Chemical Compounds:  Molecular & Ionic
A molecule is a compound composed of a group of two or more atoms held together in a definite spatial arrangement by forces called covalent bonds.

A molecular formula is a symbolic representation of the composition of a compound in terms of its constituent elements.
        Inorganic compounds do not contain carbon or carbon and hydrogen (e.g., B2O3, NH3, H2O).
        Organic compounds contain carbon and hydrogen, and can contain oxygen, nitrogen, phosphorus,
        as well as other atoms (e.g., CH4, C6H14,  HC2H3O2,  C6H12O6,  NH2CH3)

The molecular formula is the actual number of atoms in a molecule represented by whole number ratio.
An empirical formula is the simplest formula that can be written for a compound.
        C6H12O6 = molecular formula        CH2O = empirical formula

Structural formula is a chemical formula that shows how atoms are attached to one another.

Nomenclature of Binary Molecules
A binary molecule is a compound of only two elements (e.g., H2O, CH4 ).
1. First element is named.
        Element to the left in the period named first. HCl: hydrogen named first.
        Element in the period below named first. BrCl: bromine is named first.
2. The other element is named with -ide ending.
        HCl hydrogen chloride
        BrCl bromine chloride
3. When two nonmetals form more than one compound from each other, Greek prefixes are used.
        (Note: mono- is not affixed to first element of compound if there is only one atom per molecule
        (e.g., CO2 is carbon dioxide, not monocarbon dioxide).
                     Prefixes for binary molecule nomenclature

Number one two  three  four five six seven eight nine  ten 
Prefix mono- di- tri- tetra- penta-  hexa- hepta- octa- nona-  deca-

NO nitrogen monoxide;     N2O dinitrogen monoxide;      NO2 nitrogen dioxide;      N2O3 dinitrogen trioxide;
N2O4 dinitrogen tetroxide;     N2O5 dinitrogen pentoxide;     PCl5 phosphorus pentachloride; 

 P2O5 diphosphorus pentoxide; SF6 sulfur hexafluoride;      Cl2O7 dichlorine heptoxide

Many binary molecules have common names:
H2O Water;  H2O2 Hydrogen peroxide; NH3 Ammonia;  CH4 Methane;  B2H6 Diborane;  SiH4 Silane;  PH3 Phosphine;
H2S Hydrogen sulfide;
HF Hydrogen fluoride;  HCl Hydrogen chloride;  HBr Hydrogen bromide;   HI Hydrogen iodide;
NO Nitric oxide;   N2O Nitrous oxide (laughing gas)

Ionic Compounds and Ions
An ionic compound is a compound composed of ions, and is also, in some cases, referred to as salts.

An ion is an electrically charged particle obtained from an atom or chemically bonded group of atoms by adding or removing electrons.

There are two types:  cations, which are positively charged (+), and anions, which are negatively charged (-).
        Metals typically lose electrons and acquire a positive charge, becoming cations.
        Nonmetals typically gain electrons and acquire a negative charge, becoming anions.

Cation charge = number of electrons lost.
    Group 1A metals always lose one electron. Na1+
    Group 2A metals always lose two electrons. Ca2+
    Aluminum always loses three electrons. Al3+

Anion charge = number of electrons gained.
    Group 5A  (15) elements can gain 3 electrons. N3-
            Electrons gained = 8 – 5 = 3
    Group 6A  (16) elements can gain 2 electrons. O2-
            Electrons gained = 8 – 6 = 2
    Group 7A  (17) elements (halogens) can gain 1 electron. Cl1-
            Electrons gained = 8 – 7 = 1

In some periodic tables hydrogen appears in two locations.
    An atom of hydrogen can either lose or gain one electron.
    When one electron is lost, the hydrogen ion is formed, H1+.   (This ion is simply a free proton.)
    When one electron is gained, the hydride ion is formed, H1-.

Noble gases do not readily gain or lose electrons.

Transition metals form cations of various charges.
     There is no general method for predicting the charge an elements of a given group will form.

