Download binary molecular compounds

Chapter 7: Chemical Formulas
and Chemical Compounds
Section 1: Chemical Names and Formulas (Emily
Brown)
Section 2: Oxidation Numbers (Anu Ninan)
Section 3: Using Chemical Formulas (Johana Barxha)
Section 4: Determining Chemical Formulas (Justine
Baird)
7.1: Chemical Names and Formulas
Vocabulary
Hydrocarbons—molecular compounds composed solely of carbon and hydrogen
Monatomic ions—ions formed from a single atom
Binary compounds—compounds composed of two different elements
Nomenclature—naming system
Oxyanions—polyatomic ions that contain oxygen
Binary acids—acids that consist of two elements, usually hydrogen and one of the
halogens (fluorine, chlorine, bromine, iodine)
Oxyacids—acids that contain hydrogen, oxygen, and a third element (usually a
nonmetal)
Hydrochloric acid—a water solution of the molecular compound hydrogen
chloride
Salt—an ionic compound composed of a cation and the anion from an acid
7.1 The Naming System
These are the prefixes used in the stock system for naming
binary molecular compounds:
1. The element with the smaller group
number always goes first, except if both
elements have the same group number
(in which the greatest period number
goes first)
2. The second element combines a prefix
giving the number of atom contributed,
the root name, and an -ide suffix
7.1 Naming Cont.
The following is the naming system for binary ionic
compounds:
• In this method, the elements are
sorted according to cations (+)
and anions (-). These can be
determined by an atom's charge
• The positive element always
goes first, followed by the anion.
• The cation retains its original
name, but the anion drops its
ending and adds an -ide suffix
7.2 Oxidation Numbers
VOCABULARY:
• Oxidation numbers: the general distribution of electrons
among the bonded atoms in a molecular compound or a
polyatomic ion
• Oxidation states: another name for oxidation numbers
7.2 Steps Involved In Assigning
Oxidation Numbers
The main key is that the sum of the oxidation numbers of the
atoms equals 0
• In a pure element, the oxidation numbers of all the atoms in
0. (ex. pure Na, each sodium atom's oxidation number is 0,
and in P4 each phosphorus atom has an oxidation number of
0)
• In a binary molecular compound:
o the more electronegative element is given the number
equal to its negative charge it would have as an anion
o the less electronegative element is given the number
equal to its positive charge it would have as an anion
• Fluorine, F, is ALWAYS given the oxidation number of -1,
because it is the most electronegative
7.2 Steps Involved In Assigning
Oxidation Numbers
• Oxygen has an oxidation number of -2 in almost all
compounds. Exceptions: when it is in peroxides or in a
compound with fluorine
• The algebraic sum of the oxidation numbers is always equal
to 0, as long as the compound is neutral
• The algebraic sum of the oxidation numbers of all atoms in a
polyatomic ion is equal to the charge of the ion
• Oxidation numbers can also be assigned to ions
• monoatomic ions have oxidation numbers equal to the
charge of the ion (ex. Ca2+ has an oxidation number of +2)
• Hydrogen, H, has an oxidation number of +1 in compound
where the other atoms are more electronegative,but the
oxidation number is -1 in compounds where there are metals
7.2 Oxidation numbers of certain nonmetals
7.2 Steps Involved In Assigning
Oxidation Numbers
UF6
ClO3-
H2SO4
N2O5
CS2
P4O10
7.2 Stock System and
OxidationNumbers
Ionic Compound have names based on prefixes as well as
names based on the Stock System
Example
PbO2
can be written as:
Lead dioxide prefix system
Lead (IV) oxide Stock System
because oxygen has a charge of -2 and there are to oxygen so
it become (-2)(2)=-4 and lead must make the chemical formular
balance to 0, so
-4+x=0
x=4
7.2 Practice with naming using the
Stock System
As2S3
NCl3
SO3
Mo2O3
NO
PCl5
7.3: Formula Masses
• The formula mass of any molecule, formula unit, or ion is
the sum of the average atomic masses of all atoms
represented in its formula.
For example, the mass of a water molecule is found by adding
the masses of the three atoms in the molecule.
average mass of H: 1.01 amu
average mass of O: 16.00 amu
2 H atoms x 1.01 amu = 02.02 amu
H atom
1 O atom x 16.00 amu = 16.0 amu
O atom
amu.
When added, the
sum equals the
average mass,
about 18.02
7.3: Molar Masses
• The molar mass is calculated by summing the masses of
the elements in a mole of the molecules that make up the
compound.
For example, the mass of a water molecule is found by adding
the masses of the three atoms in the molecule.
2 mol H x 1.01 g H = 02.02 g H
H atom
1 mol O x 16.00 g O = 16.0 g O
O atom
When added, the sum
equals the molar
mass of H2O, about
18.02 g/mol.
A compound's molar mass is numerically equal to its
formula mass.
Molar Mass as a Conversion Factor
7.3: Percentage Composition
• The percent composition is the percentage by mass of
each element in a compound. It is calculated by dividing the
mass of the element of the sample in a compound by the
total mass of the sample. Then, multiply this value by 100.
mass of element in sample of compound x 100 = % element in
mass of sample of compound
compound
mass of element in 1 mol of compound x 100 = % element in
molar mass of
compound
compound
7.3: Percentage Composition
Find the percent composition of copper(I) sulfide, Cu2S.
7.3: Percentage Composition
Find the percent composition of Ba(NO3)2.
7.4: Determining Chemical Formulas
Empirical Formula- Is the smallest is the smallest whole
number ratio of the elements in a compound.
•
For ionic compounds the empirical formula is usually the
formula unit but...
• For molecular compounds the empirical formula does not
always represent the number of atoms in the molecule.
Practice:
What is the empirical formula for
N6H4
S8O4
P2H5
7.4 Finding the Empirical Formula
Steps:
1. If given the % composition (if
given in grams skip to step 2)
figure out the amount of each
substance in the sample by
putting the % into grams
(remember 32% of a substance
is 32 grams).
Question: if a substance
has 24% carbon and 76%
oxygen what is the empirical
formula?
1. 24%= 24 grams
84%= 84 grams
2. Then divide the amount of
each element by it's molar
mass (found on the periodic
table).
2. 24g/12g = 2 mole
76g/16g = 4.75 mole
7.4 Finding the Empirical Formula Con.
3. Next divide the numbers
from both elements ( which
are in moles now) by the
smallest number.
4.Finally round the results
to a whole number and
write them as a subscript
of the appropriate element
3. 2 mol of carbon/ 2= 1
4.75 mol of oxygen/2 = 2.4
4. Co2
7.4 Finding the Molecular Formula
The Molecular Formula, or the actual formula of a molecular
compound can be found using the empirical formula ad the
equation x(empirical formula)= the molecular formula. To solve
for the molecular formula you...
1. Dividing the molecular
What is the molecular formula for
formula mass (given) by the
a compound with the emperical
empirical formula mass (
formula HO and a formula mass of
sum of the element's molar
34 amu?
mass)
1. 34/ 17 = 2
2. Then multiply the
empirical formula by that
number rounded to a
whole number.
2. 2(HO) = H2O2
7.4 Practice
What is the empirical formula for a compound that is 64%
Sulfur, 20% Carbon and 16 % Oxygen?
If the molecular formula mass for the above empirical formula is
312 grams what is the molecular formula?
What is the difference between a molecular and an empirical
formula?