SUGGESTED ANSWER FOR SK015 TUTORIAL 1.0 Matter 2023/2024 1.1 Atoms and Molecules 1. a) i. Number of protons in the nucleus of an atom of an element. ii. ii. Total number of protons and neutrons in the nucleus of an atom of an element. iii. iii. Two or more atoms of the same element having the same number of protons but different number of neutrons. b) 56 138 3+ 13 27 14 28 2− 16 32 c) Chromium p: 24, e:24, n: 28 fluorine p: 9, e: 9, n: 10 2. a) average atomic mass Mg = (78.9 23.98504 ) + (10.0 24.98584 ) + (11.1 25.98259 ) (78.9 + 10.0 + 11.1) = 24.30 amu Relative atomic mass Mg = 24.30 1 12 12.00 = 24.30 Relative atomic mass Mg is 24.30
b) 3. 63 Cu : X 65Cu : 100-X Average atomic mass Cu is 63.5 63.5 = 63()+65(100−) 100 X = 75% Percentage abundance 63Cu: X: 75% Percentage abundance 65Cu: 100-X: 25% 1.2 Mole Concept 1. Given: 10.0 g CuSO4.5H2O a) number of moles of CuSO4.5H2O. 10.0 g n CuSO4.5H2O = = 0.0400 mol 249.7 g mol 1 b) number of moles of copper atoms. 1 mol CuSO4.5H2O ≡ 1 mol Cu atom 0.0400 mol CuSO4.5H2O ≡ 0.0400 mol Cu atom c) mass of anhydrous copper (II) sulphate. 1 mol CuSO4.5H2O ≡ 1 mol CuSO4 0.0400 mol CuSO4.5H2O ≡ 0.0400 mol CuSO4 ∴ mass of C atoms = 0.0400 mol x 159.7 g mol-1 = 6.39 g 2. a) Definition Empirical formula: formula that shows the simplest ratio of atoms of different elements present in amolecule
Molecular formula: formula that shows the actual number of atoms of different elements present in amolecule b) Element C H N O Mass (g) 0.198 0.025 0.116 0.661 Moles (mol) 0.0165 0.025 8.286 X10-3 0.0413 ratio 2 3 1 5 Empirical formula C2H3NO5 n = 121 121 = 1 Molecular formula : C2H3NO5 3. mass sample: 0.255 g mass CO2: 0.561 g moles of CO2 = 0.561 44 / = 0.0128 mol Mass H2O: 0.306 g moles of H2O = 0.306 18 = 0.017 mol Mass O = 0.255g - (0.153+ 0.034)g = 0.068 g Element C H O Mass (g) 0.153 0.034 0.068 Moles (mol) 0.0128 0.034 4.25 X 10-3 ratio 3 8 1 Empirical formula C3H8O Empirical formula is C3H8O 4. a) molality is number of moles of solute per kilogram of solvent. b) density solution: 0.954 g/mL molarity: 2.5 M
0.954 g/mL = 2.5 M = / Assume V solution = 1000 mL @1L Mass solution: 954 g Moles of solute: 2.5 mol, mass solute: 80 g Mass solvent: 954g- 80g = 874 g Molality = 2.5 (874/1000) = 2.86 mol/kg c) %w/w: 37.8% density solution: 1.19 g/mL 37.8 % = 100% 1.19 g/mL = / Assume mass solution: 1000 g Mass solute: 378 g, moles solute: 10.36 mol V solution: 840 mL @ 0.84L Molarity = 10.36 0.84 = 12.33 M d) Mass solute CaCl2: 16 g mass solvent, H2O: 64.0 g n solute = 16 1101.6 n solvent = 64.0 18 = 0.145 mol = 3.556 mol Mol fraction = 0.145 (0.145+3.556) 100% = 3.92% % w/w = 16 (16+64) 100% = 20%
1.3 Stoichiometry 1. ii. 8I -(aq) + SO4 2-(aq) + 10H + (aq) 4I 2(s) + H2S (aq) + 4H2O(l) i. C5H6O (l) + 6O2 (g) 5CO2 (g) + 3H2O (g) iii. 4Cl2 (aq) + 8OH - (aq) ClO4 - (aq) + 7Cl - (aq) + 4H2O (l) 2. 3. a) nCO = 50 28 n hydrogen= 0.8 mol available = 1.786 mol 1 mol CO ≡ 2 mol H2 1.786 mol CO ≡ 2 X 1.786 mol = 3.572 mol (needed) H2 is a Limiting reactant
b) percentage yield = ℎ 100% n CH3OH yield: 0.4 mol mass CH3OH: 0.4 mol X 32 gmol-1 = 12.8g = 11.9 12.8 100% = 92.97% 4. Given: Mass of copper(II) oxide, CuO = 7.0 gV nitric acid, HNO3 solution = 50 mL Concentration of nitric acid, HNO3 solution = 0.20 M a. a reactant that is completely consumed during a reaction and limits the amount of product(s) formed. b. CuO(s) + 2HNO3(aq) Cu(NO3)2(aq) + H2O(l) c. Given, mol of CuO = 0.0881 mol mol of HNO3 = 0.01 mol 1 mol CuO ≡ 2 mol HNO3 0.0881 mol CuO ≡ 0.1762 mol HNO3 ∴ mol of HNO3 given < needed. HNO3 is limiting reactant. d. expected mass of copper (II) nitrate formed. 1 mol HNO3 ≡ 1 mol Cu(NO3)2 0.01 mol HNO3 ≡ 5 x 10-3 mol Cu(NO3)2Mass of Cu(NO3)2 = 0.94 g e. Given: actual mass of copper(II) nitrate = 0.85 g.
% yield = actual mass x100% theoritical mass = 0.85g x100% 0.94g = 90%