom with one proton is always a hydrogen atom; one with two protons is a helium atom and so on.

There are more substances in the world than the 100 or so different kinds of atom. The other substances are made by combining atoms in various ways to make molecules.

When a chemical reaction takes place the atoms are rearranged to make different molecules but no atoms can be made or destroyed. To show this you have to be able to find a method of counting the atoms that take part in a reaction and its products.

The mass of an individual atom is very small and it is much more convenient to measure atomic masses as relative masses.

The definition of relative atomic mass Ar is:

The mass of a single atom on a scale on which the mass of an atom of carbon — 12 has a mass of 12 atomic mass units. The relative atomic mass does not have units.

The definition of Relative Molecular Mass Mr (also referred to as Molar Mass) is

The mass of a single molecule on a scale on which the mass of an atom of carbon — 12 has a mass of 12 atomic mass units.

Relative Molecular Mass of a molecule is calculated by adding together the relative atomic masses of the atoms in the chemical formulae.

Relative formula mass: in many ways this is more accurate than Relative Molecular Mass. Many salts, even in the solid state, exist as ions rather than molecules. Although the formula of sodium chloride is normally given as NaCl, it is not a simple molecule but a giant lattice and it is more accurately written as (Na+Cl–)n. Since this compound does not have molecules, it cannot have relative ‘molecular’ mass. However, the principle is the same: add the relative atomic masses of sodium (23) and chlorine (35.5) to give 58.5, the relative formula mass of NaCl.

Remember: relative atomic mass, molecular mass and formula mass have no units.

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

Examples: Calculation of Molar Mass from Relative Atomic Mass data

Before you start any of these questions make sure you read the Section 4 of this booklet (The mole on page 27).

When you carry out experiments you will weigh chemicals in grams. Molar mass has the same numerical value as the Relative Molecular Mass; it is calculated by adding together the relative atomic masses of the elements in the molecule. The total is expressed in units of grams per mol or g mol-1.

Example 1

Calculate the Molar Mass of sulphuric acid H2SO4

This molecule contains

2 atoms of hydrogen each of mass 1 = 2 x 1 = 2 g mol–1

1 atom of sulphur of mass 32 = 1 x 32 = 32 g mol–1

4 atoms of oxygen of mass 16 = 4 x 16 = 64 g mol–1

Total mass = 98 g mol–1 Example 2

Calculate the Molar Mass of lead nitrate Pb(NO3)2

Care! This molecule contains TWO nitrate groups

1 atom of lead of mass 207 = 1 x 207 = 207 g mol–1

2 atoms of nitrogen of mass 14 = 2 x 14 = 28 g mol–1

6 atoms of oxygen of mass 16 = 6 x 16 = 96 g mol–1

Total mass = 331 g mol–1 Example 3

Calculate the Molar Mass of CuSO4.5H2O

Care! This molecule has 5 molecules of water attached to each molecule of copper sulphate. Many students make the mistake of thinking that there are 10 hydrogens and only 1 oxygen.

In CuSO4

1 atom of copper of mass 63.5 = 1 x 63.5 = 63.5 g mol–1

1 atom of sulphur of mass 32 = 1 x 32 = 32 g mol–1

4 atoms of oxygen of mass 16 = 4 x 16 = 64 g mol–1

In 5H2O

5 x 2 atoms of hydrogen of mass 1 = 10 x 1 = 10 g mol–1

5 x 1 atoms of oxygen of mass 16 = 5 x 16 = 80 g mol–1

Total mass = 249.5 g mol–1

Calculations of this type are generally written as follows

CuSO4.5H2O = [ 63.5 + 32 + (4 x 16) + 5{(2x1) + 16} ] = 249.5 g mol–1

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

Exercise 1 Calculation of the Molar Mass of compounds

Calculate the Molar Mass of the following. You will find data concerning Relative Atomic Masses on the periodic table (on page 113). When you have finished this set of calculations keep the answers for reference. You will find them useful in some of the other questions in this workbook.

H2O

CO2

NH3

C2H5OH

C2H4

SO2

SO3

HBr

H2SO4

HNO3

NaCl

NaNO3

Na2CO3

NaOH

Na2SO4

KMnO4

K2CrO4

KHCO3

KI

CsNO3

CaCl2

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

Ca(NO3)2

Ca(OH)2

CaSO4

BaCl2

AlCl3

Al(NO3)3

Al2(SO4)3

FeSO4

FeCl2

FeCl3

Fe2(SO4)3

PbO

PbO2

Pb3O4

Pb(NO3)2

PbCl2

PbSO4

CuCl

CuCl2

CuSO4

ZnCl2

AgNO3

NH4Cl

(NH4)2SO4

NH4VO3

KClO3

KIO3

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

NaClO

NaNO2

CuSO4.5H2O

FeSO4.7H2O

(NH4)2SO4.Fe2(SO4)3.24H2O

Na2S2O3.5H2O

(COOH)2.2H2O

MgSO4.7H2O

Cu(NH3)4SO4.2H2O

CH3CO2H

CH3COCH3

C6H5CO2H

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

Section 2

Chemical formulae

A chemical formula is a useful shorthand method for describing the atoms in a chemical: sometimes you will see the formula used instead of the name, but you should not do this if you are asked for a name.

The chemical formula of an element or compound tells you:

• Which elements it contains: eg FeSO4 contains iron, sulphur and oxygen
• How many atoms of each kind are in each molecule: eg H2SO4 contains two atoms of hydrogen, one atom of sulphur and four atoms of oxygen in each molecule
• How the atoms are arranged: eg C2H5OH contains a group of atoms known as the ethyl group, -C2H5, and a hydroxyl group, -OH
• The masses of the various elements in a compound: eg 18 g of water, H2O, contains 2 g of hydrogen atoms and 16 g of oxygen since the relative atomic mass of hydrogen is 1 (x 2 because there two hydrogen atoms) and that of oxygen is 16.

You should not learn large numbers of chemical formulae by heart. However, it is useful to know a few of them and when you do you should be able to work out the rest. The table on page 10 shows the names, formulae and valency of the more common elements and some groups of atoms, called radicals, that you will study and you should refer to it when necessary.

Although it’s best to learn formulae by using the valency of the common parts, it is sometimes useful to be able to work out the formula of a compound. This set of rules helps you to do this using information in the table.

You can think of valency as the combining power and use it to show the simplest ratio in which the atoms of the elements and radicals combine together in the formula. The following rules can now be applied:

• Write down the symbols of the elements and radicals given in the chemical name of the compound
• Now write down the valency of each element or radical under the corresponding symbols for the element or radical
• Now cross them over as shown in the example on page 10
• The valency shows the simplest combining ratio and may be cancelled down but only the valency can be simplified in this way
• If an element has more than one valency, the name of the compound will indicate which valency is to be used.

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

Here are a few examples:

• Sodium Sulphate

NaSO4

= Na2SO4

2 1

• Calcium hydrogen carbonate

Ca(HCO3)

= Ca(HCO3)2 2 1

Note: A bracket must be placed around the radical if it is multiplied by 2 or more and composed of more than one element.

Eg MgBr2 no bracket required Ca(OH)2 bracket essential as CaOH2 is incorrect.

• Often you can cancel the numbers on the two formulae:

Ca2(CO3)2 =CaCO3

However, you should not do this for organic compounds: C2H4 has two atoms of carbon and four of hydrogen so it cannot be cancelled down to CH2.

• Copper(I) oxide means use copper valency 1, ie Cu2O: lead(II) nitrate means use lead valency 2, ie Pb(NO3)2

The periodic table can help you to find the valency of an element and hence the formula of its compounds.

Although you can use the table above to work out the formulae of many compounds it is important to realise that all formulae were originally found by experiment.

On page 11 you will find a table of the more common elements and groups that you may have met at GCSE. Also included are a few that you will meet in the first few weeks of your Advanced course or are mentioned in some of the calculations in this booklet. These are in italics.

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

Symbols and Valences of Common Elements and Radicals ELEMENTS RADICALS Symbol Valency Symbol Valency

Aluminium Al Ammonium NH4

Barium Ba Carbonate CO3

Bromine Br Chloride Cl

Calcium Ca Hydrogen-carbonate HCO3

Carbon C Hydrogen-sulphate HSO3

Chlorine Cl Hydroxide OH

Cobalt Co Nitrate NO3

Copper Cu 1&2 Nitrite NO2

Hydrogen H Sulphate SO4

Iodine I Sulphite SO3

Iron Fe 2&3

Lead Pb 2&4 Chlorate(I) ClO

Magnesium Mg Chlorate(V) ClO3

Manganese Mn 2&4 Vanadate(V) VO3

Mercury Hg 1&2 Manganate(VII) MnO4

Nitrogen N 3&5 Chromate(VI) CrO4

Oxygen O Dichromate(VI) Cr2O7

Phosphorus P 3&5

Potassium K

Silicon Si

Silver Ag

Sodium Na

Sulphur S 2,4,6

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

Exercise 2 Writing formulae from names

Use the data in the table on page 11 to write the formulae of the following. Before you start this exercise, make sure you have read Section 3 (Naming of compounds on page 19) of this booklet.

Sodium Chloride

Sodium Hydroxide

Sodium Carbonate

Sodium Sulphate

Sodium Phosphate

Potassium Chloride

Potassium Bromide

Potassium Iodide

Potassium Hydrogen Carbonate

Potassium Nitrite

Magnesium Chloride

Magnesium Nitrate

Magnesium Hydroxide

Magnesium Oxide

Magnesium Carbonate

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

Calcium Oxide

Calcium Chloride

Calcium Sulphate

Calcium Carbonate

Barium Chloride

Barium Sulphate

Aluminium Chloride

Aluminium Oxide

Aluminium Hydroxide

Aluminium Sulphate

Copper(II) Sulphate

Copper(II) Oxide

Copper(II) Chloride

Copper(II) Nitrate

Copper(I) Oxide

Copper(I) Chloride

Zinc Nitrate

Zinc Carbonate

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

Zinc Oxide

Silver Chloride

Silver Bromide

Silver Iodide

Silver Nitrate

Silver Oxide

Lead(II) Nitrate

Lead(II) Carbonate

Lead(II) Oxide

Lead(IV) Oxide

Lead(II) Chloride

Lead(IV) Chloride

Lead(II) Sulphide

Tin(II) Chloride

Tin(IV) Chloride

Iron(II) Sulphate

Iron(II) Chloride

Iron(III) Sulphate

Iron(III) Chloride

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

Iron(III) Hydroxide

Iron(II) Hydroxide

Ammonium Chloride

Ammonium Carbonate

Ammonium Hydroxide

Ammonium Nitrate

Ammonium Sulphate

Ammonium Phosphate

Phosphorus Trichloride

Phosphorus Pentachloride

Phosphorus Trioxide

Phosphorus Pentoxide

Hydrogen Phosphate (Phosphoric Acid)

Hydrogen Sulphate (Sulphuric Acid)

Hydrogen Nitrate (Nitric Acid)

Hydrogen Chloride (Hydrochloric Acid)

Carbon Tetrachloride

Silicon Tetrachloride

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

Silicon Dioxide

Sulphur Dioxide

Sulphur Trioxide

Hydrogen Sulphide

Chlorine(I) Oxide

Nitrogen Dioxide

Nitrogen Monoxide

Carbon Dioxide

Carbon Monoxide

Hydrogen Hydroxide

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

Section 3

Naming of compounds

At Advanced Level you will meet many compounds that are new to you; a lot of these will be organic compounds. In this section, we will be looking at the naming of compounds that you may already have met at GCSE level. Many of these compounds are named using simple rules. However, there are some that have ‘trivial’ names not fixed by the rules. It is important that you learn the names and formulae of these compounds. Later in the course, you will learn the rules for naming most of the organic compounds you will meet.

Naming inorganic compounds

The name must show which elements are present and, where confusion is possible, the valency of the elements concerned.

You need to remember that if there are only two elements present then the name will end in

-ide

Thus, oxides contain an element and oxygen eg Na2O is

Sodium Oxide

CaO is

Calcium Oxide

Chlorides contain an element and chlorine eg MgCl2 is

Magnesium Chloride

AlCl3 is

Aluminium Chloride

Bromides and Iodides have an element and either bromine or iodine eg KBr is

Potassium Bromide

ZnI is

Zinc Iodide

Hydrides contain an element and hydrogen and Nitrides an element and nitrogen.

eg LiH is

Lithium Hydride

Mg3N2 is

Magnesium Nitride

Other elements also form these types of compounds and the name always ends in -ide. The exceptions to this are hydroxides that have the -OH group and cyanides, which have the -CN group.

eg NaOH is

Sodium Hydroxide

Ca(OH)2 is

Calcium Hydroxide

KCN is

Potassium Cyanide

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If the elements concerned have more than one valency this may need to be shown. Thus as iron has valency 2 and 3, the name Iron Chloride would not tell you which of the two possible compounds FeCl2 or FeCl3 is being considered. In this case the valency of the iron is indicated by the use of a Roman II or III in brackets after the name of the metal. In this case Iron(II) Chloride for FeCl2 or Iron(III) Chloride for FeCl3.

eg PbCl2 is

Lead(II) Chloride

PbCl4 is

Lead(IV) Chloride

Fe(OH)2 is

Iron(II) Hydroxide

Mn(OH)2 is

Manganese(II) Hydroxide

For compounds containing two non-metal atoms the actual number of atoms of the element present are stated.

eg CO is Carbon Monoxide where mon- means one

CO2 is Carbon Dioxide where di- means two SO2 is Sulphur Dioxide. This could be called Sulphur(IV) Oxide

SO3 is Sulphur Trioxide. This could be called Sulphur(VI) Oxide PCl3 is Phosphorus Trichloride.

This could be called Phosphorus(III) Chloride

PCl5 is Phosphorus Pentachloride.

