Transcript TOPIC III: THE MOLE

TOPIC III: THE MOLE
LECTURE SLIDES
• Definition and Use
• Avogadro’s Number
• Conversion Problems
• Empirical Formula Calculations
Kotz & Treichel, Chapter 3, Sections 3.6-3.8
From PARTICULATE (“too small to touch”)
to MACROSCOPIC (amounts we can
handle):
THE MOLE
Many different items we
encounter in our daily lives
come packaged in set amounts
described by various
“counting terms”.
Let’s consider a few of them:
•shoes and socks and earrings come in pairs
(2),
• eggs come in dozens (12),
•pencils are wholesaled by the gross (144),
•donuts and sweet corn are often sold as “the
baker’s dozen” or “the farmer’s dozen” (13),
•and diet pop and beer by the 6- pack or case
(24).....
Chemists deal in atoms, molecules and ions,
which need to be counted and measured as well.
BUT:
The mass of one atom of the 19F isotope is
3.156X10-23 g.
The radius of a nucleus is about .001 pm and the
radius of an atom about 100 pm. (1012 picometers
or 1,000,000,000,000 pm = 1 m)
THEREFORE......
Chemists need their own unit for
counting and weighing amounts of
substances which come in particle size
too tiny to be seen or weighed on any
balance.
For convenience in describing amounts
of atoms, molecules, and ions , chemists
have a unique unit of measure,
THE MOLE
The MOLE
•The chemist’s counting number
•Comes from the Latin meaning “whole heap
or pile of”
•SI base unit for measuring amount of
substance
•Defined as the number of atoms in exactly 12
grams of 12C, 6.022 X 1023
1 Mole always contains Avogadro’s number of
particles or units:
1 Mole = Avogadro’s number of particles
= 6.022136736 X 1023 particles
= 602,213,673,600,000,000,000,000 particles
If one used A’s number to describe macroscopic
objects, one would be overwhelmed:
I mole of green peas would cover the entire United States
to a depth of 3 miles!
Grams, amu’s, What’s the Difference?
One mole is defined as the number of particles
in exactly 12 g of the 12C isotope of carbon.
Carbon was used as the standard for the amu scale,
where the mass of one atom of 12C was defined as 12 amu.
The mole answers the question: “How many atoms
would you have if you took the amu scale (which describes
mass of one atom) and use it as grams instead?
So, How to Get a Mole:
•We consult the periodic table, obtain the atomic
mass of an element in amu’s, the relative
mass of one atom.....
•We weigh this amount out in grams.....
•We now have one mole of atoms, A’s number,
6.02 X 10 23 atoms, a convenient “package of
atoms”.....
•We have gone into the chemist’s counting
system and can deliver not a dozen eggs
but a mole of atoms....
This system works because of the relative nature
of the atomic mass unit scale, in which all atoms
were assigned a mass relative to 12C, the mole
standard:
“One mole is the number of atoms in exactly 12
g of 12C”
The mole “pile or heap” of atoms for each
element will weigh more or less than the mole
“pile” for carbon, depending on whether the
individual atoms weigh more or less than
carbon.
If one mole of carbon atoms weighs 12.0 g, then
one mole of oxygen atoms, which weighs 1.33
times more than carbon, would be:
1.33 X 12.0 g =16.0 g = 1 mol O
Since H atom is 1/12 the mass of a carbon atom,
the matching pile of hydrogen atoms would be
1/12 X 12.0 g = 1.00 g = 1 mol H
The “Molar Mass, M”
•For any element, the molar mass, M, is the mass
in grams of a mole of atoms, “#g/mol”
•M , molar mass, is NUMERICALLY equal to the mass
of one atom in amu’s as given on your PT.
