Download Naming Binary Molecular Compounds

Chapter 7 – Chemical Formulas
and Chemical Compounds
Taken from Modern Chemistry
written by Davis, Metcalfe, Williams
& Castka
Section 7.1 – Chemical Names and Formulas
HW – Notes on section 7.1 pgs 203-215
Objectives
Students will be able to :
• Explain the significance of a chemical formula
• Determine the formula of an ionic compound
formed between two given ions
•Name an ionic compound given its formula
•Using prefixes, name a binary molecular compound
from its formula.
•Write the formula of a binary molecular compound
given its name.
Section 7.1 – Chemical Names and Formulas
Significance of a chemical formula
The chemical formula indicates the relative
number of atoms of each element in a chemical
compound.
C12H22O11
Elements’ subscripts indicate
the number of atoms in the
compound.
Al2(SO4)3
Note how in this example
parenthesis surround a
polyatomic anion and the
subscript refers to the entire
unit
Section 7.1 – Chemical Names and Formulas
Monatomic Ions
Group 1
metals lose
one e- to
give 1+
cations.
Group 2
metals lose
two e- to
give 2+
cations.
By gaining or losing electrons many
main-group elements form ions with
stable configurations.
Ions formed from a single atom
are known as monatomic ions
Section 7.1 – Chemical Names and Formulas
Monatomic Ions
(continued)
The
nonmetals in
groups 15,
16 & 17 gain
e- to form
anions.
Not all main-group elements readily
form ions, C and Si form covalent
bonds where they share electrons.
Section 7.1 – Chemical Names and Formulas
Monatomic Ions
(continued)
Elements which give up 1 or more eand take a positive (+) charge are
called cations.
Elements which gain 1 or more
e- and take a negative (-)
charge are called anions.
Section 7.1 – Chemical Names and Formulas
Naming Monatomic Ions
(continued)
Potassium
cation
Magnesium
cation
Cation naming is simple, element
name and the word cation.
For anions you drop the end of
the element name and add –ide
to the root.
Fluorine  fluoride
Nitrogen  Nitride
Section 7.1 – Chemical Names and Formulas
Binary Ionic Compounds
Compounds composed of two different
elements are known as binary compounds
Cation goes first : Mg2+, Br-, BrBalance to become electrically neutral
And you get
Section 7.1 – Chemical Names and Formulas
Naming Binary Ionic Compounds
The naming system involves combining the
names of the compound’s positive and
negative ions
Aluminum cation & oxide
And you get
Aluminum Oxide
Section 7.1 – Chemical Names and Formulas
Naming Binary Ionic Compounds
The Stock system of nomenclature
Some elements form more than one cation,
(no elements form more than one
monoatomic anion) each with a different
charge – add Roman Numerals
Iron(II) oxide
Iron(III) oxide
Section 7.1 – Chemical Names and Formulas
Naming Binary Ionic Compounds
Compounds Containing Polyatomic Ions
Oxyanions each is a polyatomic ion that
contains oxygen.
Nitrate
Nitrite
Section 7.1 – Chemical Names and Formulas
Naming Binary Molecular Compounds
1. Less electronegative element
given first, prefix only for
multiples
2. Second element named with
prefix indicating # of atoms, with
few exceptions ends with –ide
(only 2 elements)
3. The o or a at the end of the
prefix is dropped when the word
following begins with another
vowel.
1 mono2 di3 tri4 tetra5 penta6 hexa7 hepta8 octa9 nona10 deca-
Section 7.1 – Chemical Names and Formulas
Naming Binary Molecular Compounds
Example pg 212
Prefix need if more than one
Prefix indicating number
Less-electronegative element
Root name +ide
Section 7.1 – Chemical Names and Formulas
Covalent-Network Compounds
Similar to naming molecular compounds
Section 7.1 – Chemical Names and Formulas
Acids and Salts
Acids are a specific class of compound which usually
refer to a solution of water of one of these special
compounds.
Section 7.1 – Chemical Names and Formulas
Acids and Salts - (continued)
An ionic compound composed of a cation and the
anion from an acid is often referred to as a salt.
Section 7.1 – Chemical Names and Formulas
Practices
PRACTICE – pg 207 Q 1 & 2 all
Practice Naming
PRACTICE – pg 209 Q 1 & 2 all
PRACTICE – pg 211 Q 1 & 2 all
PRACTICE – pg 213 Q 1 & 2 all
– pg 215 Q 2,3 & 4 all
Section 7.1 – Chemical Names and Formulas
Quiz Break
Quiz Key
Section 7.1 – Chemical Names and Formulas - POGIL
KEY
Section 7.1 – Chemical Names and Formulas - POGIL
KEY
Section 7.1 – Chemical Names and Formulas - POGIL
KEY
Section 7.1 – Chemical Names and Formulas - POGIL
KEY
Section 7.1 – V2 – Ions and Compounds
Quiz Break
Quiz Key
Section 7.1 – Chemical Names and Formulas - POGIL
KEY
Section 7.2 - Oxidation Numbers
Objectives
HW – Notes on section 7.2 pgs 216-219
Students will be able to :
• List the rules for assigning oxidation numbers.
