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EXERCISE 5 Practice Examples first done in class-on white board

EXERCISE 5 Practice Examples first done in class-on white board. 5.2 Calculate the maximum yield problems. 2 C 4 H 10 + 13 O 2 --------  8CO 2 + 10H 2 O 58 32 44 18 g/mol. Mole-mol-mol

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EXERCISE 5 Practice Examples first done in class-on white board

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  1. EXERCISE 5 Practice Examples first done in class-on white board

  2. 5.2 Calculate the maximum yield problems 2C4H10 + 13O2-------- 8CO2 + 10H2O 58 32 44 18 g/mol Mole-mol-mol • a)0.25 moles of C4H10and 1.4 moles of O2 are reacted. • What is the maximum yield of CO2 in moles? 0.25 mol C4H10 produces 8*0.25/2 =1 mol CO2 1.4 mol O2 produces 8*1.4/13 =0.86 mol CO2

  3. 5.2 Calculate the maximum yield problems (cont.) 2C4H10 + 13O2-------- 8CO2 + 10H2O 58 32 44 18 g/mol • Weight-weight-mol • b)1 gram of C4H10 and 10 grams of O2 are reacted. • What is the maximum yield in H2O in moles ? 1 gram C4H10= 1/58=0.017 mol C4H10 0.017 mol C4H10 produces 10*0.017/2=0.085 mol H2O 10 gram O2= 10/32=0.312 mol O2= 0.085 mol H2O 0.312 mol O2 produces 10*0.312/13 =0.24 mol H2O

  4. 5.2 Calculate the maximum yield problems (cont.) 2C4H10 + 13O2-------- 8CO2 + 10H2O 58 32 44 18 g/mol Weight-weight-molecules • c) 5.8 grams of C4H10and 160 grams of O2 are reacted. • What is the maximum yield of CO2in molecule count? 5.8/58=0.1 mol C4H10 160/32=5 mol O2 0.1 mol C4H10 produces 8*0.1/2=0.4 mol CO2 5 mol O2 produces 8*5/13 =3.07 mol CO2 C4H10limits 0.4*6*1023 = molecules CO2= 2.4*1023

  5. 5.2 Calculate the maximum yield problems (cont.) 2C4H10 + 13O2-------- 8CO2 + 10H2O 58 32 44 18 g/mol Weight-molecules-weight • d)116 grams of C4H10 and 1.66*1024 molecules of O2 react. • What is the maximum yield of H2O in grams? 116/58=2 mol C4H10 1.66*1024/6*1023=2.8 mol O2 2 mol C4H10 produces 10*2/2=10 mol H2O 2.8 mol O2 produces 10*2.8/13 =2.15 molH2O mol O2limits Multiply down to calculate: grams H2O= 2.15*18=38.8g

  6. % Yield: a Familiar, non-chemical analogy Billy comes home with his math test. He got 40 out of 75 questions right. Did he `pass’ the test ? Test Score = Test Yield = 100* 40 =53.3% 75

  7. Examples of chemical % yield 5O2 + C3H83CO2 + 4H2O Given mol excess 0.250.7(obs) Exp. Product mol Theory product mol 4 x 0.25 =1 (theory)1 % yield=exp. Molx 100 theory mol. = 70% =0.7 x 1001

  8. Examples of chemical % yield (continued) 5O2 + C3H83CO2 + 4H2O MW 32 44 44 18 Given excess 1.1 g 0.5 g Mol C3H8 = 1.1=0.025 44 Exp. Mol CO2 = 0.5= 0.01136 44 Theory product CO2 moles 3 x 0.025 =0.075 1 % yield =0.01136 x 100% =15.1% 0.75

  9. Where we’ve been so far on the chemistry bus trip…

  10. Where we’ve been…. Rutherford & Bohr 1910 1) Atomic structure & dimensions 2) p+.no , e- bookkeeping 3) Crazy quantum models

  11. Where we’ve been …… Mendeleev 1865 AD 1)Organizing element behavior to predict chemistry 2)Periodicity and the Periodic Table

