THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS Volume 1 1st 20th IChO 1968 1988 Edited by Anton Sirota IUVENTA, Bratislava, 2008 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 1Editor:Anton SirotaISBN 978-80-8072-082-7 Copyright 2008 by IUVENTA IChO International Information Centre, Bratislava, SlovakiaYou are free to copy, distribute, transmit or adapt this publication or its parts for unlimited teaching purposes, however,youareobligedtoattributeyourcopies,transmissionsoradaptationswithareferenceto"The CompetitionProblemsfromtheInternationalChemistryOlympiads,Volume1"asitisrequiredinthe chemical literature. The above conditions can be waived if you get permission from the copyright holder. Issued by IUVENTA in 2008 with the financial support of the Ministry of Education of the Slovak Republic Number of copies: 250 Not for sale. International Chemistry OlympiadInternational Information CentreIUVENTABdkov 2 811 04 Bratislava 1, SlovakiaPhone: +421-907-473367 Fax: +421-2-59296123 E-mail: anton.sirota@stuba.sk Web: www.icho.sk Contents Preface. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 VOLUME 1 The competition problems of the: 1st IChO. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 2nd IChO. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 3rd IChO. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 4th IChO . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 5th IChO. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 6th IChO. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 7th IChO. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 8 th IChO. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 9th IChO. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11610th IChO. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 11th IChO. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 12th IChO. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 13th IChO. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220 14th IChO. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 15th IChO. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268 16th IChO. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296 17th IChO. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319 18th IChO. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 340 19th IChO. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 366 20th IChO. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383 1 Preface This publication contains the competition problems from the first twenty International ChemistryOlympiads(IChO)organizedintheyears19681988.Ithasbeenpublished bytheIChOInternationalInformationCentreinBratislava(Slovakia)ontheoccasionof the 40th anniversary of this international competition.Not less than 125 theoretical and 50 practical problems were set in the IChO in the mentioned twenty years. In the elaboration of this collection the editor had to face certain difficulties because the aim was not only to make use of past recordings but also to give themsuchaformthattheymaybeusedinpracticeandfurtherchemicaleducation. Consequently, it was necessary to make some corrections in order to unify the form of the problems.However,theydidnotconcernthecontentsandlanguageoftheproblems. Many of the first problems were published separately in various national journals, in different languages and they were hard to obtain. Some of them had to be translated into English.Mostofthexeroxcopiesoftheproblemscouldnotbeuseddirectlyandmany texts, schemes and pictures had to be re-written and created again. The changes concern inparticularsolutionsoftheproblemssetinthefirstyearsoftheIChOcompetitionthat wereoftenavailableinabriefformandnecessaryextentonly,justfortheneedsof members of the International Jury. Some practical problems, in which experimental results andrelativelysimplycalculationsarerequired,havenotbeenaccompaniedwiththeir solutions. Recalculations of the solutions were made in some special cases ony when the numeric results in the original solutions showed to be obviously not correct. Although the numbersofsignificantfiguresintheresultsofseveralsolutionsdonotobeythecriteria generally accepted, they were left without change.InthispublicationSIquantitiesandunitsareusedandamoremodernmethodof chemicalcalculationsisintroduced.Onlysomeexceptionshavebeenmadewhen,inan effort to preserve the original text, the quantities and units have been used that are not SI.Unfortunately, the authors of the particular competition problems are not known and due to the procedure of the creation of the IChO competition problems, it is impossible to assignanyauthor'snametoaparticularproblem.Nevertheless,responsibilityforthe scientific content and language of the problems lies exclusively with the organizers of the particular International Chemistry Olympiads. 2 Nowadays many possibilities for various activities are offered to a gifted pupil. If we wanttogainthegiftedandtalentedpupilforchemistrywehavetolookforwayshowto evokehisinterest.TheInternationalChemistryOlympiadfulfilsallpreconditionstoplay this role excellently. This review of the competition problems from the first twenty International Chemistry Olympiadsshouldservetobothcompetitorsandtheirteachersasasourceoffurther ideasintheirpreparationforthisdifficultcompetition.Forthosewhohavetakenpartin some of these International Chemistry Olympiads the collection of the problems could be of help as archival and documentary material. The edition of the competition problems will continuewithitssecondpartandwillcontaintheproblemssetintheInternational Chemistry Olympiads in the years 1989 2008. TheInternationalChemistryOlympiadhasits40thbirthday.Inthepreviousforty yearsmanyknownandunknownpeople-teachers,authors,pupils,andorganizers- provedtheirabilitiesandknowledgeandcontributedtothesuccessofthisalreadywell known and world-wide competition. We wish to all who will organize and attend the future International Chemistry Olympiads, success and happiness. Bratislava, July 2008 Anton Sirota, editor 1 11 1st st st st 4 theoretical problems 2 practical problems THE 1ST INTERNATIONAL CHEMISTRY OLYMPIAD, Prague, 1968 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 1 Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 4 THE FIRST INTERNATIONAL CHEMISTRY OLYMPIAD 1821 JULY 1968, PRAGUE, CZECHOSLOVAKIA _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ THEORETICAL PROBLEMS PROBLEM 1 A mixture of hydrogen and chlorine kept in a closed flask at a constant temperature was irradiated by scattered light. After a certain time the chlorine content decreased by 20 % compared with that of the starting mixture and the resulting mixture had the composition asfollows:60volume%ofchlorine,10volume%ofhydrogen,and30volume%of hydrogen chloride. Problems: 1.1What is the composition of the initial gaseous mixture? 1.2How chlorine, hydrogen, and hydrogen chloride are produced? ____________________ SOLUTION 1.1H2 + Cl2 2 HCl 30volumepartsofhydrogenchloridecouldonlybeformedbythereactionof15 volumepartsofhydrogenand15volumepartsofchlorine.Hence,theinitial composition of the mixture had to be: Cl2: 60 + 15 = 75 % H2:10 + 15 = 25 % 1.2 ChlorineandhydrogenareproducedbyelectrolysisofaqueoussolutionsofNaCl:NaCl(aq)Na+(aq) + Cl- (aq) anode:2 Cl- 2 e Cl2 cathode:2 Na++ 2 e2 Na THE 1ST INTERNATIONAL CHEMISTRY OLYMPIAD, Prague, 1968 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 1 Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 5 2 Na + 2 H2O2 NaOH + H2 Hydrogen chloride is produced by the reaction of hydrogen with chlorine. THE 1ST INTERNATIONAL CHEMISTRY OLYMPIAD, Prague, 1968 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 1 Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 6 PROBLEM 2 Write down equations for the following reactions:2.1Oxidation of chromium(III) chloride with bromine in alkaline solution (KOH). 2.2Oxidation of potassium nitrite with potassium permanganate in acid solution (H2SO4). 2.3Action of chlorine on lime water (Ca(OH)2) in a cold reaction mixture. ____________________ SOLUTION 2.12 CrCl3 + 3 Br2 + 16 KOH2 K2CrO4 + 6 KBr + 6 KCl + 8 H2O 2.2 5 KNO2 + 2 KMnO4+ 3 H2SO4 2 MnSO4 + K2SO4 + 5 KNO3 + 3 H2O 2.3. Cl2 + Ca(OH)2 CaOCl2 + H2O THE 1ST INTERNATIONAL CHEMISTRY OLYMPIAD, Prague, 1968 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 1 Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 7 PROBLEM 3 The gas escaping from a blast furnace has the following composition: 12.0 volume % of CO228.0 volume % of CO 3.0 volume % of H2 0.6 volume % of CH4 0.2 volume % of C2H456.2 volume % of N2

Problems: 3.1Calculatethetheoreticalconsumptionofair(inm3)whichisnecessaryforatotal combustion of 200 m3 of the above gas if both the gas and air are measured at the same temperature. (Oxygen content in the air is about 20 % by volume). 3.2Determinethecompositionofcombustionproductsifthegasisburnedina20% excess of air.____________________ SOLUTION O2 ________ 3.12 CO + O22 CO214 2 H2 + O2 2 H2O1.5 CH4 + 2 O2

CO2 + 2 H2O 1.2 C2H4 + 3 O2 2 CO2 + 2 H2O0.6 _________ 17.3 parts x 5 = 86.5 parts of the air 200 m3 of the gas ........ 2 x 86.5 = 173.0 m3 of the air+ 20 %34.6 m3 ________________ 207.6 m3 of the air 3.2 207.6:5 = 41.52 parts of O2 : 2 = 20.76 parts of O2 for 100 m3 of the gas20.76x4 = 83.04 parts of N2for 100 m3 of the gas THE 1ST INTERNATIONAL CHEMISTRY OLYMPIAD, Prague, 1968 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 1 Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 8 Balance:CO2 H2ON2O2 (volume parts)12.00 3.00 56.2020.76 28.00 1.20 83.04- 17.30 0.60 0.40 0.40___________________________________________ 41.00 4.60 139.24 3.46 Total: 41.00 + 4.60 + 139.24 + 3.46 = 188.30 of volume parts of the gaseous components. 24.60%H O 100 2.44188.30= = 2139.24%N 100 73.95188.30= = 23.46%O 100 1.84188.30= = THE 1ST INTERNATIONAL CHEMISTRY OLYMPIAD, Prague, 1968 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 1 Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 9 PROBLEM 4 Avolumeof31.7cm3ofa0.1-normalNaOHisrequiredfortheneutralizationof0.19 g of an organic acid whose vapour is thirty times as dense as gaseous hydrogen.Problem: 4.1Give the name and structural formula of the acid. (The acid concerned is a common organic acid.) ____________________ SOLUTION 4.1 a) The supposed acid may be: HA, H2A, H3A, etc. n(NaOH) = c V = 0.1 mol dm-3x0.0317 dm3=3.17 10-3 mol mol10 17 . 3(acid)3vn= where v = 1, 2, 3,......

(acid)(acid)(acid)Mmn = 13mol g 60mol 10 17 . 3g 19 . 0(acid) = = v v M (1) b) From the ideal gas law we can obtain: 112 2MM=M(H2) = 2 g mol-1 M(acid) = 30x2 = 60 g mol-1 By comparing with (1): v = 1 The acid concerned is a monoprotic acid and its molar mass is 60 g mol-1. The acid is acetic acid: CH3COOH THE 1ST INTERNATIONAL CHEMISTRY OLYMPIAD, Prague, 1968 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 1 Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 10 PRACTICAL PROBLEMS PROBLEM 1 (Practical) Therearetentesttubesintherackatyourdisposal(110)andeachtesttubecontainsoneofaqueoussolutionsofthefollowingsalts:Na2SO4,AgNO3,KI,Ba(OH)2, NH4Cl, Ag2SO4, Pb(NO3)2, NaOH, NH4I, KCl.Foridentificationoftheparticulartesttubesyoucanusemutualreactionsofthe solutions in the test tubes only. Determineinwhichorderthesolutionsofthesaltsinyourrackareandwrite chemical equations of the reactions you used for identification of the salts. PROBLEM 2 (Practical) Eachofthesixtesttubes(AF)intherackcontainsoneofthefollowing substances: benzoic acid, salicylic acid, citric acid, tartaric acid, oxalic acid and glucose.Determine the order in which the substances in the test tubes are placed in your rack and give chemical reactions you used for identification of the substances.Foridentificationofthesubstancesthefollowingaqueoussolutionsareatyour disposal: HCl, H2SO4, NaOH, NH4OH, CuSO4, KMnO4, FeCl3, KCl,and distilled water. 2 22 2nd nd nd nd 4 theoretical problems 2 practical problems THE 2ND INTERNATIONAL CHEMISTRY OLYMPIAD, Katowice, 1969 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 12 THE SECOND INTERNATIONAL CHEMISTRY OLYMPIAD 1620 JUNE 1969, KATOWICE, POLAND _______________________________________________________________________ THEORETICAL PROBLEMS PROBLEM 1 Anamountof20gofpotassiumsulphatewasdissolvedin150cm3 ofwater.The solution was then electrolysed. After electrolysis, the content of potassium sulphate in the solution was 15 % by mass. Problem: Whatvolumesofhydrogenandoxygenwereobtainedatatemperatureof20Canda pressure of 101 325 Pa? ____________________ SOLUTION Onelectrolysis,onlywaterisdecomposedandthetotalamountofpotassium sulphate in the electrolyte solution is constant. The mass of water in the solution: 1.1Before electrolysis (on the assumption that = 1 g cm-3): m(H2O) = 150 g 1.2After electrolysis:m(H2O) = m(solution) m(K2SO4) = 20 g0.15 20 g = 113.3 gThe mass of water decomposed on electrolysis: m(H2O) = 150 113.3 = 36.7 g,i. e. n(H2O) = 2.04 mol Since,2 H2O2 H2 + O2 thus,n(H2) = 2.04 mol n(O2) = 1.02 mol

