Transcript From PARTICULATE to MACROSCOPIC:

LECTURE 6
CHM 151 ©slg
TOPICS:
1. Ionic Nomenclature, Completion
2. Naming Binary Molecular Compounds
3. “The Mole”: Introduction
Tues. Group Work:
•Cr3+
CO32-
•Ni2+
CN-
•Zn2+
NO2-
•Bi3+
H2PO4
•Pb2+
N3-
Cr2(CO3)3 Chromium(III)
carbonate
-
Ni(CN)2
Nickel(II)
cyanide
Zn(NO2)2
Zinc Nitrite
Bi(H2PO4)3 Bismuth(III)
dihydrogen phosphate
Pb3N2 Lead(II) Nitride
Polyatomic Anions of O, S (6A) Cr (6B)
Oxygen:
OH-
Remember also: O2-
Sulfur:
SO42SO32HSO4 HSO3 -
hydroxide
oxide
sulfate
sulfite
hydrogen sulfate
hydrogen sulfite
Remember also: S2-, sulfide
Chromium:
CrO42-
chromate
“BASES” and “SALTS”
The “hydroxides” and “oxides” of the metallic elements
are referred to as “bases”; all other ionic combinations
are referred to as “salts”
Bases:
Salts:
Mg(OH)2
MgCl2
MgHSO4 MgCO3
NaOH
Na3PO4
NaNO2
Na2SO3
CaO
Ca(NO3)2
Ca3N2
CaSO4
Fe(OH)3
Fe(CN)2 Fe(CH3CO2)3 Fe(H2PO4)2
Polyatomic Anions of Cl, Br, I (7A) Mn (7B)
Fluorine, F forms only the monatomic anion F-;
Bromine, Br and Iodine, I form the same ions as
chlorine, Cl:
Chlorine:
ClOClO2ClO3ClO4-
hypochlorite
chlorite
chlorate
perchlorate
Remember also: Cl-, Chloride
Manganese:
MnO4-
permanganate
SUMMARY, NAMING IONIC SALTS AND BASES
State name of the cation, then name of the anion.
Cations with a variable charge are named by
adding a Roman numeral
Monoatomic anions are named by changing their
elemental name to end in “ide”
Polyatomic anions (memorized) end in “ite” or
“ate”...
GROUP WORK
FORMULA
NH4ClO
Cd(BrO2)2
Co(IO3)3
Ba(ClO4)2
KMnO4
Ag2CrO4
NAME
FORMULA
NAME
NH4ClO
Ammonium hypochlorite
Cd(BrO2)2
Cadmium bromite
Co(IO3)3
Cobalt(III) Iodate
Ba(ClO4)2
KMnO4
Ag2CrO4
Barium perchlorate
Potassium permanganate
Silver chromate
Naming Binary Molecular Compounds
All compounds beginning with a metal or ammonium
are named as “ ionic compounds.”
Compounds containing only two elements (“binary”)
in which both elements in the formula are a
non-metal or metalloid are named in a different
manner...
The change in nomenclature reflects the fact that
these compounds are “molecular” and not “ionic” in
nature!
Binary Molecular Nomenclature Method:
• Name the first element in the formula
• Name the second element in the formula to end in
“ide”:
carbide, nitride, phosphide, oxide, sulfide,
fluoride, bromide, chloride, iodide
•Add numerical prefixes to indicate more than one
atom of the element in the formula:
di (2), tri (3), tetra (4), penta (5), hexa (6), hepta (7),
octa(8)
Typical Nomenclature
•
•
•
•
•
•
•
NO2
SF6
ICl5
N2O5
CBr4
SO3
P2O3
•
•
•
•
•
•
•
nitrogen dioxide
sulfur hexafluoride
iodine pentachloride
dinitrogen pentoxide
carbon tetrabromide
sulfur trioxide
diphosphorus trioxide
COMMON NAMES, BINARY MOLECULES
ENDING IN H
• BH3
• borane
• CH4
• methane*
• SiH4
• silane
• NH3
• ammonia*
• PH3
• phosphine
From PARTICULATE (“too small to touch”)
to MACROSCOPIC (amounts we can
handle):
THE MOLE
Kotz & Treichel, Chapter 3, 3.6-3.8
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 the same number of particles
of whatever is being described, that is, Avogadro’s
number of particles:
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 of particles 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!
How do we get “a mole” of a substance ???
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?
Mass, amu,
= 1 atom
Mass, g
= ? atoms = 6.02 X 10 23 atoms (1 mole)
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/mole”
•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
PT: amu’s / 1 atom
M, g /mol, A’s # atoms
Li, 6.941 amu/atom
Li, 6.941 g/ mol
Pb, 207.2 amu/atom
Pb, 207.2 g/ mol
Zn, 65.39 amu/atom
Zn, 65.39 g/ mol
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
8.008 x 1023
atoms Al
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
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
g
mL
75.0 mL Hg
atoms
mol
13.534 g Hg
1mol Hg
6.02 x1023 atoms Hg
1 mL Hg
200.59 g Hg
1 mol Hg
= 30.46 x 1023 atoms = 3.05 X 1024 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!
Molar Mass, M, NaCl
Element
Na
+
(as Na )
Cl
(as Cl-)
Total
# of atoms M, g/mol
Total
1
22.99
22.99
1
35.45
35.45
M, NaCl = 58.44 g/mol
58.44
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
How many moles of hydrogen atoms are contained in
33.95 g CH3CH2OH?
Question:
Relationship:
33.95 g CH3CH2OH = ? mol H
46.07 g CH3CH2OH = 1 mol
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
1 mol CH3CH2OH = 6 mol H
1 mol H = 1.008 g
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
END, Lecture 6
gH
1.008 g H
1 mol H