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Ch 7 Mole JKS

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					        Chapter 7
    Chemical Quantities

     Masuk High School
      Mr. J. K. Stoelzel

1
             Section 7.1
     The Mole: A Measurement of
               Matter
     OBJECTIVES:
     –Describe how Avogadro’s
      number is related to a mole of
      any substance.
     –Calculate the mass of a mole of
      any substance.
2
              What is a Mole?

     You  can measure mass,
     or volume,
     or you can count pieces.
     We measure mass in grams.
     We measure volume in liters.

     We   count pieces in MOLES.
3
        Moles (abbreviated: mol)
     Defined   as the number of carbon
      atoms in exactly 12 grams of
      carbon-12.
     1 mole is 6.02 x 1023 particles.
     Treat it like a very large dozen

     6.02   x 1023 is called Avogadro’s
     number.
4
             What is the Mole?


       1 mole of hockey pucks would
        equal the mass of the moon!

       1 mole of basketballs would fill a
        bag the size of the earth!


       1 mole of pennies would cover
5       the Earth 1/4 mile deep!
        Representative particles
 The   smallest pieces of a substance.
    –For a molecular compound: it is
     the molecule.
    –For an ionic compound: it is the
     formula unit (ions).
    –For an element: it is the atom.
      »Remember the 7 diatomic
       elements (made of molecules)
6
            Types of questions
     How  many oxygen atoms in the
      following?
       –CaCO3
       –Al2(SO4)3
     How many ions in the following?
       –CaCl2
       –NaOH
       –Al2(SO4)3
7
              Types of questions
     How many molecules of CO2 are
     there in 4.56 moles of CO2 ?
     = (4.56 moles)(6.02 x 1023 molecules/mole)
     = 2.75 x 1024 molecules
     How   many moles of water is 5.87 x
     1022   molecules?
     = (5.87 x 1022 molecules)/(6.02 x 1023
       molecules/mole)
     = 0.0975 moles
8
              Types of questions
     How many atoms of carbon are
     there in 1.23 moles of C6H12O6 ?
     = (1.23 moles)(6 atoms of C/mole)
     = 7.38 atoms of C
     Howmany moles is 7.78 x 1024
     formula units of MgCl2?
     = (7.78 x 1024 units)/(6.02 x 1023 units/mole)
     = 12.9 moles

9
          Measuring Moles
  Remember     relative atomic mass?
  The amu was one twelfth the mass
   of a carbon-12 atom.
  Since the mole is the number of
   atoms in 12 grams of carbon-12,
  the decimal number on the periodic
   table is also the mass of 1 mole of
   those atoms in grams.
10
     Gram Atomic Mass (gam)
 Equals  the mass of 1 mole of an
   element in grams
  12.01 grams of C has the same
   number of pieces as 1.008 grams of
   H and 55.85 grams of iron.
  We can write this as
              12.01 g C = 1 mole C
  We can count things by weighing
11 them.
                   Examples
  How   much would 2.34 moles of
     carbon weigh?
     = (2.34 moles C)(12.01 g/mole)
     = 28.1 g C
  How    many moles of magnesium is
     24.31 g of Mg?
     = (24.31 g Mg)/(24.31 g/mole)
     = 1.000 moles Mg

12
                  Examples
  How    many atoms of lithium is 1.00
     g of Li?
       (1.00 g Li)(6.02 x 1023 atoms Li/mol)
     
                   (6.94 g Li/mol)
    8.67 x 1022 atoms Li
  How much would 3.45 x 1022 atoms
   of U weigh?
       (231.04 g U/mol)(3.45 x 10 22 atoms U)
     
             (6.02 x 1023 atoms U/mol)
       38.38 g U
13
       What about compounds?
  in 1 mole of H2O molecules there are
   two moles of H atoms and 1 mole of O
   atoms
  To find the mass of one mole of a
   compound
    – determine the moles of the elements
      they have
    – Find out how much they would weigh
14
    – add them up
       What about compounds?
  What is the mass of one mole of CH4?
   1 mole of C = 12.01 g
   4 mole of H x 1.01 g = 4.04g
   1 mole CH4 = 12.01 + 4.04 = 16.05g
  The Gram Molecular Mass (gmm) of
   CH4 is 16.05g
   – this is the mass of one mole of a
     molecular compound.
15
     Gram Formula Mass (gfm)
  The mass of one mole of an ionic
   compound.
  Calculated the same way as gmm.
  What is the GFM of Fe2O3?
   2 moles of Fe x 55.85 g = 111.70 g
   3 moles of O x 16.00 g = 48.00 g
   The GFM = 111.70 g + 48.00 g =
     159.70 g
16
                Section 7.2
         Mole-Mass and Mole-Volume
               Relationships
        OBJECTIVES:
         – Use the molar mass to convert
           between mass and moles of a
           substance.
         – Use the mole to convert among
           measurements of mass, volume,
           and number of particles
17
            Molar Mass
  Molar mass is the generic term for
   the mass of one mole of any
   substance (in grams)
  The same as: 1) gram molecular
   mass, 2) gram formula mass, and 3)
   gram atomic mass- just a much
   broader term.