•Give the oxidation number for each element in the
formula of a chemical compound.
•Name the binary molecular compounds using
oxidation numbers and the Stock sytem.
Section 7.2 - Oxidation Numbers
Oxidation Numbers
To indicate the general
distribution of electrons
among bonded atoms in
molecular compounds ,
oxidation numbers (or
states) are assigned to
the atoms that compose
the same.
Some are arbitrary, but they are useful in naming compounds, in writing
formulas and in balancing equations.
Section 7.2 - Oxidation Numbers
Assigning Oxidation Numbers – the rules
The following are guidelines...
1. Atoms of pure elements have an oxidation number of
zero.
All zero.
Section 7.2 - Oxidation Numbers
Assigning Oxidation Numbers – the rules
2. The more-electronegative element in a binary compound
is assigned the number equal to the negative charge it
would have as an anion. The less-electronegative is
assigned the number equal to the positive charge it
would have as a cation.
Section 7.2 - Oxidation Numbers
Assigning Oxidation Numbers – the rules
3. Fluorine has an oxidation number of -1 as it is the most
electronegative element.
Section 7.2 - Oxidation Numbers
Assigning Oxidation Numbers – the rules
4. Oxygen has an oxidation number of -2 in almost all
compounds. Exceptions peroxides is -1, compounds with
halogens +2
5. Hydrogen has an oxidation number of +1 in all
compounds containing elements that are moreelectronegative; it has an oxidation number of -1 in
compounds with metals.
Section 7.2 - Oxidation Numbers
Assigning Oxidation Numbers – the rules
6. The algebraic sum of the oxidation numbers of all atoms
in a neutral compound = zero.
7. The algebraic sum of the oxidation numbers of all atoms
in a polyatomic ion = the charge of the ion.
8. Rules 1-7 apply to covalently bonded atoms, oxidation
numbers can also be assigned to atoms in ionic
compounds.
Section 7.2 - Oxidation Numbers
Using Oxidation Numbers for Formulas and Names
Many non-metals have more than one oxidation state.
Recall the use of Roman numerals to denote charges.
Formula
Prefix system
Stock System
PCl3
phosphorus
trichloride
phosphorus(III)
chloride
NO
nitrogen
monoxide
nitrogen(II)
oxide
PbO2
lead
dioxide
lead (IV) oxide
Section 7.2
Practices
Ba(NO3)2
Is the substance elemental?
No, three elements are present.
Is the substance ionic?
Yes, metal + non-metal.
Are there any monoatomic ions?
Yes, barium ion is monoatomic.
Barium ion = Ba2+
Oxidation # for Ba = +2
Which elements have specific rules?
Oxygen has a rule....-2 in most compounds
Oxidation # for O = -2
Which element does not have a specific rule?
N does not have a specific rule.
Use rule 8 to find the oxidation # of N
Let N = Oxidation # for nitrogen
(# Ba) (Oxid. # of Ba) + (# N) (Oxid. # N) + (# of
O) (Oxid. # of O) = 0
1(+2) + 2(N) + 6(-2) = 0
N = +5
Section 7.2
Practices
NF3
Is the substance elemental?
No, two elements are present.
Is the substance ionic?
No, two non-metals.
Are there any monoatomic ions?
Since it is molecular, there are no ions present.
Which elements have specific rules?
F = -1
Which element does not have a specific rule?
N does not have a specific rule.
Use rule 8 to find the oxidation # of N
Let N = oxidation # of N
(# N) (Oxid. # N) + (# F) (Oxid. # F) = 0
1(N) + 3(-1) = 0
N = +3
Section 7.2
Practices
(NH4)2SO4
Is the substance elemental?
No, four elements are present.
Is the substance ionic?
Yes, even though there are no metals present, the
ammonium ion is a common polyatomic cation.
Are there any monoatomic ions?
No, the cation and anion are both polyatomic.
Which elements have specific rules?
H = +1 because it is attached to a non-metal (N)
O = -2
Which elements do not have a specific rule?
Neither N nor S has a specific rule.
You must break the compound into the individual ions that
are present and then use rule 9 to find the oxidation
numbers of N and S. Notice that if you try to use rule 8, you
end up with one equation with two unknowns: 2N + 8(+1) +
1S + 4(-2) = 0
The two ions present are NH4+ and SO42-.
N + 4(+1) = +1 so N = -3
S + 4(-2) = -2 so S = +6
Section 7.2
Practices
PRACTICE – pg 218 Q 1 all
– pg 219 Q 1 & 2 all
Practice Sheet
Key
Section 7.2
Quiz
Key
Section 7.3 - Using Chemical Formulas
Objectives
HW – Notes on section 7.3 pgs 221-228
Students will be able to :
• Calculate the formula mass or molar mass of any
given compound.
