Transcript Chapter 3 Chemical Compounds

Chapter 3 Chemical Compounds
When atoms approach other in a chemical reaction, the electrons of
the atoms interact to form chemical bonds. Compounds are
substances composed of more than one element, chemically
combined.
E. g. HCl, H2O, NH3
There are three fundamental kinds of chemical bonds between
atoms-covalent bonds, ionic bonds and metallic bonds.
Types of Chemical Compounds and Their Formulas
i) Molecular Compounds
~ A molecular compound is made up of discrete units called
molecules, which typically consist of two or more of nonmetal atoms
held together by covalent bonds. A covalent bond, the most common
kind of chemical bond, results when two atoms share electrons.
Even some elements exit as molecules rather than as atoms.
Hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine, iodine, sulfur
and phosphorus all exist as molecules whose atoms held together by
covalent bond. Therefore, we have to write them as H2, N2, O2, F2,
Cl2, Br2, I2, S8 and P4 when using any of these elements in a chemical
equation.
Chemical Formulas
A compound is represented by giving its chemical formula, a notation that
uses atomic symbols with numerical subscripts to convey the relative
proportion of atoms of different elements in the substance.
E.g.
The two elements present
H2O
Lack of subscript means one atom of O per formula unit
two H atoms per formula unit
The three elements present
CH2O
Lack of subscript means one atom of O per formula unit
two H atoms per formula unit
Lack of subscript means one atom of C per formula unit
Some Common Types of Formulas
i) An empirical formula is the simplest formula for a compound; it
shows the types of atoms present and their relative numbers. Compounds
with different molecular formulas can have the same empirical formulas
and such substances will have the same percentage composition. E.g.
Acetic acid (C2H4O2), formaldehyde (CH2O), and glucose (C6H12O6) all
have the empirical formula CH2O. Generally, empirical formula does not
tell us much about a molecule.
ii) A molecular formula is based on an actual molecule of a compound.
It gives the exact number of different atoms of an element in a molecule.
In some cases, the empirical formula and the molecular formula are
identical
E.g. formaldehyde CH2O. In other cases, the molecular formula is a
multiple of the empirical formula unit E.g. C6H12O6 = (CH2O)6
Compound
Molecula
formula
Empirical
formula
Ratio of atoms in compound
Carbon
dioxide
CO2
CO2
1 carbon atom : two oxygen atoms
formaldehyde
CH2O
CH2O
1 carbon atom : two hydrogen atoms : 1
oxygen atom
Acetic acid
C2H4O2
CH2O
1 carbon atom : two hydrogen atoms : 1
oxygen atom
Empirical and molecular formulas tell us the combining ratio of the atoms
in the compound, but show nothing about how the atoms are attached to
each other.
There are other types of formula that will show the connectivity of atoms
in a molecule.
Structure and Condensed Formulas
H
H
O
C
C
O
H
CH3COOH
H
structural formula
condensed formula
iii) A structural formula shows the order in which atoms are
bonded together in a molecule and by what types of bonds. The
covalent bonds in the structure formula are represented by lines ().
Each line represents one bond.
iv) A condensed structural formula: a less cumbersome way of
showing how the atoms are connected
acetic acid
name
H
CH2O
empirical formula
H
O
C
C
O
H
C 2 H 4O 2
molecular formula
CH3COOH
H
structural formula
condensed formula
Molecular Models
To show three-dimensional structures of molecules is by structural models
ball-and-stick
type
space-filling
type
Structural Models
1) ball-and stick-model, the centers of the bonded atoms are
represented by small balls, and the bonds between atoms by
sticks. Such model help us to visualize distances between the
centers of atoms (bond length) and the geometrical shapes of
molecules.
2) A space-filling model shows that the atoms in a molecule
occupy space and that they are in actual contact with one
another. This model is the most accurate representation of the
size and shape of a molecule because it constructed to scale.
Ionic Compounds
A ionic compound is made up of positive and negative ions (cation and
anion) joined together by ionic bonds.
A ionic bond result from a transfer of one or more electrons from one
atom to another then joint together by electrostatic forces of attraction.
For example,
the arrangement of Na+ions and Cl-ions in a crystal of sodium chloride.