An older system uses the Latin form of the element’s name and the prefixes –ic for the ion of higher charge and –ous for the ion of lower charge.          Fe3+ is the ferric ion and Fe2+ is the ferrous ion.

Polyatomic Ions
A polyatomic ion is a charged particle formed from more than one atom.
It is a unit of two or more covalently bonded atoms that possess an overall charge.
Cation (1+)
NH41+ Ammonium
Anion (1-)
OH1- Hydroxide
HSO41- Hydrogen sulfate (or bisulfate)
C2H3O21- Acetate
ClO1- Hypochlorite
ClO21- Chlorite
ClO31- Chlorate
ClO41- Perchlorate
NO21- Nitrite
NO31- Nitrate
MnO41- Permanganate
H2PO41- Dihydrogen phosphate
CN1- Cyanide
HCO31- Hydrogen carbonate (or bicarbonate)
Anion (2-)
CO32- Carbonate
HPO42- Hydrogen phosphate
Cr2O72- Dichromate
S2O32- Thiosulfate
SO32- Sulfite
SO42- Sulfate
C2O42- Oxalate
Anion (3-)
PO43- Phosphate
Ionic Compound Formulas
Compounds are electrically neutral.  Cation symbol written first followed by the anion symbol.  Do not write charge.
NaCl, CaCO3, K2SO4

Naming Ionic Compounds
Ionic compounds can be divided into two categories:  Binary Type I & Binary Type II.

Binary Type I Nomenclature
A binary type I ionic solid is composed of only 2 elements with the cation having only one type of charge.

1.  Cation named first and anion named last.
2.  Metal cations take name from the element.
            Group 1A metals (Alkali metals) all form cations with a 1+ charge.  Li1+, Na1+, K1+
            Group 2A (alkaline earth metals) all form cations with a 2+ charge.  Mg2+, Ca2+, Ba2+
            Aluminum forms only the 3+ cation.  Al3+
3.  Anion is named by taking the first part of the elements name and adding -ide.

KF, potassium fluoride;   MgS, magnesium sulfide;   LiH, lithium hydride;   NaH, sodium hydride;
Al2O3, aluminum oxide;   CaO calcium oxide

Binary Type II Nomenclature
A binary type II ionic solid is composed of only 2 elements but the cation can have more than one type of charge.
( In other words, the cation will not be from Group 1A, Group 2A, or will not be aluminum.)
Most transition metals can form two or more differently charged cations (e.g., Fe2+, Fe3+).

1.  Cation named first and anion named last.
2.  Metal cations take name from the element.
        Immediately following the metal a Roman numeral is used to indicate the charge.
        Parentheses are used to enclose the Roman numeral.
        There is no space between the metal and the opening parenthesis.
        There is a space between the closing parenthesis and the anion name.
3.  Anion is named by taking the first part of the elements name and adding -ide.

FeCl2, iron(II) chloride;  FeCl3, iron(III) chloride
CrO, chromium(II) oxide;  Cr2O3, chromium(III) oxide;    CrO3, chromium(VI) oxide


Properties of Ionic Compounds
A crystal lattice is the ordered array of cations and anions that make up an ionic compound.
This type of arrangement maximizes the attractions and minimizes the repulsion.
A formula unit is the group of atoms or ions explicitly symbolized in the formula.
The formula unit is "hypothetical," because it does not exist as a separate entity.
It represents only the smallest whole number ratio of ions in the compound.

The crystal lattice gives rise to two characteristic properties of ionic solids:
    1. high melting points
    2. distinctive crystalline shapes

Cleavage along definite lines results from layer shifting and bringing like charges in close proximity.
The repulsion from opposite charges causes the crystal to split along definite lines.

Ionic solids do not conduct electricity because the ions are fixed in place and cannot move.
When ionic solids melt, then they can conduct electricity because the ions can then move freely.


Ionic Compounds in Aqueous Solution: Electrolytes
Electric current is the flow of charged particles.
In solid and liquid metals, the charged particles that flow are electrons.
Molten (liquid) ionic compounds and aqueous solutions of ionic compounds are also good electrical conductors,
but in these cases the charged particles that flow are ions.

