# 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:
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
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
  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
 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

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?
–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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