Transcript Ions - Clark College

Ions
Ions: A charged particle formed when a neutral
atom or group of atoms gain or lose one or
more electrons.
Example
Na  Na+ + eF + e-  F1
Cations
Mg  Mg2+ + 2eAl  Al3+ + 3eCation: A positively-charged ion. One or more
electrons are lost from a neutral atom
 oxidation
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Anions
I + e -  IO + 2e-  O2S + 2e-  S2Anion: A negatively-charged ion. Electrons are
gained by a neutral atom
 reduction
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Ions

Ion charges can be predicted from the Periodic Table

Main Group metal (IA-IVA) ion-charges correspond to group number
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Sodium (Na) in IA  +1
Exceptions
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Fluorine (F) in VIIA  (7-8) = -1
Boron is exception  -3 not -5
All this has to do with electron configuration
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We’ll talk about non-main group metal ion-charges later
Main Group non-metal (IIIA-VIIIA) ion-charges correspond to (group# - 8)
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Tl, Sn, Pb, Sb, and Bi
More anon
See website as well:
http://web.clark.edu/aaliabadi/CHEM131ions%20to%20memorize.htm
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Compounds That Contain
Ions


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Require metal and non-metal
Form ionic bonds
Called an ionic compound
Characteristic Properties
1. Very high melting points
2. Conduct an electric current when melted or
when dissolved in water
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Ionic compound

The number of cations and anions must have a net charge of
zero.
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Compounds that Contain
Ions
Writing Formulas for Ionic Compounds
Give the formulas for the compounds that contain the following
pairs of ions:
(a) K and I
(b) Mg and N
(c) Al and O
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Types of Ionic Compounds
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Type I Compounds: The metal present forms
only one type of cation.
Examples: the main group metals
Type II Compounds: The metal present can
form two or more cations that have different
charges or oxidation states
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Oxidation state = ionic charge, if any, on species
Examples include Cr2+, Cr3+, Cu+, Cu2+, etc.
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Naming Ionic Compounds
1. The cation is always named first and the
anion second.
2. The cation takes its name from the name of
the element.
3. The anion is named by taking the first part of
the element name and adding –ide.
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Naming Ionic Compounds
Name the following Type I
compounds
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NaCl
RaBr2
Rb2O
AlI3
K3N
Cs4Si
Give the chemical formula for the
following Type I compounds
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Strontium phosphide
Calcium fluoride
Beryllium carbide
Lithium hydride
Barium sulfide
Magnesium telluride
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Naming Ionic Compounds
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Type II compounds need to be identified by a Roman numeral
 (I), (IV), etc.
Represents oxidation state of cation
Not how many cations are present in compound!
Example: NaCl  sodium (I) chloride is INCORRECT
Example: SnCl4  tin (IV) chloride is correct
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Naming Ionic Compounds
Type II Ionic Compounds
FeCl2 and FeCl3
PbO and PbO2
MnS and Mn2S7
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A Mixed Bag
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PbBr2 and PbBr4
Aluminum arsenide
FeS and Fe2S3
Thallium (III) boride
Mercury (II) carbide
Na2S
CoCl3
Cerium (IV) phosphide
ScF3
Gold (I) selenide
Vanadium (V) telluride
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Naming Compounds that
Contain Polyatomic Ions
Polyatomic Ion: An ion that contains more than one atom. They
are charged entities composed of several atoms bound
together.
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Consult my website for the list that must be
memorized:
http://web.clark.edu/aaliabadi/CHEM131ions
%20to%20memorize.htm
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Trends
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Sulfide, sulfite, sulfate
Nitride, nitrite, nitrate
Phosphide, phosphite, phosphate
Chloride, hypochlorite, chlorite, chlorate,
perchlorate
Parenthesis required if more than one
polyatomic ion present
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Ca(IO3)2 is correct
Ca(I)2 is INCORRECT
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Naming Compounds that
Contain Polyatomic Ions
Name or provide the chemical formula for each of the following
compounds:
(a) Ca(OH)2
(e) Co(ClO4)2
(b) Sodium phosphate
(f) platinum (IV) bicarbonate
(c) KMnO4
(g) Cu(NO2)2
(d) Ammonium carbonate
(h) nickel (III) sulfite
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Naming Compounds that
Contain Polyatomic Ions

Name or provide each of the following compounds:
(a) calcium carbonate
(e) MoO
(b) BaSO4
(f) Iridium (VII) acetate
(c) CsClO4
(g) ZnO2
(d) Zirconium (IV) bisulfite
(h) lithium cyanide
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Naming Acids
Acids: A substance that yields hydrogen ions
(protons, H+) when dissolved in water.