This could be called Phosphorus(V) Chloride CCl4 is Carbon Tetrachloride and

SiCl4 is Silicon Tetrachloride. Where a compound contains a metal, a non-metal and oxygen it has a name ending in -ate or -ite. You need to remember the names and formulae of the groups listed on page 11. To cover the ideas we will look at the following groups

Carbonate -CO3 Sulphate -SO4 Nitrate -NO3

Thus a compound of sodium, carbon and oxygen would be Na2CO3 and would be called Sodium Carbonate.

eg NaNO3 is

Sodium Nitrate

Mg(NO3)2 is

Magnesium Nitrate

Fe2(SO4)3 is

Iron(III) Sulphate

FeSO4 is

Iron(II) Sulphate

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Because most non-metals can have more than one valency they can also produce more than one acid upon which these groups are based. Thus sulphur can form sulphates and sulphites. The ending -ite is used when an element forms more than one such compound. In all cases the -ite is used for the compound with the lower number of oxygens atoms. Sulphate can also be referred to as sulphate(VI) and sulphite can also be referred to as sulphate(IV). In the case of nitrogen with oxygen the compounds would be nitrate and nitrite or nitrate(V) and nitrate(III).

In summary:

Common name Systematic name Formulae

Sulphate Sulphate(VI) -SO4 Sulphite Sulphate(IV) -SO3 Nitrate Nitrate(V) -NO3 Nitrite Nitrate(III) -NO2 Chlorate

Chlorate(V) -ClO3 Hypochlorite Chlorate(I) -ClO

Great care needs to be taken when using these systematic names, as they are called, because the properties of the two groups of compounds will be very different. In some cases the use of the wrong compound in a reaction could cause considerable danger. For this reason you should always read the label on a bottle or jar and make sure it corresponds exactly to what you should be using.

Other elements can form compounds involving oxygen in this way. These include Chlorate(V), Chromate(VI), Manganate(VII) and Phosphate(V).

eg KNO2 is Potassium Nitrite or Potassium Nitrate(III)

Na2SO3 is Sodium Sulphite or Sodium Sulphate(IV)

K2CrO4 is

Potassium Chromate(VI)

KMnO4 is

Potassium Manganate(VII)

KClO3 is

Potassium Chlorate(V)

When a compound is considered it is usual to put the metal down first both in the name and the formula. The exceptions to this rule are in organic compounds where the name has the metal first but the formula has the metal at the end.

eg CH3COONa is Sodium Ethanoate

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The elements nitrogen and hydrogen can join together to form a group called the ammonium group. This must not be confused with the compound ammonia, but more of that later. This ammonium group has the formula NH4

+ and sits in the place generally taken by a metal in the formula.

eg NH4Cl is

Ammonium Chloride

(NH4)2SO4 is

Ammonium Sulphate

NH4ClO3 is

Ammonium Chlorate(V)

There are a small number of simple molecules that do not follow the above rules. You will need to learn their names and formulae.

They include:

Water which is H2O

Sulphuric Acid which is H2SO4

Nitric Acid which is HNO3

Hydrochloric Acid which is HCl

Ammonia which is NH3

Methane which is CH4

Organic compounds have their own set of rules for naming but you will need to learn some of the basic rules. The names are generally based on the names of the simple hydrocarbons. These follow a simple pattern after the first four:

CH4 is Methane

C2H6 is

Ethane

C3H8 is

Propane

C4H10 is

Butane

After butane the names are based on the prefix for the number of carbons C5-pent, C6 - hex and so on.

Thus organic compounds with 2 carbons will either start with Eth- or have -eth- in their name.

eg C2H4 is

Ethene

C2H5OH is

Ethanol

CH3COOH is

Ethanoic Acid

C2H5Cl is

Chloroethane

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Exercise 3 Names from formulae

Use the notes in this section, the data in the table on page 11 and the copy of the periodic table on page 113 to write the names of the following. Before you start this exercise make sure you have read Section 2 of this booklet (Chemical formulae on page 9).

H2O

CO2

NH3

O2

H2

SO2

SO3

HCl

HI

HF

CH4

H2S

HBr

H2SO4

HNO3

NaCl

NaNO3

Na2CO3

NaOH

Na2SO4

CaCl2

Ca(NO3)2

Ca(OH)2

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CaSO4

BaCl2

AlCl3

Al(NO3)3

Al2(SO4)3

FeSO4

FeCl2

FeCl3

Fe2(SO4)3

PbO

PbO2

Pb(NO3)2

PbCl2

PbSO4

Cu(NO3)2

CuCl

CuCl2

CuSO4

ZnCl2

AgNO3

NH4Cl

(NH4)2SO4

NH4VO3 (V is Vanadium)

KClO3

KIO3

NaClO

NaNO2

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C2H6

C4H10

C8H18

(NH4)2CO3

KMnO4

K2CrO4

KHCO3

KI

Co(NO3)2

KAt

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Section 4

The mole

When chemists measure how much of a particular chemical reacts they measure the amount in grams; or they measure the volume of a gas. However, chemists find it convenient to use a unit called a mole. You need to know several definitions of a mole and be able to use them.

• The mole is the amount of substance, which contains the same number of particles (atoms, ions, molecules, formulae or electrons) as there are carbon atoms in 12 g of carbon -12
• This number is known as the Avogadro constant, L, and is equal to 6.02 x 1023 mol–1
• The molar mass of a substance is the mass, in grams, of one mole
• The molar volume of a gas is the volume occupied by one mole at room temperature and atmospheric pressure (r.t.p). It is equal to 24 dm3 at r.t.p
• Avogadro’s Law states that equal volumes of all gases, under the same conditions of temperature and pressure contain the same number of moles or molecules. If the volume is 24 dm3, at room temperature and pressure, this number, once again, is the Avogadro constant.

When you talk about moles, you must always state whether you are dealing with atoms, molecules, ions, formulae etc. To avoid any ambiguity it is best to show this as a formula.

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Example calculations involving the use of moles

These calculations form the basis of many of the calculations you will meet at A level.

Example 1

Calculation of the number of moles of material in a given mass of that material a Calculate the number of moles of oxygen atoms in 64 g of oxygen atoms. You need the mass of one mole of oxygen atoms. This is the Relative Atomic Mass in grams; in this case it is 16 g mol–1.

atoms of mass molar grams in mass atoms of moles of number

= mol g of oxygen of mass molar atoms oxygen of g oxygen of moles of number − = ∴

= 4 moles of oxygen atoms

b Calculate the number of moles of chlorine molecules in 142 g of chlorine gas.

atoms of mass molar grams in mass atoms of moles of number

=

The first stage of this calculation is to calculate the molar mass of Chlorine molecules. Molar mass of Cl2 = 2 x 35.5 = 71 g mol–1 mol g of chlorine of mass molar gas chlorine of g chlorine of moles of number − = ∴

= 2 moles of chlorine molecules

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004 c Calculate the number of moles of CuSO4.5H2O in 100 g of the solid.

The Relative Molecular Mass of CuSO4.5H2O =

[63.5 + 32 + (4 x 16) + 5{(2x1) + 16}] = 249.5 g mol–1 mol g 5. of O H 5. CuSO of mass molecular

O H 5. CuSO of g O H 5. CuSO of moles of number − = ∴

= 0.4008 moles of CuSO4.5H2O molecules Example 2

Calculation of the mass of material in a given number of moles of that material

The mass of a given number of moles = the mass of 1 mole x the number of moles of material concerned

a Calculate the mass of 3 moles of sulphur dioxide SO2

1 mole of sulphur dioxide has a mass = 32 + (2x16) = 64 g mol–1

∴ 3 moles of SO2 = 3 x 64 = 192 g b What is the mass of 0.05 moles of Na2S2O3.5H2O ?

1 mole of Na2S2O3.5H2O = [(23 x 2) + (32 x 2) + (16 x 3)] + 5[(2 x 1) +16] = 248 g mol–1

∴ 0.05 moles of Na2S2O3.5H2O = 0.05 x 248 = 12.4 g

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Example 3 Calculation of the volume of a given number of moles of a gas

All you need to remember is that 1 mole of any gas has a volume of 24 dm3 (24000 cm3) at room temperature and pressure.

∴

The volume of a given number of moles of gas = number of moles x 24000 cm3

a What is the volume of 2 mol of carbon dioxide?

Remember you do not need to work out the molar mass to do this calculation as it does not matter what gas it is.

∴ 2 moles of carbon dioxide = 2 x 24000 cm3 = 48000 cm3 = 48 dm3 b What is the volume of 0.0056 moles of chlorine molecules?

Volume of 0.0056 moles of chlorine = 0.0056 x 24000 cm3 = 134.4 cm3

Example 4

Calculation of the number of moles of gas in a given volume of that gas

cm cm in gas of volume gas of moles of number

= a Calculate the number of moles of hydrogen molecules in 240 cm3 of the gas.

moles .0 cm cm moles of number

= = b How many moles of a gas are there in 1000 cm3 of the gas?

moles .0 cm cm gas of moles of number

= =

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Example 5 Calculation of the volume of a given mass of gas

This calculation require you to apply the skills covered in the previous examples

Calculate the volume of 10 g of hydrogen gas.

This is a two-stage calculation a) you need to calculate how many moles of hydrogen gas are present, and b) you need to convert this to a volume.

mol g of ) H ( hydrogen of mass molecular

) H ( hydrogen of g ) H ( hydrogen of moles of number − = ∴

= 5 moles

∴ 5 moles of hydrogen = 5 x 24000 cm3 = 120000 cm3 = 120 dm3

Example 6 Calculation of the mass of a given volume of gas

This calculation require you to apply the skills covered in the previous examples

Calculate the mass of 1000 cm3 of carbon dioxide

Again this is a two-stage calculation a) you need to calculate the number of moles of carbon dioxide and then b) convert this to a mass.

cm of CO of mole of volume

CO of cm CO of moles of number = ∴

= 0.0147 moles

∴ 0.0147 moles of carbon dioxide = 0.0147 x 44 g = 1.833 g

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Example 7

Calculation of the molar mass of a gas from mass and volume data for the gas

Calculations of this type require you to find the mass of 1 mole of the gas, ie 24 000 cm3. This is the molar mass of the gas.

eg Calculate the Relative Molecular Mass of a gas for which 100 cm3 of the gas at room temperature and pressure, have a mass of 0.0667 g

100 cm3 of the gas has a mass of 0.0667 g cm cm g .0 of mass a have must gas the of cm × = ∴

= 16 g

∴ The molar mass of the gas is 16 g mol–1

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Exercise 4a Calculation of the number of moles of material in a given mass of that material

In this set of calculations all the examples chosen are from the list of compounds whose molar mass you calculated in exercise 1.

In each case calculate the number of moles of the material in the mass stated.

9.00 g of H2O

88.0 g of CO2

1.70 g of NH3

230 g of C2H5OH

560 g of C2H4

0.640 g of SO2

80.0 g of SO3

18.0 g of HBr

0.0960 g of H2SO4

3.15 g of HNO3

19.3 g of NaCl

21.25 g of NaNO3

2.25 g of Na2CO3

0.800 g of NaOH

17.75 g of Na2SO4

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3.16 g of KMnO4

32.33 g of K2CrO4

100 g of KHCO3

7.63 g of KI

3.90 g of CsNO3

0.111 g of CaCl2

41.0 g of Ca(NO3)2

1.48 g of Ca(OH)2

3.40 g of CaSO4

41.6 g of BaCl2

14.95 g of CuSO4

13.64 g of ZnCl2

1.435 g of AgNO3

13.76 g of NH4Cl

13.76 g of (NH4)2SO4

23.4 g of NH4VO3

10.0 g of KClO3

10.7 g of KIO3

100 g of NaClO

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1.70 g of NaNO2

50.9 g of CuSO4.5H2O

19.6 g of FeSO4.7H2O

9.64 g of (NH4)2SO4.Fe2(SO4)3.24H2O

12.4 g of Na2S2O3.5H2O

32.0 g of (COOH)2.2H2O

3.075 g of MgSO4.7H2O

40.0 g of Cu(NH3)4SO4.2H2O

6.00 g of CH3CO2H

3.10 g of CH3COCH3

0.530 g of C6H5CO2H

4.79 g of AlCl3

56.75 g of Al(NO3)3

8.35 g of Al2(SO4)3

3.8 g of FeSO4

200 g of FeCl2

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Exercise 4b Calculation of the mass of material in a given number of moles of the material

In each case calculate the mass in grams of the material in the number of moles stated.

2 moles of H2O

3 moles of CO2

2.8 moles of NH3

0.50 moles of C2H5OH

1.2 moles of C2H4

0.64 moles of SO2

3 moles of SO3

1 mole of HBr

0.012 moles of H2SO4

0.15 moles of HNO3

0.45 moles of NaCl

0.70 moles of NaNO3

0.11 moles of Na2CO3

2.0 moles of NaOH

0.90 moles of Na2SO4

0.050 moles of KMnO4

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0.18 moles of K2CrO4

0.90 moles of KHCO3

1.5 moles moles of KI

0.12 moles of CsNO3

0.11 moles of CaCl2

4.1 moles of Ca(NO3)2

0.0040 moles of Ca(OH)2

0.10 moles of CaSO4

0.21 moles of BaCl2

0.10 moles of CuSO4

0.56 moles of ZnCl2

0.059 moles of AgNO3

0.333 moles of NH4Cl

1.1 moles of (NH4)2SO4

0.025 moles of NH4VO3

0.10 moles of KClO3

0.10 moles of KIO3

10 moles of NaClO

0.0010 moles of NaNO2

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0.20 moles of CuSO4.5H2O

0.10 moles of FeSO4.7H2O

0.0050 moles of (NH4)2SO4.Fe2(SO4)3.24H2O

0.040 moles of Na2S2O3.5H2O

2.4 moles of (COOH)2.2H2O

3.075 moles of MgSO4.7H2O

0.15 moles of Cu(NH3)4SO4.2H2O

0.17 moles of CH3CO2H

0.20 moles of CH3COCH3

0.080 moles of C6H5CO2H

0.0333 moles of AlCl3

0.045 moles of Al(NO3)3

0.12 moles of Al2(SO4)3

2.0 moles of FeSO4

11 moles of FeCl2

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Exercise 4c Calculation of the volume of a given number of moles of a gas

In each case calculate the volume of the number of moles of gas stated.

