Transcript Naming Ionic and Covalent Compounds

11/20 Today you will need…



The paper from the side table
Take out the paper title “types of chemical
bonds”. You have 10 minutes to finish this.
Please do this ON YOUR OWN so you wont
be confused when we start naming/writing
formulas for these two types of compounds
Turn in when finished
Chapter 7-Chemical Formulas &
Chemical Compounds
7.1-Chemical Names & Formulas
7.2-Oxidation Numbers
7.3-Using Chemical Formulas
7.4-Determining Chemical Formulas
7.1-Chemical Names & Formulas
Pages 203-215
Significance of a Chemical Formula

Molecular formula-tells the number of each
type of atom that make up one molecule of
the compound.
C8H18
There are 8 carbon atoms in
one molecule of octane
There are 18 hydrogen atoms
in one molecule of octane
Significance of a Chemical Formula

Ionic formula-represents one formula unit
–
The simplest whole number ratio of cations to
anions in the compound
CaBr2
No subscript=implied “1”
For every calcium ion in the
compound, there are 2
bromide ions
Molecule vs. Ionic Compound
Monatomic Ions


Ions formed from a single atom
Gain or lose electrons to attain a noble gas
configuration
Monatomic Ions: Common Charges
1+
0
Chart with charges of ions on page 205
2+
3+ NA 3- 2- 1
Naming Monatomic Ions

Cations-same as element’s name
–

Example- Li+ is a lithium cation
Anions-drop the ending of the element’s
name and at –ide
–
Example- F- is a fluoride anion
Binary Ionic Compounds



Compounds composed of 2 different
elements that are ionically bonded
Charges of ions must “cancel out” so the
overall charge on the compound is 0.
Use the “crisscross method” to write the
formula.
Binary Ionic Compunds:
Crisscross method

1.
2.
3.
Find the formula for a compound formed by the ions of
aluminum and oxygen.
Write the symbols for the ions side by side, CATION FIRST.
1.
Al3+
O2Cross over the charges by using the number only from each
ions’ charge. Make this the subscript for the other ion.
1.
Al3+
2.
Al2O3
O2-
Check the subscripts to make sure that they cannot be
simplified.
Al2O3
Binary Ionic Compounds:
Crisscross Method

Write the formulas for the binary ionic compounds
formed between the following elements:
– Sodium and phosphorus Na3P
– Gallium and nitrogen
GaN
Naming Binary Ionics

Combine the names of the cation and anion,
cation first.
–
Example: Al2O3



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Cation: aluminum
Anion: oxide
Compound: Aluminum oxide
Most elements form only one ion, so you
usually don’t have to indicate the charge in
the name.
Stock System

Some elements can form more than one ion.
–

Example: Iron can form a +2 and +3 ion
When you name these compounds, you have
to indicate which ion is present.
–
If I asked you to write the formula for iron oxide
without telling you if it was Fe+2 or Fe+3, then you
could write 2 different formulas


FeO
Fe2O3
Use Roman numerals in parentheses in the
name to indicate the charge on the ion.
–
FeO is Iron (II) oxide
Fe2O3 is Iron (III) oxide
Stock System




ONLY use the Stock system for elements
that can form more than one ion
Almost always found in the d-block
Always the cation. Never use Stock system
on anions.
See Table 7-1: Some Common Monatomic
Ions on page 205
Naming Practice
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Write the formula and give the name for the
compound formed by the ions:
Cr3+ and F–

Na+ and S2–

Na2S, Sodium sulfide
Sn2+ and S2–

CrF3, Chromium (III) fluoride
SnS, Tin (II) sulfide
Pb2+ and Cl–
PbCl2, Lead (II) chloride
Naming Practice
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Name the following compounds
CuO
–

LiBr
–
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Copper (II) oxide
Lithium bromide
CoF3
–
Cobalt (III) fluoride
Compounds Containing Polyatomics
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
1.
2.
Chart of polyatomics on page 210.
Compounds containing polyatomics are
named the same way at binary ionics.
Name the cation
Name the anion

Example: Name the compound AgNO3


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Cation: silver
Anion: nitrate
Compound: silver nitrate
Polyatomic Examples