HCl(aq)  H+(aq) + Cl-(aq)

H3PO4(aq)  3H+(aq) + PO43-(aq)
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Rules for naming acids
If the formula does not contain oxygen the prefix of the acid
is hydro and the suffix –ic is attached to the root name for
the element.
Ex: HCl = hydrochloric acid, H2S = hydrosulfuric acid
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When the anion contains oxygen, the acid name is formed
from the anion name. The suffix –ic or –ous is added.
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When the anion ends in –ate, the suffix –ic is used.
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H2CO3 = carbonic acid
When the anion ends in –ite, the suffix –ous is used.
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H2SO3 = sulfurous acid
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More
Anion
ClO4- ______________
Acid
HClO4 _______________
ClO3- ______________
HClO3 _______________
ClO2- ______________
HClO2 _______________
ClO- _______________
HClO ________________
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Naming Compounds that Contain
Only Nonmetals: Type III
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Type III compounds contain only nonmetals.
Form covalent bonds
 share electrons
Rules for Naming Type III Binary Compounds
1. The first element in the formula is named first, and the full
element name is used.
2. The second element is named as though it were an anion.
3. Prefixes are used to denote the numbers of atoms present.
4. The prefix mono- is never used for naming the first element.
5. Drop the “a” when it is followed by an “o”
 Tetraoxide should be tetroxide
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Naming Compounds that Contain
Only Nonmetals: Type III
Prefixes Used to Indicate Numbers in Chemical Names
Prefix
Number Indicated
mono1
di2
tri3
tetra4
penta5
hexa6
hepta7
octa8
nona9
deca10
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Practice
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CCl4
Silicon dioxide
NO2
Sulfur trioxide
P2O5
Iodine pentafluoride
Dinitrogen tetroxide
SeI2
Xenon hexafluoride
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The Name Game
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Extra credit opportunity
Mix-and-match
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Remember the mole?
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Mole = amt that contains as many “things” as
there are atoms of 12 g of C-12
1 mole = 6.022 x 1023 particles
Molar mass (MM) = mass in grams per 1 mole
of particle (g/mol)
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Molecular Mass
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Summation of molar masses from Periodic
Table based on molecular formula
NaCl
I2
V2O5
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Solutions
g
NaCl : 22.98977
g
I 2 : 2 126.9045
mol
g
 35.4527
mol
g
 58.4425
mol
g
 253.8090
mol
mol
V2O5 : 2  50.9415 g
 5 15.9994 g
 181.8800 g
mol
mol
mol
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Molar Mass
Example
Calculate the mass of 30.0 moles of polyvinyl chloride (PVC),
C2H3Cl.
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Solution
C2 H 3Cl : 2 12.011 g
mol
 3 1.00794 g
mol
 35.4527 g
mol
 62.499 g
62.499 g
30.0mol 
 1874.97  1870 g
mol
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mol
Molar Mass
Example
A sample of Na2SO4.with a mass of 300.0 grams
represents what number of moles of Na2SO4?
Example
Calculate the number of grams of caffeine,
C8H10N4O2, in 8.13 x 1023 molecules.
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Solution
Na2 SO4 : 2  22.98977 g
300.0 g 
mol
142.043 g
mol
 32.066 g
mol
 4 15.9994 g
mol
 142.043 g
mol
 2.112mol
mol
C8 H10 N 4O2 : 8 12.011 g
mol
 10 1.00794 g
 4 14.0067 g
 2  15.9994 g
mol
mol
1mol
194.193g
8.13 1013 molecules 