(Assume that all volumes are measured at room temperature and pressure and that 1 mole of gas has a volume of 24 000 cm3 under these conditions).

1 mole of CO2

0.1 moles of NH3

0.5 moles of C2H4

2 moles of SO2

0.12 moles of SO3

3.4 moles of HBr

0.11 moles of Cl2

0.0040 moles of CH4

10 moles of H2

0.45 moles of O2

0.0056 moles of C2H6

0.0090 moles of C3H8

0.040 moles of C2H2

0.123 moles of NO

0.0023 moles of HCl

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8.0 moles of HBr

0.000010 moles of HI

6.0 moles of NO2

0.0076 moles of F2

3.0 moles of N2

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Exercise 4d

Calculation of the number of moles of gas in a given volume of that gas

In each case calculate the volume of the number of moles of gas stated.

(Assume that all volumes are measured at room temperature and pressure and that 1 mol of gas has a volume of 24 000 cm3 under these conditions).

200 cm3 of CO2

500 cm3 of NH3

1000 cm3 of C2H4

2000 cm3 of SO2

234 cm3 of SO3

226 cm3 of HBr

256 cm3 of Cl2

200 cm3 of CH4

2000 cm3 of H2

2400 cm3 of O2

700 cm3 of C2H6

5600 cm3 of C3H8

2200 cm3 of C2H2

210 cm3 of NO

800 cm3 of HCl

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80 cm3 of HBr

2 cm3 of HI

20 000 cm3 of NO2

420 cm3 of F2

900 cm3 of N2

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Exercise 4e Calculation of the mass of a given volume of gas

Calculate the mass of the volume of gases stated below.

(Assume that all volumes are measured at room temperature and pressure and that

1 mol of gas has a volume of 24 000 cm3 under these conditions).

200 cm3 of CO2

500 cm3 of NH3

1000 cm3 of C2H4

2000 cm3 of SO2

234 cm3 of SO3

226 cm3 of HBr

256 cm3 of Cl2

200 cm3 of CH4

2000 cm3 of H2

2400 cm3 of O2

700 cm3 of C2H6

5600 cm3 of C3H8

2200 cm3 of C2H2

210 cm3 of NO

800 cm3 of HCl

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80 cm3 of HBr

2 cm3 of HI

20 000 cm3 of NO2

420 cm3 of F2

900 cm3 of N2

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Exercise 4f Calculation of the volume of a given mass of gas

In each case calculate the volume on cm3 of the mass of gas given.

(Assume that all volumes are measured at room temperature and pressure and that

1 mol of gas has a volume of 24 000 cm3 under these conditions).

2 g of CO2

5 g of NH3

10 g of C2H4

20 g of SO2

2.34 g of SO3

2.26 g of HBr

10 g of Cl2

20 g of CH4

200 g of H2

240 g of O2

70 g of C2H6

56 g of C3H8

22 g of C2H2

20 g of NO

8 g of HCl

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8 g of HBr

2 g of HI

23 g of NO2

42 g of F2

90 g of N2

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Exercise 4g

Calculation of the Relative Molecular Mass of a gas from mass and volume

data for the gas

In each case you are given the mass of a certain volume of an unknown gas. From each set of data calculate the Relative Molecular Mass of the gas.

(Assume that all volumes are measured at room temperature and pressure and that

1 mol of gas has a volume of 24 000 cm3 under these conditions).

0.373 g of gas occupy 56 cm3

0.747 g of gas occupy 280 cm3

0.467 g of gas occupy 140 cm3

0.296 g of gas occupy 100 cm3

0.0833 g of gas occupy 1000 cm3

0.175 g of gas occupy 150 cm3

0.375 g of gas occupy 300 cm3

0.218 g of gas occupy 90 cm3

0.267 g of gas occupy 200 cm3

1.63 g of gas occupy 1400 cm3

0.397 g of gas occupy 280 cm3

0.198 g of gas occupy 280 cm3

0.0602 g of gas occupy 38 cm3

0.0513 g of gas occupy 44 cm3

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0.0513 g of gas occupy 28 cm3

1.33 g of gas occupy 1000 cm3

8.79 g of gas occupy 1000 cm3

0.0760 g of gas occupy 50 cm3

0.338 g of gas occupy 100 cm3

0.667 g of gas occupy 125 cm3

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Section 5

Using the idea of moles to find formulae

You can find the formula of copper(II) oxide by passing a stream of hydrogen over a known mass of copper oxide and weighing the copper formed.

• A known mass of copper(II) oxide is used.
• A stream of hydrogen from a cylinder is passed over the copper until all the air has been swept out of the apparatus.
• It is heated to constant mass (until two consecutive mass determinations at the end of the experiment are same) in a stream of dry hydrogen.
• The mass of the copper is finally determined.

Note:

• Excess hydrogen must not be ignited until it has been tested (by collection in a test tube) to make sure that all the air has been expelled from the apparatus. If the hydrogen in the test tube burns quietly, without a squeaky pop, then it is safe to ignite it at the end of the tube.
• The combustion tube is tilted to prevent the condensed steam from running back on to the hot part of the tube.
• When the reduction process is complete, ie after heating to constant mass, the tube is allowed to cool with hydrogen still being passed over the remaining copper. This is to prevent the copper from being oxidized to copper(II) oxide.

The working on the next page shows you how to calculate the results:

Typical results

Mass of copper (II) oxide = 5 g Mass of copper = 4 g Mass of oxygen = 1 g

Dry hydrogen

Boat containing copper oxide

HEAT

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÷ by relative atomic mass (r.a.m)

÷ by smallest Ratio of atoms

Moles of copper atoms .0 4 = .0

.0 =

Moles of oxygen atoms .0 1 = .0

.0 =

Therefore, the simplest (or empirical) formula is CuO.

The apparatus may be modified to determine the formula of water. Anhydrous calcium chloride tubes are connected to the end of the combustion tube and the excess hydrogen then ignited at the end of these tubes. Anhydrous calcium chloride absorbs water; the mass of the tubes is determined at the beginning and end of the experiment. The increase in mass of the calcium chloride tubes is equal to the mass of water produced.

Typical results

Mass of water = 1.125 g Mass of oxygen (from previous results) = 1.000 g Mass of hydrogen = 0.125 g

÷ by r.a.m ÷ by smallest Ratio of atoms

Moles of hydrogen atoms .0 .0 = .0

.0 =

Moles of oxygen atoms .0 1 = .0

.0 =

Since the ratio of hydrogen to oxygen is 2:1 the simplest (or empirical) formula is H2O.

In examination calculations of this type the data is often presented not as mass, but as percentage composition of the elements concerned. In these cases the calculation is carried out in an identical fashion as percentage composition is really the mass of the element in 100 g of the compound.

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Example 1

Sodium burns in excess oxygen to give a yellow solid oxide that contains 58.97% of sodium. What is the empirical formula of the oxide?

N.B. This is an oxide of sodium. It must contain only Na and O. Since the percentage of Na is 58.97 that of O must be 100 – 58.97 = 41.03%

÷ by r.a.m ÷ by smallest Ratio of atoms

Moles of sodium atoms .2 . = .2

.2 =

Moles of oxygen atoms .2 . = .2

.2 =

Therefore the empirical formula is NaO.

The result of the above calculation does not seem to lead to a recognisable compound of sodium. This is because the method used only gives the simplest ratio of the elements - but see below.

Consider the following series of organic compounds:

C2H4 ethene, C3H6 propene, C4H8 butene, C5 H10 pentene. These all have the same empirical formula C H2.

To find the Molecular Formula for a compound it is necessary to know the Relative Molecular Mass (Mr).

Molecular Formula Mass = Empirical Formula Mass x a whole number (n)

In the example above the oxide has an Mr = 78 g mol–1.

Thus

Molecular Formula Mass = 78

Empirical Formula Mass = (Na + O) = 23 + 16 = 39

∴ 78 = 39 x n

∴ n = 2

The Molecular Formula becomes (NaO)2 or Na2O2

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Example 2

A compound P contains 73.47% carbon and 10.20% hydrogen by mass, the remainder being oxygen. It is found from other sources that P has a Relative Molecular Mass of 98 g mol-1. Calculate the molecular formula of P.

NB It is not necessary to put in all the details when you carry out a calculation of this type. The following is adequate.

C H O

73.47 10.20 (100 – 73.47 – 10.20)

= 16.33

By r.a.m . . .

= 6.1225 = 10.20 = 1.020

By smallest .1

.6 .1

. .1

.1

Ratio of atoms

Therefore the empirical formula is C6H10O.

To find molecular formula:

Molecular Formula Mass = Empirical Formula Mass x whole number (n)

98 = [(6x12) + (10x1) + 16] x n = 98 x n

∴ n =

The molecular formula is the same as the empirical formula C6H10O.

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A warning

In calculations of this type at GCE Advanced level you may meet compounds that are different yet have very similar percentage composition of their elements. When you carry out a calculation of this type you should never round up the figures until you get right to the end. For example NH4OH and NH2OH have very similar composition and if you round up the data from one you may well get the other. If you are told the Relative Molecular Mass and your Empirical Formula Mass is not a simple multiple of this you are advised to check your calculation.

Example 3

Calculate the empirical formula of a compound with the following percentage composition:

C 39.13%; O 52.17%; H 8.700%

C O H

By r.a.m . . .8

= 3.26 = 3.25 = 8.70

Divide by smallest 2.66

It is clear at this stage that dividing by the smallest has not resulted in a simple ratio. You must not round up or down at this stage. You must look at the numbers and see if there is some factor that you could multiply each number by to get each one to a whole number. In this case if you multiply each by 3 you will get:

C O H

Thus C3H8O3 is the empirical formulae not C1H2.66O1

You need to watch carefully for this, the factors will generally be clear and will be 2 or 3. What you must not do is round 1.33 to 1 or 1.5 to 2.

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Calculations involving the moles of water of crystallization

In calculations of this type you need to treat the water as a molecule and divide by the Relative Molecular Mass.

Example 4

24.6 grams of a hydrated salt of MgSO4 . xH2O, gives 12.0 g of anhydrous MgSO4 on heating. What is the value of x ?

Your first job is to find the mass of water driven off .

Mass of water evolved = 24.6 – 12.0 = 12.6 g

MgSO4 H2O

12.0 12.6

Divide by Mr 0. 6.

= 0.100 = 0.700

Ratio of Atoms

Giving a formula of MgSO4.7H2O

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Exercise 5 Calculation of a formula from experimental data

In Section a. calculate the empirical formula of the compound from the data given. This may be as percentage composition or as the masses of materials found in an experiment. For Section b. you are given the data for analysis plus the Relative Molecular Mass of the compound, in these cases you are to find the empirical formula and thence the molecular formula. Section c. is more difficult, the data is presented in a different fashion but the calculation of the empirical formula/ molecular formula is essentially the same.

Section a

Ca 40%; C 12%; O 48%

Na 32.4%; S 22.5%; O 45.1%

Na 29.1%; S 40.5%; O 30.4%

Pb 92.8%; O 7.20%

Pb 90.66%; O 9.34%

H 3.66%; P 37.8%; O 58.5%

H 2.44%; S 39.0%; O 58.5%

C 75%; H 25%

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C 81.81%; H 18.18%

H 5.88% ; O 94.12%

H 5%; N 35%; O 60%

Fe 20.14%; S 11.51%; O 63.31%; H 5.04%

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Section b

A hydrocarbon with a Relative Molecular Mass (Mr) of 28 g mol–1 has the following composition: Carbon 85.7%; Hydrogen 14.3%. Calculate its molecular formula.

A hydrocarbon with a Relative Molecular Mass (Mr) of 42 g mol–1 has the following composition: Carbon 85.7%; Hydrogen 14.3%. Calculate its molecular formula.

P 10.88%; I 89.12%. Mr = 570 g mol–1

N 12.28%; H 3.51%; S 28.07%; O 56.14%. Mr = 228 g mol–1

P 43.66%; O 56.34%. Mr = 284 g mol–1

C 40%; H 6.67%; O 53.3%. Mr = 60 g mol–1

Analysis of a compound with a Mr = 58 g mol–1 shows that 4.8 g of carbon are joined with 1.0 g of hydrogen. What is the molecular formula of the compound?

3.36 g of iron join with 1.44 g of oxygen in an oxide of iron that has a Mr = 160 g mol–1 What is the molecular formula of the oxide?

A sample of an acid with a Mr = 194 g mol–1 has 0.5 g of hydrogen joined to 16 g of sulphur and 32 g of oxygen. What is the molecular formula of the acid?

Analysis of a hydrocarbon showed that 7.8 g of the hydrocarbon contained 0.6 g of hydrogen and that the Mr = 78 g mol–1. What is the formula of the hydrocarbon?

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Section c

22.3 g of an oxide of lead produced 20.7 g of metallic lead on reduction with hydrogen. Calculate the empirical formula of the oxide concerned.

When 1.17 g of potassium is heated in oxygen 2.13 g of an oxide is produced. In the case of this oxide the empirical and molecular formulae are the same. What is the molecular formula of the oxide produced?

A hydrocarbon containing 92.3% of carbon has a Relative Molecular Mass of 26 g mol–1. What is the molecular formula of the hydrocarbon?

When 1.335 g of a chloride of aluminium is added to excess silver nitrate solution 4.305 g of silver chloride is produced. Calculate the empirical formula of the chloride of aluminium.

Hint; you will need to work out how much chlorine there is in 4.305 g of AgCl. This will be the amount of chlorine in the initial 1.335 g of the aluminium chloride.

