Book on Moles / Stoichiometry for O'Levels and A'levels (Grade 9/10/11/12)

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AII:i&hts, ... ""ed.Noparlof,h,ispublic
..,ionmaybereproduoed.>toredin.
retrltV.!syStem or ,ran.mttttd,n
any rOfTfIOJ by.oy me.ns,electronie.
me.:h.niraJ,phororopying.
",.ording orothe,wi$t."';,hou,
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. ... 'SlOnl.yThornes(Publi>h.rs)L'd.OldS
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tAcckhampron. CHELTENH.y.I GL5J ODN. Englond.
Firstpublisllcdinl981
SecDndedit;onpubli.l>edinI981
rhin!edit;onpubli>hedinl993by:
SllUIley Thomcs (Publishers) LId
Elknborou811 Hou~e. Wellington Street. CHELTENHAM

GLSO J YW

book i.,vailable

1211
from 'he Bri,ish Ubrary

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A calalogue n:con!ofthis

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ACKNOWU;OGEMENTS
~'~~k~~:,~~l~~;;~~

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1'11. A"oci .. td h.minin~ Boa,d, The joint M.. ';cul.r.ion Boa,d, The Nurthern
1",land 5£hool." E~""in .. ;on, .nd A ..... m.n' CouMiI;The Od",d and Cambridg.
5<:hool. h.m,n.tlon
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Un;~";'y <>f Camhridge Sd"",,)o Loal Examination. Syndicate, The Uni~<rsi'y of
London School Ex.min.,ion'Council;Thc
""el 'h join, EducationComm in ..

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Many hum.rinlvalue,h
.. ebun Ukcnf,om
th<Cb ...";<rryl),,, .. B,,oi by
J GS.>rk and HG W.lla"" (publi,hcd by Juhn Murr.y). forh.lp wi,h ddini'ion •
• nd physical COnl. n ... ,dcfI'n"" h•• b<:.n mad. '0 Pby,;co-cbn .. icai Q"",,!j,j~,
.
~mlUn'-" by M L McGI.,h.n (published by The Roy.1 In"itut< of Chem""yl
Ihovebeenforluna,einn:
.. i .ingcx«llent.dvicef,oml'rof
.. ..,rRII.B :t1dwin,
Prof • .."r II. P nell. fRS. Dr GH D.~i ••. P,of • ..",WCE
Higgin.on. Dr KA
Holbrook. Dr RBMoy ... ndD,j RShorter. I ,hankth.",
chemi ... for 'he
l>eJp.heyhovegiv<nme.J.mindobtedloMrJPDT.ylo,forcll«kingthe
Ul_"" to the problem> in the First Edit;on .nd for m.ny v,tu.hle commentS
.oodco,,",,«ion.
Ith.nkSunleyThorn
,he;,.neuu",sem<nl

•• (Publi.h .... lfor'''.i'coll.bontion.ndmyfamilyfor
E/I."Ram,d."
Oxford 1993

Typesc:tbyTc-ch-S.t.C.,eshc.d.TyncIlr.W
PrintedinG,,,,,,BrilainbyRedwoodRoob.

•• ,
Trowbridge. Wiltshi",

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Contents
Acknowledemcms

ForewordbyProfes,orRl>

een r-es

vii

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Furmulae and egualions
Formulae 17 cguatioos18

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e 1l~rions2-a trian )efor rearran in'
4 - eft ell anons on ratio 5 - wor 'In
ro enter ex onenrs on a
-10 arithms 7-antilo
-choice of a calc ulator10ap s 1 -

ellcnl;]fjoD o f molar m ass 29

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percentage

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The mole

relative molecular mass 21

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Relative atomic mass 21
composition 23

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Eqll~tionsandthcmolc
32
Calculation based on chemical equations 32 using:the masses of the
reactants to work out ,he equation for a reaction 36 percent:lg<>
yield38
Jimiriog.cactant39

(;

Findin fomlulac
Empirical formub~ 41

7

Kcacringvolumcsofpscs
Gas molar volume 46 finding formulae by combustion 49

8

VolumclricAnal-sis
Concentration 59 acid-base nnaticns 62 oxjdMjon=f!'dllctjon
rCJC[iom69 potassium man~anate(V!!)tirmtions 76 potassium
dichromate(Vl} titrarions 79 sodium thiosulphatc rilratiOQS 80
cotnplexomerrictitrations88
precipitltiontitrations90
Mass Sp~ClrOmetlyI 04

molccubr formulae 42

nuclear reactions 111

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Materi~1

Th~gaslawsJl5

Boy1c'sla",

1 15

Char1cs'law115

of state lor an ideal gas 115
Graham's
law ofdiffusiof]
gas molar volume
119
Dahan's
law of panial
pressures
ideal g"-' ~quatiolll22
the kin~tic 'heory
of gases 123
II

Liuuids
Drrcrmmarinn o t mpl;)r m.ss 127

lhctqualion
117
120

the
[he

127
aOprn.!p1s '''<l]I, trom m ea" re-

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156-cakuIation oftne degree of dissociationand the dissociation
constant of a weak electrolyte from ~onductanC<'measurements158
- calculation of pH, pOH and pK 159 - buffer solutions 167solubility and solubility product 169-the common ion effect 171
-electrode potentials 175 - galvaniccdb 176

Readinn Kiuetic,
Rcoct"ou rare 223
the effect ofCDun'otrarj"o
00 wrof!!'3ct;OQ
224 Qait'r of (radon 225 firu-ordrr r eaerinns 225 half_He

225 pseudo-fir:5t-orderreactions226 second~rdtrreactions.with
reactants of equal concentration 227 zero-order reactions 229
the effect oftemperarure on reaction rates 229
15

Euilinri ..
Chemical equilibrium 248

the equillbrium law 248

16 OrganieCbemistry

260

Answers to Exercises
Table of Relative Atomic Masses
Periedic Table cf the Elemcnts

294
-2~

~a_

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List of Exercises
15
20
23
25
30
lJ

ppctjcewjrbCalclIlaljons

Practice with Eguatiolls
Problems 011Relative Molecular M=
Problems on Perl'entage Composition

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40

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of Gases
Combustion

120

29 Problems on Partial Pressures of Gases
30 Problemson the KineticTheory and rhe Id~alGas Equation
31 Quesrions from A·levd Papers
32 zmhlsrns Oil Molar M~s<e~ofV"lati!e Substances
13 rrcblcms on Associarjona"dDissociation
34 Prol>lemson Vapour Pressures of Liquids
ll.......fwbll:lllullI ..li.morj·Prc5S!rr
i
36 Prohlems on Vapour Pr<'ssuresof Solutions of Two Liquids
37

121
12)

124
30

ur

112

13.
IU

137

proh kmsQnSt,...mDj5Ji)!ariQIl

38
39
40
41

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54
61
66
74
85
89
9J
92
105
0.
111
2
117
IJB

ProblemsonPartirion
Qucsrion.fromA-levdPapcrs
Problemson Electrolysis
Problems on Molar Conductivity

140
141
151
59
164

~:~~rt~jfm

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49
50
51
52
5~
54
55
56
57
58

Problems on Solubility Products
172
Problems on Standard E1cctrodePoteorials
178
Question, from A·lcvcl Papers
179
ProblemsonStandardEnthalpyofNcutralisauon
192
Problems on Standard Emhalpy of ncaction and A~'cragcStandard
Bond Enthalpies
102
Problems on Standard Entropy Change and Standard Frc~ Energy
Change
209
Questiol1sftom A-!cvelPapers
Problems on Finding the Order of Reaction
232
Problems on First-order
Problems on.Second-order Reactions
234
Problems on Radioactive Decay
2J5
Problems on;Rates of Reaction
236
Problems on Activation Energy
237
QuestionsfromA-leveIPapen;
217
Problems on Equilibria
m
254
:~brg~j;~:ea~7s7ry
262
Questions from A-level Paper,;

nsasncns

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46
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Some of the exercises are divided into an easier section (Section I)
and a more advanced section (Section 2). questions from A-level
papers are on the immediately preceding topic{s). Each qUe5tion is
appended with the name of the Examination Board and the year
(90= 1990 etc). p indicates a pan question, S an Sqevel question,
AS an Advanced Supplementary question and N a Nuffield syllabus.
The most difficult (often S·level) questions are also denoted by an
asterisk.

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ABBREVIATI0NS OF EXAMINATION BOARDS
AEB
C
JMB
L
NI

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0& C
WJEC

Associated Examining Board
University of Cambridge Schools Local Examinations
Syndicate
Joint Matriculation Board
University of London Schools Examinations Council
Northern Ireland Schools Examinations and Assessment
Council
Oxford Delegacy of Local Examination
Oxford and Cambridge Schools Examinnions Soard
WelshJoint Educa~ion Comminee

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Foreword

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It is a common complaint of university and coUegeteachers of physical
sd~cesthatma!lyoftheirinoomingstudentsareunabletocarryout
even simple calculations, although they may appear to have a satisfactory grasp of the underlying subject matter. Moreover- this is by
no mews a trivial complaint, since inability to solve numerical
problems nearly always stems from a failure to understand fundamental principles, rather than from mathematical or computational
diffirulties. This situation is more likely to arise in chemistry man in
physics, since in the latter subject it is much more difficult to avoid
quantitative problems and at the same time produce some semblance
of understanding.

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In auemptingto remedy this state of affairsteachers in schocls often
feel the lack of a single source of well..::hosen calculations covering
all branches of chemistry. This g~p is admirably filled by Dr J:l.amsden's
collection of problems. The brief mathematical introduction serves
to remind the student of some general principles, and the remaining
sections cover the whole range of chemistry. Each section contains
a theoretical introduction, followed by worked examples and a large
n~mber of problems, some of them from past exarninaticn papers.
Since answers are also given, the book willbe equally useful in schools
and in home study. It should make a real contribution towards
~mproving the facility and understanding of students of chemistry
m their last years at school and inthe~rlypanofthelruniversiry
or college courses.
RPBellFRS
Honorary Research Professor, Universiry of'Leeds.end
formerly Professor of Chemistry, University of Stirling

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Preface
Many topics in O!e!~try involve numerical problems. Textbooks are
not long enough to include sufficient problems 10 gIVcsrudcnts the
practice which they need in order to acquire a thorough mastery of
calculations. This book aims to fill that need

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Chapter! is a quick revision of mathemarical rechniqucs, with special
reference to the use of the calculator, and some hints on how to
tackle chemical calculations. With each topic. a theoretical background is given, leading to worked examples and followed by a large
~umber of problems and a i>Clectio~ questions from past examinaof
tion papers. The theeretica! section IS not intended as a fuU treatment,
rorcplace.atcxtbook,butisindudedromakciteasierforthcsrudent
to use the book for individual study as well as for class work. The
indusion of answers is also an aid to private study.
The material will take students up to GeE A- and S-Ievelexaminations
It will also serve the needs of students preparing for the Ordinary
National Diploma. A few of the topics covered arc not in the A·level
syllabuses of aU the Examination Boards, and it is expected that
students will be sufficiently familiar with the syl1abu~ they are
following to omit material outside their course if they wah. S-Icvel
topics and the more difficult calculations are marhd with an
asterisk.

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Many students are now starting A-level work after GCSF. Science:
Double Award. The coverage of chemica! calculations in the syllabuses
for double award science is slighter than it was in the syllabuses for
GeE Odevel Chemistry and GCSE dlemistry. I have taken the; into
account in the Third Edition. In previous editions, r assumed that
students had mastered wme topics before and only an extension was
needed for Aievel. Since students who have done Double Award
Science ha.ve spent little time OHquantitative work, I have taken a
more gradual approach in the Third Edition. Ihopclhatthisapproach
will also suit students with no previous experience of chemical calcul.1lions.ln the Third Edition I have includ..d mor.. Foundation work on
formulae, the mole, calculations based on chemical e'luation~ and
volumetric analysis than in earlier edition ... A number of topics which
are no longer included in A"level ~l1abuseshavebecndropped,and
the selection of questions from past papers ha,beeH updated
lo'NlIam,dell
0.dord1993

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Basic Mathematics
INTRODUCTION
Calculation,s ar~ a part of your chemistry course. The time you sJ:lcnd
on calculations will be richly rewarded. Your perception of chemistry
will become at the same time deeper and more precise. No one can
come to an undcr;tanding of science without acquiring the sharp,
logicalapproachthatisnecdcdforsolving!lurncricalprobkm~

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To succeed in solving numerical problems you need two things. The
first is an understanding of the chemistry involved. The second is
some facility in ~mple mather:mtics. Ca.lcul~tioru are a perfectly
straightforward matter. A numerical problem gives you some data and
asks you to obtain some other numerical values. The connection
between t~e data you arc given and the information youareas.ked for
is a chemical relationship. You will need to know your chcnusvy rc
recognisewhatthatrelatiomhipis.

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This introduction is a reminder of some of the mathematics which you
studied earlier in your school career. !t is included for the sake of
students who are not studying mathematics concurrently with their
chemistry course. A few problems are included to help you to brush
up your mathcmancal skills before you go on to tackle the chemical
problems.