Each Na+ion is surrounded by six neighboring Cl-ions, and each Cl-ion is
surrounded by six neighboring Na+ions.
Formula unit of an ionic compound
There is no discrete “molecule” of NaCl. Instead, the entire crystal is
an ionic solid.
The formula unit of an ionic compound is the smallest
electrically neutral collection of ions. The ratio of atoms (ions)
in the formula unit is the same as in the chemical formula
A sodium atom
Na + Cl
A chlorine atom
A sodium cation
_
lose/ gain e+
Na + Cl
A chlorine anion
NaCl
Organic and Inorganic Compounds
• Chemical compounds also can be classified as organic or inorganic.
Organic compounds are those formed by carbon and hydrogen
(hydrocarbon) or carbon and hydrogen together with oxygen, nitrogen,
and a few other elements.
• Inorganic compounds are compounds composed of elements other than
carbon. Except a few simple compounds of carbon, including carbon
monoxide, carbon dioxide, carbonates and cyanides are generally
considered to be inorganic.
Formula Mass of Compounds
i) Formula mass
Formula mass(weight): is the mass of a formula unit in atomic mass unit.
F.M.(or F. W.) = Sum of atomic mass of all the atoms in a formula
unit of the compound, whether molecular or not.
Molecular mass: is the mass of a molecule in atomic mass unit.
M. M.(or M. W.)
= Sum of atomic mass of all the atoms in a molecular.
For a molecular compound, the formula unit an actual molecule and
FM =MM.
E.g. Calculate the formula mass (weight) of each of the following
(a) Sodium chloride
FM NaCl = 1 atomic mass Na + 1 atomic mass Cl
= 22.99 amu + 35.45amu
= 58.44 amu
(b) Fe(III) sulfate, Fe2(SO4)3
FM = 2(atomic mass Fe) + 3(atomic mass S)+ 12(atomic mass O)
= 2(55.85amu) + 3(32.07amu) + 12(16.00amu)
= 399.91amu
Mole of Compound
A mole is an amount of substance that contains the same number
of elementary entities (atoms, molecules or formula units) as the
number of atoms in exactly 12g of carbon-12. (the quantity of a
substance whose mass in gram is numerically equal to the formula
mass of the substance).
A mole of compound is an amount of compound containing
Avorgadro’s number of formula units or molecules.
The molar mass is the mass of one mole of compound.
Formula
Units
related by
Avogadro's
number
Mole
relate to
grams by
Molar mass
For elements H2, N2, O2, F2, Cl2, Br2, I2, S8 and P4, we speak of
atomic mass or molecular mass, and the molar mass can be
expressed in two ways.
E.g. atomic mass of hydrogen = 1.008 amu and a molecular mass
= 2.016 amu; its molar mass can be expressed as 1.008g H / mol
H or 2.016g H2/ mol H2.
E. g. Both empirical and molecular formula of halothane is C2HBrClF3;
(a) What is its molecular mass and its molar mass?
(b) How many F atoms are present per mole of halothane?
From the formula, one mole of halothane contains two moles of C atoms,
one mole each of H, Br and Cl atoms, and three moles of F atoms.
• E.g. How many H2O molecules are there in 1.00 L of water ? Density of
water is 1.00 g/cm3
Determining Chemical Formulas
When a chemist has discovered a new compound, the first question to
answer is, What is the formula?
To determine this, we begin by analyzing the compound into amount of
the elements for a given amount compound.
This is expressed as percentage composition-that is the mass percentages
of each element in the compound. We then determine the formula from
this percentage composition
Mass Percentages from the Formula
We define the mass percentage of A as the parts of A per hundred
parts of the total, by mass. That is
Mass % A =
mass of A in the whole compound x 100%
mass of whole compound
E.g. Calculate the percentage composition from formula
(a) What is the mass percent composition of formaldehyde, CH2O?
1 Check the accuracy of the computations by ensuring that the
percentages do total 100%.
2 Determine the percentages of all the elements but one. Obtain that
one by difference (subtraction). The above example
%O = 100.00% - % C -% H
= 100.00% - 39.99% -6.73% = 53.28%
(b) Calculating the mass of an element in a given mass of compound
How Many grams of carbon are there in 83.5g of formaldehyde? use
the percentage composition obtain in (a)
Establishing Empirical Formulas from the Experimentally
Determined Percent Composition of Compounds
Elemental Analysis: percentage of Carbon, Hydrogen, and Oxygen
Percent composition establishes the relative proportion of the elements
in a compound on a mass basis. A chemical formula requires these
proportions to be on a mole basis, that is in terms of # of atoms.