16 g of a hydrocarbon burn in excess oxygen to produce 44 g of carbon dioxide. What is the empirical formula of the hydrocarbon.

Hint; you will need to work out what mass of carbon is contained in 44 g of CO2. This is the mass of C in 16 g of the hydrocarbon.

When an oxide of carbon containing 57.1% oxygen is burnt in air the percentage of oxygen joined to the carbon increases to 72.7%. Show that this data is consistent with the combustion of carbon monoxide to carbon dioxide.

When 14.97 g of hydrated copper(II) sulphate is heated it produces 9.60 g of anhydrous copper(II) sulphate. What is the formula of the hydrated salt?

When the chloride of phosphorus containing 85.1% chlorine is heated a second chloride containing 77.5% chlorine is produced. Find the formulae of the chlorides and suggest what the other product of the heating might be.

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Section 6

Chemical equations

Chemical equations do much more than tell you what reacts with what in a chemical reaction. They tell you how many of each type of molecule is needed and are produced and so they also tell you what masses of the reactants are needed to produce a given mass of products.

Often you will learn equations that have been given to you. However, if you are to interpret equations correctly you must learn to write them for yourself.

Equations in words

Before you can begin to write an equation, you must know what the reacting chemicals are and what is produced in the reaction. You can then write them down as a word equation. For instance, hydrogen reacts with oxygen to give water, or, as a word equation:

Hydrogen + Oxygen

→

Water

Writing formulae

When you have written the equation in words you can write the formula for each of the substances involved; you may know them or you may have to look them up. In our example:

Hydrogen is represented as H2 ; Oxygen is represented as O2; and Water is H2O.

So we get:

H2 + O2

→

H2O

However this will not do as a full equation as you will discover if you read on!

Balancing the equation

One of the most important things you must understand in chemistry is that atoms are rearranged in chemical reactions: they are never produced from ‘nowhere’ and they do not simply ‘disappear’. This means that in chemical equation you must have the same number of atoms of each kind on the left-hand side of the equation as on the right. Sometimes you need to start with two or more molecules of one of the reactants and you may end up with more than one molecule of one of the products.

Let us look at two very simple examples:

Carbon + Oxygen → Carbon dioxide

C + O2 → CO2

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It so happens that carbon dioxide has one atom of carbon and two atoms of oxygen in one molecule: carbon is written as C (one atom) and oxygen molecules have two atoms each: written as O2.

This equation does not need balancing because the number of atoms of carbon is the same on the left as on the right (1) and the number of atoms oxygen is also the same (2) ie it is already balanced.

Now let us try one that does not work out:

Magnesium + Oxygen

→

Magnesium oxide

Magnesium is written as Mg (one atom just like carbon) and oxygen is, of course, O2, but magnesium oxide has just one atom of oxygen per molecule and is therefore written as MgO.

So we might write:

Mg + O2

→

MgO

The magnesium balances, one atom on the left and one on the right, but the oxygen does not as there are two atoms on the left-hand side of the equation and only one on the right hand side. You cannot change the formulae of the reactants or products.

Each ‘formula’ of magnesium oxide has only one atom of oxygen: each molecule of oxygen has two atoms of oxygen, so you can make two formulae of magnesium oxide for each molecule of oxygen. So we get:

Mg + O2

→

2MgO

Even now the equation does not balance, because we need two atoms of magnesium to make two formulae of MgO, and the final equation is:

2Mg + O2

→

2MgO

Sometimes, you will need to show in the equation whether the chemicals are solid, liquid or gas. You do this by putting in state symbols: (aq) for aqueous solution, (g) for gas, (1) for liquid and

• for solid or precipitate:

2Mg(s) + O2(g) → 2MgO(s)

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Exercise 6a Balancing equations

Balance the following equations. To get you started _ indicates the first six questions where numbers need to be inserted to achieve the balance. In one or two difficult examples some of the numbers have been added. You will not need to change these. Also remember all the formulae are correct!

_ H2

+

O2

→

_ H2O

BaCl2

+

_ NaOH

→

Ba(OH)2

+

_ NaCl

H2SO4

+

_ KOH

→

_ K2SO4

+

H2O

K2CO3

+

_ HCl

→

_KCl

+

H2O

+

CO2

CaCO3

+

_HNO3

→

Ca(NO3)2

+

H2O

+

CO2

Ca

+

_H2O

→

Ca(OH)2

+

H2

Pb(NO3)2

+

NaI

→

PbI2

+

NaNO3

Al2(SO4)3

+

NaOH

→

Al(OH)3

+

Na2SO4

Al(OH)3

+

NaOH

→

NaAlO2

+

H2O

Pb(NO3)2

→

PbO

+

NO2

+

O2

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FeSO4

→

Fe2O3

+

SO2

+

SO3

NH4NO3

→

N2O

+

H2O

NaNO3

→

NaNO2

+

O2

CH4

+

O2

→

CO2

+

H2O

C4H10

+

O2

→

CO2

+

H2O

PCl3

+

H2O

→

H3PO3

+

HCl

8HNO3

+

3Cu

→

Cu(NO3)2

+

NO

+

H2O

4HNO3

+

Cu

→

Cu(NO3)2

+

NO2

+

H2O

H3PO4

+

NaOH

→

NaH2PO4

+

H2O

H3PO4

+

NaOH

→

Na3PO4

+

H2O

H3PO4

+

NaOH

→

Na2HPO4

+

H2O

6NaOH

+

Cl2

→

NaClO3

+

NaCl

+

H2O

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N2

+

H2

→

NH3

NaBr

+

H2SO4

→

Na2SO4

+

HBr

HBr

+

H2SO4

→

H2O

+

SO2

+

Br2

C2H5OH

+

PCl3

→

C2H5Cl

+

H3PO3

Fe3O4

+

H2

→

Fe

+

H2O

Fe2O3

+

CO

→

Fe

+

CO2

C2H5OH

+

CH3CO2H

→

CH3CO2C2H5

+

H2O

2KMnO4

+

HCl

→

MnCl2

+

Cl2

+

8H2O

+

KCl

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Exercise 6b What’s wrong here?

The following equations have one or more mistakes. These may be in a formula, in the balancing, in the state symbols or even in the chemistry. Your job is to identify the error and then write a correct equation.

Na(s) + H2O(l) → NaOH(aq) + H(g)

PbNO3(aq) + NaCl(aq) → PbCl(s) + NaNO3(aq)

CaOH2(aq) + 2HCl(aq) → CaCl2(aq) + 2H2O(l)

C2H4(g) + 2O2(g) → 2CO2(g) + 2H2(g)

MgSO4(aq) + 2NaOH → Ca(OH)2(s) + Na2SO4(aq)

Cu(NO3)2(s) + CuO(s) → 2NO(g) + O3(g)

Cu(s) + H2SO4(aq) → CuSO4(aq) + H2(g)

AlCl2(s) + 2KOH(aq) → Al(OH)2(s) + 2KCl(aq)

NaCO3(s) + 2HCl(aq) → NaCl2(aq) + CO2(g) + H2O(l)

2AgNO3(aq) + MgCl2(aq) → Mg(NO3)2(s) + 2AgCl(aq)

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Exercise 6c Writing equations in symbols from equations in words

In the following examples you will need to convert the names of the materials into formulae and then balance the resulting equation. In some cases more than one experiment is described. In these cases you will need to write more than one equation.

Zinc metal reacts with copper sulphate solution to produce solid copper metal and zinc sulphate solution.

Solid calcium hydroxide reacts with solid ammonium chloride on heating to produce solid calcium chloride, steam and ammonia gas.

When lead(II) nitrate is heated in a dry tube lead(II) oxide, nitrogen dioxide gas and oxygen are produced.

Silicon tetrachloride reacts with water to produce solid silicon dioxide and hydrogen chloride gas.

When a solution of calcium hydrogen carbonate is heated a precipitate of calcium carbonate is produced together with carbon dioxide gas and more water.

When octane (C8H18) vapour is burnt with excess air in a car engine carbon dioxide and water vapour are produced.

All the halogens, apart from fluorine, react with concentrated sodium hydroxide solution to produce a solution of the sodium halide (NaX) the sodium halate (NaXO3) and water.

The elements of Group 1 of the periodic table all react with water to produce a solution of the hydroxide of the metal and hydrogen gas.

The last two examples in this section will need a lot of thought as they involve changes in the valency of the elements concerned. Before you start to balance the equations check with your teacher that you have the formulae correct.

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Tin(II) chloride solution reacts with mercury(II) chloride solution to produce a precipitate of mercury(I) chloride and a solution of tin(IV) chloride. This precipitate of mercury(I) chloride then reacts with further tin(II) chloride solution to produce liquid mercury and more tin(IV) chloride solution.

Concentrated sulphuric acid reacts with solid potassium iodide to produce solid potassium hydrogen sulphate, iodine vapour, water and hydrogen sulphide gas.

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Section 7

How equations are found by experiment

Although equations are often printed in books for you to learn, you must remember that they have all been found originally by someone doing experiments to measure how much of each chemical reacted and how much of each product was formed.

Below are set out some of the methods you could use:

• Direct mass determinations, eg the reaction of magnesium with oxygen

A known mass of magnesium is heated in a crucible to constant mass and hence the mass of magnesium oxide is found. Supposing 0.12 g of magnesium produce 0.20 g of magnesium oxide. By subtraction, the mass of oxygen combined with the magnesium is 0.080 g.

Each of these masses is then converted to moles and it is found that every 2 moles of magnesium react with one mole oxygen molecules and produce two moles of magnesium oxide: hence

2Mg + O2

→

2MgO

• Reacting volumes in solution: usually you have to calculate concentrations of acids or alkalis by reaction with the appropriate standard solution and use the chemical equation for the reaction

However, you can calculate the ratio of reacting moles from experimental data, in order to construct the equation. To do this you use solutions, both of whose concentrations you know. You then do a titration in the usual way and use the volumes used in the titration to find the number of moles of each reagent which react.

These are then used in the equation straight away, just as in the magnesium oxide example above.

• Measurement of gas volumes: the molar volume of a gas is taken as 24 dm3 at room temperature and atmospheric pressure (r.t.p.)

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Examples

In an experiment a solution containing 3.31 g of lead(II) nitrate reacts with a solution containing 1.17 g of sodium chloride to produce 2.78 g of lead(II) chloride solid and leave a solution that contains 1.70 g of sodium nitrate. What is the equation for the reaction?

In this type of question you are required to calculate the ratio of the reacting moles and then use these to write the equation.

Pb(NO3)2 Mr = 331

NaCl Mr = 58.5

PbCl2 Mr = 278

NaNO3 Mr = 85

∴ 3.31 g of Pb(NO3)2 = (3.31/331) mol = 0.010 mol

1.17 g of NaCl = (1.17/58.5) mol = 0.020 mol

2.78 g of PbCl2 = (2.78/278) mol = 0.010 mol

1.70 g of NaNO3 = (1.70/85) mol = 0.020 mol

∴ 0.01 mol of Pb(NO3)2 reacts with 0.02 mol of NaCl to give 0.01 mol of PbCl2 and 0.02 mol of NaNO3 ie 1 mol of Pb(NO3)2 reacts with 2 mol of NaCl to give 1 mol of PbCl2 and 2 mol of NaNO3

Pb(NO3)2 + 2NaCl

→

PbCl2 + 2NaNO3

It is not necessary to write all of this out each time.

When 5.175 g of lead are heated at 300oC the lead reacts with the oxygen in the air to produce 5.708 g of an oxide of lead. This is the only product. What is the equation for this reaction? In a question of this type you seem to be short of information but in fact you know the mass of oxygen reacting. Remember this is oxygen molecules that are reacting not oxygen atoms.

Mass of oxygen used is 5.708 – 5.175 g = 0.533 g

Moles of lead reacting = (5.175/207) mol = 0.025 mol

Moles of oxygen reacting = (0.533/32) mol = 0.0167 mol

∴ 0.025 mol of Pb react with 0.0167 mol of O2 to give product

∴ 1.5 mol of Pb react with 1 mol of O2 to give product

∴3 mol of Pb react with 2 mol of O2 to give product

3Pb + 2O2

→

Pb3O4

You do not have the information to write the full equation but as you know there is only one product and this has 3 lead atoms and 4 oxygen you can suggest a formula.

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25 cm3 of 2M sulphuric acid solution react with 50 cm3 of 2M sodium hydroxide solution to produce sodium sulphate and water. Construct the equation for this reaction.

You will need to look at the start of chapter 11 before you can follow this question.

25 cm3 of 2M H2SO4 contain (25 x 2/1000) mol of H2SO4 = 0.050 mol

50 cm3 of 2M NaOH contain (50 x 2/1000) mol of NaOH = 0.10 mol

∴ 0.05 mol of H2SO4 react with 0.10 mol of NaOH to give Na2SO4 plus H2O

∴ 1 mol of H2SO4 react with 2 mol of NaOH to give Na2SO4 plus H2O

H2SO4 + 2NaOH → Na2SO4 + 2H2O

2 cm3 of nitrogen gas react completely with 6 cm3 of hydrogen gas to produce 4cm3 of ammonia gas. Use the data to write the equation for this reaction.

2 cm3 of nitrogen = (2/24000) mol = 8.33 x 10-5 mol

6 cm3 of hydrogen = (6/24000) mol = 2.50 x 10-4 mol

4 cm3 of ammonia = (4/24000) mol = 1.67 x 10-4 mol

∴ ratios are (8.33 x 10-5/8.33 x 10-5) of nitrogen = 1

(2.50 x 10-4/8.33 x 10-5) of hydrogen = 3

(1.67 x 10-4/8.33 x 10-5) of ammonia = 2

N2 + 3H2 → 2NH3

1 g of CaCO3 reacts with 10 cm3 of 2M HNO3 to produce 1.64 g of Ca(NO3)2, 240 cm3 of CO2 and water.

In practice the acid will be in water and it is almost impossible to measure the amount of water produced by the reaction.

