BASIC IDEAS INCHEMISTRY

by Michael Clark

Success in studying Chemistry depends upon the familiarity of students with

a few basic ideas, conventions, and methods upon which later studies are built.

This small book presents these basic ideas, conventions and methods. When

a student has achieved mastery of them, further studies can be pursued with

greater confidence. Without mastery of them, students are likely to find higher

levels of study in Chemistry difficult.

Three basic areas are developed:

1. use of chemical symbols and formulae, (with a simple

introduction to bonding)

2. writing chemical equations,

3. calculations involving moles (solids, gases, and solutions).

There is no reference to laboratory activities in this book. This is not to

suggest, however, that laboratory experience with chemical substances and

their reactions is not a vital part of learning Chemistry.

While theoretical rigour is a desirable objective in Chemistry courses, it is not

always appropriate for beginning courses. In some of the pages that follow,

theoretical rigour has been sacrificed to the need for simplicity. A more

rigorous treatment of some of these topics can be developed after a student has

mastered the essential basics.

BASIC IDEAS IN CHEMISTRY is not intended to stand alone as a text in

its own right, but rather to provide a supplement to other text-books, to which

students must return for more detailed and theoretical development of the

topics covered in this book.

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CONTENTS

Page

1. Chemical symbols and formulae

a) Elements and symbols

b) Compounds and formulae

c) Writing names and formulae of ionic compounds

d) List of names and formulae of common ions

e) Practice exercises for writing formulae

f) Introduction to Bonding Theory and explanation

of rules for writing formulae.

3

3

5

7

8

11

2. Chemical equations

a) What balancing an equation means

b) How to balance equations by inspection

c) Summary and practice exercises in balancing

simple chemical equations

d) Molecular and ionic equations, and spectator

ions

17

18

22

23

3.

4.

Redox reactions and redox equations

a) Definitions

b) Identifying redox reactions

c) Oxidation states

d) Writing redox equations : oxidation state method

e) Writing redox equations : half-equation method

Moles and mole calculations

a) Introduction

b) Definition of a mole and of molar mass

c) Percentage composition

d) Use of molar mass to predict reacting masses

e) Calculations involving gases

f) Calculations involving solutions

g) Calculations combining masses, gas volumes,

and concentrations of solutions

h) Answers to exercises

25

25

26

28

30

35

35

37

38

42

44

52

57

3

CHEMICAL SYMBOLS AND FORMULAE

ELEMENTS AND SYMBOLS

All chemical substances are made up of atoms. Substances made of only one kind

of atom are called elements. There are about ninety different chemical elements that

occur naturally on Earth. Of these, some are very rare. About twenty-five to thirty

elements are regarded as "common" or "well-known".

Each chemical element is known by its SYMBOL, comprised of one or two letters.

The symbols of common elements should be memorised thoroughly, and less

common ones might also be learnt as they come to attention.

The following list of symbols and the elements they represent should be

memorised. They are listed by atomic number, (which is the number of protons

present in the nucleus of an atom of the element: see page 11 ):

1. H = hydrogen

6. C = carbon

7. N = nitrogen

8. O = oxygen

11. Na = sodium

12. Mg = magnesium

13. Al = aluminium

14. Si = silicon

15. P = phosphorus

16. S = sulfur

17. Cl = chlorine

19. K = potassium

20. Ca = calcium

22. Ti = titanium

24. Cr = chromium

25. Mn = manganese

26. Fe = iron

29. Cu = copper

30. Zn = zinc

35. Br = bromine

47. Ag = silver

50. Sn = tin

53. I = iodine

56. Ba = barium

79. Au = gold

80. Hg = mercury

82. Pb = lead

A chemical symbol represents the name of the element. It is used also to represent

one atom of the element.

COMPOUNDS AND FORMULAE

Atoms of elements can join together. Sometimes two identical atoms join together,

but more often different kinds of atoms form compounds. A compound is made of

at least two different elements.

MOLECULES

If two or more atoms join together, they form a molecule.

A formula shows what kinds of atoms, and how many of each, join together when

a molecule is formed. A small (subscript) number after a symbol shows the number

of atoms of that element that are present in one molecule of the compound. If there

is no number, it means that there is one atom of that element.

4

For example:

2CO means that one atom of carbon is joined to two atoms of oxygen.

2H O means that two atoms of hydrogen are joined to one atom of oxygen.

2 4H SO means that two atoms of hydrogen, one atom of sulfur, and four atoms of

oxygen are joined in one molecule.

IONIC COMPOUNDS

In some compounds, atoms or groups of atoms have an electrical charge, and are

then called ions. A group of atoms with an electrical charge is called a compound

ion. Ions with opposite electric charges attract each other, but do not usually

become permanently joined together.

Name of

compound

Names of ions

in compound

Formulae

of ions

Formula of

compound

Ratio of

positive to

negative ions.

Sodium iodide sodium

iodide

Na+

I-

NaI 1:1

Silver oxide silver

oxide

Ag+

O2-2Ag O 2:1

Zinc sulfate zinc

sulfate

Zn2+

4SO 2-4ZnSO 1:1

Aluminium

phosphate

aluminium

phosphate

Al3+

4PO 3-4AlPO 1:1

Lead nitrate lead

nitrate

Pb2+

3NO -3 2Pb(NO ) 1:2

Iron(III) chloride iron(III)

chloride

Fe3+

Cl-3FeCl 1:3

The formula of an ionic compound shows how many of each kind of ion are

attracted to each other in the compound.

5

WRITING NAMES AND FORMULAE OF IONIC COMPOUNDSThe first requirement for students needing to master the writing of formulae is

to memorise - and memorise thoroughly - the formulae of common ions.

While writing of formulae is being learnt and practised, a list of the common ions

and their formulae should be kept close at hand for ready reference. On the

following page is a list of ions that students may encounter. Also provided are three

sheets of examples of ionic compounds for practising writing formulae.

Basic rules for writing names and formulae are provided here. Explanations of the

rules are presented starting on page 11.

1. Clarity and accuracy are of greatest importance. Upper case (capital) letters must

be clearly written as capital letters, lower case (small) letters must be written

clearly as small letters, subscript numbers (small numbers after a symbol) must

be written accurately and clearly.

2. The name of an ionic compound has two parts: the first part is the positive ion,

usually a metal, but may also be ammonium, a positive compound ion containing

nitrogen and hydrogen. The second part of the name is the negative ion, either

the name of a non-metal with the end of its name changed to -ide, or the name

of a negative compound ion.

EXAMPLES:

Name of compound Positive ion Negative ion

Calcium iodide calcium, Ca iodide, I2+ -

4Copper phosphate copper, Cu phosphate, PO2+ 3-

4Aluminium sulfate aluminium, Al sulfate, SO3+ 2-

4Ammonium chloride ammonium, NH chloride, Cl+ -

3. An acid contains hydrogen joined with a negative ion, and has "acid" as the

second word of its name. The first word is usually the name of the negative ion,

with the end of its name changed: -ate changes to -ic, -ite changes to -ous.

EXAMPLES OF ACIDS:

Name of acid Positive ion Negative ion

3Nitric acid hydrogen, H nitrate, NO+ -

3Carbonic acid hydrogen, H carbonate, CO+ 2-

3Sulfurous acid hydrogen, H sulfite, SO+ 2-

But note an important exception: hydrochloric acid, HCl, is hydrogen chloride.

6

4. Accuracy in reading and clarity in writing names is essential. Names of different ions maydiffer by only one letter, so any error alters the meaning of a name or formula.

The names of negative ions end in -ide, -ite, or -ate.

The ending "-ide" means that the ion contains only one atom (except hydroxide, OH , and-

cyanide, CN ). The endings "-ite" and "-ate" indicate that the ion is a compound ion, with-

"-ite" ions containing less oxygen than "-ate" ions. If only one compound ion with oxygenexists for a particular element, the "-ate" ending is used. Some compounds have special prefixes, such as "per-", "hypo-", and "thio-". "Per-" meanseven more oxygen than "-ate", "hypo-" means less oxygen than "-ite". "Thio-" meanssome sulfur is present instead of oxygen

EXAMPLES

3 4sulfide, S sulfite, SO sulfate, SO2- 2- 2-

2 3nitride, N nitrite, NO nitrate, NO3- - -

3carbonate, CO 2-

3 4phosphide, P phosphite, PO phosphate, PO3- 3- 3-

2 3 4chloride, Cl- hypochlorite, ClO chlorite, ClO chlorate, ClO perchlorate, ClO- - - -

4 2 3 4sulfate = SO ; thiosulfate = S O (think of one O from SO being replaced by S).2- 2- 2-

5. Formulae are written by the following steps:

a) Identify the ions indicated in the name.Common examples are listed on the nextpage, but these should be memorised as amatter of high priority!

b) Check the charge values on the ions. Positiveand negative values must balance, so thenumbers of positive and negative ions usedmust be chosen so that positive and negativecharges cancel out.

c) Combine the ions into a single formula,leaving out charge values. The number of eachkind of ion used in the formula is shown by asubscript (small number at the bottom afterthe symbol of each ion involved), except thatthe number 1 is not shown. If there is two oror more of a compound ion, its formulashould be enclosed in brackets, with thesubscript outside.

EXAMPLESZinc chloride : Zn and Cl 2+ -

4 4Ammonium sulfate: NH and SO+ 2-

3Aluminium nitrate: Al and NO3+ -

3Carbonic acid: H and CO+ 2-

One Zn requires two Cl 2+ -

4 4Two NH require one SO+ 2-

3One Al requires three NO3+ -

3Two H require one CO+ 2-

2ZnCl

4 2 4(NH ) SO

3 3Al(NO )

2 3H CO

The following are lists of common ions and their formulae. Note that a few metals have morethan one possible charge number.

7

Positive ions are called CATIONS, negative ions are called ANIONS.

"Former names of ions" are not officially used today, but are still frequently encountered inold books and chemical labels, so should be recognised.

SIMPLE IONS

Modernname of ion

Formername of

ion

Modernname of

ion

Formername of

ion

Name ofion

aluminium Al3+ magnesium Mg2+ bromide Br-

barium Ba2+ mercury(I) mercurous Hg+ chloride Cl-

calcium Ca2+ mercury(II) mercuric Hg2+ fluoride F-

chromium(III) chromic Cr3+ nickel Ni2+ iodide I-

cobalt(II) cobaltous Co2+ potassium K+ nitride N3-

copper(I) cuprous Cu+ silver Ag+ oxide O2-

copper(II) cupric Cu2+ sodium Na+ phosphide P3-

iron(II) ferrous Fe2+ tin(II) stannous Sn2+ sulfide S2-

iron(III) ferric Fe3+ tin(IV) stannic Sn4+

lead(II) plumbous Pb2+ zinc Zn2+

COMPOUND IONS

Name of compound ion Name of compound ion

acetate 3CH COO- hypochlorite ClO-

carbonate 3CO 2- nitrate 3NO -

chlorate 3ClO - nitrite 2NO -

chromate 4CrO 2- oxalate 2 4C O 2-

cyanide CN- permanganate 4MnO -

dichromate 2 7Cr O 2- phosphate 4PO 3-

dihydrogenphosphate 2 4H PO - sulfate 4SO 2-

hydrogencarbonate (bicarbonate) 3HCO - sulfite 3SO 2-

hydrogensulfate (bisulfate) 4HSO - thiosulfate 2 3S O 2-

hydrogenphosphate 4HPO 2-

hydroxide OH- ammonium 4NH +

8

WRITE FORMULAE FOR EACH IONIC COMPOUND IN THE SPACES PROVIDED.

potassium zinc lead silver ammonium

carbonate

bromide

hydroxide

chromate

iodide

nitrate

sulfite

acetate

cyanide

sulfide

sulfate

nitrite

oxide

chloride

phosphate

9

WRITE FORMULAE FOR EACH IONIC COMPOUND IN THE SPACES PROVIDED.