THE 2ND INTERNATIONAL CHEMISTRY OLYMPIAD, Katowice, 1969 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 13 -1 -122(H ) 2.04 mol8.314 JmolK293.15 K(H )101325 Pan RTVp = = 0.049 m3, resp. 49 dm3

V(O2) = V(H2) 0.0245 m3 24.5 dm3 THE 2ND INTERNATIONAL CHEMISTRY OLYMPIAD, Katowice, 1969 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 14 PROBLEM 2 A compound A contains 38.67 % of potassium, 13.85 % of nitrogen, and 47.48 % of oxygen.Onheating,itisconvertedtoacompoundBcontaining45.85%ofpotassium, 16.47 % of nitrogen, and 37.66 % of oxygen. Problem: 2.1What are the stoichiometric formulas of the compounds? 2.2Write the corresponding chemical equation. ____________________ SOLUTION 2.1Compound A: KxNyOz 1647.481413.8539.138.67z : y : x = = = = 0.989 : 0.989 : 2.968 = 1 : 1 : 3 A : KNO3 Compound B: KpNqOr1637.661416.4739.145.85r : q : p = = = = 1.173 : 1.176 : 2.354 = 1 : 1 : 2 B : KNO2 2.2Equation:2 KNO32 KNO2 + O2 THE 2ND INTERNATIONAL CHEMISTRY OLYMPIAD, Katowice, 1969 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 15

PROBLEM 3 A10cm3 sampleofanunknowngaseoushydrocarbonwasmixedwith70cm3 of oxygen and the mixture was set on fire by means of an electric spark. When the reaction wasoverandwatervapourswereliquefied,thefinalvolumeofgasesdecreasedto65 cm3.Thismixturethenreactedwithapotassiumhydroxidesolutionandthevolumeof gases decreased to 45 cm3. Problem: What is the molecular formula of the unknown hydrocarbon if volumes of gases were measured at standard temperature and pressure (STP) conditions? ____________________ SOLUTION The unknown gaseous hydrocarbon has the general formula:CxHy mol22.40.010mol dm 22.4dm 0.010) H (C1 33y x= =nBalance of oxygen: -Before the reaction: 70 cm3,i. e. 0.07022.4mol -After the reaction: 45 cm3,i. e. 0.04522.4mol Consumed in the reaction: 0.02522.4mol of O2 According to the equation: CxHy + (x +y4) O2 = x CO2 + y2H2O Hence, 0.02022.4molofO2 reactedwithcarbonand 0.02022.4molofCO2 wasformed(C + O2 = CO2), 0.00522.4molO2 combinedwithhydrogenand 0.01022.4molofwaterwasobtained(2 H2 + O2 = 2 H2O). THE 2ND INTERNATIONAL CHEMISTRY OLYMPIAD, Katowice, 1969 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 16 3 n(C) = n(CO2) = 0.02022.4moln(H2) = 2 n(H2O) = 0.02022.4 mol x : y = n(C) : n(H2) = 0.020 : 0.020 = 1 : 1 FromthepossiblesolutionsC2H2,C3H3,C4H4,C5H5.onlyC2H2 satisfiestotheconditions given in the task, i. e. the unknown hydrocarbon is acetylene. THE 2ND INTERNATIONAL CHEMISTRY OLYMPIAD, Katowice, 1969 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 17 PROBLEM 4 Calcium carbide and water are the basic raw materials in the production of: a)ethanol b)acetic acid c)ethylene and polyethylene d)vinyl chloride e)benzene Problem: Givebasicchemicalequationsforeachreactionbywhichtheabovementioned compounds can be obtained. SOLUTION Basic reaction:CaC2 + 2 H2O = Ca(OH)2 + C2H2 From acetylene can be obtained: a) ethanol CH CHCH2CH3CH3HgSO4(catalyst)diluted H2SO4+H2O CHOHvinyl alcoholrearrangement CHOreductionCH2OHacetaldehyde ethanol b)acetic acid CH CHCH2CH3CH3COOHacetic acidHgSO4(catalyst)diluted H2SO4+H2O CHOHvinyl alcoholrearrangement CHOoxidationacetaldehyde THE 2ND INTERNATIONAL CHEMISTRY OLYMPIAD, Katowice, 1969 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 18 c)ethylene, polyethylene CH CH CH2catalyst+H2O CH2ethylene

CH2CH2pressure, temperaturecatalyst( - CH2 - CH2 - )n polyethylene d) vinyl chloride CH CH CH2+HCl CHCl vinyl chloride e) benzene

CH CH 3400 -500 C THE 2ND INTERNATIONAL CHEMISTRY OLYMPIAD, Katowice, 1969 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 19 PRACTICAL PROBLEMS PROBLEM 1 (Practical) a)Threenumberedtest-tubes(1-3)containmixturesoftwosubstancesfromthe following pairs (4 variants): 1. ZnSO4 - NaBr NaCl - Ca(NO3)2MgSO4 - NH4Cl 2. AlCl3 - KBr CaCl2 - NaNO3ZnCl2 - (NH4)2SO4 3. KNO3 - Na2CO3 KCl - MgSO4NH4Cl - Ba(NO3)2 4. MgCl2 - KNO3 K2CO3 - ZnSO4 Al(NO3)3 - NaCl b)Each of the test-tubes numbered 4 and 5 contains one of the following substances: glucose, saccharose, urea, sodium acetate, oxalic acid. Problem: Bymeansofreagentsthatareavailableonthelaboratorydeskdeterminethe contentoftheindividualtest-tubes.Givereasonsforboththetestsperformedandyour answers and write the chemical equations of the corresponding reactions. Note: Fortheidentificationofsubstancesgivenintheabovetask,thefollowingreagents wereavailabletocompetingpupils:1NHCl,3NHCl,1NH2SO4,concentratedH2SO4, FeSO4, 2 N NaOH, 20 % NaOH, 2 N NH4Cl, 2 N CuSO4, 2 N BaCl2, 0,1 N AgNO3, 0,1 % KMnO4,distilledwater,phenolphtalein,methylorange.Inaddition,furtherlaboratory facilities, such as platinum wire, cobalt glass, etc., were available. THE 2ND INTERNATIONAL CHEMISTRY OLYMPIAD, Katowice, 1969 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 20 PROBLEM 2 (Practical) Allow to react 10 cm3 of a 3 N HCl solution with the metal sample (competing pupils weregivenpreciselyweighedsamplesofmagnesium, zincoraluminium) and collect the hydrogen evolved in the reaction in a measuring cylinder above water. Perform the task by means of available device and procedure. Inordertosimplifytheproblem,calculatethemassofyourmetalsamplefromthe volume of hydrogen on the assumption that it was measured at STP conditions.