 2.62 108 g
23
6.022 10 molecules
mol
mol
 194.193 g
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mol
Percent Composition
of Compounds: NaCl
mass percent for a given element = (
part
) 100%
whole
22.99 g
mol
35.45g
Cl =
mol
Na =
Thus, NaCl =
22.99 g 35.45g 58.44g


mol
mol
mol
22.99g 58.44g
Na: [(1 
)
] 100% = 39.34%
mol
mol
35.45g 58.44g
Cl: [(1 
)
] 100%  60.66%
mol
mol
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Percent Composition
of Compounds
Example
Compute the mass percent of each element in
sodium sulfide, Na2S.
Example
Compute the mass percent of each element in
nitric acid, HNO3(aq).
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Solution
Na : 2  22.98977 g
S : 32.066 g
mol
mol
 32.066 g
45.97954 g
78.046 g
32.066 g
And, S =
mol
mol
Na2 S : 45.97954 g
Thus, Na =
 45.97954 g
78.046 g
mol
 78.046 g
mol
mol 100%  58.913%
mol
mol 100%  41.086%
mol
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Solution
H :1.00794 g
mol
N :14.0067 g
mol
O : 3 15.9994 g
 47.9982 g
mol
mol
HNO3 :1.00794 g
 14.0067 g
 47.9982 g
 63.0128 g
mol
mol
mol
mol
1.00794 g
mol 100%  1.59958%
Thus, H =
63.0128 g
mol
14.0067 g
mol 100%  22.2283%
And, N =
63.0128 g
mol
47.9982 g
mol 100%  76.1721%
And, O =
63.0128 g
mol
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Empirical Formulas
Empirical Formula: or the simplest formula; the smallest
whole-number ratio of the atoms present.
Molecular Formula: the actual formula of a compound. It gives
the composition of the molecules that are present.
Empirical Formula
CH
CH2O
H2O
Molecular Formula
C6H6
C6H12O6
H2O
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Calculation of Empirical
Formulas
Steps for Determining the Empirical Formula of a Compound
1. Obtain the mass of each element present (in grams).
2. Determine the number of moles of each type of atom present.
3. Divide the number of moles of each element by the smallest
number of moles to convert the smallest number to 1. If all
of the numbers are integers (whole numbers), these are the
subscripts in the empirical formula. If one or more of these
numbers are not integers, go to step 4.
4. Multiply the numbers derived in step 3 by the smallest integer
that will convert all of them to whole numbers. This set of
whole numbers represents the subscripts in the empirical
formula.
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Calculation of Empirical
Formulas
Experiment:
Suppose we weigh out 6.50 grams of Cr. We decide to heat this
Cr in the air so that the Cr can react with O to form CrxOy .
After the sample cools, we weigh it again and find its mass to
be 9.50 grams. How do we find the mass of oxygen?
What is the empirical formula of this compound?
Let’s work on this together.
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Calculation of Empirical
Formulas
Example
Nylon-6 consists of 63.68% C, 12.38% N, 9.80% H, and
14.14%O.
Calculate the empirical formula for Nylon-6.
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Solution
mol
 5.302mol  0.8838mol=5.999  6
12.011g
mol
N: 12.38g 
 0.8839mol  0.8838mol=1.000  1
14.0067g
mol
H: 9.80g 
 9.72mol  0.8838mol=11.00  11
1.00794g
mol
O: 14.14g 
 0.8838mol  0.8838mol=1.000  1
15.9994g
 C6 H11 NO
C: 63.68g 
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Calculation of
Molecular Formulas
We need to know the empirical formula and molar mass of the
molecular compound.
Molecular Formula = (empirical formula)n
where n is a small whole number.
Molecular Formula = n x Empirical Formula
Molar Mass = n x Formula Weight
 n = Molar Mass/Formula Weight
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Calculation of
Molecular Formulas
Example
A compound shows the following percentage compostion:
71.65% Cl
24.27% C 4.07% H
The molar mass is known to be 98.96 g/mol. Determine the
empirical formula and the molecular formula for this
compound.
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Solution
71.65% Cl
24.27% C
4.07% H
The molar mass is known to be 98.96 g/mol. Determine the empirical formula and the molecular formula
mol
 2.021mol  2.021mol  1
35.4527g
mol
C: 24.27g 
 2.021mol  2.021mol  1
12.011g
mol
H: 4.07g 
 4.038mol  2.021mol  1.998  2
1.00794g
Cl: 71.65g 
empirical formula = CH 2Cl
empirical formula mass = 12.011 g
mol
 2 1.00794 g
mol
 35.4527 g
mol
 49.476 g
mol
98.96 g
molar mass
mol  2

empirical formula mass 49.476 g
mol
molecular fomrula = C2 H 4 Cl2
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Calculation of
Molecular Formulas
Example
Vitamin C consists of 40.92% C, 4.58% H, and 54.50% O on a
mass basis, and has a molar mass of 176.12 g/mol. Determine
the molecular formula of the compound.
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Solution
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