An electrode is an electrical conductor (wires, plates, rods) partially immersed in a solution and connected to a source of electricity.   The anode is  the electrode connected to the positive pole of the source of electricity.   The cathode is the electrode connected to the negative pole of the source of electricity.   An ion is a carrier of electricity through a solution. (Ion is derived from Greek and means "wanderer.")
The cation is positively charged  (+) and is attracted to the negatively charged cathode (-).   The anion is negatively charged (-) and is attracted to the positively charged anode (+).

Theory of Electrolytic Dissociation
An electrolyte is a compound that conducts electricity when dissolved in water or melted.  Strong electrolytes are solutes that exist in solution predominantly in the form of ions.  A weak electrolyte is a solute that exists in solution predominantly in the form of molecules.  In other words, if the solute is a strong electrolyte it is essentially 100% dissociated in solution while a solute that is a weak electrolyte will hardly be dissociated.  A nonelectrolyte is a substance that does not conduct an observable amount of electricity when dissolved in water.  "Strong" and "weak" refer to the extent to which an electrolyte produces ions in solution.
Strong Electrolyte
Sodium chloride dissolved in water.
NaCl(s) + H2O(l) ---> NaCl(aq)

How it actually exists in solution.
NaCl(aq) ---> Nal+(aq) + Cll-(aq)

Weak Electrolyte
Oxalic acid dissolved in water.
H2C2O4(s) + H2O(l) ---> H2C2O4(aq)

How it actually exists in solution
H2C2O4(aq) <---> 2Hl+(aq) + C2O42-(aq)

Nonelectrolyte
Glucose dissolved in water. 
C6H12O6(s) + H2O(l) ---> C6H12O6(aq)

Glucose does not dissociate, but exists in solution exclusively as the hydrated molecule, C6H12O6(aq).


Amount of a Compound: The Mole
Molar mass is the sum of the atomic weights of the atoms in a molecule of the substance.  It is sometimes referred to as molecular weight (MW).  If the compound is ionic, it is sometimes referred to as formula weight (FW).

Example Mass
Calculate the molar mass of water, H2O.
2 H @ 1.01 =     2.02
1 O @ 16.00 = 16.00
                         18.02 g/mol
Example Molar Mass
Calculate the molar mass of phosphoric acid, H3PO4.
3 H @ 1.01 =    3.03
1 P @ 30.97 = 30.97
4 O @ 16.00 = 64.00
                          98.00 g/mol

 

Example Molar Mass
Calculate the molar mass of potassium carbonate,  K2CO3.
2 K @ 39.10 = 78.20
1 C @ 12.01 = 12.01
3 O @ 16.00 = 48.00
                         138.21 g/mol

Example Molar Mass
Calculate the molar mass aluminum sulfate, Al2(SO4)3.

2 Al @ 26.98 =    53.96
3 S @ 32.07 =      96.21
12 O @ 16.00 = 192.00
                           342.17 g/mol

Hydrates
A hydrate is a crystalline compound with a fixed number of water molecules weakly bound within the crystal.
The water in the compound is referred to as waters of hydration.

Some examples of hydrates are:
CuSO4• 5 H2O Copper(II) sulfate pentahydrate
NiSO4• 6 H2O Nickel(II) sulfate hexahydrate
AlK(SO4)2• 12 H2O Aluminum potassium sulfate dodecahydrate

Formula Weight of a Hydrate:
BaCl2 • 2H2O
1 Ba @   137.33 = 137.33
2 Cl @      35.45 =   70.90
2 H2O @ 18.02  =   36.04
                               244.27 g/mol

Example
How many moles of water are contained in a 355.0 gram sample?
[ 355.0 grams H2O][1 mol H2O/18.02 g H2O]  = 19.70 mol H2O

Example
How many grams of Ca(OH)2 are required to provide 4.0102 x 10-1 moles of Ca(OH)2?
[4.0102 x 10-1 mol Ca(OH)2][74.10 g Ca(OH)2/mol Ca(OH)2]  = 29.716 grams Ca(OH)2

