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Q1.????????? The picture shows a painting which was painted in a cave in France about 17 000 years ago.????????? ????????? ???????? By Carla Hufstedler [CC-BY-SA-2.0], via Wikimedia CommonsOne of the pigments in this painting contains:70 % of iron (Fe) and 30 % of oxygen (O)Calculate the simplest (empirical) formula of this substance.Relative atomic masses: O = 16; Fe = 56.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(4)(Total 4 marks)??Q2.?????????Cosmetic powders were widely used in ancient Egypt.Cosmetic powders that may have been used in face paints have been analysed. These powders contained compounds that are rare in nature. The compounds must have been made by the ancient Egyptians using chemical reactions.One of these compounds is called phosgenite.Analysis of this compound shows that it contains:76.0% lead (Pb)?? 13.0% chlorine (Cl)?????? 2.2% carbon (C)?????????? 8.8% oxygen (O)Calculate the empirical formula of this compound.To gain full marks you must show all your working.Relative atomic masses: C = 12 ; O = 16 ; Cl = 35.5 ; Pb = 207........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(Total 4 marks)??Q3.????????? Lead compounds have been used for thousands of years as colours in paint.Johannes Vermeer [Public domain], via Wikimedia Commons(a) ????A sample of a red oxide used in paint was found to contain 6.21 g of lead and 0.64 g of oxygen.Calculate the empirical (simplest) formula of this compound.You must show all your working to gain full marks.Relative atomic masses: O = 16; Pb = 207.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(4)(b) ????A problem with lead compounds is that they slowly react with hydrogen sulfide in the air. This produces lead sulfide which is black.(i)????? Hydrogen sulfide has the formula H2S. The bonding in a molecule of hydrogen sulfide can be represented as:?H–S–HComplete the diagram below to show the arrangement of the outer electrons of the hydrogen and sulfur atoms in hydrogen sulfide.Use dots (●) and crosses (x) to represent the electrons.You need only show the outer shell electrons.(Atomic numbers: H = 1; S = 16.)(1)(ii)?????Hydrogen sulfide has a low boiling point.Explain why.............................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(iii)????Lead white is also used in paint. The white colour slowly darkens when lead sulfide is produced.The painting can be restored with hydrogen peroxide. This converts the black lead sulfide into white lead sulfate.Balance the equation for the reaction between lead sulfide and hydrogen peroxide (H2O2).?PbS(s)+............H2O2(aq)→PbSO4(s)+4H2O(l)(1)(Total 8 marks)??Q4.???????? Liquefied petroleum gases such as propane and butane are used as heating fuels for caravans, boats and barbecues.(a)???? Propane and butane have no smell, so for safety reasons a very small amount of thioethanol – the smelliest substance known – is added, even though it is toxic in large concentrations.Suggest one safety reason why thioethanol is added to propane and butane...........................................................................................................................................................................................................................................................................(1) (b)???? Suggest how mass spectrometry could be used to distinguish between propane (C3H8) and butane (C4H10)...........................................................................................................................................................................................................................................................................(1) (c)???? When 0.4 g of a hydrocarbon gas was completely burned in oxygen, 1.1 g of carbon dioxide and 0.9 g of water were the only products.Relative formula masses: CO2 = 44; H2O = 18.Use this information to calculate the number of moles of carbon dioxide and of water produced in this reaction. Use your answers to calculate the empirical formula of this hydrocarbon.You must show all your working to gain full marks..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................Empirical formula is ..............................(4)(Total 6 marks)Q5.????????? Perfumes contain a mixture of chemicals.??????????????????????????????????????????????????????????????????????? ????????? The main ingredients of perfumes are a solvent and a mixture of fragrances. (a)???? A sample of the solvent used in one perfume contained 0.60 g of carbon, 0.15 g of hydrogen and 0.40 g of oxygen.????????? Relative atomic masses: H = l; C = 12; O = 16.????????? Calculate the empirical (simplest) formula of the solvent.????????? You must show all of your working to gain full marks for this question..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(4) (b)???? Solvent molecules evaporate easily.????????? Explain why substances made of simple molecules evaporate easily..