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Chemical Compounds
Molecular and Ionic compounds
●Mass percent composition
●Empirical formula determination
●Combustion analysis
●Oxidation states
●Compound Nomenclature
●Examples
●
Chemical Compounds
Molecular Compounds:
Discrete aggregates of atoms.
●Held together by covalent bonds consisting
of shared pairs of electrons
●
Water
Ammonia
Methane
© 2012 K. Brown
Chemical Compounds
Ionic Compounds:
Groups of oppositely charged atoms or
polyatomic species held together by
electrostatic forces. No discrete,
identifiable molecules.
Sodium Chloride
© 2012 K. Brown
Chemical Compounds
Molecular Compounds
Empirical Formula: shows the type of atoms
in the molecule and their relative numbers.
Molecular Formula: shows the type of atoms
in the molecule and their actual numbers.
Hydrogen peroxide:
Hydrazine:
empirical: HO
molecular: H2O2
empirical: NH2
molecular: N2H4
Chemical Compounds
Molecular Compounds
Structural Formula: shows the type of atoms
in the molecule, their actual numbers and
how they are connected.
Water:
© 2012 K. Brown
Chemical Compounds
Ratio of number
of atoms
Exact number of
atoms
Empirical
formula

Molecular
formula


Structural
formula


© 2012 K. Brown
How atoms are
connected

Chemical Compounds
Ionic Compounds
Formula unit: shows the types of atoms and
their relative numbers very similar to an
empirical formula but ... it does not imply
the existence of discrete units of ionic
molecules.
© 2012 K. Brown
Chemical Compounds
Ions with predictable charges
Li + 1
N
Al + 3
+1
Na Mg + 2
K
+1
+1
Ca
Rb Sr
Cs
+1
+2
+2
Ba
−3
−2
F
−2
Cl
−1
Se−2 Br
−1
O
S
−2
Te
+2
© 2012 K. Brown
−1
I −1
Chemical Compounds
Mixed Molecular/Ionic Compounds
Many compounds are a mixture of ions and
covalently bonded species.
Ammonium hydroxide
NH4+
OH-
© 2012 K. Brown
Chemical Compounds
Molar Mass
The mass of 1 mol of a substance in grams
is defined as it’s molar mass.
For a monatomic element it is simply a
matter of taking its atomic mass in u/atom
and changing the unit to grams/mol.
Chemical Compounds
Molar Mass
For compounds consisting of two or more
elements we can find the molar mass (or the
formula mass) by using the molecular
formula or the formula unit and simply
adding up the relative atomic masses of
each atom in the formula.
Chemical Compounds
Molar Mass
What is the molar mass of methane (CH4)?
Molar mass methane=12.01g/ mol+4×1.008g/ mol
=16.04 g/ mol
Chemical Compounds
Molar Mass
What is the molar mass of sodium chloride,
NaCl?
NaCl is not actually a molecular formula
since sodium chloride is an ionic compound,
as previously discussed. However, we can
treat it as such for the purpose of
calculating a formula mass:
Formula mass NaCl=22.99g/ mol+35.45g/ mol
= 58.44g/ mol
Chemical Compounds
Mass Percent Composition
A relative measure of the masses of the
elements in a compound.
●
Lavoisier's original estimate of the
composition of water was 85% oxygen and 15%
hydrogen by mass.
Chemical Compounds
Percent Composition
What is the mass percent composition of
carbon dioxide, CO2?
Molar mass CO2=12.01g/ mol+2×16.00 g/ mol
= 44.01g/ mol
12.01 g/ mol
Mass %C=
×100=27.29%
44.01 g/ mol
2×16.00g/ mol
Mass % O=
×100=72.71%
44.01g/ mol
Chemical Compounds
Determining Empirical Formulas
Recall that the empirical formula gives us
only the ratio of atoms to each other in a
compound.
The problem in finding empirical formulas
is then generally “how do we determine the
ratios of atom A to atom B (to atom C ...
etc.) in our compound”.