1/ 100 mol of CaCO3

+ (10x2)/1000 mol of HNO3

→

1.64/164 mol of Ca(NO3)2

+ 240/24000 mol of CO2

+ H2O

∴ 0.01 mol of CaCO3

+ 0.02 mol of HNO3

→

0.01 mol of Ca(NO3)2

+ 0.01 mol of CO2

+ H2O

CaCO3 + 2HNO3 → Ca(NO3)2 + CO2 + H2O

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Exercise 7 Writing chemical equations from experimental data

Use the data below to write the equations for the reactions listed. In some cases you may not be able to calculate the moles of all the materials involved. In these cases you should indicate that you have ‘balanced’ this part yourself.

In examples involving gases you should assume 1 Mole of gas occupies 24 000 cm3 at room temperature and pressure.

In an experiment a solution containing 6.675 g of aluminium chloride reacted with a solution containing 25.50 g of silver nitrate. 21.52 g of silver chloride was produced together with a solution of 10.65 g of aluminium nitrate, Al(NO3)3. What is the equation for the reaction taking place?

100 cm3 of a solution of potassium chromate(VI), containing 97.05 g dm–3, reacts with 50 cm3 of a solution, containing 331 g dm–3 of lead nitrate, to produce 16.15 g of a precipitate of lead(II) chromate and 150 cm3 of a solution of potassium nitrate, which gives 10.1 g of solid when the water is evaporated off from the solution. Write the equation for the reaction.

1.133 g of silver nitrate was heated in an open tube. The silver residue weighed 0.720 g. During the reaction 0.307 g of nitrogen dioxide was also produced. The rest of the mass loss was due to oxygen. Use the data to write the equation for the reaction.

In a titration using methyl orange as an indicator 25.0 cm3 of a solution of 0.1 M sodium hydroxide reacted with 25.0 cm3 of 0.1 M phosphoric acid, H3PO4, solution. If the experiment is repeated using phenolphthalein in place of the methyl orange as the indicator the volume of the sodium hydroxide used to cause the indicator to change colour is 50.0 cm3.

i Use the data to calculate the number of moles of sodium hydroxide that reacts with one mole of phosphoric acid in each case ii Suggest the formula of the salt produced in each case iii Write the equations iv What volume of the alkali would be needed to produce the salt Na3PO4?

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50 cm3 of a solution of citric acid, Mr = 192, containing 19.2 g dm–3 reacted with 50 cm3 of a solution of sodium hydroxide containing 12 g dm–3. Citric acid can be represented by the formula HxA, where x represents the number of hydrogen atoms in the molecule. Use the data above to calculate the number of moles of sodium hydroxide that react with one mole of citric acid and hence find the value of x.

When 12.475 g of hydrated copper(II) sulphate, CuSO4.xH2O, was heated 7.980 g of anhydrous salt were produced. Use the data to find the value of x and hence write the equation for the reaction.

When 20 cm3 ammonia gas is passed over a catalyst with excess oxygen 20 cm3 of nitrogen monoxide (NO) and 30 cm3 of water vapour are produced. Use this data to write the equation for the reaction.

10 cm3 of a hydrocarbon CaHb reacted with 50 cm3 of oxygen gas to produce 30 cm3 of carbon dioxide and 40 cm3 of water vapour. Use the data to calculate to reacting moles in the equation and suggest value for a and b.

When 8.4 g of sodium hydrogen carbonate are heated 5.30 g of solid residue are produced 1200 cm3 of carbon dioxide are produced and 0.900 g of water are evolved. Show that this data is consistent with the following equation.

2NaHCO3

→

Na2CO3 + CO2 + H2O

10 When 13.9 g of FeSO4.xH2O is heated 4 g of solid iron (III) oxide is produced together with the loss of 1.6 g of sulphur dioxide and 2.0 g of sulphur trioxide. The rest of the mass loss being due to the water of crystallization being lost. Use the data to write the full equation for the action of heat.

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Section 8

Amounts of substances

Equations can also tell you how much of a chemical is reacting or is produced. The equation in Section 7 tells us that 2 moles of (solid) magnesium atoms react with 1 mole of (gaseous) oxygen molecules to produce 2 moles of (solid) magnesium oxide molecules.

We know that the relative atomic mass of magnesium is 24, and that of oxygen is 16, (see periodic table on page 113). And from the equation we balanced in Section 6 we can suggest that 48 g of magnesium react with 32 g of oxygen (because an oxygen molecule contains two atoms) to give 80 g of magnesium oxide.

Since we know the ratio of reacting masses (or volumes in the case of gases) we can calculate any reacting quantities based on the equation.

Example 1

a What mass of magnesium oxide would be produced from 16 g of magnesium in the reaction between magnesium and oxygen?

i Write the full balanced equation

2Mg(s) + O2(g)

→

2MgO(s) ii Read the equation in terms of moles

2 moles of magnesium reacts to give 2 moles of magnesium oxide iii Convert the moles to masses using the Mr values

∴ (2 x 24g) of magnesium gives 2 x (24+16) = 80 g of Magnesium oxide

∴ 16 g of magnesium gives ×

× =

26.7 g of Magnesium oxide

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004 b What volume of oxygen would react with 16 g of magnesium in the above reaction?

In this case the oxygen is a gas so the volume of each mole is 24000 cm3 at room temperature and pressure and you do not have to worry about the molecular mass of the gas.

From the equation:

2 moles of Mg reacts with 1 mole of O2

∴2 x 24 g of Mg reacts with 1 x 24000 cm3 of O2(g)

∴ 16 g of Mg reacts with g g cm

×

× × =

8000 cm3 of oxygen Example 2

What mass of lead(II) sulphate would be produced by the action of excess dilute sulphuric acid on 10 g of lead nitrate dissolved in water ?

Pb(NO3)2(aq) + H2SO4(aq) → PbSO4(s) + 2HNO3(aq)

∴ 1 mole of lead nitrate gives 1 mole of lead sulphate

∴ 331 g of lead nitrate gives 303 g of lead sulphate

∴ 10 g of lead nitrate gives g sulphate lead of g g 303 × =

9.15 g of lead sulphate Example 3

What is the total volume of gas produced by the action of heat on 1 g of silver nitrate?

2AgNO3(s) → 2Ag(s) + 2NO2(g) + O2(g)

2 moles of silver nitrate give 2 moles of nitrogen dioxide gas plus 1 mole of oxygen gas = 3 moles of gas

∴ 2 x 170 g of silver nitrate give 3 x 24000 cm3 of gas

∴1 g of silver nitrate gives g gas of g cm

×

× × = 211.8 cm3 of gas

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Example 4

When excess carbon dioxide is passed into sodium hydroxide solution, sodium carbonate solution is formed. This can be crystallised out as Na2CO3.10H2O . What mass of crystals would be produced from 5 g of sodium hydroxide in excess water.

Care. You need the water as moles in the equation.

2NaOH(aq) + CO2(g) + 9H2O → Na2CO3(aq) + 10H2O(l) → Na2CO3.10H2O(s)

∴ 2 moles of sodium hydroxide give 1 mole of the crystals of sodium carbonate

∴ 2 x 40 g of sodium hydroxide give 286 g of crystals

∴ 5 g of sodium hydroxide give ×

× =

17.88 g of crystals Example 5

What mass of ethanoic acid and what mass of ethanol would be needed to produce 100 g of ethyl ethanoate assuming the reaction went to completion?

Care! In this question you know how much you want to get and are asked how much you will need to start with. In these cases you must read the equation from the other end ie 1 mole of the ethyl ethanoate is produced from 1 mole of acid and 1 mole of alcohol.

CH3CO2H(l) + C2H5OH(l)

→

CH3CO2C2H5(l) + H2O(l)

(12+3+12+32+1) (24+5+16+1) (12+3+12+32+24+5)

= 60 = 46 = 88

∴ 88 g of ethyl ethanoate are produced from 60 g of ethanoic acid and 46 g of ethanol

∴ 100 g of ethyl ethanoate are produced from g g g 60 × =

68.2 g of ethanoic acid

and g g g 46 × =

52.3 g of ethanoic acid

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Exercise 8 Calculations of products/reactants based on equations

In this exercise the equations you need are given in the question, unless they were included in Exercise 6a.

What mass of barium sulphate would be produced from 10 g of barium chloride in the following reaction?

BaCl2 + H2SO4

→

BaSO4 + 2HCl

What mass of potassium chloride would be produced from 20 g of potassium carbonate?

What masses of ethanol and ethanoic acid would need to be reacted together to give 1 g of ethyl ethanoate?

What mass of iron(III) oxide would need to be reduced to produce 100 tonnes of iron in a blast furnace?

What mass of silver nitrate as a solution in water would need to be added to 5 g of sodium chloride to ensure complete precipitation of the chloride?

AgNO3(aq) + NaCl(aq)

→

AgCl(s) + NaNO3(aq)

A solution of copper sulphate reacts with sodium hydroxide solution to produce a precipitate of copper hydroxide according to the following equation:

CuSO4(aq) + 2NaOH(aq) → Cu(OH)2(s) + Na2SO4(aq)

What mass of sodium hydroxide would be needed to convert 15.95 g of copper sulphate to copper hydroxide and what mass of copper hydroxide would be produced?

What volume of ammonia gas would be needed to produce 40 g of ammonium nitrate in the following reaction?

NH3(g) + HNO3(aq)

→

NH4NO3(aq)

In the reaction between calcium carbonate and nitric acid what mass of calcium nitrate and what volume of carbon dioxide would be produced from 33.3 g of calcium carbonate?

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What would be the total volume of gas produced by the action of heat on 33.1 g of lead(II) nitrate ?

Magnesium reacts with sulphuric acid to produce a solution of magnesium sulphate. If this is allowed to crystallise out the solid produced has the formula MgSO4.7H2O.

Write the equation for this reaction and calculate the mass of magnesium sulphate heptahydrate that could be produced from 4 g of magnesium.

Copper(II) oxide reacts with sulphuric acid to produce copper(II) sulphate. If this is allowed to crystallise the formula of the crystals is CuSO4.5H2O.

What mass of copper oxide would be needed to produce 100 g of crystals?

Sulphur dioxide can be removed from the waste gases of a power station by passing it through a slurry of calcium hydroxide. The equation for this reaction is:

SO2(g) + Ca(OH)2(aq) → CaSO3(aq) + H2O(l)

What mass of calcium hydroxide would be needed to deal with 1000 dm3 of sulphur dioxide?

In a fermentation reaction glucose is converted to alcohol and carbon dioxide according to the following equation:

C6H12O6 → 2C2H5OH + 2CO2

What mass of alcohol and what volume of carbon dioxide would be produced from 10 g of glucose?

In the following reactions calculate the mass of precipitate formed from 20 g of the metal salt in each case.

• ZnSO4(aq) + 2NaOH

→

Zn(OH)2(s) + Na2SO4(aq)

• Al2(SO4)3(aq) + 6NaOH

→

2 Al(OH)3(s) + 3Na2SO4(aq)

(iii) MgSO4(aq) + 2NaOH

→

Mg(OH)2(s) + Na2SO4(aq)

What volume of hydrogen would be produced by 1 g of calcium in its reaction with water?

What mass of magnesium would be needed to produce 100 cm3 of hydrogen?

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Chlorine reacts with sodium hydroxide as follows:

Cl2(g) + 6NaOH(aq) → 5NaCl(aq) + NaClO3(aq) + 3H2O(l)

What mass of sodium chloride and what mass of sodium chlorate(V) be produced from 240 cm3 of chlorine gas?

When nitrogen reacts with hydrogen in the Haber Process only 17% of the nitrogen is converted to ammonia. What volume of nitrogen and what volume of hydrogen would be needed to produce 1 tonne of ammonia? (1 tonne = 1 x 106 g)

Nitric acid is produced by the following series of reactions:

4NH3 + 5O2 → 4NO + 3H2O

4NO + O2 → 4NO2

4NO2 + O2 + 2H2O → 4HNO3

What mass of nitric acid would be produced from 17 tonnes of ammonia and what volume of oxygen would be needed in the reaction?

Hardness in water is caused by dissolved calcium compounds. When heated some of these break down and deposits calcium carbonate as follows:

Ca(HCO3)2

→ CaCO3

+ H2O + CO2

This builds up as ‘fur’ on the inside of boilers. It can be removed by reaction with hydrochloric acid.

What mass of calcium carbonate would be produced from 10000 dm3 of water containing 0.356 g of calcium hydrogen carbonate per dm3 of water and what volume of 10 mol dm-3 hydrochloric acid solution would be needed to remove the solid calcium carbonate from the inside of the boiler?

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Section 9

Reactions involving gases

Whenever gases are involved in a reaction you need to remember that they have both mass and volume and that 1 mole of any gas has the same volume, 24 000 cm3, at room temperature and 1 atmosphere pressure as 1 mole of any other gas. (See Section 4 for more details)

This means:

2 g of hydrogen, H2, has a volume of 24 000 cm3

32 g of oxygen, O2, has a volume of 24 000 cm3

81 g of hydrogen bromide, HBr, has a volume of 24 000 cm3

The effect of this is to make calculations involving gas volumes much easier than you might expect.

Consider the following reaction:

2NO(g) + O2(g)

→

2NO2(g)

This says:

2 moles of NO(g) react with 1 mole of O2(g) to give 2 moles of NO2(g)

∴ (2 x 24 000) cm3 of NO react with (1 x 24 000) cm3 of oxygen to give (2 x 24 000) cm3 of NO2

2 cm3 of NO react with 1 cm3 of oxygen to give 2 cm3 of NO2 ie for gases only the reacting volume ratios are the same as the reacting mole ratios in the equation.

Example 1

What volume of sulphur trioxide would be produced by the complete reaction of 100 cm3 of sulphur dioxide with oxygen? What volume of oxygen would be needed to just react with the sulphur dioxide?