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(c)???? Most companies claim that their perfumes have been tested on skin. A study was made of the tests they used. The study found that each company used different tests.????????? The perfumes were tested in the companies’ own laboratories and not by independent scientists.????????? Some companies did not give any information about the tests that they had used.(i)????? Suggest why companies test their perfumes on skin..................................................................................................................................................................................................................................................................................................................................................................................(1)(ii)???? Did the study show that the tests made by the different companies were valid and reliable????????? Explain your answer........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(Total 9 marks)?Q6.????????? The diagram shows the main parts of an instrumental method called gas chromatography linked to mass spectroscopy (GC-MS).This method separates a mixture of compounds and then helps to identify each of the compounds in the mixture. (a)???? In which part of the apparatus:(i)????? is the mixture separated? ...................................................................(1) (ii)???? is the relative molecular mass of each of the compounds in the mixture measured?...............................................................................................................(1) (b)???? (i)????? Athletes sometimes take drugs because the drugs improve their performance. One of these drugs is ephedrine.Ephedrine has the formula:C10H15NOWhat relative molecular mass (Mr) would be recorded by GC-MS if ephedrine was present in a blood sample taken from an athlete?Show clearly how you work out your answer.Relative atomic masses: H = 1; C = 12; N = 14; O = 16.............................................................................................................................................................................................................................................................................................................................................................................................................................................................Relative molecular mass = .....................................(2)(ii)???? Another drug is amphetamine which has the formula:C9H13NThe relative molecular mass (Mr) of amphetamine is 135.Calculate the percentage by mass of nitrogen in amphetamine.Relative atomic mass: N = 14..............................................................................................................................................................................................................................Percentage of nitrogen = ..................................... %(2)(c)???? Athletes are regularly tested for drugs at international athletics events.An instrumental method such as GC-MS is better than methods such as titration.Suggest two reasons why.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(d)???? When a blood sample is taken from an athlete the sample is often split into two portions. Each portion is tested at a different laboratory.Suggest why.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(Total 10 marks)???M1.????????? 70/56??????30/16division by atomic mass1= 1.25?????? = 1.875proportion12?????? 3ratio (accept 1:1.5 / 4:6 / etc)allow e.c.f from proportion if sensible attempt at step 11Fe2O3formula allow e.c.f from ratio if sensible attempt at step 1allow correct formula with no working = 1 mark1[4]?M2.???????? Pb????? Cl????????? C????????? O76?????? 13??????? 2.2??????? 8.8207??? 35.5????? 12???????? 161 mark for dividing one mass by Arallow upside down ratio to lose this mark only1= 0.367???? =???? 0.366????? = 0.183???? = 0.551 mark for one correct proportion – accept to one d.p. or rounded up to 1 d.p.11 mark for all four correct proportions correctly rounded12???????????? 2??????????? 1???????????? 3or Pb2Cl2CO31 mark for correctly written formulaorcorrect whole number ratiocorrect formula without working gets only 1 mark.e.c.f. can be allowed from incorrect proportions to formula or ratio1[4]??M3.????????? (a)???? 1 mark for dividing mass by Ar max 2 if Ar divided by mass1= 0.03????????????? = 0.041 mark for correct proportions13????????????????????????????? 41 mark for correct whole number ratio (allow multiples) can be awarded from correct formula1?????Pb3O41 mark for correct formulaecf allowed from step 2 to step 3 and step 3 to step 4 if sensible attempt at step 1correct formula with no working gains 2 marks 1(b)?????