1. Basic mole-mass-count conversions ( divide up, multiply down) Mole ratio conversion within a compound (`body parts’ relationships) The road trip through mole land so far …

2. % composition problems and combustion analysis (pp. 94-103) : moles level 2b Reaction balancing (pp. 105-108) Reaction stoichiometry predictions, limiting yields and % yields ( pp. 108-123): moles level 3 The mole road ahead...

3. Why bother with all of this ????

4. ` stoichiometric’ studies in the 18 th and 19 th century are foundation to things like…

5. Doped semiconductors

6. Combustion analysis

7. also leads to… Optical fiber Without which there is no….

8. Sample % composition problem #1 A compound of N and O contains 63.63 wt % N and 36.36 wt % O. What is the empiric formula of the compound ? % composition problems: using mole concept to convert weights to formulas

9. Definition: Example: CH 2 O = empiric formula of glucose (sugar we metabolize) empiric formula= mole ratio of elements in a compound expressed in lowest whole numbers

10. CH 2 O = empiric formula of glucose (sugar we metabolize) (CH2O)6(CH2O)6 Sugar =Carbo hydrate Actual molecular formula (what it really has in atom count): C 6 H 12 O 6 = Empiric formulas (continued)

11. Which formulas below are empiric (= lowest common denominator form) ?? C3H6O3C3H6O3 NOT EMPIRIC ÷ 3 CH 2 O EMPIRIC

12. Which formulas below are empiric (continued) ?? N3F7N3F7 EMPIRIC…3 & 7 have no shared factors

13. Which formulas below are empiric (continued) ?? H 3 P 9 O 12 S 5 EMPIRIC ( 3,9,12 divisible by 3…but 5 is not)

14. A compound of N and O contains 63.63% N and 36.36 %O. What is the empiric formula of the compound? % composition Problem 1: empiric formula determination let’s do it on the board…

15. Sample problem #1; SUMMARIZED element N O w(g) 63.63 36.36 14 16 63.63/14= 4.545 36.36/16 = 2.273 4.545 = 2 2.273 2.273 =1 2.273 AW=Atomic wt (g/mol) n=w/AW moles n n min =>N 2 O =Laughing gas OFFICIAL NAME… Dinitrogen monoxide

16. http://www.youtube.com/watch?v=_Ha-ZrUPJ_E Joseph Priestley 1772

17. An oxide of nitrogen contains 46.7 wt. % N and 53.3 % O. What is it’s empiric formula ? A.NO 3 B.NO 2 C.N 2 O D.N 3 O 7 E.NO F.None of the above

18. The previous empiric compound (C 2 H 4 O 3 ) has a molecular weight of 228 g/mol. What is the molecular formula of the compound ? % composition Problem 3: empiric + molecular formula determination Let’s do this on the board…

19. A compound composed of C,H and O is 48.64% C and 8.16 % H and 43.2 % O by mass. What is the empiric formula ? Answer: C 1.5 H 3 O 1  C 3 H 6 O 2 % composition Problem 4: empiric formula determination with a twist elementMass,gAtomic wt(g/mol) n=mass Atomic wt n/n min C 48.6412 H 8.161 O 43.216 48.64/12= 4.05 8.16/1= 8.16 43.2/16= 2.70 4.05/2.7 =1.5 ? 8.16/2.7 =3.0 2.7/2.7= 1 x FACTOR 1.5*2= 3 3.0*2= 6 1*2= 3