Edexcel Jun 2012 Paper 5 Q14 (with explained solutions) |

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Edexcel Jun 2012 Paper 5 Q14

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14 An organic compound X contains carbon, hydrogen, oxygen and nitrogen. 0.132 g of X is burned completely in oxygen to produce 0.072 g of water, 0.176 g of carbon dioxide and 24.0 cm3 of nitrogen. [Molar volume of nitrogen under the conditions of the experiment = 24000 cm3 mol1] (a) Calculate the empirical formula of compound X. (b) The molar mass of X is 132 g mol1. Deduce the molecular formula for X. Show how you arrived at your answer. 14 *P39309A01420* (5) (1)<br />
(c) When X is refluxed with concentrated hydrochloric acid for several hours, cooled and neutralized, there is only one organic product, Y, which has the molecular formula C2H5O2N. One mole of Y will react with either one mole of hydrochloric acid or one mole of sodium hydroxide solution. When Y is sprayed with a solution of ninhydrin and heated, a purple colour is observed. (i) Use all the information above to deduce the functional groups present in Y and to classify the type of compound it is. Justify your answer. (3)(ii) Deduce the displayed formula for Y and give its name. (2) Name of Y(iii) Hence give the structural formula for X. (1) (Total for Question 14 = 12 marks) TOTAL FOR SECTION B = 48 MARKS *P39309A01520* 15 Turn over<br />

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Reject Mark Acceptable Answers Route 1 by mol of H, C and N 0.072 = 0.004 mol water 18 OR 0.008 mol H(atoms) And 0.176 = 0.004 mol carbon (dioxide) (1) 44 24.0 = 0.001 mol nitrogen N2 24000 OR 0.002 mol N(atoms) (1) Mass of H + mass of C + mass of N = 0.008 + 0.004 x 12 + 0.028 (1) = 0.084 g mass of oxygen = 0.132(0.008 + 0.004 x 12 + 0.028) = 0.048 g amount of oxygen = 0.048 = 0.003 mol (1) 16 empirical formula is C4H8O3N2 (1) Route 2 by mass of H, C and O calculated in one step mass of H = 2/18 x 0.072 = 0.008 g (1) mass of C = 12/44 x 0.176 = 0.048 g (1) mass of N = 24/24000 x 28 = 0.028 g (1) mass of O = 0.132(0.008 + 0.048 + 0.028) = 0.048 g moles of O = 0.003 (1) moles of H = 0.008 Question Number 14(a)<br />
moles of C = 0.004 moles of N = 0.002 empirical formula is C4H8O3N2 (1) Route 3 Percentage by mass of each element in 0.132 g First three marks by either method above. Then percentages are: H6.06 C36.36 N21.21 So O is 100(6.06 + 36.36 + 21.21) = 10063.63 = 36.37 Mole ratios O2.27allow = or0.02 (1) H6.06, C3.03, N1.515 Dividing by smallest gives H4, C2, N1, O1.5 empirical formula is C4H8O3N2 (1) The following transferred errors are allowed: If nitrogen gas taken as N, first two marks can still be awarded for all methods Then mass of nitrogen is 0.014 g This gives mass of oxygen as 0.062 g and amount of oxygen as 0.003875 mol (1) now empirical formula is C4H8O4N (1) OR percentage method: N10.61% O46.97%<br />
Mole ratio N0.7575 O2.935 (1) empirical formula is C4H8O4N (1) Transferred error for hydrogen Two from first three marks still awarded Then amount of hydrogen is 0.004 mol This gives 0.003125 mol oxygen empirical formula is C4H4O3N2 (1) Both the above nitrogen and hydrogen errors Award 1 mark for correct mass of carbon or correct moles of carbon Then mass of nitrogen is 0.014 g Then mass of hydrogen is 0.004 g This gives 0.004125 mol oxygen (1) Empirical formula is C4H4O4N (1)<br />
Reject Mark Acceptable Answers (12 x 4 + 1 x 8 + 16 x 3 + 14 x 2)n = 132 n = 1 So molecular formula is C4H8O3N2 Some element of working must be shown TE from (a) of nitrogen error can be given only if: (12 x 4 + 1 x 8 + 16 x 4 + 14)n = 132 n = 0.98 (which is approximately 1) TE from (a) of hydrogen error can be given only if: (12 x 4 + 1 x 4 + 16 x 3 + 14 x 2)n = 132 n = 1.03 (which is approximately 1) TE from (a) of nitrogen and hydrogen error can be given only if: (12 x 4 + 1 x 4 + 16 x 4 + 14)n = 132 n = 1.015/1.02 (which is approximately 1) Question Number 14(b)<br />
Acceptable Answers Reject Mark Question Number 14(c)(i) + (1) - (1) Y reacts with HCl/acid so it is an amine /contains NH2 /CO2 It reacts with alkali/NaOH so it is a carboxylic acid/contains CO2H/NH3 It forms a purple colour/reacts with ninhydrin so it is an amino acid (1) OR As it is an amine/contains NH2 /CO2 react with HCl/acid (1) + As it is a carboxylic acid/contains CO2H/NH3 it will react with alkali/NaOH (1) As it is an amino acid so it forms a purple colour/reacts with ninhydrin (1) Each marking point is independent and requires both the functional group and the test NOTE: It is an amino acid so it reacts with acid and alkali (with neither of first two points) (1) - it will Justit is a base Justit is an acid it is amphoteric (alone) Mark Question Number 14(c)(ii) H H O-H Acceptable Answers | | | H-N-C-C=O ALLOW OH | H OR H H O- | | | H-N+-C-C=O | | H H (1) 2-aminoethanoic acid/ aminoethanoic acid/glycine (1) Mark independently Reject C-H-O if bond is clearly to H 1- aminoethanoic acid<br />
Acceptable Answers H2NCH2CONHCH2CO2H Or NH2CH2CONHCH2CO2H Or HOCOCH2NHOCCH2NH2 ALLOW H Reject Mark O H Or reversed displayed formula ALLOW ionic formulae with H3N+ and CO2 - TOTAL FOR SECTION B = 48 MARKS Question Number 14 (c)(iii)<br />

Question 12 13 14 15

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