3 33 3rd rd rd rd 6 theoretical problems 2 practical problems THE 3RD INTERNATIONAL CHEMISTRY OLYMPIAD, Budapest, 1970 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1 Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 22 THE THIRD INTERNATIONAL CHEMISTRY OLYMPIAD 15 JULY 1970, BUDAPEST, HUNGARY _______________________________________________________________________ THEORETICAL PROBLEMS PROBLEM 1 An amount of 23 g of gas (density = 2.05 g dm-3 at STP) when burned, gives 44 g of carbon dioxide and 27 g of water. Problem: What is the structural formula of the gas (compound)? ____________________ SOLUTION The unknown gas : X 1(X)From the ideal gas law : (X) 46 g molR TMp= = 123 g(X) 0.5 mol46 g moln= =mol 1mol g 44g 44) (CO12= =n n(C) = 1 mol m(C) = 12 g mol 1.5mol g 18g 27O) (H12= =nn(H) = 3 mol m(H) = 3 g THE 3RD INTERNATIONAL CHEMISTRY OLYMPIAD, Budapest, 1970 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1 Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 23 The compound contains also oxygen, since m(C) + m(H) = 12 g + 3 g = 15 g < 23 g m(O) = 23 g - 15 g = 8 g n(O) = 0,5 mol n(C) : n(H) : n(O) = 1 : 3 : 0,5 = 2 : 6 : 1 The empirical formula of the compound is C2H6O. C2H6OC2H5OHCH3CH3Oethanoldimethyl ether Ethanol is liquid in the given conditions and therefore, the unknown gas is dimethyl ether. THE 3RD INTERNATIONAL CHEMISTRY OLYMPIAD, Budapest, 1970 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1 Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 24 PROBLEM 2 A sample of crystalline soda (A) with a mass of 1.287 g was allowed to react with an excess of hydrochloric acid and 100.8 cm3 of a gas was liberated (measured at STP). Anothersampleofdifferentcrystallinesoda(B)withamassof0.715gwas decomposed by 50 cm3 of 0.2 N sulphuric acid. Aftertotaldecompositionofsoda,theexcessofthesulphuricacidwasneutralized whichrequired50cm3 of0.1Nsodiumhydroxidesolution(bytitrationonmethylorange indicator). Problems: 2.1 How many molecules of water in relation to one molecule of Na2CO3 are contained in the first sample of soda? 2.2 Have both samples of soda the same composition? Relative atomic masses: Ar(Na) = 23; Ar(H) = 1; Ar(C) = 12; Ar(O) = 16. ____________________ SOLUTION 2.1Sample A: Na2CO3 . x H2O m(A) = 1.287 g 2(CO ) 0.0045 mol (A)p Vn nRT= = = 1mol g 286mol 0045 . 0g 287 . 1) A (= = MM(A) = M(Na2CO3) + xM(H2O) 10mol g 18mol g ) 106 286 () O H () CO Na ( ) A (x1123 2===MM M Sample A:Na2CO3 .10 H2O THE 3RD INTERNATIONAL CHEMISTRY OLYMPIAD, Budapest, 1970 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1 Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 25 2.2Sample B:Na2CO3 . x H2O m(B) = 0.715 g H2SO4+2 NaOH=Na2SO4+2 H2O n(NaOH) = c V = 0.1 mol dm-3 0.05 dm3 = 0.005 mol Excess of H2SO4:n(H2SO4)=0.0025 mol Amount of substance combined with sample B: n(H2SO4) = 0.0025 mol = n(B) -1-10.715 g(B) = = 286 g mol0.0025 g molM Sample B:Na2CO3 .10 H2O THE 3RD INTERNATIONAL CHEMISTRY OLYMPIAD, Budapest, 1970 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1 Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 26 PROBLEM 3 Carbon monoxide was mixed with 1.5 times greater volume of water vapours. What will bethecomposition(inmassaswellasinvolume%)ofthegaseousmixtureinthe equilibrium state if 80 % of carbon monoxide is converted to carbon dioxide? ____________________ SOLUTION CO + H2O CO2 + H2 Assumption: n(CO) = 1 mol n(H2O) = 1.5 mol After reaction: n(CO)= 0.2 mol n(H2O) = 0.7 mol n(CO2) = 0.8 mol n(H2)= 0.8 mol (CO) 0.2 mol(CO) 0.08 i.e. 8 vol. % of CO2.5 molVV = = =22 2(H O) 0.7 mol(H O) 0.28 i. e. 28 vol. % of H O2.5 molVV = = =22 2(CO ) 0.8 mol(CO ) 0.32 i.e. 32 vol. % of CO2.5 molVV = = =22 2(H ) 0.8 mol(H ) 0.32 i.e. 32 vol. %of H2.5 molVV = = =Before reaction: m(CO) = n(CO)M(CO)=1 mol 28 g mol-1=28 g m(H2O) = 1.5 mol 18 g mol-1=27 g THE 3RD INTERNATIONAL CHEMISTRY OLYMPIAD, Budapest, 1970 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1 Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 27 After reaction: m(CO)=0,2 mol 28 g mol-1=5.6 g m(H2O)=0.7 mol 18 g mol-1=12.6 g m(CO2)=0.8 mol44 g mol-1=35.2 g m(H2)=0.8 2 g mol-1=1.6 g CO of % mass 2 . 10 . e . i 102 . 0g 0 . 55g 6 . 5 ) CO () CO ( = = =mmw 22 2(H ) 1.6 g( ) 0.029 i.e. 2.9 mass % of H55.0 gmw Hm= = = 222CO of % mass 0 . 64 . e . i 640 . 0g 0 . 55g 2 . 35 ) CO () CO ( = = =mmwO H of % mass 9 . 22 . e . i 229 . 0g 0 . 55g 6 . 12 ) O H () O H (222= = =mmwTHE 3RD INTERNATIONAL CHEMISTRY OLYMPIAD, Budapest, 1970 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1 Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 28 PROBLEM 4 An alloy consists of rubidium and one of the other alkali metals. A sample of 4.6 g of the alloy when allowed to react with water, liberates 2.241 dm3 of hydrogen at STP. Problems: 4.1Which alkali metal is the component of the alloy? 4.2What composition in % by mass has the alloy? Relative atomic masses:Ar(Li) = 7; Ar(Na) = 23; Ar(K) = 39; Ar(Rb) = 85.5; Ar(Cs) = 133 ____________________ SOLUTION 4.1M - alkali metal Reaction: 2 M + 2 H2O2 MOH+H2 n(H2)=0.1 mol n(M)=0.2 mol Mean molar mass: -14.6 g= 23 g mol0.2 molM= 4.2Concerningthemolarmassesofalkalimetals,onlylithiumcancomeinto consideration, i.e. the alloy consists of rubidium and lithium. n(Rb)+n(Li)=0.2 mol m(Rb)+m(Li)=4.6 g n(Rb) M(Rb)+n(Li) M(Li)=4.6 g n(Rb) M(Rb)+(0.2 n(Rb)) M(Li)=4.6 n(Rb) . 85.5 + (0.2 n(Rb)) 7 = 4.6 n(Rb) = 0.0408 mol n(Li) = 0.1592 mol 76 100g 6 . 4mol g 5 . 85 mol 0408 . 0Rb %1= = THE 3RD INTERNATIONAL CHEMISTRY OLYMPIAD, Budapest, 1970 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1 Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 29 24 100g 6 . 4mol g 7 mol 1592 . 0Li %1= = THE 3RD INTERNATIONAL CHEMISTRY OLYMPIAD, Budapest, 1970 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1 Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 30 PROBLEM 5 An amount of 20 g of cooper (II) oxide was treated with a stoichiometric amount of a warm 20% sulphuric acid solution to produce a solution of copper (II) sulphate. Problem: Howmanygramsofcrystallinecopper(II)sulphate(CuSO4 .5H2O) have crystallised when the solution is cooled to 20 C? Relative atomic masses: Ar(Cu) = 63.5; Ar(S) = 32; Ar(O) = 16; Ar(H) = 1 Solubility of CuSO4 at 20 oC: s = 20.9 g of CuSO4 in 100 g of H2O. SOLUTION CuO +H2SO4 CuSO4+H2O -1(CuO) 20 g(CuO) = 0.2516 g(CuO) 79.5 g molmnM= = n(H2SO4)=n(CuSO4)=0.2516 mol Mass of the CuSO4 solution obtained by the reaction: m(solution CuSO4) = m(CuO) + m(solution H2SO4) = -12 4 2 42 4(H SO ) (H SO ) 0.2516 mol98 g mol(CuO) 20 g +(H SO ) 0.20n Mmw = + = m(solution CuSO4) = 143.28 g Mass fraction of CuSO4: a) in the solution obtained:4 4 444 4(CuSO ) (CuSO ) (CuSO )(CuSO ) 0.28(solution CuSO ) (solution CuSO )m n Mwm m= = = b) in saturated solution of CuSO4 at 20oC: 173 . 0g 9 . 120g 9 . 20) CuSO (4= = w THE 3RD INTERNATIONAL CHEMISTRY OLYMPIAD, Budapest, 1970 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1 Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 31 c) in crystalline CuSO4 . 5 H2O: 639 . 0) O H 5 . CuSO () CuSO () CuSO (2 444= =MMwMass balance equation for CuSO4: 0.28 m=0.639 m1+0.173 m2 m - mass of the CuSO4 solution obtained by the reaction at a higher temperature. m1 - mass of the crystalline CuSO4 . 5H2O. m2 - mass of the saturated solution of CuSO4 at 20 oC. 0.28 143.28 = 0.639 m1 + 0.173 (143.28 - m1) m1 = 32.9 g The yield of the crystallisation is 32.9 g of CuSO4 . 5H2O. THE 3RD INTERNATIONAL CHEMISTRY OLYMPIAD, Budapest, 1970 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1 Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 32 PROBLEM 6 Oxideofacertainmetalcontains22.55%ofoxygenbymass.Anotheroxideofthe same metal contains 50.48 mass % of oxygen. Problem: 1. What is the relative atomic mass of the metal? ____________________ SOLUTION Oxide 1: M2Ox ) O () O (:) M () M (x : 2r rAwAw= ) M (95 . 54162255 . 0:) M (7745 . 0x : 2r rA A= = (1) Oxide 2: M2Oy 0.4952 0.5048 15.6952 : y :(M) 16 (M)r rA A= =(2) When (1) is divided by (2): 5 . 3695 . 1595 . 54xy= = 27xy=By substituting x = 2 into equation (1): Ar(M) = 54.95 M = Mn Oxide 1 = MnO Oxide 2 = Mn2O7 ) O () O (:) M () M (y : 2r rAwAw=THE 3RD INTERNATIONAL CHEMISTRY OLYMPIAD, Budapest, 1970 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1 Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 33 PRACTICAL PROBLEMS PROBLEM 1 (Practical) An unknown sample is a mixture of 1.2-molar H2SO4 and 1.47-molar HCl. By means of available solutions and facilities determine: 1. the total amount of substance (in val) of the acid being present in 1 dm3 of the solution, 2. the mass of sulphuric acid as well as hydrochloric acid present in 1 dm3of the sample. PROBLEM 2 (Practical) Bymeansofavailablereagentsandfacilitiesperformaqualitativeanalysisofthe substances given in numbered test tubes and write down their chemical formulas. Give 10 equations of the chemical reactions by which the substances were proved: 5 equations for reactions of precipitation, 2 equations for reactions connected with release of a gas, 3 equations for redox reactions. 4 44 4th th th th

6 theoretical problems 2 practical problems THE 4TH INTERNATIONAL CHEMISTRY OLYMPIAD, Moscow, 1972 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 35 THE FOURTH INTERNATIONAL CHEMISTRY OLYMPIAD 110 JULY 1972, MOSCOW, SOVIET UNION _______________________________________________________________________ THEORETICAL PROBLEMS PROBLEM 1 Amixture of two solid elements with a mass of 1.52 g was treated with an excess of hydrochloric acid. A volume of 0.896 dm3 of a gas was liberated in this process and 0.56 g of a residue remained which was undissolved in the excess of the acid. Inanotherexperiment,1.52gofthesamemixturewereallowedtoreactwithan excessofa10%sodiumhydroxidesolution.Inthiscase0.896dm3 ofagaswerealso evolved but 0.96 g of an undissolved residue remained. In the third experiment, 1.52 g of the initial mixture were heated to a high temperature without access of the air. In this way a compound was formed which was totally soluble in hydrochloric acid and 0.448 dm3 of an unknown gas were released. All the gas obtained was introduced into a one litre closed vessel filled with oxygen. After the reaction of the unknown gaswithoxygenthepressureinthevesseldecreasedbyapproximatelytentimes(T= const). Problem: 1.1Writechemicalequationsfortheabovereactionsandprovetheircorrectnessby calculations. In solving the problem consider that the volumes of gases were measured at STP and round up the relative atomic masses to whole numbers. ____________________ SOLUTION 1.1a) Reaction with hydrochloric acid: 1.52 g 0.56 g = 0.96 g of a metal reacted and 0.896 dm3 of hydrogen (0.04 mol) were formed.THE 4TH INTERNATIONAL CHEMISTRY OLYMPIAD, Moscow, 1972 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 36 0.96combining mass of the metal:11.2 = 12 g0.896 Possible solutions: Relative atomic mass of the metal Oxidation number ElementSatisfying? 12 ICNo 24 IIMgYes 36 IIIClNo Reaction:Mg+ 2 HCl MgCl2+H2 b) Reaction with sodium hydroxide: 1.52 g 0.96 g = 0.56 g of an element reacted, 0.896 dm3 (0.04 mol) of hydrogen were formed. 0.56combining mass of the metal:11.2 = 7 g0.896 Possible solutions:Relative atomic mass of the element Oxidation number ElementSatisfying? 7 ILiNo 14 IINNo 21 IIINeNo 28 IVSiYes Reaction:Si+2 NaOH+ H2ONa2SiO3+2 H2 c)Combining of both elements: 0.96 g Mg+0.56 g Si=1.52 g of silicide MgxSiy 37 . 0g 52 . 1g 56 . 0) Si ( 63 . 0g 52 . 1g 96 . 0) Mg ( = = = = w w THE 4TH INTERNATIONAL CHEMISTRY OLYMPIAD, Moscow, 1972 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 37 1 : 22837 . 0:2463 . 0y : x = =silicide: Mg2Si d)Reaction of the silicide with acid: Mg2Si+4 HCl 2 MgCl2+SiH4 mol 02 . 0mol g 76g 52 . 1) Si Mg (12= =n mol 02 . 0mol dm 4 . 22dm 448 . 0) SiH (1 334= =n e)Reaction of silane with oxygen: SiH4+2 O2SiO2+2 H2O V=1 dm3 On the assumption that T=const: 1122pnnp = mol 0446 . 0mol dm 4 . 22dm 1) O (1 332 1= =nConsumption of oxygen in the reaction:n(O2)=0.04 mol The remainder of oxygen in the closed vessel: n2(O2)=0.0446 mol 0.04 mol=0.0046 mol 1 1 21 . 0mol 0446 . 0mol 0046 . 0p p p = THE 4TH INTERNATIONAL CHEMISTRY OLYMPIAD, Moscow, 1972 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 38 PROBLEM 2 Amixtureofmetallicironwithfreshlypreparediron(II)andiron(III)oxideswas heatedinaclosedvesselintheatmosphereofhydrogen.Anamountof4.72gofthe mixture when reacted, yields 3.92 g of iron and 0.90 g of water. Whenthesameamountofthemixturewasallowedtoreactwithanexcessofacopper(II) sulphate solution, 4.96 g of a solid mixture were obtained. Problems: 2.1 Calculate the amount of 7.3 % hydrochloric acid ( = 1.03 g cm-3) which is needed for a total dissolution of 4.72 g of the starting mixture. 2.2 What volume of a gas at STP is released? Relative atomic masses:Ar(O) = 16; Ar(S) = 32; Ar(Cl) = 35.5; Ar(Fe) = 56; Ar(Cu) = 64 ____________________ SOLUTION 2.1a) Reduction by hydrogen: FeO+H2 Fe+H2O n(Fe)=n(FeO);n(H2O) = n(FeO) Fe2O3 + 3 H2 2 Fe + 3 H2O n(Fe) = 2 n(Fe2O3);n(H2O) = 3 n(Fe2O3) The mass of iron after reduction: 3.92 g The total amount of substance of iron after reduction: mol 07 . 0mol g 56g 92 . 3) O Fe ( 2 ) FeO ( ) Fe (13 2= = + +n n n(1) b) Reaction with copper(II) sulphate: Fe + CuSO4Cu + FeSO4 Increase of the mass: 4.96 g 4.72 g = 0.24 g After reaction of 1 mol Fe, an increase of the molar mass would be:M(Cu) M(Fe) = 64 g mol-1 56 g mol-1 = 8 g mol-1 THE 4TH INTERNATIONAL CHEMISTRY OLYMPIAD, Moscow, 1972 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 39 Amount of substance of iron in the mixture: mol 03 . 0mol g 8g 24 . 0) Fe (1= =n(2) c) Formation of water after reduction: 0.90 g H2O, i.e. 0.05 mol 0.05 mol = n(Fe) + 3 n(Fe2O3) (3) By solving equations (1), (2), and (3): n(FeO) = 0.02 mol n(Fe2O3) = 0.01 mol d) Consumption of acid: Fe + 2 HClFeCl2 + H2 FeO + 2 HClFeCl2 + H2O Fe2O3 + 6 HCl2 FeCl2 + 3 H2O n(HCl) = 2 n(Fe) + 2 n(FeO) + 6 n(Fe2O3) = =0.06 mol + 0.04 mol + 0.06 mol=0.16 mol A part of iron reacts according to the equation: Fe + 2 FeCl3 3 FeCl2 n(Fe) = 0.5 n(FeCl3) = n(Fe2O3) n(Fe) = 0.01 mol It means that the consumption of acid decreases by 0.02 mol. The total consumption of acid: n(HCl) = 0.14 mol 331cm 68cm g 03 . 1 073 . 0mol g 5 . 36 mol 14 . 0) HCl % 3 . 7 ( == = wM nV2.2Volume of hydrogen: Fe + 2 HClFeCl2 + H2 Iron in the mixture: 0.03 mol Iron reacted with FeCl3: 0.01 mol Iron reacted with acid: 0.02 mol Hence, 0.02 mol of hydrogen, i.e. 0.448 dm3 of hydrogen are formed. THE 4TH INTERNATIONAL CHEMISTRY OLYMPIAD, Moscow, 1972 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 40 PROBLEM 3 Avolumeof200cm3 ofa2-normalsodiumchloridesolution(=1.10gcm-3)was electrolysedatpermanentstirringinanelectrolyticcellwithcopperelectrodes.Electrolysis was stopped when 22.4 dm3 (at STP) of a gas were liberated at the cathode. Problem: 3.1Calculate the mass percentage of NaCl in the solution after electrolysis. Relative atomic masses:Ar(H) = 1;Ar(O) = 16; Ar(Na) = 23; Ar(Cl) = 35.5; Ar(Cu) = 64. ____________________ SOLUTION 3.1 Calculations are made on the assumption that the following reactions take place: 2 NaCl2 Na+ + 2 Cl- cathode: 2 Na+ + 2 e2 Na anode: 2 Cl- 2 eCl Cl2 + CuCuCl2 Becausetheelectrolytesolutionispermanentlybeingstirredthefollowingreaction comes into consideration:CuCl2 + 2 NaOHCu(OH)2 + 2 NaCl Ontheassumptionthatallchlorinereactswithcopper,themassofNaClinthe electrolyte solution remains unchanged during the electrolysis. m(NaCl)=n M=c V M=2 mol dm-3 0.2 dm3 58.5 g mol-1 =23.4 g V(H2) = 22.4 dm3 , i. e. n(H2) = 1 mol The amount of water is decreased in the solution by:n(H2O) = 2 mol m(H2O) = 36 g Before electrolysis: m(solution NaCl) = V = 200 cm3 1.10 g cm-3 = 220 g 64 . 10 100g 220g 4 . 23NaCl % = = THE 4TH INTERNATIONAL CHEMISTRY OLYMPIAD, Moscow, 1972 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 41 After electrolysis: m(solution NaCl) = 220 g 36 g = 184 g 72 . 12 100g 184g 4 . 23NaCl % = = THE 4TH INTERNATIONAL CHEMISTRY OLYMPIAD, Moscow, 1972 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 42 PROBLEM 4 Amount of 50 g of a 4 % sodium hydroxide solution and 50 g of a 1.825 % solution of hydrochloricacidweremixedinaheatinsulatedvesselatatemperatureof20C.The temperature of the solution obtained in this way increased to 23.4 C. Then 70 g of a 3.5 % solution of sulphuric acid at a temperature of 20 C were added to the above solution. Problems: 4.1 Calculate the final temperature of the resulting solution. 4.2 Determine the amount of a dry residue that remains after evaporation of the solution. In calculating the first problem use the heat capacity value c = 4.19 J g-1 K-1. Relative atomic masses:Ar(H) = 1; Ar(O) = 16; Ar(Na)= 23; Ar(S) = 32; Ar(Cl) = 35.5. ____________________ SOLUTION 4.1 a) NaOH + HClNaCl + H2O 1(solution NaOH) (NaOH) 50 g 0.04(NaOH) 0.05 mol(NaOH) 40 g molm wnM = = =