Copper(II) calcium iron(II) tin(II) aluminium

carbonate

bromide

hydroxide

chromate

iodide

nitrate

sulfite

acetate

cyanide

sulfide

sulfate

nitrite

oxide

chloride

phosphate

10

WRITE FORMULAE FOR EACH IONIC COMPOUND IN THE SPACES PROVIDED.

sodium nickel iron(III) barium magnesium

carbonate

bromide

hydroxide

chromate

iodide

nitrate

sulfite

acetate

cyanide

sulfide

sulfate

nitrite

oxide

chloride

phosphate

11

EXPLANATION OF RULES FOR WRITING FORMULAE

All matter is made of atoms. Atoms are very small: the radius of an atom can be

measured in picometres. A picometre (pm) is one million-millionth (10 ) of a metre.-12

The mass of a single atom ranges from about 1.6 x 10 g for a hydrogen atom to-24

about 4 x 10 g for an atom of uranium.-22

PROTONS, NEUTRONS and ELECTRONS

Atoms are made up of three kinds of particles: neutrons and protons that are located

in the nucleus, and electrons that revolve around the nucleus in orbits. Protons have

a positive electric charge, so the nucleus has a positive charge. This positive charge

attracts the electrons, keeping them near the nucleus. Neutrons have no electric

charge; they provide part of the mass of the nucleus, and help to hold the protons

together (otherwise the repulsion of their positive charges would cause the nucleus

to break apart). Protons and neutrons have equal mass. An electron has a negative

electric charge. Its mass is slightly more than of the mass of a proton or neutron.

ATOMIC NUMBER

Atoms of different elements have different numbers of protons in their nuclei. Atoms

of the same element always have the same number of nuclear protons. The number

of protons in the nucleus of an atom is called its atomic number. On page 3 of these

notes, some common elements are listed by their atomic numbers. For example, any

atom with 11 protons in its nucleus must be sodium. Every sodium atom has 11

protons in its nucleus.

NEUTRON NUMBERS

The number of neutrons in the nuclei of atoms of any particular element can vary. For

example, atomic nuclei of the element hydrogen can have one proton and no neutrons

(ordinary hydrogen), or one proton and one neutron (called deuterium, or heavy

hydrogen), or one proton and two neutrons (called tritium). These different kinds of

atoms of the same element are called isotopes.

Not every atom of any particular element will have the same mass, since atoms with

more neutrons will have greater mass than those with fewer neutrons.

ELECTRON MOVEMENTS AND NUMBERS

Electrons move around the nucleus in "orbits". The movement of electrons around the

nucleus is likened sometimes, for simplicity, to the movement of planets around the

Sun, and in the diagrams that follow, they will be represented like that.

Electrons really move in much less regular paths; each electron tending to move

freely anywhere within a defined region surrounding the nucleus. The region where

each electron can move is called a shell, or energy level. In these notes, the term shell

will be used. The rules about the numbers of electrons that can move within each

shell are explained below.

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ELECTRON NUMBERS and CHEMICAL REACTIONS

In an atom of a pure element, the number of electrons in orbit around the nucleus

equals the number of protons in the nucleus. When atoms are involved in chemical

reactions, and form compounds, the numbers of electrons in orbit change, as electrons

move from atoms of one element to atoms of another element. In a compound, the

number of protons in the nucleus does not equal the number of electrons in orbit

around the nucleus.

Most (but not all) chemical reactions between compounds also involve the shifting

of electrons from one atom to another.

RULES ABOUT SHELLS AND NUMBERS OF ELECTRONS IN THEM

Shells can be thought of like layers of space around a nucleus. The one closest to the

nucleus is called the first shell, the second one from the nucleus is the second shell,

and so on.

Electrons tend to fill a shell closer to the nucleus before starting to fill one further

out. There are, however, some rules that must be observed by electrons in shells:

there is a maximum number of electrons allowed in each shell, and as each shell is

filled, electrons must start to fill the next one, further out from the nucleus.

Maximum number of electrons allowed in each shell:

SHELL NUMBER MAXIMUM NUMBER OFELECTRONS ALLOWED

MAXIMUM NUMBER OFELECTRONS ALLOWED IFIT IS THE OUTSIDE SHELL

1 2 2

2 8 8

3 18 8

4 32 8

5 50 8

OCTET RULE

Atoms combining to form compounds tend to swap and share electrons between

themselves so that every atom has eight electrons in its outside shell (or two electrons

if the outside shell is the first shell).

There are some exceptions to the Octet Rule, but they will not be considered here.

DIAGRAMS on the next four pages show the arrangement of electrons in atoms, and

how electrons are rearranged when compounds are formed. It is emphasised that these

are very simplified diagrams, showing electrons as if they were moving in simple

orbits, like planets around the Sun, instead of freely within the region of a shell.

13

When atoms combine, electrons are rearranged so that each atom has eight

electrons in its outermost shell (or two electrons if the outermost shell is the

first). This can happen in two main ways: electrons move completely from

one atom to another, or electrons are shared between pairs of atoms.

1. ELECTRONS MOVE COMPLETELY FROM ONE ATOM TO ANOTHER.

The sodium ion still has 11 protons (this cannot change because it is sodium) but now

has only ten electrons. Overall it has a positive charge of one, so is symbolised Na .+

The fluoride ion still has 9 protons (this cannot change because it is fluorine) but now

has ten electrons. Overall it has a negative charge of one, so is symbolised F .-

14

Each magnesium atom has two outer shell electrons. When a magnesium atom

combines with chlorine, it loses its two outer electrons, and is left as a magnesium ion

with eight electrons in its outer shell. With 12 protons (because it is still magnesium)

and ten electrons, the ion has two positive electric charges. Its formula is Mg .2+

A chlorine atom has seven outer electrons, so needs to gain only one electron to have

eight outer-shell electrons. Each magnesium atom that reacts with chlorine can give

one electron to each of two chlorine atoms. With an extra electron, the atoms are now

called chloride ions. With 17 protons and 18 electrons, the ion has one negative

charge. Its formula is Cl .-

Since one magnesium atom can give electrons to two chlorine atoms, the formula for

2magnesium chloride is MgCl .

Refer again to the rules on page 6 for writing formulae.

Exercises: The arrangements of electrons for several elements are listed below.Lithium: 2,1 Nitrogen: 2,5 Oxygen: 2,6Fluorine: 2,7 Sodium: 2,8,1 Magnesium: 2,8,2Aluminium: 2,8,3 Sulfur: 2,8,6 Chlorine: 2,8,7Potassium: 2,8,8,1 Calcium: 2,8,8,2

Using only the information about electron shells, write formulae fora) calcium oxide d) potassium sulfide g) lithium sulfideb) lithium fluoride e) aluminium chloride h) aluminium nitridec) sodium nitride f) calcium nitride i) aluminium oxide

15

2. ELEMENTS WITH FOUR OR MORE OUTER-SHELL ELECTRONS

COMBINE BY SHARING ELECTRONS, RATHER THAN BY

TRANSFERRING THEM FROM ONE ATOM TO ANOTHER.

For example, sulfur (2,8,6) can combine with chlorine (2,8,7) by sharing electrons so

that atoms of both sulfur and chlorine have eight outer-shell electrons.

Compounds like magnesium chloride, in which electrons are transferred completely

from one atom to another, are called IONIC COMPOUNDS. The ions are attracted to

each other by the attraction between the positive and negative charges, yet are not

actually joined together.

2Compounds like sulfur dichloride, SCl , in which electrons are shared between atoms,

are called COVALENT COMPOUNDS. A pair of shared electrons is called a

COVALENT BOND. The atoms are actually joined together, to form a MOLECULE.

Atoms can share more than one pair of electrons. In carbon dioxide, carbon and

oxygen atoms are bound together by two pairs of electrons, or two covalent bonds.

16

3 3In compound ions such as carbonate, CO , or nitrate, NO , the atoms are held2- -

together by covalent bonds. The atoms take up one or more extra electrons as they

join, so that the compound ion has a negative charge, and can attract to positive ions.

The electron shells for the nitrate ion

are shown in the diagram below:

Uncombined oxygen atoms have

six outer-shell electrons, while

uncombined nitrogen atoms have

five outer shell electrons.

One extra electron is taken up by

the atoms as they unite, so that the

nitrate ion has a negative charge.

17

Copper nitrate does not exist as molecules; copper and nitrate are both ions in the1

compound, and are attracted to each other, but do not join together. The formulaindicates that in a sample of copper nitrate, there will be two nitrate ions for everycopper ion. In equations, it is sometimes convenient to show such compounds asif the ions were joined together.

CHEMICAL EQUATIONS

WHAT BALANCING AN EQUATION MEANS

WHAT IS A CHEMICAL EQUATION?A chemical equation is a way of representing a chemical reaction in symbolic form.

For example, when hydrochloric acid is added to sodium hydroxide, the equation reads

The symbols for the reactants (the substances that are mixed) are written on the left, whilethe symbols for the products (the substances that are formed in the reaction) are written onthe right. The arrow indicates that the materials shown on the left are changes to thematerials shown on the right.

A chemical equation shows not only what the reactants and products are, but it shows alsohow much of each reactant combines to form how much of each product. For example:

This equation can be interpreted to read that three atoms of copper react with eightmolecules of nitric acid to form three "molecules" of copper nitrate, two molecules of nitric1

oxide (NO), and four molecules of water. It can also be interpreted to read that three molesof copper react with eight moles of nitric acid to form three moles of copper nitrate, twomoles of nitric oxide (NO), and four moles of water. Moles will be explained in a latersection; the important thing to understand now is that the correct number (coefficient) infront of each formula is critically important.

In the first example above, the formulae have no coefficients. The absence of a coefficientmeans 1 (one). Thus, one mole of sodium hydroxide reacts with one mole of hydrochloricacid to produce one mole of sodium chloride and one mole of water.

Other information that may be included in an equation is the state (solid, liquid, gas, oraqueous solution) of each reactant and product. This additional information is not alwaysincluded in an equation, but sometimes it may be useful, or even vitally necessary.

Symbols used are: (s) = solid, (l) = liquid, (g) = gas, (aq) = aqueous (dissolved in water).The equation shown above may be rewritten:

18

BALANCED CHEMICAL EQUATIONSA balanced chemical equation is one in which for each kind of atomic symbol, the numberon the left side of the equation equals the number on the right side.

There is one Na, one Cl, one O, and two H on each side. Notice with the H, on the left side

2there is one H in NaOH and one H in HCl. On the right, they are shown as two H in H O.

3On the left there are three Cu, eight H, eight N, and twenty four O. Each HNO has one H,one N, and three O. The coefficient, 8, multiplies everything in the formula. On the right

3 2there are three Cu; six N in 3Cu(NO ) plus two N in 2NO, total eight N; eighteen O in

3 2 33Cu(NO ) (three in NO , doubled to six by the subscript 2 outside the brackets), then

2tripled to eighteen by the coefficient 3) plus two O in 2NO plus four O in 4H O, totaltwenty four.

Sometimes it is easier to check the balancing of an equation by counting the numbers ofcompound ions on each side of the equation, provided that none of the compound ions hasbeen changed (as in the example above, where some of the nitrate ions change to NO).For example:

2 4On the left there are three sulfate ions (one sulfate in each H SO , times the coefficient 3).

4On the right there are three sulfate ions ((SO ) times the subscript 3; there is no coefficient

2 4 3to the Al (SO ) , so the value of the coefficient is 1).

HOW TO BALANCE EQUATIONS BY INSPECTIONThere are three basic approaches to balancing chemical equations:

a) by inspectionb) by half-equationsc) by oxidation states.