Example
What is the mass of 2.50 x 10-3 mols of barium chloride dihydrate, BaCl2 • 2H2O?
[2.50 x 10-3 mol BaCl2 • 2H2O][244.27 g/mol]  =  0.611 g

Example
How many mmol are in 1.750 x 10-2 g of aluminum sulfate, Al2(SO4)3?
[1.750 x 10-2 g Al2(SO4)3][1mol/342.17 g][1 x 103 mmol/1 mol]  =  5.11 x 10-2 mmol

SAMPLE PROBLEMS:
1.  Calculate the molar mass of each of the following.
    a.  sodium nitrate, NaNO3 [85.00 g/mol]
    b.  dinitrogen tetroxide, N
2O4 [92.02 g/mol]
    c.  sulfuric acid, H
2SO4 [98.09 g/mol]
    d.  Ferric ammonium oxalate, Fe(NH
4)3(C2O4)3 [374.06 g/mol]

2.  What is the mass of 3.500 moles of sodium nitrate? [297.5 g]

3.  How many moles are in 50.00 g of dinitrogen tetroxide? [0.5434 mol]

4.  How many molecules are in 25.00 g of sulfuric acid? [1.535 x 1023]

5.  How many atoms of oxygen are in 1.00 g of ferric ammonium oxalate, Fe(NH
4)3(C2O4)3?
[1.93x1022 O atoms]


Percent Composition
Mass percent composition describes the proportions of the constituent elements in a compound as the number of grams of each element per 100 g of the compound.

Percent composition =       mass of atom A            x 100
                                          total mass of sample

Example
What is the percent composition of sulfuric acid,  H2SO4  ?  The molar mass of sulfuric acid is  98.08 g/mol.
        %H = [2.02/98.08][100] = 2.06%
        %S = [32.06/98.08][100] = 32.69%
        %O = [64.00/98.08][100] = 65.25%

Example
How many grams of oxygen are there in 36.45 g of hydrogen peroxide?  Hydrogen peroxide is 94.06% oxygen.
        (36.45 g H2O2)(94.06 g O/100 g of H2O2) = 34.30 g O

Example
What mass of oxygen is contained in a 1 gallon of water?
The molar mass of water is 18.02 g/mol and the density is 1.00 g/mL.
        %O = (16.00g /18.02 g)(100) = 88.79%

       (1 gallon)(3.7854 L/1 gallon)(1000 mL/1 L)(1.00 g/mL) = 3785.4 g

        (3785.4 g H2O)(88.79 g O/100 g H2O)= 3361.06 g O


Determining Empirical and Molecular Formulas
Empirical formula is the formula of a substance written with the smallest integer (whole number) subscripts.
        The molecular formula is the actual number of atoms in a molecule represented by whole number ratio.
        An empirical formula is the simplest formula that can be written for a compound.
                C6H12O6 = molecular formula        CH2O = empirical formula

Converting Percent Composition to Empirical Formula
Step 1   If percent composition is given, convert the percent of each element to a mass.
Step 2   Convert the mass of each element to an amount in moles.
Step 3   Use the number of moles of the elements as subscripts in a tentative formula.
Step 4   Attempt to get integers as subscripts by dividing each of the subscripts by the smallest subscript.
Step 5   If any subscripts obtained after Step 4 are fractional quantities, multiply each of the subscripts by the
              smallest integer that will convert all the subscripts to integers.

Example
What is the empirical formula of benzene? The percent composition is 92.2%C and 7.8%H.
Step 1 Convert the percent of each element to a mass.
                    92.2 g C and 7.8 g H
Step 2   Convert the mass of each element to an amount in moles.
                    (92.2g C)(1 mol C/12.0 g C) = 7.68 mol C
                    (7.8g H)(1 mol H/1.01 g ) = 7.73 mol H
Step 3   Use the number of moles of the elements as subscripts in a tentative formula.
                C7.68H7.73
Step 4   Attempt to get integers as subscripts by dividing each of the subscripts by the smallest subscript.
                    C7.68/7.68H7.73/7.68 = CH        Therefore, the empirical formula is: CH.