(i)????? allow all dots or all crosses or e or e– ignore inner shells and any inner electronsallow 4 non-bonded electrons anywhere on shell as long as not in overlap – need not be paired1(ii)?????forces of attraction / bonds between molecules are weak (owtte)do not accept intramolecular forces / covalent bonds are weakdo not accept reference to ionsorintermolecular forces / bonds are weak (owtte)orit is made of small molecules with weak forces of attractionif 2 marks not awardedmade of small molecules / simple molecular gains 1 markforces of attraction are weak (without specifying between molecules / intermolecular) gains 1 mark(accept easily broken / not much energy needed to break instead of weak)bonds are weak without specifying intermolecular would not gain a mark and would be ignored2(iii)????41[8]?M4.???????? (a)???? (smell) warns of a leak / gas escapeaccept leak / gas escape by implicationignore smell alone1(b)???? eg (mass spectrometry gives)different molecular ions / Mr / formula mass orshows that one has mass 44 and the other 58‘mass of butane is more than mass of propane’ is insufficientaccept different fragmentation / patterndo not accept Ar / RAMaccept references to butane deflects less or converse1(c)???? CO2????????????????????? 2 H2O1.1??????????????????????? 0.9–––????????????????????? –––44???????????????????????? 181= 0.025???????????????? = 0.0511 (mole) CO2???????? 2 (moles) H2O11?????????????????????????? 4orCH41or alternative methodMass of C = (1)Mass of H =(1)C : Hproportions 0.025 : 0.1 (1)whole number 1 : 4 (1)or CH4correct formula with no working is only 1 markM3 can be awarded from the formula if steps one and two are clearcorrect formula from their incorrect ratio gets 1 markif fraction is wrong way around e.g. Mr / mass, then lose M1 and M2 but accept ecf for M3 and M4 max 4[6]??M5.????????? (a)????????????????? C???????????????????? H??????????????? O???????????????????? 0.60????????????????? 0.15???????????? 0.401????????????????????? 12???????????????????? 1??????????????? 16?????????????????? = 0.05?????????????? = 0.15???????? = 0.0251?????????????????????? 2????????????????????? 6???????????????? 11?????????????????????????????????????????? C2H6O????????????? 1 mark for dividing the correct amount or multiples of correct amount by Ar1 mark for proportions1 mark for whole number ratio – accept any multiple1 mark for correctly written simplest formulacorrect formula without working gets only 2 markscorrect formula gains full marksprovided steps 1 and 2 are correct.ecf can be allowed from step 2 to 3 or step 3 to 4formula can be in any order eg OH6C21(b)???? intermolecular forces / bonds1????????? are weak(covalent) bonds are weak = 0????????? or????????? forces between molecules or bonds between molecules (1)(attractive) forces are weak = 1????????? are weak (1)if no marks awarded, allow low boiling point or small Mr for 1 mark1(c)???? (i)????? to check the safety of the perfume (owtte)accept references to possible harmful / dangerous effects of perfume or possible reactions on skineg to show it does not damage skin / cause cancer etc.allow to see what it smells like on the skinallow so the company do not have to test on animals1(ii)???? any two from:???????? idea from text linked with an explanation???????? the company claim to have tested the product:but we cannot be certain they have or how thorough theyare or how accurately reported???????? companies did not disclose how they did their tests:so they could not be checked or so they could not beshown to be reliable / valid or so they could not be repeatedor converseeg companies should disclose how they did their tests so that results can be checked etc.???????? companies may not have repeated their tests:so they may not be reliable???????? companies do their own tests:so they may be biased or so they may not be truthful about their results or so they may not be reliableor converseeg independent tests should be done so as to ensure there is no bias etc.???????? the companies are using different tests:so the results cannot be compared or so results will be different or so results will not be fair / valid / reliableor converseeg companies should do the same tests so that the results will be fair etc.???????? companies would not give false information because of damage to reputation or it might lead to litigation2[9]?M6.????????? (a)???? (i)????? column1(ii)???? mass spectrometer1(b)???? (i)????? 165if answer is not correct then evidence of correct working gains one mark.e.g. (10 × 12) + 15 + 14 + 162(ii)???? 10.37%accept 10 / 10.4 / 10.37...............if answer is not correct then evidence of correct working gains one mark.e.g. minimum evidence would be 14/1352(c)???? any two from:???????? faster???????? more accurate???????? detects smaller amounts2(d)???? to avoid biasaccept to check / compare the result1to improve reliability1[10]???E1.????????? A reasonable proportion of students gave completely correct calculations. A substantial number of students were able to complete the first 2 steps of the calculation successfully, but were unable to produce a ratio from their answers. Common errors included dividing Ar by % and incorrect rounding to produce the ratio. Weaker students made attempts at adding 16 and 56 and dividing by 72.??E2.???????? This question was a challenging question aimed at A to A* students. The question sought to give the students a