OCR A Jun 2013 Paper 2 Q6 (with explained solutions) |

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OCR A Jun 2013 Paper 2 Q6

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6 Mass spectrometry and infrared spectroscopy are used in analysis. 15 (a) The element sulfur exists as molecules, Sn. The mass spectrum that would be given by a sample of sulfur is shown below. All the sulfur atoms are the same isotope. abundance 32 64 96 128 160 192 224 256 m /z (i) State the m/z value of the molecular ion.[1] (ii) Suggest the formula for a molecule of sulfur.[1] (iii) What is the formula for the fragment ion with m/z = 128?[1]OCR 2013 Turn over<br />
(b) A sample of an element, L is analysed using mass spectrometry. The mass spectrum is shown below. 16 percentage abundance (%) 40 30 20 10 33% 34% 25% 194 195 196 m /z 8% 198 Calculate the relative atomic mass of L. Give your answer to one decimal place. relative atomic mass of L =[2] (c) Give an everyday use for infrared spectroscopy.[1]OCR 2013<br />
(d) The solvent, M, is an organic compound used in paints. The solvent M was analysed. 17 M has a relative molecular mass of 72.0. The percentage composition by mass of M is C, 66.7%; H, 11.1%; O, 22.2%. The infrared spectrum of M is shown below. 100 transmittance (%) 50 0 4000 3000 2000 1500 1000 500 wavenumber / cm1 The analysis produces several possible organic structures. Suggest, with reasons, two possible structures for M.[5]OCR 2013 [Total: 11] Turn over<br />

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F322/01 Mark Scheme June 2013 Answer Marks Question 6 (a) (i) 256 (cid:57) (ii) S8 (cid:57) (iii) S4 + (cid:57) (b) FIRST, CHECK THE ANSWER ON ANSWER LINE IF answer = 195.2, award 2 marks. IF answer = 195.16 award 1 marks. = (19433)+(19534)+(19625)+(1988) 195.2 (cid:57) Monitor air pollution OR breathalysers (cid:57) 100 (cid:57) (c) Guidance 16 S8 OR 32 16 S8 etc ALLOW 32S8 OR 32 DO NOT ALLOW 33S8 OR 30 Positive ion must be present ALLOW 32S4 16 S4 DO NOT ALLOW 33S4 + 195 on its own with no working scores 0 marks OR 30 16 S4 etc ALLOW measure the concentration or abundance of atmospheric pollutants ALLOW measure concentration of named atmospheric pollutant ALLOW monitoring of gases in car exhaust fumes ALLOW drug detection or drug identification IGNORE night vision goggles, identifying gases on distant planets / ice samples 20<br />
F322/01 Mark Scheme June 2013 Question 6 (d) Answer : 11.1 1.0 OR 5.56 :11.1 : 1.39 (cid:57) mole ratio C : H : O : 22.2 66.7 12.0 16.0 4 : 8 : 1 OR C4H8O (cid:57) contains a C=O or carbonyl because of absorbance at about 1710 cm1 (cid:57) Any two from: Marks Guidance PLEASE LOOK AT THE SPECTRA AND ABOVE THE SPECTRA FOR POSSIBLE ANSWERS ALLOW two marks for 72 x 66.7/100 = 48/12 = 4 (C) 72 x 11.1/100 = 8 = 8 (H) 72 x 22.2/100 = 16 = 1 (O) ALLOW C=O or carbonyl since has absorbance within the range 1640 to 1750 cm1 ALLOW ketone OR aldehyde linked to correct absorbance ALLOW could be aldehyde, ketone, carboxylic acid, ester (or amide) because of absorbance between range 1640 to 1750 cm1 (ie direct quote from the data book) DO NOT ALLOW reference to M being a carboxylic acid, ester or amide unless they are included in a list with aldehyde/ketone in which case IGNORE carboxylic acid/ester/amide IGNORE reference to CO / absence of OH DO NOT ALLOW has OH ALLOW correct structural OR displayed OR skeletal formula OR mixture of the above (as long as unambiguous) eg CH3CH2CH2CHO, CH3COCH2CH3 OR (CH3)2CHCHO DO NOT ALLOW C3H7CHO IGNORE incorrect name correct name on its own is not sufficient CH3 CH2 CH2 C CH3 CH3 CH CH3 CH2 CH3 (cid:57)(cid:57) Total 11 21<br />

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