1. CHEMICAL QUANTITIES The Mole

2. If we had to count the atoms of Silicon and Oxygen in this sand sculpture? What unit would we use? Gram? No! Kilograms? No! The Mole

3. The “MOLE” is a specialized unit for measuring amounts of matter as molecules, formula units, atoms, ions, and subatomic particles. 1 mole is defined as 6.0221415×10 23 For the purposes of this class 1 mole = 6.02 × 10 23

4. The “MOLE” is counting unit like “1 dozen = 12” or “1 gross = 12 dozen”…1 mole = 6.02 x 10 23 1 mole of bagels = 6.02 x 10 23 bagels 1 mole of bicycles = 6.02 x 10 23 bicycles 1 mole of people = 6.02 x 10 23 people

5. Some Other Specialized Units 1 dozen = 12 Used to measure eggs, bagel, & donuts 1 carat = 200 milligrams Used to measure gems & precious stones or metals 1 acre = 1/640 of a mile 2 Used to measure parcels of property 1 hogshead = 1 kegs = 63 gallons Used to measure liquids: e.g. molasses, beer (see “Huck Finn”) 1 grain = 0.06479891 gram Used for precise measurements of metals, e.g. bullets 1 Ǻngstrom(Ǻ) = 10 -10 meters Used to measure atomic radii

6. Why this number? 6.02 x 10 23 Recall that: 1 a.m.u. (atomic mass unit) is defined as 1/12 of the mass of a Carbon-12, ( 12 C)atom; therefore the mass of one 12 C atom equals 12 a.m.u. The mass of an Copper-63 ( 63 Cu) equals 63 a.m.u. and the Average Atomic Mass of all isotopes on Copper equals: a.m.u. (–look it up on your Chemhelper) 63.55

7. Why this number? 6.02 x 10 23 Our scales measure mass in grams not a.m.u., wouldn’t it be nice if instead of 63.546 a.m.u. of Copper, we could talk about 63.546 grams of Copper. One number that works with both units: a.m.u. and grams.

8. 6.02 x 10 23 Avogadro’s Number But, how many atoms of Copper are present in a sample that has a mass of 63.55 grams?

9. NOTE: 6.02 x 10 23 is known as Avogadro’s Number in honor of Amedeo Avogadro di Quaregna (1776- 1856), who was a pioneer in the properties of gases Remember 1 mole = 6.02 x 10 23 so…1 mole a.m.u. s = 1 gram 6.02 x 10 23 a.m.u. s = 1 gram

10. If you don’t already have it out, get out your Sargent-Welch PTOE for immediate use.

11. How do we used the mole? GAM  gram atomic mass 79 Au Gold 196.97 1 gold atom has a mass of 196.97 a.m.u. 1 mole of gold atoms have a gram atomic mass (GAM) of 196.97 grams See Chemhelper

12. What is the Gram Atomic Mass of… 88 Ra Radium

13. What is the Gram Atomic Mass of… 88 Ra Radium 226.02 grams per mole (g/mol)

14. What is the Gram Atomic Mass of… 88 Ra Radium 226.02 grams per mole (g/mol) As Arsenic

15. What is the Gram Atomic Mass of… 88 Ra Radium 226.02 grams per mole (g/mol) As Arsenic 74.92 grams per mole (g/mol)

16. Gram Formula Mass (GFM) Vs. Gram Molar Mass (GMM) What is the difference? Recall: There are two types of compounds that you learned to name: Molecular and Ionic. Molecular Compounds form Molecules, which have a Gram Molar Mass Ionic Compounds form a Formula Units, which have a Gram Formula Mass

17. Calculating an Ionic Compounds’ Gram Formula Mass (Ex 1- Lithium Fluoride) GFM for LiF = ElementsGAM Number of Atoms in the Formula Unit Mass for each Element in the Compound

18. Calculating an Ionic Compounds’ Gram Formula Mass (Ex 1) GFM for LiF = ElementsGAM Number of Atoms in the Formula Unit Mass for each Element in the Compound Li F

19. Calculating an Ionic Compounds’ Gram Formula Mass (Ex 1) GFM for LiF = ElementsGAM Number of Atoms in the Formula Unit Mass for each Element in the Compound Li 6.94g/mol F TOTAL

20. Calculating an Ionic Compounds’ Gram Formula Mass (Ex 1) GFM for LiF = ElementsGAM Number of Atoms in the Formula Unit Mass for each Element in the Compound Li 6.94g/mol F 19.00g/mol TOTAL

21. Calculating an Ionic Compounds’ Gram Formula Mass (Ex 1) GFM for LiF = ElementsGAM Number of Atoms in the Formula Unit Mass for each Element in the Compound Li 6.94g/mol1 F 19.00g/mol1 TOTAL

22. Calculating an Ionic Compounds’ Gram Formula Mass (Ex 1) GFM for LiF = ElementsGAM Number of Atoms in the Formula Unit Mass for each Element in the Compound Li 6.94g/mol16.94-g/mol F 19.00g/mol119.00-g/mol TOTAL

23. Calculating an Ionic Compounds’ Gram Formula Mass (Ex 1) GFM for LiF = ElementsGAM Number of Atoms in the Formula Unit Mass for each Element in the Compound Li 6.94g/mol16.94-g/mol F 19.00g/mol119.00-g/mol TOTAL 25.94-g/mol