Write the formula for tin (IV) sulfate
–
–
–

Cation: Sn+4
Anion: SO42Formula using crisscross method: Sn2(SO4)4
If you need more than one polyatomic ion,
put it in parentheses to avoid confusion!
–
Wrong: Sn2SO44
Right: Sn2(SO4)4
Polyatomic Examples


Write formulas for the following ionic
compounds:
Sodium iodide
–

Calcium nitrite
–

NaI
Ca(NO3)2
Copper (II) sulfate
–
CuSO4
Polyatomic Examples


Give the names for the following compounds
Ag2O
–

KClO3
–

Silver oxide
Potassium chlorate
FeCrO4
–
Iron (II) chromate
Naming Binary Molecular Compounds

Named using a prefix
system to indicate
how many atoms of
each element are
present.
Table 7-3: Numerical Prefixes (pg. 212)
1
Mono-
2
Di-
3
Tri-
4
Tetra-
5
Penta-
6
Hexa-
7
Hepta
8
Octa-
9
Nona-
10
Deca-
Rules for Naming
Binary Molecular Compounds
1.
2.
The less-electronegative element is given first. It is
given a prefix only if it contributes more than one
atom to a molecule. (Names will never start with
mono-.)
 Order of nonmetals: C, P, N, H, S, I, Br, Cl, O, F
The second element is named by combining a prefix
indicating the number of atoms, the root of the name
of the element, and the ending –ide.
 You may drop an o or a at the end of the prefix if the
name of the element starts with a vowel. Example:
pentoxide, not pentaoxide.
Naming Binary
Molecular Compounds Example

Name the following compound: P4O10.
P4O10
Prefix needed if less
electronegative
element contributes
more than one atom
+
tetraphosphorus
Prefix indicating
Name of
number of atoms
less-electronegative contributed by moreelement
electronegative
element
+
decoxide
Root name of
more-electronegative
element + -ide.
Naming Binary
Molecular Compounds Practice


Name the following compounds:
As2O5
–

XeF4
–

Diarsenic pentoxide
Xenon tetrafluoride
CCl4
–
Carbon tetrachloride
Naming Binary
Molecular Compounds Practice


Write formulas for the following compounds:
Carbon diselenide
–

Sulfur pentachloride
–

CSe2
SCl5
Dihydrogen monoxide
–
H2O
7.2-Oxidation Numbers
Pages 216-219
What are oxidation numbers?

In ionic compounds, charges on ions reflect the
electron distribution
–

Negative ions have a greater electron density in the
compound.
Oxidation numbers or oxidation states reflect to
distribution of electrons among the bonded
atoms in a molecular compound
–
–
Do not have an actual physical meaning
Useful in naming compounds, writing formulas, and
balancing chemical equations.
How do you assign oxidation numbers??

Rules are on page 591. You will need to know
them! (rules on page 216 are not complete!)
1.
Atoms in a pure compound have an oxidation
number of zero.
•
Example: atoms in Na, O2, P4, and S8 all
have oxidation numbers of 0.
Oxidation Number Rules
2.
More electronegative-element in a binary
molecular compound is assigned the oxidation
number equal to the negative charge it would
have as an anion.
•
Example: The oxidation number of O
in NO is -2.
Oxidation Number Rules
Fluorine always has an oxidation number of
-1 in all compounds because it is the most
electronegative element.
3.
•
Example: The oxidation number of F
in LiF is -1.
Oxidation Number Rules
Oxygen has an oxidation number of -2 unless
it is combined with F, when it is +2, or it is in a
peroxide, such as H2O2, when it is -1.
4.
•
Examples:
• In H2SO4, O -2
• In H2O2 (a peroxide due to O22-), O -1
• In OF2, O +2
Oxidation Number Rules
5.
The oxidation state of hydrogen in most
compounds is +1 unless it is combined with a
metal, in which case it is -1.
•
Examples:
• In KH, H’s oxidation number is -1 because
K is a metal.
• In HF, H’s oxidation number is +1.
Oxidation Number Rules
In compounds, Group 1 and 2 elements and
aluminum have oxidation numbers of +1, +2,
and +3, respectively.
6.
–
Example: The oxidation number of Ca
in CaCO3 is +2.
Oxidation Number Rules
The algebraic sum of the oxidation numbers of
all atoms in a neutral compound is equal to
zero.
7.
•
Example:
•
In H2SO4
• H 2(+1)=2
• O 4(-2)=-8
• Subtotal: 2-8=-6
• Therefore, S must be +6 to give a sum of
zero.
Oxidation Number Rules
The sum of the oxidation numbers of all atoms
in an polyatomic ion is equal to the charge of
the ion.
8.
•
Example:
•
In SO42-:
• O=4(-2)=-8
• Therefore, S must be +6 to give a sum of -2
(equal to the charge of the ion)
Oxidation Number Rules
Oxidation numbers can also be assigned to
atoms in ionic compounds. A monatomic ion
has an oxidation number equal to the charge
on the ion.
9.
•
•
Na+ has an oxidation number of +1
N-3 has an oxidation number of -3
Examples
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Cl2
PI3
NO2As2S3
ClO2-
•0
•-1, +3
•+3, -2
•+3, -2
•+3, -2
Using oxidation numbers: Stock
System


