1. Chapter 8 The Mole

2. 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. 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. 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. 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. 7 The Chemist’s Dozen - A Mole Avogadro’s Number of “items” 6.022 141 99 x 10 23 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 thousands millions billions trillions quadrillions quintrillions s e x t i l l i o n s

8. 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: CCl 4 ✓ 1 mole C = 12.01 g ✓ 4 mole Cl = 4 (35.45 g) = 141.8 g ✓ 12.01 g + 141.8 g = 153.81 g/1 mol of CCl 4 molecules

9. 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 78.01 g/1mole of Al(OH) 3 formula units aka ionicules

10. 10 Molar Mass of Large Compounds Determine the MM (molar mass) of iron(III) sulfate: Fe 2 (SO 4 ) 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 = 399.91 g/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 ~2000 g of iron(III) sulfate

11. 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. 12 So what is the unit label on Avogadro’s number? 6.02 x 10 23 items/mol

13. 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 3.09 moles of Mg atoms What is the mass of 0.732 mole of Cl 2 molecules? ✓ 1 mole Cl atoms = 35.45 g ✓ 1 mole Cl 2 molecules = 70.90 g ✓ (0.732 mole)(70.90 g/1mole) = 51.8988 g 51.9 g of Cl 2 molecules

14. 14 Moles ↔ Items How many atoms are in 4.7 moles of Mg? ✓ 1 mol Mg = 6.02 x 10 23 atoms ✓ (4.7 mole)(6.02 x 10 23 atoms/1mol) = 2.8294 x 10 24 atoms 2.8 x 10 24 atoms of Mg How many mole of oxygen molecules in 5.83 x 10 22 molecules? ✓ 1 mole O 2 = 6.02 x 10 23 molecules ✓ (5.83 x 10 22 molecules)(1 mol/6.02 x 10 23 molecules) 0.096843853 mole 0.0968 mole of O 2 molecules

15. 15 Mass ↔ Moles ↔ Items What is the mass of 2.8 x 10 24 atoms of Mg? ✓ 1 mole Mg = 6.02 x 10 23 atoms, 1 mole Mg = 24.31 g ✓ This is a two step process: ✓ Change to moles, then change to grams ✓ (2.8 x 10 24 atoms)(1 mole/6.02 x 10 23 atoms)(24.31 g/1mole) = 113.0697674 g 110 g of Mg atoms How many ammonia molecules are in 34 g of NH 3 ? ✓ 1 mole NH 3 = 17.04 g, 1 mole = 6.02 x 10 23 molecules ✓ This is a two step process: ✓ Change to moles, then change to molecules ✓ (34 g)(1 mole/17.04 g)(6.02 x 10 23 molecules/1mole) = 1.201173709 x 10 24 molecules 1.2 x 10 24 molecules of NH 3

16. Practice Problem What would be the mass of 3.75 x 10 21 atoms of iron? ✓ Change to moles, then change to grams ✓ (3.75 x 10 21 atoms)(1 mole/6.02 x 10 23 atoms)(55.85 g/1mole) = 0.3479028 g 0.348 g of Fe atoms 16

17. Practice Problem 17 How many water molecules would be found in a 54 gram sample of water? ✓ 1 mole H 2 O= 18.02 g, 1 mole = 6.02 x 10 23 molecules ✓ This is a two step process: ✓ Change to moles, then change to molecules ✓ (54 g)(1 mole/18.02 g)(6.02 x 10 23 molecules/1mole) = 1.80459 x 10 24 molecules 1.8 x 10 24 molecules of H 2 O

18. 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 MgF 2 ) Mg = 24.31 g/1mol, F = 19.00 g/1mol MgF 2 = 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: 19 1.CO 2 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% 2.C 6 H 12 O 6 MM = 12.01(6) +1.01(12) +16.00(6) =180.g/mol So % C = (72.06/180)(100) = 40.0%

20. 20 Elemental Analysis