Internal assessment resource number Chem/2/3 – E version 1 PAGE FOR TEACHER USE 2005 Internal Assessment Resource Subject Reference: Chemistry 2.3 Internal assessment resource reference number: Chem/2/3 – E version 1 “Quantitative Chemistry 2” Supports internal assessment for: Achievement Standard 90763 Version 1 Solve simple quantitative chemical problems Credits: 2 Date version published: November 2004 Ministry of Education quality assurance status For use in internal assessment from 2005. © Crown 2004 1 Internal assessment resource number Chem/2/3 – E version 1 : Quantitative Chemistry 2 PAGE FOR STUDENT USE 2005 SUBJECT REFERENCE: CHEMISTRY 2.3 Internal assessment resource reference number: Chem/2/3 “Quantitative Chemistry 2” Achievement Standard 90763 Version 1 Solve simple quantitative chemical problems Credits: 2 Time allowed: 1 hour This is a closed book activity. Student Instructions Sheet For some of the following calculations you will need to refer to the Periodic Table on page 5 to find the required molar masses of elements. Give answers to 3 significant figures where appropriate. 1 What is the mass of 0.126 moles of oxalic acid, H2C2O4? M(H2C2O4) = 90.0 g mol1 2 19.4 g of baking soda, NaHCO3, is dissolved in water to make exactly 1 litre of solution. What is the concentration in mol L1? M(NaHCO3) = 84.0 g mol1 3 4 How many moles of sucrose are present in 14.0 mL of 0.450 mol L 1 sucrose olution? Allicin is the compound responsible for the characteristic smell of garlic. Its formula is C6H10S2O. What is the % of carbon in this compound? © Crown 2004 1 Internal assessment resource number Chem/2/3 – E version 1 : Quantitative Chemistry 2 PAGE FOR STUDENT USE 5(a) To carry out a colorimetric analysis a set of standard solutions are prepared from a stock solution of 0.00500 mol L1 FeCl3.6H2O. What mass of FeCl3.6H2O is required to make up 500 mL of this solution? (b) 20.0 mL of a 0.00500 mol L1 solution from part (a) above was pipetted into a volumetric flask and diluted to 100 mL. What is the concentration of the diluted solution? 6 The porphryn molecule forms an important part of the haemoglobin structure. This porphryn molecule is 78.5% carbon, 3.2% hydrogen and 18.3% nitrogen. (i) Calculate the empirical formula of porphryn. (ii) If the molar mass of porphryn is 306 g mol1 calculate the molecular formula. 7 A precipitate of 1.34 g of Fe2O3 was obtained by treating a vitamin supplement dissolved in water with NaOH and then heating. What is the mass of iron in the tablet? © Crown 2004 2 Internal assessment resource number Chem/2/3 – E version 1 : Quantitative Chemistry 2 PAGE FOR STUDENT USE 8 Hydrated magnesium chloride is heated in a crucible. The following data is collected. mass of crucible and lid = 26.47 g mass of crucible, lid and hydrated magnesium chloride = 28.62 g mass crucible, lid and magnesium chloride after first heating = 27.49 g mass of crucible, lid and magnesium chloride after second heating = 27.48 g Use these results to determine the formula of hydrated magnesium chloride. M(MgCl2) = 95.0 g mol1 M(H2O) = 18.0 g mol1 9 Wine makers convert sugars