Only the first of these will be treated in this section: the other two will be explained in latersections. Balancing a chemical equation by inspection means no more than to look at theequation and see by looking what coefficients are needed.

For a balanced equation to be written, it is necessary that the reactants and products beknown. This requires that the writer of the equation have a sufficient knowledge ofDescriptive Chemistry, that is, of that branch of Chemistry that describes the properties andreactions of various elements and their compounds.

The purpose of this section is to explain the balancing of equations. A little DescriptiveChemistry will be presented, however, to cover a few common kinds of chemical reactions.

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i). NEUTRALISATION OF ACIDS AND BASES.An acid is a substance with "acid" in its name, and its formula usually starts with H.

2 4 3 4Examples: sulfuric acid, H SO , hydrochloric acid, HCl, phosphoric acid, H PO .

2(Water, H O, is sometimes considered to be an acid, but here will be taken as an"exception" to the rule about acids.)

A base is defined here as a compound with a metal ion, and either oxide or hydroxide.

2Examples: sodium hydroxide, NaOH, calcium hydroxide, Ca(OH) , copper oxide, CuO,

3zinc oxide, ZnO. Ammonia gas, NH , is also a base: when it dissolves in water, it reactswith the water; the solution is sometimes called ammonium hydroxide. The ammonium

4ion, NH , has a positive charge, and resembles a metal in some of its properties.+

An acid plus a base always gives a salt and water. A salt is any compound of a metal (orammonium) ion with a negative ion other than oxide or hydroxide. Examples: zinc

2 3 4 3 2iodide, ZnI , sodium phosphate, Na PO , copper nitrate, Cu(NO ) , calcium carbonate,

3CaCO , sodium chloride, NaCl. Sodium chloride is one of the most common salts in theworld, so is often called common salt, or just salt, for short.

When an acid and base are mixed, the metal (or ammonium) of the base and the negativeion from the acid form the salt, while the hydrogen of the acid combines with the oxide

2or hydroxide from the base to make hydrogen oxide (H O) or hydrogen hydroxide(HOH), which are both water.

EXAMPLES:

Inspection of this equation (that is, just looking at it) indicates that it is not balanced. Thereis only one H and one Cl on the left, but two of each on the right. It can be balanced byinserting a coefficient = 2 in front of HCl, so that there are two H and two Cl on both sides.

3It is easier to count nitrate (NO ) ions than N and O separately. Writing HOH instead of-

2 2 3H O shows that the OH from the Cu(OH) has combined with the H from the HNO . There

2 3are two OH in Cu(OH) , so two H are needed to form 2HOH. Two NO are needed on- -

3 3 2the left to allow two NO in Cu(NO ) . Balancing can be achieved by writing 2 in front of-

3 2both HNO and HOH. In the final writing, HOH can be rewritten as H O.

It might be noticed that in the paragraph above, the symbols for the negative ions were

3written with the charges shown, OH and NO . The charges are not shown in the full- -

3 3 2formula, as in HNO and Cu(NO ) .

It is absolutely essential that all formulae be written correctly. If an equation cannotbe balanced, it may be that a formula is written wrongly. It is NEVER allowable tomake an equation balance by altering a formula that is already correct.

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ii). REACTION OF ACIDS AND CARBONATES.

3A carbonate is a compound with "carbonate" in its name and CO in its formula.2-

3 3Examples: lead carbonate, PbCO , iron(II) carbonate, FeCO , potassium carbonate,

2 3 2 3K CO , silver carbonate, Ag CO .

3When a carbonate is mixed with an acid, the H from the acid joins with the CO from+ 2-

2 3the carbonate, forming H CO , and the metal from the carbonate joins with the negative

2 3 2 2ion from the acid to form a salt. The H CO then splits up to make H O and CO .

To summarise: an acid and a carbonate make a salt, water, and carbon dioxide.

EXAMPLES:

The coefficient 2 was placed before the HCl because two Cl are needed on the right to

2 2 3 2 3make CaCl , and two H are needed to make H CO . By writing H CO first, then altering

2 2it to H O + CO , it is easier to see how the equation can be balanced just by rearrangingions.

To balance the folowing equation:

2 2 2 3a) rewrite the H O + CO as H CO .b) observe that an even number of H is needed on the right, whatever the coefficient of

2 3H CO may become. An even number of H must also be on the left: one way to achieve

3 4this is to put a 2 in front of the H PO . This would give 6H on the left, so 3 in front of

2 3H CO on the right will balance the H.

4 3 4c) There are now 2 PO on the left, so 2 should be placed in front of K PO . To balance3-

3 2 3both K and CO , 3 should be written in front of K CO .+ 2-

2 3 2 2.d) The 3H CO should now be separated into 3H O + 3CO

iii). REACTION OF ACIDS AND METALS.Some common acids react with certain metals to produce a salt and hydrogen gas.Whether or not a particular acid and a particular metal actually do this must be decidedusing other sources of information. Examples used here are all true examples.

It should be evident that all that is needed to balance this is 2 in front of HCl.

21

An example involving aluminium and sulfuric acid is shown on page 18. What aboutaluminium and hydrochloric acid?

3Three Cl are needed on the left to balance the three Cl in AlCl . If 3 is written in front ofHCl, there will be an odd number of H on the left and an even number on the right. Tomake H even on the left, and still be able to provide 3Cl for each Al, a coefficient of 6

3should be tried. This will allow 2AlCl on the right. It will then be necessary to have 2Al on

2the left and 3H on the right to achieve a full balance.

iv). REACTION OF SALTS TO FORM A PRECIPITATE.Some salts are insoluble in water. If solutions of two soluble salts are mixed, the ionsmay simply rearrange to form a precipitate of an insoluble salt.In the examples below, it should be recalled that (aq) means dissolved in water, while(s) means solid.

Silver chloride is insoluble in water. Silver nitrate, calcium chloride, and calcium nitrateall dissolve easily in water.

3To balance this, a 2 should be placed before both AgNO and AgCl.

v). OTHER REACTIONS.Another common kind of reaction whose equation can usually be balanced by inspectionis combustion, that is, burning of a substance so that its elements combine with oxygen.For example, when magnesium metal burns in oxygen, it forms magnesium oxide, MgO.

A coefficient 2 in front of Mg and MgO will balance the equation. It is also permissible

2to balance it by placing in front of O .

Fuels containing carbon and hydrogen burn in oxygen to produce carbon dioxide and

4 3 8water. Two such fuels are methane, CH , and propane, C H .

22

SUMMARY:1. A chemical equation is a means of representing the nature and amounts of reactants and

products involved in a chemical reaction.

2. To write a chemical equation it is necessary to know what the reactants and productsare, and to be able to write their chemical formulae correctly. It is essential thatchemical symbols, and methods of reading and writing formulae, are well-learned.

3. Equations must be balanced: this means that the numbers of each kind of symbol on theleft-hand and right-hand sides of the equation must be equal.

4. It is often possible to balance an equation by inspection, that is, by just looking at it.(Methods for balancing equations that cannot be balanced by inspection will beexplained later.)

PRACTICE EXERCISES:Write formulae for the substances in the word-equations below, then balance thechemical equations by inspection: (see page 7 for formulae of ions)

1. Barium hydroxide + sulfuric acid produce barium sulfate + water

2. Lead oxide + nitric acid produce lead nitrate + water

3. Zinc carbonate + hydrochloric acid produce zinc chloride + carbon dioxide + water

4. Sodium carbonate + nitric acid produce sodium nitrate + carbon dioxide + water

5. Magnesium + phosphoric acid produce magnesium dihydrogenphosphate + hydrogen

6. Iron + hydrochloric acid produce iron(II) chloride + hydrogen

7. Calcium chloride(aq) + silver nitrate(aq) produce silver chloride (s) + calcium nitrate(aq)

8. Aluminium sulfate(aq) + potassium hydroxide(aq) produce aluminium hydroxide(s) +potassium sulfate(aq)

2 5 9. Alcohol (C H OH) burns in oxygen to produce carbon dioxide and water

10. Iron (as powder) burns in oxygen to produce iron(III) oxide.

23

MOLECULAR AND IONIC EQUATIONS, AND SPECTATOR IONS

Any solution of an ionic substance contains two different kinds of ions, which exist quiteseparately from one another. The only effect that one may have on the other is to balanceits electric charge.

2 4For example, in a pure solution of sodium sulfate, Na SO , there are sodium ions, Na , and+

4 4sulfate ions, SO , in a ratio two Na to one SO .2- + 2-

In solutions of potassium sulfate or sodium chloride, the sulfate ions and sodium ions areexactly the same kinds of particles as they would be in sodium sulfate, and would haveexactly the same role in a chemical reaction as sodium and sulfate ions in sodium sulfate.

If solutions containing equal concentrations of sodium nitrate and potassium chloride weremixed, the resulting solution would be exactly the same as if solutions (of the sameconcentration) of sodium chloride and potassium nitrate had been mixed.

Consider the mixtures of solutions whose equations are shown below. Each mixtureinvolves two clear solutions which are mixed to produce the same white precipitate.

zinc chloride + sodium carbonate produce zinc carbonate (solid) + sodium chloride

zinc sulfate + potassium carbonate produce zinc carbonate (solid) + potassium sulfate

zinc nitrate + ammonium carbonate produce zinc carbonate (solid) + ammonium nitrate

The equations above are written as if the substances involved are molecules: these aremolecular equations.

24

The equations can be rewritten, to show that the substances are ions:

Equations written like this are called complete (or full) ionic equations.

Inspection of these equations shows that only the zinc and carbonate ions take part in thereaction. The other ions, chloride, sulfate, nitrate, sodium, potassium, ammonium, are allspectator ions. It should be possible, therefore, to write the equation for the actual reaction

This last equation is called a net ionic equation. It shows only the ions or molecules thatactually react, and excludes the spectator ions, which are present but which have no partin the reaction, except to keep positive and negative electric charges in the solution equal.

All three kinds of equations can be used, the value of each depending upon thecircumstances. Molecular equations and net ionic equations are used more commonly, andare generally more useful, than complete ionic equations.

Further examples are presented below, with molecular equations and a net ionic equation,for the reaction of metallic magnesium with three different acids.

The molecular, complete ionic, and net ionic equations for the reaction of NaOH and HCl:

The net ionic equation for all neutralisation reactions is the same as the one shown abovefor HCl and NaOH.

All reactions between acids and carbonates also have the same net ionic equation:

25

REDOX REACTIONS AND REDOX EQUATIONS

DEFINITIONSMost of the reactions and their equations considered so far have been based upon arearrangement of ions between the components of a mixture.

Another very common kind of chemical reaction is the redox reaction. Redox stands forreduction and oxidation.

The original definition of oxidation is "a reaction in which a substance combines withoxygen", while reduction is "a reaction in which oxygen is removed from a substance".These definitions were extended to cover reactions involving hydrogen gas: if a substancecombined with hydrogen, it was reduced, while if it gave off hydrogen it was oxidised.

OXIDATION REDUCTION

Substance combines with oxygen OR

Substance gives off hydrogen

Substance gives off oxygen OR

Substance combines with hydrogen

Both of these old definitions continue to be useful, and should be used.

Modern definitions describe oxidation and reduction as the movement of electrons from theouter shell of one kind of atom, ion, or molecule to the outer shell of another kind of atom,ion or molecule. This movement usually involves rearrangement of atoms within the ionsor molecules in the reaction.

When an atom, ion, or molecule loses one or more electrons, it is oxidised.When an atom, ion, or molecule gains one or more electrons, it is reduced.

IDENTIFYING REDOX REACTIONSHow can a chemical reaction be identified as a redox reaction?