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USINGEQUATIONS

Scientists are often concerned with measuring quantities such as
pre5Sure, volume, electric current and electric potential difference
Sometimes they find that when one quantity ch.angesanother quantity
changes as a result. The related quantities ate described es dependent
oari(1bic.'. The rdationship between the variables can be written in the
form of a mathematical equation. For "xample, when a rna.. of gl..
expands, its volume increases and its density decreases:
Density ex IIVolume
·lhesign exmeans 'is proportional ro', so the expression means 'densiry
is proportional to l!volume' or 'density is invl'I":5elyroportional to
p
volume'.
The relationship between the volume and density of a mass of gasis:
Density"'--

Mass
Volume

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CslclJl:.riomfo,A-I_IO!/IffJ/,,1Y

Thi,; ~quation c= alsobe written as:
Density .. Mass';- Volum~
and as:

Density .. MassNolume

If you are given two quantities, mass and volume, you can use the
equation to calculate the third, density.
~o;.;;~~:~~~

i~S: d!r:t~~f~f~~~ni~a~;s

21.4g, and its volume

Using the equation
Density = MasslVolume
Density "" 21.4gl7.92cmJ
= 2.70gcm-J

REARRANGING EQUATIGNS

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Notice the units. Since mass has the unit gram (g) and volume has the
unit cubic centimetre (cmJ), density has the unit gram per cubic
cenrimetre Ig cm").

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You might want to use t.he above equation II_! find the mass of an
object when you know us volume and density. It would help to
rearrange the above equation to put mass by itself on one side of the
equation, that is in the form

In the equation

Density=--

Mass
Volume

mass is divided by volum<:,so to obtain mass by itself you must mulriply the right-hand side of the equation by volume. Naturally, YOIl
must do the same to the left-hand side. Then
DensityXVolurne
that is

= ~~XJlD.lume-

Mass .. Density x Volume

Pethapsyou need to use the equation to find the volume of an object
when you know its mass and den.~ity.Then you rearrange the equation
toputvolume~loneononesideofthecquarion_lnrheequHion
Mall = DensityxVolume
the term volume is multiplied by density. To obtain volume on its

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own you must divide by density. Doing the same on both sides of the
equation gives
~",,~il:y"XVolume
Density
-DeMityVolume=~

Density

A TRIANGLE FOR REARRANGING EQUATIONS
A short cut for rearranging equations is to put the quantities into a
triangle. For the equation,
=

YXZ

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the triangle is

PRACTice1

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Cover up the letlcryou want; then what you see is the eGuation you
need to use. If you want F. cover up Y; then you read XIZ, so you
know that V = X/Z. !fyou cover 7., you read XIV, so you know that
Z '" XI}" is the equation you need

1. Draw a triangle

to show

the equation

Mass = Density X Volume
Cover up Density. and write the equstion for Density = ?
Cover up Volume, and write the equation for Volume

=

?

2. The equation relating potential difference, V, resistance, N, and
current,l,is V= R Xt. The equation means just the same written
without the multiplication sign as V = RI. Rearrange the equation
a) in the form I=?, b) in the form R =?
3. Draw a triangk to show the relationship V = RJ
a) Covet up R. Complete the equation R = ?
h) Cover up 1. CompJeTethe "Guarion 1 =?

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Cair:u!ations

forA-!81'fJ1 Olemitlrv

4. Theconccntrationofasolutioncanbeexpressed
Concentraticn

M3!;$ofsolutc
"vo"',o"'m,"'o"', "'0"""-0"
00'

e=

Rearrange the equation a) into the fonn Mass of solute
b) into the form Volume of solution =?
!L ReQlTIUIge equation P = QR
the
b) inro the form R =?

=

? and

into the form Q =? and

a)

CRess·MULTIPLYING

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Once you have understood the ideas behind rearranging equations,
you can uy the method of cross-multiplying. If

H

then by cross-multiplying, you obtain

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ad"" be

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How can you find out whether this is correct? First multiply both
sides of the first eqcation by d.

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Nextmultipl}'bnthsidesbyb

That is

=

b

5!!_ ""

c

"'

7=I>C
ad = be

which is the equation you obtained by cross-multiplying. This shows
that cross'mukiplyingonly PUtSinto practice the method of multiply .
ing or dividing both sides of the equation by the same quantity.
Now that you have the equation

ad

=

be

to obtain an equation fora, divide both sides by ui then
a = beld
to obtain an equation for d. divide by a; then
and similarly,
b

=

adle

c

=

ad/b

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PRACTICE

2

1, The pressure.wolurue and tempcrarurc ofa gas are related to the
gas constant, R,by the equation
/'

R

T"'j!
Rearrange the equation hy crass-multiplying ta obtain equations
far a) Tand b) v_
2. Theresistanceofanc:1ectricalconducrorisgivenby
I< '"

pXI/A

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where R =resistance, o e resistiviry, ''''length and A = crosssectional area. Rearrange the equation to give a) an equation of
the form {J"'? and b) an equation of the form II"'?

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3. Rearrangethe equation

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CALCULATIONSON RATIO

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10give aj an equaticn for p and h) an equauon for q

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Many of the calcularions ,'ou meet involve ratios. You have met this
type of problem in your maths lessons, do not forget how to solve
them when you meet them in chemistry!
EXAMPLE Nancy pal's 78p for two toffee bars. How much does Nina have to
1
p~yforfive of rhc same bars?
You can tackle this problem by the unitary method
If2toffeebarscost78p,
1 toffee bar costs 78/2p
and 5 toffee bar~cost 5 X78/2p '" 195p

=<

£1.95.

EXAMPlE Zinc reacts with dilute acids to give hydrogen. If O.0400g of hydrogen
2
is formed when 1.30 g of zinc reacts with an excess of acid, what maSS
of zinc is needed to produce 6_00g of hydrogen?
Again, the unitary method will help you.
IfO_0400goihydrogenisproducedbyl.30gofzinc,
then 1.OOg of hydrogen is produced by 1.30/0.0400gofzinc
and 6.00g of hydrogen are produced by 6.00X 1_30l0.0400gofzinc
1959ofzinc

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c>/cu/ilriam farA'/.wI

Cilemistry

PRACTICE3
1. If O.020g of a gas has a volume of 150cmJwhat is the volume of
32gof[hegas{atthe~ametemperJtureandpresrure)?
2

SSg of iron{1I) sulphide is the maximum quantity that can be
~btained from the ~action of an excess of sulphur with 56gof
What is the maximum quantity of iron(lI) sulphide that can be
obtained from Lnug ofircn?

3. A firm obtains 80 eonnes of pure calcium carbonate from 100
ronnes of limestone. What mass of limesroue must be quarried
to yield 240 tonnes of pure calcium carbonate?

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WORKINGWITH NUMBERSIN STANDARD FORM

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You arc accustomed to writing numbers in decimal notation, for
example 123677.54 and 0.001678. In working witb lug<' numbers
and small numbers, you will fmd it convenient to write tbem in >I.
different way, known as scientific "alation or standard foml. Tbis

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~=:,

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~:i~irn':Jmp':i~tasc~m~~~ft~r
~~erc::srfad~~' '~eth;c:J
factor is a multiple of ten. For example, 2123 = 2.123 X 10' and
O.OOO167=1.67Xl0-4.
t<P means 10XI0XI0, and 10-" means
If(10X lOX lOX 10). The number 3 or -4 is called the exponent,
and the number 10 is the base. 10l is referred to as '10 to the power 3'
or '10 ro the third power'. You will have noticed that, if the exponent
isincreasedby:,thedecimalpointmustbemovedoneplacetotile
left.
= 250XlOI = 2S00XlOO

2.SXJOJ =.0.25 X lcr' = 25X10'

Since 10°= 1,thislast factor is normaUy omitted
When you m!ltiply numbers in standard form, the exponcms are
added. The product of2X 104and6X 10-1 is given by
(2X lO')X(6X 10-1) = (2X6)X(104X

10-')

= 12X 10' = 1.2XlOJ
In division, the exponents are subtracted
1.44XlO"
4.50Xl0-1

=

!.:±ix~=

4_5010-2

O.J20XI0!

= 3.20XIO'

In addition and subtraction, it isconvcnitnttoexptessnumbcrsusin
the same expo¥Uts. An example of addition is
(6.~OOX102) + (4.00 X io-')

g

= (6.HlOX W') + (0.00400 X to!)
" 6.304 X 10'

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An example

of subtraction

is

1O-l)-(4.20x

10-4)

'"

(3.60 X1O-l)-(0.420X 10-3)

=

(3.60x

3.18X 10-'

Howtoentllrllxponentsonaealeulator
To enter 1.44xI0·, you enter 1.44;then press the EXP key, then
the 6 key
To enter 4.50XlO-1, you cmer4.5; thenpressthc
the 2 key, and lastly the +I-kcy.

EXPkey, then

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To enter 10-', you enter 1; then press the EXP key, then the 3 key,
andlastly the+f-key.

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ESTIMATING YOUR ANSWER

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One advantage of standard form is that """ry large and very small
numbers can be enter ed on a calculator Another advantage is that you
can easily estirnarc the answer to a calculationtothecorrectord~rof
magnitude (i.e. the correct power of 10)

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For example,

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2456xO.0123xO.00414
5223X60.7XS.51

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Purring the numbers into standard form gives
2.456xlO'xl.23xIW'x4.14xlO-l
------s.inXIO'X 6.07xIOx851

This is approximately
~xlOjXlO-'XlO-J=..!.XlO_6=3X10_3
5X6X8
10 ' X10
30
Dy purring the numbers into standard form, you can estimate the
answer very quickly. A complete calculation gives the answer
4.64 X 1O-"_ The rough estimate is sufficiently close to this to reassure you that you have not made any slips with exponents of ten.

LOGARITHMS
The logarithm (or 'log') ofanumbcr
be raised to give the number

Nisthepowertowhich

to must

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Ca!cularionslofA-/BvraIClwmtfrry

IfN
If N
IfN'"

=

1,
100,

then since
then since

10° = 1,
102 = 100,

0.001,

=

then since

10-3 = 0.001, 19N

19N = 0
IgN = 2.
= -~.

W~say that the logarithm of 100 to the base 10 is 2 or IglOO = 2
There is another widely used set of logarithms to the base e, They aTe
called natural logarithms as e is a significant quantity in mathematics
It has tlJe value 2.71828..
Natural logarithms are written as inN.
The relationship between the two systems is
InN

1n10XIgN

=

Since Inl0 = 2.3026, fOf mcsr purposes it is sufficiently accurate
to write
inN

=

2.303lgN

AD

Whenever scientific IVorkgives an equation in which InN appears, you
can substirure 2.303 times the valueoflgN.

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To obtain the logofa numher, enter the number on your calculator
and press the log key. The value of the log will appear in the display.
This will happen whether you enter the number in srandard fonn or
anotber form: For example, Ig12345 = 4.0915, whether you enter
the number as 12345 or as 1.2345 X 104• However, there is a limit
to the number of digits your calculator will accept. and you need to
enter very large and very small numbers in standard notation.

AD

Opttations on logarithms are:

FA

H

MultiplicatioR The logs of the numbers are added
Ig(AXB) = IgA+lgR

Divi$ion. The logs are subtracted:
Ig(1'IQ)

= IgP -lgQ

Powers. This is a special case of multiplication.
19A3 = 19l1 + Ig/l

21gA

=

IgA-l = -31gA
Routs. It is easy to show that IgyB= t 19B.
Since
19B = IgBI/3+lg8113
19B"3 '" ~lgB
Similarly,

Ig-«8'" llgB

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ANTILOGARITHMS

Your calculator will give you the antilo g of a llumber. You should
consult the manual to find Out th~ procedure for your own model
of calculator.
Most calculators will give you reciprocals. squares and other powers,
square roots and other rootsditeccly. If you have a simpler form of
calculator, you canohtain powers and roots by usinglogarithms
ROUNDINGOFF NUMBERS

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Often your calculator will disp.layan answer containing more digits
than (h~ numbers rou fed mto, it. Suppo~ you are given the
informatIOn that 18.6cm3 of sodium hydroxide solution exactly
neutralise 2S.0cm'ofa solutiollofbydroch.loricaddofconcemranon O.100moldm-l. You want to find the concentration of sodium
hydroxide solution, and you put the numbers (25.0XO.l00)/18.6
into your calculator and obtain a value ofO.1344086moldm-J. The
concentration of the solution is not known as accurately as this,
however. because YOllcannot read the burette as accurately as this.

FA

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Since you read the burette tothreefigures,youquOtcyouransw~rto
three figures. In the numberO.1344086,the figures you are sure of
are termed the significant figures. The significant figuresMCretained,
and me insignificant figures arc dropped. This operation is called
rounding off. If the fim number had been 0.1347086, it would have
beenroundedofftoO.135_lfthefirstofthcinsignificantfigures
being dropped is 5 or greater, the last of the significant figures is
rounded up to the next digit. If the first of the dropped figuresis less
than 5, the last significant figure is left unalten~d
Some calculations invo!vcscveraln;lgcs.lt is sound practice to give
one more significant figure in your answer at each stage than the
number of significant figures in the data_ Then, in the final smge,the
answer is rounded off.
If the calculation were (25.0XO.lOO)/26.2= 0.09541984moldm-3,
would you still round off to 3 significant figures? This would make
th~ ~swer 0.0954moldm-' .. Stated in this way. the answer is
c!alffimgan accuracy of 1 pan 10 954 - about 1 pan in 1000. Since
the hydrochloric add concentration is known to about 1 pan in 100,
the answer cannot be stated to a higher degree of accuracy. You have
to usc the 3"'5ignificant-figureule sensibly, and say that an error of
r
±I in 95 is about as significant as an errorof±l in 134. T he answer
should therefore he quoted as 0.095 moldm "-',
The number of significant figures is the number of figures which is
accurately known, The llumbt:r 123 has 3 significan't figures. The
number 1.23XlO<113S 3 significant figures, but 12300 has Ssignificant figures because the final zeros mean mat each of these digits is

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Ca/c<Jlar!(K1$forA-feWJI

Olsml,;rry

known to be zero and not some other digit. The number 0.00123 has
3 significant figuresvThe number 25.1 has 35ignificamfigures,and
the number 25.10has4 significant figures as the final o states that the
value of this number is known to an accuracy of 1 part in 2500
In addition, the sum is known with the accuracy of the least reliable
numbers in the sum. For example, the sum of
1.4167g
+100.5
g

~

109.0367g

AD

Since 1 figure is known to only 1 place after the decimal point, the
sum also is known to 1 place after decimal point and should be
written as 109.0g. The same guideline is used for subtraction.
In multiplication and division the producr or quotient is rounded off
the same' number ofsignifieant figures as the number with the
fewest significant figures. For example, 12 340X2.7XO.00~65 '"
121.6107 The product is rounded off to 2 significant figures,
1.2 X 10'.