Oxidation numbers can be used in naming
molecular compounds as an alternative to
the prefix system.
Example: lead dioxide
–
–
–
PbO2
+4, -2
Lead(IV) oxide
Practice

Phosphorus trichloride
–
–

PCl3
+3, -1
Phosphorus(III) chloride
Dimolybdenum trioxide
–
–
Mo2O3 +3, -2
Molybdenum(III) oxide
7.3-Using Chemical Formulas
Pages 221-228
Formula Masses
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The average atomic mass of an atom is
given on the periodic table.
Like atoms, compounds have average
masses.
Formula mass-the sum of the average
atomic masses of all the atoms represented
in it formula.
–
Round masses from the periodic table to the
hundredths position.
Formula Mass Example

Find the formula mass of potassium chlorate, KClO3.
1 atom K X 39.10 amu  39.10 amu
1 atom K
1 atom Cl X 35.45 amu  35.45 amu
1 atom K
3 atoms O X 16.00 amu  48.00 amu
1 atom K
Formula mass of KClO  39.10 amu  35.45 amu  48.00 amu 122.55 amu
3
Molar Mass


Molar mass-the mass of one mole of a
compound
A compound’s molar mass is numerically equal
to is formula mass
–
Just as one atom of oxygen has a mass of 16.00
amu and one mole of oxygen (6.022 X 1023 atoms)
has a mass of 16 grams.
Molar Mass Example

Don’t forget to distribute
the “2” to each atom in
the nitrate ion!
Find the molar mass of barium nitrate, Ba(NO3)2.
137.33 g Ba  
14.01 g N  
16.00 g O 

1 mol Ba X
   2 mol N X
   6 mol O X

1 mol Ba  
1 mol N  
1 mol O 

261.35 g/mol
Molar mass practice

Find the molar mass of each of the following
compounds:
–
Al2S3

–
NaNO3

–
150.17 g/mol
85.00 g/mol
Ba(OH)2

171.35 g/mol
Molar Mass as a Conversion Factor

Molar mass can be used to relate the amount of a
compound in moles to a mass in grams.
molar mass of compound
moles of compound X
 grams of compound
1 mole
1 mole
mass of compound X
 moles of compound
molar mass of compound
Molar Mass Conversion Example 1

What is the mass in grams of 2.50 mol of
oxygen gas, O2?
–
–
Given: 2.50 mol O2
Unknown: ? g O2
32.00 g O 2
2.50 mol O 2 X
 80.0 grams O 2
1 mol O 2
Molar Mass Conversion Example 2

How many moles of sulfur dioxide are in 3.82 g
of the substance?
–
–
Given: 3.82 g SO2
Unknown: ? mol SO2
1 mol SO 2
3.82 g SO 2 X
 0.0596 mol SO 2
64.07 g SO 2
Percentage Composition

Percentage by mass of each element in a compound
% compositio n 

mass of element in 1 mol of the compound
molar mass of the compound
X 100
Percentages should always total very close to 100!
Always check!
Percentage Composition Example 1
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

