1. (I have now prefaced the slides for my talk on Sept. 5 with the solution to the atomic mass problem from Sept. 3. The lecture on moles follows those slides.)

2. Problem: Given the exact masses, 35 Cl = 34.969mu and 37 Cl = 36.966 u, and the average atomic mass of chlorine = 35.453 u, find the abundances of the two isotopes.

3. Problem: Given the exact masses, 35 Cl = 34.969 u and 37 Cl = 36.966 u, and the average atomic mass of chlorine = 35.453 u, find the abundances of the two isotopes. Ave. atomic mass = a 1 m 1 + a 2 m 2 +...

4. Problem: Given the exact masses, 35 Cl = 34.969 u and 37 Cl = 36.966 u, and the average atomic mass of chlorine = 35.453 u, find the abundances of the two isotopes. Ave. atomic mass = a 1 m 1 + a 2 m 2 +... 35.453 u = a 1 (34.969 u) + a 2 (36.966 u)

5. Problem: Given the exact masses, 35 Cl = 34.969 u and 37 Cl = 36.966 u, and the average atomic mass of chlorine = 35.453 u, find the abundances of the two isotopes. Ave. atomic mass = a 1 m 1 + a 2 m 2 +... 35.453 u = a 1 (34.969 u) + a 2 (36.966 u) Also: a 1 + a 2 = 1

6. Problem: Given the exact masses, 35 Cl = 34.969 u and 37 Cl = 36.966 u, and the average atomic mass of chlorine = 35.453 u, find the abundances of the two isotopes. Ave. atomic mass = a 1 m 1 + a 2 m 2 +... 35.453 u = a 1 (34.969 u) + a 2 (36.966 u) Also: a 1 + a 2 = 1 So 35.453 u = (1 – a 2 )(34.969 u) + a 2 (36.966 u)

7. 35.453 u = 34.969 u – 34.969 u(a 2 ) + 36.966 u(a 2 ) 35.453 u = 34.969 u + 1.997 u(a 2 ) a 2 0.2424 and a 1 = 1 – 0.2424 =.7576

8. How to Count Atoms (when they are really really small) Review: 1 12 C atom = 12.0000 u = 1.9926x 10 -23 g (from mass spectroscopy experiments) So

9. Goal: a number that is capable of expressing numbers of atoms in convenient terms. H and O react in simple numbers of atoms to form water (H 2 O): 2 H : 1 O There is not a simple relation between the masses that react. Yet – masses are what we can readily measure in a laboratory. To get a quantity capable of expressing numbers of atoms, Define: 1 mole = number of atoms in 12.0000 g of 12 C.

10. We now have enough information to calculate the number of atoms in one mole of carbon: But this is also the number of atoms in one atomic mass, expressed in grams, of any element. This important number is known as Avogadro’s Number, N A.

11. Both a mole and a dozen express quantities of things by a collective number. A dozen eggs weighs more than a dozen ping-pong balls because each individual unit is heavier.

12. 12.0 amu C 144.0 amu in carton 16.0 amu O 192.0 amu in carton Take enough cartons to have 6.02 x 10 23 atoms The total mass will be 12.0 gram Take the same number of atoms as at the left. The total mass will be 16.0 grams (A dozen oxygen atoms weigh more than a dozen carbon atoms) Moral: a mole of oxygen weighs more than a mole of carbon because each individual oxygen atom weighs more.

13. So now, atomic masses have two interpretations: 1 atom of C weighs 12.0000 u 1 mole of C weighs 12.0000 g Some conversion factors: 1 mole of C = 12.00 g 1 mole of C = 6.02 x 10 23 atoms C

14. Now we can calculate the mass of any atom: 1 atom of U = 238 u

15. Now we can calculate the mass of any atom: 1 atom of U = 238 u Find the number of carbon atoms in 3.0 g C.

16. A mole road-map Moles gramsatoms NANA Atomic mass

17. This can be applied to compounds, which have a molecular mass (in amu) or a molar mass (in grams). For instance: In one molecule of H 2 O, there are 2 H atoms = 2 x 1 amu = 2 u 1 O atom = 1 x 16 amu = 16 u 18 u 1 mole of H 2 O has 1 mole of O atoms and 2 moles of H atoms. 1 mole of water weighs 18 g and has 6.02 x 10 23 molecules

18. For Ca(OH) 2 : 1 Ca = 40.1 u 2 O = 32.0 u 2 H = 2.0 u 74.1 u 1 mole of Ca(OH) 2 has a mass of 74.1 g Ca(OH) 2 is an ionic compound so we can’t talk about molecules. Sometimes the simplest formula – which indicates only the ratio of the ions – is referred to as a formula unit.

19. How many atoms of O are there in 5.0 g of CO 2 ?