AD

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to

CHOICE OF A CALCULATOR

•

FA

H

The functions which you need in a calculator for the problems in
this book are'
Addition,Subuaction,MultiplicarionandDivision
Squares lind other powers (x1andxY keys)
Square rcocs end other roots

/.VX

and ...,lfy keys)

Reciprocals
•

Wg,oandanrilog,o(lO'")

•

NaturaJlogaothms,ln,andantiln.(e")

•

Exponent key and-tv+key

•

Brackets

•

Memory

A variety cf scieerific calculators have these functions and others
(such as sin.eos, tan and l:x) which wiUbe useful to you in physics
and mathernarics problerns

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UNITS

There are tWOsets of units currently employed in scientific work. One
is the CGS system, based On the centimetre, gram and second. The
other is the Systeme Internaticnale (51) which is based on the metre,
kilogram, second and ampere. SI units were introduced in 1960, and
in 1979 the Association for Science Education published a booklet
called Chemical Nomenclature, Symbols and Tensinoiogy for Use
ill School Science that recommended that schools and colleges adopt
this system.
Listed below arc the SI units for the seven fundamental physical
quantities on which the system is based and also a number of derived
quamitics and their units

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Chemists are still using some of the CGS units. You will find mass in g;
volume in cm3 and dm3; concentrations in mol dmP or mollirre-';
conductivity in ,n-'cm as well as rr'm.
Pressure is sometimes
given i" mm mercur}' and temperatures in -c

.A

Basic 51 Units

H

FA
PhY!iiC(1IQulIllriry
Energy
Force
Electric charge
Electric potential
difference
Electric resistance
Ac~
Volume
Density
Pressure
Molar mass

Symbol

kilogmm
second
ampere
kelvin
mole
candela

k,

H

NamcofUllit

AD

Pbysical Quamity
Length
Mass
Time
Electric current
Temperature
Amount of substance
Light intensity

A
K
mol
cd

Derivcd Sl Units
Name of Unit
j:c~~on
coulomb
volt
ohm
square metre
cubic metre
kil:-::pcrcubic

J

Symbol

Dcfinitio"
kgm1.s-1
Jm-l
A,

V

JA-'s-'
VKI

N
C

a

m'
m'
kgm-3

newton per square
metre or pascal

Nm-10r
p,
kgmol-'

kilogram per mole

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Caleulallons for A·IIIWI Cilemiwy

wlrh all these units, the following prefixes (and others) may be used
Prefix
deci
centi
milli
micro

Symbol
d

kilo

Meaning
ro-'
10-2
10-3
10-6
10-9
W'
10'
10'
10"

k
M
G
T

nRT

H

=

.A

PV

M

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It is n:ry important whenp~ttingva1ues iorphysica! quantities into
an equation to be consistenr in the use of umts. If you arc,tben the
units can be treated as factors in the same way as numbers. Suppose
you are asked to calculate the volume occupied by O.OllOkg of
carbon dioxide at 27°C end a pressure of 9.80Xl04Nm-2.
You
know that the gas constant is 8.31J mol"! K-' and that the molar
mass ot carbcn dioxide is 44.0gmol-'. Use the ideal gas equation:

p = 9.80X10"Nm-1

The constant

H '" 8.31)K-'mol-'

H

The pressure

= 300K

AD

Thetempetatyre T = 27+273

FA

H

The number of moles
II =

Mass/Molarmass

=

O.OllOkg/(44.0X10-Jkgmol-')

=

O.250moi

V = O.2S0moiXS.31JK-'mol-1X300K
9.S0X104Nm-1
=

Since

J

6.34X 1O-3J N-'m'

= Nm

(l

joule == 1 newton metre)

V = 6.34XlO-JNmN-'m'
= 6.34 X 10-'mJ
Volume has the unit of cubic metre. This calculation illustrates what
people mean when they say that SI units form a cohcre_m. ystem of
s
unns. You ca.n convert from one unit to another by multiplication and
division, without introducing any numerical factors.

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SOLUTION

OF QUADRATIC

EQUATIONS

A quadmtic equation is the name for an equation of the type
x is the unknown quantity, a and b are the coefficients of:.:, and c is
a cnnstant. The solution of this equation is given by
x

-b±~

2a

=

AD

There are two solution.s In thc equation. Often you will be al~le eo
decide that one solution IS mathematically correct but physically
impossible. You may be calculating some physical quantity that
cannot possibly be negative sOlharyou will ignore a ncgativesolulion
andadoptapositi"csolution.

M

DRAWINGGRAPHS

H

Here are some binrs for drawing graphs.

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a) Whenever possible, data should be planed in a form that gives a
straight line graph. it is easier to draw the best straight line through a
set of points than to draw a CUl"le.

FA

H

AD

if the dimension~ x and y are related by th~ e""pressiony = (1.-'; + b,
then a straight line will result wben expcnmental values of yare
plotted against the corresponding values of x. The values of x are
plotted against the horizontal axis (the x-axis or abscissa), and the
corres~onding values ofy are plotted along t?e vertical axis (the y-axis
or ordinate}. The gradient of the straight line obtained =a, and the
intercept on the y-<l."'tis b (sec Fig. 1.1)
=

14
r;~~,
------,.
Fig. 1.1

Ploning.graph

b) Choose a scale which will allow the graph to cover as much of the
piece of graph paper as poo;sible.There is no need to starr a! ZerO. If
the points lie between 80 and 100,tosnmatzerowouldcrampyour
graph into a small section at the tOp of the page (see Fig. 1.2)

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Ca/cu/ar;r.msfarA·/8111l1rCi>INnisrry

Choose a scale which wilt make plotting the data and reading the
graph as simpleas possible.
c) Label theax~swith the dimensions and the units. Make the scale
units as simple as possible. Instead of plotting as scale units
IX 10-' mol dm-', 2X 1O-'moldm-l, s x IO-J mol dm'", etc., plot
1,2 and 3, erc., and label the axis as (Concenrranon/moldrnvj x 10'
(seeFig.U).
The solidus (f) is used because it means 'divided by'. The numbers, 1,
2 and 3,erc. {see Fig. 1.3) are the va lues of the physical quantity,
concentration, divided by the unit,moldm-',and multiplied hy lhe
factor 10'

M

AD

d) When you come to draw a straight line through the points, draw
the best straight line you can, to pass th rough,or close to, as many
points as possible (see Fig. 1.4). Owing to experimental error, not all
the points wi11 fall on the line. A graph of experimental results gives
you a better accuracy than calculating a value from just one point. If
you are drawing a curve, draw a smooth curve (see Fig. 1.5). Do not
join up the points with straight lines. The curve may not pass through
every point, bur it is more reliable than any one of the points

H
.A

H

'~. !"mi
::

AD

:

ro

H

ec

FA

w

c

5

1 0 15

'

20 25

•

0

Fig.l.2

Don"te,ampyou, graph!

Fig,1.3

Fig. 1.4

Drawing
the

Fig.I.5

be't line

1~.~,r.o,;!'mo:dm"~X1D'
Labellh •• x.,

Dr.wing.,moothcu".

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j

A FINAL

POINT

Always look critically at your answer. Ask yourself whether it is a
reasonable answer. Is it of the right order of magnitude for the data?
ts it in the right units? Many errOTScan be detected by all a5S(ssmem
of this kind

EXERCISE1

Practice with Calculations

1. Convert the following numbers into standard form'
a) 23678
b) 437.6
c}0.O I69
d) 0.000345
c) 672891

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2. Convert each of the following numbers into standard fonn, cnter
into your calculator, and multiply by 237.Givcthe answers in standard
form.
l)246.8
b) 11230
c) 267831
d) 0.051
e) 0.567
Find the fol!owing quotients
a) 236010.00071
b) 28780/0.106
d) 58/900670
e) 0.0008810.144

4.

Find the following sums and differences·
a) (2.000X104) + (0.10X10~)
b) 48.0+ (5.600Xl0J)
c) (L23 X10')+ (6.00X10')
d) (4.80X1O-4)-(l.6X 10-')
e) (6.300X104)-(4.8 X102)

c) 85.42/460000

FA

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3.

5. Make an approximate estimate of the answers to the foUowing:
a) 4.0 X 18~~:
~~_: ~~-.~: ~~JX1O~
b) 567X4183 XO.OOI27xO.I07
c) 49~;:01~:; :30~~~7~4. 7
d) ~.~~OXXO~:~~8XXO~1::8:~~~~I:
e)

560x2~~C:;SI2~:~49

6. find the logarithms of the following numben'
a) 4735
b) 5.072XIO'
c) 0.001327
d) 10.076
e )2,314XIO-'

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2

Formulae and Equations
Calculations are based on formulae and on equations. In order to
tackk the calculations in this buok}'ou "'ill havc to be quite sure you
can work out the formulae of compounds correctly, and that you can
balance equations_ This section i., a revision of work on formulae and
equations.

AD

FORMULAE

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M

Elerrrovalenr compounds consist of oppositely charged ions. The
compound formed is neutral because the charge on the positive ion
(or ions) is equal rc me charge on the negative ion (or ions). ln sodiu m
chloride, NaCl, one sodium ion, Na+, is balanced in charge by one
chloride ion, a-

FA

H

AD

H

Tbis i, bow the jommlac of e!~(/"J<'idenl
compounds CIf" be worked
out
Z;ncchlu,ide
Compound
Zn:+ andCl"
Ions present are
On~ 2n1+ ion needs two a- ions
Now balance the charges
Znl+ and2C1Ions needed are
The formula is
ZnClI
CompuulIIl
Ions pr~sent arc
Ions needed are
The formllb i<;

.)·<}dilJm5ulph~lc
Na+andSO.1Two Na+ balance one 504'2N:l:andSO.1NalS04

Compound
tens present are
Ions needed are
The formula is

Alu"'inil,msu/piJare
A!;+ and 504'Two AlJ+balance three 50.12All+and3S04'Al1(50.h

Compound
Ions presenr are
Now balance lhe charges
ions needed are
The formula is

Iron(ll)su/phMC
Fr>+and SO/One Fel+ balances one S04'FeH and SO/FeSO.

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Calculstinl/s(<>rA-IevtJIChsmi<try

Compollnd
Ions present are
Ions needed are
The formula is

lron(lll)sulphilte
Fe'·andSO.'Two Fe'+ balance three SO.z2Fc:3+and 350.'Fe:,(SD.h

You need to know the: charges of the ions in the table below, Then
you can work out the formula of any electrcvalent compound.
You will notice that the compounds of iron are:named iwnO!) sulphate
and iton(IlI) sulphate: to show which of its valencies iron is using in
the compound. This is always done with the: compounds of elements
of variable valency. For valency and oxidation number, see:Chapter 8,
p.72.
Symbol

Charge

W
NH.+
K·

+1
+1

Ag"

H
+1
+1

Barium
Calcium
Capper(l!)
Iron(ll)
Lead
Magnesium
Zinc
Aluminium
1ron(lII)

B3'•
CaH
Culo
Fe"
Pb"
Mf·
ZnH

AD
AI'·

FeJ+

+2
+2
+2
+2
+2
+2
+2
+J
+J

Symbol

Ch_

Hydroxide
Nitrate
Chloride
Bromide
Iodide:
Hydrogencarhonat"
Oxid"
Sulphide
Sulphite
Sulphate
Carbonate

OH
NO,CIB,1-

1
-,
-,
-,
-,
-,

CO,'-

-2
-2
-2
-2
-2

Phosphate

PO/-

-3

M
H

H
.A

co-

H

FA

+r

N,·

Name

AD

Hydrogen
Ammonium
Potassium
Sodium
Silver
Coppcrfl)

HCO)0'S'SOl-

so,>

EQUATI9NS
Having symbols for elements and formulae for compounds gives us a
way of representing chemical reactions
£XAMl'lE Instead of writing 'Coppe:r(II) carbonate forms copper(I!) oxide and
1
carbondioxide:',wecanwrite:
CuCD, _
CuO + CO,
The atoms we finish with are the same in num]J"r and kind as the
atolll5 we Start with_ w~stan with one atom of copper, On~atom of
carbon and three: atoms of oxygen, and we:finish with the same. This
make:sthetW05idesoftheexptession~ua!,andwe:c:!Il;tanequa!inll
A simple way of conveying a lot more information is to include:Slate

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FomJUI""""dEquatlon.
symbols
(aq) =

in the equation. These are (s) = solid, (I) = liquid, (g) = gas,
in solution in water. Theequation
CuO(s} + Co,{g)

CuCOJ(S} ---..

tells you that snliJ copp~r(l!) carbonate dissociates to form solid
copper(JJ) oxide and the gas carbon dioxide.
EXAMPLE2
Tbe equaticn

Zn(s) + H,SOiaq)

__,._

ZnSOhq)

+ H,{g}

tells you that solid zincreacls with a solution of sulphuric acid to give
a solution of zinc sulphate and hydrogen g;L5. Hydrogen is written as
H" since each mokcule of hydrogen gas comains tWOatoms.
Sodium carbonate reacts with dilute hydrochloric acid to give carbon
dioxide and a solution of sodium chloride. The equation could be
Na,COJ(s)

+ HO(a'l)

AD

EX P
...... lE3

CO,(g) + Nad(aq) + H,O{)

---..