If either of the elements oxygen or hydrogen is a reactant or a product in the reaction, thenthe reaction must be redox, according to the old definitions of oxidation and reduction. Thefollowing equations represent examples of redox reactions: oxygen or hydrogen is involved

2 in each. NB Hydrogen is H , not H .+

Many reactions do not involve elemental oxygen or hydrogen. To decide if a reaction isredox, involving transfer of electrons from one atom, ion, or molecule to another, oxidationstates may be used. The idea of oxidation state is explained on the following pages.

26

OXIDATION STATESWhen elements combine to form compounds, they either transfer electrons completely fromatoms of one kind to atoms of another, or they transfer electrons partially by sharing thembetween different atoms. This was explained in pages 11 to 16.

Oxidation state is a way to describe the number of electrons that have been transferred orshared between atoms of different kinds.

RULE ONE states that uncombined elements have an oxidation state = zero, sincethere are no electrons being shared with atoms of other elements.

2 2In simple ionic compounds, such as Na S or CaI , each atom has either gained or lost oneor more electrons, forming ions such as Na , Ca , S , I .+ 2+ 2- -

RULE TWO states that the oxidation state of a simple ion is the same as the chargeon the ion. It is usually written with the sign before the number.

2For example, in Na S, the oxidation state of sodium is +1, of sulfur is -2; in

2CaI , the oxidation state of calcium is +2, of iodide is -1.

Oxygen and hydrogen are found in very many different compounds. When oxygen, whichhas electron shells 2,6, forms a compound, it gains two electrons (either by completetransfer or by sharing) to achieve full electron shells 2,8. Hydrogen has a single electron inits first shell, which it may lose to other atoms, to form an H ion.+

RULE THREE states that in its compounds, oxygen has an oxidation state = -2,while hydrogen has an oxidation state = +1. (In just a few compounds

2this rule is not observed: in peroxides, containing the peroxide group O ,2-

and in metal hydrides, in which hydrogen is combined directly with a

2metal, for example, calcium hydride, CaH .)

RULE FOUR: the total oxidation states of an ion equals the charge of the ion. Thetotal oxidation states of an uncharged molecule = zero.

EXAMPLES: Oxidation states of elements in compounds containing three or moredifferent elements can be determined in the following way:

a) If the compound is a molecule (see page 15), give hydrogen a value = +1,oxygen a value = -2. Give the other element a value such that the total of all theoxidation states = 0.

For example, it is evident that the oxidation stateof nitrogen in nitric acid = +5.

27

b) If the compound is ionic, divide it into its ions and deal with each ionseparately. Apply the same rules as above in part a), but let the total of alloxidation states equal the charge of the ion. Forexample, the oxidation state of sulfur in calcium

4sulfate, CaSO , is the same as the oxidation state of

4sulfur in the sulfate ion, SO , by itself. The oxidation2-

state of sulfur in sulfate = +6.

Further examples:What are oxidation states of nitrogen and phosphorus in ammonium dihydrogenphosphate?

4 2 4 4 2 4The formula is NH H PO , which can be separated into the ions NH and H PO . Treating+ -

each kind of ion separately, according to the rules above:

The oxidation states are: nitrogen = -3, phosphorus = +5

Oxidation states may have fractional values: for example,

3 8the oxidation state of carbon in propane, C H , is - .

The same element may have different oxidation states in different compounds: using sulfuras an example:

Name of ion Formula Oxidationstate

Name of ion Formula Oxidationstate

3sulfide S -2 sulfite SO +42- 2-

2 3 4thiosulfate S O +2 sulfate SO +62- 2-

4 6 2 8tetrathionate S O +2 persulfate S O +72- 2-

In addition, sulfur as the uncombined element has oxidation state = 0.

If the same element can exist at different oxidation states, then it must be possible for it tochange from one oxidation state to another.

Returning to the original question on page 25: How can a chemical reaction be identifiedas a redox reaction?

If two elements in a reaction change in oxidation state, one increasing, the otherdecreasing, then the reaction is a redox reaction.

28

WRITING REDOX EQUATIONS : OXIDATION STATE METHOD

The rules for writing balanced redox equation by this method are set out with an example.

Balance the equation :

1. Check that the equation is a redox equation: write oxidation states under key elements,and show that one element increases in oxidation state while the other decreases.

2. Insert coefficients so that the total decrease in oxidation state of one element equals thetotal increase in oxidation state of the other element:

23. Balance oxygen by adding water: 4H O on the right will balance oxygen.

4. Balance hydrogen by adding H : 6H on the left will balance hydrogen. The balanced+ +

equation is

35. Spectator ions (nitrate in this example) can be added if required: 6NO on each side:-

Further example:Balance :

1. Balance any elements other than oxygen or hydrogen, then write in the oxidation statesof those elements that change in oxidation state:

2. The loss of oxidation state by chromium = 2 x (-3) = -6. The gain by one sulfur = +2.Three sulfite ions will therefore increase total oxidation state by +6, to balance the lossby chromium. Sulfite and sulfate should have coefficients = 3:

29

3. Balance oxygen by adding water:

4. Balance hydrogen by adding H :+

5. Check the balancing, including the balancing of charges. In the equation above, chargeson left = (2-) + 3(2-) + 8(+) = 0. Charges on right = 2(3+) + 3(2-) = 0.

6. Add spectator ions if required.

FURTHER EXAMPLES TO BE BALANCED:Summary of procedures:

a) Write the unbalanced equation, excluding spectator ions, ensuring that all formulaeare correct.

b) Check that the equation is redox by writing oxidation states under the elementspresent.

c) Balance all elements other than oxygen and hydrogen by inserting appropriatecoefficients.

d) Calculate changes in oxidation states, and insert coefficients so that the total increasein oxidation state of one element is balanced exactly by the total loss of oxidationstate by the other element.

e) Balance oxygen by adding a suitable amount of water.f) Balance hydrogen by adding a suitable amount of H .+

g) Check the balancing, including the balancing of charge.h) Add spectator ions if required.

1. Permanganate ions + iodide ions give manganese(II) ions + iodine.

2. Iron(III) chloride + sodium sulfide give iron(II) chloride, sodium chloride, and sulfur.

3. Iron(II) ions + silver ions give silver metal + iron(III) ions.

4. Silver ions + methanal (HCHO) give silver metal + formic acid (HCOOH).

2 2 5. Tin(II) ions + hydrogen peroxide (H O ) give tin(IV) hydroxide (s)

6. Nitric acid + zinc metal give zinc nitrate + nitrogen dioxide

7. Potassium dichromate + oxalic acid give chromium(III) ions + carbon dioxide.

8. Permanganate ions + hydrogen sulfide give manganese dioxide + sulfur.

9. Copper(II) ions + potassium iodide give copper(I) iodide(s) + iodine.

10. Potassium chlorate + hydrochloric acid give chlorine gas + potassium chloride.

30

WRITING REDOX EQUATIONS : HALF-EQUATION METHOD

Comparing different methods for balancing redox equations:If the objective is simply to balance an equation, and the equation can be balanced easily BYINSPECTION, then it may be balanced that way. Balancing by inspection is described onpages 18 to 23. Most of the examples given involve only exchanges of ions, and only thereaction of acids and metals on page 20, and combustion reactions on page 21, areexamples of redox. Among the redox equations listed for balancing on page 29, some canbe balanced quite easily by inspection.

The OXIDATION-STATE METHOD, described in pages 28 - 29, is a useful and quick methodfor balancing redox equations that cannot be balanced by inspection. It does not have agood theoretical basis, and occasionally it may be difficult to apply.

The HALF-EQUATION METHOD has a strong theoretical basis, and is therefore a moreimportant method, even though not always the quickest method to balance an equation. Itis important that students know how to use half-equations, since they have otherapplications besides as a way of balancing equations, notably in situations involving cellsand electrolysis.

Some background theory:On page 25, it was stated that oxidation is a loss of one or more electrons by an atom, ion,or molecule, whereas reduction is a gain of one or more electrons by an atom, ion ormolecule.

There are two ways to remember which is loss, which is gain of electrons.a) When an atom is oxidised, its oxidation state becomes more positive; it increases.

When an atom is reduced, its oxidation state becomes less positive or morenegative, that is, it is reduced. Gain of (negative) electrons effectively makes theatom less positive or more negative, that is, gain of electrons reduces the atom.The same idea applies, of course, to molecules and ions: reduction involves gainof electrons.

b) Think of a simple example of an element combining with oxygen, for example,magnesium plus oxygen forms magnesium oxide.

It should be clear that electrons have been transferred from magnesium to oxygen.Magnesium has lost electrons in the process of combining with oxygen, that is,while being oxidised. Oxygen has been reduced by magnesium by gaining electronsfrom it.

If there is ever any hesitation about whether oxidation is a loss or a gain ofelectrons, use this simple example (or any like it) as a way of remembering.

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The equations written to show electrons (written as e ) are called half-equations. -

Note that it is not usual to write a half equation with "- e ”, as -

Such an equation is properly written with a positive sign before the electrons, as

Every redox reaction is comprised of two separate components:1) an oxidation half-reaction, which concerns the atom, molecule or ion that is

oxidised, that is, which loses electrons, and2) a reduction half-reaction, which concerns the atom, molecule or ion that is

reduced, that is, which gains electrons.

Both of these must be involved, and each can be expressed by a half-equation. For example,consider the reaction between zinc and sulfuric acid, that produces hydrogen gas and zincsulfate in solution:

Written without the sulfate spectator ions, the equation is:

The reaction is redox, because zinc is oxidised, while hydrogen ions are reduced. Theequation can be separated into two half equations:

OXIDATION HALF-EQUATION:

REDUCTION HALF-EQUATION:

Any redox equation can be separated into two half-equations. For example,

If the second half-equation is simplified, the two half-equations are

It should be noticed that electrons are always shown in a redox half-equation, but that theelectrons cancel out when the two half-equations are added together. The "simplified" half-equations shown above cannot be added together unless the second one has been multipliedby three, so that the electron numbers are the same in both half-equations.

32

It should also be noticed that when the half-equations are properly added together, it maybe necessary to simplify the equation with respect to water and hydrogen ions:

How to write balanced redox equations:EXAMPLE:

Manganese dioxide + hydrochloric acid produces manganese(II) ions + chlorine gas.Step one: write the equation in unbalanced form:

Step two: separate the unbalanced equation into two unbalanced half-equations:

Step three: balance either or both equations for any elements other than oxygen orhydrogen. (NB this is a step often overlooked by students! Be careful!).

Step four: balance oxygen by adding water:

Step five: Balance hydrogen by adding H :+

Step six: Add electrons to balance charge:

NB the electrons should be on the left-hand-side of one equation and on the right-hand-sideof the other. If this is not so, then either the substances shown on the left of the half-equations will not react with each other in the way that the half-equations show, or else anerror has been made in writing and balancing the half-equations.

33

Step seven: If the electrons numbers in the two equations are not equal, multiply one orboth equations by factors so that the numbers of electrons become equal. Thisstep is not necessary in the example being used, since both equations have 2e .-

A second example below shows the application of step seven.

Step eight: Add the two equations, so that the electrons are cancelled out.

Step nine: Simplify the equation so that water and H each appear in only one place:+

Step ten: Add spectator ions if required: in this example, chloride ions.

Second example:Potassium hydroxide solution heated with iodine produces potassium iodate and iodide.

Step one:

Iodine is reduced from oxidation state 0 to -1, and is also oxidised fromoxidation state 0 to +5. This is a redox reaction.

Step two: NB spectator ions are not written.

Step three:

Step four:

Step five: :

Step six:

Step seven:

Step eight:

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Step nine: Simplify the equation:

Step ten: Add spectator ions if required: in this example, potassium ions.

This equation is correct and balanced, but it does not agree with the originalunbalanced equation at step one, which has KOH as a reactant.

If 6OH is added to each side of the equation, then 6KOH appears on the left,-

2and 6HOH (or 6H O) on the right. The water cancels from left and right, leaving

23H O on the right.