2SO2(g) + O2(g) → 2SO3(g)

Ratios ie ½

100 cm3 50 cm3 100 cm3

Thus 100 cm3 of sulphur dioxide will need 50 cm3 of oxygen and produce 100 cm3 of sulphur dioxide.

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Example 2

What would be the composition of the final product in Example 1 if 100 cm3 of oxygen had been used rather than 50 cm3?

Since 100 cm3 of the sulphur dioxide needs only 50 cm3 of oxygen there must be 50 cm3 of oxygen unused. Thus the final volume is:

100 cm3 of sulphur dioxide plus 50 cm3 of excess oxygen = 150 cm3 Example 3

What volume of ammonia would be produced if 10 cm3 of nitrogen was reacted with 20 cm3 of hydrogen?

N2(g) + 3H2(g)

→

2NH3(g)

You need to think before you start this question. The reacting volumes given in the question are not the same as those in the reaction. You must have excess of one of the gases.

From the equation:

10 cm3 of nitrogen needs 30 cm3 of hydrogen. You only have 20 cm3 of hydrogen so the nitrogen is in excess.

In this case you will need to use the hydrogen volume in the calculation.

N2(g) + 3H2(g) → 2NH3(g)

Ratios

1/3 2/3

1/3 x 20 2/3 x 20

6.67 cm3 20 cm3 13.33 cm3

Thus 20 cm3 of hydrogen will react to give 13.33 cm3 of ammonia and there will be 3.33 cm3 of hydrogen left over.

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Exercise 9 Calculations based on equations involving only gases Section a

In Section a. assume that you have 10 cm3 of the first named reactant and then calculate the volumes of all the gases involved in the equation. In these examples the reactions are being carried out at above 100°C and you should assume the water is present as a gas and therefore has a volume.

CH4

+ 2O2 → CO2 + 2H2O

C2H4

+ 3O2 → 2CO2 + 2H2O

2C2H2

+ 5O2 → 4CO2 + 2H2O

2C8H18

+ 25O2 → 16CO2 + 18H2O

N2

+ 3H2 → 2NH3

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Section b

In Section b. you are asked to find the total volume of gas produced at room temperature and pressure. You should ignore the volume of water produced as this will have condensed as a liquid. Be careful in some cases, as there is an excess of one of the reactants.

What volume of oxygen would be needed to convert 1000 cm3 of nitrogen monoxide, NO, to nitrogen dioxide, NO2? (Assume all volumes are measured at the same temperature and pressure.)

In the production of sulphuric acid sulphur dioxide is converted to sulphur trioxide by reaction with the oxygen in the air. What volume of air (assume 20% of the air is oxygen) would be needed to produce 150 cm3 of sulphur trioxide? Assume complete conversion of sulphur dioxide to sulphur trioxide.

In equation for the oxidation of ammonia to nitrogen monoxide is:

4NH3 + 5O2 → 4NO + 6H2O

What volume of ammonia would be required to produce 2500 cm3 of nitrogen monoxide and what volume of air would be used in the conversion? Again assume that air is 20% oxygen by volume.

What volume of oxygen at room temperature and pressure would be needed to completely burn 1 mole of butane?

What volume of hydrogen at room temperature and pressure would be needed to convert 1 mole of ethene, C2H4, to ethane, C2H6?

What is the final volume of gas produced at room temperature when 10 cm3 of methane is burnt with 30 cm3 of oxygen?

What is the final volume of gas produced at room temperature if 5 cm3 of octane are burnt with 100 cm3 of oxygen?

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In a reaction between methane and oxygen 60 cm3 of methane was burnt with 60 cm3 of oxygen. What is the composition of the gas mixture produced?

What volume of ammonia would be produced if 10 cm3 of nitrogen was reacted with 60 cm3 of hydrogen?

What would be the final volume of gas produced in the reaction between 10 cm3 of hydrogen and 10 cm3 of oxygen?

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Section 10

Ions and ionic equations Ionic theory

Many of the chemicals which you will use at GCE Advanced are ionic, that is the chemical bonds which hold the atoms together are ionic bonds. When you melt these compounds the ions are free to move and this gives them some special properties. Often, but not always, these chemicals are soluble in water and when they dissolve the ions separate to produce a solution containing positive and negative ions.

A few covalent substances also form ions when they dissolve in water. Some of these are extremely important: hydrogen chloride and sulphuric acid are examples.

Structures of ionic compounds

In your course you will study bonding and structure, but at a of the most important ideas are set out below.

• Ions are atoms or groups of atoms, which have a positive or negative electric charge.
• Positive ions are called cations (pronounced cat-ions) and negative ions are called anions (pronounced an-ions).
• Positive ions attract negative ions all around them and are firmly held in a rigid lattice; this is what makes ionic compounds solids.
• When an ionic compound is solid it is crystalline, but when it melts or is dissolved in water the ions become free and can move around.
• Ions have completely different properties from the atoms in them; chlorine is an extremely poisonous gas, but chloride ions are found in sodium chloride, which is essential to human life.

Ionic equations and spectator ions

Many of the chemicals, which you study are ionic, and in these cases it is the ions which react, not the molecules. For instance, copper(II) sulphate is usually written as CuSO4 but it is more often the ion Cu2+ which reacts. When you write an ionic equation you include only the ions which actually take part in the reaction.

Let us look at molecular equation and see how it may be converted into an ionic equation. For example, look at the reaction between iron(II) sulphate solution and aqueous sodium hydroxide.

FeSO4(aq) + 2NaOH(aq) → Fe(OH)2(s) + Na2SO4(aq)

In water, the iron (II) sulphate and the sodium hydroxide are in the form of freely moving ions. When the two solutions are mixed together, we see a green precipitate of iron (II) hydroxide solid. Remaining in solution will be a mixture of sodium ions and sulphate ions.

Fe2+(aq) + 2OH–(aq) → Fe(OH)2(s)

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Also when silver nitrate solution reacts with sodium chloride solution the changes do not involve the nitrate ion from the silver nitrate or the sodium ion from the sodium chloride. These are referred to as ‘spectator ions’. The equation for this reaction can be written

Ag+(aq) + Cl–(aq) → AgCl(s)

This equation represents the reaction between any aqueous solution containing silver ions and any aqueous solution containing chloride ions. This is the equation for the test for a chloride ion in solution.

You can work out an ionic equation as follows using the example of the reaction of iron(II) sulphate solution with excess sodium hydroxide solution.

Write down the balanced molecular equation

FeSO4(aq) + 2NaOH(aq) → Fe(OH)2(s) + Na2SO4(aq)

Convert those chemicals that are ions in solution into their ions

Fe2+(aq) + SO4

2–(aq) + 2Na+(aq) + 2OH–(aq) → Fe(OH)2(aq) + 2Na+(aq) + SO4 2–(aq)

Cross out those ions that appear on both sides of the equation as they have not changed during the reaction. They started in solution and they finished in the solution. To give the ionic equation:

Fe2+(aq) + 2OH–(aq) → Fe(OH)2(s)

Check that the atoms and the charges balance.

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Exercise 10 Ionic equations

In questions 1–5 you are required to balance the equations, in questions 6–10 you are required to complete the equation and then balance it. For questions 1–17 you are required to write the full, balanced ionic equation. Questions 18–20 involve more complex ions again you are just asked to balance the equation.

Pb2+ (aq)

+

OH–(aq)

→

Pb(OH)2(s)

Al3+(aq)

+

OH–(aq)

→

Al(OH)3(s)

Al(OH)3(s)

+

OH–(aq)

→

AlO2

2–(aq)

+

H2O(l)

Cl2(g)

+

OH–(aq)

→

ClO3

3–(aq)

+

Cl–(aq)

+

H2O(l)

S2O3

2–(aq)

+

I2(s)

→

S4O6

2–(aq)

+

2I–(aq)

Cu2+(aq)

+

OH–(aq)

→

CO3

2–(s)

+

H+(aq)

→

Zn(s)

+

H+(aq)

→

Zn(s)

+

Pb2+(aq)

→

H+(aq)

+

OH–(aq)

→

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Write an ionic equation for the reaction of magnesium with sulphuric acid.

Write an ionic equation for the reaction of sodium carbonate solution with nitric acid.

Write an ionic equation for the reaction of copper oxide with hydrochloric acid.

Write an ionic equation for the reaction of barium chloride solution with sodium sulphate solution.

Write an ionic equation for the reaction of silver nitrate solution with potassium chloride solution.

Write an ionic equation for the reaction of zinc with silver nitrate solution.

Write ionic equations for the reactions of sodium hydroxide and potassium hydroxide with hydrochloric acid.

Write ionic equations for the reactions of sodium hydroxide and potassium hydroxide with nitric acid.

Write ionic equations for the reactions of sodium hydroxide and potassium hydroxide with sulphuric acid.

What do you notice about the answers to questions 17, 18 and 19?

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Section 11

Calculations involving chemicals in solution

These are often referred to as Volumetric Analysis. The name should not worry you, the basis of the calculations is the same as all the rest ie moles and equations.

Many reactions take place in solution involving solutions of known concentration.

Concentration in solution is generally measured as moles per 1000 cm3 of solution. For example the sodium chloride on the bench may be labelled as 1M NaCl. This means that each 1000 cm3 of the solution contains 1 Mole of NaCl (58.5 g).

It does not mean that 58.5 g of NaCl have been added to 1000 cm3 of water.

The solution will have been made up by measuring out 58.5 g of the solid, dissolving it in about 500 cm3 of water and then adding water to make the total volume of the mixture up to 1000 cm3. (1 dm3)

Concentration in mol dm-3 is called molarity.

dissolved material the for M cm per grams in ion concentrat molarity

r

=

) cm ( volume molarity volume given a in material of moles of number

× =

M ) cm ( volume molarity solution of volume given a in material of mass r 3 × × =

In reactions in solution it is often more convenient to use molarity rather than g dm–3.

There are two ways you can approach calculations involving solutions. The first method (A) detailed below is really a short cut way of using the more detailed method B. Most of the straight forward calculations you will meet at the start of your course and the ones in this booklet can be carried through using method A.

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Method A

Consider the following reaction between two solutions aA(aq) + bB(aq) → cC(aq) + dD(aq)

In this reaction a moles of substance A react with b moles of substance B

Let us suppose that Va cm3 of the solution of A react with Vb cm3 of the solution of B. If this is an acid/alkali reaction we could find these volumes out using an indicator.

∴ Number of moles of A in Va cm3 of solution = = a

M V a a

∴ Number of moles of B in Vb cm3 of solution = = b

M V b b

∴ If we divide equation (i) by equation (ii) we get b a M V

M V b b a a =

This relationship will hold good for any reaction between two solutions.

Examples

2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)

V M

V M

SO H SO H

NaOH NaOH = × ×

BaCl2(aq) + H2SO4(aq) → BaSO4(s) + 2HCl(aq)

V M

V M

SO H SO H

BaCl BaCl = ×

×

MnO4

-(aq) + 5Fe2+(aq) + 8H+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)

V M

V M

Fe Fe

MnO MnO = ×

×

+ +

− −

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Calculation examples

What is the molarity of a solution of NaOH which contains 4 g of NaOH in 250 cm3 of solution?

Mr NaOH = 40 g mol–1

4 g per 250 cm3 = 16 g per 1000 cm3

∴ molarity = =

0.040 mol dm-3

This can also be written molarity = =

0.040 mol dm-3

×

× ×

) (

What mass of KMnO4 would be needed to prepare 250 cm3 of 0.020 mol dm–3 KMnO4 solution? (Mr = 158)

1000 cm3 of 0.020 mol dm–3 KMnO4 will need 158 x 0.02 g

∴ 250 cm3 will need =

0.79 g

× × .

How many moles of H2SO4 will be contained in 25 cm3 of 0.10 mol dm–3 H2SO4?

number of moles =

0.0025 moles

× .

25 cm3 of 0.10mol dm–3 NaOH react with 50 cm3 of a solution of H2SO4. What is the molarity of the H2SO4?

2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)

V M

V M

SO H SO H

NaOH NaOH = × ×

∴

M

2SO H

= × × .

∴ =

0.025 mol dm-3

M

2SO H ×

× = .

NB If you are required to calculate the concentration in g dm–3 at this stage you need to multiply by the Mr of the material. In this case 98 g mol–1

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What volume of 0.02 mol dm–3 KMnO4 solution will be needed to react with 25 cm3 of 0.1 mol dm–3 Iron(II) ammonium sulphate ?

NB in Iron(II) ammonium sulphate only the iron(II) ions react with the manganate(VII) ions

MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)

V M

V M

Fe Fe

MnO MnO = ×

×

+ +

− −

∴

V MnO = ×

× −

.

.

∴

MnO

cm .0 1.0 V

= × × = −

25 cm3 of a solution of 0.05 mol dm–3 silver nitrate react with 10 cm3 of a solution of NaCl. What is the concentration of NaCl in g dm–3 in the solution?

NaCl(aq) + AgNO3(aq) → NaNO3(aq) + AgCl(s)

V M

V M

AgNO AgNO

NaCl NaCl = × ×

∴ M NaCl = × ×

.

∴ NaCl dm mol M − = × = . .

∴

concentration of NaCl dm g .7

5.

dm mol .0

− − = × =

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In the reaction between an acid HxA and 0.1 mol dm–3 NaOH solution. 25 cm3 of a solution of 0.1 mol dm–3 HxA react with 50 cm3 of the 0.1 mol dm–3 NaOH. What is the value of x?

This is not as difficult as it looks. You need to think what the equation for the reaction would be.

HxA(aq) + xNaOH(aq) → NaxA (aq) + xH2O(l)

∴ x V M

V M

NaOH NaOH

A H A H x x = ×

×

∴ x 1.0 1.0 = × ×

∴ x = 2

Thus the acid is H2A.

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Method B

In this method the actual amounts of materials in the volumes involved are calculated rather than the ratios.

Example

25 cm3 of 0.10 mol dm–3 NaOH react with 50 cm3 of a solution of H2SO4.