Chemical Compounds
Determining Empirical Formulas
A modern elemental analysis of water gives:
%H = 11.19%
%O = 88.81%
both by mass. What is the empirical formula
of water?
In100 g of water
mass H=11.19g
mass O=88.81g
Chemical Compounds
Determining Empirical Formulas
In100 g of water
1
moles H=11.19 g×
=11.10mol
1.008 g/ mol
1
moles O=88.81g×
=5.551mol
16.00g/ mol
Chemical Compounds
Determining Empirical Formulas
Mol ratio of H to O in water:
moles H/ molesO=11.10 mol/ 5.551 mol
= 2.000
empirical formula of water is H2 O
Chemical Compounds
Write empirical formulas for the compounds
represented by the molecular formulas.
Workbook example 3.1
C4 H8
C H2
B2 H 6
BH 3
CCl 4
CCl 4
Chemical Compounds
Determining Empirical Formulas
1. Use the given % composition, or
determine what the %composition is and then
find the mass of each element in a 100 g
sample of the substance.
2. For each elemental mass, determine the
number of mols of the element in the 100 g
sample using the corresponding atomic mass
from the periodic table.
Chemical Compounds
Determining Empirical Formulas
3. If necessary, divide each mol quantity
by the smallest of the numbers of mols to
get whole numbers. If any fractions result,
multiply all results by the appropriate
number to obtain all whole numbers.
4. Use the whole numbers from step 3 to
generate an empirical formula.
Chemical Compounds
Combustion Analysis
© 2012 K. Brown
Chemical Compounds
Combustion Analysis
Vitamin C has only elements C, H and O in
it. A 6.49 g sample is burned and 9.74 g of
CO2 and 2.64 g of H2O are formed. What is
the empirical formula of vitamin C?
12.01 g/ mol
Fraction of C in CO2 =
=0.2729
44.01 g/ mol
mass C in CO2 sample=0.2729×9.74 g=2.66 g C
1
mols C in CO2 sample=2.66 g×
12.01g/ mol
= 0.221mol C
Chemical Compounds
Combustion Analysis
2.016g/ mol
=0.1119
18.01g/ mol
mass H in H2 O sample=0.1119×2.64 g=0.296 g H
1
mols H in H2 O sample=0.296g×
1.008g / mol
= 0.293 mol H
Fraction of H in H2 O=
Chemical Compounds
Combustion Analysis
Mols C=0.221 mol
Mols H=0.293 mol
Mols O=?
Mass O=mass original sample−massC−mass H
= 6.49 g−2.66g−0.296g=3.53 g
1
Mols O=3.53g×
16.00g/ mol
= 0.221mol O
Chemical Compounds
Combustion Analysis
Mols C=0.221 mol
Mols H=0.293 mol
Mols O=0.221 mol
Dividing by the smallest:
Mols C=1
Mols H=1.33
Mols O=1
Chemical Compounds
Combustion Analysis
Multiplying by 3:
Mols C=3
Mols H=4
Mols O=3
Empirical formula of vitamin C: C3 H4 O3
Chemical Compounds
Oxidation States
The idea of the oxidation state or
oxidation number comes about from
considering “what if all compounds were
ionic”. There would then be an identifiable
charge on each atom in the molecule ... it
is this 'what if' charge that is referred
to as the oxidation state or oxidation
number.
Chemical Compounds
Oxidation States
1. The oxidation state of an atom in it's
elemental form is zero.
2. For ions composed of only one atom the
oxidation number is equal to the charge on the
ion.
3. The oxidation state of oxygen in most
compounds is -2. A notable exception is in
peroxides such as hydrogen peroxide in which
the oxidation state of oxygen is -1.
Chemical Compounds
Oxidation States
4. The oxidation state of hydrogen is generally
+1 except when it is bonded to metals such as
sodium (NaH) in which case it's oxidation
number is -1.
5. Fluorine has an oxidation number of -1 in
its compounds … always. Group 1 elements have
an oxidation number of +1 in their compounds …
always. Group 2 elements have an oxidation
number of +2 in their compounds … always.