A redox reaction of this kind, in which the same reactant is both oxidised and reducedsimultaneously (it oxidises and reduces itself) is called disproportionation.

For practice exercises, the equations on page 29 can be balanced by the half equationmethod.

Further examples for balancing:

1. Silver metal + nitric acid give silver nitrate + nitric oxide (NO)

2. Calcium hydroxide + chlorine give calcium chloride + calcium hypochlorite

3. Copper + (concentrated) sulfuric acid give copper sulfate + sulfur dioxide

4. Nitric acid + iodide ions give nitric oxide + iodine

5. Nitric acid + sulfide ions give nitrogen dioxide + sulfur

6. Iron(III) chloride + iodide ions give iron(II) chloride + iodine.

4 6 7. Thiosulfate ions + iodine give tetrathionate ions (S O ) + iodide ions.2-

8. Ammonium ions + nitrite ions give nitrogen gas + water

9. Manganese dioxide + hydrochloric acid give manganese(II) ions + chlorine gas

10. Potassium dichromate + sulfur dioxide give chromium(III) sulfate + potassium sulfate

35

The definitions given here do not allow for the existence of different isotopes of1

elements. This omission does not affect the validity of the methods described.Detailed treatment of this topic should consider the presence of isotopes.

MOLES AND MOLE CALCULATIONS

INTRODUCTIONThe purpose of this section is to present some methods for calculating both how much ofeach reactant is used in a chemical reaction, and how much of each product is formed.

In this section theoretical details are kept to the minimum necessary for a mastery ofcalculation methods to be achieved. It is desirable, however, that at some stage studentslearn of the historical development of the ideas applied here, and of the underlying theory.

It is assumed that students have a good mastery of basic arithmetic and algebra,especially of calculations involving ratios and proportions.

It is assumed also that students are able to both read and write chemical formulae andequations, accurately and with fair confidence. Until such knowledge and skills have beenacquired, it is difficult to achieve success with mole calculations.

SCOPE OF CALCULATIONS COVERED IN THIS SECTIONThere are three basic requirements to be mastered in this section:

1) to calculate masses of reactants that will be consumed in a reaction, and the massesof products likely to be formed from a given mass of reactants.

2) to calculate what volume of gas may be either consumed or produced in a reaction.

3) to calculate what volumes and concentrations of solutions of reactants should bemixed to produce predicted concentrations of products in the solution.

DEFINITION OF A MOLE AND OF MOLAR MASSAtoms of different elements have different masses. It is possible both to measure the massof individual atoms, and to measure the relative masses of different kinds of atoms. Actualmasses of atoms are very small (an atom of carbon has mass = 2 x 10 g approx), so the-23

masses are given in atomic mass units (amu).

The mass of an atom of carbon has been selected as a standard for comparing the masses

of atoms: an atom of carbon has been assigned a mass of 12.00 amu. The relative atomic1

mass of an atom of any other element is given in proportion to the mass of an atom of

carbon. For example, the relative atomic mass of an oxygen atom is given as 16.00 amu,

which indicates that an atom of oxygen is times more massive than an atom of carbon.

The relative atomic mass of hydrogen = 1.00 amu; an atom of hydrogen has of the

mass of an atom of carbon.

36

The relative atomic masses (ram) of some common elements are listed below:

Element ram Element ram Element ram

Aluminium 27.0 Gold 197.0 Oxygen 16.0

Barium 137.3 Hydrogen 1.0 Phosphorus 31.0

Bromine 79.9 Iodine 126.9 Potassium 39.1

Calcium 40.1 Iron 55.8 Silicon 28.1

Carbon 12.0 Lead 207.2 Silver 107.9

Chlorine 35.5 Magnesium 24.3 Sodium 23.0

Chromium 52.0 Manganese 54.9 Sulfur 32.1

Cobalt 58.9 Mercury 200.6 Tin 118.7

Copper 63.5 Nickel 58.7 Titanium 47.9

Fluorine 19.0 Nitrogen 14.0 Zinc 65.4

The mass of one mole of an element is the relative atomic mass of an

element, stated in grams.

A mole is defined as the number of atoms in 12.00 g of pure carbon. The

Avalue of this number is 6.023 x 10 . It is called the Avogadro Number, symbol N .23

AOne mole of any element also contains 6.023 x 10 , or N , atoms.23

WHAT ABOUT COMPOUNDS?The mass of one mole of any compound can be calculated by adding the relative atomicmasses of all atoms in the formula, and stating the total in grams.

Examples:

Compound Formula Sum of masses of atoms in

formula

Mass of

one mole

2Water H O (1.0 x 2) + 16.0 18.0 g

2Carbon dioxide CO 12.0 + (16.0 x 2) 44.0 g

Sodium chloride NaCl 23.0 + 35.5 58.5 g

4Calcium sulfate CaSO 40.1 + 32.1 + (16.0 x 4) 136.2 g

3 2Lead nitrate Pb(NO ) 207.2 + 2(14.0 + 16.0 x 3) 331.2 g

4 3 4Ammonium phosphate (NH ) PO 3(14.0 + 1.0 x 4) + 31.0 + (16.0 x4)

149.0 g

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The mass of one mole of a substance is the MOLAR MASS, for which the symbol is M.

AOne mole of any substance contains N , or 6.023 x 10 , of each particle present.23

Examples:

One mole of withformula

contains the followingnumbers of moleculesor ions

and the followingnumbers of atoms

2 A AWater H O N molecules of water 2 x N atoms of hydrogen,

AN atoms of oxygen

2 ACarbon dioxide CO N molecules of carbondioxide

AN atoms of carbon,

A2 x N atoms of oxygen

ASodium chloride NaCl N sodium ions and

AN chloride ionsAN atoms of sodium,

A N atoms of chlorine

4 ACalcium sulfate CaSO N calcium ions and

AN sulfate ionsAN atoms of calcium,

AN atoms of sulfur,

A 4 x N atoms of oxygen

3 2 ALead nitrate Pb(NO ) N lead ions and

A2 x N nitrate ionsAN atoms of lead,

A2 x N atoms of nitrogen,

A6 x N atoms of oxygen.

4 3 4 AAmmonium phosphate (NH ) PO 3 x N ammonium ions

Aand N phosphate ionsA3 x N atoms of nitrogen,

A12 x N atoms of hydrogen,

AN atoms of phosphorus,

A4 x N atoms of oxygen

PERCENTAGE COMPOSITIONThe proportion by mass of an element in a compound can be calculated easily by usingmolar masses.For example, the proportion of sodium present in sodium chloride, NaCl, is the ratio of themass of one mole of sodium to the mass of one mole of sodium chloride, expressed as apercentage. Proportion of sodium in sodium chloride = = 0.393

Percentage composition of sodium in sodium chloride = 39.3%.

4 3What is the percentage composition of nitrogen in ammonium nitrate, NH NO ?Molar mass of ammonium nitrate = 80.0 gMass of nitrogen in one mole of ammonium nitrate = 2 x 14.0 g = 28.0

Percentage composiition of nitrogen in ammonium nitrate = = 35.0%.

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USE OF MOLAR MASS TO PREDICT REACTING MASSESA balanced chemical equation is a way of describing the relative quantities of reactants andproducts that are involved in a reaction. The coefficients may be read to indicate the relativenumbers of atoms, ions, or molecules involved in the reaction.

It is more useful to read the coefficients as indicating the numbers of moles of eachsubstance involved.

The following equation, for magnesium metal burning in oxygen, can beread "two atoms of magnesium combine with one molecule of oxygen toform two 'molecules' (actually ions Mg and O ) of magnesium oxide."2+ 2-

A better way to read it is "two moles of magnesium metal combine with onemole of oxygen gas to form two moles of magnesium oxide".

From this, the relative masses of the reactants and products can be predicted:

2 moles 1 mole 2 moles2 x 24.3 g 32.0 g 2 x 40.3 g

If the relative masses of reactants and products in a reaction can be predicted from abalanced equation and knowledge of molar masses, then the actual mass of any reactant orproduct can be calculated by the use of ratios.

Example one:What mass of sodium carbonate will be obtained if 3.36 g of pure sodium hydrogen-carbonate is heated? (The other products of the reaction are carbon dioxide and water.)1) Write a balanced equation:

2) Write mole ratios underneath equation: 2 moles 1 mole3) Calculate molar masses and write them under: 2 x 84.0 g 106.0 g4) Write in known value, and "x" for unknown: 3.36 g x g

Use ratio to calculate x: = x = = 2.12 g

The mass of sodium carbonate formed by heating 3.36 g of sodium hydrogencarbonate = 2.12 g.

39

Example two:What mass of carbon will be converted to carbon monoxide in reducing 1000 g of iron(III)oxide to iron metal? What masses of iron and carbon monoxide should be formed?

Note that with three "unknowns" in this problem, three algbraic symbols, x, y, z, are used.The values of x, y, and z are calculated by simple ratio:

= = =

Mass of carbon converted = x = 226 gMass of iron formed = y = 700 g

Mass of carbon monoxide formed = z = 526 g

The total mass of reactants should equal the total mass of products: 1000 g + 226 g = 700 g + 526 g

Example three:What mass of lead can be extracted by heating 120 g of solid lead sulfide in air, forminglead oxide and sulfur dioxide, then heating the lead oxide with carbon, to form metallic leadand carbon monoxide?

This problem can be solved as above, by writing the equations and carrying out all ratiocalculations. An alternative method uses percentage composition: the problem can besummed up as "how much lead can be separated from 120 g of lead sulfide?"

Molar mass of PbS = (207.2 + 32.1) g = 239.3 g.

Percentage of lead in lead sulfide = = 86.6%

86.6% of 120 g = 104 g = mass of lead that can be extracted from 120 g of lead sulfide.

Exercises:1. What mass of copper can be extracted from 5.0 g of copper(II) sulfate by dissolving the

copper sulfate in water and adding zinc metal? (The other product is zinc sulfate).2. What mass of potassium iodide is needed to react exactly with 8.0 g of lead nitrate, to

form lead iodide? (The other product is potassium nitrate).3. When calcium carbonate is heated strongly, it forms calcium oxide and carbon dioxide.

What mass of calcium carbonate is needed to make 50.0 g of calcium oxide?4. Sodium carbonate reacts with hydrochloric acid to form sodium chloride, water, and

carbon dioxide. Some hydrochloric acid was added to some sodium carbonate: 6.0 g ofsodium chloride were formed. What mass of carbon dioxide was produced?

5. What mass of lead oxide would need to be reacted with nitric acid to produce 10.0g oflead nitrate?

40

CALCULATIONS INVOLVING EXCESS OF ONE REACTANTChemical reactants combine in definite proportions which can be predicted from theequation for the reaction. A word used to describe this is STOICHIOMETRY.

Consider the reaction

2 4 2 4"The stoichiometric ratio of KOH to H SO is two moles of KOH to one mole of H SO ."

2 4"The stoichiometry of this reaction requires 98.1 g of H SO for 112.2 g of KOH."

2 4"If the reactants, H SO and KOH, are mixed in stoichiometric proportion, then theproduct will contain only potassium sulfate and water."

The situation often exists, however, where reactants are not mixed in stoichiometric

2 4proportions. For example, if 12.0 g of KOH is mixed with 10.0 g of H SO , how muchpotassium sulfate will be present when the reaction is completed?

2 4It should not be assumed that 12.0 g of KOH and 10.0 g of H SO is in stoichiometricproportion. If they are not in stoichiometric proportion, then at the end of the reaction,

2 4there will be either an excess of KOH or an excess of H SO mixed with the potassiumsulfate product. How can this be checked by calculation?

2 4Step one Assume H SO to be excess. This means that all of the KOH will be used up.

2 4Use the mass of KOH present to calculate the mass of H SO used.

= x = 10.5 g

2 4The mass of H SO needed to react with 12.0 g of KOH is greater than the 10.0 g provided.

