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The Mole

The Mole. Dimensional Analysis Review. How many seconds are in 5.0 hours? 5.0 hr 5.0 hr x 60 min x 60 sec = 18000 sec 1 hr 1 min. Dimensional Analysis Review. Calculate the number of inches in 26 yards 26 yards

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The Mole

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  1. The Mole

  2. Dimensional Analysis Review • How many seconds are in 5.0 hours? • 5.0 hr • 5.0 hr x 60 min x 60 sec = 18000 sec 1 hr 1 min

  3. Dimensional Analysis Review • Calculate the number of inches in 26 yards • 26 yards • 26 yards x 3 ft x 12 inches = 940 inches 1 yd 1 ft

  4. Stoichiometry • Stoichiometry is just a long word for changing units in chemistry • Just remember to ALWAYS start with your given! • If you can do Dimensional Analysis, you can do stoichiometry

  5. Steps • Start with your given • Use conversion factors and cross out until you get what you wanted • Check sig. figs

  6. The Mole • Chemists need a convenient method for counting accurately the number of atoms, molecules, or formula units in a sample of a substance. • The mole, commonly abbreviated mol, is the SI base unit used to measure the amount of a substance.

  7. The Mole • A mole of anything contains 6.02 x 1023 representative particles. • A representative particle is any kind of particle such as atoms, molecules, formula units, electrons, or ions. • 6.02 x 1023 is called Avogadro’s number

  8. Conversion Factor #1

  9. Representative Particles • Anything - Representative particles • Elements – Atoms • Covalent Compounds – Molecules • Ionic Compounds – Formula Units • Ions - Ions

  10. Mole – Representative Particle Calculations • Calculate the number of atoms in 3.50 moles of copper • Start with your given • 3.50 mol Cu • Draw your line • 3.50 mol Cu x _________

  11. Mole – Representative Particle Calculations • Place the conversion factors • 3.50 mol Cu x 6.02 x 1023 atoms 1 mol Cu • Work the problem • 2.11 x 10 24 atoms Cu

  12. Another Example • Calculate the number of molecules in 2.6 moles of H2O • 2.6 mol H2O x 6.02 x 1023 molec = 1.6 x 1024 1 mol H2O molec H2O

  13. Mole – Representative Particle Calculations • How many moles of MgO are in 9.72 x 10 23 molecules of MgO? • 1.61 mol MgO

  14. Another Example • How many moles are in 4.50 x 1024 atoms of Zinc? • 7.48 mol Zn

  15. Mass & the Mole • The mass in grams of 1 mole of a substance is called the molar mass • It can also be called molecular mass, molecular weight, and formula mass • To get the molecular weight you just add up all of the masses of all of the elements in a compound

  16. Molecular Weight • Calculate the molecular weight of the following: • Ca • 40.08 g/mol • Na • 22.99 g/mol

  17. Molecular Weight • Calculate the molecular weight of the following: • MgO • 40.31 g/mol • NaCl • 58.44 g/mol • H2O • 18.02 g/mol • Fe2O3 • 159.70 g/mol

  18. Conversion Factor # 2 1 mole Molecular weight (g) The molecular mass comes from the periodic table!

  19. Mole – Mass Calculations • What is the mass of 4.21 moles of iron (III) oxide? • Start with your given: • 4.21 mol Fe2O3 • Draw your line • 4.21 mol Fe2O3 x _____________

  20. Mole – Mass Calculations • Place conversion factors • 4.21 mol Fe2O3 x 159.70 g Fe2O3 1 mol Fe2O3 • Cross out units & work the problem • 672 g Fe2O3

  21. Another Example • Calculate the mass of 1.630 moles of Na • 37.47 g Na

  22. Mole – Mass Calculations • How many moles of Ca(OH)2 are in 325 grams? • 4.39 moles Ca(OH)2

  23. Mass – Particle Conversions • How many atoms of gold are in 25.0 g of gold? • 25.0 g Au x 1 mol Au x 6.02 x 1023 atoms Au 196.79 g Au 1 mol Au • 7.65 x 1022 atoms of Au

  24. Mass – Particle Conversions • How many grams of He are in 5.50 x 1022 atoms of He? • 0.366 g He

  25. Moles of Compounds • Freon has the formula CCl2F2 • This means that • 1 mol CCl2F2 = 1 mol C • 1 mol CCl2F2 = 2 mol Cl • 1 mol CCl2F2 = 2 mol F • This now gives us new conversion factors