Example
A compound contains only nitrogen and oxygen.  It is 30.4% N by mass. Calculate the empirical formula.
Step 1   If percent composition is given, convert the percent of each element to a mass.
                30.4 g of N and 69.6 g of O
Step 2   Convert the mass of each element to an amount in moles.
                (30.4 g N)(1 mol N/14.01 g N) = 2.17 mol N
                (69.6 g O)(1 mol O/16.00 g O) = 4.35 mol O
Step 3   Use the number of moles of the elements as subscripts in a tentative formula.
                N2.17O4.35
Step 4   Attempt to get integers as subscripts by dividing each of the subscripts by the smallest subscript.
                N2.17/2.17O4.35/2.17 = NO2

Example
What is the empirical formula of a hydrocarbon with 93.71%C and 6.29%H?
Step 1   If percent composition is given, convert the percent of each element to a mass.
                 93.71 g C and 6.29 g H
Step 2   Convert the mass of each element to an amount in moles.
                (93.71 g C)(1 mol C/12.0 g C) = 7.803 mol C
                (6.29 g H)(1 mol H/1.01 g ) = 6.23 mol H
Step 3   Use the number of moles of the elements as subscripts in a tentative formula.
                 C7.803H6.23
Step 4   Attempt to get integers as subscripts by dividing each of the subscripts by the smallest subscript.
                C7.803/6.23H6.23/6.23 = C1.25H
Step 5   If any subscripts obtained after Step 4 are fractional quantities, multiply each of the subscripts by the
              smallest integer that will convert all the subscripts to integers.
                Multiplying subscripts by 4 gives: C5H4.


The molecular formula is the actual number of atoms in a molecule represented by whole number ratio.
An empirical formula is the simplest formula that can be written for a compound.
        C6H12O6 = molecular formula        CH2O = empirical formula
Compound Molecular Formula Molecular Weight % Composition Empirical
Formula
Empirical
Weight
water H2O 18.02 11.21% H
88.79% O
H2O 18.02
hydrogen
peroxide
H2O2 34.02 5.94% H
94.06% O
HO 17.02
acetylene C2H2 26.04 92.2% C
7.8% H
CH 13.02
benzene C6H6 78.12 92.2% C
7.8% H
CH 13.02
ethyl butyrate C6H12O2 116.18 62.04% C
10.41% H
27.55% O
C3H6O 58.09
caproic acid C6H12O2 116.18 62.04% C
10.41% H
27.55% O
C3H6O 58.09
glucose C6H12O6 180.18 39.99% C
6.73% H
53.28% O
CH2O 30.03
Relating Molecular Formulas to Empirical Formulas
Molar Mass (MW) = n x Empirical Formula Mass (EFW)
n = the number of empirical formula units in the molecular formula.

        MW/EFW  = n

Example
What is the molecular formula of benzene if the molecular weight is 78.12 and the empirical formula is CH?
        MW/EFW = n
        78.12/13.02 = 6         Therefore, the molecular formula is, C6H6.

Example
A compound with an empirical formula of NO2 has a MW of 92.02. What is the molecular formula?
        MW/EFW = n
        92.02/46.01 = 2 Therefore: N2O4

Example
Nicotine is 74.03% C, 8.70% H, and the remainder is N.
If the molecular weight of nicotine is found to be 162.26 g/mol, what is the molecular formula?

Step 1   If percent composition is given, convert the percent of each element to a mass.
                74.03 g of C,  8.70 g of H, and 17.27 g N
Step 2   Convert the mass of each element to an amount in moles.
                74.03 g of C = 6.164 mol C
                8.70 g of H = 8.61 mol H
                17.27 g of N = 1.233 mol N
Step 3   Use the number of moles of the elements as subscripts in a tentative formula.
                C6.164H8.61N1.233
Step 4   Attempt to get integers as subscripts by dividing each of the subscripts by the smallest subscript.
                C6.164/1.233H8.61/1.233N1.233/1.233 = C5H7N1

The empirical formula of C5H7N1 gives a EFW of  81.13.
        MW/EFW = n
         162.26/81.13 = 2, therefore the molecular formula of nicotine is, C10H14N2.