Chemical Compounds
Oxidation States
6. The sum of the oxidation numbers of all of
the atoms in a molecule or ion must equal the
net charge on the molecule or ion.
Chemical Compounds
Oxidation States
What is the oxidation number of each atom in
the compound KMnO4?
K=+1
4Oxygens=−2×4=−8
Mn=?
Total=0
Mn=+7
Chemical Compounds
Nomenclature of Compounds
Metal-containing Binary Compounds
Compounds containing only two different
elements are binary compounds. If one of
the elements is a metal they are named by
using the element name of the metal
followed by the name of the second element
modified with the suffix 'ide'.
Chemical Compounds
Nomenclature of Compounds
Metal-containing Binary Compounds
exceptions:
NaCl
Al2O3
sodium chloride
aluminum oxide
NaOH
KCN
sodium hydroxide
potassium cyanide
Chemical Compounds
Nomenclature of Compounds
Metal-containing Binary Compounds
Many metals can exist in more than one
oxidation state and can form two or more
compounds with a given anion:
Fe+2
Fe+3
iron(II) ion
iron(III) ion
FeCl2
FeCl3
iron(II) chloride
iron(III) chloride
Chemical Compounds
Nomenclature of Compounds
Binary Acids
When binary compounds with acidic
properties are discussed in the context of
their acidity we alter their names by using
the prefix 'hydro' and suffix 'ic':
HCl
HF
HCN
hydrochloric acid
hydrofluoric acid
hydrocyanic acid
Chemical Compounds
Nomenclature of Compounds
Polyatomic Ions
Most polyatomic ions are anions rather than
cations and the most common modifiers used
as suffixes are 'ate' and 'ite', depending
primarily on oxidation numbers. Also, it is
most common for polyatomic anions to have
oxygen in them and are called oxoanions.
Chemical Compounds
Nomenclature of Compounds
Polyatomic Ions
NH4+
CO3-2
NO3NO2-
ammonium
carbonate
nitrate
nitrite
The 'ate' suffix is applied to the anion which
has the primary atom in the higher oxidation
state and 'ite' to the anion with the primary
atom in the lower oxidation state.
Chemical Compounds
Nomenclature of Compounds
Oxo Acids
H2CO3
H3BO3
carbonic acid
boric acid
H3PO4
H3PO3
phosphoric acid
phosphorous acid
H2SO4
H2SO3
sulfuric acid
sulfurous acid
Chemical Compounds
Nomenclature of Compounds
Oxo Acids
HClO
HClO2
HClO3
HClO4
hypochlorous acid
chlorous acid
chloric acid
perchloric acid
Chemical Compounds
Nomenclature of Compounds
Binary Compounds of Nonmetals
The nomenclature of these compounds is
similar to the binary compounds mentioned
earlier except for those in which the
primary element can exist in multiple
oxidation states.
Chemical Compounds
Nomenclature of Compounds
Binary Compounds of Nonmetals
HCl
hydrogen chloride
PCl5
PCl3
CO2
CO
P4O6
phosphorous pentachloride
phosphorous trichloride
carbon dioxide
carbon monoxide
tetraphosphorus hexaoxide
Chemical Compounds
Nomenclature of Compounds
Hydrates
Some compounds have a fixed number of water
molecules associated with them:
CuSO4.5H2O
copper(II) sulfate pentahydrate
Chemical Compounds
Nomenclature of Compounds
Workbook Examples 3.6, 3.7 and 3.8
Name the compounds:
PbCl 4
Lead ( IV ) chloride
N I3
Nitrogen Triiodide
P 4 S10
Tetraphosphorus DecaSulfide
HIO3 ( aq)
Iodic acid
HF( aq)
Hydrofluoric acid
Chemical Compounds
Chemical Equations
The chemical equation is a shorthand
description of a reaction that gives the
molecular formulas of all of the reactants
and products:
2H2(g) +O2(g) → 2H2 O(g)
reactants
products
Chemical Compounds
Chemical Equations
The equation must be consistent with
Lavoisier's law of conservation of mass.