•If, however, one weighs out the molar mass, one has
6.022 x 10 23 atoms every time
MOLES of ATOMS: the MOLAR MASS, M
1 atom
1 mole, A’s # atoms
Li, 6.941 amu
Li, 6.941 g
Pb, 207.2 amu
Pb, 207.2 g
Zn, 65.39 amu
Zn, 65.39 g
24
Cr
51.9961
atomic mass,
one atom,
atomic mass units,
relative to C
molar mass, 6.022 x 1023 atoms, in grams
ONE MOLE
M,
molar mass
in grams
A's NUMBER
23
6.02 x 10 UNITS
The molar mass, “g/mol”, like density, “g/cm3”, is a
convenient conversion factor:
For any element:
1mole = atomic weight, grams = 6.022 X 1023 atoms
Using this knowledge, the chemist can interconvert
grams, moles, and atoms of any element.
Suppose you weighed out 35.89 g of aluminum metal.
How many moles and how many atoms of aluminum
would be contained in this sample?
Question: 35.89 g Al = ? mol Al = ? atoms Al
Relationships: 1 mol Al = 26.98 g Al
= 6.022 X 1023 atoms Al
Setup and Solve: g ---> mol
#1
35.89 g Al
=_ ?____mol Al
g
#1
35.89 g Al
1 mol Al
=_ ?____mol Al = 1.330 mol Al
26.98 g Al
ans.
g
35.89 g Al
#2
mol
mol
atoms
=_ ?____atoms Al
g --------> mol ---------> atoms
35.89 g Al
#2
ans
6.022 x 1023 atoms Al
1 mol Al
=_ ?____atoms Al
26.98 g Al
=
1 mol Al
8.011 x 1023
atoms Al
Group Work 3.1
Suppose you weighed out 15.00 g each of V and
Mo metal. How many moles and how many atoms
of each do you now have?
15.00 g V = ? Mol =? Atoms
15.00 g Mo= ? Mol =? Atoms
What would 9.00 x 1024 atoms of mercury weigh
in grams?
Question: 9.00 x 1024 atoms Hg = ? g Hg
Relationships: 1 mol Hg = 200.59 g Hg
= 6.022 X 1023 atoms Hg
Setup and Solve: atoms -----> mol -----> g
9.00 x 1024 atoms Hg
=
g Hg
mol
atoms
9.00 x 1024 atoms Hg
1 mol Hg
6.022 x 1023 atoms Hg
=
2998
g
200.59 g Hg
1 mol Hg
g Hg = 2.998 X 103 = 3.00 x 103 g Hg
GROUP WORK 3.2
Mercury is a liquid metal with a density of 13.534
g/cm3. If you measured out 75.0 mL of Hg into a
graduated cylinder, how many atoms of Hg would be
in the sample?
Question: 75.0 mL Hg = ? Atoms Hg
Relationships:
13.534 g Hg = 1cm3 or mL Hg
200.59 g Hg = 1 mol Hg
1 mol Hg = 6.022 x 1023 atoms Hg
Setup and solve: mL---> g ---> mol --->atoms
75.0 mL Hg
= ? atoms Hg
Molecules, Compounds, and the Mole
Let us now extend the use of molar mass, M, to
include all particles chemists need to measure:
not just atoms but also especially ions and
molecules....
The basic principle is this: whenever you weigh
out the“formula weight” of any substance or
species in grams, you have A’s number of
particles of that species, and the molar mass of
that species...
Molar Mass of Molecules
The formula of any molecule describes the number of
atoms making up one unit of that molecule:
Br2 The diatomic bromine molecule, as bromine
is found in nature: the formula tells us that 2 atoms
of bromine are contained in every molecule.
By extension, 2 moles of bromine atoms are contained
in every 1 mole of bromine molecules. The
calculation of the molar mass of molecular bromine
then looks like this:
The atomic weight of Br, from the PT, is 79.904 amu’s.
Therefore:
2 moles of Br = 2 X 79.904 g = 159.808 g
And the molar mass, M, of Br2 is 159.808 g/mol
Now let’s try the molar mass of CH3CH2OH, ethyl
alcohol:
Molar Mass, M, CH3CH2OH
Element
# of atoms M, g/mol
C
2
12.01
H
6
1.008
O
1
16.00
Total
M, CH3CH2OH, 46.07 g/mol
total
24.02
6.048
16.00
46.068
Molar Mass of Ionic Compounds
The formula of an ionic compound indicates the
simplest ratio of ions present in any sample of the
compound. It is this “formula unit” that we use for
calculating the molar mass.
