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Elements

Elements. Compounds. Identical. atoms. Atoms. Molecules. ± e -. ± n. ± e -. ± n. Ions. Isotopes. (Chemical) Atomic Mass: The average of the isotope masses of an element, weighted to reflect their relative natural abundances.

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Elements

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  1. Elements Compounds Identical atoms Atoms Molecules ± e- ± n ± e- ± n Ions Isotopes

  2. (Chemical) Atomic Mass: The average of the isotope masses of an element, weighted to reflect their relative natural abundances. • Example: Chlorine has two naturally occurring isotopes, 35Cl • (A = 34.97 a.m.u.) and 37Cl (A = 36.96 a.m.u.). • The respective natural abundances of these • isotopes are 75.5% and 24.5%. (34.97 x 75.5) + (36.96 x 24.5) 100 Atomic mass of Cl = = 35.45 a.m.u. • Molecular Mass: The sum of the atomic masses of all of the • constituent atoms in a molecule. Example: Molecular mass of H2? (A.M. of H = 1.008 a.m.u.) M.M. of H2 = 2 x 1.008 = 2.016 a.m.u.

  3. Example 2: M.M. of carbon monoxide, CO? (A.M. of C = • 12.01 a.m.u., A.M. of O = 15.99 a.m.u) M.M. of CO = 12.01 + 15.99 = 29.01 a.m.u. • Example 3: M.M. of lead nitrate, Pb(NO3)2? (Pb =207.2, • N = 14.01, O = 15.99) • M.M. of Pb(NO3)2 = 207.2 + (14.01 x 2) + (15.99 x 6) • = 331.16 a.m.u.

  4. Chemical Accounting 1: The Mole Relating Number of Chemical Entities to their Mass

  5. Consider the atomic mass table: • ElementAtomic mass (of one atom) • H 1.008 a.m.u. • He 4.003 a.m.u. • Li 6.941 a.m.u. • Be 9.012 a.m.u. • C 12.000 a.m.u. • ...etc. ...etc. • Gives: Relative mass of one atom of a given element w.r.t. to • mass of one atom of one element (carbon)

  6. Practical work requires a table which gives: • (a) relative mass of a given large number of atoms of each • element w.r.t. mass of the same number of C atoms; • (b) a value of 12 g for this given large number of C atoms. • Thus this “practical” table should look like............... • ElementMass (of n atoms) • H 1.008 g • He 4.003 g • Li 6.941 g • Be 9.012 g • C 12.000 g • ...etc. ...etc. • where n is the given large number of atoms.

  7. What value of n will allow us to “scale up” from the a.m.u. scale (single atoms) to the gram scale (large numbers of atoms)? n = 6.022 x 1023 entities = Avogadros Number (N) = 1 mole (mol) Now have a system for relating large numbers of atoms to their total mass, and for comparing the relative masses of any given large number of atoms of different elements: • Element Atomic Mass Scale “Mole Mass Scale” • No. of MassNo. of Mass • Atoms (a.m.u.)Atoms (g) • H 1 1.008 1 mole 1.008 • He 1 4.003 1 mole 4.003 • Li 1 6.941 1 mole 6.941 • Be 1 9.012 1 mole 9.012 • ...etc.. ...etc.. ...etc.. 6.022 x 1023 atoms

  8. This system can also be used to determine the masses of large numbers of other chemical entities (e.g. molecules, ions, etc.). Example 1: Mass of 0.5 mole of plutonium (Pu) atoms? From table of atomic masses: Pu = 244. This means that 1 atom of Pu has a mass of 244 a.m.u. and that 1 mole of Pu has a mass of 244 g. Since 1 mole Pu = 244 g, then 0.5 mole Pu has a mass of = 244 x 0.5 g. Answer: 122 g

  9. Example 2: Mass of 1 mole of silicon carbide (SiC)? A molecule of SiC consists of one atom of silicon and one atom of carbon bonded together. Thus 1 mole SiC consists of 1 mole Si + 1 mole C. From atomic mass table, Si = 28.086 and C = 12.011. So the mass of 1 mole SiC = 28.086 + 12.011 g • Answer: 40.097 g

  10. Example 3: Mass of 2 moles of SO42- ions? (O = 15.999 and • S = 32.06) • 1 SO42- ion contains 1 S and 4 O’s, thus: • 2 mol SO42- contains 2 mol sulphur and 4 x 2 mol oxygen • Mass of 2 mol S = 2 x 32.06 g = 64.12 g • Mass of 4 x 2 mol O = 4 x 2 x 15.99 g = 127.92 g Mass of 2 moles of sulphate ions = 192.04 g

  11. Molarity Molarity = moles of solute litres of solution Example: What is the molarity of a solution made by dissolving 2.355 g of sulfuric acid (H2SO4) in water and diluting to a final volume of 50.0 ml? Molecular weight of H2SO4 = 2.0 x 1 + 32.1 + 16.0 x 4 = 98.1 a.m.u. Molar mass of H2SO4 = 98.1 g/mol 2.355 x 1/98.1 = 0.024 Units g x mol g-1 = mol => 0.024 mol H2SO4 Molarity = moles of solute/ Litres of solution = 0.024 mol/ 0.05 L = 0.48 M

  12. Example: What is the concentration of water in 1 litre of water? Molecular weight of H2O = 2.0 x 1 + 16.0 = 18.0 a.m.u. Density of water = 1.0 g/cm3 or 1.0 kg/L Mass of H2O in 1 L = 1 kg molarity = moles of solute/ litres of solution = 1000/18.0 g L -1 g-1 mol = 55.5 mol L-1 = 55.5 M

  13. Example: The molecular mass of sugar (C12H22O11) How many molecules of sugar are there in 1 spoon (0.025 kg) and what is the concentration of sugar in 1 cup of coffee (100 ml)? Molecular mass of sugar = 12 x 12.00 + 22 x 1.01 + 11 x 16.00 = 342.22 a.m.u. Mass of 1 mole of sugar = 342.22 g # of moles of sugar = 0.025 x 1000/342.22 = 0.073 mole # of molecules of sugar = = 0.073 x 6.022 x 1023 mole molecule mole-1 = 4.4 x 1022 molecules

  14. 2000 Exam How many moles of copper atoms are there in a copper coin of mass 2.5 g given that the atomic mass of copper is 63.55 a.m.u. and that the coin contains 55% copper by mass? (a) 0.04 (b) 0.02 (c) 0.01 (d) 0.06 (e) 0.03 Mass of copper = 0.55 x 2.5 g = 1.38 g # of moles of copper = 1.38/ 63.55 (g/ g mol-1) = 0.02 moles

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