Find the percentage composition of sodium nitrate.
Formula of sodium nitrate: NaNO3
Molar mass of NaNO3=85.00 g/mol
% Na 
22.99 g Na
X 100  27.05 % Na
85.00 g NaNO3
%N
14.01 g Na
X 100  16.48 % N
85.00 g NaNO3
%O
316.00 g O 
X 100  56.47 % O
85.00 g NaNO3
Total
100%
Percentage Composition Example 2
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

Find the percentage composition of silver sulfate.
Formula of silver sulfate: Ag2SO4
Molar mass of Ag2SO4=311.81 g/mol
% Ag 
2107.87 g Ag 
X 100  69.19 % Ag
311.81 g Ag 2SO 4
%S
32.07 g S
X 100  10.29 % S
311.81 g Ag 2SO 4
%O
416.00 g O 
X 100  20.53 % O
311.81 g Ag 2SO 4
Total
100.01%
Percentage Composition Example 3



How many grams of copper are in a 38.0 g sample of copper(I)
sulfide?
Formula of copper(I) sulfide: Cu2S
Molar mass of Cu2S=159.17 g/mol
263.55 g Cu 
% Cu 
X 100  79.85% Cu
159.17 g Cu 2S
0.798538.0 g Cu 2S  30.34 g Cu
February
 Pick
nd
2
up an “Empirical Formula”
worksheet from side lab table.
Turn in word search from
yesterday if you haven’t done so
already and then have a seat.
7.4-Determining Chemical
Formulas
Pages 229-233
Why % composition?


When a new compound is made or
discovered, it is analyzed to determine its %
composition.
From this info, its empirical formula can be
determined.
Empirical Formula

Simplified formula for a compound
–
For an ionic compound, empirical formula is the
compound’s formula unit because we always
simplify the formula

–
Calcium sulfide: Ca2S2CaS
For a molecular compound, may not represent
actual number of atoms present in a molecule


Molecular formula: B2H6
Empirical formula: BH3
Empirical Formula

Smallest whole
number ratio
of atoms
in a compound
C2H6
reduce subscripts
CH3
How do you calculate empirical formula?
1. Find mass (or %) of each element.
2. Convert grams to moles of each element.
3. Divide moles by the smallest # to find
subscripts.
4. When necessary, multiply subscripts by
2, 3, or 4 to get whole #’s.
Empirical Formula Example 1

A compound consist of 75% carbon and 25% hydrogen by mass.
Determine the empirical formula for this compound
75 g 1 mol
= 6.24 mol N
6.24 mol
12.01 g
25 g
1 mol
1.01 g
=1C
= 24.75 mol O = 4 H
6.24 mol
Empirical Formula Example 2
Find the empirical formula for a sample
of 25.9% N and 74.1% O.
25.9 g 1 mol
= 1.85 mol N
=1N
1.85 mol
14.01 g

74.1 g
1 mol
= 4.63 mol O
16.00 g
1.85 mol
= 2.5 O
Empirical Formula Example
N1O2.5
Need to make the subscripts whole
numbers  multiply by 2
N2O5
Molecular Formula

“True Formula” - the actual number of
atoms in a compound
empirical
formula
CH3
?
molecular
formula
C2H6
How do you calculate molecular formula?
1. Find the empirical formula.
2. Find the empirical formula mass.
3. Divide the molecular mass by the
empirical mass.
4. Multiply each subscript by the answer from
step 3.
MF mass
n
EF mass
EF n
Molecular Formula Example

The empirical formula for ethylene is
CH2. Find the molecular formula if the
molecular mass is 28.1 g/mol.
empirical mass (CH2) = 14.03 g/mol
28.1 g/mol
14.03 g/mol
= 2.00
(CH2)2  C2H4
General pathway to solve problems

Mass of each element

Percent composition
of compound

Empirical formula
for compound

Molecular formula
for compound
Molecular Formula Example

A sample of a compound with a molar mass
of 34.00 g/mol is found to consist of 0.44 g H
and 6.92 g O. Find its molecular formula.
–
–
–
5.98 % H, 94.02 % O
Empirical formula: HO
Molecular formula: (HO)2=H2O2