question in which the determination of an empirical formula was relevant to a research problem. It is pleasing to note, however, that a fair number of candidates gave perfect answers to this difficult question. The mark scheme was also eased slightly, compared to previous years, to allow more credit to be given to those who completed part of the calculation.?????????A common error was to divide the Ar by the percentage mass in the first step of the calculation.???E3.????????? (a)???? Empirical formulae calculations can be very challenging, but it was pleasing to see that a majority of candidates gained 2 or more marks, and most were able to make a realistic start to the calculation. Having successfully started the calculation by dividing the mass by the Ar for each element, a significant number of candidates found obtaining a whole number ratio challenging, and so rounded the ratio of 1:1.3 to give 1:1, and hence PbO as the formula, gaining credit for the correct working, though not for the incorrect answer. Even lower scoring candidates were aware that division was needed to start the calculation, but some were unsure about what to divide. Other common errors included Pb4O3 (obtained by confusing and swapping the elements part way through the calculation, or by dividing the Ar by the mass for each element), and Pb3O2 ( obtained by dividing 0.64/32 for oxygen). It was pleasing to see that many answers were clearly written with logically presented working, enabling examiners to award the credit due. A small minority of candidates continue to give a jumble of numbers, or no working at all. If no correct working was given or discernable, then 2 marks were awarded if a correct answer is given. Correct working had to be shown to access the remaining marks.(b)?????(i)????? The majority of candidates gained credit in this question. The most common errors included putting more electrons in the outer shell of sulfur than 2 lone pairs, adding an extra electron to the hydrogen shell, or giving only 2 lone electrons on the sulfur shell. A minority of candidates put 3 bonding electrons in the overlap between the hydrogen and sulfur shells. If crossings out are necessary, candidates should be encouraged to make the diagram clear.(ii)?????This question revealed a number of misconceptions, and discriminated well. Candidates showed some confusion between covalent bonds and intermolecular forces, with many stating that covalent bonds are weak. There were some excellent responses, but others showed that candidates had learnt the phrase ‘weak intermolecular forces’, while the rest of their answer revealed that they did not understand it, for example stating that covalent bonds have weak intermolecular forces. Examiners found occasional references to ionic bonds or delocalised electrons. There were many references to unspecified weak bonds. (iii)????The majority of candidates gained this mark with 2 often seen as an incorrect response. ??E4.???????? Part (a) was well attempted by the majority of the candidates who realised that the smell would alert people to the presence of the gas if there was a leak.???????? Part (b) was the worst attempted question on the Paper. Mass spectrometry was poorly understood by most of the candidates, and there were many references to atomic mass (rather than molecular/formula mass), isotopes and elements.???????? This part seemed to discriminate well. About half of the candidates in part (c) were able to gain some credit on this question. The first two steps in the calculation were generally done well. However, the translation that 1 mole of carbon dioxide is equivalent to 2 moles of water and hence 4 moles of hydrogen proved challenging for many.??E5.????????? This part was designed to discriminate between the A* and A candidates. The performance of the candidates was slightly better this year than for the equivalent calculation last summer. About one quarter of the candidates were able to complete the calculations and gain all four marks. These candidates usually set out their calculation neatly and in a logical way. A number of candidates lost marks due to untidy working and careless processing of numbers. Candidates should be encouraged to check their calculations using a calculator. A number of candidates worked out the number of moles of each element correctly, worked out the ratio correctly but then did not write the formula of the compound so that they only gained three of the four marks.????????? A fairly large number of candidates could not convert the number of moles of each element into a correct ratio. This was sometimes caused by difficulty in deciding which number, 0.05, 0.15 or 0.025, or was the smallest number of moles.????????? As in previous years some candidates incorrectly divided the relative atomic mass by the mass of the element. Error carried forward was allowed so that they could still gain two of the four marks.????????? About 8% of the candidates were unable to make any attempt at this question.?? ................
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