Actually, we needn’t ask what kind of compound we
are getting the M for; we simply calculate for all
atoms found in the formula of any species!
Cl2 (2 Cl)
Fe(CN)2 (1Fe 2C 2N)
(NH4)2CO3 (2N 8H 1C 3O)
M, CH3CH2OH, 46.07 g/mol, use in
problems:
Given a mass, or volume and density,
solve for:
a) moles of compound or individual atoms
b) grams of individual atoms
c) number of molecules or atoms
How many moles of ethyl alcohol are contained in a
sample that weighs 33.95 g? (CH3CH2OH, 46.07 g/mol).
Question:
33.95 g CH3CH2OH = ? mol CH3CH2OH
Relationship:
46.07 g CH3CH2OH = 1 mol
Setup and Solve: ( g ---> mol)
33.95 g CH3CH2OH
= ? mol CH3CH2OH
g ----------> mol
33.95 g CH3CH2OH
1 mol
46.07 g
= ? mol CH3CH2OH
= .7369 mol CH3CH2OH
H
H
H
C
H
C
H
O
H
1 molecule CH3CH2OH,
contains:
1 O atom
2 C atoms
6 H atoms
1 mole CH3CH2OH,
6.022x1023 molecules of CH3CH2OH
contains:
2 mole C atoms
6 moles H atoms
1 mole O atoms
23
6 X 6.022x1023
Hydrogen atoms
1 X 6.022x1023
Oxygen atoms
2 X 6.022x10
Carbon atoms
How many moles of hydrogen are contained in
33.95 g CH3CH2OH?
Question:
33.95 g CH3CH2OH = ? mol H
Relationship:
46.07 g CH3CH2OH = 1 mol CH3CH2OH
1 mol CH3CH2OH = 6 mol H
Setup and Solve: ( g ---> mol CH3CH2OH ---> mol H)
33.95 g CH3CH2OH
= ? mol H
g
33.95 g CH3CH2OH
mol alcohol
mol H
1 mol CH3CH2OH
6 mol H
46.07 g CH3CH2OH
1 mol CH3CH2OH
= 4.422 mol H
= ? mol H
How many grams of hydrogen are contained in
33.95 g CH3CH2OH?
Question:
33.95 g CH3CH2OH = ? g H
Relationship:
46.07 g CH3CH2OH = 1 mol CH3CH2OH
1 mol CH3CH2OH = 6 mol H
1 mol H = 1.008 g H
Setup and Solve:
( g CH3CH2OH ---> mol CH3CH2OH ---> mol H -----> g H)
33.95 g CH3CH2OH
= ? gH
g alcohol
33.95 g CH3CH2OH
= ? g H
mol alcohol
mol H
1 mol CH3CH2OH
6 mol H
46.07 g CH3CH2OH
1 mol CH3CH2OH
= 4.457 g H
gH
1.008 g H
1 mol H
Group Work 3.3
How many grams of carbon are contained in 33.95 g
CH3CH2OH?
Question:
33.95 g CH3CH2OH = ? g C
Relationship:
46.07 g CH3CH2OH = 1 mol CH3CH2OH
1 mol CH3CH2OH = 2 mol C
1 mol C = 12.01 g C
Setup and Solve:
( g CH3CH2OH ---> mol CH3CH2OH ---> mol C -----> g C)
How many atoms of carbon are contained in
33.95 g CH3CH2OH?
Question:
33.95 g CH3CH2OH = ? atoms C
Relationship:
46.07 g CH3CH2OH = 1 mol CH3CH2OH
1 mol CH3CH2OH = 2 mol C
1 mol C = 6.02 x 1023 atoms C
Setup and Solve:
( g CH3CH2OH ---> mol CH3CH2OH ---> mol C
-----> atoms C)
