1. Inorganic Nomenclature
lithium nitrate
lead (II) sulfide
lithium nitride
barium sulfide
Chemistry
lithium nitrite
sulfur dioxide
Unit 4 Part 2: Bonding and
Inorganic Nomenclature
NaClO3
NO2
Fe(ClO3)2
N2O4
Fe(ClO3)3
N2O5
PowerPoint Presentation by Mr. John Bergmann

2. atoms that stay together
Chemical Bonding
Ionic Bonds: atoms give up or gain e– and
are attracted to each other by
coulombic attraction
loses e–
gains e– Na
Na1+ Cl
Cl1–
Na Cl1–
NaCl
ionic compounds = salts
K NO31–
KNO3
where NO31– is a polyatomic ion:
a charged group of
atoms that stay together

3. Properties of Salts
1. very hard –
each ion is bonded
to several oppositely
-charged ions
2. high melting points –
many bonds must be
broken
3. brittle –
with sufficient force,
like atoms are
brought next to
each other and repel
calcite

4. Covalent Bonds
…atoms share e– to get a full valence shell C
1s2 2s2 2p2
(4 v.e–) F
1s2 2s2 2p5
(7 v.e–)
both need 8 v.e– for a full outer shell (octet rule)
Lewis structure:
a model of a covalent molecule that
shows all of the valence e–
1. Two shared e– make a single covalent bond,
four make a double bond, etc.
2. unshared pairs: pairs of unbonded valence e–
3. Each atom needs a full outer shell, i.e., 8 e–.
Exception: H needs 2 e–

5. carbon tetrafluoride (CF4)x F x F x F o C x F o C x F x F x F x F
methane (CH4) H x H x x H o C H x o C x H H x H x x H

6. nitrogen triiodide (NI3)x I x I o
N o x I x I o
N o x I x I
carbon dioxide (CO2) o C x O x O o C x O
O = C = O
x x x O

7. covalent compounds = molecular compounds
-- have lower melting points
than do ionic compounds
(consist of two
nonmetal elements)
butter

8. In metals, valence shells of atoms overlap, so v.e–
Metallic Bonds
In metals, valence shells of atoms overlap, so v.e–
are free to travel between atoms through material.
In insulators (like wood),
the v.e– are attached
to particular atoms.
Not so in metals.

9. All due to free-moving v.e–.
Properties of Metals
conduct heat and electricity
ductile
malleable
All due to free-moving v.e–.

10. Other Types of Bonds
DNA
boiling H2O
dipole-dipole forces
hydrogen bonds
London dispersion forces
ion-dipole forces
These are much weaker than ionic, covalent, or metallic bonds, but very important in determining states of matter, boiling and melting points, and molecular shape (among other things).

11. Writing Formulas of Ionic Compounds
chemical formula:
has neutral charge;
shows types of atoms
and how many of each
To write an ionic compound’s formula, we need:
1. the two types of ions
(i.e., “pink” and “blue”)
2. the charge on each ion
Na1+ and F1–
NaF
Ba2+ and O2–
BaO
Na1+ and O2–
Na2O
Ba2+ and F1–
BaF2

12. charge on cation / anion “becomes” subscript of anion / cation
criss-cross rule:
charge on cation / anion
“becomes” subscript of anion / cation
** Warning:
Reduce to lowest terms.
Al3+ and O2–
Ba2+ and S2–
In3+ and Br1– 3 2 2 2 3 1
Al O
Ba S
In Br
Al2O3
BaS
InBr3

13. Writing Formulas w/Polyatomic Ions
Parentheses are required only when you need more
than one “bunch” of a particular polyatomic ion.
Ba2+ and SO42–
BaSO4
Mg2+ and NO21–
Mg(NO2)2
NH41+ and ClO31–
NH4ClO3
Sn4+ and SO42–
Sn(SO4)2
Fe3+ and Cr2O72–
Fe2(Cr2O7)3
NH41+ and N3–
(NH4)3N

14. Inorganic Nomenclature
sodium
hydroxide
NaOH
potassium
nitrate
KNO3
copper (II)
sulfate
Cu2SO4
dinitrogen monoxide
N2O

15. Ionic Compounds (cation/anion combos)
Single-Charge Cations with Elemental Anions
i.e., “pulled off the
Table” anions
The single-charge cations are:
groups 1, 2, 13, and Ag1+, Cd2+, and Zn2+

