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Ch. 3 Notes

by: Stephanie Belo

Ch. 3 Notes CHM2045

Marketplace > Chemistry > CHM2045 > Ch 3 Notes
Stephanie Belo
GPA 4.0

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About this Document

These cover week 3 of Mitchell's lecture notes.
Chemistry 1
John Mitchell
Class Notes
25 ?




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This 3 page Class Notes was uploaded by Stephanie Belo on Saturday January 23, 2016. The Class Notes belongs to CHM2045 at a university taught by John Mitchell in Spring 2016. Since its upload, it has received 9 views.


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Date Created: 01/23/16
CHM  2045  Ch.  3  Notes   • Know:  mole  concept  and  mole  calculations,  percent  composition,   empirical  formula  &  molecular  formula  determination   • Key  points  about  the  MOLE  concept   o The  mole  maintains  the  same  mass  relationships  between   macroscopic  samples  as  exists  between  individual  chemical   entities   o The  mole  relates  the  number  of  chemical  entities  to  the  mass  of   a  sample  of  those  entities   o So  we  count  chemical  entities  by  weighing  them   23 o We  use  Avogadro’s  number  (6.022  *  10 )  to  say  that  one  mole   contains  6.022  *  10  Avogadro’s  number  of  ions,  molecules,   particles,  etc.  (You  must  know  more  than  just  the  number!)   o A  mole  represents  a  fixed  number  of  chemical  entities  and  has  a   fixed  but  different  mass  for  each   Mole  and  Molar  Mass   Aspirin,  (C 9 O8) 4  What  is  aspirin’s  molar  mass?  (12.01  g  C  /  mol  C,  1.008  g  H  /  mol  H,  16.00   g  O  /  mol  O)   9  (12.01)  +  8  (1.008)  +  4(16.00)  =  108.09  +  8.064  +  64  =  180.15  g/mol  of   Aspirin,  C 9 O8 4 You  must  be  able  to  do  calculations  like  these  shown  here!   g  à  mol  à  molecules  à  atoms   If  you  have  12.3  g  of  aspirin  (ASA)  how  many  oxygen  atoms  do  you  have?   12.3  g  ASA  *  (1  mol  ASA  /  180.15  g  ASA)  *  (6.022*10  molecules  /  1  mol   23 ASA)  *  (4  oxygen  atoms  /  1  ASA  molecule)  =  1.64  *  10  oxygen  atoms   %  Composition   9  C  =  108.09  /  180.15  =  0.5999  *  100%  =  59.99%  C   8  H  =  8.064  /  180.15  =  0.04476  *  100%  =  4.476%  H   4  O  =  64  /  180.15  *  100%  =  35.53%  O   59.99  g  C  *  (1  mol  C  /  12.01  g  C)  =  4.995  mol  C   4.476  g  H  *  (1  mol  H  /  1.008  g  H)  =  4.44  mol  H   35.53  g  O  *  (1  mol  O  /  16.00  g  O)  =  2.22  mol  O   then  divide  by  the  smallest  quantity   4.995  mol  C  /  2.22  =  2.25   4.44  mol  H  /  2.22  =  2   2.22  mol  O  /  2.22  =  1   then  multiply  each  by  4  (to  make  2.25  a  whole  number)   C 9 O8 4   A  hydrate  is  an  ionic  compound  that,  when  it  crystallizes,  sucks  water  in  with   it   Hydrate  problem  (**will  be  on  exam!  Practice!!**)   Discussion  Problem   You dry 4.450 g of hydrated lithium iodide, LiI • xH O in an oven. When 2 you remove and cool the anhydrous salt it has a mass of 3.170 g. What is the value of x? LiI is the anhydrous salt of H O 2 4.450 g “wet” –3.170 g “dry = 1.28 g H O (subtract to ge2 the mass of water!) 1.28 g H2O * (1 mol H2O / 18.0 g H2O) = 0.0711 mol H2O 3.170 g LiI * (1 mol LiI / 133.84 g LiI) = 0.0237 mol LiI 0.0711 mol H2O / 0.0237 mol LiI = 2.95 à x = 3 (round bc coefficients must be whole numbers) 2   **practice balancing equations** **practice naming compounds** 3 Sn + 2 NO 31-+ 8 H à 3 Sn + 2 NO + 4 H O 2 (If you balance just the mass, the charge will not be balanced! Both the mass and the charge must be balanced.) Sn: goes through oxidation—loss of 2 electrons NO à3 NO : goes through reduction: gains 3 electrons The loss/gain must be equal so we must cross multiply. Mass Relationships in Chemical Reaction Equation, is it balanced? 4 NH (3) + 5 O (g)2à 4 NO (g) + 6 H O (l) 2 Yes, there are equal amounts of N, H, and O on both sides. If you start w only 4 moles of NH how much oxygen do we need? 5 3 *Check slides and practice the following qs* 3  


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