g alcohol
mol alcohol
33.95 g CH3CH2OH
mol C
atoms C
= ? atoms C
g alcohol
33.95 g CH3CH2OH
mol C
mol alcohol
1 mol CH3CH2OH
2 mol C
46.07 g CH3CH2OH
1 mol CH3CH2OH
= ? atoms C = 8.873 x 1023 atoms C
atoms C
6.02x1023 atoms C
1 mol C
GROUP WORK 3.4
What mass of ethyl alcohol, CH3CH2OH, would contain
2.06 X 1024 atoms of carbon?
Question: 2.06 X 1024 atoms C = ? g CH3CH2OH
Relationship: 6.02 x 1023 atoms C = 1 mol C
2 mol C = 1 mol CH3CH2OH
1 mol CH3CH2OH = 46.07 g CH3CH2OH
Setup and Solve:
(atoms C-----> mol C ---> mol CH3CH2OH
---> g CH3CH2OH )
New TOPIC:
% Composition from Formula of the
Compound
The formula of a compound can be used to determine
the mass % of each element by using the molar
relationships we have learned...
Consider the following:
What is the % by mass of Carbon, Hydrogen and Oxygen
in a sample of ethyl alcohol, CH3CH2OH?
Mass % of Elements in a Compound
This is done using the formula weight calculations and
the approach:
% by mass = total mass of one element (the part) X 100 %
total mass of compound (the whole)
based on one
mole of the
compound
% Composition of Ethyl Alcohol, CH3CH2OH
Molar Mass Calculation:
2 C = 2 X 12.011 = 24.022
6 H = 6 X 1.008 = 6.048
1 O = 1 X 15.999 = 15.999
46.069
gC
gH
gO
g/mol CH3CH2OH
%C =
24.022 g C
X 100 = 52.144% C
46.069 g CH3CH2OH
%H =
6.048 g H
X 100 = 13.13% H
46.069 g CH3CH2OH
%O =
15.999 g O
X 100 = 34.728% O
46.069 g CH3CH2OH
CHECKING.....
52.144% C
13.13 % H
34.728% O
100.002 % CH3CH2OH
= 100.00% (2 digits allowed after decimal)
Formula from % Composition
If one can go from formula to % composition,
one should be able to go from % composition to the
formula of the compound.
This is quite true (almost):
We can take % composition back to the “empirical
formula”, which describes the simplest mole ratio of
atoms in the formula... That is not always the same
thing...
Empirical Vs Molecular Formulas
Name
Molecular
Formula
Empirical “n”
Formula
Acetylene
C2H2
(CH)n
2
Benzene
C6H6
(CH)n
4
Vinyl
Acetylene
C4H4
(CH)n
6
For “organic” or “molecular carbon containing
compounds”, there exist a list of compounds
which share almost every conceivable empirical
formula.
To determine which compound one has from
analytical data, one needs the molar mass as
well, as we will see...
For most ionic compounds and simple molecular
compounds, the empirical and molecular formulas
are identical.
Empirical Formula from % Composition:
Let’s take our ethyl alcohol compound back to its
formula from its percent composition.The trick is to
use 100 grams of sample whenever one is
calculating from mass %. Then we have:
52.144% C
13.13 % H
34.728% O
= 52.14 g carbon
= 13.13 g hydrogen
= 34.73 g Oxygen
Our procedure will be:
% element ---> g element ---> mol element ---> simplest
mole ratio
Method of choice: make a chart:
element
grams
List each
Use based
separately on 100 g
when %
given
Exp data
if
provided
Molar
mass
#moles
Use mass
of 1 mol
of atoms
#g X 1 mol Divide all
#g
moles by
smallest #
(calculate
of moles
and insert)
“atomic
weight
only”
Simplest
mole ratio
Chart out your information in this fashion to determine
formula:
grams
Molar
mass
Moles:
#g/ M
Simplest
mole ratio
C
52.14 g
12.01
4.341
4.341/2.171 =2.000
H
13.13 g
1.008
13.03
13.03/2.171= 6.002
O
34.73 g
16.00
2.171
2.171/2.171= 1.000
Moles: 52.14 g C X 1 mol C = 4.341 mol C
12.01 g C
Your empirical formula is the simplest ratio of moles
of the elements in the formula, for which you have
determined that:
For every 1.00 mole of O atoms, you have 6.002 mole
of H atoms and 2.000 mole of C atoms
Empirical formula:
C2H6O
GROUP WORK 3.5
Cumene is an organic hydrocarbon containing
only C and H. It is 89.94% C and 10.06% H. What is its
empirical formula?