150 g 0.01825(HCl) 0.025 mol36.5 g moln= =unreacted: n(NaOH) = 0.025 mol b)When 1 mol of water is formed, neutralization heat is: 1 11neutr2100 g 4.19 J g K 3.4 K57 000 J mol(H O) 0.025 molm c tHn = = = c) NaOH + H2SO4NaHSO4 + H2O The temperature of the resulting solution is calculated according to the equation: m1 c1 t1 + m2 c2 t2 = m c t c1 = c2 = c THE 4TH INTERNATIONAL CHEMISTRY OLYMPIAD, Moscow, 1972 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 43

m1 t1 + m2 t2 = m t

1 1 2 2(100 23.4) (70 20.0)22 C170m t m ttm+ + = = =

d) The temperature increase due to the reaction of NaOH with H2SO4 is as follows: -12 neutr-1 -1(H O) 0.025 mol 57000 J mol= 2 K170 g 4.19 J g Kn Htm c= = The final temperature of the solution: t = 22 + 2 = 24 oC 4.2e) Whenthesolutionhasevaporatedthefollowingreactionisassumedtotake place:NaCl + NaHSO4 Na2SO4 + HCl Na2SO4 is the dry residue. m(Na2SO4) = n M = 0.025 mol 142 g mol-1 = 3.55 g THE 4TH INTERNATIONAL CHEMISTRY OLYMPIAD, Moscow, 1972 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 44 PROBLEM 5 Onlyoneproductwasobtainedbythereactionofbrominewithanunknown hydrocarbon. Its density was 5,207 times as great as that of the air. Problem: 5.1 Determine the structural formula of the unknown hydrocarbon. Relative atomic masses: Ar(H) = 1;Ar(C) = 12;Ar(Br) = 80. ____________________ SOLUTION 5.1 Relative molecular mass of the initial hydrocarbon can be calculated from the density value: Mr(RBr) = 29 5.207 = 151 Monobromoderivativecanonlycomeintoconsiderationbecausetherelative molecular mass of dibromo derivative should be greater: Mr(RBr2)>160 Mr(RH) = 151 - 80 + 1 = 72 The corresponding summary formula: C5H12 The given condition (the only product) is fulfilled by 2,2-dimethyl propane: CCH3CH3CH3CH3 THE 4TH INTERNATIONAL CHEMISTRY OLYMPIAD, Moscow, 1972 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 45 PROBLEM 6 OrganiccompoundAis41.38%carbon,3.45%hydrogenandtherestisoxygen. Compound A when heated with ethanol in the presence of an acid yields a new substance B which contains 55.81 % carbon, 6.97 % hydrogen, and oxygen. TheinitialcompoundAwhenallowedtoreactwithhydrobromideyieldsproductC which on boiling in water gives substance D containing 35.82 % carbon, 4.48 % hydrogen, andoxygen.Anamountof2.68gofsubstanceDrequiredreactingwith20cm3 ofa2N solution of potassium hydroxide. Problems: 6.1 Determine structural formulas of all the above mentioned substances A, B, C and D. Use the finding that compound A splits off water when heated. 6.2 Write chemical equations for the above reactions. Relative atomic masses: Ar(H) = 1;Ar(C) = 12;Ar(O) = 16;Ar(K) = 39. ____________________ SOLUTION 6.1 Stoichiometric formulas of compounds: A :CxHyCz 1 : 1 : 11617 . 55:145 . 3:1238 . 41z : y : x = =B :CmHnOp 1 : 3 : 21622 . 37:197 . 6:1281 . 55p : n : m = =D :CaHbOc 20cm3 of2NKOHcorrespond0.04/vmolofsubstanceDanditcorrespondsto2.68 g of substance Dv = 1, 2, 3, ... 1 mol of compound D = v 67 g 5 : 6 : 41670 . 59:148 , 4:1282 . 35c : b : a = =THE 4TH INTERNATIONAL CHEMISTRY OLYMPIAD, Moscow, 1972 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 46

Mr(D)=67or134or201,etc. Due to both the stoichiometric formula and relative molecular mass of compound D, its composition is C4H6O5. Then molecular formulas for compounds A, B, and C are as follows: A: C4H4O4B: C8H12O4C: C4H5O4Br 6.2 Equations: CH - COOHCH - COOCH2CH3 + 2 CH3CH2OH 2 H2O+ CH - COOHCH - COOCH2CH3 AB CH - COOH CH2 - COOH CH2 - COOH HBr O H2 CH - COOH CHBr - COOHCH(OH) - COOH A C D CH2 - COOH CH2 - COOK + 2 KOH2 H2O+ CH(OH) - COOHCH(OH) - COOK CH - COOH CH - CO heatingO CH - COOHCH - CO Compound A:maleic acid THE 4TH INTERNATIONAL CHEMISTRY OLYMPIAD, Moscow, 1972 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 47 PRACTICAL PROBLEMS PROBLEM 1 (Practical) Determine unknown samples in ten numbered test tubes using reagents and facilities available on the laboratory desk. Write chemical equations for the most important reactions that were used to identify each substance. In case that the reactions take place in solutions, write equations in a short ionic form. PROBLEM 2 (Practical)On June 10th, a mixture of formic acid with an excess of ethanol was prepared. This mixture was kept in a closed vessel for approximately one month. Determine quantitatively thecompositionofthemixtureonthedayofthecompetition,usingonlyreagentsand facilities available on the laboratory desk. Calculate the amounts of the acid and ethanol in per cent by mass which were initially mixed together. 5 55 5th th th th 6 theoretical problems 3 practical problems THE 5TH INTERNATIONAL CHEMISTRY OLYMPIAD, Sofia, 1973 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 49 THE FIFTH INTERNATIONAL CHEMISTRY OLYMPIAD 110 JULY 1973, SOFIA, BULGARIA _______________________________________________________________________ THEORETICAL PROBLEMS PROBLEM 1 In nitrating a hydroxy derivative of benzene a compound is formed which contains 49.0 % by mass of oxygen. A charge of 4350 C is required for a total electroreduction of 0.458 g of the compound, efficiency being 80 %. Problem: 1.1Determinethestoichiometricaswellasstructuralformulasofthecompoundifthe product of the electrochemical reduction is an aromatic hydroxy amino derivative. F (Faraday's charge) = 96 500 C mol-1 ____________________ SOLUTION 1.1a) Formula of the compound: C6HxOyNz The compound is a hydroxy nitroderivative of benzene: C6H6-(y-2z)-z(OH)y-2z(NO2)z b) Equation of the reduction: R-NO2 + 6 H R-NH2 + 2 H2O Combining mass of the compound: (compound)6rMEz=(1)An amount of charge which is required for the electrochemical reduction: Q = 4350 C 0.8 = 3480 C Combining mass of the compound: THE 5TH INTERNATIONAL CHEMISTRY OLYMPIAD, Sofia, 1973 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 50 7 . 12C 3480C 96500458 . 0C 3480= = =FmEIn relation to (1):Mr (compound) = 76.2 z (2) c) y (O) 100% O(compound)rrMM = y 16 10049(compound)rM =Mr(compound)=32.7 y d) Mr(compound) = 6 Mr (C) + x Mr (H) + y Mr (O) + z Mr (N) Mr (compound) = 6 12 + x + 16 y + 14 z Taking into consideration the general formula of the unknown hydroxy derivative of benzene: x=6 (y 2 z) z+ y 2 z x=6 z(4) Then: Mr(compound)=72 + 6 z + 16 y + 14 z Mr(compound)=78 + 16 y + 13 z (5) By solving equations (2), (3), (4), and (5) we obtain: Mr(compound) = 229 x = 3 y = 7 z = 3 The molecular formula of the compound is:C6H3O7N3 or C6H2(OH)(NO2)3. The compound is 2, 4, 6-trinitrophenol THE 5TH INTERNATIONAL CHEMISTRY OLYMPIAD, Sofia, 1973 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 51 PROBLEM 2 A mixture of a gaseous hydrocarbon and oxygen is in a vessel of a volume of 1 dm3 at a temperature of 406.5 K and a pressure of 101 325 Pa. There is twice as much oxygen in themixtureasisneededforthereactionwiththehydrocarbon.Aftercombustionofthe hydrocarbon the pressure in the vessel (at the same temperature) is increased by 5 %. Problem: 2.1 Whathydrocarbonwasinthemixturewhenthemassofwaterformedbythe combustion was 0.162 g. ____________________ SOLUTION 2.1Amounts of substances of reactants and reaction products: Equation:CxHy+(x + y4)O2=x CO2+y2 H2O mol 0.009mol g 18g 0.162O) (HO) (HO) (H1222= = =Mmnmoly018 . 02ymol 009 . 0) H C (y x= = n (1) mol 018 . 0y4yx2ymol 009 . 0)4yx ( ) O (2+= + = n (2) mol 018 . 0yx2ymol 009 . 0x ) CO (2 = = n (3) Before reaction: mol 0.03K 406.5 K mol J 8.314dm 1 kPa 101.325(mixture)1 13== = T RV pn n(CxHy)+2 n(O2)=0.03 mol (4) THE 5TH INTERNATIONAL CHEMISTRY OLYMPIAD, Sofia, 1973 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 52 After reaction: p = 101. 325 kPa 1.05 = 106.4 kPa mol 0.0315K 406.5 K mol J 8.314dm 1 kPa 106.4(mixture)1 13== = T RV pn n(CO2) + n(O2) + n(H2O)=0.0315 mol n(CO2) + n(O2)=0.0225 mol (5) When(1),(2),and(3)aresubstitutedin (4) and (5), an equation of two unknowns is obtained which when solved yields x = 3; y = 6 The stoichiometric formula of the unknown hydrocarbon is:C3H6. THE 5TH INTERNATIONAL CHEMISTRY OLYMPIAD, Sofia, 1973 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 53 PROBLEM 3 Equalvolumes(10cm3)of0.01-molarsolutionsofCH3COOHandHClOweremixed and then diluted to a total volume of 100 cm3. Ionisation constant of CH3COOH is equal to 1.8 10-5 and that for HClO is 3,7 10-8. Problems: Calculate: 3.1 degree of ionisation for each of the acids in the solution, 3.2degree of ionisation of HClO if the diluted solution would not contain CH3COOH, 3.3pH value for the solution containing at the same time CH3COOH and HClO. ____________________ SOLUTION CH3COOH: K1, 1, c1 HClO: K2, 2,c2 c1 = c2 = 1 10-3mol dm-3=c 3.1 -3 3 1 2 1 1 2 113 1 1[H O ] [CH COO ]( ) ( )(1)[CH COOH] (1 ) 1c c cKc ++ += = = -3 1 2 122[H O ] [ClO ]( )(2)[HClO] 1cK ++= = K1 >> K2, therefore also 1 >> 2 and1 + 2 1 K1 (1 - 1) = 12 c c 12 + K1 1 K1=0 1=0,125 When (2) is divided by (1):

