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Homework 6 posted and due Monday 3/5

Learn mole-mass-count conversions, reaction balancing, and % composition problems with practical examples and molecular formula determinations. Includes an engaging group project task.

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Homework 6 posted and due Monday 3/5

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  1. Homework 6 posted and due Monday 3/5 Sigh, another Monday…

  2. The road trip through mole land so far… • Basic mole-mass-count conversions ( divide up, multiply down): =Moles level 1 • Mole ratio conversion within a compound (`body parts’ relationships) = Moles level 2

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

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

  5. % composition problems: using mole concept to convert weights to formulas 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 ?

  6. Definition: empiric formula= mole ratio of elements in a compound expressed inlowestwhole numbers Example: CH2O = empiric formula of glucose (sugar we metabolize)

  7. Empiric formulas (continued) CH2O = empiric formula of glucose (sugar we metabolize) Actual molecular formula(what it really has in atom count): C6H12O6= (CH2O)6 Sugar =Carbo hydrate

  8. Which formulas below are empiric (= lowest common denominator form) ?? ÷ 3 NOT EMPIRIC EMPIRIC C3H6O3 CH2O

  9. Which formulas below are empiric (continued) ?? EMPIRIC…3 & 7 have no shared factors N3F7

  10. Which formulas below are empiric (continued) ?? H3P9O12S5 EMPIRIC (3,9,12 divisible by 3…but 5 is not)

  11. Which compound in the list is not empiric: NO2 , N2O3 ,N3O9 , N2O • NO2 • N2O3 • N3O9 • N2O

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

  13. Sample problem #1; SUMMARIZED OFFICIAL NAME… Dinitrogen monoxide =>N2O =Laughing gas n nmin AW=Atomic wt (g/mol) n=w/AW moles w(g) 4.545 = 2 2.273 63.63/14= 4.545 63.63 14 36.36/16= 2.273 2.273 =1 2.273 36.36 16

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

  15. % composition problem 2: An oxide of nitrogen contains 46.7 wt. % N and 53.3 % O. What is it’s empiric formula ? • NO3 • NO2 • N2O • N3O7 • NO • None of the above

  16. % composition Problem 3: empiric formula determination with a twist 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: C1.5H3O1 C3H6O2 x FACTOR 1.5*2= 3 48.64/12= 4.05 4.05/2.7 =1.5 ? 8.16/1= 8.16 3.0*2= 6 8.16/2.7 =3.0 43.2/16= 2.70 2.7/2.7= 1 1*2= 3

  17. Compound X is composed of 39.34 wt % C, 8.20 wt% H and 52.56 wt% O. What is compound X’s empiric formula ? • C4H8O5 • CH2O • C2H5O2 • CH3O

  18. Naughty Kitty Group Project (% composition problem 5) • Using the % composition data for your team: • Determine the compound on your assigned • kitty’s anatomical parts • 2) Decide what your kitty’s naughty act was.

  19. Tigger Socks Bosco Buddy Socks

  20. Table 1: Percent Composition Data of the Compounds Found on my 4 Naughty Kitties Naughty Kitty C H N O anatomical site Bosco 20.00 6.72 46.70 26.66 paws and tail Buddy 57.14 6.16 9.52 27.18 whiskers, saliva, urine Socks 63.15 5.30 - 31.55 paws 46.66 4.48 31.1 17.76 paws Tigger 60.00 4.48 - 35.53 tail 68.5 6.66 4.44 20.30 paws. nose Note: C=12 H=1 N=14 O=16 (atomic masses)

  21. Table 2: Possible Compounds Identity Molecular formula Notes Aspirin C9H8O4 Over-the-counter pain killer Codeine C­18H21NO4 Over-the-counter pain killer Aspartame C­14H18N2O5 Artificial sweetener Urea CH4N2O component in urine Vanilla C8H8O3 Flavoring in cakes, pies, cookies Theobromine C7H8N4O2 Chocolate flavoring

  22. Examining Buddy’s whiskers, saliva and urine sample C14H18N2O5 =Aspartame (artificial sweetener) n=mol =w/AW n/nmin X 2 4.76 7 14 18 2 5 6.16 9 1 0.68 2.5 ? 1.698

  23. Buddy Bosco Whiskers etc. => asparatame Paws, tail => urea Took my diet cookies Peed on the living room floor

  24. Socks Tigger Paws => vanilla Paws => chocolate Tail => aspirin Paws/nose=> codeine Walked on kitchen counter and cake mix Made mess in bathroom

  25. One more twist on composition problems… empiric  molecular formula determination The empiric compound (C2H4O3) has a molecular weight of 228 g/mol. What is the molecular formula of the compound ? Let’s do this on the board…

  26. Burn, Baby, Burn !

  27. Chemical Compositions from Combustion Data 1) A 0.30 gram sample of carbon black is burned in a covered crucible. The collected gas weighs 0.70 grams. What is the empiric formula of the gas ?(C =12, O =16) ANSWER: CO

  28. Chemical Compositions from Combustion Data (continued) 2) A hydrocarbon sample (Cx Hy) is burned in oxygen producing 1.0 gram of CO2 and 0.4086 g H2O. What is the empiric formula for the hydrocarbon ? (C=12, O=16, H=1) ANSWER: CH2

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