16. Na
A. To name, given
the formula: Ba
1. Use name of cation.
2. Use name of anion (it has the ending “ide”).
NaF
sodium fluoride
BaO
barium oxide
Na2O
sodium oxide
BaF2
barium fluoride

17. Zn Ca Ag
B. To write formula,
given the name:
1. Write symbols for the two types of ions.
2. Balance charges to write formula.
silver sulfide
Ag1+
S2–
Ag2S
zinc phosphide
Zn2+
P3–
Zn3P2
calcium iodide
Ca2+
I1–
CaI2

18. Multiple-Charge Cations with Elemental Anions
i.e., “pulled off the
Table” anions
The multiple-charge cations are:
Pb, Sn, and the transition elements
(but – of course! – not Ag, Cd, or Zn)

19. A. To name, given the formula:Cu Fe
Figure out charge on
cation.
2. Write name of cation.
3. Write Roman numerals
in ( ) to show cation’s charge.
Stock System
of nomenclature
4. Write name of anion.
FeO
iron oxide
Fe?
Fe2+
O2–
iron (II) oxide
Fe2O3
Fe?
iron oxide
Fe3+
Fe?
Fe3+
O2–
O2–
O2–
iron (III) oxide
CuBr
copper bromide
Cu?
Cu1+
Br1–
copper (I) bromide
CuBr2
Cu?
copper bromide
Cu2+
Br1–
Br1–
copper (II) bromide

20. B. To find the formula, given the name:
1. Write symbols for the two types of ions.
2. Balance charges to write formula. Co Sn
cobalt (III) chloride
Co3+
Cl1–
CoCl3
tin (IV) oxide
Sn4+
O2–
SnO2
tin (II) oxide
Sn2+
O2–
SnO

21. Compounds Containing Polyatomic Ions
Insert name of ion
where it should go
in the compound’s
name.
Write formulas:
iron (III) nitrite
iron (III) nitrite
Fe3+
NO21–
Fe(NO2)3
ammonium phosphide
ammonium phosphide
NH41+
P3–
(NH4)3P
ammonium chlorate
ammonium chlorate
NH41+
ClO31–
NH4ClO3
zinc phosphate
zinc phosphate
Zn2+
PO43–
Zn3(PO4)2
lead (II) permanganate
lead (II) permanganate
Pb2+
MnO41–
Pb(MnO4)2

22. Write names: (NH4)2SO2 (NH4)2SO2 ammonium hyposulfite AgBrO2 AgBrO2
silver bromite
(NH4)3N
(NH4)3N
ammonium nitride
Mn(CrO4)2
Mn(CrO4)2
Mn4+
Mn?
CrO42–
manganese (IV) chromate
CrO42–
Cr2(SO2)3
Cr2(SO2)3
Cr3+
Cr?
SO32–
chromium (III) hyposulfite
Cr?
Cr3+
SO32–
SO32–

23. Use Greek prefixes to indicate how many atoms of each element, but
Covalent Compounds
-- contain two types of
nonmetals
nonmetals
** Key:
FORGET CHARGES!
What to do:
Use Greek prefixes to indicate how
many atoms of each element, but
don’t use “mono” on first element.
1 –
2 –
3 –
4 –
5 –
6 –
7 –
8 –
9 –
10 –
mono
hexa di
hepta
tri
octa
tetra
nona
penta
dec

24. EXAMPLES:
carbon dioxide
CO2 CO
carbon monoxide
dinitrogen trioxide
N2O3
N2O5
dinitrogen pentoxide
carbon tetrachloride
CCl4
NI3
nitrogen triiodide

25. A Tale of Danger and Irresponsibility
Dihydrogen Monoxide:
A Tale of Danger and Irresponsibility
-- major component of acid rain
-- found in all cancer cells
-- inhalation can be deadly
-- excessive ingestion results in acute physical symptoms:
e.g., frequent urination,
bloated sensation,
profuse sweating
-- often an industrial byproduct of chemical
reactions; dumped wholesale into rivers and lakes

26. Traditional System of Nomenclature
(i.e., NOT the Stock System)
…used historically (and still some today) to name
compounds w/multiple-charge cations Fe Cu Sn Au Pb
To use:
1. Use Latin root of cation.
2. Use -ic ending for higher charge;
-ous ending for lower charge.
3. Then say name of anion, as usual.

