1 / 69

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

rex
Download Presentation

The Mole

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related