+ Het(aq)

.A

Na,CO,(s)

H

M

but, when you add up the atoms on the right, you find that they are
not equal 10 thearoms on the left. The equation is not 'bal,mced'. so
the next step is to balance it. Multiplying NaCI by tWOgives
__,._

CO,(g)

+ 2NaCl(aq) + H,O(1)

Na,CO.(s)

AD

H

This makes the number of sodium atoms Onthe right-hand side equal
to the number on the left-hand side. But there are two chlorine atoms
ontheright·han<lside,thereforetheHClmustbcmultipliedbytwo:

+ 2Hd{aq)

_

CO,(g)

+ 2NaO(aq) + H,O(l)

H

The equation is now ba/anced

FA

When you are balancing a chemical equation, the only way you do it is
ro multiply the number of atoms ormo!ecules. You never try to alter
a formula. In the above example, you gOt two chlorine atoms by
multiplying HCI by two, not by altering the fonnula to HCI" which
doe~ nor exist
The steps in writing an equation are'
1. Write a word equation
2. PUt in the symbols and formulae (symbolsfor elements.forrnulae
for compounds and state symbols).
3. Balance the equation.
EXAMPLE"
When

methane burns,
Methane + Oxygen _
CHig)

+ O,(g)

_

Carbon dioxide

+ Water

CO,{g) + H,O(g)

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There is one carbon
atom on the left-hand
side and one carbon atom
on the right-hand
side. There are fOUT hydrogen atoms on the leftband side,and therefore we need to put four hydrogen atoms on the

right-hand side. PUtting 2H~Oon the right-hand side will accomplish
this:
CH.(g) + Oig) _
C01{g) + 2H:0(g)
There is one molecule of 0, on the left-hand side and four 0 atoms on
the right-hand side, We can make the two sides equal by putting 201
on the left-hand side
CHig)

+ 20lg)

__..

C01(g) + 2H,0(g)

EXERCISE 2

AD

This is a balanced equation. The numbers of atoms of carbon, hydrogen and oxygen on the left-hand side arc ~qual to the numbers of
atoms of carbon, hydrogen and oxygen on the right-hand side
Practice with Equations

H
.A

Now try writlog balanced equations for the reactions
a) Calcium + Water _
Hydrogen + Calcium hydroxide
solution
b) Copper + Oxygen _
Copper(1I) oxide
c) Sodium + Oxygen _
Sodium oxide
d)lron+Hydrochloricacid_
Iron(11)chloridesolulion
e) Jroo+Chlorine_
Iron(1l1) chloride

FA

H

2.

AD

+

H

M

1. For practice, try writing the equations for the reactions:
a) Hydrogen + Copper oxide _
Copper + Water
h) Carbon Carbon dioxide __..
Carbon monoxide
~) Carbon';': Oxygen __..
Carbon dioxide
d) Magnesium + Sulphuric acid __..
Hydrogen + Magnesium
sulphate
e) Copper + Chlorine _
Copper(ll) chloride

3. Balance these equations.
a) NalO(S) H,O(l) _
+
b) KC!O)(s) _
KO(s)

NaOH(aq)

+ O,(g)

c) HlO~(aq) H100) + 0l(g)
d) Fe(s)+ Dig) _
Fe)O.(s)
e) Mg(s) + N,(g) _
MglNl(s)
NHJ(g)+O,(g) N,(g)+H,O(g)
g) Fe(s) + HiO(g) _
Fe,O.(s) + H,(g)
h) HlS(g) + O~(g) _
H,O(g) + SOl(g)
i) H,S(g) + SO::(g) _
H10(1) + SIs)

o

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3

Relative Atomic Mass

RELATIVE ATOMIC MASS
Atoms are tiny: one atom of hydrogen has a mass of 1.66x IO-z"g;
une alom of caruon has a mas, of 1.99 x lO-ng. Numbers as small as
this are awkw:ll"dto handle, and, imread of the actual masses, we use
relative atomic masses. Since hydrogen aromsarethesmalIes!ofaU
atoms, one atom of hydrogen was originally taken as the mass with
which all ether atoms would be compared. Then,

:1.;:0:

O:foollne"

~f~;:::;~nt

AD

Original relative atomic mass '" ~:

M

Thus, on this scale, the relative atomic mass of hydrogen is I, and,
since one atom of carbon is 12 times as heavy asonearomofhydrogen. the relative atomic mass of carbon is 12.

H

I"
Relative aromic mass

FA

I

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H

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H

The modern method of finding relative arornic masses is to use a mass
spectrometer. The most accurate meaouremcnrs are made with volatile compounds of carbon, and it was therefore convenient to change
the standard of reference to carbon. There are three isotopes of
carbon (with the same number of protons and electrons but different
numbers of neutrons, and therefore different masses). It was decided
to use the most plentiful carbon isotope, carbon-12. Thus,

=

Mass of one atom of an element
~1112)M"'SSOfnneawmofcarbnn-12

I

On this sc:1.le,rarbon-12 has a rdative atomic mass of 12.000000,
<:arbon has a relative atomic massof12.01l11,andhydrogenhasa
relative atomic mass 1.00797. Since relative atomic masses are ratios
of two masses, they have no units. As this value for hydrogen is very
close to one, rhe value of H = I is used in most calculations. A table
of approximate relative atomic ma.. eS is givcn on page 293. The
symbol for relative atomic mass iS/I,.
RELATIVEMOLECULAR
MASS
You can find the mass of a molecule by adding up the masses of all
the atoms in it. You can find the relative molecular mass of a compound by adding the relative atomic masses of all the atoms in a molecule of the compound. For example, you can work OUtthe relative
molecularmas~ofearbondioxidcasfollows:

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Calculariol1$forA.(,wlo,emisuy

The formula is COl'
1 atomofC, relative atomic mass 12 '" 12
2 atoms of 0, relative atomic mass 16 '" 32
Total== 44
Relative molecular mass of CO! '" 44
The symbol for relative molecular mass is AI,

H

M

AD

A vast num~r of compounds consist of ions, not molecules. The compound sodium chloride, for example, consists of sodium ions and
chloride ions. You cannot correctly refer to a 'molecule of sodium
chloride'. For ionic compounds, the term formula unit is used to
describe the ions which make lip the compound. A formula unit of
sodium chloride i, NaCl. A formula utlit ofcopper(1I) sulphate·Swater is CuS04·SHlO. It is still correct to use the term relative molecular mass for ionic compounds'

H

.A

.
ilassofone formula unit
Relative UJ?lecular mass"" (1/12) Mass of one atom of carbon-12

AD

We work out the relative molecul,1r mass of calcium chloride as
follows'

H

The formula is CaCll.

FA

1 atom ofCa, relative atomic mass qu '" 40

2 atoms ofCI,relativeatomic

mass 35.5 '" 71
Total

=

111

Relative molerular mass ofCaCll '" 111
We work
follows

OUt

the relative molecular mass of aluminium sulphate as

The formula is AI,(SO.h.
2 atoms of AI, relative atomic mass 27

=

54

3 atoms ofS, relative atomic mass 32

=

96

12 atoms of 0, relative atomic mass 16

=

192

Total

=

342

Relative molecular mass of Al,(S04h

=

342

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EXERCISE

3

Problems RelativeMolecularMass
on

Workoutth~reJatj'enloletll!armassesofthesecompounds,
Na OH
PbCI,
Zn(O H),
H NO)
Ph,O.
Ca(OH),
Ca(HCO,),
Na,CO,"10H,O

KNOJ
MgCl,
Z nSO.
MgSO."7H,0
P ,O,
CuCO,
CUS04"5H,0
feS0 4"7H,0

AD

SO,
MgCO"
Mg(NO,),
H,S04
CaSO.
Na,COJ

PERCENTAGECOMPOSITION

H

M

From the formula of a compound, we can workout the percentage
by mass of each element present in the compound.

H
.A

Calculate the percentage of _~i!icon
and oxygen in
(silica)

siliccntlv)

oxide

AD

First, work Out the relative molecular mass. The formula is SiO,

H

1atom of silicon, relative atomic mass 28 = 28
2 atoms of oxygen, relative atomic mass 16 = 32
Total = Relative molecular mass = 60

FA

EXAMPLE 1

Percentageofsilicon

=

~XIOO

=

fsXlOO

= 7X20 = 47%
3
=

aa
8
6oX100 = i5X100

=

Percentageof oxygen

8~20

=

53%

Silicon(1V) oxide contains 47% ~ilicon and 53%oxygen by mass
Since every formula unit ofsilicon(lV) oxide is 47% silicon, and all
formula units are identical, bulk samples of pure silioon{IV)oxide all
contain 47% silicon. This is true whether you m~
talking about
silicon(IV) oxide found as quartz, or amethyst Or crystoballiteor sand

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d

~:~c:::: ofdement

A ~

"R,,,,,,,,,tiv,,,,,,;:,o,"~,:,'.C"m'':' 'CCo,,f":N",o,~ofc;"~om=,~of,,:A'Ci,,,f~mm~",,,bX 100
A,;,'
Relative molecular mass of compound

EXAMPLE Find the percentage by mass of magnesium, nXyg<"ll and mlphur
2
magnesium sulphate.

in

Fir~t calculate the rdath·e molecular mass. The formula is MgSO.

=

24
32
64
120

A,(Mg),~,7~~;~~g x 100
atoms

H

Percentage of magnesium

==
=
...
==

M

AD

1 atom of magnesium, relative atomic mass 24
1 atom of sulpburvrelative atomic mass 32
4 atoms of oxygen, relative atomic mass 16
Total '" Reiativemolecularmass,IH,

AD

H

.A

24
=-x100
120

FA

H

Perccntageofsulphur

=

A,(S)X No. ofS atoms
M,(MgSO.>
XIOO

==

UoX100

J2

Percentage of oxygen = A,(O),~j:~~;~~

atoms X 100

16x4
=-xIOO
120
=

53%

Magnesium 20%; sulphur 27%: oxygen 53%. You can check on the
calculation by adding up the percentages to see whether they add up
to 100.In this case 20+27 + S3 = 100
EXAMPlE3Cakub.te the percent:J.geof water in copper sulphate crystals.

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Find

the rdative

molecular

mass.

The formula

is CuSO.'

1 awmof
copper,
relativentcmic mass e-t ""
1 atom ofsulphm,relative atomic mass 32 =
4 atoms of oxygen, relative atomic mass 16 =
5 rnoleculcs of water, 5X [{2X 1}+16] =
Total = Relative molecular mass""
Mass of water =

Percentage of water

=

SH,O.

64 (approx.)
32
64
5XI8 = 90
250
90

Ma~ of wat~r in formula
Rda!ivc molecular ma55 X 100
90
150

=-XIOO

36%

AD

=

EXERCISE4

.A

H

M

The percentage of water in copper sulphate crystals Is 36%

Problems on Percentage Composition

AD

H

S~cTION
1
Calculators are not needed for these problems

FA

H

1. CaleuJatc thc pcrccmagcs by ma5Sof
a} carbon and hydrogen in ethane,C,1i6
b) sodium, oxygen and hydrogen in sodium hydroxide, N30H
c) sulphur and oxygen in sulphur trioxide. SO,
d) earhon and hydrogen in propyne,C,H,
2. Calculate the percentages by mass of
a) carbon and hydrogen in heptane,C,H,.
b) magnesium ~nd nitrogen in magnesium nitride, Mg)N~
c) sodium and iodine in sodium iodide, Nal
d) calcium and bromine in caicium bromide, CaBr,.

1

CalcuJatethepercentagebyma~of
a) ClIrbonand hydrogen in penrene,C,Hw
b) nitrogen, hydrogen and oxygen in ammonium nitrate
c) il'on, oxygen and hydrogen in iron(ll) hydroxide
d) carbon. hydrogen and oxygen in ethanedioicacid,ClO,lil.

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Calculatloo;

forA·/evei

Cheminry

2. Calcutare the percentages of
a) carbon, hydrogen and oxy~n in propanol, C,H,OJ-l
b) carbon, hydrogen and oxygen in ethancic acid, CH,CO,H
c) carbon, hydrogen and oxygen in methyl rnerbanoate, HC02CHj
d) aluminium and SUlphurin aluminium sulphide, AI,S,.
3. Haemoglobin contains 0.33% by mass of iron. Tbere are 2 Fe atoms in
1 molecule of haemoglobin. What is the relative molecular mass of
haemoglobin?

FA

H

AD

H

.A

H

M

AD

4. An adult's bones wcigh about 11 kg, and 50% of this mass is calcium
phosphate, Ca,(PO.,),. What is the mass of phosphorus in the bones of
an average adult?

