Transcript Chapter One

The Mole
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1 atom or 1 molecule is a very small entity not
convenient to operate with
The masses we usually encounter in chemical
experiments vary from milligrams to kilograms
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Just like one dozen = 12 things
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One mole = 6.022 x 1023 things
Avogadro’s number:
NA = 6.022 x 1023
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The Mole
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NA = 6.022 x 1023
Why 6.022 x 1023 ?
This is the number of carbon atoms found in
12 g of the carbon-12 isotope
Molar mass – mass of one mole of atoms,
molecules, ions, etc.
Numerically equal to the atomic or molecular
weight of the substance in grams:
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m (1 mole H2) = Mr(H2) =
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m (1 mole Fe) = Mr(Fe) =
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The Mole: Example 1
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Example: Calculate the mass of a single
Mg atom in grams to 3 significant figures.
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The Mole: Example 2
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Example: How many C6H14 molecules are
contained in 55 ml of hexane (d = 0.78 g/ml).
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The Mole: Example 3
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Example: Calculate the number of O atoms
in 26.5 g of lithium carbonate, Li2CO3.
Solution:
1) Calculate the molar mass of Li2CO3:
Mr(Li2CO3) = 73.89 g/mol
2) Calculate the amount of Li2CO3 in moles:
n (Li2CO3) = 0.359 mol
3) Calculate the amount of O atoms in moles:
n (O) = (0.359x3) mol = 1.077 mol
4) Multiply by the Avogadro’s number to obtain the answer:
N (O) = 1.077 mol x 6.022·1023 atom/mol =
= 6.50·1023 atoms
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Percent Composition and
Formulas of Compounds
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If the formula of a compound is
known, its chemical composition can be
expressed as the mass percent of
each element in the compound
(percent composition), and vice versa.
When solving this kind of problems, we
can use masses expressed in a.m.u. or
in g/mol
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Percent Composition: Example 1
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What is the percent composition of
each element in sodium chloride, NaCl?
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Percent Composition: Example 2
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Calculate the percent composition of iron(III)
sulfate, Fe2(SO4)3, to 3 significant figures
Solution:
1) Calculate the molar mass of Fe2(SO4)3:
Mr(Fe2(SO4)3) = 399.88 g/mol
2) Calculate the percent composition for the elements:
2  55.85 g/mol
2  Mr (Fe)
ω(Fe) =
 100% 
 100%  27.93%
Mr (Fe2 (SO4 )3 )
399.88 g/mol
3  32.07 g/mol
3  Mr (S)
ω(S) =
 100% 
 100%  24.06%
Mr (Fe2 (SO4 )3 )
399.88 g/mol
12  16.00 g/mol
12  Mr (O)
ω(O) =
 100% 
 100%  48.01%
Mr (Fe2 (SO4 )3 )
399.88 g/mol
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Simplest (Empirical) Formula
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The smallest whole-number ratio of
atoms present in the compound
Molecular formula, on the other hand,
indicates the actual number of atoms
present in a molecule of the compound
water
hydrogen peroxide
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Empirical Formula: Example 1
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The first high-temperature superconductor,
prepared by Bednorz and Müller in 1986,
contained 68.54% lanthanum, 15.68% copper,
and 15.79% oxygen by mass. What was the
simplest formula of this compound?
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Empirical Formula: Example 2
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A sample of a compound contains 6.541g of Co
and 2.368g of O. What is its empirical formula?
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Elemental Composition
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A combustion train for carbon-hydrogen analysis
 percent composition is determined experimentally
magnesium
perchlorate
sodium
hydroxide
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Empirical Formula: Example 3
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0.1172 g of a pure hydrocarbon was burned in a C-H
combustion train to produce 0.3509 g of CO2 and 0.1915 g
of H2O. Determine the masses of C and H in the sample,
the percentage of these elements in this hydrocarbon, and
the empirical formula of the compound.
Solution:
1) First we calculate the molar weights of CO2 and H2O:
Mr(CO2) = 44.01 g/mol;
Mr(H2O) = 18.02 g/mol
2) Now we can calculate the amount of both CO2 and H2O in moles:
0.3509g
m(CO2 )
n (CO2 ) =
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 0.007973mol  7.973 103 mol
Mr (CO2 ) 44.01 g/mol
n (H2O) =
0.1915g
m(H2O)
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 0.01063mol  1.063 102 mol
Mr (H2O) 18.02 g/mol
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Example 3 (continued)
3) Notice that 1 mole of CO2 molecules contains 1 mole of C atoms
and
1 mole of H2O molecules contain 2 moles of H atoms
Therefore,
n (C) = 7.973·10-3 mol; n (H) = 2 x 1.063·10-2 mol = 2.126·10-2 mol
4) Now we can answer the first question – determine the masses of
carbon and hydrogen in the starting compound – using the formula:
n=
m
Mr
m = n  Mr
m (C) = 7.973·10-3 mol x 12.01 g/mol = 0.09576 g
m (H) = 2.126·10-2 mol x 1.008 g/mol = 0.02143 g
Now, if you add these masses, they should give you the mass of
the starting compound: 0.1172 g
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Example 3 (continued)
5) Now we can easily find the mass percentage of the elements
in the hydrocarbon by dividing the mass of the element by
the total mass of the hydrocarbon and multiplying by 100%:
0.09576g
ω(C) =
 100%  81.71%
0.1172 g
ω(H) =
0.02143g
 100%  18.28%
0.1172 g
We can also find the percentage of hydrogen by subtracting
the percentage of carbon from 100%:
ω(H) = 100% - ω(C)  18.29%
The slight difference between two results in caused by
rounding errors.
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Example 3 (continued)
6)
C:
Finally, to find the empirical formula we use the amounts
of C and H in moles found in step 3 and divide them by
the smallest of both numbers:
7.973 10-3 mol
 1.000
-3
7.973 10 mol
H:
2.126  10-2 mol
 2.666
-3
7.973 10 mol
We know that the fraction like .666 is in fact 2/3.
Therefore, to convert both results to whole numbers,
we need to multiply them by 3:
C: 3.000
H: 7.998
These result tells us that the empirical formula for the
hydrocarbon in question is
C3H8
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Empirical Formula: Example 4
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0.1014 g sample of purified glucose was burned
in a C-H combustion train to produce 0.1486 g
of CO2 and 0.0609 g of H2O. An elemental
analysis showed that glucose contains only
carbon, hydrogen, and oxygen. Determine the
empirical formula of the compound.
This example is worked out in your textbook
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Molecular Formula
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Indicates the actual number of atoms present in a
molecule of the compound
To determine the molecular formula for a molecular
compound, both its empirical formula and its
molecular weight must be known
The molecular formula for a compound is either the
same as, or an integer of, the empirical formula
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Molecular Formula: Example
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A compound is found to contain 85.63% C
and 14.37% H by mass. In another
experiment its molar mass is found to be
56.1 g/mol. What is its molecular formula?
I’ll work on this example in the
beginning of the next lecture.
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Other Examples
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What mass of ammonium phosphate,
(NH4)3PO4, would contain 15.0 g of N?
You do it !
The answer is:
53.2 g
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Reading Assignment
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Go through Lecture 3 notes and learn
again the problems we solved in class
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Read Chapter 2 to the end
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Learn Key Terms (p. 82)
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Take a look at Lecture 4 notes
(already posted on the web)
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If you have time, read Chapter 3
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Homework #1 due by 9/13
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