When the % mass composition of a sample is known, consider the
following five steps
1 assume the masses of the sample exactly 100g
Then the example of CH2O above: 39.99g C, 6.73g H and 53.28g O
2 Convert the masses of the elements in the 100.0g sample to amounts in
moles.
mol C = 39.99g/12.01(g/mol)=3.330mol
mol H = 6.73g/1.01(g/mol)=6.66 mol
mol O = 53.28g/(16.00g/mol)= 3.330mol
3 Write a tentative formula based on the numbers of mole just determined.
C3.330H6.66O3.330
4 Attempt to convert the subscript in the tentative formula to small whole #.
C 3.330H 6.66O 3.330
3.330
3.330
3.330
5 If the subscripts at this point differ only slightly from whole #, round up
them off to whole # to obtain the final formula. If one or more subscripts is
not a whole #, multiply all of them by a small whole # that will make all
subscript integral.
CH2O
Combustion Analysis
One of the most common methods used to determine empirical formula
of a compound. In combustion analysis, a weighted sample of a
compound is burned in a stream of oxygen gas. The water vapor and
carbon dioxide gas produced in combustion are absorbed by appropriate
substance. The increases in mass of these absorbers correspond to the
masses of water and carbon dioxide.
E.g. Combustion of a sample of a 1.152g of isobutyl propionate, a
carbon-hydrogen-oxygen compound, yields 2.726g CO2 and 1.116g H2O.
What is the empirical formula of isobutyl propionate?
Molecular Formula from Empirical Formula
The molecular formula of a compound is a multiple of its empirical
formula. For any molecular compound, we can write
Moleculas weight = n x empirical formula weight.
n=
molecular weight____
empirical formula weight
Where n is the number of empirical formula units in the molecule.
E.g. The percentage composition of benzene is 92.3% C and 7.70%H
and its molecular weight is 78.12amu. What is the molecular formula of
benzene?
First, determine the empirical formula from percentage composition
E.g. A 0.2612g sample of a hydrocarbon produces 0.8661g CO2 and
0.2216g H2O in combustion analysis. Its molecular mass is found to be
106amu. For this hydrocarbon, determine:(a) its mass percentage
composition (b) its empirical formula (c) its molecular formula
(b) its empirical formula
(c) its molecular formula
IV Oxidation States (O. S.)
~ is related to the number of electrons that an atom loses, gain or
otherwise appear to use in joining with other atoms in compounds.
~ is defined to be the charge an atom in a substance would have if the
pairs of electrons in each bonded belonged to the more electronegative
atom.
Whenever two rules appear contradict each other, follow the rule that
appears higher on the list.
1 The oxidation state (number) of an individual atom in a free element
is 0.
E.g the O. S. of a Cl atom in Cl2 or an O atom in O2 is zero.
2 The total of the oxidation states of all the atoms in
(a) neutral species, such as isolated atoms, molecules, and formula
units, is 0
(b) an ion is equal to the charge on the ion
E.g. the sum of the O. S. of all atoms in MnO4- is –1
3 In their compounds, the group 1 metals have an O.S. of +1 and the
group 2 metals have O.S. of +2.
4 The O.S. of fluorine is –1 in all of its compounds.
5 H has an O.S. 0f +1 in most of its compounds. The exceptions are
hydrides, compounds such as NaH in which H is bonded to metallic
elements, where the O. S. of H is –1.
6 The usual O. S. of oxygen in a compound is –2. The exceptions are
peroxides, such as H2O2 and Na2O2, in which the O. S. of oxygen is –1.
7 In binary compounds with metals, group 17 elements have an O. S.
of –1; group 16 elements, -2; and group 15 elements, -3.
E.g. What is the oxidation state of the underlined element in each of
the following?
(a) HClO4
(b) MnO4-
(c ) KO2
Nomenclatures- systematic naming of chemical compounds
Will be go over during your lab section. See your lab manual