18
               Molar Mass
      The  number of grams of 1 mole
       of atoms, ions, or molecules.
      We can make conversion factors
       from these.
        –To change grams of a
         compound to moles of a
         compound.

19
         B. Molar Mass Examples
        water
         – H 2O
         – 2(1.01) + 16.00 = 18.02 g/mol

        sodium chloride
          – NaCl
          – 22.99 + 35.45 = 58.44 g/mol

20
         B. Molar Mass Examples
        sodium bicarbonate
       – NaHCO3
       – 22.99 + 1.01 + 12.01 + 3(16.00)
                         = 84.01 g/mol
      sucrose
       – C12H22O11
       – 12(12.01) + 22(1.01) + 11(16.00)
                         = 342.34 g/mol
21
                   Examples
      Calculate the molar mass of the
       following and tell what type it is:
      Na2S
        2 mole of Na x 22.99 g = 45.98g
        1 mole of S = 32.07 g
        1 mole Na2S = 45.98 + 32.07 =
        78.05 g/mol

22
                   Examples
      Calculate the molar mass of the
       following and tell what type it is:
      N2O4
        2 mole of N x 14.01 g = 28.02 g
        4 mole of O x 16.00 g = 64.00 g
        1 mole N2O4 = 28.02 + 64.00 =
        92.02 g/mol

23
                      Examples
      Calculate the molar mass of the
       following and tell what type it is:
      Ca(NO3)2
        1 mole of Ca = 40.08 g
        2 mole of N x 14.01 g = 28.02 g
        6 mole of O x 16.00 g = 96.00 g
         1 mole Ca(NO3)2 = 40.08 + 28.02 + 96.00 =
         164.10 g/mol; ionic compound
24
                  For example
        How many moles is 5.69 g of NaOH?




25
                  For example
        How many moles is 5.69 g of NaOH?
                     
     5.69 g          
                     




26
                  For example
        How many moles is 5.69 g of NaOH?
                 mole 
      5.69 g          
                    g 
      need to change grams to moles




27
                  For example
        How many moles is 5.69 g of NaOH?
                 mole 
      5.69 g          
                    g 
      need to change grams to moles
      for NaOH




28
                  For example
        How many moles is 5.69 g of NaOH?
                  mole 
      5.69 g           
                     g 
      need to change grams to moles
      for NaOH
      1mole Na = 22.99g 1 mol O = 16.00 g
       1 mole of H = 1.01 g

29
                  For example
        How many moles is 5.69 g of NaOH?
                  mole 
      5.69 g           
                     g 
      need to change grams to moles
      for NaOH
      1mole Na = 22.99g 1 mol O = 16.00 g
       1 mole of H = 1.01 g
      1 mole NaOH = 40.00 g
30
                  For example
        How many moles is 5.69 g of NaOH?
                1 mole 
      5.69 g            
               40.00 g 
      need to change grams to moles
      for NaOH
      1mole Na = 22.99g 1 mol O = 16.00 g
       1 mole of H = 1.01 g
      1 mole NaOH = 40.00 g
31
                  For example
        How many moles is 5.69 g of NaOH?
                1 mole 
     5.69 g             = 0.142 mol NaOH
               40.00 g 
      need to change grams to moles
      for NaOH
      1mole Na = 22.99g 1 mol O = 16.00 g
       1 mole of H = 1.01 g
      1 mole NaOH = 40.00 g
32
                    Examples
      Howmany moles is 4.56 g of
      CO2?
                   1 mole      
            12.01  2 *16.00 g  = 0.104 mol CO 2
     4.56 g                    
                               
      Howmany grams is 9.87 moles of
      H2O?
                   1 mole      
             16.00  2 *1.01 g  = 0.548 mol H 2 O
      9.87 g                   
                               

33
                    Examples
      Howmany molecules is 6.8 g of
      CH4?
                  1 mole      
            12.01  4 *1.01 g  = 0.399 mol CH 4
      6.8 g                   
                              
                           6.02 x 10 molecules 
                                     23
      0.399 mol CH 4 
                                                 
                                                  
                                 1 mole          
                         23
              = 2.40 x 10 molecules CH 4

