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


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Excellent book on Moles / Stoichiometry, which covers the following topics for O'Levels and A'levels

-Writing and Balancing Equations
-Moles and Mass/Solids
-Moles and Solutions/Concentration/Volume
-Moles and Gases
-Empirical and Molecular Formula
-Reaction Kinetics

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

  1. 1. FA H AD H .A H M AD Contact: +92 323 509 4443
  2. 2. 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, ,h . prior w,;tre n ."h5<hl of 1b.",pyrigh,ho!4.1l. Appl;catiansfo,,,,,.h!"",missionshoul db e .dd"'.,..d,othepuhlish . ... 'SlOnl.yThornes(Publi>'d.OldS .. 'ionO';" •. 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 H M A calalogue n:con!ofthis AD 00010203040511019181716151413 bo"d, for p<rmi"';on to prin. question. from H u.min.,;on .A ACKNOWU;OGEMENTS ~'~~k~~:,~~l~~;;~~ H AD 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 Boord, Th. Oxford ~I.g.<y of Loc.1 Exam;n .. ;on.; The 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 .. FA 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", Contact: +92 323 509 4443 Urheberre<;hlliGh geschlitlles ""'aleri~1
  3. 3. Contents Acknowledemcms ForewordbyProfes,orRl> een r-es vii .A Furmulae and egualions Formulae 17 cguatioos18 H 2 H M AD 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 H n percentage FA The mole relative molecular mass 21 AD Relative atomic mass 21 composition 23 5 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 Contact: +92 323 509 4443 U,hebcfrochtlichgcscllutztes Materi~1
  4. 4. 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 127 lhctqualion 117 120 the [he 127 aOprn.!p1s '''<l]I, trom m ea" re- If FA H AD H .A H M AD 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_ Contact: +92 323 509 4443 Urhellerr6Ch!li~hgeschli!z!es Ma!erin!
  5. 5. List of Exercises 15 20 23 25 30 lJ ppctjcewjrbCalclIlaljons Practice with Eguatiolls Problems 011Relative Molecular M= Problems on Perl'entage Composition " 18 40 FA H AD H .A H M AD 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·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 " 47 H 5 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 J69 Contact: +92 323 509 4443 Urhebermchllichge"chutltesMatmiBI
  6. 6. 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 AD 44 45 46 47 48 M !~ ~~~I~::s H .A H '" H AD 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. FA 61 ABBREVIATI0NS OF EXAMINATION BOARDS AEB C JMB L NI o 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 Contact: +92 323 509 4443 U,hdm,mchllichgc5chlitzlesMaleriBI
  7. 7. Foreword AD 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. FA H AD H .A H M 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 Contact: +92 323 509 4443 Urheberre<;htlichgcschutltcsMaterial
  8. 8. 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 AD H .A H M AD 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. FA H 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 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 Contact: +92 323 509 4443 Urheb~r",chtliGhge"GhutltcsMaleri"i
  9. 9. 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~ H M AD 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. H AD H .A 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. FA 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 Contact: +92 323 509 4443 Urheberre<;htlichgcschlitltcsMa!erial
  10. 10. 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 H .A H M AD 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"). FA H AD 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 Contact: +92 323 509 4443 Urhebmre<;htlichgcschlitltcsMatmiol
  11. 11. 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 AD x H .A H M the triangle is PRACTice1 FA H AD 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 =? Contact: +92 323 509 4443 Urheb~r",chtliGhge"GhutltcsMaleri"i
  12. 12. 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 M AD 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 .A ad"" be H How can you find out whether this is correct? First multiply both sides of the first eqcation by d. AD ~ FA H 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 Contact: +92 323 509 4443 Urheberre<;htliGh geschlitltes ""'ateri~1
  13. 13. 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 AD 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"'? H M 3. Rearrangethe equation AD CALCULATIONSON RATIO H .A 10give aj an equaticn for p and h) an equauon for q FA H 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 Contact: +92 323 509 4443 UrheberrcchtlicilgeschutltosMa!m
  14. 14. 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? AD WORKINGWITH NUMBERSIN STANDARD FORM H M 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 H .A ~=:, FA H AD ~: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' Contact: +92 323 509 4443 Urhebmre<;htlichgcschlitltcsMatmiol
  15. 15. 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 AD To enter 10-', you enter 1; then press the EXP key, then the 3 key, andlastly the+f-key. M ESTIMATING YOUR ANSWER .A H 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) H For example, AD 2456xO.0123xO.00414 5223X60.7XS.51 FA H 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 Contact: +92 323 509 4443 Urheb~r",chtliGhge"GhutltcsMaleri"i
  16. 16. 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. H .A H M 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 Contact: +92 323 509 4443 Urheberre<;htlichgcschlitltcsMa!erial
  17. 17. 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 H M AD 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 H AD H .A 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; 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 Contact: +92 323 509 4443 Urheberre<;htlichgeschutltesMa!eriol
  18. 18. 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 H .A H M 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 Contact: +92 323 509 4443 Urhet>erre<;htlichgeschutltesMateriol
  19. 19. 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 H M AD 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 Contact: +92 323 509 4443 Urhebmre<;htlichgcschlitltcsMatmiol