such as glucose to ethanol and carbon dioxide: C6H12O6 2 C2H5OH + 2 CO2 M / g mol1: C2H5OH = 46.0, C6H12O6 = 180 Starting with 540 g of glucose: (a) what is the mass of ethanol that can be obtained by this process? (b) what is the volume of ethanol that can be obtained by this process? density ethanol = 0.789 g mL1 © Crown 2004 3 Internal assessment resource number Chem/2/3 – E version 1 : Quantitative Chemistry 2 PAGE FOR STUDENT USE 10 10.0 g of iron is reacted with excess chlorine gas to produce iron chloride. The resulting precipitate has a mass of 29.0 g. Use this information to decide whether the product is iron(II) chloride FeCl2 or iron(III) chloride FeCl3. 11 In an experiment sodium hydroxide was used to prepare a sample of Glauber's salt, Na2SO4.10H2O, by neutralisation with sulfuric acid followed by recrystallisation from aqueous solution. H2SO4 + 2 NaOH Na2SO4 + 2 H2O Calculate the maximum mass of Glauber's salt which could be made from 5.00 g of sodium hydroxide. © Crown 2004 4 Internal assessment resource number Chem/2/3 – E version 1 : Quantitative Chemistry 2 PAGE FOR STUDENT USE PERIODIC TABLE OF THE ELEMENTS 18 Atomic Number 1 1 2 3 4 Li 6.9 11 Na 23.0 19 K 39.1 37 Rb 85.5 55 Cs 133 87 Fr 223 Be 9.0 12 Mg 24.3 20 Ca 40.1 38 Sr 87.6 56 Ba 137 88 Ra 226 13 Atomic Mass 5 B 10.8 13 Al 3 4 5 6 7 8 9 10 11 12 27.0 21 22 23 24 25 26 27 28 29 30 31 Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga 45.0 47.9 50.9 52.0 54.9 55.9 58.9 58.7 63.5 65.4 69.7 39 40 41 42 43 44 45 46 47 48 49 Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In 88.9 91.2 92.9 95.9 98.9 101 103 106 108 112 115 71 72 73 74 75 76 77 78 79 80 81 Lu Hf Ta W Re Os Ir Pt Au Hg Tl 175 179 181 184 186 190 192 195 197 201 204 103 104 105 106 107 108 109 Lr Rf Db Sg Bh Hs Mt 262 57 Lanthanide Series Actinide Series © Crown 2004 2 H 1.0 58 59 La Ce Pr 139 140 141 89 90 91 Ac Th Pa 227 232 231 60 61 62 63 Nd Pm Sm Eu 144 147 150 152 92 93 94 95 U Np Pu Am 238 237 239 241 14 15 16 17 6 7 8 9 C 12.0 14 Si 28.1 32 Ge 72.6 50 Sn 119 82 Pb 207 N 14.0 15 P 31.0 33 As 74.9 51 Sb 122 83 Bi 209 O 16.0 16 S 32.1 34 Se 79.0 52 Te 128 84 Po 210 F 19.0 17 Cl 35.5 35 Br 79.9 53 I 127 85 At 210 64 65 66 67 Gd Tb Dy Ho 157 159 163 165 96 97 98 99 Cm Bk Cf Es 247 249 251 254 He 4.0 10 Ne 20.2 18 Ar 40.0 36 Kr 83.8 54 Xe 131 86 Rn 222 68 69 70 Er Tm Yb 167 169 173 100 101 102 Fm Md No 257 258 255 5 Internal assessment resource number Chem/2/3 – E version 1 PAGE FOR TEACHER USE AS90763 Version 1 Assessment Schedule: Quantitative Chemistry 2 Chem 2.3 E version 1 Q Judgement for Achievement Evidence Judgement for Merit 1 11.3 g Correct 2 0.231 mol L1 Correct 3 6.30 x 103 mol Correct 4 M(C6H10S2O) 162 g mol1 Correct molar mass of C6H10S2O. Correct %. Amount of FeCl3.6H2O correct. Correct mass determined. Amount of FeCl3.6H2O correct. Correct determination of concentration of diluted solution. Amount in moles of each element correct. Empirical formula correct. Correct process, but incorrect empirical formula used. Molecular formula correct. % C = 44.4% 5(a) n(FeCl3.6H2O) = 5.00 x 103 mol L1 x 0.500 L = 2.50 x 103 mol