What is the molarity of the H2SO4?

2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)

∴ 2 mol of NaOH react with 1 mol of H2SO4

In this case you know the concentration of the sodium hydroxide so

∴ 1 mol of NaOH reacts with 0.5 mol of H2SO4 always put the reactant you know as ‘1 mol’

In this reaction you have used 25 cm3 of 0.10 mol dm–3 NaOH

NaOH of mol .0 25× = mol 5.2 − × =

This will react with 0.5 x 2.5 x 10–3 moles of H2SO4

= 1.25 x 10–3 moles of H2SO4

∴ 1.25 x 10–3 moles of H2SO4 will be found in 50 cm3 of the solution

∴ In 1000 cm3 of the acid the same solution there will be

SO H of moles ) .1( − × × =

= 0.0250 moles

∴ The concentration of the sulphuric acid is 0.025 mol dm–3.

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Exercise 11a Calculations based on concentrations in solution

Calculate the number of moles of the underlined species in the volume of solution stated.

25 cm3 of 1.0 mol dm-3 HCl

50 cm3 of 0.5 mol dm-3 HCl

250 cm3 of 0.25 mol dm-3 HCl

500 cm3 of 0.01 mol dm-3 HCl

25 cm3 of 1.0 mol dm-3 NaOH

50 cm3 of 0.5 mol dm-3 KOH

50 cm3 of 0.25 mol dm-3 HNO3

100 cm3 of 0.1 mol dm-3 H2SO4

25 cm3 of 0.05 mol dm-3 KMnO4

25 cm3 of 0.2 mol dm-3 FeSO4

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Calculate the mass of material in the given volume of solution

25 cm3 of 1 mol dm-3 HCl

50 cm3 of 0.5 mol dm-3 NaCl

100 cm3 of 0.25 mol dm-3 NH4NO3

100 cm3 of 0.1 mol dm-3 AgNO3

25 cm3 of 1 mol dm-3 BaCl2

50 cm3 of 0.2 mol dm-3 H2SO4

20 cm3 of 0.1 mol dm-3 NaOH

50 cm3 of 0.1 mol dm-3 K2CrO4

25 cm3 of 0.02 mol dm-3 KMnO4

25 cm3 of 0.1 mol dm-3 Pb(NO3)2

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What is the concentration in moles dm-3 of the following?

3.65 g of HCl in 1000 cm3 of solution

3.65 g of HCl in 100 cm3 of solution

6.62 g of Pb(NO3)2 in 250 cm3 of solution

1.00 g of NaOH in 250 cm3 of solution

1.96 g of H2SO4 in 250 cm3 of solution

1.58 g of KMnO4 in 250 cm3 of solution

25.0 g of Na2S2O3.5H2O in 250 cm3 of solution

25.0 g of CuSO4.5H2O in 250 cm3 of solution

4.80 g of (COOH)2.2H2O in 250 cm3 of solution

10.0 g of FeSO4.(NH4)2SO4.6H2O in 250 cm3 of solution

240 cm3 of NH3(g) dissolved in 1000 cm3 of solution

480 cm3 of HCl(g) dissolved in 100 cm3 of solution

120 cm3 of SO2(g) dissolved in 250 cm3 of solution

24 cm3 of HCl(g) dissolved in 200 cm3 of solution

100 cm3 of NH3(g) dissolved in 10 cm3 of solution

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Exercise 11b Simple volumetric calculations

In this series of calculations you should start by writing the equation for the reaction taking place then generate the molarity/volume ratio. In some cases you will need to calculate the molarity of the solutions before you start the main part of the question.

For questions 1–10 calculate the molarity of the first named solution from the data below.

25 cm3 of sodium hydroxide reacts with 21.0 cm3 of 0.2 mol dm–3 HCl

25 cm3 of sodium hydroxide reacts with 17.0 cm3 of 0.1 mol dm–3 H2SO4

20 cm3 of hydrochloric acid reacts with 23.6 cm3 of 0.1 mol dm–3 NaOH

20 cm3 of hydrochloric acid reacts with 20.0 cm3 of a solution of NaOH containing 40 g dm–3 of NaOH

25 cm3 of nitric acid reacts with 15 cm3 of a solution of 0.2 mol dm–3 NH4OH

25 cm3 of a solution of barium chloride reacts with 20 cm3 of a solution of 0.05 mol dm–3 sulphuric acid

25 cm3 of a solution of NaCl reacts with 10 cm3 of a 0.02 mol dm–3 silver nitrate

10 cm3 of a solution of AlCl3 reacts with 30 cm3 of 0.01 mol dm-3 silver nitrate

25 cm3 of HxA reacts with 25 cm3 of 0.2 mol dm–3 NaOH to give Na2A

25 cm3 of H3PO4 reacts with 100 cm3 of 0.1 mol dm–3 NaOH to give NaH2 PO4

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25 cm3 of a solution of 0.1 mol dm–3 NaOH reacts with 50 cm3 of a solution of hydrochloric acid. What is the molarity of the acid?

25 cm3 of a solution of 0.2 mol dm–3 KOH reacts with 30 cm3 of a solution of nitric acid. What is the concentration of the acid in moles dm–3?

In a titration 25 cm3 of ammonia solution react with 33.30 cm3 of 0.1 mol dm–3 HCl. What is the concentration of the ammonia solution in g dm–3?

In the reaction between iron(II) ammonium sulphate and potassium manganate(VII) solution. 25 cm3 of the Fe2+ solution reacted with 24.8 cm3 of 0.020 mol dm–3 KMnO4 solution. What is the molarity of the iron(II) ammonium sulphate solution?

10 cm3 of a solution of NaCl react with 15 cm3 of 0.02 mol dm–3 silver nitrate solution. What is the concentration of the NaCl solution in g dm–3?

25 cm3 of a solution of an acid HxA containing 0.1 mol dm–3 of the acid in each 1000 cm3 of solution reacts with 75 cm3 of a solution of 0.1 mol dm–3 NaOH. What is the value of x?

25 cm3 of a solution of sodium carbonate react with 10 cm3 of a 0.1 mol dm–3 HCl. What is the concentration of the sodium carbonate?

What volume of 0.1 mol dm–3 HCl will be needed to react with 25 cm3 of 0.2 mol dm–3 NaOH?

What volume of 0.05 mol dm–3 H2SO4 will be needed to react with 25 cm3 of 0.2 mol dm–3 NaOH?

What volume of 0.02 mol dm–3 KMnO4 will be needed to react with 25 cm3 of 0.1 mol dm–3 FeSO4 solution?

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The last five questions will require you to use the skills you have learnt in this section, together with those from other sections.

What weight of silver chloride will be produced if 25 cm3 of 0.1 mol dm–3 silver nitrate is added to excess sodium chloride solution?

What weight of calcium carbonate will dissolve in 100 cm3 of 0.2 mol dm–3 HCl?

What volume of carbon dioxide will be produced if 100 cm3 of 0.2 mol dm–3 HNO3 is added to excess sodium carbonate solution?

What weight of magnesium will dissolve in 10 cm3 of 1 mol dm–3 HCl and what volume of hydrogen will be produced?

What volume of ammonia gas will be produced in the following reaction if 50 cm3 of 0.5 mol dm–3 sodium hydroxide is boiled with 50 cm3 of 0.4 mol dm–3 ammonium chloride solution? (Care: one of these is in excess.)

NaOH(aq) + NH4Cl(aq) → NaCl(aq) + H2O(l) + NH3(g)

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Section 12

Data — The periodic table

H

Li

Na

Be

Mg

K Ca

Rb

Cs

Fr

Sr

Ba

Ra

Sc Ti

Y Zr

La

Ac

Hf

Unq

V

Nb

Ta

Unp Unh

W

Mo

Cr Mn Fe Co

Tc

Re Os

Rh Ru

Ir

Ni Cu Zn

Pd Ag Cd

Pt Au Hg

B C N O F Ne

Al Si P S Cl Ar

Ga Ge As Se Br Kr

In Sn Sb Te I Xe

Tl Pb Bi Po At Rn

He

Ce Pr Nd Pm Sm Eu Gd

Th Pa U Np Pu Am Cm

Tb Dy Ho Er Tm Yb Lu

Bk Cf Es Fm Md No Lr

Hydrogen

Lithium

Berylium

Sodium Magnesium

Potassium Scandium Calcium Titanium Vanadium Chromium

Rubidium Strontium Yttrium Zirconium Molybdenum Niobium Technetium

Iron Manganese

Caesium

Francium

Barium

Radium

Lanthanum

Actinium

Hafnium

Unnilquadium Unnilhexium Unnilpentium

Tantalum Tungsten Rhenium

Ruthenium

Osmium

Rhodium

Iridium

Cobalt Nickel Copper

Silver Palladium

Platinum Gold Mercury Thallium

Cadmium

Zinc

Boron Carbon

Aluminium

Gallium

Indium

Lead

Tin Antimony

Bismuth

Silicon Phosphorus

Arsenic

Nitrogen

Helium

Oxygen Fluorine

Chlorine Sulphur

Selenium Germanium Bromine

Argon

Neon

Krypton

Tellurium Iodine Xenon

Polonium Astatine Radon

Cerium Praseo- dymium Neodymium Promethium

Thorium Protactinium

Samarium

Uranium Neptunium

Europium

Curium Berkelium Californium Einsteinium

Gadolinium Terbium Holmium Erbium Thulium Ytterbium

Fermium Mendelevium Nobelium Lawrencium

Lutetium

35.5

(210) (210) (222)

(99)

(223) (226) (227) (261) (262) (263)

63.5 65.4

(147)

(231) (237) (242) (243) (247) (245) (251) (254) (253) (256) (254) (257)