2 4The assumption that H SO is present in excess is false. KOH is in excess.

2 4Step two KOH is in excess, which means that all the H SO will be used up. Use the mass

2 4 2 4of H SO present to calculate the mass of KOH used and the mass of K SOformed in the reaction. The mass of unused KOH can also be calculated.

= = x = 11.4 g y = 17.8 g

2 4Mass of KOH used = 11.4 g Excess KOH = 0.6 g Mass of K SO made = 17.8 g

41

It may be possible, by looking carefully at the quantities of reactants, to decide which oneis in excess without having to do a trial-and-error calculation like the one above.

For example: What mass of zinc sulfate will be obtained by treating 20.0 g of zinc oxidewith 1.0 g of sulfuric acid?

It should be obvious at a glance that zinc oxide is in excess, and that it is the sulfuric acidthat will be completely used up. The equation can be rewritten, therefore:

= = x = 1.6 g y = 0.83 g

2 4Only 0.83 g of zinc oxide reacts with 1.0 g of H SO , leaving 19.2 g of ZnO unreacted.1.6 g of zinc sulfate is formed.

In some situations, it is stated clearly that an excess of one reactant is used, so thecalculation can be very straightforward.For example: 5.0 g of magnesium metal is dissolved in excess hydrochloric acid. What

mass of magnesium chloride will be formed?

= x = 19.6 g Mass of magnesium chloride = 19.6 g

Exercises:

21. Manganese dioxide, MnO , reacts with concentrated HCl to form manganese(II)chloride, chlorine gas, and water. What mass of chlorine gas will be formed if excess HClis added to 6.0 g of manganese dioxide?

2. Iron(III) oxide reacts with carbon at very high temperatures to form iron metal andcarbon monoxide gas. If 2000 g of iron oxide is heated with 500 g of carbon in a furnace,which reactant will be in excess, and by how much?

4 23. 50.0 g of copper(II) sulfate crystals (CuSO .5H O) are dissolved in water, and zinc metalis added. The zinc displaces the copper: metallic copper and zinc sulfate are formed.Will 12.0 g of zinc be sufficient to displace all the copper?

42

This statement about gases is, in fact, only approximately true. In this book, it is1

accepted without qualification that the molar volume of any gas = 22.4 L at STP.

The actual pressure of the atmosphere varies with weather conditions. 2

The standard atmosphere is a unit used in chemistry: it is exactly 101.3 kPa. The unit is usually named simply as atmospheres (atm).

Methods for calculating volumes of gases at various temperatures and pressures3

can be found in many text-books, under such headings as Boyle's Law, Charles'Law, Combined Gas Law, Universal (or Ideal) Gas Equation.

CALCULATIONS INVOLVING GASESMany chemical reactions involve gases either as reactants or as products. It is often moreuseful to calculate quantities of gases as volumes rather than as masses.

It is a fact about gases that the same number of moles of any gas occupies the same volume,provided that the volumes are measured at the same temperature and pressure.1

At Standard Temperature and Pressure, the volume of 1.0 mole of any gas is 22.4 litres.This volume, 22.4 L at STP, is called the molar volume of a gas.

Standard Temperature and Pressure is commonly written as STP.

Standard temperature = 0 C, standard pressure = 101.3 kilopascals or 1.0 atmospheres .o 2

The volume of a gas at temperatures and pressures other than STP can be calculated easily .3

For current purposes, however, all calculations will assume that a gas is at STP.

Calculations may be set out in the same way as calculations involving masses, except thatwhen volumes of gases are involved, for every mole of gas write 22.4 L under the equation.

Example one: What volume of hydrogen at STP will be produced when 5.0 g of iron istreated with excess hydrochloric acid?

When calculating using ratios, it is important to ensure that the units used in each ratio arethe same. In the calculation below, 5.0 grams are divided by 55.8 grams, and x litres aredivided by 22.4 litres.

= x = 2.0 L Volume of hydrogen formed = 2.0 litres at STP

43

Example two: Hydrogen chloride gas and ammonia gas combine readily to form solidammonium chloride. What volume of each gas (at STP) will combine toform 25.0 g of solid ammonium chloride?

= x = 10.5 L

3 410.5 litres each of HCl and NH (at STP) are needed to make 25.0 g of NH Cl.

Excess reactants may have to be considered in reactions involving gasesExample three: 1.5 L (at STP) of carbon dioxide gas is passed into a solution containing

5.0 g of sodium hydroxide. Will all the sodium hydroxide be converted tosodium carbonate?

= x = 1.4 L

2 2 3Only 1.4 L at STP of CO are needed to convert 5.0 g of NaOH to Na CO . Since 1.5 L

2 2 3of CO are passed through the solution, all the NaOH will be converted to Na CO .

2(In this situation, the excess CO will form hydrogencarbonate ions at room temperature.)

Exercises:

2 21. If manganese dioxide is added to hydrogen peroxide, the H O decomposes to form

2 2oxygen gas and water. What mass of H O has decomposed to produce 1.5 L of oxygengas (at STP)?

2. What volumes (at STP) of hydrogen and oxygen gas are produced by decomposition of1.0 g of pure water?

3. What volume of ammonia gas (at STP) and what mass of pure sulfuric acid must reactto produce 20.0 g of ammonium sulfate?

4. Carbon monoxide gas reacts with heated copper(II) oxide to produce copper metal andcarbon dioxide gas. Excess CO gas is allowed to react with 20.0 g of CuO until it has

2been converted completely to copper. What volume of CO (at STP) is produced?

5. Lead sulfide, heated in air, produces lead oxide and sulfur dioxide gas. Oxygen makesup one-fifth of the volume of pure air (most of the rest is nitrogen gas).a) What volume of oxygen will react with 1.0 tonne (= 1 000 000 g) of lead sulfide?b) Assuming that all the oygen in the air used combines with sulfur, what volume of air

is needed to react with 1.0 tonne of lead sulfide?

44

One litre = 1 000 mL. Since 1.0 mL = 1.0 cm , then 1.000 litre = 1000 cm .1 3 3

One decimetre (dm) = metre = 10 cm.

(1.0 dm) = (10 cm) .3 3

1.0 dm = 1000 cm = 1.000 L.3 3

CALCULATIONS INVOLVING SOLUTIONSINTRODUCTION AND DEFINITIONS

Many chemical reactions take place in aqueous (water) solution. Quantities of suchsolutions are measured as volumes, while the amounts of solute present in a given volumeof solution are the concentrations of the solutions.

The concentrations of solutions are commonly stated in moles (of solute) per litre (ofsolution)(mol. L ), or as Molarity (M), which means the same. In practical terms, however,-1

amounts of solutes have to be measured as mass, for example in grams. It is not possibleto measure the number of moles of a solid directly with a balance or any other instrument.Concentrations of solutions can also be expressed, therefore, in grams (of solute) per litre(of solution) (g. L ).-1

Grams per litre is therefore a practical way to express concentration of solute in a solution.Moles per litre, or molarity, is more important in making calculations and predictions aboutchemical reactions.

Volumes of solutions are defined in litres, (L), (which is sometimes expressed as dm , or3

cubic decimetres ). Millilitres (mL) are also used widely, but may have to be converted to1

litres for some calculations.

FORMULAE USEFUL FOR CALCULATIONS:It is important, therefore, to be able to convert grams to moles and moles to grams for anysubstance.

Number of moles = or n =

This can also be written:Actual mass (grams) = Number of moles x Molar mass, or m = n x M

Concentration of a solution can be expressed as a formula:

Concentration (in mol. L ) = or C = -1

This can also be written:Number of moles of solute = Concentration (mol. L ) x Volume of solution (litres)-1

or n = C x V

It is suggested strongly that these formulae be remembered as word formulae, rather thanas algebraic formulae. With word formulae, the student is remembering the meanings of theparts of each formula. It is easy, with the algebraic formulae, to confuse n and m and M.

45

Example one:A solution of sodium hydroxide has a concentration = 0.50 mol. L (also written as-1

0.50M). How many grams of sodium hydroxide are dissolved in 1.000 L of solution?

Calculate first the number of moles in the 1.000 L of solution.

Concentration = 0.50M = =

Number of moles of solute = 0.50 x 1.000 = 0.50 molSodium hydroxide: formula = NaOH, molar mass = (23.0 + 16.0 + 1.0) g = 40.0 g

One mole of NaOH = 40.0 g, so 0.50 mol of NaOH = 20.0 g.20.0 g of solid sodium hydroxide dissolved in 1.000 L solution makes a solution of

concentration = 0.50M.

Example two:What mass of lead nitrate should be dissolved in 250 mL of water to make a solutionof concentration 0.040M?

3 2Lead nitrate: formula = Pb(NO ) , molar mass = 331.2 g 250 mL = 0.250 L

Concentration = 0.040M = =

Number of moles of solute = 0.040 x 0.250 = 0.010 molMass of lead nitrate required = (0.010 x 331.2) g = 3.31 g.

Example three:What is the concentration of the solution if 10.0 g of sodium carbonate is dissolvedin 200 mL of solution?

2 3Sodium carbonate: formula = Na CO , molar mass = 106.0 g 200 mL = 0.200 L

2 3Number of moles of Na CO in 10.0 g = = 0.0943 mol.

Concentration = = = 0.47 M

Example four:What volume of solution must be made if 5.10 g of silver nitrate is to be dissolved tomake a solution of concentration = 0.050M?

3Silver nitrate: formula = AgNO , molar mass = 169.9 g

Number of moles of silver nitrate = = = 0.0300 mol

Concentration = 0.050 =

Volume = L = 0.600 L = 600 mL

Calculations of this kind should be practised until they can be done quickly and easily.

46

Exercises: 1. How many moles of calcium hydroxide are needed to make 2.000 L of 0.001M

solution? What mass of calcium hydroxide would be required?

2, What volume of 0.05M potassium sulfate contains 50.0 g of potassium sulfate? (Hint:first convert the number of grams to number of moles.)

4 2 3. What is the concentration of magnesium sulfate if 20.0 g of crystals (MgSO .7H O) aredissolved in 500 mL of solution?

4. What is the mass of hydrogen chloride in 2.50L of 10.5M hydrochloric acid solution?(Hydrochloric acid is a solution of hydrogen chloride gas in water. "Concentratedhydrochloric acid" is approximately 10.5M).

5. What mass of iron(II) sulfate is present in 25.00 mL of 0.10M solution?

CALCULATIONS INVOLVING PARTICULAR IONS IN A SOLUTIONConsider a solution made by dissolving one mole of solid calcium chloride in one litre ofwater to make a solution that is 1.00M in calcium chloride.

2The one litre of solution contains 1.00 mol of CaCl . It can also be said to contain 1.00 molof calcium ions and 2.00 mol of chloride ions.

2This can be shown symbolically [CaCl ] = 1.00M [Ca ] = 1.00M [Cl ] = 2.00M2+ -

(The square brackets, [ ], mean "concentration of ".)

Further examples:1. What is the concentration of ammonium ions in 0.25M ammonium sulfate solution?

4 2 4Formula for ammonium sulfate is (NH ) SO . In a given volume of solution, there aretwice as many moles of ammonium ions as there are moles of ammonium sulfate or ofsulfate alone.

4 4 4 2 4[NH ] = 2[SO ] = 2[(NH ) SO ]+ 2-

4Therefore [NH ] = 0.50M+

2. What is the concentration of nitrate ion in 0.40M aluminium nitrate solution?

3 3Formula for aluminium nitrate is Al(NO ) .

3Therefore [NO ] = 1.20M-

3. What are the concentrations of iron(III) and of sulfate ions in 0.10M iron(III) sulfate?

2 4 3Formula for iron(III) sulfate is Fe (SO )

4Therefore [Fe ] = 0.20M and [SO ] = 0.30M3+ 2-

4. In a solution of silver sulfate in water, the concentration of silver ions is 0.0004M. Whatis the concentration of sulfate ions in the solution?