The total number of atoms of each element
must be the same in the products and
reactants. When this condition is satisfied
we say the equation is balanced.
CH4(g)+O2(g) → CO2(g) +H2 O(g) (unbalanced!)
CH4(g)+2O2(g) → CO2(g)+2H2 O(g)
(balanced)
Chemical Compounds
Balancing Chemical Equations
1.Write the unbalanced equation showing
each reactant and product compound
molecular formula.
2.Balance the equation.
Start with the more complex compounds
●Balance atoms in elemental substances last
●Remove fractional coefficients by
multiplying with an appropriate factor
●
Chemical Compounds
Balancing Chemical Equations
3. Indicate the states of matter of the
reactants and products.
Example: Tin oxide is heated with hydrogen
gas to form tin metal and water vapor.
Write the balanced equation that describes
this reaction.
Chemical Compounds
Balancing Chemical Equations
SnO2+H2 → Sn+H2 O
SnO2+H2 → Sn+2H2 O or SnO2+H2 → Sn+H2 O2 ?
SnO2+2H2 → Sn+2H2 O
SnO2( s)+2H2(g) → Sn(s) +2H2 O(g)
Chemical Compounds
Valence
Valence is the bonding capacity of a particular
element. For the second period of the table the
valences are:
Li – 1
Be – 2
B - 3
C - 4
N - 3
O - 2
F - 1
Ne – 0
Chemical Compounds
Valence
We can use valences to predict formulas. For
example, the combination of nitrogen and
fluorine might be expected to give a molecular
formula of NF3.
Chemical Compounds
Example Calculations
It was found that 56g of iron combined with
32g of sulfur. Calculate the empirical
formula of the compound formed.
1
moles Fe=56g×
55.85g/ mol
= 1.0 mol
1
moles S=32g×
32.07g/ mol
= 1.0 mol
Chemical Compounds
Example Calculations
1.0
1.0
empirical formula is FeS
mole ratio Fe to S=
Chemical Compounds
Example Calculations
A pure compound was found on analysis to
contain 31.9% potassium, 28.9% chlorine and
39.2% oxygen. Calculate its empirical
formula.
In a 100g sample there will be 31.9g K,
28.9g chlorine and 39.2g oxygen.
Chemical Compounds
Example Calculations
1
moles K=31.9 g×
39.10g / mol
= 0.815mol
1
moles Cl=28.9 g×
35.45g / mol
= 0.815mol
1
moles O=39.2 g×
16.00g/ mol
=2.45 mol
Chemical Compounds
Example Calculations
Dividing by the smallest number, 0.815,
gives a K:Cl:O ratio of 1:1:3
empirical formula: KClO3
Chemical Compounds
Example Calculations
A metal sulfide, MS2, is 40.06% by mass
sulfur. What is the identity of the metal?
One mole of MS2 will have two mols of sulfur
atoms in it and one mol of the metal, M.
Two mols of sulfur has a mass of 32.07 x 2
= 64.14 g, which represents 40.06% of the
mass of one mole of MS2.
Chemical Compounds
Example Calculations
We can now determine what the mass of one
mol of MS2 will be:
64.14g 40.06
=
mol
x g/mol
100
100×64.14
x=
g/mol
40.06
= 160.1 g/mol
Chemical Compounds
Example Calculations
mass of M in one mole of MS2 =160.1g−64.14 g
= 95.96g
Referring to the periodic table we find
that the element with this molar mass is
molybdenum.
Chemical Compounds
Example Calculations
Name the compounds:
Cr2O3:
MoS2:
K2SO3:
Fe2(SO4)3:
SiCl4:
P4O10:
KHCO3:
chromium(III) oxide
molybdenum(IV) sulfide
potassium sulfite
iron(III) sulfate
silicon tetrachloride
tetraphosphorus decaoxide
potassium hydrogen carbonate
Further Reading
1.N.J. Tro, Chemistry: A Molecular
Approach, Pearson-Prentice Hall, 2008.
(your text!)
2.Extra problems at
http://chem4823.usask.ca/chem112