24. Calculating an Ionic Compounds’ Gram Formula Mass (Ex 1) GFM for LiF = ElementsGAM Number of Atoms in the Formula Unit Mass for each Element in the Compound Li 6.94g/mol16.94-g/mol F 19.00g/mol119.00-g/mol TOTAL 25.94-g/mol

25. Calculating an Ionic Compounds’ Gram Formula Mass (Ex 2-Sodium Oxide) GFM for Na 2 O = ElementsGAM Number of Atoms in the Formula Unit Mass for each Element in the Compound Na O TOTAL

26. Calculating an Ionic Compounds’ Gram Formula Mass (Ex 2) GFM for Na 2 O = ElementsGAM Number of Atoms in the Formula Unit Mass for each Element in the Compound Na 22.99-g/mol O 16.00g/mol TOTAL

27. Calculating an Ionic Compounds’ Gram Formula Mass (Ex 2) GFM for Na 2 O = ElementsGAM Number of Atoms in the Formula Unit Mass for each Element in the Compound Na 22.99-g/mol2 O 16.00g/mol1 TOTAL

28. Calculating an Ionic Compounds’ Gram Formula Mass (Ex 2) GFM for Na 2 O = ElementsGAM Number of Atoms in the Formula Unit Mass for each Element in the Compound Na 22.99-g/mol245.98-g/mol O 16.00g/mol116.00-g/mol TOTAL

29. Calculating an Ionic Compounds’ Gram Formula Mass (Ex 2) GFM for Na 2 O = ElementsGAM Number of Atoms in the Formula Unit Mass for each Element in the Compound Na 22.99-g/mol245.98-g/mol O 16.00g/mol116.00-g/mol TOTAL 61.98-g/mol

30. Calculating an Ionic Compounds’ Gram Formula Mass (Ex 3-Chromium(III) Sulfate) GFM for Cr 2 (SO 4 ) 3 = ElementsGAM# of AtomsElement Mass Cr S O TOTAL

31. Calculating an Ionic Compounds’ Gram Formula Mass (Ex 3-Chromium(III) Sulfate) GFM for Cr 2 (SO 4 ) 3 = ElementsGAM# of AtomsElement Mass Cr 52.00-g/mol S 32.06-g/mol O 16.00-g/mol TOTAL

32. Calculating an Ionic Compounds’ Gram Formula Mass (Ex 3-Chromium(III) Sulfate) GFM for Cr 2 (SO 4 ) 3 = ElementsGAM# of AtomsElement Mass Cr 52.00-g/mol 2 S 32.06-g/mol 3 O 16.00-g/mol 12 TOTAL

33. WHOA! Cr 2 (SO 4 ) 3 : think “Distributive Property” 2 Cr 3+ ions: chromium (III) ions 3 S 2- ions: sulfide ions 12 O atoms: oxygen atoms Cr 2 (S 1x3 O 4x3 ) =

34. Calculating an Ionic Compounds’ Gram Formula Mass (Ex 3-Chromium(III) Sulfate) GFM for Cr 2 (SO 4 ) 3 = ElementsGAM# of AtomsElement Mass Cr 52.00-g/mol2104.0 S 32.06-g/mol396.18 O 16.00-g/mol 12 192.0 TOTAL

35. Calculating an Ionic Compounds’ Gram Formula Mass (Ex 3-Chromium(III) Sulfate) GFM for Cr 2 (SO 4 ) 3 = ElementsGAM# of AtomsElement Mass Cr 52.00-g/mol2104.0 S 32.06-g/mol396.18 O 16.00-g/mol 12 192.0 TOTAL 392.1-g/mol

36. (Ex 4-Ammonium Sulfite) GFM for (NH 4 ) 2 SO 3 = ElementsGAM# of AtomsElement Mass TOTAL

37. (Ex 4-Ammonium Sulfite) GFM for (NH 4 ) 2 SO 3 = ElementsGAM# of AtomsElement Mass N 14.01-g/mol228.02-g/mol H 1.01-g/mol88.08-g/mol S 32.06-g/mol 1 O 16.00-g/mol 3 48.00-g/mol TOTAL 116.16-g/mol

38. That was GFM—Gram Formula Mass… now Here’s GMM—Gram Molar Mass

39. Calculating a Molecular Compounds’ Gram Molar Mass GMM (Ex 1: S 2 O 4 ) GMM for S 2 O 4 = ElementsGAM# of AtomsElement Mass S 32.062 O 16.00 4 TOTAL

40. Calculating a Molecular Compounds’ Gram Molar Mass GMM (Ex 1: S 2 O 4 ) GMM for S 2 O 4 = ElementsGAM# of AtomsElement Mass S 32.06264.12 O 16.00 4 64.00 TOTAL

41. Calculating a Molecular Compounds’ Gram Molar Mass GMM (Ex 1: S 2 O 4 ) Disulfur Tetroxide GMM for S 2 O 4 = ElementsGAM# of AtomsElement Mass S 32.06264.12-g/mol O 16.00 4 64.00-g/mol TOTAL 128.12-g/mol

42. What’s the difference? There is no difference in the process of calculating GMM--Gram Molar Masses-- and GFM--Gram Formula Masses. The differing terminology reflects the two different types of compounds: molecular compounds and ionic compounds.

43. THE END HaHa