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Chapter 8

Chapter 8. The Mole. I say, “I have a dozen….”. You say.... A dozen what ? eggs, donuts, pencils, dogs, siblings....whatever. What is a dozen? 12 items How would you label a dozen? 12 items/1 dozen. I’ll give you a dozen rice…. You might say, “Big deal.”

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Chapter 8

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  1. Chapter 8 The Mole

  2. I say, “I have a dozen….” • You say.... • A dozen what? • eggs, donuts, pencils, dogs, siblings....whatever. • What is a dozen? • 12 items • How would you label a dozen? • 12 items/1 dozen

  3. I’ll give you a dozen rice… • You might say, “Big deal.” • This would not be useful because rice grains are so small. • You would still be hungry.

  4. OK, I’ll give you a dozen gold atoms… • You might again say, “So what.” • This would not be valuable because a dozen atoms is so small, so in chemistry we need a much LARGER dozen. • Introducing...

  5. The really big dozen The Mole

  6. Using the Mole Just Like the Dozen • If you had 1 dozen bicycles, you’d have… • how many bicyles, dozen frames, dozens of wheels, frames, wheels? • 1 dozen bikes*12 bikes/1doz = 12 bikes • 1 dozen bikes*1frame/1bike = 1 dozen frames • 1 dozen bikes*2 wh/1 bike = 2 dozen wheels • 1 dozen bikes*12 bikes/1doz*1frame/1bike • 12 frame • 1 dozen bikes*12 bikes/1doz*2wheels/1bike • 24 wheels

  7. The Chemist’s Dozen - A MoleAvogadro’s Number of “items” • 6.022 141 99 x 1023 items/mol • Atoms or molecules or ions or units or whatever (dogs, donuts, eggs, bicycles, etc) • 602,214,199,000,000,000,000,000 items/mol • The average atomic mass of any element is the mass of a mole of atoms of that element. • H = 1.01 g/mol • C = 12.01 g/mol • Fe = 55.85 g/mol quintrillions sextillions quadrillions thousands trillions billions millions

  8. Molar Mass - The Average Mass in the Periodic Table • The mass (in grams) of 1 mole of a substance. • Determine the MM (molar mass) of carbon tetrachloride: CCl4 • 1 mole C = 12.01 g • 4 mole Cl = 4 (35.45 g) = 141.8 g • 12.01 g + 141.8 g = 153.81g/1 molof CCl4 molecules

  9. Molar Mass, again • Determine the MM (molar mass) of aluminum hydroxide: Al(OH)3 • 1 mole Al = 26.98 g • 1 mole O = 16.00 g * 3 = 48.00 g • 1 mole H = 1.01 g * 3 = 3.03 g • 26.98 g/1mol + 48.00 g/1mol + 3.03 g/1mol 78.01g/1moleof Al(OH)3 formula units aka ionicules

  10. Molar Mass of Large Compounds • Determine the MM (molar mass) of iron(III) sulfate: Fe2(SO4)3 • 2 mole Fe = 2 (55.85 g) = 111.70 g • 3 mole S = 3 (32.07 g) = 96.21 g • 12 mole of O = 12 (16.00) = 192.00g • 111.70 g + 96.21 g + 192.00 g = 399.91g/1mol of iron(III) sulfate ionicules • So if you had 5 moles of iron(III) sulfate, how much would it weigh? • 5 moles * ~400 g/1mol = ~2000 g ~2000gof iron(III) sulfate

  11. So what is the unit label on the molar masses that you look up in the periodic table? • g/mol • (or g/1 mole)

  12. So what is the unit label on Avogadro’s number?6.02 x 1023 • items/mol

  13. Mass ↔ Moles • How many moles of Mg in 75.0 g of Mg? • 1 mole Mg = 24.31 g • (75 g)(1 mol/24.31g) = 3.085150144 moles 3.09molesof Mg atoms • What is the mass of 0.732 mole of Cl2 molecules? • 1 mole Cl atoms = 35.45 g • 1 mole Cl2 molecules = 70.90 g • (0.732 mole)(70.90 g/1mole) = 51.8988 g 51.9gof Cl2 molecules

  14. Moles ↔ Items • How many atoms are in 4.7 moles of Mg? • 1 mol Mg = 6.02 x 1023 atoms • (4.7 mole)(6.02 x 1023 atoms/1mol) = 2.8294 x 1024 atoms 2.8 x 1024atoms of Mg • How many mole of oxygen molecules in 5.83 x 1022 molecules? • 1 mole O2 = 6.02 x 1023 molecules • (5.83 x 1022 molecules)(1 mol/6.02 x 1023 molecules) 0.096843853 mole 0.0968moleof O2 molecules

  15. Mass ↔ Moles ↔ Items • What is the mass of 2.8 x 1024 atoms of Mg? • 1 mole Mg = 6.02 x 1023 atoms, 1 mole Mg = 24.31 g • This is a two step process: • Change to moles, then change to grams • (2.8 x 1024 atoms)(1 mole/6.02 x 1023 atoms)(24.31 g/1mole) = 113.0697674 g 110gof Mg atoms • How many ammonia molecules are in 34 g of NH3? • 1 mole NH3 = 17.04 g, 1 mole = 6.02 x 1023 molecules • This is a two step process: • Change to moles, then change to molecules • (34 g)(1 mole/17.04 g)(6.02 x 1023 molecules/1mole) = 1.201173709 x 1024 molecules 1.2 x 1024moleculesof NH3

  16. Practice Problem • What would be the mass of 3.75 x 1021 atoms of iron? • Change to moles, then change to grams • (3.75 x 1021 atoms)(1 mole/6.02 x 1023 atoms)(55.85 g/1mole) = 0.3479028 g 0.348gof Fe atoms

  17. Practice Problem • How many water molecules would be found in a 54 gram sample of water? • 1 mole H2O= 18.02 g, 1 mole = 6.02 x 1023 molecules • This is a two step process: • Change to moles, then change to molecules • (54 g)(1 mole/18.02 g)(6.02 x 1023 molecules/1mole) = 1.80459 x 1024 molecules 1.8 x 1024moleculesof H2O

  18. Sometimes in chemistry it is useful to report the percentages of elements in a compound. This is called:Percent Composition (by mass)aka Elemental Analysis • Determine the % composition by mass • (aka an elemental analysis of MgF2) • Mg = 24.31 g/1mol, F = 19.00 g/1mol • MgF2 = 62.31 g/1mol • Remember that % is • always part out of total • Mg = (24.31 / 62.31)(100) = 39.01% • F (all of it) = (38 / 62.31)(100) = 60.99%

  19. Determine the Percent Composition of carbon in the following: • CO2 • C = 12.01 g/1mol, O = 16.00 g/1mol • MM = 12.01+16.00(2)= 44.01 g/mol • So % C = (12.01 / 44.01)(100) = 27.29% • C6H12O6 • MM = 12.01(6) +1.01(12) +16.00(6) =180.g/mol • So % C = (72.06/180)(100) = 40.0%

  20. Elemental Analysis

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