  20. 20. 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 AD 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. Contact: +92 323 509 4443 Urheberre<;htliGh geschlitlles ""'aleri~1
  21. 21. 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. H .A 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) Contact: +92 323 509 4443 Urhcbcrr",hllichgcschulllcsMaleriol
  22. 22. 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". Contact: +92 323 509 4443 UrhebGrrc<;hllichgcschulzlesMater;ill j
  23. 23. 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 H M AD 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 H AD H .A 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-' Contact: +92 323 509 4443 Urheberre<;htlichgeschulltesMateriol
  24. 24. 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 .A H 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 Now balance the charges Ions needed are The formllb i<; .)·<}dilJm5ulph~lc Na+andSO.1Two Na+ balance one 504'2N:l:andSO.1NalS04 Compound tens present are Now balance the charges 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. Contact: +92 323 509 4443 Urneberre<;htlichge5chl'it;:lesMateri"1
  25. 25. Calculstinl/s(<>rA-IevtJIChsmi<try Compollnd Ions present are Now balance the charges 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 Contact: +92 323 509 4443 Urhebmre<;htlichgcschlitltcsMatmiol
  26. 26. 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) Contact: +92 323 509 4443 Urhebmre<;htlichgcschlitltcsMatmiol
  27. 27. 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 Contact: +92 323 509 4443 Urheberre<;htlichgcschlitltcsMaterial
  28. 28. 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 AD H .A 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: Contact: +92 323 509 4443 Urhebmre<;htlichgcschlitltcsMatmiol
  29. 29. 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 Contact: +92 323 509 4443 Urheberre<;htliGh geschlitlles ""'aleri~1
  30. 30. 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 Contact: +92 323 509 4443 Urhcberre<;htlichgcschutltcsMaterial
  31. 31. 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. Contact: +92 323 509 4443 Urheb~r",chtliGhge"GhutltcsMaleri"i
  32. 32. 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. Contact: +92 323 509 4443 UrheberrcchtlicilgeschutltosMa!m
  33. 33. 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? Contact: +92 323 509 4443 Urneberre<;htlichge5chl'it;:tesMateri"1
  34. 34. 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 Contact: +92 323 509 4443 Urhebmre<;htlichgcschlitltcsMatmiol
  35. 35. 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 Contact: +92 323 509 4443 Urheberre<;htliGh geschlitlles ""'aleri~1
  36. 36. 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. Contact: +92 323 509 4443 Urheberre<;htliGh geschlitlles ""'aleri~1
  37. 37. 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? Contact: +92 323 509 4443 Urheberre<;htlichgcschutltcsMaterial
  38. 38. 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? Contact: +92 323 509 4443 Urhcbcrre<;htlichgcschlitltcsMatcrial
  39. 39. €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? Contact: +92 323 509 4443 Urhcbcrre<;htlichgeschutltesM"tc
  40. 40. 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"~lu~ Fluorideion(F-) + Charcoal (Cl _ Carbon tetrafluoride (CF.) The product, CF4, is harmless. The firm claims that I kg of charcoal Contact: +92 323 509 4443 Urheberre<;htliGh geschlitltes "__'ateri~1
  41. 41. 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 Contact: +92 323 509 4443 Urhebmre<;htlichgcschlitltcsMatmiol
  42. 42. 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 Contact: +92 323 509 4443 Urheberre<;htlichgeschutltesM"te
  43. 43. 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 Contact: +92 323 509 4443 Urheberre<;htliGh geschlitlles ""'aleri~1
  44. 44. Equafiomlll1drheMoie 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 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 Contact: +92 323 509 4443 Urhebmre<;htlichgcschlitltcsMatmiol
  45. 45. OllculfJtloM fOf A-"vel OU/miW'I 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. Contact: +92 323 509 4443 Urheberre<;htliGh geschlitlles ""'aleri~1
  46. 46. 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 Contact: +92 323 509 4443 UrheberrcchtlicilgeschutltosMa!mial
  47. 47. 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 •. Contact: +92 323 509 4443 Urneberre<;htlichge5chl'it;:tesMateri"1
  48. 48. 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 Contact: +92 323 509 4443 UrhebermchtliGhgeschlitltosMaterial
  49. 49. 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 Contact: +92 323 509 4443 Urheberre<;htlichgeschutltesM"terial
  50. 50. 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; Contact: +92 323 509 4443 Urheberre<;htlichgeschulltesMateriol
  51. 51. 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, 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<: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 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 Contact: +92 323 509 4443 Urheb~r",chtliGhge"GhutltcsMaleri"i
  52. 52. ReactingVolrJfflO'$afG_ 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.) Contact: +92 323 509 4443 UrhebermchtliGhgeschlitltosMateriat
  53. 53. 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) Contact: +92 323 509 4443 Urheb~r",chtliGhge"GhutltcsMaleri"i
  54. 54. R(J1!cringVo'umefofGases 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 Contact: +92 323 509 4443 Urhebmre<;htlichgcschlitltcsMatmiol
  55. 55. 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 Contact: +92 323 509 4443 Urheberre<;htliGh geschlitlles ""'aleri~1
  56. 56. 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.? Contact: +92 323 509 4443 Urheberre<;htlichgcschutltcsMaterial
  57. 57. 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? Contact: +92 323 509 4443 Urheb~r",chtliGhge"chutltcs Maleri"'
  58. 58. 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) Contact: +92 323 509 4443 UrnebermchtliGilgeschlitltosMaterial
  59. 59. 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) Contact: +92 323 509 4443 Urheberre<;htliGh geschlitlles ""'aleri~1