27. plumbic phosphide plumbous phosphide stannic chloride
Element Latin root -ic -ous
gold, Au aur- Au3+ Au1+
lead, Pb plumb- Pb4+ Pb2+
tin, Sn stann- Sn4+ Sn2+
copper, Cu cupr- Cu2+ Cu1+
iron, Fe ferr- Fe3+ Fe2+
Write formulas:
Write names:
P3–
P3–
cuprous sulfide
cuprous sulfide
Pb3P4
Pb3P4
Pb4+
Pb?
Pb4+
Pb?
Pb?
Pb4+
P3–
P3–
plumbic phosphide
Cu1+
S2–
Cu2S
auric nitride
auric nitride
Pb3P2
Pb3P2
Pb?
Pb2+
Pb?
Pb2+
Pb?
Pb2+
P3–
P3–
plumbous phosphide
Au3+
N3–
AuN
Cl1–
ferrous fluoride
ferrous fluoride
SnCl4
SnCl4
Sn?
Sn4+
Cl1–
Cl1–
Cl1–
stannic chloride
Fe2+
F1–
FeF2

28. and type of atoms in a m’cule
Empirical Formula and Molecular Formula
lowest-terms
formula
shows the true number
and type of atoms in a m’cule
Compound
Molecular Formula
Empirical
Formula
glucose
C6H12O6
propane
C3H8
butane
C4H10
naphthalene
C10H8
sucrose
C12H22O11
octane
C8H18
CH2O
C3H8
C2H5
C5H4
C12H22O11
C4H9