• Use molar mass to convert between mass in grams
and amount in moles of a chemical compound.
• Calculate the # of molecules, formula units, or ions
in a given molar amount of a chemical compound.
•Calculate the % composition of a given chemical
compound.
Section 7.3 - Using Chemical Formulas
Formula Masses
The formula mass of any molecule,
formula unit, or ion is the sum of the
average atomic masses of all the atoms
represented in the formula.
Book example
Average atomic mass of H : 1.01 amu
Average atomic mass of O : 16.00 amu
H2O  (2 x 1.01 amu) + 16 amu = 18.02 amu
Section 7.3 - Using Chemical Formulas
Molar Masses
A compound’s molar mass is numerically
equal to its formula mass.
Book example
Formula mass of H2O = 18.02 amu
Which is also the molar mass of
water 18.02 g/mol.
Section 7.3 – Using Chemical Formulas
Molar Mass as a Conversion Factor – Remember our old
friends. . .
There are 3 mole equalities. They are:
1 mol = 6.02 x 1023 particles
1 mol = g-formula-mass (periodic table)
1 mol = 22.4 L for a gas at STP*
These become. . .
1 mol
[-------------] [-------------]
6.02 x 1023 particles
or
6.02 x 1023 particles
1 mol
[-------------] [-------------]
[----]
[----]
1 mol
g-formula-mass (periodic table)
g-formula-mass (periodic table)
1 mol
22.4 L
or
OR
1 mol
22.4 L
1 mol
Section 7.3 - Using Chemical Formulas
Percentage Composition
The percentage by mass of each
element in a compound is known as
the percentage composition of the
compound.
Mass of X in sample
of compound
Mass of sample of
compound
X 100 %
=
Mass % X in compound
Section 7.3 - Using Chemical Formulas
Percentage Composition - example
Percent Composition Example:
Calculate the percent composition of
Mg(NO3)2
1 Mg = 1 x 24 = 24
2 N = 2 x 14 = 28
6 O = 6 x 16 = 96
148g/mole
Double check do they total 100%?
% Mg = 24/148 x 100 = 16.2%
% N = 28/148 x 100 = 18.9%
% O = 96/148 x 100 = 64.0%
Section 7.3 - Using Chemical Formulas
Percentage Composition - practice
Quiz Break
Key
Section 7.4 - Determining Chemical Formulas
Objectives
HW – Notes on section 7.4 pgs 229-233
Students will be able to :
•Define empirical formula, and explain how the terms
applies to ionic and molecular compounds.
•Determine an empirical formula from either a
percentage or a mass composition.
•Explain the relationship between the empirical
formula and the molecular formula of a given
compound.
•Determine a molecular formula from an empirical
formula
Section 7.4 - Determining Chemical Formulas
Calculation of Empirical Formulas
An empirical formula consists of the symbols for
the elements combined in a compound, with
subscripts showing the smallest whole-number mole
ratios of the different atoms in the compound.
Section 7.4 - Determining Chemical Formulas
Calculation of Empirical Formulas - example
Let’s Determine the empirical formula for a compound which
is 54.09% Ca, 43.18% O, and 2.73% H
1) Divide each percent by that element's atomic weight.
Ca = 54.09/40 = 1.352
O = 43.18/16 = 2.699
H = 2.73/1 = 2.73
1.352/1.352 = 1
2.699/1.352 = 2
2.73/1.352 = 2
2) To get the answers to whole numbers, divide through by the
smallest one.
This gives us  CaO2H2
better yet  Ca(OH)2
Section 7.4 - Determining Chemical Formulas
Calculations of Molecular Formulas
An empirical formula may or may not be
a correct molecular formula.
Book example , diborane1’s empirical
formula is BH3, any multiple of that
equals the same ratio – B2H6,B3H9,B4H12 etc
It is a colorless gas at room temperature with a repulsively sweet odor.
Diborane mixes well with air, easily forming explosive mixtures.
Diborane will ignite spontaneously in moist air at room temperature.
Section 7.4 - Determining Chemical Formulas
Calculations of Molecular Formulas – (continued)
The relationship between an empirical
formula and a molecular formula is
seen below:
X(empirical formula) = molecular formula
X is a whole-number multiple indicating the
factor that you need to multiply the empirical
formula by to get the molecular formula.
Section 7.4 - Determining Chemical Formulas
Calculations of Molecular Formulas – (continued)
X = molecular formula
empirical formula
Formula mass of diborane = 27.67 amu
Empirical mass of diborane 13.84 amu
The molecular formula of diborane
is therefore B2H6
(BH3)2 = B2H6
Section 7.4 - Determining Chemical Formulas
PRACTICE
Empirical Formula Practice
Molecular Formula Practice