  26. Example • How many moles of F are in 5.50 mol of CCl2F2? • 5.50 mol CCl2F2 x 2 mol F = 10.0 mol F 1 mol CCl2F2

  27. Example • How many moles of Al are in 1.25 mol of Al2O3? • 2.50 mol Al2O3

  28. Example • How many Cl- ions are in 35.6 g of AlCl3? • 4.82 x 10 23 ions Cl- • How many Al +3 ions? • 1.61 x 10 23 ions Al+3

  29. % Composition, Empirical Formulas, & Molecular Formulas

  30. % Composition • % = (part / whole ) x 100 • When calculating the % composition, you are calculating the % of each element in a compound

  31. % Composition • Calculate the % Composition of MgO Mg = 24.31g O = 16.00g Total = 40.31g / 40.31 / 40.31 X 100 X 100 = 60.31 % Mg = 39.69% O

  32. % Composition • Calculate the % Composition of iron (III) oxide • % Fe = 69.94% • % O = 30.06%

  33. Empirical & Molecular Formulas • Empirical formula – the smallest whole number ratio of elements • Molecular formula – the true number of elements in a compound

  34. Empirical Formula • What is the empirical formula for H2O2? • HO • What is the empirical formula for C6H12O6? • CH2O

  35. Steps for Calculating the Empirical Formula • List your givens • Change % to grams • Change grams to moles • Divide everything by the smallest number of moles • Write your formula

  36. Empirical Formula Problem • Calculate the empirical formula of a compound containing 40.05 % S and 59.95 % O. 1 mol S = 32.07 g S 1 mol O = 16.00 g O 1.249 mol 3.747 mol / / 1.249 mol = 1 3.747 mol = 3 40.05 g S x 59.95 g O x SO3

  37. Empirical Formula Problem • Calculate the empirical formula for a compound containing 48.64 g C, 8.16 g H, and 43.20 g O. • C3H6O2

  38. Steps for Calculating Molecular Formula • Calculate the empirical formula • Get the molecular mass of the empirical formula that you just determined • Divide the experimentally determined molecular mass (given) by the molecular mass of the empirical formula • You will get a whole number • Multiply everything in the empirical formula by this number

  39. Molecular Formula Problem • Calculate the molecular formula of a compound containing 40.68%C, 5.08%H, and 54.25%O with an experimentally determined molecular weight of 118.1 g/mol

  40. Molecular Formula Problem 40.68g C x 1 mol C = 3.387 / 3.387 = (1)2 = 2 12.01 g C 5.08g H x 1 mol H = 5.04 / 3.387 = (1.5)2= 3 1.01 g H 54.25 g Ox 1 mol O = 3.390 / 3.387 = (1)2 = 2 16.00 g O C2H3O2

  41. Molecular Formula Problem • C2H3O2 • Molecular Mass = 59.04 • EDMM / EFMM = 118.1 / 59.04 = 2 • Molecular Formula • C4H6O4

  42. Molecular Formula Problem • Calculate the molecular formula of a compound containing 57.84 g C, 3.64 g H, and 38.52 g O with an experimentally determined molecular mass of 249.21 g/mol • C12H9O6

  43. Stoichiometry

  44. Stoichiometry • Using the methods of stoichiometry, we can measure the amounts of substances involved in chemical reactions and relate them to one another.

  45. Steps • Write the chemical equation • Balance the chemical equation • Start with your given • Cross out until you get what you want • Check sig. figs, units, and circle your answer

  46. Conversion Factor # Moles A # Moles B The #’s in from if A & B MUST come from the balanced chemical equation

  47. Mole – Mole Relationship • 4 Fe + 3O2 2Fe2O3 • 4 mol Fe / 3 mol O2 • 4 mol Fe / 2 mol Fe2O3 • 3mol O2 / 2 mol Fe2O3

  48. Mole – Mole Relationship • How many moles of Fe2O3 will I form from 5.0 mol of Fe? • 5.0 mole Fe x 2 mol Fe2O3 4 mol Fe • 2.5 mol Fe2O3

  49. Mass – Mole Relationship • How many g of NaCl will be produced from 1.25 mol of chlorine gas reacting with sodium? • Write the reaction • Na + Cl2 NaCl • Balance the reaction • 2Na + Cl2 2NaCl

  50. Mass – Mole Relationship • Work the problem • 1.25 mol Cl2 x 2 mol NaCl x 58.44 g NaCl • 1 mol Cl2 1 mol NaCl • 146 g NaCl

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