34
                          Examples
      49molecules of C6H12O6 weighs
       how much?
                     1 mole          6 *12.01  12 *1.01  6 *16.00 g 
49 molecules                                                          
              6.02 x 10 molecules                                     
                         23
                                                   1 mole

                          1          180.18 
                  49           23            g
                      6.02 x 10       1     


                      = 1.47 x 10 -20 g C 6 H12 O 6

35
                 Gases
  Many  of the chemicals we deal with
   are gases.
    –They are difficult to weigh.
  Need to know how many moles of gas
   we have.
  Two things effect the volume of a gas
    –Temperature and pressure
  We need to compare them at the same
   temperature and pressure.
36
         Standard Temperature and
                 Pressure
      0ºC and 1 atm pressure
      abbreviated STP
      At STP 1 mole of gas occupies
       22.4 L
      Called the molar volume
      1 mole = 22.4 L of any gas at STP

37
                   Examples
      Whatis the volume of 4.59 mole
      of CO2 gas at STP?
                         22.4 L 
              4.59 mole           103 L CO 2
                         1 mole 

      How many moles is 5.67 L of O2
      at STP?
                      1 mole 
              5.67 L           0.253 mole O 2
                      22.4 L 
38
                   Examples
      Whatis the volume of 8.8 g of
      CH4 gas at STP?

                1 mole         22.4 L 
            12.01  4 *1.01 g  1 mole   12 L
     8.8 g                    
                                      




39
           Density of a gas
    D=m/V
     –for a gas the units will be g / L
   We can determine the density of any
    gas at STP if we know its formula.
   To find the density we need the mass
    and the volume.
   If you assume you have 1 mole, then
    the mass is the molar mass (from PT)
40 At STP the volume is 22.4 L.
                     Examples
      Find   the density of CO2 at STP
          12.01  2 * 16.00 g  1 mol    1L     
                                              
                 1 mol         22.4 L  1000 ml 

         1.96 x 10 -3 g/ml

      Find   the density of CH4 at STP.
           12.01  4 * 1.01 g  1 mol     1L    
                                              
                 1 mol         22.4 L  1000 ml 

          7.17 x 10 -4 g/ml

41
                The other way
      Given  the density, we can find the
       molar mass of the gas.
      Again, pretend you have 1 mole at
       STP, so V = 22.4 L.
     m = D x V
      m is the mass of 1 mole, since you
       have 22.4 L of the stuff.


42
                      Examples
      What   is the molar mass of a gas with
         a density of 1.964 g/L?
                           22.4 L 
               1.964 g/L           44.0 g/mol
                           1 mol 


        2.86 g/L?
                           22.4 L 
                2.86 g/L           64.1 g/mol
                           1 mol 


43
                Summary
      These four items are all equal:
      a) 1 mole
      b) molar mass (in grams)
      c) 6.02 x 1023 representative
       particles
      d) 22.4 L at STP
      Thus, we can make conversion
       factors from them.
44
                  Section 7.3
            Percent Composition and
              Chemical Formulas
        OBJECTIVES:
         – Calculate the percent composition
           of a substance from its chemical
           formula or experimental data.
         – Derive the empirical formula and
           the molecular formula of a
           compound from experimental data.
45
 Calculating Percent Composition of
            a Compound
      Like all percent problems:
         Part
                  x 100 %
         whole
      Find the mass of each
       component,
      then divide by the total mass.

46
                   Example
      Calculate the percent
      composition of a compound that
      is 29.0 g of Ag with 4.30 g of S.
              29.0 g        
           
            29.0  4.30 g   100 %  87.1 % Ag
                             
                            
              4.30 g        
           
            29.0  4.30 g   100 %  12.9 % S
                             
                            

47
            Getting it from the formula
      Ifwe know the formula, assume
       you have 1 mole.
      Then you know the mass of the
       pieces and the whole.




48
                   Examples
      Calculate   the percent composition
       of C2H4?
      How about Aluminum carbonate?
       –Sample Problem 7-11, p.191
      We can also use the percent as a
       conversion factor
       –Sample Problem 7-12, p.191

49
                      Examples
      Calculate    the percent composition
      of C2H4?

                 2 * 12.01 g         
         
                                     100 %  83.99 % C
                                      
            2 * 12.01  4 * 1.01 g   
                4 * 1.01 g           
          
           2 * 12.01  4 * 1.01 g    100 %  16.01 % H
                                      
                                     


50
                         Examples
               the percent composition
      Calculate
      of Aluminum carbonate, Al(CO3 ) 3
                                         
