1. Happy Birthday J.J. Abrams (1966)
HOMEWORK – DUE Wednesday 6/28/17
HW 7: CH 7 #’s 11, 12, odd, 31, 33, 35, 39, 40, 53, 59, 67, 70, all, 77, 82, 84, 88, 89
Chapter 7 Extra Credit: DUE Wednesday 6/28
Lab Wednesday: Dry lab, no prelab
Beacon with Chris
Monday and Wednesday from 3:00 – 5:00 in LRC 107
One-on-one tutoring with Chris

2. Words that Represent Numbers
mole
things
6.022x1023 things
JUST A NUMBER!!!
1 dozen things = 12 things
1 mole things = 6.022x1023 things
dozen
12 things
gross
144 things
ream
500 things

3. One mole of ANYTHING = 6.022x1023 things
One dozen of ANYTHING = 12 things
Let’s assume that there is an omnipotent being with a pile in front of him which contains 1 mole of carbon atoms, and let’s say he has a pair of tweezers capable of picking up a single atom at a time (omnipotence has its benefits). Scientist say that the universe is about 15 billion years old (1.5x1010 years old). That is a LONG time, but there are 6.022x1023 atoms of carbon sitting in front of this guy, so let’s use the shortest time frame possible that we can comprehend, the second. In the 15 billion years since the Big Bang, there have been 4.7x1017 seconds. Now, let’s say that our being has been picking up one carbon atom with his tweezers EVERY SECOND of the 15 billion years since the big bang. By today, he is barely 1 MILLIONTH of the way through his pile of carbon… that is how big one mole is.

4. One mole of ANYTHING = 6.022x1023 things
One dozen of ANYTHING = 12 things
Or, try this one on… one mole (6.022x1023) chocolate chips would burry the entire surface of the Earth by 590 METERS (close to 2000 feet)!

5. This number, 6.022x1023 is called Avogadro’s number.

6. Using Avogadro’s number
6.022x1023 [things] = 1 mole [things]
6.022x1023 atoms K = 1 mole K
6.022x1023 molecules H2O = 1 mole H2O

7. Using Avogadro’s number
8.33x1022 atoms of Cl is equal to how many moles?
1 mol Cl = 6.022x1023 atom Cl

8. Using Avogadro’s number
How many molecules are there in mol of carbon tetrachloride?
1 mol CCl4 = 6.022x1023 molecules CCl4

9. Molar Mass
Using the following information, calculate the mass (in grams) of ONE mole of Ca
= amu (if one atom)
= g/mol (if one mole of atom)

10. Molar Mass
molar means “per 1 mole”
molar mass means mass per 1 mole

11. Molar Mass = 40.08 amu/atom (if one atom) 1 atom Ca = 40.08 amu Ca
= g/mol (if one mole of atom)
1 mol Ca = g Ca

12. Molar Mass What is the MOLAR MASS of calcium oxide? CaO = Ca2+ + O2-
= g/mol
= g/mol
= molar mass CaO
1 mol CaO = g CaO
56.08 g CaO = 1 mol CaO = 6.022x1023 formula units CaO
1 mol CaO = 6.022x1023 formula units CaO

13. Molar Mass MgCl2 Mg = 24.31 g/mol 2 x Cl Cl = 35.45 g/mol
= 2 x (35.45 g/mol) = g/mol
= molar mass MgCl2

14. Molar Mass MgCl2 95.21 g/mol = molar mass MgCl2
1 mol MgCl2 = g MgCl2

15. Molar Mass Sr(OH)2 Sr = 87.62 g/mol 2 x O O = 16.00 g/mol
= 2 x (16.00 g/mol) = g/mol
2 x H H
= 1.01 g/mol
= 2 x (1.01 g/mol) = 2.02 g/mol
= molar mass Sr(OH)2

16. Molar Mass Sr(OH)2 121.64 g/mol = molar mass Sr(OH)2
1 mol Sr(OH)2= g Sr(OH)2

17. Molar Mass Fe Fe(C2H3O2)3 C H O = 55.85 g/mol 55.85 g/molx 3 x 2 x 3 3 3 x x x
= g/mol
55.85 g/mol
= 6 x (12.01 g/mol) = g/mol
= g/mol
72.06 g/mol
= 1.01 g/mol
= 9 x (1.01 g/mol) = 9.09 g/mol
9.09 g/mol
= g/mol
= 6 x (16.00 g/mol) = g/mol
96.00 g/mol
molar mass Fe(C2H3O2)3 =
233 g/mol
g/mol

18. Molar Mass Fe(C2H3O2)3 Fe = 55.85 g/mol 3 x (2 x C) 2 x C) C)
= 6 x (12.01 g/mol) = g/mol
3 x (3 x H)
3 x H) H)
= 9 x (1.01 g/mol) = 9.09 g/mol
= 1.01 g/mol
3 x (2 x O)
2 x O) O)
= g/mol
= 6 x (16.00 g/mol) = g/mol
molar mass Fe(C2H3O2)3 =