2 4Formula for silver sulfate is Ag SO

4 4[Ag ] = 2[SO ] Therefore [SO ] = 0.0002M+ 2- 2-

47

Exercises:1. What is the concentration of

a) lead ions in a 0.02M solution of lead nitrate?b) nitrate ions in a 1.0M solution of zinc nitrate?c) ammonium ions in a 0.40M solution of ammonium chloride?

3d) iron(III) ions in a solution of iron(III) nitrate where [NO ] = 0.45M?-

e) iodide ions in a solution of magnesium iodide where [Mg ] = 0.20M2+

2. What mass of solid would need to be dissolved in 250 mL of water to make a solutionin which a) [Cl ] = 0.50M, using potassium chloride?-

4b) [SO ] = 0.50M, using sodium sulfate?2-

4 2c) [Fe ] = 0.05M, using iron(II) sulfate-7-water (FeSO .7H O)?2+

4d) [NH ] = 0.02M, using ammonium hydrogenphosphate?+

e) [Cl ] = 0.60M, using iron(III) chloride?-

3. What is the molarity of chloride ions in each of the following solutions?a) 5.85 g of sodium chloride are dissolved in 500 mL of solution.b) 1.1 g of calcium chloride is dissolved in 100.0 mL of solution.c) 13.3 g of aluminium chloride dissolve in 600 mL of solution.d) 9.5 g of magnesium chloride and 0.74 g of potassium chloride are dissolved together

in 250 mL of solution.e) 1.47 g each of zinc chloride, potassium chloride, and iron(III) chloride are dissolved

together in 750 mL of solution.

4. To make a solution that is 0.10M in chloride, what volume of solution should be madea) if 10.7 g of ammonium chloride is dissolved to make the solution?b) if 22.2 g of calcium chloride is dissolved to make the solution?

CALCULATIONS INVOLVING CONCENTRATIONS OF SOLUTES INCHEMICAL REACTIONSThe basic procedure here is similar to that for other calculations:1) Write a balanced equation.2) Write mole ratios under the equation, to show the ratio of moles of reactants to moles

of products. 3) Write in the given data for concentrations and volumes, writing x for unknown values.4) Convert values of concentration and volume to moles. 5) Use ratios to calculate x. The ratio between the actual numbers of moles of the reactants

(line 4 in example one on the next page) and the number of moles shown in the equation(line two in example one on the next page) are equal (line 5 in example one on the nextpage).

(Examples next page)

48

Example one:What volume of 0.25M hydrochloric acid will exactly neutralise 40.0 mL of 2.00M sodiumhydroxide solution?

1) Write a balanced equation:2) Write mole ratios under the equation.3) Write in the given data for concentrations

and volumes, writing x for unknown values.4) Convert values of concentration and volume

to moles.

5) Use ratios to calculate x: =

The factor "/1000" converts the given volumes from millilitres to litres. Since it is acommon factor, it is cancelled from the next step.

x = = 320 mL

Example two:Silver nitrate solution of concentration 0.025M is added to 50.0 mL of sodium sulfidesolution, causing silver sulfide to precipitate. 21.5 mL of the silver nitrate solution was justsufficient to precipitate all the sulfide ions present. What was the concentration of sulfideions in the sodium sulfide solution?1) Write a balanced equation.2) Write mole ratios under the equation.3) Write in the given data for concentrations and volumes, writing x for unknown values.4) Convert values of concentration and volume to moles.

When the factor 1000 has been cancelled, as in the previous example, the product ofthe concentration and volume in millilitres can be stated in millilmoles (mmol).

5) Use ratios to calculate x. The ratios are . If the units used in both ratios are

the same, the ratio can be calculated in the usual way.

=

2Molarity of Na S = x = = 0.0054M = 5.4 × 10 M-3

NB the concentrations of sodium ions and of nitrate ions in the mixture are equal.

3[Na ] = 2 × ( ) × 5.4 × 10 M = 0.0075M, [NO ] = ( ) × 0.025M = 0.0075M+ -3 -

49

Example three:When iron(II) sulfate solution is added to a solution of potassium cyanide, it reacts to form

6the hexacyanoferrate(II) ion, Fe(CN) . What is the largest volume of 0.50M iron(II) sulfate4-

solution that would react completely with 120.0 mL of 0.40M KCN solution? "Largest volume that would react" is the volume that is needed to react exactly. If asmaller volume were used, there would be excess of cyanide solution.

= x = = 16.0 mL

Example four: This example varies from the previous ones in that it involves an excess of one reactant.30.0 mL of 0.25M sodium hydroxide solution is added to 12.5 mL of 0.40M sulfuric acidsolution. What is the concentration of unreacted sodium hydroxide or sulfuric acid in thesolution after the solutions are mixed?

= 7.5 mmol = 5.0 mmol

2 4The equation shows that 2 mol of NaOH react with 1mol of H SO , so 7.5 mmol of NaOH

2 4require only 3.75 mmol of H SO for complete reaction. Sulfuric acid is in excess; the

2 4excess of H SO is (5.0 - 3.75)mmol = 1.25 mmol, or 1.25 x 10 mol.-3

Students uncomfortable with using millimoles can work with moles by inserting the factor1000 to convert millilitres to litres wherever necessary: for example, the calculation abovewould appear as

= 7.5 × 10 mol = 5.0 × 10 mol-3 -3

2 4 2 4To calculate the concentration of H SO in the final solution: the number of moles of H SO isknown, as is the volume of the solution : (30.0 + 12.5) mL = 42.5 mL = 42.5 × 10 L.-3

The relationship concentration = can be used:

2 4 4Concentration of H SO = = 0.024M (Actually, HSO ions are formed.)-

50

Example five: This is a more difficult problem.What volume of 0.10M silver nitrate solution should be added to 25.00 mL of 0.10Mcalcium chloride solution to leave a final chloride concentration = 0.05M?

The student must be aware of several things. Silver ions form insoluble AgCl when chlorideis present. The initial concentration of calcium chloride is given as 0.10M, but the chlorideconcentration is 0.20M. It is inferred from the wording of the question that calciumchloride will be present in excess.

A chemical problem, however complex it may seem, should be started by writing achemical equation and filling in the data under the equation in the way demonstratedpreviously. Unknown quantities, such as, in this case, the volume of silver nitrate solution,should be written as an algebraic symbol (such as x). Additional information may need tobe inserted in the space under the equation.

In problems such as this, it is often easier to use an ionic equation than a molecularequation (see pages 23 - 24).

Total volume of solution after mixing = (25.00 + x) mL

Concentration of chloride, [Cl ], after mixing = = 0.05M-

The equation in the previous line can be solved to give x = 23.0 mL

Exercises:1. What volume of 0.12M potassium hydroxide will exactly neutralise 40.0 mL of 0.15M

nitric acid?

2. What volume of 0.010M calcium hydroxide solution ("lime water") is neutralised exactlyby 25.00 mL of 0.026M hydrochloric acid?

3. What is the concentration of lead ions in a solution if they are completely precipitatedfrom 24.0 mL of solution by 32.0 mL of 0.28M potassium iodide. (Lead iodide isinsoluble in cold water.)

4. What is the smallest concentration of sulfate ions that must be present for 22.0 mL ofsodium sulfate solution to precipitate all the barium in 20.0 mL of 0.15M bariumchloride solution? (Barium sulfate is insoluble in water).

51

5. 25.0 mL of 0.065M sodium hydroxide solution is mixed with an equal volume of0.030M sulfuric acid solution. What is the concentration of sodium hydroxide in themixed solution?

4 6. Mercury(II) and iodide ions react to form an ion HgI (called tetraiodomercury(II)).2-

What is the smallest volume of 0.40M potassium iodide solution that should be mixedwith 10.0 mL of 0.10M mercury(II) chloride to convert all the mercury(II) ions totetraiodomercury(II) ions?

7. Potassium dichromate solution is mixed with acidified sodium oxalate solution. Theproducts of the reaction are chromium(III) ions and carbon dioxide. (See page 7 forformulae, page 28 or page 30 for methods of balancing equations.) 20.0 mL of 0.025Mpotassium dichromate solution was reduced completely by 30.0 mL of sodium oxalatesolution. What is the concentration of chromium(III) in the final solution?NB there is a quick way to calculate the answer to this question.

8. Equal volumes of 0.020M sodium carbonate and 0.030M calcium nitrate solutions aremixed. What are the concentrations in the final mixture of a) sodium ions? b) nitrate ions? c) calcium ions? d) carbonate ions?NB calcium carbonate is insoluble in water.

9. If hydrogen peroxide solution is added to acidified potassium permanganate solution,the permanganate ions are reduced to Mn and oxygen gas is produced. What volume2+

of 0.040M hydrogen peroxide will exactly reduce 20.00 mL of 0.0080M potassiumpermanganate solution? (See page 28 or page 30 for method to balance equation).

10. Excess solid zinc oxide is reacted with 40.0 mL of 1.0M hydrochloric acid until all ofthe acid has reacted. What is the concentration of zinc ions in the final solution? Howmany moles of zinc chloride will have been formed?

The following exercises are more difficult. Remember that even if you cannot see howto solve a problem, start by writing an equation (molecular or ionic - see page 23) andthen entering mole ratios and given data under the equation, using algebraic symbolsfor thm quantities that are unknown and have to be found.

11. Both lead chloride and lead bromide are insoluble in cold water. 20.0 mL of 0.12Msodium chloride solution are added to 50.0 mL of 0.05M lead nitrate solution. Whatvolume of 0.20M potassium bromide solution would be needed to precipitate theremaining lead from the solution?

12. Barium sulfate and zinc hydroxide are both insoluble in water. Both substances will beprecipitated if barium hydroxide solution and zinc sulfate solution are mixed. If 20.0 mLof barium hydroxide solution reacts exactly with 60 mL of zinc sulfate solution, whatis the ratio of the concentrations of the two solutions?

13. Barium sulfate is insoluble in water. What volume of 0.10M barium hydroxide solutionshould be added to 20.0 mL of 0.15M sulfuric acid solution so that the concentrationof sulfuric acid in the final solution is 0.10M?

52

CALCULATIONS COMBINING MASSES, GAS VOLUMES, AND CONCENTRATIONS OF SOLUTIONS

Calculations about the amounts of reactants and products in chemical reactions ofteninvolve some combination of mass, gas volume, and volume and concentration of solutions.There may be more than one method to obtain a solution to a problem, but it is still alwaysnecessary to start with a balanced equation.

In the first line under the equation, write the mole ratio from the equation.

Then, in the second line,

a) where masses are involved, write in the molar masses of the materials

concerned, multiplied by the number of moles shown in the equation.

b) where gas volumes are involved (at STP), write in 22.4 L, multiplied by the

number of moles of gas shown in the equation.

c) where solution volumes and concentrations are involved, do not enter any

values in the second line.

In the third line, enter the values given in the question, using algebraic symbols for

unknown values that must be calculated. Where solution volumes and

concentrations are involved, calculate and enter the number of moles present.

Ratios can then be written, using , , and .

It should be checked that units of mass and volume used are the same.

Example one:What volume of carbon dioxide gas (at STP) can be obtained by heating 20.0 g of solid zinccarbonate?

= x = L = 3.57 L

Example two:Calculate the mass of magnesium that will dissolve in 100.0 mL of 0.50M hydrochloric acid.What volume of hydrogen will be produced?

