1. Body parts Mole calculations: Moles level 2(continued)

2. `body’ parts example 2 : Weight  moles first, then mol  Mol Octane = C 8 H 18 How many moles of octane contain 24 grams of C (at. wt. = 12 g/mol) 24 g = 2 moles C 12 g/mol C Step 1) Convert grams C to moles C (all roads lead through moles divide up)

3. `body’ parts example 2 : (continued) Octane = C 8 H 18 How many moles of octane contain 24 grams of C (at. wt. = 12 g/mol) Step 2) 2 moles C => how many moles octane? 1 mol octane = x 8 mol C 2 x= 2/8=0.25 moles octane (body parts relationship ?)

4. `body’ parts example 3 : Weight  moles, then mol  Mol  weight Octane = C 8 H 18 (MW=114) How many grams of H are in 10 grams of Octane ? (1 mole H=1 g H) Step 1) Convert grams octane to moles octane All roads lead through moles; divide up 10 g = 0.0877 mol octane 114 g/mol

5. `body’ parts example 3 (continued) : Octane = C 8 H 18 (MW=114) How many grams of H are in 10 grams of Octane ? (1 mole H=1 g H) Step 2) Relate moles H to moles octane (body parts relationship ?) 14 mol H = x 1 octane 0.0877 x= 14*0.0877=1.228 mol H

6. `body’ parts example 3 (continued) : Octane = C 8 H 18 (MW=114) How many grams of H are in 10 grams of Octane ? (1 mole H=1 g H) Step 3) convert moles H to grams H (multiply down) 1.228 mol H * 1 g H/mol=1.228 mol H

7. Solving Mole body parts problems: A SUMMARY  Step 1: Convert any given weight or molecule count to moles (“all roads lead through moles” DIVIDE UP)  Step 2: Relate the computed moles above to desired moles (set up ratio and solve x)  If necessary –Step 3: convert desired moles to target grams or molecule count (MULTIPLY DOWN)

8. Isohexane has the formula C 6 H 14. How many grams of H (at. Wt. = 1 g/mol) are combined in 300 grams of C 6 H 14 (MW=86 g/mol) A.36.84 g H B.1.32 g H C.0.249 g H D.0.020 g H E.48.84 g H

9. Isohexane has the formula C 6 H 14. How many moles of isohexane are formed with 24 grams of C (1 mol C=12 g) A.0.333 moles isohexane B.2 moles isohexane C.0.1667 moles isohexane D.3 moles isohexane E.Stoichiometry blows F. None of the above

10. Mole Body Part Calculations level 2: extra problems for you to try at home if you want 1)mole to mole: how many moles of O are present in 0.1666 mole of C 6 H 12 O 6 ? 2)mole to mole how many moles of C 6 H 12 O 6 can be made with 12 mole of O ? 1 mol O 2 moles C 6 H 12 O 6

11. 3) weight to moles how many moles of C 6 H 12 O 6 in a sample containing 216 g C ? 3 mol 4) moles to weight how many grams of C 6 H 12 O 6 are formed with 0.2666 mol H? = 4 g Mole Calculations: part 2 (cont.)

12. 5) moles to molecules how many molecules of O are present in 0.8333 mol of C 6 H 12 O 6 ? =5*10 23 6) molecules to moles how many moles of C 6 H 12 O 6 are formed from 4.32*10 25 atoms of H ? 6 moles Mole Calculations: part 2 (cont.)

13. 7) mass to molecules: how many molecules of C 6 H 12 O 6 form from 84 g of C ? 7 *10 23 molecules of C 6 H 12 O 6 8) atoms to mass: how many grams of H are combined with 2.4*10 24 atoms of O in C 6 H 12 O 6 ? 8 grams H Mole Calculations: part 2 (cont.)

14. 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 …

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

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

17. 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

18. 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

19. 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)

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

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

22. 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)

23. 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…

24. 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

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

26. 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

27. 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 2: 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