19. Molar Mass Fe(C2H3O2)3 233.00 g/mol = molar mass Fe(C2H3O2)3
1 mol Fe(C2H3O2)3 = g Fe(C2H3O2)3

20. Net Ionic Equations
Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone:
I.E.
2 Fe+3(aq) + 3 SO4–2(aq) + 3 PbCl2(s)  2 Fe+3(aq) + 6 Cl–(aq) + 3 PbSO4(s)
Cross out every thing that is EXACTLY the same on both sides:
2 Fe+3(aq) + 3 SO4–2(aq) + 3 PbCl2(s)  2 Fe+3(aq) + 6 Cl–(aq) + 3 PbSO4(s)
N.I.E.
3 SO4–2(aq) + 3 PbCl2(s)  2 Cl–(aq) + 3 PbSO4(s)
3 SO4–2(aq) + 3 PbCl2(s)  6 Cl–(aq) + 3 PbSO4(s)

21. Net Ionic Equations
Start with a balanced chemical equation (including subscripts):
M.E.
2 HC2H3O2(aq) + Ba(OH)2(aq)  Ba(C2H3O2)2(aq) + 2 H2O(l)
2 HC2H3O2(aq) + Ba(OH)2(aq)  Ba(C2H3O2)2(aq) + 2 H2O(l)
2 HC2H3O2(aq) + Ba(OH)2(aq)  Ba(C2H3O2)2(aq) + 2 H2O(l)
2 HC2H3O2(aq) + Ba(OH)2(aq)  Ba(C2H3O2)2(aq) + 2 H2O(l)
2 HC2H3O2(aq) + Ba(OH)2(aq)  Ba(C2H3O2)2(aq) + 2 H2O(l)
W.E.
S.E.
S.E.
N.E.
Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone:
I.E.
2 HC2H3O2(aq) + Ba2+(aq) + 2 OH–(aq)  Ba2+(aq) + 2 C2H3O2–(aq) + 2 H2O(l)
2 HC2H3O2(aq) + Ba2+(aq) + 2 OH–(aq)  Ba2+(aq) + 2 C2H3O2–(aq)
2 HC2H3O2(aq)
2 HC2H3O2(aq) + Ba2+(aq) + 2 OH–(aq)

22. Net Ionic Equations
Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone:
I.E.
2 HC2H3O2(aq) + Ba2+(aq) + 2 OH–(aq)  Ba2+(aq) + 2 C2H3O2–(aq) + 2 H2O(l)
Cross out every thing that is EXACTLY the same on both sides:
2 HC2H3O2(aq) + Ba2+(aq) + 2OH–(aq)  Ba2+(aq) + 2C2H3O2–(aq) + 2 H2O(l)
N.I.E.
2 HC2H3O2(aq) + 2OH–(aq)  2C2H3O2–(aq) + 2 H2O(l)
HC2H3O2(aq) + OH–(aq)  C2H3O2–(aq) + H2O(l)

23. Net Ionic Equations
Start with a balanced chemical equation (including subscripts):
M.E.
H2SO4(aq) + 2 NaOH(aq)  Na2SO4(aq) + 2 H2O(l)
H2SO4(aq) + 2 NaOH(aq)  Na2SO4(aq) + 2 H2O(l)
H2SO4(aq) + 2 NaOH(aq)  Na2SO4(aq) + 2 H2O(l)
H2SO4(aq) + 2 NaOH(aq)  Na2SO4(aq) + 2 H2O(l)
H2SO4(aq) + 2 NaOH(aq)  Na2SO4(aq) + 2 H2O(l)
S.E.
S.E.
S.E.
N.E.
Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone:
I.E.
2 H+(aq) + SO4–2(aq) + 2 Na+(aq) + 2 OH–(aq)  2 Na+(aq) + SO4–2(aq) + 2 H2O(l)
2 H+(aq) + SO4–2(aq)
2 H+(aq) + SO4–2(aq) + 2 Na+(aq) + 2 OH–(aq)  2 Na+(aq) + SO4–2(aq)
2 H+(aq) + SO4–2(aq) + 2 Na+(aq) + 2 OH–(aq)

24. Net Ionic Equations
Break ALL strong electrolytes into ions. Leave non and weak electrolytes alone:
I.E.
2H+(aq) + SO4–2(aq) + 2Na+(aq) + 2OH–(aq)  2Na+(aq) + SO4–2(aq) + 2H2O(l)
Cross out every thing that is EXACTLY the same on both sides:
2 H+(aq) + SO4–2(aq) + 2 Na+(aq) + 2 OH–(aq)  2 Na+(aq) + SO4–2(aq) + 2 H2O(l)
N.I.E.
H+(aq) + OH–(aq)  H2O(l)
2 H+(aq) + 2 OH–(aq)  2 H2O(l)