  12. Where we’ve been…. In the land of the spectroscopists: 1s2 2s2 2p6 3s23p4 …… Describing atoms by electronic configurations implied by observation and predicted by Periodic table spectroscopy `fugetabout theory….let the line spectra rule or I break your face

  13. Where we’ve been… Dalton 1805AD 1)Law of constant proportions 2)Law of multiple proportions Moles till we drop…..

  14. The Americans land… Lewis 1925 AD: using the Periodic table to predict ionic and covalent compound formulas without breaking a sweat…Father of an American-style Chemistry that looks at the big picture and uses simple, intuitive models . (sans calculations …)* * We know how to use the fancy math…but not as a crutch

  15. Legacies of the American chemist:GilbertNewtonLewis Gilbert Newton LewisAmericanChemist UC Berkeley “at work” ~1930 Before Lewis: UC Berkeley Chemistry Building ~1915

  16. Chemistry `Quad’ UC Berkeley 2013 + more `on the hill’ at L-L Labs

  17. As young chemistry nerd …Young Doc Fong sits in here staring every day at a 4’x 6’ photo of… The House that LewisBuilt:Lewis Hall of Chemistry UC Berkeley

  18. Reading: • Chapter 4 • pp. 154-157 ionic model • pp. 168-182 covalent model • Chapter 5 • pp. 193-205 VSEPR structures

  19. Trends in binary ionic and near-ionic formulas LiCl Li2O Li3P Trends down a column MaXb …all M’scombine the same for a given X NaCl Na2O Na3P KCl K2O K3P

  20. Trends in ionic and not ionic compound formulas HCl NaCl Na2O Na3P CH4 NH3 H2O MgCl2 MgO Mg3P2 Al2O3 AlP Is there a simple pattern here ?? Trends across columns

  21. The rule of 8 ( early version of the `octet rule’ for building compound formulas) The sum of column numbers of elements combined in many simple1binary compounds add to 8. Col 26 CaO 2+ 6=8 Example: 1simple ionic compounds ~formed of column 1- 2 elements (cations) with columns 5-7 (anions)

  22. More examples of the rule of 8 Col 1 7 NaCl 1 + 7 = 8 Col 1 6 Li + Se?? 2*1+6=8 Li2Se

  23. You predict formulas for…. H+ N H + O K + P 3*1+5=8=>H3N=NH3 2*1+6=8=>H2O 3*1+5=8=>K3P

  24. Is it safe to assume the rule of 8 works all the time ????

  25. …what about… col2 5 Mg + N ??? Actual formula Mg3N2 Rule of 16 ???3*2 +2*5=16 =2x8 (ICKY…too much math for us Americans) 

  26. Thinking about the curious case of column 8 1 2 3 4 5 6 7 8

  27. What’s so curious ?? 8 The “Noble” Elements are utterly stable … Noble gas chemistry 0 None Nada …And what’s with this number 8 ?? Noble…because like all aristocrats they have nothing to do with any of the lower class elements.

  28. The crazy stability of the noble gas electronic configuration is the key to a better model for predicting ionic and near-ionic compound formulas (see also: p. 155-6) Example: CaO O=[He]2s22p4 Ca= [Ar]4s2 Lose 2 e- gain 2 e- Ca2+=[Ar] O2- =[He]2s22p6=[Ne]

  29. Lewis dot pix for atoms +2 -2 Ca O O Ca Ar Complete octets (ns2 np6) Ne [CORE] VALENCE Ca= [Ar]4s2 O= [He]2s22p4

  30. A quick visual way to know element charges Cations: count left to col 8 for + charge Anions:count right to col 8 for - charge +2 -3 …but don’t do this with transition metals

  31. Knowing cation and anion charges lets us predict binary compound formulas. EXAMPLE: Predict compound made of Na and S 1)FIND PREFERRED CHARGES +1 -2 Na S 2 1 2)“CROSS” CHARGE MAGNITUDES

  32. In-Class practice: Deducing formulas for… Ca + Cl CaCl2 Al2O3 Al + O B + N BN K + S K2S Mg+ As Mg3As2

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