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The Mole
Looking at equations tells us a great deal about chemical reactions.
For-example.
Fe(s)+S(s)

-----+-

FeS(s)

.A

H

M

AD

tells us tblt iron and sulphur combine to form iron(ll) sulphide, and
that one atom of iron combines with one atom of sulphur. Chemists
are interested in the exact quantities of substanceswhichreact tcgether
in chemical reactions. For example, in the reaction between iron and
sulphur, if you want to measure ourjust enough ironto combine with,
say, 109 of sulphur, bow do you go about it? What you need to do is
to coum out equal numbersof atoms of iron aod sulphur.111;5
sounds
a formidable task,and ir puzzled J chemisl called Avogadro, working
in Italy early in the nineteenth century. He managed to solve this
problem with a piece of dear thinkingwhich makesthe problem look
very simple once you have followed his argument.
Avogadro reasoned in this way

AD

H

We know from their fdati.-!' atomic masses that an atom of carbon is
12 times as heavy as anatom of hydrogen. Thcrefore, we can say

H

If 1 atom of carbon is
then I dozen C atoms are
and I hundred Caroms are
and 1 million C atoms are

FA

4

12timesasheavyaslatomofhydro~n,
12 times as heavy as 1 dozen H atoms,
12 times as heavy as 1 hundred H atoms,
12 times as heavy as 1 million H atoms,

and it follows thn when we see a mass of carbon which is 12 times as
heavy as a mass of hydrogen, the tWO masses must contain equal
numbers of atoms. If we have 12gofcarbonandlgofhydrogen,we
know that we have the same number of atoms ofca:rbonand hydrogen
The same argument applies to any element. When we take the rebti"e
aromic mass of an element in grams:

I4Q;I ["24,1 ["32;1 ~

~

~~~~~

all these ma~ contain the same number of atoms. This number is
6.022X 10". The amount of an element which contains this numher
of atoms is called one mole of the element, (The symbol for mole is
mol.) The ratio 6.022x lOll/mol is called the Avogadro constant

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The mole is defined
as many
e1~mentary
carbon-12

as the amount
of a substance
enriries
as there
are atoms

which contains
in 12 gratru of

count out 6x 1013 atoms of any dement by weighmg OUtits
relative atom~c mass in grams. If we want to react i~otland sulphur so
that there is an atom of sulphur for every atom of Iron, we can count
out 6X 1023 atoms of sulphur by weighing out 32g of sulphur and we
can count out 6x 1013atoms of iron by weighing out S6g of iron.
Since one atom of iron reacts with one atom of sulphur to form one
formula unit of iron(lI) sulphide, one mole of iron reacts with one
mole of sulphur to form one mole of iron(lI) sulphjd~:
We can

Fe(s)+S(s)_

FeS(s)

AD

and 56g of iron react with 32g of sulphur to form 88g ofiron(Il)
sulphide.

.A

H

M

Just as one ~Ie of an element is the_relative atomic mass in grams,
one mole of a compound is the relative molecular mass in grams. If
you want to weigh out one mole of sodium hydroxide, you firs! work
out its relative molecular mass.

H

The fonnula is NOlOH

AD

Relative molecular mass

=

23 + 16 + 1

=

40

FA

H

If you weigh our 40g of scdiuru hydroxide, you have one moc of
sodium hydroxide. The quantity 40gmol-1 is the molar mass of
sodium hydroxide. The molar mass of a compound is the relative
molecular mass in graffi5 p~r mole, The molar mass of an element is
the relative atomic mass in grams per mole. The molar mass of sodium
hydroxide is ang mol'", and the molar mass of sodium is 23 g mol"
Remember that most gaseous elements consist of molecules. not
atoms. Chlorine exists as el, molecules, oxygen as 0, molecules,
hydrogen :IS H, molecules, and so on. To work out the mass of a mole
of chlorine molecules, you must use the relative molecular mass of

ct,
Relative atomic mass of chlorine = 35.5
Relatlve molecular muss of Cl,

=

2x3S.5

Mass of 1 mole of chlorine, Cl,

=

71 grams.

=

71

The noble gases, helium, neon, argon, krypton and xenon. exist as
atoms. Since the relative atomic mass of helium is 4, the mass of 1
mole of helium is 4g

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What is rhe molar mass of glucose?
Formula = C.,H120.
Relative mow mass = (6 X 12) + (12 X I) + (6 X 16) = 180
The relative molar mass is 180, and the molar mass is 180gmol-1

EXERCise 5

Problems on the.Mole

SECTION1

AD

1. State the mass of each elementin.
a) 05molchromium
b) 117moliron
c) 1/3 mol carbon
d) 1/4molmagnesium
e) 117molnitrogenmolecules
f) 1/4moloxygenmolecule~.
Remember that nitrogen and oxygen exist as diatomic molecules, Nl
and 0,.
c) 109 calcium

.A

H

M

2. Calculate the; amoum of each element in
a) 46gsodium
b) 130gzinc
d} 2.4gmagnesium
e) 13gchromium.

H

AD

H

3. Find the mass of each element in:
a) JOmolleW
c) 0.1 mol iodine.molecules
e) 0.25 mol calcium
g) ~moliron
i) tmochiorinemolccuie!5

b}
d)
f)
h)
j)

1/6molcoppcr
10molhydrogenmolecuies
0.25 mol bromine molecules
0.20 mol zinc
0.1 mol neon.

FA

4. State the amount of substance (mol) in:
a) 58.5gsodiumchloridc
b) 26.S g anhydrous sodium carbonate
c) 50.0gcalciumC:l.rbonate
d}I5.9gcopper(JJ)oxide
e) 8.00gsodiumhydroxide
() 303gpotllSSiumnitrate
g)9.8gsulphuricacid
h) 499g oopper(ll) sulphate+watcr.
5. Given Avogadro's constant Is e x lO'lmol-', calculate the number of
aroms m.
a) 35.5gchlorine
b) 27galuminium
c) 3.1gphosphorus
d) B6giron
e) 48gmagncsium
f) 1.6goxygen
g) o.4goxygt'n
h)216gsilver.

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6. How many

gralTl5 of zinc contain

'3atoms

h) 6XO,Oatoms?

a) 6xtO
7.

How many grams of aluminium contain:
a) 2X102Jatoms
b) 6xtOlOatoms?

8. what mass of carbon contains
a) 6xlO"atoms

b) 2XIOlJatoms?

Write down:
a) the mass of eakium which has the same number of atoms as t2g
ofmngnesium
b) the mas.";of silver which has the same number of atoms as 3 g of
aluminium
c) the mass of zinc with the same number of atoms as 1g of helium
d) the mass of sodium which has 5 times the number of atoms in 39g
of potassium

M

AD

9.

H

Use Avogadro constant e o x lO"mo]-J

Ethanol, C,H60, is the alcohol in alcoholic drinks. If }'OU hav~ 9.2g of
ethanol.How many moles do you have of
a) ethanolmolecules
b) carbon atoms
c) hydrogen atoms
d) oxygen atoms?

H

AD

2.

H
.A

1. Imaginc a hardware store is having a sale. Theknock-dolVtl price of
titanium is one billion (109) atoms for tp. How much would you have
to pay for 1milligram(1X lO-Jg)oftitaniurn?

FA

3. A car releases about 5g of nitrogen oxide, NO, into the air for each
mile driven. How many molecules of NO are emitted pcr mild
4. How many moles of H20 are there in 1.00 litre of water?
5. How many moles of Fe,O,
6.

art:

there in 1.00kg of rust?

What is the mass of one molecule of water?

7, Wha, is fhe amount.(mol) of sucrose,C"H"O", i n a one kilogrJ.m
bagofsuga:r?

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5

Equations and the Mole
You will find that the mole concept, which you studied in Chapter 4.
helps with all your chemical calculations. In chemisny, caJcuLuions
are related to the equations for chemical reactions. Thequanriries of
substa.ncesthat react together are expressed in moles.

CALCULATIGNS BASE£I0N CHEMICAL eOUATIONS

AD

Equations tell us not only what substances react togetller but aim
what amounts of substances react together. The equation for the
action of heat on sodium hydrogencarbonate
ZNaHCOJ(s) _

NalCOl(s)

+

COig)

+

l'I,O(g)

H

M

tells us that 2moles of NaHCO ,giv,", mole of Na,CO). Since the
I
molar masses are NaHCO, = S4gmol-1 and Na,CO, = 106gmol-1,
it follows that 168g of NaHCO, give l06g of Na,CO,

H

AD

H
.A

The amounts of substances undergoing reaction, as given by the
balanced chemical equation, are called the stoichiometric amounts.
Stoicbiomnry is the felation,J,ip between the amounts ofrcactILnts
and products in a chemical reaction. If one reactant is present in
excess of the srcichicmetnc amount required for reaction with
another of the reactants, then the excess oi osv: reactant will be left
unuscdattheendoftherCiction.

FA

EXAWLE How many moles of iodine can be obtained from i mole of potassium
1
iodate(V)?
The equation
Potassium + Potassium + Hydrogen
iodate(V) iodide
ion
KIOJ(aq) + 5Kl(aq)

+ 6W(aq)

_

iodine

+ Potassium
ion
+ Water

31,(aq)+ 6K+(aq)+ 3H,O(l)

tells us that 1 mol of KID) gives 3 mol of I,. Therefore
~ mol of KID] gives ~X 3 mol of I, == t mol of I,
E)lA"'PL~2What is the maximum mass of ethyl ethanoate that can k obtained
from 0.1 mol of ethanol?

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€qlNltiom,."d rtre Mole

Wrirethe equation.

+ Ethanok acid
C,HsOH(I) + CH,CO,H(I)
Ethanol

_

Ethyl ethancare

-

CH,CO,C,H,(I)

+ Water
+ H,O(I)

1 mol of C,H,OH gives 1mol CH,CO,C,Hs
0.1mol ofC,H,OH gives 0.1 mol CH,CO,C,H,
The molar mass of CH,CO,C,H5 is 8f!g mol-I, Th~refore:
0.1 mol of ethanol gives 8.8goferhylerhanoate.
EXAMPLE

a A mixture of 5.00g of sodium carbonate and sodium hydrogencarbonate is heated. The loss in mass is OJ 1 g. Calculate the percentage by mass of sodium carbonate in the mixture.
On heating the mixture, the reaction
_

Na,CO,(s)

+

CO,(gl

AD

2NaHCOis)

+

H,O(g)

.A

H

M

takes place. The loss in mass is due to the decomposition of NaHCO,.
Since 2mol NaHCO, form 1mol CO, + 1mol H,O
2 X 84g NaHCO, form 44g CO, and 18g H,O
168 g NaHCO, lose 62 g in mass.
The observed loss in mass of 0.31 g is due to the decomposition of

H

%¥X168gNaHco,

'" 0.84g

H

AD

The mixture comains 0.84 g NaHCO,
The difference, 5.00-0.84
'" 4.16gNa,CO,.

EXERCISE6

FA

Percentage of Na,CO, eo

Wo

X 100 = 83.2%,

Prohlerrn on Reacting Massesof Solids

1. A sulphuric acid plant usc52500 tonnes of sulphur dioxide each day
What rna.. of sulphur mUSt be burned to produce thi.quantity of
sulphur dioxide?
2. An antacid tablet COntains O.lg of magnesium hydrogencarbonate,
Mg(HCO,),. What mass of stomach acid, HCI, will it neutralise?
3. Aspirin,C9H.O., is made by the reacrion:
Salicylicacid+Erbanoicanhydride
_

Aspirin+EThanoica~id

C,HoO. + C,HoO,
C.Hs04 + ClH.Ol
How many grams of salicylic acid. C7H60). are needed ro make One
aspirin tablet. which contains 0.33g of aspirin?

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CaI"tJlafjo"sforA-I~ .. ICiUlmilfry

4. Aluminium sulphate is used to treat sewage. It can be made by the
reaction:
Aluminium
hydroxide

+ Sulphuric

Aluminium

acid

+ Water

-sulphate

a} Balance this unbalanced equation for the reaction
Al(OHMs) + H,SO.(aq} _
Al,(SO~),(aq)

+ H,O(I)

b) Say what masses of (i) aluminium hydroxide and (ii) sulphuric
acidareneededtomakel.OOkgofaluminiumsulphate
5. Washing soda, Na,CO,'lOH,O, loses some of its water of crystallisalion if it is not kept ill an air-tight container to form Na,CO,·H,O.

AD

A grocer buys a 10kg bag of washing soda at 30p/kg. While it is
standing in his storeroom. the bag puncrures,aodthccrystalsrurn
into a powder, Na,COJ'H,O. The grocer sells this powder at 50p/kg.
Does he makea profit or a loss?

Nitrogen monoxide, NO, is a pollutaru gas which comes OUf of vehicle
exhausts.Onctechniqueforreducingthequantityofnitrogenmonoxide
in vehicle exhausts is to inject a stream of ammonia. NH" into the
exhaust. Nitrogen monoxide is converted into the harmless products
nitrogen and water
4NH,(g) + 6NO(g) _
5N,(gl + 6HP(I)

FA

H

7.

AD

H

.A

H

M

6. when you take a warm bath, the power starion has to burn about
1.2 kg of coal to provide enough electricity to heat the warer.
a) Iftbe coal contains 3%sulphur, what massofsulpburJioxide does
the power station emit as a resu!t?
b) Multiply your answer by the number of warm baths you take in a
year
c) This is only a part of your contribtlri<;mttl air pollution. What can
be dOfletoreducethi~sourceofpollution-apartfromtakingcold
baths?

An average vehicle emits 5 g of nitrogen monoxide per mile. Assuming a mileage of 10000 miles a year, what mass of ammonia would
t>eneeded to dean up the exhaust;
8. Some industrial plants, for example aluminium smelters, emit
fluorides. In the past, there have been cases of fluoride pollution
affecting the teeth and joints of cattle. The Union Carbide Corpora"
rion has invented ~ process for removing fluorides from waste gases.
lrinvolves the reaction.
high
l.mp"~lu~

Fluorideion(F-) + Charcoal (Cl _

Carbon tetrafluoride (CF.)

The product, CF4, is harmless. The firm claims that I kg of charcoal

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"__'ateri~1

EqIJarionsa"dlheMole

will remove 6.3 kg of fluoride ion. Do you believe this claim, Explain

your answer.
9.

A factory m~kes a detergent of formula CuH",SO.,Na from !aury!
alcohol, C"H'60. To manufactu~ 1J tonnes of detergent daily, what
mass oflauryl alcohol is needed?

10. A mass of 0.65g of zinc powder was added to a beaker containing
silver nitrate solution. Whcnallthezinchadreacted,2.16gofsilvcr
were obtained.

AD

Calculate
a) the amount of zinc used
b} me amount of silver formed
c) the amount of silver produced by 1mol of zinc.
d) write a balancedionic equatiou forrhereacrjon.

M

SeCTIO'l2

.A

H

1. The element X has a relative atomic mass of 35.5. It reacts with a
~lutionofthesodiumsaltofYacrordinglotheequarion
Xt+2NaY _
Y,+2NaX

H

If 14.2g of X, displace 50.8g of Y" what is the relative atomic
mass of Y?

AD

2. TNT is an explosive. The name stands for trinitrotoluene. The compound is made by thercact:ion

FA

H

Toluene + Nitric acid
C,H,(l) + 3HI'.:03(1) _

TNT + Water
C,H,N306{S) + 3H,O(l)

Calculate the masses of aj toluenc and b}nitrie acid that must he
used to make 10.00ronnes of TNT. (1 tonne 0= 1000kg.)
J.

A large power plant produces about 500 ronnes of sulphur dioxide in
a day. One way of removing this pollutant from the waste gases is to
inject limestone. This converts sulphur dioxide into calcium sulphate.
~;mestone + ~~~~du: + Oxygen
2CaCO){s) + 2S01(g) + O,(g)

Calcium + Carb.on
sulphate
dioxide
2CaSO.(s) + 2CO,(g}

Another mcthod of removing mlphur dioxide is to 'scrub' me waste
gases with ammonia. The product is ammonium sulphalc.
~%monja + ~~~~du: + Oxygen + Water
_
~;~~t~iUm
4NH){g) 2S0,(g) 0l(g) + 2HP{1)
+
+
2(N H4hSOiaq)
a) Calculate the mass of calcium sulphate produced in a day by

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Fq<Jillion8ar1dII~MoJe

EXAMPLE Iron burns in chlorine 10 form iron chloride. An experiment showed
1
that S.60g of iron combined with 1O_65gof ch1or;I1~.Deduce th~
equation for the r..action.
5,60gofironcombinewithl0.65gofchlorine
Relative atomic masses are: Fe = 56, CI = 35.5
Amount (mol) of iron = 5,60/56 = 0.10
Amount (mol) of chlorine molecules = 10.65/71.0
The equation must be

=

0.15

Fe+l.5Ci,2Fe+3Ci,_

Tobalancetheequation,therighl-handsidemustre~d
Therefore,

2FeCiJ•

AD

2Fe{s)+30,{g)_2FeCl,{s)

H

M

EXAMPlE217.0g of sodium oitrate react wirh 19.6g of sulphuric add 10 give
12.6gofniuicacid.
Deduce theequarion for the reaclioo.

.A

Relative molecular massesare: NaND, '" 85, H,S04= 98, HND, "" 63

H

Number of moles of NaND, = 17.0/85 = 0.2
Number of moles of H,S04 '" 19_6198 = 0.2
Number of moles ofHNO, = 12.6/63 = 0.2

H

AD

0,2 mol NaND, rC3ct~with 0,2 mol H,S04 to form 0.2 mol of HNOJ
I mol NaN~, reacts with 1 mol H,S04 10 form I mol of HNO)

FA

The "'<.luationmust be balanc~,,1by inserting NaHSO.
hand sIde:

EXERCISE7

00

the tight-

Problems on Deriving Equations

1. To a solution containing 2.975 g of sodium persulpbste, Nal-Sl-Oe,s
i
added an excess of potassium iodide solution. A reaction occurs, in
which sulphate ions are formed and 3_175g of iodine are formed
Dcducetheequarionforthereaction.
2. Aminosulphuric arid, H,NSO,H, reacts wirh warm sodium hydroxide
solution to give ammonia and asoiurion which contains sulphate ions
Q,540g of the acid, when treated with an excess of alkali, gave
153cml of ammonia at 60°C and 1 atm. Deduce the equation for the
reaction

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C1lcularlons forA-Ie..t

ChlNllistry

3. A solution containing S.OOXIO-'mol of sodium thiosulphate was
sbakenwith1gofsilverchioride.O.717Sgofsilverchloridedissolved,
and analysis showed that 5.00X IO-'mol of chloride ions were
preientin the resulting solution. Derive an equation for the reaction.
4. An unsaturated hydrocarbon of molar mass 80gmol-1 reac~ with
bromine. If 0.250g of hydrocarbon reacts with 1.00g of bromine,
wbatis theequacion for the reaction?
5. Civen that 1,90g of phenylamine, C6HsNH" reacts with 5.16g of
bromine, derive an equation for the n:action

AD

PERCENTAGEYIELD

=

Actual mass of product
X 100
Calculated mass of product

.A

Percentage yield

H

M

There are many re.acrions which do not go 10 completion. Reactions
between orgapic compounds do not often give a 100% yield of
product. The actual yield is compared with the yield calculated from
the molar masses of the reactants. The equation

H

is used to give rbe percentage yield.

H

AD

from 23gofctbanolare
obcaincd36gofcthylcthanoolebyelterificanon with ethanoie acid in the presence ofconccntrated ,ulphuric
acid. What is the percentage yield of the reaction?

FA

Write the equation:
CH,COlH(I) + ClHjOH(l)

*

_

CHJC01C1H5(l)

+

H10(1)

46 g of ClHsOH forms 8Hg ofCH,C01CzHl
23g ofClHjOH should give

x 88g

=

44g ofCH,CO!C!Hj

Actual mass obtained = 36g
Percentage yield == Ca~:~~aal~l::s~f:rr;~:;~c

x 100

percentage yield = ~ x 100 = 82%
EXERCISE8

Problems on Percentage Yield

1. Phenol, C"H,OH, is convened into trichlorophenol, C.H,Cl.,OH. If
488g of product are obtained from 2S0gofphenol,
calculate the
per~e!ll;llgeyield

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2. 29.5 g of ethancic acid, CHJCOlH, are obtained from the oxidation
of 25.0g of ethanol, C,H,OH. What percentage yield does this
represent?
3. 0.8500g of hexanone, C6H"O,is converted into its 2,4-dinitrophenylhydrazone. After isolation and purification, 2.11BOg of product,
C'lH,sN"404,are obtained. What percentllge yield does this represent?
4. Benzaldehyde, C,H60, forms a hydrogensulphke compound of
formula C,H7SO.Na. From 1.21Og of benzaldehyde, a yield of 2.1S1 g
of the product was obtained. Calculate the percentage yield.

H

M

AD

5, 100 ern- of barium chloride solution of concentration 0.0500 mol dm-l
were treated with an excess ofsu!phate ions in solution. The precipitate of barium sulphate formed was dried and weighed. A mass of
1.1S5Sgwasrccorded.Whatperccmageyie1ddocsthisrepresem?

.A

LIMITING REACTANT

FA

H

AD

H

In a chemical reaction, the reactants are often added in amounts
which are notstoichiomctric. One or more of the reactants is in excess
and is not completely used up in the reaction. The amount of product
is determined by the amount of the reactant that is not in excess and
i.~used up completely in the reaction. This is called the limitillg
rcaClarlt,Youfim
have to decide which is rhelimiting reacranr before
you can calculOl.tc amount of product formed.
the
5.00g of iron and 5.00g of suiphur arc heated together to form
iron(ll) sulphide. Which reactant is present in excess? What ma~s of
product is formed?
Write the equation
Fe(s) +S(s)_

resoi

ImoleofFe+
l mcle of S form 1 mole of FeS
S6gFeand32gSformS8gFcS
S.OOgFeis 5156 mol = 0.0893molfe
5.00gSis 5/32mol = 0.156mo1S
There is insufficient Fe to react with 0,156moIS;ironisthelimiting
reactant
0.0893 mol Fe forms 0.0893 mol FeS '" 0.0893X88g = 7.86g
Mass formed '" 7,S6g

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EXERCISE 9

Problems on Limitin9 Reactant

1. In the blast furnace, the overall reaction is
2FelO,(s) + 3e(s) _
3C01(g) + 4h(s)
What is the maximum mass of iron that can be obtained from 700
tonncsofiron(lIl)oxideand
70tonncsofcokc?(1 tonne = 1000kg.)
2. In the manufacrurc of calcium carbide
oo» + 3C{s} _
c:aC,(s} + CO(g}
Whar is the maximum mass of calcium carbide that can be obtained
from 40kg of quicklime and 40 kg of coke?

AD

3. In the manufacture of the fertiliser ammonium sulphate
H,SO.(aq) + 2NH,(g) (NH.)lSOiaq)
What is the maximum mass of ammonium sulphate that can be
obtainedflOm2.0kgofsulphuricacidandl.Okgofnmmonia?

H

6. In the Thermit reaction

.A

H

M

4. In the Solvay process, ammonia is recovered by the reaction
2NH.Cl(s) + CaO(s) _
CaCl,{s) + H,O{gl + 2NHJ(gl
What is the maximum mass of ammonia that can be recovered from
2.00 X lO'kg of ammonium chloride and 500 kg of quicklime?

FA

H

AD

Calculate the percentage yield when 180g of chromium are obtained
from a reaction between 100g of aluminium and 400g of
chromium(lll) oxide.

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6

Finding Formulae

EMPIRICAL FORMULAE
The formula of a compound is determined by finding the mass of each
dcmcnr present in a certain mass of th~ compound.
Remember,
Amount (in
moles) of substance

AD

MassofsubstaoL'e
Molarmassofthesubstanet:

OX.ygC71
0
32g
16
32
16
= 2mol
lmnl

.A

Copper
co
127g
63.5
127
63.5
'" 2 mol

AD

H

Elements
Symbols
Masses
Relative atomic masses

H

M

E)(AMPlElGiven that 117g of copper combine with 32g of oxygen, what is the
formula of cnpper oxide?

H

Divide through by 2
Ratio of atoms

'"

FA

CoO

We,divide_through by lWO to obtain, the simplest formula for copper
OXIdewhIch ""II fit the data, .,he "'''plcst [ormll,", ",b,eb represents
Ibrc()lIlposilitJ>lo[JCOmpuuIJdj,c"f/edrheempiricaljimnll/"

When 127g of copper combine with oxygen, 143g of an oxide are
formed. What is the ~mpirical formula of the oxide?
You will.notice here that the ma~s of ox}'gen is not given
You obtain it by subtraction
Mass of copper

=

Massofoxide""
Massofox}'gen

to

you

127g
143g

=

143-128

=

16g

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O!lcIJlariom fur A.Jewl Chemistry

Now you C<l1I