= =

2x = mass of Mg used = = 0.61 g, y = volume of H produced =

= 0.56 L

53

Example three:If carbon dioxide gas is passed into a solution of calcium hydroxide (lime water), calciumcarbonate settles as an insoluble precipitate. Carbon dioxide was passed into 500 mL of limewater until all the calcium hydroxide had been turned into calcium carbonate. The calciumcarbonate formed was separated, dried, and weighed. There was 0.40 g of dried calciumcarbonate. What was the molarity of the original calcium hydroxide solution?NB that in this question, an excess of carbon dioxide is used, so the gas volume need notbe considered in the calculation.

= x = M = 0.008M = 8 x 10 M-3

Molarity of calcium hydroxide = 0.008M = 8 x 10 M-3

Example four:When hydrogen sulfide gas is passed into a solution containing copper(II) ions, insolublecopper(II) sulfide is formed. a) Will 200 mL of hydrogen sulfide gas (at STP) be sufficient to remove all copper ions

from 50 mL of 0.10M copper(II) sulfate solution? b) What mass of copper(II) sulfide will be formed if all the copper is precipitated?

2It is not clear from inspection of the data whether there is enough H S present to turn all

2copper(II) to CuS. It is suggested, therefore, that the volume of H S that will precipitate

20.0050 mol of Cu as CuS be calculated. Let the volume of H S needed be y L.2+

= y = (0.0050 x 22.4)L = 0.112 L = 112 mL

2200 mL of H S provides an excess of gas, and all the copper will form CuS.

Now the mass of CuS, which is x g, can be calculated:

= x = (95.6 x 0.0050) g = 0.48 g

54

Example five: This is a more difficult problem.Potassium permanganate solution reacts with sodium oxalate solution (for oxalate ion, seepage 7), in the presence of excess sulfuric acid, to produce Mn ions and carbon dioxide.2+

Enough 0.040M sodium oxalate solution must be made to react with 1.000 L of 0.020Mpotassium permanganate solution.What mass of sodium oxalate solid should be dissolved, in how much water, to make therequired solution?As usual, it is necessary to start with an equation, and to write the mole ratios of reactants and/orproducts under the equation. If data are entered under the equation in the usual way, a method ofsolution may become apparent.There are however, two choices: should the volume and concentration of the sodium oxalatesolution be written under the equation, or the required mass of sodium oxalate? Either course maybe followed, as is demonstrated below.Methods for writing a balanced redox equation are explained on pages 28 and 30.

Method one: using volume and concentration of sodium oxalate:

=

x = ( )L = 1.25 L = volume of 0.040M sodium oxalate solution needed.

Number of moles of sodium oxalate needed (see page 44) = 0.040 x 1.25 = 0.050 mol

2 2 4Molar mass of sodium oxalate, Na C O = 134.0

Mass of sodium oxalate required (see page 44) = (0.050 x 134.0) g = 6.7 g

Method two: using mass of sodium oxalate:

4 2 4 2 22MnO (aq) + 5C O (aq) + 16H (aq) 6 2Mn (aq) + 10CO (g) + 8H O(l)- 2- + 2+

2 mol 5 mol 5×134g1.000L0.020M= 0.020mol x g

=

x = = 6.7 g = mass of sodium oxalate needed to make the solution.

The number of moles of sodium oxalate = = 0.050 mol.

Concentration = , so 0.040M =

2 2 4 2 4V = L = 1.25 L, which will dissolve 6.7 g of Na C O to make [C O ] = 0.040M2-

55

Example six: Another more difficult problem.It is not possible to make a solution of iodine of exactly known concentration by weighinga sample of solid iodine and dissolving it in a known volume of solvent.One way to make a solution of iodine of exactly known concentration is to add dilute

3 3sulfuric acid and bromate ions (BrO ), as potassium bromate (KBrO ), to a solution-

containing excess potassium iodide. The bromate oxidises iodide to iodine, and is itselfreduced to bromide.

What mass of potassium bromate should be used, and what is the minimum mass ofpotassium iodide needed, to make 1.000L of 0.0100M iodine solution?

It is easier to write a balanced ionic equation first, but this should then have spectator ionsinserted to make it a molecular equation, since masses have to be calculated for both

3KBrO and KI. (See pages 23, 24, 28, 30 for methods of writing equations).The question asks for "the minimum mass of potassium iodide needed". In the calculationbelow, the stoichiometric (see page 40) mass of potassium iodide is calculated: this is the"minimum mass needed". More than the minimum is then used to make the solution.

= =

x = mass of potassium bromate needed = ( ) = 0.557 g

y = minimum mass of potassium iodide needed = ( ) = 3.32 g

The main difficulty with a question like this is in interpreting the question. Once theequation has been written and the data written in under the equation, it is no more difficultthan any of the preceding ones.

Summing up, therefore: for problems involving stoichiometry (see page 40):1. WRITE THE EQUATION2. WRITE IN MOLE RATIOS UNDER FORMULAE IN EQUATION.3. WRITE IN MOLAR MASSES OR MOLAR GAS VOLUME (22.4 L) UNDER

FORMULAE OF MATERIALS INVOLVED IN THE QUESTION.4. WRITE IN GIVEN DATA ON MASS, GAS VOLUME, SOLUTION VOLUME AND

CONCENTRATION, USING x, y, etc FOR UNKNOWN QUANTITIES.5. CALCULATE x, y, etc BY USING RATIOS

56

Exercises on stoichiometry: mass, gas volume, and solution volume andconcentration are involved together in these exercises.Assume that all gas volumes are given at STP (see page 42).

1. Ammonia gas can be produced by heating a mixture of solid sodium hydroxide withsolid ammonium sulfate. What mass of ammonium sulfate should be heated with excesssodium hydroxide to produce 2.00 L of ammonia gas?

2. Hydrochloric acid is a solution in water of hydrogen chloride gas. What will be theconcentration of the hydrochloric acid if 400 mL of hydrogen chloride are dissolved in100 mL of water?

3. Will 10.0 mL of 0.20M calcium chloride solution be sufficient to precipitate all of thesilver from 60.0 mL of 0.020M silver nitrate solution?

4. Excess solid zinc carbonate is treated with 40.0 mL of 2.0M hydrochloric acid. Whatvolume of carbon dioxide is likely to be formed? What will be the concentrations of zincand chloride ions in the final solution? (NB zinc chloride is very soluble in water.)

5. Potassium dihydrogenphosphate is going to be made by reacting solid potassiumcarbonate with 100.0 mL of 1.00M phosphoric acid (see page 7 for formulae, page 5for names and formulae of acids). What mass of potassium carbonate is needed?

6. A 25.00mL sample of 0.1012M hydrochloric acid is exactly neutralised by 23.47 mLof sodium hydroxide solution. It needed 27.33 mL of the same sodium hydroxidesolution to neutralise a second 25.00 mL sample of hydrochloric acid. What is theconcentration of the second sample of hydrochloric acid?(HINT: there is a short way to solve this problem, that does not require that theconcentration of the sodium hydroxide solution be calculated.)

7. To make 10.0 g of lead iodide, lead nitrate is going to be mixed with 100.0 mL of0.50M potassium iodide solution. What mass of solid lead nitrate is required? Howmuch excess potassium iodide is being used?

8 18 8. Petrol is mainly octane, C H . The mass of 1.00 L of octane is 698 g. What volume ofcarbon dioxide, and what mass of water, are formed when 1.00 L of petrol is burnedto form carbon dioxide and water only?

9. When potassium chloride is oxidised, chlorine gas is formed. What volume of chlorine would be formed if 20.0 g of potassium chloride is oxidised? (HINT: is a full equationnecessary? See page 30).

410. Carbon dioxide and water only are formed when 2.00L of methane gas (CH ) is burnt.If all of the carbon dioxide formed is passed into a solution containing excess bariumhydroxide, what mass of barium carbonate should be formed?

11. A 20.00 g sample of a mixture of calcium carbonate and calcium sulfate was treatedwith excess hydrochloric acid. The carbon dioxide produced was all recovered and itsmass was found to be 3.30 g. What mass of calcium sulfate was present in the sample?

57

ANSWERS TO EXERCISESPage 15: writing formula:

3 2 3 3 2 2 2 3a) CaO b) LiF c) Na N d) K S e) AlCl f) Ca N g) Li S h) AlN i) Al O

Page 29: equations for balancing:

Page 34: equations for balancing:

Page 39:1. 2.0 g 2. 8.0 g 3. 89.2 g 4. 2.3 g 5. 6.7 g

Page 41:1. 4.9 g 2. C is in excess by 49 g 3. No: 13.1 g of zinc are needed

Page 43:

2 2 3 2 41. 4.6 g 2. H = 1.24 L O = 0.62 L 3. NH = 6.8 L H SO = 14.8 g4. 5.6 L 5. a) 1.4 x 10 L of oxygen b) 7.0 x 10 L of air.5 5

Page 46:1. 0.002 mol, 0.15 g 2. 5.74 L 3. 0.162M 4. 958 g 5. 0.38 g

58

Page 47:1. a) 0.02M b) 2.0M c) 0.40M d) 0.15M e) 0.40M 2. a) 9.3 g b) 17.8 g c) 3.5 gd) 0.33 g e) 8.1 g 3. a) 0.20M b) 0.20M c) 0.50M d) 0.84M e) 0.091M4. a) 2.0 L b) 4.0 L

Page 50:1. 50.0 mL 2. 32.5 mL 3. 0.19M 4. 0.14M 5. 2.5 x 10 M 6. 10.0 mL-3

7. 0.020MThe "quick way" mentioned recognises that one mole of dichromate ions produces two

2 7moles of Cr ions. As the solution volume increases from 20 mL to 50 mL, [Cr O ]3+ 2-

2 7decreases from 0.025M to 0.010M. However, at the same time, Cr O ions are2-

2 7changed to Cr ions, so [Cr O ] = 0.010M becomes [Cr ] = 0.020M 3+ 2- 3+

8. a) 0.020M b) 0.030M c) 0.0050M d) zero Explanation: initially, since equal volumes are mixed, the volume doubles so the con-centations of the ions is halved, so [Na ] = 0.020M (0.040M in the initial solution+

2 3 3 3because of two sodium ions in Na CO ), [CO ] = 0.010M, [Ca ] = 0.015M, [NO ]2- 2+ -

= 0.030M (initially doubled like the sodium ions). The amount of calcium ions exceedsthe amount of carbonate ions, so in the reaction all the carbonate precipitates, leaving0.005 mol L of calcium ions in the solution. Sodium and nitrate ions are spectator-1

ions, so are unaffected by the reaction.

2 49. 10.0 mL 10. [Zn ] = 0.5M, 0.040 mol 11. 13.0 mL 12. [Ba(OH) ] = 3 x [ZnSO ]2+

13. This problem resembles Example five on page 50. The method of solution suggestedbelow is a little different from that used on page 50.

2Let volume of Ba(OH) solution = x

2 4Initial amount of H SO = 20.0 × 0.15 = 3.00 mmolAmount of barium hydroxide added = 0.10x mmol = amount of sulfuric acid reacted.

2 4Final [H SO ] = (3.00 - 0.10x) mmolFinal volume of mixture = (20.0 + x) mL

2 4Final [H SO ] = = 0.10

This solves to give x = 5.0 mLPage 56:1. 5.9 g 2. 0.18M 3. Yes: there are 4 mmol of Cl to 1.2 mmol of Ag- +

24. 0.90 L of CO , [Zn ] = 1.0M, [Cl ] = 2.0M 5. 6.91 g 6. 0.1178M2+ -

7. 7.2 g of lead nitrate; excess KI = 0.0067 mol or 13 mL of the solution used.

2 28. Volume of CO = 1097 L at STP, mass of H O = 992 g9. 3.0 L Explanation about use of a half-equation: the question does not say what oxidises

the chloride, so only a half equation can be written. Potassium is a spectator ion, andmust be included in the equation since the question involves a mass of KCl. (It isassumed in the equation that KCl is a solid, although the question does not say so.)

410. 17.6 g 11. 12.49 g of CaSO