                                          100 %  13.87 % Al
                    28.98 g
        
         28.98  3 *12.01  9 *16.00 g   
                                         

                                         
                                          100 %  17.24 % C
                  3 *12.01 g
        
         28.98  3 *12.01  9 *16.00 g   
                                         

                                         
                                          100 %  68.90 % O
                  9 *16.00 g
        
         28.98  3 *12.01  9 *16.00 g   
                                         

51
             The Empirical Formula
      The  lowest whole number ratio of
       elements in a compound.
      The molecular formula = the actual
       ratio of elements in a compound.
      The two can be the same.
      CH2 is an empirical formula
      C2H4 is a molecular formula
      C3H6 is a molecular formula

52    H2O is both empirical & molecular
          Calculating Empirical
 Just   find the lowest whole number
   ratio
  C6H12O6
  CH4N
  It is not just the ratio of atoms, it is
   also the ratio of moles of atoms.
  In 1 mole of CO2 there is 1 mole of
   carbon and 2 moles of oxygen.
  In one molecule of CO2 there is 1
53 atom of C and 2 atoms of O.
            Calculating Empirical
      We  can get a ratio from the
       percent composition.
      Assume you have a 100 g.
      The percentages become grams.
      Convert grams to moles.
      Find lowest whole number ratio
       by dividing by the smallest.

54
                   Example
      Calculate the empirical formula of a
       compound composed of 38.67 % C,
       16.22 % H, and 45.11 %N.
      Assume 100 g so
      38.67 g C x 1mol C      = 3.220 mole C
                    12.01 gC
      16.22 g H x 1mol H      = 16.09 mole H
                    1.01 gH
      45.11 g N x 1mol N = 3.219 mole N
55                  14.01 gN
                   Example
      The ratio is 3.220 mol C = 1 mol C
                    3.219 molN    1 mol N
      The ratio is 16.09 mol H = 5 mol H
                    3.219 molN    1 mol N
      = C1H5N1




56
                   Example
     A  compound is 43.64 % P and 56.36 %
       O. What is the empirical formula?
      Assume 100 g so
      43.64 g P x 1mol P      = 1.409 mole P
                     30.97 g P
      56.36 g O x 1mol O      = 3.523 mole O
                     16.00 g O
      The ratio is 1.409 mol P = 2 mol P
                    3.523 mol O 5 mol O
      = P2O5
57
                     Example
      Caffeine is 49.48% C, 5.15% H, 28.87% N and
       16.49% O. What is its empirical formula?
      Assume 100 g so
      49.48 g C x 1mol C = 4.120 mole C
                12.01 gC
      5.15 g H x 1mol H     = 5.010 mole H
                1.01 gH
      28.87 g N x 1mol N = 2.061 mole N
                14.01 gN
      16.49 g N x 1mol N = 1.031 mole O
                16.00 gN
58
                    Example
      The ratio is 4.120 mol C = 4 mol C
                        1.031 mol O 1 mol O
      The ratio is 5.010 mol H = 5 mol H
                        1.031 mol O 1 mol O
      The ratio is 2.061 mol N = 2 mol N
                        1.031 mol O 1 mol O
      = C4 H 5 N 2 O 1



59
          Empirical to molecular
      Since the empirical formula is the
       lowest ratio, the actual molecule
       would weigh more.
      By a whole number multiple.
      Divide the actual molar mass by the
       empirical formula mass.
      Caffeine has a molar mass of 194 g.
       what is its molecular formula?
60
             Empirical to molecular
      Caffeine has a molar mass of 194 g.
      what is its molecular formula?
      C4H5N2O1 empirical formula mass is

       = 4*12.01 + 5*1.01 + 2*14.01 + 16.00
       = 97.11 g/mol
      Molarmass/empirical formula mass =
      194/97.11 = 2.00
      So,   actual formula is C8H10N4O2
61
                Example
     Acompound is known to be
     composed of 71.65 % Cl, 24.27%
     C and 4.07% H. Its molar mass is
     known (from gas density) to be
     98.96 g. What is its molecular
     formula?



62
                   Example
      Assume   100 g so
      71.65 g Cl x 1mol Cl = 2.021 mole Cl
                    35.45 g Cl
      24.27 g C x 1mol C      = 2.021 mole C
                    12.01 g C
      4.07 g H x 1mol H       = 4.030 mole H
                    1.01 g H


63
                         Example
      The ratio is 2.021 mol C = 1 mol C
                    2.021 mol Cl 1 mol Cl
      The ratio is 4.030 mol H = 2 mol H
                    2.021 mol Cl 1 mol Cl
      = C1H2Cl1
      C1H2Cl1 empirical formula mass is

         = 12.01 + 4*1.01 + 35.45
         = 51.50 g/mol
      Molar mass/empirical formula mass =
       98.96/51.5 = 1.92 = 2
      So, actual formula is C2H4Cl2

64

				
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