2 1 21 2 1(1 )(1 )KK = After substitution of 1:2=2.94 . 10-4 THE 5TH INTERNATIONAL CHEMISTRY OLYMPIAD, Sofia, 1973 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS,Volume 1Edited by Anton Sirota, IChO International Information Centre, Bratislava, Slovakia 54 3.2 22221cK= 2 2.4 Cu2OcanbeobtainedbythereductionofCu2+inacidorbasicmedia,e.g.by Fehling's solution or reducing sugars. 2.8 [Cu(NH3)2]+ Cu+ + 2 NH3 KD = 233 2[Cu ] [NH ][Cu(NH ) ]++ ( = 110-11 Knowing E0(Cu+/Cu) = 0.52 V, the E0([Cu(NH3)2]+ / Cu+) becomes:Ef1 = 0.52 0.06 pKD = 0.14 V THE 22ND INTERNATIONAL CHEMISTRY OLYMPIAD, Paris, 1990 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 436 2.9 ThestandardemfofaCu2+/Cucellisthus:E0=(0.5+0.16)/2=0.33VandE30 = 0.33 0.03 pK2. Thereout: pK2 = (0.33 E30) / 0.03 = (0.33 ( 0.02)) / 0.03 = 12 [Cu(NH3)4]2+ + 2 eCu + 4 NH3E0 = 0.02 V [Cu(NH3)2]+ +eCu + 2 NH3E0 = 0.14 V ______________________________________________ [Cu(NH3)4]2+ + e[Cu(NH3)2]+ + 2 NH3 Since only G0 is additive and from G0 = n F E0 it follows: Ef2 = 2 ( 0.02) ( 0.14) = 0.10 V 2.10[Cu(NH3)2]+ + eCu + 2 NH3Ef1 = 0.14 V [Cu(NH3)4]2+ + e[Cu(NH3)2]+ + 2 NH3 Ef2 = 0.10 V SinceEf1 0.95 to the continuous plug reactor has the same loss of benzene as already indicated in the Fig 1 below. 7.4Intheplugreactorthepercentageofbenzenelostis2%(thebestmeanyieldis 0.98), therefore the amount of benzene annually lost is 2000 tons. In a MP installation, the yield of hydrogenation is 0.99 (except for 0.95 < p < 1 where itslightlydecreasesto0.985,butthiscanbeneglected)andtherefore,theoverall mean yield of hydrogenation is also 0.99. The amount of benzene annually lost is therefore 1000 tons. THE 22ND INTERNATIONAL CHEMISTRY OLYMPIAD, Paris, 1990 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 457 Figure 1

THE 22ND INTERNATIONAL CHEMISTRY OLYMPIAD, Paris, 1990 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 458 PRACTICAL PROBLEMS PROBLEM 1(Practical) Synthesis involving Carbanions Chalcone Theaimofthisexperimentistocondenseacetophenoneandbenzaldehydeina mixedaldolcondensation. After spontaneous dehydratation, an , - unsaturated ketone is obtained: the chalcone, 1,3-diphenylprop-2-en-1-on. a) Starting the condensation reaction Ina250cm3groundneckErlenmeyer(conical)flask,dissolveabout5gof potassium hydroxide, KOH, (50 pellets) in 30 cm3 of water, then add slowly under stirring 20cm3ofethanol.Fromtheautomaticdispensers,add9.6g(0.08mol=9.5cm3)of acetophenone and 8.5 g (0.08 mol = 8.5 cm3) of benzaldehyde. Set up a condenser in the verticalrefluxpositionandrefluxvigorouslywhilestirringthemixturewiththemagnetic stirrer for an hour. b) Isolation of the crude chalcone After refluxing, cool the contents of the Erlenmeyer flask in an ice bath. The chalcone should crystallize. If crystals do not appear, scratch the inside wall of the flask with a glass rod. Collect the crystals in a Bchner funnel, wash them with a little ice cold ethanol, then air dry and weigh the crystals. c) Recrystallization of the chalcone Recrystallizethiscrudechalconefromethanol,usinga100cm3beaker.Usethe heatingplatelocatedinthehood(fumechamber).Whenthecrystallizationisover(wait long enough and scratch if necessary), collect the crystals in a Bchner funnel, and air dry them.

THE 22ND INTERNATIONAL CHEMISTRY OLYMPIAD, Paris, 1990 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 459 Giveyourentireproducttothesupervisorwhowillmeasureitsmassand melting-point. Donotbreathethevapourwhilerecrystallizingandmeasuringthemeltingpoint.It contains irrigating chalcone!!! Questions: 1.1Write the mechanism for this reaction. 1.2Give the mass of the crystals of the crude chalcon obtained and calculate the yield of crude product. 1.3Calculate the yield of recrystallized product based on starting material and calculate the efficiency of the recrystallization process. _______________ SOLUTION 1.1

OOH(-)(-)+O+H2O (-)O+O(-)OO H2O(-)OO+OOH+ OH(-) THE 22ND INTERNATIONAL CHEMISTRY OLYMPIAD, Paris, 1990 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 460 OOHHOH(-)O+ OH(-)+ HOH THE 22ND INTERNATIONAL CHEMISTRY OLYMPIAD, Paris, 1990 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 461 PROBLEM 2(Practical)Qualitative Analysis Equipment: Five flasks are labelled 2.1 through 2.5. Each contains an aqueous solution of a colourless metalnitrate.Exceptforsilvernitrate,theconcentrationsofthenitratesusedtoprepare these solutions, are all 0.1 mol dm-3. Moreover,availablearepHindicatorpaperandthefollowingthreereagentsofa concentration of about 5 mol dm-3, contained in flasks labelled 2.6 through 2.8. 2.6 : hydrochloric acid, 2.7 : aqueous ammonia solution, 2.8 : sodium hydroxide solution. Beware: these solutions are all concentrated and corrosive. Tasks: 2.1 Carryoutreactionsbetweeneachreagentandeachsolution.Foreachofthe solutions 2.1 through 2.5, record your observations for each reaction observed. 2.2 Write the name of the cation contained in each of the solutions 2.1 through 2.5 in the corresponding space on the answer sheet. 2.3For each cation identified, write the equation for each reaction observed. _______________ SOLUTION 2.1Aluminium(III) nitrate - Al3+ + HClno reaction - Al3+ + 3 NH3 + 3 H2OAl(OH)3 + 3 +4NH- Al3+ + 3 OHAl(OH)3 The solutions are:aluminium(III) nitrate calcium(II) nitrate lead(II) nitrate silver(I) nitrate zinc(II) nitrate THE 22ND INTERNATIONAL CHEMISTRY OLYMPIAD, Paris, 1990 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 462 - Al(OH)3 + OH[Al(OH)4] (aq) calcium(II) nitrate - Ca2+ + HClno reaction - Ca2+ + 2 OHCa(OH)2 lead(II) nitrate -Pb2+ + 2 ClPbCl2 (in cold solutions) -Pb2+ + 2 NH3 + 2 H2OPb(OH)2 + 2 +4NH-Pb2+ + 2 OHPb(OH)2 Pb(OH)2 + 2 OH[Pb(OH)4]2 (aq) silver(I) nitrate -Ag+ + ClAgCl -Ag+ + NH3 + H2OAgOH + +4NH(or Ag2O) AgOH + 2 NH3[Ag(NH3)2]+ (aq) + OH -Ag+ + OHAgOH 2 AgOHAg2O + H2O zinc(II) nitrate - Zn2+ + HClno reaction - Zn2+ + 2 NH3 + 2 H2OZn(OH)2 + 2 +4NHZn(OH)2 + 4 NH3[Zn(NH3)4]2+(aq) + 2 OH - Zn2+ + 2 OHZn(OH)2 Zn(OH)2 + 2 OH[Zn(OH)4] (aq) THE 22ND INTERNATIONAL CHEMISTRY OLYMPIAD, Paris, 1990 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 463 PROBLEM 3(Practical) Titration of Oxygen Dissolved in Water (Winkler's method) Inaqueousalkalinesolution,dissolvedoxygenoxidizesmanganese(II)hydroxideto hydratedmanganese(III)oxide(writtenasMn(OH)3forsimplification).Inacidicsolution, manganese(III)ionsoxidizeiodideionstoiodine.Theiodineformedistitratedwitha sodium thiosulphate solution. Data: Redox coupleElectrode potentials (in V) at pH = 0at pH = 14 Mn3+ / Mn2+1.51 Mn(OH)3 / Mn(OH)2 0.13 O2 / H2O1.230.39 I2 / I-0.620.62 S4O62- / S2O32-0.090.09 Solubility products:

Ksp(Mn(OH)2) = 110-13

Ksp(Mn(OH)3) = 110-36 Gas constant: R = 8.315 J K-1mol-1 Procedure: Preliminary remarks: To reduce volume variations, the reagents are added either as solids (sodium hydroxide pellets ...), or as concentrated solutions (sulphuric acid). 1. Thewatertobetestedisstoredinalargecontainerlocatedonthegeneral-use bench. Place two glass beads into a 250 cm3 ground top Erlenmeyer (conical) flask. Fill it to the rim with the water to be tested. At this stage of the manipulation the THE 22ND INTERNATIONAL CHEMISTRY OLYMPIAD, Paris, 1990 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 464 Erlenmeyer flask should be standing in the basin provided to avoid overflow of water onto the bench. Measure the temperature of the water. 2. Add to the water, avoiding any loss of reagent: a)2gofmanganese(II)chloride,preweighedwithinadecigramandcontainedin thebag. b)about8pelletsofsodiumhydroxide(yieldingabasicmedium,pH=14).The pellets will be found on the general-use bench. 3. StoppertheErlenmeyerflask,avoidingairbubbles,andswirlituntilcomplete dissolution of the sodium hydroxide and of the manganese chloride has occurred. A brown precipitate forms. 4. Let the flask stand for at least 30 minutes. 5. Open the Erlenmeyer flask, add concentrated sulphuric acid dropwise, stirring with a glassroduntilthesolutionisdefinitelyacidic(checkwiththepHindicatorpaper); make sure that nearly all the precipitate has disappeared. Sulphuric acid will be found on the general-use bench. 6. Add to the Erlenmeyer flask 3 g of potassium iodide, preweighed within a decigram and contained in a bag. Stopper the flask and shake it until the potassium iodide has dissolved. The solution should now be clear. 7. Removea50cm3sampleofthesolutionandtitrateitwithaXmoldm-3sodium thiosulphate solution (the numerical value of X will be shown on the board). For this titration,theendpointcanbedeterminedusingtheindicatorthiodenewhichis equivalent to starch. A small quantity of this solid indicator should be dissolved in the solution being titrated just prior to the endpoint. Questions: 3.1Justifythattheoxidationofmanganese(II)bydissolvedoxygenispossibleonlyin alkaline solution. 3.2Write the equation of the reaction between: - dissolved oxygen and manganese(II) hydroxide (in alkaline solution), - manganese(III) ions and iodide ions, - iodine and thiosulphate. 3.3Record the volume of thiosulphate required to reach the endpoint. THE 22ND INTERNATIONAL CHEMISTRY OLYMPIAD, Paris, 1990 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 465 3.4Deriveanexpressionthatrelatesthedissolvedoxygenconcentrationinwater (expressed as mol dm-3) to the volume in cm3 of thiosulphate added at the endpoint. Determine that concentration for the water tested. Deduce the volume of oxygen (in cm3), determined at 0 oC and at a pressure of 101,325 Pa,containedinalitreofwater(measuredatambienttemperature).Givethetemperature of the water. _______________ SOLUTION 3.1E0(O2/H2O)>E0(Mn(III)/Mn(II) at pH = 14 3.2O2 + 4 e + 2 H2O4 OH

Mn(OH)2 + HOMn(OH)3 + e

O2 + 4 Mn(OH)2 + 2 H2O4 Mn(OH)3 2 II2 + 2 e

Mn3+ + eMn2+ 2 I + 2 Mn3+I2 + 2 Mn2+ I2 + 2 e2 I

2 2-2 3S O 2-4 6S O+ 2 e I2 + 2 2-2 3S O

2 I + 2-4 6S O 3.41 mol O24 mol Mn(II)4 mol Mn(III) 4 mol Mn(III)4 mol I2 mol I2 2 mol I24 mol 2-2 3S O c(2-2 3S O ) V(2-2 3S O )=4 c(O2) 50 THE 22ND INTERNATIONAL CHEMISTRY OLYMPIAD, Paris, 1990 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 466 PROBLEM 4(Practical) Kinetics of an SN1 Reaction Inanaqueousethanolicsolution,tert-butylchloride(2-chloro-2-methylpropane) undergoesanSN1nucleophilicsubstitutionreactionleadingtotheformationof 2-methylpropan-2-ol and hydrochloric acid: tert-BuCl + H2O = tert-BuOH + HCl The rate of the reaction is given by1[HCl][ BuCl]dk tertd t=The aim of this experiment is to determine the rate constant k1 at ambient temperature. Procedure: 1. Usingapipettetransfer2.0cm3oftert-butylchloride(flask4.1onthegeneral-use bench) into a clean and dry 250 cm3 ground neck Erlenmeyer (conical) flask. 2. Assoonaspossibleadd148cm3(measurewiththegraduatedcylinder)ofthe aqueous ethanol solution standing on the general-use bench (flask 4.2). 3. Stopper the Erlenmeyer flask and stir vigorously using the magnetic stirrer. Start your timer. Note carefully the temperature, T(0), of the solution. 4. Afterapproximately5,15,25, 35, 45, 55 min (determined accurately), transfer 10.0 cm3samplesusingapipetteinto20cm3ofamixtureoficeandacetone (propanone). Add2dropsofbromothymolblueandtitratetheliberatedacidwithasolutionofY mol dm-3 aqueous sodium hydroxide (Y will be shown on the board). Questions: 4.1 The concentrations of tert-BuCl at time t = 0 and at time t are linked by the following relationship: 01t[tertBuCl]ln[tertBuCl]= k .t Establish the theoretical expression: 1lnV= k tV V THE 22ND INTERNATIONAL CHEMISTRY OLYMPIAD, Paris, 1990 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 467

whereVstandsforthevolumeofsodiumhydroxideusedat time t and V(8) for the corresponding value at t(8). (8) (8) = t ; V Vt =4.2 CalculateV(8)(densityoftert-BuClat20 oC:850kgm-3,molarmassoftert-BuCl: 92.5 g mol-1). 4.3Fill in the table on the answer sheet whose columns will be: 3, ln(min), (cm ),V VV, ,t V VV V V V Draw the curve lnV= f (t)V V 4.4Determine the value of k1 (give the value of t(0)). _______________ SOLUTION 4.1t-BuCl + H2Ot-BuOH + HCl t = 0(t-BuCl)0

t(t-BuCl)(t-BuCl)0 (t-BuCl) t(8)(t-BuCl)0 1(HCl) ( BuCl)( BuCl)d tk td t d t= = ln0( )( )t BuClt BuCl=k1 t t: c(HCl) = c0(t-BuCl) c(t-BuCl) = (NaOH) (NaOH)(HCl)V cV= 2Y 1010V t(8) c(HCl) = c0(t-BuCl) = 10aY c(t-BuCl) = ( ) Y10a V THE 22ND INTERNATIONAL CHEMISTRY OLYMPIAD, Paris, 1990 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 468 01(t-BuCl)ln ln(t-BuCl)c ak tc a V= = 4.2n0(t-BuCl) = 1.7092.5M(t-BuCl) = 92.5 g mol-1 Y a 10-3 = 010(t-BuCl)150n a(ml) = 3170Y 10 15 92.5

23 23 23 23rd rd rd rd 6 theoretical problems 2 practical problems THE 23RD INTERNATIONAL CHEMISTRY OLYMPIAD, Lodz, 1991 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 470 THE TWENTY-THIRDINTERNATIONAL CHEMISTRY OLYMPIAD 715 JULY 1991, LODZ, POLAND_________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ THEORETICAL PROBLEMS PROBLEM 1 1.1Show that 0.1 mol of Tl2S dissolves in a 1 M solution of any strong monoprotic non-coordinating acid. 1.2 Show that 0.1 CuS dissolves in a 1 M HNO3 but not in a 1 M HCl solution. Information: AssumethatCu2+ionsdonotformstablecomplexeswithchlorideionsinaqueous solutions. E0(S/S2-) = 0.48 VE0(-3NO / NO(aq)) = 0.96 V pKa(H2S) = 7pKa(HS) = 13 Ksp(Tl2S) = 110-20Ksp(CuS) = 110-35 Solubility of NO in water (298 K): 2.5310-2mol dm-3 Solubility of H2S in water (298 K):0.1 mol dm-3 R = 8.314 J K-1 mol-1F = 96 487 C mol-1 _______________ SOLUTION 1.1 Solubility condition:[Tl+]2 [S2-] 110-20 [Tl+] = c(Tl+) = 0.2 mol dm-3 c(S2-) = [S2-] + [HS-] + [H2S] = [S2-] 22 1 2H H1K K K+ +| | (( |+ + |\ = 0.1 mol dm-3 THE 23RD INTERNATIONAL CHEMISTRY OLYMPIAD, Lodz, 1991 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 471 2-+ + 22 1 20.1[S ][H ] [H ]1K K K=+ + For a strong monoprotic acid (1 mol dm-3) and [H+] 1. Then 1 + 1013 [H+]1.5)sothatcompleteseparationis obtained.ForcolumnA,R=0.91( xgivesx = 6.5 . 10-4 , not valid Quadratic formula:x = 5.810-4 , [H3O+] = 5.810-4 , pH = 3.24 1.21: HL + -3HCOH2CO3 + L-:K1 2: HL + H2O H3O+ + L-:K2 = KHL 3: -3HCO+ H3O+ H2CO3 + H2O:K3 = a11K Reaction 1 = 2 + 3,K1 = K2 . K3 =311 (3.1102) Alternative:K1 = -2 3-3[H CO ] [L ][HL] [HCO ] +3+3[H O ][H O ] = + -3[H O ][L ][HL]2 3+3 3[H CO ][HCO ][H O ] 1.3i) 3HCOis amphoteric, pH 1 21( )2a apK pK + = 8.34 ii) HL + -3HCO H2CO3 + L- , "reaction goes to completion" Before:0.00300.02400 After :0 0.0210.00300.0030 Buffer:pH pKa1 + log 0.0210.0030= 6.35 + 0.85 = 7.20 (Control:HL+3[H O ]K= [L-][HL]= 2.2103, assumption is valid) 1.4 A:pH = 7.40; [H3O+] = 4.010-8;[-3HCO ]A = 0.022.From Ka1:[H2CO3]A = 0.0019; (1)[-3HCO ]B + [H2CO3]B = 0.0239 (0.024) B:pH = 7.00; -32 3[HCO ][H CO ]= 4.5;

THE 26TH INTERNATIONAL CHEMISTRY OLYMPIAD, Oslo, 1994 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 542 (2)[-3HCO ]B = 4.5 [H2CO3]B From (1) and (2):[-3HCO ]B = 0.0196 [H2CO3]B = 0.0043 n(HL) = n(H2CO3) = c( H2CO3) 1.00 dm3 =2.410-3 mol 1.5[OH-] = 8.910-5 [H2CO3] of no importance Reactions:A:CaCO3(s)Ca2+ +23CO c0c0 B: 23CO+ H2O -3HCO+ OH-K = Kb = 2.110-4 c0 - xx x From B: [-3HCO ] = [OH-] = 8.910-5

[23CO] =-3b[HCO ][OH ]K=3.810-5

[Ca2+] = [-3HCO ] + [23CO] = 1.310-4

c0(Ca2+) = 1.310-4 mol dm-3 = solubility 1.6Ksp = [Ca2+] [23CO] = 1.310-4 3.810-5 = 4.910-9 = 5 10-9 From Ka2 :[23CO] = 2-3+3[HCO ][H O ]aK = 2.610-5

Q = [Ca2+] [23CO]; Precipitation when Q > Ksp = 510-9 No precipitation when Q < Ksp

