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The Mole

The Mole. Diatomic elements. – elements that do not exist on their own, they must be in a covalent bond with themselves to be more stable. N 2 O 2 F 2 Cl 2 Br 2 I 2 H 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

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The Mole

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  1. The Mole

  2. Diatomic elements • – elements that do not exist on their own, they must be in a covalent bond with themselves to be more stable.

  3. N2 • O2 • F2 • Cl2 • Br2 • I2 • H2

  4. 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

  5. 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

  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. A mole is used to represent the amount of a substance or compound. • H2 means 1 mole of H2 • NaCl means 1 mole of NaCl. • Hydrates – compounds that contain water. Written as NaCl • 2H2O

  8. 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

  9. Molecular or Formula Mass Calculations • You have to add up the masses of every element found in the compound or molecule. • Ex. (NH4)3PO4 Ammonium Phosphate

  10. Make a list of how many of each element you have. • N = 3 • H = 12 • P = 1 • O = 4

  11. Multiply how many of each element you have by their mass. (Look it up on the periodic table.) Then add up the total. The units always grams.

  12. N = 3 x 14 = 42 • H = 12 x 1 = 12 • P = 1 x 31 = 31 • O = 4 x 16 = 64 • 149 g

  13. example • NaCl • Na = 23g • Cl = 35 g • Total 58 g

  14. Hydrates- • are just a molecule or compound that has a certain number of water molecules attached to it. • They are written like NaCl • 5H2O.

  15. When you are asked to calculate the mass you have to understand that there are 5 H2O molecules attached to 1 NaCl. • That means: 10 H & 5 O in addition to the Na & Cl.

  16. H = 10 x 1 = 10 • O = 5 x 16 = 80 • Na = 1 x 23 = 23 • Cl = 1 x 35 = 35 • Total 148 amu or grams

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

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

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

  20. Another Example • Calculate the mass of 1.630 moles of Na • 1.630 moles x 22.9g • 1 mole • 37.47 g Na

  21. Mole – Mass Calculations • How many moles of Ca(OH)2 are in 325 grams? • 325 g x 1 mole • 74 g • 4.39 moles Ca(OH)2

  22. Another Example • How many moles are in 62.17 g of sodium chloride? • 62.17g x 1 mole • 58 g • 1.064 mol NaCl

  23. Conversion Factor #2

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

  25. 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 _________

  26. 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

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

  28. Another Example • Calculate the number of formula units in 5.23 moles of NaCl • 5.23 mole x 6.02 x 1023 f. units • 1 mole • 3.15 x 10 24 formula units NaCl

  29. Mole – Representative Particle Calculations • How many moles of MgO are in • 9.72 x 10 23 molecules of MgO? • 9.72 x 10 23 mlcs x 1 mole • 6.02 x 10 23 mlcs • 1.61 moles MgO

  30. Another Example • How many moles are in 4.50 x 1024 atoms of Zinc? • 4.5 x 1024 atoms x 1 mole • 6.02 x 10 23 atoms • 7.48 moles Zn

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

  32. Mass – Particle Conversions • How many grams of He are in 5.50 x 1022 atoms of He? • 5.50 x 10 22 atomsx1 mole x 4 g • 6.02 x 10 23 atoms1 mole • 0.366 g He

  33. example • How many formula units are in 35.6 g of NaCl? • 3.70 x 10 23 formula units

  34. moles • Al2S3 means you have 1 mole of the compound Al2S3. • It is composed of2 moles of Al atoms and 3 moles of S atoms

  35. Ions in a compound • How many Cl- ions are there in 35.6 g of AlCl3? • 35.6 g AlCl3 x 1 mole AlCl3 x 3 mole Cl3 x 6.02 x 10 23 ions • 132 g 1 mole AlCl3 1 mole Cl3 • Remember ions is just a word, atoms particles, representative units. Anything works

  36. % Composition, Empirical Formulas, & Molecular Formulas

  37. % composition • Mass of the element x 100 • Mass of the compound

  38. % composition- • gives the relative mass of each atom in a compound. • Steps • 1. Calculate formula mass of compound • 2. Determine % of each atom • Ex: Calculate the % comp of N in NH4NO3 • N x 2 = 28.0 • H x 4 = 4.0 • O x 3 = 48.0 • 80.0 g • % N = 28.0 g/ 80.0 g x 100 = 35%

  39. Calculate the % composition of each atom in • a. Fe2O3 b. Ag2O c. HgO

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

  41. Molecular formulas- Used to tell exactly how many of each atom is present in the compound. • Empirical Formulas- The simplest formulas possible indicating the ratio of atoms found in the molecule. Think of it as reducing a fraction to the least common denominator.

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

  43. C6H6 • C2H2 • C6H12O6 • C4H10 • P4O10 • SO3 • N2O4 • NO2 • Ag2C4H4O6

  44. Empirical Formulas-calculations • The smallest whole number ratio of elements in a compound. It is determined by calculating the: • Calculate the moles present of each element in the compound. • Set up ratios for each element by dividing each mole amount by the smallest mole amount. • Use the ratio from step # 2 as the subscripts to write out the empirical formula.

  45. Ex: Determine the empirical formula for a compound with 36.5 % Na, 25.4% S, 38.1% O • Since each elements % adds up to 100% treat each % as a gram amount

  46. Na = 36.5 g Na x 1 mole Na = 1.59 mole Na • 23.0 g Na • S = 25.4 g S x 1 mole S = .791 mole S • 32.1 g S • O = 38.1 g O x 1 mole O = 2.23 mole O • 16.0 g O

  47. Na S O • 1.59 = 2.01 .791 = 1 2.23 = 3.01 • .791 .791 .791 • Ratio = 2:1:3 • Na2SO3

  48. 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 1.249 mol = 3 40.05 g S x 59.95 g O x SO3

  49. Ratio problems • Occasionally one part of your ratio will not be a whole number ( 1.5 : 1) It has to be!!!!! • Since H1.5O can’t exist you have got to fix it. • When this occurs you need to multiply by factor that will make it a whole # (like 2) • * what you do to one you must do to all, so every mole amount gets multiplied by 2!!

  50. Empirical Formula Problem • Calculate the empirical formula for a compound containing 48.64 g C, 8.16 g H, and 43.20 g O. • 48.64g x 1 mole C = 4.04 mole/ 2.7 =1.5 x 2 = 3 • 12 g • 8.16 g x 1 mole H = 8.16 mole/2.7 = 3.02 x 2 = 6 • 1 g • 43.20 x 1 mole O = 2.7 mole/2.7 = 1x 2 = 2 • 16 g • C3H6O2

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