29. Empirical Formulas Empirical Formulas
How can you calculate the empirical formula of a compound?
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30. Empirical Formulas
The empirical formula of a compound gives the lowest whole-number ratio of the atoms or moles of the elements in a compound.
An empirical formula may or may not be the same as a molecular formula.
For example, the lowest ratio of hydrogen to oxygen in hydrogen peroxide is 1:1. Thus, the empirical formula of hydrogen peroxide is HO.
The molecular formula, H2O2, has twice the number of atoms as the empirical formula.
Notice that the ratio of hydrogen to oxygen is still the same, 1:1.
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31. Empirical Formulas For carbon dioxide, the empirical and molecular
are the
same—
CO2.
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32. Empirical Formulas
The figure below shows two compounds of carbon and hydrogen having the same empirical formula (CH) but different molecular formulas.
Ethyne (C2H2), also called acetylene, is a gas used in welders’ torches.
Styrene (C8H8) is used in making polystyrene.
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33. Empirical Formulas
The percent composition of a compound can be used to calculate the empirical formula of that compound.
The percent composition tells the ratio of masses of the elements in a compound.
The ratio of masses can be changed to ratio of moles by using conversion factors based on the molar mass of each element.
The mole ratio is then reduced to the lowest whole-number ratio to obtain the empirical formula of the compound.
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34. Determining the Empirical Formula of a Compound
Sample Problem 10.12
Determining the Empirical Formula of a Compound
A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula of the compound?
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35. Analyze List the knowns and the unknown.
Sample Problem 10.12
Analyze List the knowns and the unknown. 1
The percent composition gives the ratio of the mass of nitrogen atoms to the mass of oxygen atoms in the compound. Change the ratio of masses to a ratio of moles and reduce this ratio to the lowest whole-number ratio.
KNOWNS percent by mass of N = 25.9% N percent by mass of O = 74.1% O UNKNOWN empirical formula = N?O?
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36. Calculate Solve for the unknown.
Sample Problem 10.12
Calculate Solve for the unknown. 2
Convert the percent by mass of each element to moles.
Percent means “parts per 100,” so g of the compound contains 25.9 g N and 74.1 g O.
25.9 g N ×
1 mol N
14.0 g N
= 1.85 mol N
74.1 g O ×
1 mol O
16.0 g O
= 4.63 mol O
The mole ratio of N to O is N1.85O4.63.
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37. Calculate Solve for the unknown.
Sample Problem 10.12
Calculate Solve for the unknown. 2
Divide each molar quantity by the smaller number of moles to get 1 mol for the element with the smaller number of moles.
1.85 mol N
1.85
= 1 mol N
4.63 mol O
1.85
= 2.50 mol O
The mole ratio of N to O is N1O2.5.
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38. Calculate Solve for the unknown.
Sample Problem 10.12
Calculate Solve for the unknown. 2
Multiply each part of the ratio by the smallest whole number that will convert both subscripts to whole numbers.
1 mol N × 2 = 2 mol N
2.5 mol O × 2 = 5 mol O
The empirical formula is N2O5.
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39. Evaluate Does the result make sense?
Sample Problem 10.12
Evaluate Does the result make sense? 3
The subscripts are whole numbers, and the percent composition of this empirical formula equals the percents given in the original problem.
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40. Molecular Formulas Molecular Formulas
How does the molecular formula of a compound compare with the empirical formula?
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41. Comparison of Empirical and Molecular Formulas
Interpret Data
Ethyne and benzene have the same empirical formula—CH.
Comparison of Empirical and Molecular Formulas
Formula (name)
Classification of formula
Molar mass (g/mol) CH
Empirical 13
C2H2 (ethyne)
Molecular
26 (2 × 13)
C6H6 (benzene)
78 (6 × 13)
CH2O (methanol)
Empirical and molecular 30
C2H4O2 (ethanoic acid)
60 (2 × 30)
C6H12O6 (glucose)
180 (6 × 30)
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42. Comparison of Empirical and Molecular Formulas
Interpret Data
Methanal, ethanoic acid, and glucose have the same empirical formula—CH2O.
Comparison of Empirical and Molecular Formulas
Formula (name)
Classification of formula
Molar mass (g/mol) CH
Empirical 13
C2H2 (ethyne)
Molecular
26 (2 × 13)
C6H6 (benzene)
78 (6 × 13)
CH2O (methanal)
Empirical and molecular 30
C2H4O2 (ethanoic acid)
60 (2 × 30)
C6H12O6 (glucose)
180 (6 × 30)
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43. Comparison of Empirical and Molecular Formulas
Interpret Data
Notice that the molar masses of the compounds in these two groups are simple whole-number multiples of the molar masses of the empirical formulas, CH and CH2O.
Comparison of Empirical and Molecular Formulas
Formula (name)
Classification of formula
Molar mass (g/mol) CH
Empirical 13
C2H2 (ethyne)
Molecular
26 (2 × 13)
C6H6 (benzene)
78 (6 × 13)
CH2O (methanal)
Empirical and molecular 30
C2H4O2 (ethanoic acid)
60 (2 × 30)
C6H12O6 (glucose)
180 (6 × 30)
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44. Molecular Formulas
Methanal (formaldehyde), ethanoic acid (acetic acid), and glucose have the same empirical formula—CH2O.
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45. Molecular Formulas
The molecular formula of a compound is either the same as its experimentally determined empirical formula, or it is a simple whole-number multiple of its empirical formula.
Once you have determined the empirical formula of a compound, you can determine its molecular formula, if you know the compound’s molar mass.
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46. Molecular Formulas
You can calculate the empirical formula mass (efm) of a compound from its empirical formula.
This is simply the molar mass of the empirical formula.
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47. Molecular Formulas
You can calculate the empirical formula mass (efm) of a compound from its empirical formula.
Then you can divide the experimentally determined molar mass by the empirical formula mass.
This quotient gives the number of empirical formula units in a molecule of the compound and is the multiplier to convert the empirical formula to the molecular formula.
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48. Finding the Molecular Formula of a Compound
Sample Problem 10.13
Finding the Molecular Formula of a Compound
Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH4N.
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49. Analyze List the knowns and the unknown.
Sample Problem 10.13
Analyze List the knowns and the unknown. 1
Divide the molar mass by the empirical formula mass to obtain a whole number.
Multiply the empirical formula subscripts by this value to get the molecular formula.
KNOWNS empirical formula = CH4N molar mass = 60.0 g/mol
UNKNOWN molecular formula = C?H?N?
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50. Calculate Solve for the unknown.
Sample Problem 10.13
Calculate Solve for the unknown. 2
First calculate the empirical formula mass.
efm of CH4N = 12.0 g/mol + 4(1.0 g/mol) g/mol
= 30.0 g/mol
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51. Calculate Solve for the unknown.
Sample Problem 10.13
Calculate Solve for the unknown. 2
Divide the molar mass by the empirical formula mass.
molar mass
efm =
60.0 g/mol
30.0 g/mol
= 2
Multiply the formula subscripts by this value.
(CH4N) × 2 = C2H8N2
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52. Evaluate Does the result make sense?
Sample Problem 10.13
Evaluate Does the result make sense? 3
The molecular formula has the molar mass of the compound.
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