carry on as before:
Elements
Symbols
Masses
Relative atomic masses

Copper
Cu
127g
63.5
127

Amounts

Oxygen

63.5

o

16,
16
16

i6

= 2
= 2

Ratio of aroms

= I
I

Empirical formula Culo

AD

Find n in theformulaMgS04·I1H,O. A sample of 7.38g of magnesium
sulphate crystals lost 3.7Sgofwateronheating.
Water
3.7Sg
1Sgmol-'
3.7S/18
= 0.21 mol
0.21
0.030
= 7

H

AD

H

=

M

=

Ratio of' amounrs

Magnesium sulphate
3.60g
120gmol-'
3.60/120
0.030mol
0.030
0.030
I

H

Compounds present
M~
Molar mass
Amount

.A

EXA"'PlE3

FA

MOLECULAR FORMULAE
The molecular formula is a simple multiple of the empirical formula.
If the empirical formula is CH,O, the molecular formula may be
CH,O or C,H40, or C,H60l and so on. You can tell which molecular
formula is correct by finding out which gives the ~orrectmolar mass.
For methods of fmding molar masses, see Chapters 9, to and 11. The
molar mass is a multiple of the empirical formula mass
A compound has the empirical formula CH,O and molar mass
180gmol-', What is its molecular formula?
Empirical formula mass = 30gmo]-'
Molar mass = 180grnol-)
The molar mass is 6 times the empirical formula mass. Therefore the
molecular formula is 6 times the empirical formula. Therefore:
The empirical formula is C4H"O •.

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F;nd;ngF(Jfmu/.."

EXERCISE10

Problems on Formulae

1. 0.72 g of magnesium combine with 0.28 g of nitrogen.
How many moles of magnesium does this represent?
How many moles of nitrogen atoms combind
How many moles of magnesium combine with one mole of nitrogen
atoms?
What is the formula of magnesium nitride?
2.1.68gofironcombinewithO.64gofoxygen.
How many moles of iron docs this mass represent?
How maoy moles of oxygen amIDScombine?
How many mole~ of iron combine with one mole of oxygen atoms?
What is the formula of this oxide of iron?

H
.A

H

M

AD

3. Calculate the empirical formula of the compound formed when 2.70g
of aluminium fonn 5.10gofitsoxidc.
What is the mass of aluminium?
What is the mass of oxygen (not oxide)?
How many moles of aluminium combine?
How many moles of oxygen atoms combine?
Whal is the ratio of moles of aluminium to moles of oxygen atoms?
What is the formula of aluminium oxide?

FA

H

AD

4. Barium chloride forms a hydrate which contains 85.25% barium
chloride and 14.75% water of crystallisation. What is the formula of
this hydrate?
whatis the mass of barium chloride in lOOg of the hydrate?
What is the mass of water in lOOg of the hydrate?
What is the relative mo!ecular mass of barium chloride?
What is the rdative molecular mass of water?
How many moles of barium chloride are pr"~nt in IOOg of the
hydrate;
How many moles of water are present in 100 g of the hydrate?
What is the ratio of moles of barium cbloride to moles of water?
What is the formula of barium chloride hydrate?
5. Calculate the empirical formula of the compound formed when 414g
of lead form 418gofa lead oxide.
What mass of lead is presenr?
How mmy moles of lead areprescm?
What mm ofO}.:ygen(nor oxide) is present?
How many moles of oxygen atoms are present?
What is the formula of this oxide of lead?
6. Calculate the empirical formulae of the following compounds
a) O.62gofpbospboruscombinedwitbO.48gofoxygcn
b) 1.4gofnitrogencombined1th0.30gofhydrogen
c) O.62gofkadcombincdwitbO.064gofox)'gen

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d)

3.5gofsi1iconcombjn~dwith4.0gofoxygen

e)

4.2gofnitrogcncombincdwith12.0gofoxygen

g)
7.

1.10gofmanganesc

f)

2.6 g of chromium

combined

combin~d

with

with

0.64g

of oxygen

53 g of chlorine

Find the molecular formula for each of the following compounds
from the empirical formula and the relative molecular mass:
Empirkalformula

M,

CF,

CH,

42

C1f40

CHlO

62
99

Empiricnlformub

M,

30

CH20

CH

78

C1HN01

AD

CH,

A porcelain boat was weighed. After a sample of the oxide of a metal
M, of A, = 1l<J, was placed in the boat, the boat was reweighed.
Then the boat was placed in a reduction tube, and heated while a
stream of hydrogen was passed over it. The oxide was reduced to the
metal M. The boat was allowed to cool with hydrogen still passing
OVer it, and then reweighed. Then it was reheated in hydrogen, cooled
againandrew~ighed.Thefollowingresul!swereoLrained.

FA

H

AD

H

9.

.A

H

M

B. A meral A1 forms a chloride of formula Mel1 and relative molecular
mass 127. The chloride reacts with sodium hydroxide solution to form
a precipitate of the metal hydroxide. What is the relative molecular
mass of the hydroxide?
a56
b71
c73
d90
c 146

a)
b)
c)
d)

Massof boat

=

6.lOg

=

9.67g

Massofboat+metaloxide'"
Mass of boar + metal (l)

10.63g

Massofboat+m~tal(2)
= 9.67g
Explain why hydrogen was passed over the boat while it was
cooling.
Explain why the boat + metal was reheated
Find theempirirnl formula of the metal oxide.
Write an equation for the reaction of the oxide with hydrogen.

,. Find the empirical formulae of the compounds formed in the reactions
described below:
a) 10.800gmagnesiumform18.000gofanoxide

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Finding FOfmu/ae

b) 3.400gcalcium form 9.43Sgofa chloride
c) 3.528gironformlO.237gofachloride

~?!:~~~ffO:: ~.~~~ S~~~~:1;
~
li~~~% ~
~ff~

2. CaJcuJatetheempiricalformulaeofrhccompoundswithrhefoliowing
percentage composition.
a) 77.7% Fe 22.3%0
b) 70.0% Fe 30.0%0
c) 72.4% Fe 27.6%0
d) 4O.2%K 26.9%Cr 32.9%0
e) 26.6%K 35.4%Cr 38.0%0 f) 92.3%C 7.6%H
g) 81.8%C 18.2%H

H

M

AD

3. Samples ofthcfoliowinghydratesarcWl'!ighed,he.atedtodriveoffth~
water of crystallisation, cooled and reweighed. From the results
obtained.calculate the values of a-j"ln the formulae of the hydrates
a) 0.869 g of CuSO. 'aH,O gave a residue of 0.556 g
b) 1.173 g ofCoCI,'bHP gave a residue of 0.641 g
c) 1.886g ofeaSO.'cHlO gave a residue of IA92g
d) 0.904 g of P,?(C,H,O,h'dH,O gave a residue of 0.774g
e) 1.144g ofNlSO.·cH,O gave a residue of 0.673 g
f) 1.175 g of KAI(SO.)'·jHlOga"e a residue ofO.639g.

AD

H
.A

4. An organic compound, X, which contains only carbon, hydrogen
and oxygen, has a molar masS of abour 85gmol-'. When 0.43g of X
isburntincxccssoxygcn,l.10gofcarbondioxidcandO.4Sgofwatcr
are formed
a) What is the empirical formula of Sf
b) What is the molecular formula of X?

FA

H

S. A liquid, Y, of molar mass 44 g mol-J contains 54.5% carbon, 36.4%
oxygenand9.1%hydrogcn
a) CalculatCrheempiricalformulaofl',and
b) deduce its molecular formula
6. An organic compound conuins 58.8% carboo, 9.8% hydrogen and
31.4% oxygen. The molar mass is 102gmol-'.
a) Calculate the empirical formula,and
b) deduce the molecular formula ofthc compound.
7. An organic compound has molar mass lSOgmol-' and contains 72.0%
carbon, 6.67% hydrogen and 21.33% oxygen. What is its moleculsr
formula;

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7

Reacting
Gases

Val urnes of

GAS MOLAR VOLUME

M

AD

A surpming feature of reactions between gases was noticed by a
Fr~nch chemist called Gay·LusStIl:in 1808. Gay-Llmac'. taw statex
that wben gllses combine they do so in zoiumes whicb bear a simple
ratio to one {mother and to tbe volume of tbe product if it isgaseous,
provided all.the volumes lire measured at the seme temperature and
pressure, For example, when hydrogen and chlorine combine, 1 dml
(or !itre) of hydrogen will combine exactly with ldmlofchlorineto
form2dmJofhydrogen
chloride, When nirrogen and hydrogen combine, a cen:ain volume of nitrogen will combine with lb~e times that
volume to form twice its volume of ammonia.

H

AD

H

.A

H

The Italian chemist Avogadro gave an explanation in 1811. His suggestion, kncwn.as Avogadro's hypothesis, is that, Equal volumes o[all
gase>(at the 'sametemperafure and pre55u,e) contain the S.1I11e
number
o[ molecules, It follows from Avogadro's l1yporhesis tbar, whenever
we>e<: an ",!uation "'presenting a reaction betWc~ngases, w~ can
substitute volumes of gases ill the same ratio as numbers of molecules,
Thus

FA

means that since
a rnulecule of nitrogen + 3 molecules of hydrogen form 2 molecules
of ammonia
then I volume of nitrogen + 3 volumes of hydrogen form 2 volumes
of ammonia.
For example, 1 dm' of nitrogen + 3 dm' of hydrogen form 2dm'
of ammonia.
Since equal volumesofga.ses(atthe~metemperatur<:and
pressure)
contain the same number of molecules, if you consider the Avogadro
constant {L),L molecules of carbon dioxide, L molcculcs of bydrcgen,
t molecules of oxygen and 50 on, then all the gases will occupy the
same volume. The volume occupied by L molecules of gas, which is
one mole of each gas, is called thegasI//()41'lJolullle.
In stating the volume ofa gas, one needs to state the temperature
and pressure at which the volume was me:lSl1red.lt is usual to give
the volume at OoC and 1 atmosphere pressure (273K and 1.OlX
lO'Nm-l). These conditions are called standard temperature and

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pressure (s.t.p.). Chapter 9 dells with calculating the volume of a
gas at s.t.p. from the 'Dlum!!measured under experimental conditions.

lone

mole uf

gas uccupie~ 22.4 dm'

at DOC
and 1atmosphere.

Since one rnole of gas occupies 22.4dml at s.t.p.,tbeg"s >Ila/"rvo lume
i~
22.4Jml a[ s.rp. This makes calculations on reacting volumes of
gases very 'imple. An equation which shows how many moles of
different gases react together also shows the ratio of the volumes of
the different gases that reacttogetheL For example, the equation
2NO(g)

+

O,(g) _

2NOlg)

M

Wbat is the volume of oxygen needed for the complete combustion
of 2dmlof propane?
Write the equation:

+

50,(g) -

H
.A

CJHig)

H

EXAMPLE
1

AD

tells us that 2 moles of NO + 1 mole of O2form 2 moles of NO,
44.8dm' of NO + 22.4dm' of 0, form 44.8dma of NO,
In general, 2 volumes of NO + 1 volume of 0, form 2 volumes of NO,.

+

3CO,(g)

4H,0(g)

1 mole of C.I·I"needs 5 moles of 0,
1 volume of CJIa needs 5 volumes of 0,. Therefore

AD

2dmJofpropancnecdlOdm'ofoxygen.

Wh'lt volume of hydrogen is obtained when 3.00g of zinc react with
an excess of dilute sulphuric acid at s.t.p.P

H

E)(AMl'LE2

FA

Write ilie equatiofl"
Zn(s) + H1S04(aq) _

H,(g)

+

ZnsOiaq)

1 mole of Zn forms 1 mole of H,
6Sgof Zn form 22.4dm30fH, (at s.t.p.)
3.00gofZnform

!tfX22.4dllll

'" l.D3dm'H,

3.00g of zinc give t.cj dms ct hydrogen ar s.r.p.

EXERCISE 11

Problems on Reacting Volumes of Gases

1. The problem is to find th~ p~rcentage by volume composition of a
mixture of hydrogen and ethane. When 75 em' of the mixture was
humed inan excess of oxygen, the volume of carbon dioxide produced
was 6DcmJ (allvolumes at s.t.p.)

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C.kuiBrlonsforA-.fetfiIOtemirtry

11)WrireaneqWltionforrhecomhustionofcmane
b) Say what volume of ethane would give 60 emJ of carbon dioxide.
c) Calculate the percentage of ethane in the mixture
2. 25 em> of carbon monoxide were ignited with 25 em' of oxygen. All
gas volumes were measured at the same temperature and pressure.
The reduction in the total volume was
a 2.5cm'
b 1O.Ocm' c I2.5cmJ
d IS.Oem'
e 25.0cm'
3. Ethene reacts with oxygen according ro [he equation
C,Hig) + 30,(g) __
2CO,(g) + 2H,O(l)

The table gives the formulae and relative molecular masses of some
gases.

M

4,

AD

IS.Oem' of ethene were mixed with 60.0eml of oxygen and the
mixture was sparked to complete the reaction. If all the volumes were
measured ar s.t.p., tbevolurru: of the products would be
a 15cm'
b 30em'
c 45cm'
d 60cm'
e 7Scm'

N, CoHo 0,

H

Formula

J2

NO,
40

SO,

S0,

46

64

80

4115

H

Volume (cml) occupied
bylgofgasats.t.p

.A

M,

J;O

FA

H

AD

a) Plot a graph of volume (on rhe vertical axis) againstM,(on the
horizontal axis).
b) Usethegrophto predict the volumes occupied at s.r.p. by
i) 19offluorine,F,
ii) 19ofd,.
c) What is the relative molecubt mass of a gas which occupies 508 em·
per gram ars.t.p.? If the gas contains only carbon and oxygen,
what is irs fonnula?
nCT!ON2

1, Ethene, H,C"'CH" and hydrogen react in the presence of a nickel
catalyst to form ethane.
a) Write a halanced equation for the reaction.
b) If a mixture of 30 em) of erhene and 20 em) of hydrogen is passed
over a nickel catalyst, what is the composition of the final mixture?
(Assume that the reaction is complete and that all gas volumes are
at s.t.p.)
2. Whatvolume~foxygen(ats.t,p.)isrequiredtoburnexactly:
a) ldmlofmcthane,accordingtothereaction
CH.(g) + 20,{gl
CO:(g) + 2HzO(g)

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b) 500cm' of hydrogen sulphide, according to the reaction
2HlS(g) + 30,(g) _
2S0,(g) + 2HP(g)
c)

2S0cm30fethync,accordiogtotheequation
2C,H,(g) + Se,ig) _
4co,(gl

d) 7S0cm' of ammonia, according
4NHl(g) + Se,ig) _

to

+

2H,O(gl

the reaction
4NO(g) + 6H,O(g)

e) Idm'ofphosphine,accordingwthereaction
PH,(g) + 20,(g) _
HJ'04(S)?

AD

3. 1dm' of H,S a:nd 1dm- of SO, were allowed to react, according 10
the equation
2H,S(g) + SO,(g) _
2H,O(l) + 3S(s)
WhatvoJume ofgaswiU remain after the reaction?

M

4. lOOcm30fa mixture of ethane and ethene at s.t.p. were treated with
bromine. OJ57g of bromine was used up. Calculate the percL'ntage
by volume of ethene in the mixture

H

AD

H

.A

H

5. Hydrogen sulphide bums in o:..ygen in accordance with thefoUowing
:
equation
2H,S(g) + 30,(g) _
2H,O(g) + 2S0,(g)
[f 4dm' of H,S ate burned in lOdm' of oxygen at 1 atmosphere
pressure and 120°C, what is the final volume of the mixture?
a 6dm'
b 8dm'
c lOdm'
d 12dm'
c) 14dm'

FA

FINDING FORMULAEBYCOMBUSTION
The formula of a hydrocarbon can be found from the results of a
combustion experiment. A hydrocarbon in the vapour phase is burned
ill an excess of o")'gcn to fonn carbon dioxide and water vapour
When The mixTur~ of gases is cooled [0 room temperature, water
vapour condenses ro occupy a very small volume. The gaseous mixture
consists of carbon dioxide and unused oxygen. The volume of carbon
dioxide can be found by absorbing it in an alkali. From the volumes
of gases, the equation for the reaction and [he formula of the hydro
carbon can be found.
The combustion method can be used for other compoundsalso,c,g.
ammonia (sec Examp!c~).
~XAMPlE When lOOcm' of a hydrocarbon X bum in SOOcm'of oxygen, SOcmJ
1
~6ooc~fe~f ~:a~n~:dfo~:!~~~~Ju~:~~~
and the forrnula of the hydrocarbon.

e~~~ri~n

f~rili~nr:~~t~~
f

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CaicuiatiOl1$forA-/eIl(lJChomisuy

x +

01(g)
450emJ

lOOemJ

+

C01(g)
300emJ

The volumes of gases reacting tell us thar
X + 4tOl(g) _
3CO,(g)

H,O(gl
300 em)

+

3H,O(g)

+
+

Hl,O(g)
6H,O(g)

To balance the equation, X must be C,H6' Then,
C,H6(g)
2CIHh)

+ 4!0,(g)
+ 90,(g)

_
_

3C01(g)
6CO,(g)

H

M

Write the equationC.Hb(g) + (a + bI4)O,(g) _

AD

I'XAMl'LE 10 cml of a hydrocarbon, C.H., arc exploded wi~ban excessof oxygen.
2
A contraction of 35 cml occurs,all volumes beIng measured at room
temperature and pre&5ure.On treatment of the products with sodium
hydroxide sOlution, a contraction of 40em' occurs. Deduce the
formula of the hydrocarbon.

aCO/g)

+

»a HP(!)

H

AD

H

.A

Volume of hydrocarbon '" lOem'
Volume of CO, '" a X Volume ofC.Hb
From reaction with NaOH, volume of CO, '" 40 eml
Therefore" = 4
Let [he volume of unused oxygen be cern'.
Final volume = Initial volume - 35 em'

FA

Note that H,O(l) is a liquid at room temperature and pressure, and
does not contribute to the final volume of gas.
40+c

'" lO+40+SbI2+c-35

2S = Sbl2
b =

}O

The formula is CJi,o'
EXAMf'L(l lOcm' of ammonia arc burned in an excess of oxygen at 110°C.
lOcm' of nitrogen and ,Ocm' of steam are formed. Deduce the
formula for ammonia, given that the formula of nitrogen is N1, and
the formula of steam is H,O
Let the formula of ammonia be N.H •.
The equation for combustion is
N.H.

+

bl40,

_

al2 N,

+

bl2 H,O

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Reac/ingVol"m.nofGMeS

VolumeofN.Ho = 20emJ
Volume nf N, = ,1/2.X 2.0 = lOa em' = 10eml
a = 1
VolumeofH,O(g) = bl2 X 2.0 = IObcml,., 30cml :. b= 3
The formula is NH,.
EXERCISE 12

Problems on Finding Formulae by Combustion

1. IOemJ of a hydrocarbon CxHJ. were exploded with an excess of
oxygen. There was a contraction of 30emJ. when the product was
treated with a solution of sodium hydroxide, there was n further
contraction of 30cmJ. Deduce the formula of the hydrocarbon. All
gas volumcs arc at s.t.p.

H

M

AD

2. 10 em" of a hydrocarbon CaHb were exploded with excess oxygen.
A contraction of 25cm' occurred. On treating the product with
sodium hydroxide, a further contraction of 4Qcm3 occurred. Deduce
the valucs of e and b in the formula of the hydrocarbon. All measure·
rucrus of gas volumes arc at s.t.p

H
.A

3. IOcm' of a hydTOcarboll C..Hs were explod<"d with an excess of
oxygen. A contraction of a cm-' occurred. On adding sodium
hydroxide solution, a further contraction ofbcm'occurred.
What
are the vclurnes.n and ej All gas volumes are at s.t.p.

FA

H

AD

4. When North Sea gas burns completely, it forms carbon dioxide and
water and no other products. When 250cm' of North Sea gas burn,
they need 500cm.l of oxygen, and they form 250cm" of carbon
dioxide and 500 em' of steam (the volumes being measured under the
Same conditio:!s). Deduce the equatinn for the reaction and the
formula of North Sea gas.
EXERCISE 13

Problems on Reactions Involving Solids and Gases

1. Inthereactionhetweenmarhleandhydrochloricacid,thcequarionis
CaCO,(s) + 2HCI(aq) _
C~C1iaq) + COI(g) + H,O(])
What mass of marble wouid be needed eo give I1.00g of carbon
dioxide?
Whatvolumewouldthisgasoccnpyats.t.p.?
2. Zinc reacts with aqueous hydrochloric acid to give hydrogen
Zn(s) + 2HC1(aq) _
H,(g) + ZnCl,(aq)
What mass of zinc would be needed tc give lOOg of hydrogen? What
vulumc would this gas occupy at s.t.p.?

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Calcu/ar/omfofA-I,wlChemiSlry

3. Sodiumhydrcgencarbonate
carbon dioxide
2NaHCOl(s) _

decomposes on heating, with evolution of
Na,COJ(s) + CO,(g)

+ H,O(g)

What volume of carbon dioxide (ar s.r.p.) am be obtained by heating
4.20g of sodium hydrogellcaTbonatl.'? If 4.2g of sodium hydrogen·
carbonate react with an excess of dilute hydrochloric acid, what
volume of carbon dioxide (at s.t.p.) is evolved?

M

AD

4. Many yean ago, bicycle lamps used to bum the ga.' ethyne, C,H,. The
gas was produced by allowing WltCfto drip on to calcium carbide. The
unbalanced equation for the reaction is
Calcium carbide + Water _
Ethyne + Calcium hydroxide
oc,«i + H,O(l) C,H,(g) + Ca(OH),(s)
a) Balance the equation.
b) Calculate the mass of calcium carbide which would he needed to
produC<'467 em' of cthyne (ats.t.p.)

.A

H

5. Dinirrogen oxide, N,O, is commonly called /Ilugbing gas. It can be
made by heating ammonium nitrate, NH.,NO). The unbalanced
equation for this reaction is,

to

AD

H

a) Balallce the equation
b) Calculate the mass of ammonium nitrate that must be heated
give

FA

H

i) 8.8goflaughinggas
iil 11.1 drul of laughing gas (at s.t.p.)
SECT'ON2

1. a) Analysis of an oxide of potassium shows that 1.42g of this oxide
contains O.64g of oxygen. What is irs empirical formula'
b) This oxide reacts with carbon dioxide to form oxygen and pctassium carbonate, K,COl. Write an equation for the reaction.
c) This reaction is sometimes used as a means of regenerating oxygen
ill submarines. What volume of oxygen (at s.e.p.) could be obtained
from 1.OOkgofthis oxide of potassium?
2. A cook is making a small cake. It needs 500cmJ (at s.t.p.) of carbon
dioxide to make the cake rise. The cook decides to add baking powder,
which contains sodium hydrogenC"Jrbonate. This generales carbon
dioxidebyrhermaldecomposition:
2NaHCO,{s) _
CO,(g) + Na,COJ(s) + H,O(l)
Wbll mass of sodium hydrogen carbonate must the cook add to the
cake mixture?

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Maleri"'

R/UlctingVolume$of G~

3. A certain industrial plant emits 90 tonnes of nitrogen monoxide. ~O.
daily through its chimneys. The firm decides to remove nitrogen
monoxide from its waste gases by means of the reaction
Methane