Max. concentration of "free" Ca2+ ions: [Ca2+]max = 2-3[CO ]spK = 1.910-4 THE 26TH INTERNATIONAL CHEMISTRY OLYMPIAD, Oslo, 1994 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 543 PROBLEM 2 NitrogeninagriculturalmaterialsisoftendeterminedbytheKjeldahlmethod.The method involves a treatment of the sample with hot concentrated sulphuric acid, to convert organicallyboundnitrogentoammoniumion.Concentratedsodiumhydroxideisthen added,andtheammoniaformedisdistilledintohydrochloricacidofknownvolumeand concentration. The excess hydrochloric acid is then back-titrated with a standard solution of sodium hydroxide, to determine nitrogen in the sample. 2.10.2515gofagrainsamplewastreatedwithsulphuricacid,sodiumhydroxidewas then added and the ammonia distilled into 50.00 cm3 of 0.1010 M hydrochloric acid. The excess acid was back-titrated with 19.30 cm3 of 0.1050 M sodium hydroxide. Calculate the concentration of nitrogen in the sample, in percent by mass. 2.2CalculatethepHofthesolutionwhichistitratedin2.1when0cm3,9.65cm3, 19.30 cm3and28.95cm3ofsodiumhydroxidehavebeenadded.Disregardany volumechangeduringthereactionofammoniagaswithhydrochloricacid.Kafor ammonium ion is 5.710-10. 2.3Draw the titration curve based on the calculations in b). 2.4WhatisthepHtransitionrangeoftheindicatorwhichcouldbeusedfortheback titration. 2.5TheKjeldahlmethodcanalsobeusedtodetermine themolecularweightofamino acids.Inagivenexperiment,themolecularweightofanaturallyoccurringamino acidwasdeterminedbydigesting0.2345 gofthepureacidanddistillingammonia releasedinto50.00cm3of0.1010 Mhydrochloricacid.Atitrationvolumeof17.50 cm3 was obtained for the back titration with 0.1050 M sodium hydroxide. Calculatethemolecularweightoftheaminoacidbasedononeandtwonitrogen groups in the molecule, respectively. _______________ SOLUTION 2.1[(50.00 0.1010) (19.30 0.1050)] 14.0110001000.2515= 16.84 % N 2.20 cm3 added:[H+] = 19.30 . 0.105050= 0.04053pH = 1.39 THE 26TH INTERNATIONAL CHEMISTRY OLYMPIAD, Oslo, 1994 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 544 9.65 cm3 added:[H+] == 0.01699pH = 1.77 19.30 cm3 added:[H+] = . .. 1050.000 101019 300 10505.71050 19.30 +

pH = 5.30 28.95 cm3 added:pH = pKa + log 34[NH ][NH ]+ = 9.24 + log 1.012.01 = 8.94 2.3 2 46810pH% titrated50 100 1500 2.4Indicator pH transition range: pH 5.3 1 2.5[(50.00 0.1010) (17.50 0.1050)] 14.011000 1000.2345= 19.19 % N 1 N:Mr = 73.012 N:Mr = 146.02 THE 26TH INTERNATIONAL CHEMISTRY OLYMPIAD, Oslo, 1994 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 545 PROBLEM 3 Sulphur forms many different compounds with oxygen and halogens (sulphur as the central atom). These compounds are mainly molecular, and many are easily hydrolysed in water. 3.1Write Lewis structures for molecules SCl2, SO3, SO2ClF, SF4, and SBrF5. 3.2Carefullydrawthegeometriesofthe5molecules.(Disregardsmalldeviationsfrom "ideal" angles.) 3.3Acompound,consistingofsulphur(oneatompermolecule),oxygenandoneor moreatomsoftheelementsF,Cl,Br,andI,wasexamined.A small amount of the substance reacted with water. It was completely hydrolyzed without any oxidation or reduction,andallreactionproductsdissolved.0.1Msolutionsofaseriesoftest reagentswereaddedtoseparatesmallportionsofadilutedsolutionofthe substance. Which ions are being tested for in the following tests? i)Addition of HNO3 and AgNO3. ii)Addition of Ba(NO3)2. iii)Adjustment to pH = 7 with NH3 and addition of Ca(NO3)2. Write the equations for the possible reactions in the tests: iv)Addition of KMnO4 followed by Ba(NO3)2 to an acid solution of the substance. v)Addition of Cu(NO3)2. 3.4In practice, the tests in 3.3 gave the following results: i)A yellowish precipitate. ii)No precipitate. iii) No visible reaction. iv)Themainfeatureswerethatthecharacteristiccolourofpermanganate disappeared, and a white precipitate was formed upon addition of Ba(NO3)2. v)No precipitate. Write the formulas of the possible compounds, taking the results of these tests into account. 3.5Finally, a simple quantitative analysis was undertaken: 7.190 g of the substance was weighed out and dissolved in water to give 250.0 cm3 of a solution. To 25.00 cm3 of this solution, nitric acid and enough AgNO3 was added THE 26TH INTERNATIONAL CHEMISTRY OLYMPIAD, Oslo, 1994 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 546 tosecurecompleteprecipitation.Afterwashinganddryingtheprecipitateweighed 1.452 g. Determine the formula of the compound. 3.6Write the equation describing the reaction of the substance with water.If you have not found the formula for the compound, use SOClF. _______________ SOLUTION 3.1 3.2

SCl2 SO3 SO2ClF SF4 SBrF5 THE 26TH INTERNATIONAL CHEMISTRY OLYMPIAD, Oslo, 1994 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 547 3.3i)Cl-, Br-, I- ii) 2-4SOiii)F- iv)2 -4MnO + 5 -3HSO + H+ 5 2-4SO + 2 Mn2+ + 3 H2O Ba2+ + 2-4SOBaSO4 (s) v)2 Cu2+ + 4 I-2 CuI(s) + I2 3.4SOClBr and SOBr2 3.5SOClBr [SOClBr: 1.456g, and SOBr2: 1.299g] 3.6SOClBr + 2 H2O-3HSO + Cl- + Br- + 3 H+ SOClF + 2 H2O-3HSO+ Cl- + HF + 2 H+ THE 26TH INTERNATIONAL CHEMISTRY OLYMPIAD, Oslo, 1994 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 548 PROBLEM 4 Platinum(IV) oxide is not found in the nature, but it can be prepared in a laboratory. Solidplatinum(IV)oxideisinequilibriumwithplatinummetalandoxygengasat1atm(= 1.01325105 Pa) and 650 C. 4.1ThissuggeststhattheconditionsontheEarth,whenthemineralsweknowwere formed, were: [1] p(O2) = 1 atm,t = 650 C;[2] p(O2) < 1 atm,t < 650 C;[3] p(O2) > 1 atm, t < 650 C;[4] p(O2) < 1 atm,t > 650 C;[5]p(O2) > 1 atm,t > 650 C Markthemostprobablealternative[1][5]ontheanswersheet.Pleasenotethatthe marking of only one alternative will be accepted. 4.2WhatareGandKpfortheformationofplatinum(IV)oxideatoxygenpressureof1 atm and temperature of 650 C? Thepreparationofplatinum(IV)oxideinvolvesboilingofasolutionwhichcontains hexachloroplatinate(IV) ions with sodium carbonate. In this process PtO2 . n H2O is formed andthisisinturnconvertedtoplatinum(IV)oxideuponsubsequentfilteringandheat treatment. In the following we assume n = 4. PtO2 . 4 H2O or Pt(OH)4 . 2 H2O can be dissolved in acids and strong bases. 4.3Writethebalancedequationsforthepreparationofplatinum(IV)oxideaccordingto the procedure given above. 4.4Write the balanced equations for the dissolution of PtO2 . 4 H2O in both hydrochloric acid and sodium hydroxide. Platinumismainlyfoundinthenatureasthemetal(inmixtureorinalloyingwith otherpreciousmetals).Platinumisdissolvedinaquaregiaundertheformationof hexachloroplatinate(IV)ions.Aquaregiaisamixtureofconcentratedhydrochloricand nitricacidsinproportion3:1,andof the nitrosylchloride (NOCl) and the atomic chlorine whichareformeduponthemixing.Thelatterisbelievedtobetheactivedissolving component.THE 26TH INTERNATIONAL CHEMISTRY OLYMPIAD, Oslo, 1994 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 549 Thehexachloroplatinate(IV)ionscanbeprecipitatedasdiammoniumhexachloro-platinate(IV)andbythermaldecompositionofthiscompound,finelypowderedplatinumand gaseous products are formed. 4.5Writethebalancedequationsfortheformationofaquaregiaanditsreactionwith platinum. 4.6Writethebalancedequationofthethermaldecompositionofdiammonium hexachloroplatinate(IV) at elevated temperature. From diammonium hexachloroplatinate(IV) we can prepare Pt(NH3)2Cl2 which occurs in cis (0fH = 467.4 kJ mol-1, 0fG = 228.7 kJ mol-1) and trans (0fH = 480.3 kJ mol-1, 0fG = 222.8 kJ mol-1) form. 4.7The occurrence of the isomers shows that Pt(NH3)2Cl2 has geometry: [ 1 ] linear,[ 2 ] planar, [ 3 ] tetrahedral,[ 4 ] octahedral geometry. Mark the correct alternative of [ 1 ] [ 4 ] on the answer sheet.4.8Is the cis form or trans form thermodynamically more stable? Platinumisusedasacatalystinmodernautomobiles.Inthecatalystcarbon monoxide(0fH =110.5kJmol-1, 0fG =137.3kJmol-1)reactswithoxygentocarbon dioxide (0fH = 393.5 kJ mol-1, 0fG = 394.4 kJ mol-1). 4.9Is the reaction spontaneous at 25 C?[ 1 ] yes, or[ 2 ] no. Is the reaction:[ 3 ] endothermic, or[ 4 ] exothermic? Calculate Sfor the reaction.Establish whether the entropy of the reaction system[5] increases, or[6] decreases. THE 26TH INTERNATIONAL CHEMISTRY OLYMPIAD, Oslo, 1994 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 550 4.10Establish an expression for the temperature dependence of the equilibrium constant in this case. Theoverallcatalyticreactionissimple,whereasthereactionmechanisminthe homogeneousphaseisverycomplicatedwithalargenumberofreactionsteps,andthe courseisdifficulttocontrolowingtoadistinct chaincharacter.Withplatinum as catalyst thesignificantreactionstepsare:(i)AdsorptionofCOandadsorption/dissociationofO2 (H = 259 kJ per mol CO + O), (ii) their activation (105 kJ per mol CO + O) and (iii) the reaction and the desorption of CO2 (H = 21 kJ per mol CO2). Aone-dimensionalenergy-diagramfortheplatinumcatalyzedoxidationofcarbon monoxide to dioxide can be represented as:

4.11Mark the correct alternative of [ 1 ] [ 4 ] on the answer sheet. _______________ SOLUTION 4.1Correct answer is No 4. 4.2G = 0 kJ and Kp = 1 according to the chemical equation Pt(s) + O2(g)PtO2(s) THE 26TH INTERNATIONAL CHEMISTRY OLYMPIAD, Oslo, 1994 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 551 4.3 2-3CO (aq) + H2O(l) -3HCO (aq)+OH-(aq) 2-6PtCl (aq)+4 OH(aq)+2 H2O(l)Pt(OH)4 . 2 H2O(s)+6 Cl(aq) Alternative I: PtO2 . 4 H2O(s) + 6 Cl(aq) Alternative II: (n2) H2O PtO2 . n H2O(s) + 6 Cl(aq) PtO2 . 4 H2O(s)PtO2(s) + 4 H2O(g) [PtO2 . 4 H2O(s)Pt(OH)4 . 2 H2O(s)] 4.4In hydrochloric acid: PtO2 . 4 H2O(s) + 4 H+(aq) + 6 Cl(aq)2-6PtCl (aq) + 6 H2O In sodium hydroxide: PtO2 . 4 H2O(s) + 2 OH(aq)2-6Pt(OH) (aq) + 2 H2O 4.53 HCl(sol) + HNO3(sol)NOCl(sol) + 2 Cl(sol) + 2 H2O(sol) Pt(s) + 4 Cl(sol) + 2 HCl(sol)26PtCl(sol)+2 H+(sol) 4.6(NH4)2PtCl6(s)Pt(s) + 2 NH3(g) + 2 HCl(g) + 2 Cl2(g) 4.7Correct is No 2. 4.8The cis form is thermodynamically more stable. 4.9[1]Yes.(G= 257.1 kJ for CO(g) + 1/2 O2(g) CO2(g)) [4]The reaction is exothermic. (H= 283.0 kJ forCO(g) + 1/2 O2(g)CO2(g)) [6] is correct.S= 0.0869 kJ K-1 for CO(g) + 1/2 O2(g)CO2(g);As seen from the sign for Sas well as for the reaction enthalpy the entropyof the system decreases. THE 26TH INTERNATIONAL CHEMISTRY OLYMPIAD, Oslo, 1994 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 552 4.10 ln Kp = 34037 / T 10.45 for CO(g) + 1/2 O2(g)CO2(g) Alternative:Kp = exp (34037 / T 10.45) 4.11 No 2 is correct. THE 26TH INTERNATIONAL CHEMISTRY OLYMPIAD, Oslo, 1994 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 553 PROBLEM 5 There is only one correct answer to each question 5.1What is the correct systematic name (IUPAC name) for the compound below? (CH3)2CHCH(CH2CH3)(CH2CH2CH3) 13-Isopropylhexane 22-Methyl-3-propylpentane 3Ethyl isopropyl propyl methane 43-Hexylpropane 53-Ethyl-2-methylhexane 5.2Howmanyisomers,includingstereoisomers,containingonlysaturatedcarbon atoms, are there for C5H10? 14 isomers25 isomers36 isomers 47 isomers58 isomers 5.3Whichoneofthefollowingcompoundshasadipolemomentsignificantlydifferent from zero? HO OH CN NCCN ClCH 2 ClCH 2 CH 2 Cl CH 2 Cl C Br H 3 C C Br CH 3 HN NH 1 23 5 4 THE 26TH INTERNATIONAL CHEMISTRY OLYMPIAD, Oslo, 1994 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 554 5.4Which of the following is a pair of structural isomers? 3CH 3Cl Cl Cl Cl O 3 O CH 3CH 3 CH 2 CH 3 H2 C andand1 23 and4andand H 3 C CH 35 H 3C CH 3CHCH 5.5Which of the following five options is the correct order of relative stabilities of cations a, b and c as written below (most stable first)?