Judgement for Excellence M(FeCl3.6H2O) = 270.4.g mol1 m(FeCl3.6H2O) = 270.4 g mol1 x 2.50 x 103 mol = 0.676 g (b) n(FeCl3.6H2O) = 5.00 x 103 mol L1 x 0.0200 L = 1.00 x 104 mol c(FeCl3.6H2O) = 1.00 x 103 mol L1 6(a) n(C) = 78 .5 g 12 .0 g mol n(N) = 1 18.3 g 14.0 g mol 1 = 6.54 mol n(H) = 3.20 g 1.0 g mol 1 = 3.20 mol = 1.31 mol empirical formula C10H5N2 (b) M(empirical formula) = 153 g mol1 molecular formula C20H10N4 © Crown 2004 6 Internal assessment resource number Chem/2/3 – E version 1 PAGE FOR TEACHER USE 7 method A (i) n(Fe2O3) = 1.34 g /159.8 g mol1 Correct process used in either step (i) or (iii). Correct answer for mass of Fe produced. eg An incorrect molar mass or wrong mole ratio used. = 8.39 x 103 mol n(Fe) = 2 x 8.39 x 103 mol (ii) Correct process with one error. = 1.68 x 102 mol (iii) m(Fe) = 1.68 x 102 mol x 55.9 g mol1 = 0.938 g OR method B (i) % Fe in Fe2O3 = 55.6 x 2 / 159.8 = 70.0 % (ii) 70.0% of 1.34 g = 0.938 g 8 mass of anhydrous MgCl2 = 1.01 g mass of H2O = 1.14 g amount of MgCl2 = 1.01 g / 95 g mol1 = 1.06 x 102 mol Correct determination of masses of MgCl2 and H2O. Correct determination of either amount of MgCl2 or H2O. Correct formula for hydrated salt. Correct process used in either step (i) or (iii). Correct determination for mass of ethanol. Correct determination of answer for volume of ethanol produced. amount H2O = 1.14 g / 18.0 g mol1 = 6.33 x 102 mol ratio n(H2O) / n(MgCl2) = 6 formula is MgCl2.6H2O 9(a) (i) n(glucose) = 540 g / 180 g mol1 = 3.00 mol (ii) n(ethanol) = 2 x n(glucose) = 6.00 mol (iii) m(ethanol) = 6.00 mol x 46.0 g mol1 = 276 g (b) volume ethanol = 276 g / .789 g mL1 = 350 mL © Crown 2004 7 Internal assessment resource number Chem/2/3 – E version 1 PAGE FOR TEACHER USE 10 (i) n(Fe) = 10.0 g / 55.9 g mol1 = 0.179 mol (ii) M(FeCl2) = 126.9 g mol-1 and M(FeCl2) = 162.3 g mol1 Correct amount of Fe determined. Correct process with one error. eg Incorrect molar mass. (iii) If product is FeCl2 Correct identification of product with appropriate calculations. m(FeCl2) = 0.179 mol x126.9 g mol1 = 22.7g but if product is FeCl3 m(FeCl3) = 0.179 mol x 162.3 g mol1 = 29.1 g therefore product is FeCl3 This calculation can also be done by determining the ratio of n(Fe) to n(Cl) in the 29.0 g of product. 11 (i) n(NaOH) = 5.00 g / 40.0 g mol1 = 0.125 mol (ii) n(Na2SO4) = 1/2 x n(NaOH) = 0.0625 mol Amount of NaOH correctly determined. Correct process with one error. Correct determination of mass of salt. (iii) n(Na2SO4.10H2O) = 0.0625 mol M(Na2SO4.10H2O) = 322 g mol1 (iv) m(Na2SO4.10H2O) = 0.0625 mol x 322 g mol1 = 20.1 g Correct use of significant figures and units is required for answers at excellence level. Sufficiency Statement In the application of the assessment schedule, the student's response to each question is only awarded one level of A, M or E. The requirements for sufficiency are as follows: Achievement 7 answers at achievement level. Merit 6 answers at merit level PLUS 3 other answers at achievement level. Excellence 3 answers at excellence level PLUS 4 other answers at merit level AND 3 other answers at achievement level.. © Crown 2004 8