+ ~~~~~~~c _

Nitrogen

+ ;;:;;~: + Wm:r

2N1(g) + CD1(g) + 2H,O(l)

CH.(g) + 4NO(g)

If methane (North Sea gas) costs 0.50p per cubic metre, what will this
dean""lp reaction cost the firm to run? (ignore the cost ofinstalling
the process. which will in reality be high.)
(/lint Tonnes of NO .,. moles of NO .. moles of CH•...
of CH•... cost of CH. (I m" = IOOOdm'.)
In thc Solvay process

+

N.CI('q)

NIi,(g)

+

Il,O{I)

+ CO,(g)

AD

4.

volume

______..

N&llCO,{s)

+

NIl.CI(oq)

H

M

what volume of carbon dioxide (at s.r.p.) is required to produce
1.00kgofsodiumhydrogel(:arbonate?

H
.A

S. Find-the volume ofethyne (at s.t.p.) that can be prepared from 10.0g
ofcakium carbide by the reaction
Ca(OH),(aq)

+ C,H,(g)

AD

CaC,(s) + 2H,O(1) _

Find the mass of phosphorus required fO!lhepreparationof200cm
of phosphine (at s.t.p.l by the reaction

+ 3NaOH(aq) +

3H,O(l) _

3NaH,P04(aq)

'

+ Pth(g)

FA

p.(s)

H

6.

7.

Calculate the mass of ammonium chloride required to produce
1.00dm30fammonia(ats.t,p.)jnther~accion
2NH.O(s) + Ca(OHh(s) _
2NHh) + CaCI1{s) + 2H,0(g)

8.

What mass of potassium chtcraterv) must be heated to give 1.00dm3
of oxygen at s.r.p.> The reaction is
2KC10,(s) _
2KO(s) + 30h)

9.

Whar volume of chlorin~ (at s.t.p.) can [Ko, obtained from the
elecrrolysis of a soluticn containing 60,Ogof sodium chloride?

10. What volume of oxygen (at s.r.p.j is needed for the complete ccrnbustionofl.OOkgofoctane?Thereartionis
2CsH,s{l) + 250,(g) _
16CO;o{g) + 18H,0(g)

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Calc"lariomforA·hNela.emi.rry

EXERCISE 14

Questions from A·level Papers

1. a) Describe the chemistry involved in tberustingofiron,andc"plain
two·.vays(different in principle) by whicb it maybe prevented.
b) Aerials in portable radios arc made of a mixed oxide of calcium
and iron known as 'Ferrite'. It contains 18.5% calcium and 51.9%
iron by mass. Calculate the empirical fomula of 'Ferrite' and
hence deduce the oxidation number of the iron itcomains. (C92)

AD

·2. When chlorine is passed over heated sulphur in the absence of air, an
orange liquid A that contains 47.5% of sulphur, by mass, is formed.
Its relative molecular mass is about 135. When ,1 is treated with
ammonia dissolved in benzene, an explosive orange-yellow solid II is
obtained. Compound B contains 30.4% of nitrogen by mass, the
remainder being sulphur. Its relative molecular mass is about 185.
B}' x-ray diffraction, it has been shown that all the bonds in this
compound a"" the same length

H

.A

H

M

When B is reduced by rin(II) chloride in a metbanol/henzene solurioo.
a compound C that contains 29.8% of nitrogen and 68.1% of sulphuf,
bymass,isform~d.
a) What is the likely fonnula of subnance A? Draw a diagram to show
thelikely shape ofits molecule.
b) Suggest structures for substanccs s and C, explaining c1elrly how
)'OUarrived at your condusions.
e) Write an equatjonforth~con'ersionofsubstanceB
into substance

AD

c.

(C91S)

H

d) Suggestat~asanwhysubstancellisexplo,ivc

FA

'3. A bright red salidA dissolves in water to give a highly acidic solution
When a solution of 11is made alkaline with aqueous sodium hydroxide.
the solution rums yellow. On heating l.Og of A, O.76g of a green
powder H is fonn~d and 168cm3 of oxygen (measured at s.r.p.) are
giv~n off.
An orange solid C may he obtained from the solution made by dissolv·
lng l.OgofA in 5.0cmJ of 2.0moldm-l ammonia
When 1.26g ofCarewarmed,avio]enrreaedon
takes place, giving off
112 em! (measured at u.p.) of an inert gas, as well as steam, and
leaving behind the same green powder Il that was mad~ by h~ating 11.
!fA is dissolved in cold, concentratedbydrochloricacid,andconcentrated sulphuric acid is tben grad~aUy added, a dark, ~ed·brown oil./)
separates out. /) has a boilingpomt of 117°C and rapldlyreactswnh
water.D COOlliins 5.8% of chlorine by mass.
4
Identify A. B. C andD, give equations for all the reactions involved
and show that these equ"tions arc consistent with the quantitative
data,
(Cns)

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Book on Moles / Stoichiometry for O'Levels and A'levels (Grade 9/10/11/12)

Book on Moles / Stoichiometry for O'Levels and A'levels (Grade 9/10/11/12)

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