Atomic Number

Symbol

Name

Molar mass in g mol -1 key

Group Period THE PERIODIC TABLE

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Exercise 1

159.5

161.4

53.5

117.0

122.5

166.0

74.5

69.0

58.5 162.5 249.5

99.0

134.5

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Exercise 2

NaCl 21 BaSO4 41 PbCO3 61 PCl3

NaOH 22 AlCl3 42 PbO 62 PCl5

Na2CO3 23 Al2O3 43 PbO2 63 P2O3

Na2SO4 24 Al(OH)3 44 PbCl2 64 P2O5

NO3PO4 25 Al2(SO4)3 45 PbCl4 65 H3PO4

KCl 26 CuSO4 46 PbS 66 H2SO4

KBr 27 CuO 47 SnCl2 67 HNO3

KI 28 CuCl2 48 SnCl4 68 HCl

KHCO3 29 Cu(NO3)2 49 FeSO4 69 CCl4

10 KNO2 30 Cu2O 50 FeCl2 70 SiCl4

11 MgCl2 31 CuCl 51 Fe2(SO4)3 71 SiO2

12 Mg(NO3)2 32 Zn(NO3)2 52 FeCl3 72 SO2

13 Mg(OH)2 33 ZnCO3 53 Fe(OH)3 73 SO3

14 MgO 34 ZnO 54 Fe(OH)2 74 H2S

15 MgCO3 35 AgCl 55 NH4Cl 75 Cl2O

16 CaO 36 AgBr 56 (NH4)2CO3 76 NO2

17 CaCl2 37 AgI 57 NH4OH 77 NO

18 CaSO4 38 AgNO3 58 NH4NO3 78 CO2

19 CaCO3 39 Ag2O 59 (NH4)2SO4 79 CO

20 BaCl2 40 Pb(NO3)2 60 (NH4)3PO4 80 HOH/H2O

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Exercise 3

Water 31 Iron(III) chloride

Carbon dioxide 32 Iron(III) sulphate

Ammonia 33 Lead(II) oxide

Oxygen 34 Lead(IV) oxide

Hydrogen 35 Lead(II) nitrate

Sulphur dioxide (or IV oxide) 36 Lead(II) chloride

Sulphur trioxide (or VI oxide) 37 Lead (II) sulphate

Hydrogen chloride 38 Copper(II) nitrate

Hydrogen iodide 39 Copper(I) chloride

10 Hydrogen fluoride 40 Copper(II) chloride

11 Methane 41 Copper(II) sulphate

12 Hydrogen sulphide 42 Zinc chloride

13 Hydrogen bromide 43 Silver nitrate

14 Sulphuric acid 44 Ammonium chloride

15 Nitric acid 45 Ammonium sulphate

16 Sodium chloride 46 Ammonium vanadate(V)

17 Sodium nitrate 47 Potassium chlorate(V)

18 Sodium carbonate 48 Potassium iodate

19 Sodium hydroxide 49 Sodium chlorate(I)

20 Sodium sulphate 50 Sodium nitrite

21 Calcium chloride 51 Ethane

22 Calcium nitrate 52 Butane

23 Calcium hydroxide 53 Octane

24 Calcium sulphate 54 Ammonium carbonate

25 Barium chloride 55 Potassium manganate(VII)

26 Aluminium chloride 56 Potassium chromate(VI)

27 Aluminium nitrate 57 Potassium hydrogencarbonate

28 Aluminium sulphate 58 Potassium iodide

29 Iron(II) sulphate 59 Cobalt(II) nitrate

30 Iron(II)chloride 60 Potassium astatide

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Exercise 4a

0.50 26 0.10

2.0 27 0.10

0.10 28 0.0085

5.0 29 0.26

30 0.104

0.010 31 0.20

1.0 32 0.082

0.22 33 0.050

0.0010 34 1.34

10 0.050 35 0.025

11 0.33 36 0.204

12 0.25 37 0.071

13 0.021 38 0.010

14 0.020 39 0.050

15 0.125 40 0.254

16 0.020 41 0.0125

17 0.167 42 0.152

18 1.0 43 0.10

19 0.046 44 0.053

20 0.020 45 0.0043

21 0.0010 46 0.036

22 0.25 47 0.266

23 0.02 48 0.024

24 0.0025 49 0.025

25 0.20 50 1.574

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Exercise 4b

36 g 26 14.95 g

132 g 27 76.2 g

47.6 g 28 10.03 g

23 g 29 17.82 g

33.6 g 30 145.2 g

40.96 g 31 2.925 g

240 g 32 12.25 g

81 g 33 21.4 g

1.152 g 34 745 g

10 9.45 g 35 0.069 g

11 26.3 g 36 49.9 g

12 59.5 g 37 27.8 g

13 11.66 g 38 4.82 g

14 80.0 g 39 9.92 g

15 127.8 g 40 302.4 g

16 7.9 g 41 756.5 g

17 34.92 g 42 39.53 g

18 90 g 43 10.2 g

19 249 g 44 11.6 g

20 23.4 g 45 9.76 g

21 12.2 g 46 4.34 g

22 672.4 g 47 9.59 g

23 0.296 g 48 41.0 g

24 13.6 g 49 304 g

25 43.68 g 50 1397 g

Exercise 4c

24000 cm3 11 134.4 cm3

2400 cm3 12 216 cm3

12000 cm3 13 960 cm3

48000 cm3 14 2952 cm3

2880 cm3 15 55.2 cm3

81600 cm3 16 192000 cm3

2640 cm3 17 0.24 cm3

96 cm3 18 144000 cm3

240000 cm3 19 182.4 cm3

10 10800 cm3 20 72000 cm3

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Exercise 4d

0.0083 mol 11 0.0292 mol

0.0208 mol 12 0.2333 mol

0.0416 mol 13 0.0917 mol

0.0533 mol 14 0.0088 mol

0.0098 mol 15 0.0333 mol

0.0094 mol 16 0.0033 mol

0.0106 mol 17 0.000080 mol

0.0033 mol 18 0.8333 mol

0.0833 mol 19 0.0175 mol

10 0.10 mol 20 0.0375 mol

Exercise 4e

0.367 g 11 0.875 g

0.354 g 12 10.27 g

1.166 g 13 2.38 g

5.333 g 14 0.263 g

0.78 g 15 1.217 g

0.763 g 16 0.270 g

0.757 g 17 0.011 g

0.233 g 18 38.33 g

0.167 g 19 0.683 g

10 3.20 g 20 1.05 g

Exercise 4f

1091 cm3 11 56000 cm3

7059 cm3 12 30545 cm3

8571 cm3 13 20308 cm3

7500 cm3 14 16000 cm3

702 cm3 15 5260 cm3

670 cm3 16 2370 cm3

3380 cm3 17 375 cm3

30000 cm3 18 12000 cm3

2400000 cm3 19 26526 cm3

10 180000 cm3 20 77143 cm3

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Exercise 4g

11 34

12 17

13 38

14 28

2.0 15 44

16 32

17 211

18 36.5

19 81

10 28 20 128

Exercise 5 Section (a)

CaCO3 N2H4S2O8

Na2SO4 P4O10

Na2S2O3 C2H4O2 - CH3COOH

PbO C4H10

Pb3O4 Fe2O3

H3PO3 H2S2O8

H2SO3 10 C6H6

CH4

Section (c)

C3H8 PbO

10 HO (giving H2O2) KO2

11 H4N2O3 (NH4NO3) C2H2

12 FeSO11H14 (FeSO4 - 7H2O) AlCl3

Section (b)

CH4

C2H4 yes

C3H6 CuSO4 5H2O

P2I4 PCl5, PCl3, Cl2

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Exercise 6a

2H2 + O2 → 2H2O

BaCl2 + 2NaOH → Ba(OH)2 + 2NaCl

H2SO4 + 2KOH → K2SO4 + 2H2O

K2CO3 + 2HCl → 2KCl + H2O + CO2

CaCO3 + 2HNO3 → Ca(NO3)2 + H2O + CO2

Ca + 2H2O → Ca(OH)2 + H2

Pb(NO3)2 + 2NaI → PbI2 + 2NaNO3

Al2(SO4)3 + 6NaOH → 2Al(OH)3 + 3Na2SO4

Al(OH)3 + NaOH → NaAlO2 + 2H2O

2Pb(NO3)2 → 2PbO + 4NO2 O2

2FeSO4 → Fe2O3 + SO2 SO3

NH4NO3 → N2O + 2H2O

2NaNO3 → 2NaNO2 + O2

CH4 + 2O2 → CO2 + 2H2O

2C4H10 + 13O2 → 8CO2 + 10H2O

PCl3 + 3H2O → H3PO3 + 3HCl

8HNO3 + 3Cu → 3Cu(NO3)2 + 3NO + 4H2O

4HNO3 + Cu → Cu(NO3)2 + 2NO2 + 2H2O

H3PO4 + NaOH → NaH2PO4 + H2O

H3PO4 + 3NaOH → Na3PO4 + 3H2O

H3PO4 + 2NaOH → Na2HPO4 + 2H2O

6NaOH + 3Cl2 → NaClO3 + 5NaCl + 3H2O

N2 + 3H2 → 2NH3

2NaBr + H2SO4 → Na2SO4 + HBr

2HBr + H2SO4 → 2H2O + SO2 + Br2

3C2H5OH + PCl3 → 3C2H5Cl + H3PO3

Fe3O4 + 4H2 → 3Fe + 4H2O

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

Fe2O3 + 3CO → 2Fe + 3CO2

C2H5OH + CH3CO2H → CH3CO2C2H5 + H2O

2KMnO4 + 16HCl → 2KCl + 2MnCl2 + 8H2O + 5 Cl2

Exercise 6b

Hydrogen is not H but H2, which gives

2Na(s) + 2H2O(aq) → 2NaOH(aq) + H2(g)

Since the valency of lead is 2 not 1 lead nitrate is not PbNO3 but Pb(NO3)2 and also lead chloride is PbCl2

Pb(NO3)2(aq) + 2NaCl(aq) → PbCl2(s) + 2NaNO3(aq)

Calcium hydroxide is Ca(OH)2

Ca(OH)2(aq) + 2HCl(aq) → CaCl2(aq) + 2H2O(l)

This does not balance.

C2H4(g) + 3O2(g) → 2CO2(g) + H2O(l)

A magnesium compound cannot give a calcium compound!

Ozone O3 is not produced by heating a nitrate O2 is.

2Cu(NO3)2(s) → 2CuO(s) + 4NO2(g) + O2(g)

This reaction does not take place and so no equation can be written.

Aluminium has a valency of 3 not 2 as in this equation.

AlCl3(s) + 3KOH(aq) → Al(OH)3(s) + 3KCl(aq)

Sodium has a valency of 1 not 2 as in this equation

Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)

10 Silver chloride is not soluble in water. Thus the AgCl needs a (s) symbol

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

Exercise 6c

Zn(s) + CuSO4(aq) → Cu(s) + ZnSO4(aq)

Ca(OH)2(s) + 2NH4Cl(s) → CaCl2(s) + H2O(g) + NH3(g)

2Pb(NO3)2(s) → 2PbO(s) + 4NO2(g) + O2(g)

SiCl4(l) + 2H2O(l) → SiO2(s) + HCl(g)

Ca(HCO3)2(aq) → CaCO3(s) + H2O(l) + CO2(g)

2C8H18(g) + 25O2(g) → 16CO2(g) + 8H2O(l)

6NaOH(aq) + 3Cl2(g) → NaClO3(aq) + 5NaCl(aq) + 3H2O(l)

6NaOH(aq) + 3Br2(g) → NaBrO3(aq) + 5NaBr(aq) + 3H2O(l)

6NaOH(aq) + 3I2(g) → NaIO3(aq) + 5NaI(aq) + 3H2O(l)

2M(s) + 2H2O(l) → 2MOH(aq) + H2(g)

Where M = Li, Na, K, Rb or Cs

SnCl2(aq) + 2HgCl2(aq) → 2HgCl (s) + SnCl4(aq)

9H2SO4 + 8KI → 4I2 + H2S + 8KHSO4 + 4H2O

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

Exercise 7

AlCl3 + 3AgNO3 → Al(NO3)3 + 3AgCl

K2CrO4 + Pb(NO3)2 → PbCrO4 + 2KNO3

2AgNO3 → 2Ag + 2NO2 + O2 i) 1 mole ii) 2 moles iii)

H3PO4 + NaOH → NaH2PO4 + H2O

H3PO4 + 2NaOH → Na2HPO4 + 2H2O iv) 75 cm3 x = 3 x = 5

CuSO4.5H2O → CuSO4 + 5H2O

4NH3 + 5O2 → 4NO + 6H2O

C3H8 + 5O2 → 3CO2 + 4H2O

It is

FeSO4.7H2O → Fe2O3 + SO2 + SO3 + 14H2O

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

Exercise 8

11.2 g

21.6 g

0.682 g of ethanoic acid and 0.523 g of ethanol

143 tonnes

14.5 g

8.0 g of sodium hydroxide, 9.75 g of copper hydroxide

12000 cm3

• 7 g of calcium nitrate, 8.0 dm3 of carbon dioxide

6 dm3 total (4.8 dm3 of nitrogen dioxide and 1.2 dm3 of oxygen)

Mg + H2SO4 + 7H2O → Mg SO4.7H2O + H2 41.0 g

31.9 g

324.3 g

5.11 g of ethanol, 2.67 dm3 of carbon dioxide

• 12.30 g of zinc hydroxide
• 9.12 g of aluminium hydroxide

(iii) 9.67 g of magnesium hydroxide

0.600 dm3

0.100 g

2.94 g of sodium chloride, 1.065 g of sodium chlorate(v)

4.15 x 106 dm3 of nitrogen, 12.5 x 106 dm3 of hydrogen

63 tonnes of nitric acid, 4.8 x 107 dm3 of oxygen

2198 g of calcium carbonate, 4.395 dm3 of 10M HCl

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

Exercise 9 Section (a)

20cm3 O2 10cm3 CO2 20cm3 H2O (g)

30cm3 O2 20cm3 CO2 20cm3 H2O (g)

25cm3 O2 20cm3 CO2 10cm3 H2O (g)

125cm3 O2 80cm3 CO2 90cm3 H2O (g)

30cm3 H2 20cm3 NH3

Section (b)

500cm3 O2 (2NO + O2 →2NO2)

375cm3 air (2SO2 + O2 →2SO3)

2500cm3 NH3 needed 5/4 x 2500 = 3125cm3 O2 → 15625cm3 air

6.5 x 24000cm3 = 156m3

24000cm3

Final volume = 20cm3 (10cm3 CO2 + 10 cm3 unused O2)

Final volume = 77.5cm3 (40cm3 CO2 + 37.5cm3 used O2)

This time the CH4 is in excess. We must assume that CO2 is produced (not CO or C)!

Final volume = 60 cm3 (30 cm3 CO2 + 30 cm3 CH4)

N2 + 3H2 → 2NH3 10 cm3 30 cm3 20 cm3 + 30 cm3excess 20 cm3 NH3 produced + 30 cm3 excess H2

10 2 H2 + O2 → 2H2O 10 cm3 5 cm3 + 5 cm3 excess

Final volume = 5 cm3 (all excess O2)

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

Exercise 10

Pb2+ (aq) + 2OH– (aq) → Pb(OH)2(s)

Al3+(aq) + 3OH–(aq) → Al(OH)3(s)

Al(OH)3(s) + OH–(aq) → AlO2

–(aq) + 2H2O (l)

Cl2(g) + 6OH–(aq) → ClO3

–(aq) + 5Cl–(aq) + 3H2O(l)

2S2O3 2–(aq) + I2(s) → S4O6

2–(aq) + 2I–(aq)

Cu2+ (aq) + 2OH–(aq) → Cu(OH)2(s)

CO3

2–(s) + 2H+(aq) → H2O(l) + CO2(g)

Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

Zn(s) + Pb2+ (aq) → Pb(s) + Zn2+(aq)

H+(aq) + OH– (aq) → H2O(l)

Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g)

CO3

2–(s) + 2H+(aq) → H2O(l) + CO2(g)

CuO(s) + 2H+(aq) → Cu2+(aq) + H2O(l)

Ba2+(aq) + SO4

2–(aq) → BaSO4(s)

Ag+(aq) + Cl–(aq) → AgCl(s)

Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s)

17–20

20H+(aq) + OH–(aq) → H2O(l)

In every case the reaction is the same

634rp271004LT\PD\A LEVEL TEACHERS HANDBOOK\UA008883 GCE CHEM MOLES WKBK ISS3.DOC.1-150/5

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UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue 3 – October 2004

Exercise 11a

0.025 moles 19 0.079 g

0.025 moles 20 0.828 g

0.0625 21 0.1 mol dm–3

0.005 moles 22 1.0 mol dm–3

0.025 moles 23 0.03 mol dm–3

0.025 moles 24 0.1 mol dm–3

0.0125moles 25 0.03 mol dm–3

0.01 moles 26 0.04 mol dm–3

0.00125 moles 27 0.40 mol dm–3

10 0.005 moles 28 0.40 mol dm–3

11 0.9125 g 29 0.152 mol dm–3

12 1.463 g 30 0.0102 mol dm–3

13 2 g 31 0.01 mol dm–3

14 1.70 g 32 0.2 mol dm–3

15 5.2 g 33 0.02 mol dm–3

16 0.98 g 34 0.005 mol dm–3

17 0.08 g 35 0.417 mol dm–3

18 0.97 g

Exercise 11b

0.168 mol dm–3 16 3.0

0.136 mol dm–3 17 0.02 mol dm–3

0.118 mol dm–3 18 50 cm3

1.0 mol dm–3 19 50 cm3

0.12 mol dm–3 20 25 cm3

0.040 mol dm–3 21 0.359 g

0.0080 mol dm–3 22 1.0 g

0.010 mol dm–3 23 240 cm3

0.10 mol dm–3 24 0.12 g Mg 120 cm3 H2 10 0.40 mol dm–3 25 480 cm3

11 0.050 mol dm–3

12 0.167 mol dm–3

13 2.26 g dm–3

14 0.099 mol dm–3

15 1.755 g dm–3

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