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2 Atoms, Molecules, and Ions Solutions to Exercises (b) ³H has 1 proton, 2 neutrons, and 1 electron. ³He: 2(1.672621673 10⁻²⁴ g) + 1.674927211 g + 2(9.10938215 10⁻²⁸ g) = 5.021992 10⁻²⁴ g ³H: 1.672621673 10⁻²⁴ g + 2(1.674927211 10⁻²⁴ g) + 9.10938215 10⁻²⁸ g = 5.023387 10⁻²⁴ g Tritium, ³H, is more massive. (c) The masses of the two particles differ by 0.0014 10⁻²⁴ g. Each particle loses 1 electron to form the +1 ion, so the difference in the masses of the ions is still 1.4 10⁻²⁷. A mass spectrometer would need precision to 1 10⁻²⁷ g to differentiate and ³H. 2.83 (a) Calculate the mass of a single gold atom, then divide the mass of the cube by the mass of the gold atom. 197.0 amu 1g gold atom 6.022 10²³ amu = 3.2713 10⁻²² = 3.271 10⁻²² g/gold atom 19.3g cube 3.271 gold atom = 5.90 Au atoms in the cube (b) The shape of atoms is spherical; spheres cannot be arranged into a cube so that there is no empty space. The question is, how much empty space is there? We can calculate the two limiting cases, no empty space and maximum empty space. The true diameter will be somewhere in this range. No empty space: volume cube/number of atoms = volume of one atom V = 4/3π r³; r = (3π d = 2r vol. of cube = (1.0 1.0 1.0) = 5.90 1.0 cm³ Au atoms = 1.695 10⁻²³ = 1.7x10⁻²³ r = [π (1.695 X 10⁻²³ = 3.4 cm; d = 2r = 6.8 10⁻⁸ cm Maximum empty space: assume atoms are arranged in rows in all three directions so they are touching across their diameters. That is, each atom occupies the volume of a cube, with the atomic diameter as the length of the side of the cube. The number of atoms along one edge of the gold cube is then (5.90 1/3 = 3.893 10⁷ = 3.89 10⁷ atoms/1.0 cm. The diameter of a single atom is 1.0 cm/3.89 10⁷ atoms = 2.569 = 2.6 cm. The diameter of a gold atom is between 2.6 10⁻⁸ cm and 6.8 10⁻⁸ cm (2.6 - 6.8 Å). 31