Empirical Formula using Experimental Data
A new compound weighing 0.678 g, containing xenon
and fluorine, was made from a mixture of the gases in
sunlight. If the xenon showed a mass of 0.526 g, what
is the empirical formula of the compound?
Note:
0.678 g compound - 0.526 g Xe = .152 g F2
Although the element itself is diatomic, when
combined into a compound it exists as individual
atoms, and the atomic weight of F, 19.0 g/mol, is
used for formula calculation, not 38.0 g/mol F2.
Grams
M
Moles: #g/ M
Simplest mole ratio
Xe .526
131.29
.00401
.00401/.00401 = 1
F
19.00
.00800
.00800/.00401 = 2
.152
EMPIRICAL FORMULA: XeF2
Molecular and Empirical Formula
Nicotine, a poisonous compound found in tobacco
leaves, is 74.0% C, 8.65% H, and 17.35% N. Its molar
mass is 162 g/mol.
What are the empirical and molecular formulas
of this compound?
Note: When the molar mass is included in the problem,
the exact molecular formula can be determined...
grams Molar mass Moles
C 74.0 g
H 8.65 g
N 17.35 g
12.01
1.008
14.01
6.16
8.58
1.238
Simplest mole ratio
6.16/1.238 = 4.98
8.58/1.238 = 6.93
1.238/1.238= 1.000
Empirical Formula: C5H7N
In order to obtain the molecular formula, you must
divide the molar mass (mass of molecule) by the
empirical formula mass (the mass of the simplest ratio
of atoms...)
Empirical Formula Mass:
5C = 5 X 12.01 = 60.05
7H = 7 X 1.008 = 7.056
1N = 1 X 14.01 = 14.01
81.12 g/mol
Molar mass
= 162
= 2.00 = “n”
Emp. Form. mass
81.12
(C5H7N)n
= (C5H7N)2 :
C10H14N2
Hydrated Compounds
Crystalline ionic solids (salts!) are frequently found in
nature or are produced from aqueous solutions with a
specific number of water molecules associated with
each set of formula ions:
CuSO4. 5H2O
NiCl2.6H2O
CaSO4.2H2O
Frequently the color of the salt depends on the
presence of these “waters of hydration.”
Hydrated Compounds
Crystalline ionic solids (salts!) are frequently found in
nature or are produced from aqueous solutions with a
specific number of water molecules associated with
each set of formula ions:
CuSO4. 5H2O
NiCl2.6H2O
CaSO4.2H2O
Frequently the color of the salt depends on the
presence of these “waters of hydration.”
The number of water molecules associated with a
particular salt is characteristic but not easy to
predict: therefore the value is determined
experimentally...
The hydrated salt is weighed, heated carefully to
drive off the water, and reweighed.
The mass of the water driven off is calculated,
converted to moles and compared to moles of
the parent, anhydrous salt to determine the
formula of the hydrate...
Naturally occurring hydrated copper(II) chloride
is called eriochalcite. If 0.235 g of CuCl2.xH2O is heated
to drive off the water, 0.185 g residue remains. What is
the value of x?
0.235 g hydrate - 0.185 g parent salt = 0.050 g H2O
1Cu= 1 X 63.55 = 63.55
2Cl = 2 X 35.45 =70.90
134.45 g/mol CuCl2
2H= 2 x 1.008= 2.016
1O= 1 x 16.00= 16.00
18.02 g/mol H2O
grams Molar mass Moles
Simplest mole ratio
CuCl2 0.185 g 134.45g/mol .00138 .00138/.00138= 1
H2O 0.050 g 18.02 g/mol .00278 .00278/.00138= 2
Formula of eriochalcite: CuCl2. 2H2O
Note: The correct answer will always be a whole
number, with one mole of parent compound and
(usually) several moles of water....
GROUP WORK 3.6
If 1.023 g of a hydrated compound, CuSO4. xH2O
shows a mass of 0.654 g when dehydrated, what is the
formula of the compound? (CuSO4, 159.6 g/mol;
H2O, 18.02 g/mol).