+ + CH 2 H 2 CCHCH 2 CH H 3 CC CH 3 CH 3 CH CH 3 ab c + 2 1a>b>c2 b>c>a3 c>a>b4 a>c>b5 b>a>c 5.6What is the correct stereochemical descriptor of the optically active compound drawn below? H3CCH3C Br 11R,3R,4R21R,3R,4S31R3S,4R41S,3S,4R51S,3S,4S THE 26TH INTERNATIONAL CHEMISTRY OLYMPIAD, Oslo, 1994 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 555 5.7All the molecules drawn below are neutral compounds. Which one does not contain a formal positive charge and a formal negative charge? 1(CH3)3N-B(CH3)32(CH3)2N-O-CH33CH2=N=N 4(CH3)3N-O5F3B-O(CH3)2 _______________ SOLUTION 12345 5.1X X 5.2 X 5.3X 5.4 X 5.5X 5.6 X 5.7 X THE 26TH INTERNATIONAL CHEMISTRY OLYMPIAD, Oslo, 1994 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 556 PROBLEM 6 AnopticalactivecompoundA(C12H16O)showsamongstotherastrongIR-absorptionat30003500cm-1,andtwomediumsignalsat1580and1500cm-1.The compounddoesnotreactwith2,4-dinitrophenylhydrazine(2,4-D).Upontreatmentwith I2/NaOH, A is oxidized and gives a positive iodoform reaction. Ozonolysis of A (1. O3; 2. Zn, H+) gives B (C9H10O) and C (C3H6O2). Both B and C give precipitation when treated with 2,4-D, and only C gives positive reaction with Tollens reagent. Nitration of B (HNO3/H2SO4) may give two mono-nitro compounds D and E, but in practical work only D is formed. Acidification followed by heating of the product formed by the Tollens reaction on C gives compound F (C6H8O4). The compound gives no absorption in IR above 3100 cm-1. 6.1Basedontheaboveinformationdrawthestructureformula(e)forthecompoundsA F and give the overall reaction scheme, including the (2,4-D) and the products of the Tollens and iodoform reactions. 6.2DrawCinanR-configuration.TransformthisintoaFischerprojectionformulaand state whether it is a D or L configuration. _______________ SOLUTION (See the next page.) THE 26TH INTERNATIONAL CHEMISTRY OLYMPIAD, Oslo, 1994 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 557 6.1 CH3H3C CHOOHOOOOCH3CH3NO2NH2O2NCDEAa)O3b)Zn, H++H3COHI2/NaOHCO2H3CCH3+ CHI3CH3OBH3CCH3+Ag(NH3)2H3C CO2OH1)H2)OFH3CCH3NO2OH3CCH3NO2+B and CNO2NO2NRHNO3/H2SO4ROR1R1 6.2 CHOH3COHHCHOCH3OH H R-configurationD-configuration THE 26TH INTERNATIONAL CHEMISTRY OLYMPIAD, Oslo, 1994 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 558 PROBLEM 7 7.1Whenanideal,monatomicgasexpandsreversiblyfromavolumeV1toavolume V2, a work

21VVw p dV = is performed on the system by the surroundings. In this equation, w is the work and p is the pressure of the gas. Determine the performed work when one mole ideal gas expands isothermally from V1 = 1.00 dm3 to V2 = 20.0 dm3 at the temperature T = 300.0 K. Given: The gas constant R = 8.314 J K-1 mol-1. 7.2Determine how much heat must be added to the gas during the process given under 7.1. 7.3Thegaswillperformlessworkinanadiabaticexpansionthaninanisothermal expansion.Isthisbecausetheadiabaticexpansionischaracterizedby(checkthe square you think is most important). 1The volume of the gas is constant 2The expansion is always irreversible 3No heat is supplied to the gas 7.4ThecyclicprocessshownschematicallyinFigure1showsthefourstepsina refrigeration system with an ideal gas as working medium. Identify the isothermal and adiabatic steps in the process. Here, TH and TC represent high and low temperature, respectively. Specify for each step whether it is adiabatic or isothermal. THE 26TH INTERNATIONAL CHEMISTRY OLYMPIAD, Oslo, 1994 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 559 _______________ SOLUTION 7.1Work performed on the gas isw = 21VVp dV = RT21VVdVV = RT ln 21VV = 8,314 J K-1 mol-1 300 K ln 20.001.00= 7472 J mol-1 = 7.47 kJ mol-1 7.2Becausethisisanisothermalexpansionofanidealmonatomicgas,thereisno changein internal energy. From the first law of thermodynamics, we then have that U = q + w = 0 whereqistheamountofsuppliedheatandwisperformedwork.Thisleadsto q = w = 7.47 kJ mol-1. 7.3(3)No heat is supplied to the gas. 7.4 isotherm1-22-33-44-1 adiabat1-22-33-44-1 THE 26TH INTERNATIONAL CHEMISTRY OLYMPIAD, Oslo, 1994 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 560 PROBLEM 8 Avogadro's Number:6.022 . 1023 8.1 An atom of 238U disintegrates by a series of -decays and --decays until it becomes 206Pb, which is stable. i)Howmany-decaysandhowmany--decaysdoesanatomstartingas 238U undergo before it becomes stable? ii) Oneofthefollowingtennuclidesisformedfromaseriesofdisintegrations starting at 238U. Which one ? 235U, 234U,

228Ac, 224Ra, 224Rn, 220Rn, 215Po, 212Po, 212Pb, 211Pb. 8.2In a thermal neutron-induced fission process, 235U reacts with a neutron and breaks up into energetic fragments and (normally) 2-3 new neutrons.We consider one single fission event: 235U + n 137Te + X + 2 n Identify the fragment X. 8.3Thehalf-lifeof 238Uis4.5109years,thehalf-lifeof 235Uis7.0108years.Natural uranium consists of 99.28 % 238U and 0.72 % 235U. i) CalculatetheratioinnaturalUbetweenthedisintegrationratesofthesetwo uranium isotopes. ii) A mineral contains 50 weight percent uranium. Calculate the disintegration rate of 238U in 1.0 kg of this mineral.8.4We have the following radioactive sequence: 97Ru 97Tc 97Mo (stable). Halflives: 97Ru: 2.7 days;97Tc: 2.6106 years At t = 0 a radioactive source containing only 97Ru has a disintegration rate of 1.0109 Bq. i) What is the total disintegration rate of the source at t = 6.0 days? ii) What is the total disintegration rate of the source at t = 6000 years? SOLUTION 8.1i) 8'sand6-'s(only's gives 206Os, to come from Os to Pb requires 6 -'s).ii) 234U, all other answers are incorrect. THE 26TH INTERNATIONAL CHEMISTRY OLYMPIAD, Oslo, 1994 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 561 8.2 97Zr 8.3i)D = N, i.e. D1 / D2 = 1 N1 / 2 N2 = abund.(1)T1/2.(2) / abund.(2)T1/2(1) = (99.28 7.0108) / (0.72 4.5109) = 21.4 (0.047 is also of course correct) ii) N = (m/AW(U)) abundance(238) NA = (500 / 238.01) 0.9928 6.0221023

=1.261024 D = N ln2 / T1/2 = 1.261024 ln2 / (4.5109 (y) 3.16107 (s/y)) = 6.1.106 Bq 8.4i) = ln 2 / 2.7(d) = 0.26 d-1 D = D0 e- t= 1.0109 e (0.26 6.0) = 2.1108Bq ii)Number of 97Ru atoms in the source:N = D T1/2(97Ru) / ln 2 = 1.0109 (Bq) 2.7 (d) 24 (h/d) 3600 (s/h) / 0.6931 ==3.41014 atoms Whenall 97Ruhasdisintegrated,theseatomshaveallbecome 97Tc,andthe disintegration rate of this nuclide isD = Nln 2 / T1/2(97Tc) = (3.41014 0.6931) / (2.6.106 y 3.16107 s y-1) == 2.9 Bq THE 26TH INTERNATIONAL CHEMISTRY OLYMPIAD, Oslo, 1994 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 562 PRACTICAL PROBLEMS PROBLEM 1(Practical) Determination of Fatty Acids Amixtureofanunsaturatedmonoproticfattyacidandanethylesterofasaturated monoproticfattyacidhasbeendissolvedinethanol(2.00cm3ofthissolutioncontaina totalof1.00gacidplusester).Bytitrationthe acidnumber1),thesaponificationnumber2) andtheiodinenumber3)ofthemixtureshallbedetermined.Theacidnumberandthe saponification number shall be used to calculate the number of moles of free fatty acid and esterpresentin1.00gofthesample.Theiodinenumbershallbeusedtocalculatethe number of double bonds in the unsaturated fatty acid. Note:Thecandidatemustbeabletocarryoutthewholeexambyusingthedelivered amount of unknown sample (12 cm3). There will be no supplementation. 1)Acid number: The mass of KOH in milligram that is required to neutralize one gram of the acid plus ester. 2)Saponification number: The mass of KOH in milligram that is required to saponify one gram of the acid plus ester. 3)Iodinenumber:Themassofiodine(I)ingthatisconsumedby100gofacidplus ester. Relative atomic masses: Ar(I)=126.90Ar(O)=16.00 Ar(K)=39.10Ar(H)=1.01 1) Determination of the Acid Number Reagents and Apparatus Unknownsample,0.1000MKOH,indicator(phenolphthalein),ethanol/ether(1:1 mixture), burette (50 cm3), Erlenmeyer flasks (3 x 250 cm3), measuring cylinder (100 cm3), graduated pipette (2 cm3), funnel. THE 26TH INTERNATIONAL CHEMISTRY OLYMPIAD, Oslo, 1994 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 Edited by Anton Sirota IChO International Information Centre, Bratislava, Slovakia 563 Procedure: Pipette out aliquots (2.00 cm3) from the unknown mixture into Erlenmeyer flasks (250 cm3). Add first ca. 100 cm3 of an ethanol/ether mixture (1:1) and then add the indicator (5 drops). Titrate the solutions with 0.1000 M KOH. Calculate the acid number. 2) Determination of the Saponification Number Reagents and ApparatusUnknownsample,0.5000MKOHinethanol,0.1000MHCl,indicator(phenolphthalein), volumetricflask(50cm3),roundbottomflask(250cm3),Liebigcondenser,burette(50 cm3), Erlenmeyer flasks (3 x 250 cm3), volumetric pipette ( 25 cm3), volumetric pipette (10 cm3),graduatedpipette(2cm3),funnel,glassrod.TheroundbottomflaskandLiebig condenser are to be found in the fume hoods. Procedure Pipette out a 2.00 cm3 aliquot of the unknown sample into a round bottom flask (250 cm3) andadd25.0cm30.5000MKOH/EtOH.Refluxthemixturewithaheatingmantlefor30 mininthefumehood(starttheheatingwiththemantlesetto10,thenturnitdownto5 after7min.).Bringtheflaskbacktothebenchandcoolitundertapwater.Transfer quantitativelythesolutiontoa50cm3volumetricflaskanddilutetothemarkwitha1:1 mixture of ethanol/water. Take out aliquots of 10 cm3 and titrate with 0.1000 M HCl using phenolphthalein as indicator (5 drops). Calculate the saponification number. 3) Determination of the Iodine Number In this experiment iodobromine adds to the double bond. CC +IBr C CI Br The Hanus solution (IBr in acetic acid) is added in