Writing & Naming Compounds
Chapter 7
1
Chemical Formulas – what they mean
• Shows the relative number of atoms of each
kind in a chemical formula
• C8H18 has 8 carbon atoms & 18 hydrogen
atoms.
• Their ratio is C:H = 8:18 or 4:9.
• Question: is this a molecule or a formula
unit?
• A covalently bonded molecule.
2
Al2(SO4)3
•
•
•
•
•
The cation is:
Al +3
The anion is
SO4 -2
It contains:
2 Aluminum ions, 3 sulfate ions
Inside the parenthesis there are 3 sulfur atoms
and 12 oxygen atoms
3
Monatomic Ions
• We spoke about how ions were formed when
atoms lost or gained electrons to become
stable
• Group IA (1) loses 1 e to become +1
• Group IIA (2) loses 2 e to become ___
• Group 16 gains 2 to become ___
• Group 17 gains 1 to become ___
4
Common Monatomic Ions from Main
Group Elements
1+
2+
3+
Lithium Li+
Beryllium Be2+
Aluminum Al 3+
Sodium Na+
Magnesium Mg2+
Potassium K+
Calcium Ca2+
Rubidium Rb+
Strontium Sr2+
Cesium Cs+
Barium Ba2+
1-
2-
3-
Fluoride F-
Oxide O2-
Nitride N3-
Chloride Cl-
Sulfide S2-
Phosphide P3-
Bromide BrIodide I-
5
d-block elements and others with
multiple ions
1+
2+
2+
3+
4+
Copper(I) Cu+ Vanadium(II) V2+
Chromium(II)
Cr2+
Vanadium(III)
V3+
Vanadium(IV)
V4+
Silver Ag+
Manganese(II)
Mn2+
Iron(II) Fe2+
Chromium(III)
Cr3+
Tin (IV) Sn4+
Cobalt(II) Co2+
Nickel(II) Ni2+
Iron(III) Fe3+
Lead(IV) Pb4+
Copper (II) Cu2+
Zinc Zn2+
Cobalt(III) Co3+
Cadmium Cd2+
Tin(II) Sn2+
Mercury(II) Hg2+ Lead(II) Pb2+
See page 221 or handout
6
Binary Ionic Compounds
• Composed of 2 elements
• Positive charges plus negative charges must
equal zero
• Ion charges are written as superscripts
• Number of atoms in a compound are written
as subscripts.
7
Writing binary ionic compound
formulas
1.
Write symbols for the ions side by side. Cation comes first.
Ca2+ P32. Cross over charges by using absolute value of each ion’s charge as
the subscript for the other ion
Ca2+ P33
2
3. Check subscripts and divide them by their largest common factor to
give smallest whole-number ratio of ions; then write formula:
Ca3P2
(For 3 Ca 3x2 charge = 6 and for 2 P 2x (-3) charge = -6
8
Practice-- Write the formulas for
compounds formed between:
• Potassium and iodine
• Magnesium and chlorine
• Sodium and sulfur
• Aluminum and nitrogen
9
Naming Binary Ionic Compounds
• Name of cation is first, followed by part of the
name of the anion ending in –ide.
• Lithium oxide, Li2O
• Sodium nitride, Na3N
10
Practice – name these binary
compounds
•
•
•
•
•
•
AgCl
ZnO
CaBr2
SrF2
BaO
CaCl2
11
Stock System of Nomenclature
(the new way)
• For elements with 2 or more cation forms the
stock system is used with Roman numerals
• Fe2+ is shown as iron (II)
• Fe3+ is shown as iron (III)
• Those with only one form do not need a Roman
numeral (Na+, Ba2+, Al3+)
Iron(II) chloride is FeCl2
Iron(III) chloride is FeCl3
NaCl is sodium chloride and BaO is barium oxide
12
The Old Way
Cuprous nitrate means copper (I) nitrate
CuNO3
Cupric nitrate means copper (II) nitrate
Cu(NO3)2
-ous ending would mean the lower charge for
an ion having more than one charge
-ic ending would mean the higher charge
Example: Ferrous Fe+2 Ferric Fe+3
13
Practice
• Write the formulas and give names:
• Cu2+ and BrFe2+ and O2• Pb2+ and ClSn2+ and F• Name these:
• CuO
• CoF3
SnI4
FeS
14
See page 226 in book
or your handout.
Polyatomic ions are
also called oxyanions.
If more than one
polyatomic ion is
needed, parenthesis
are used.
Source: great10hosting.com
Practice: Write formulas for the following ionic
compounds:
sodium iodate
iron(II)nitrate
hydrogen cyanide
calcium phosphate
15
Try all of these then check
your answers
on the next frame.
16
17
Binary Molecular Compounds
Numerical Prefixes
Binary Compounds of
nitrogen and oxygen
Number
Prefix
1
Mono-
2
Di-
Formula
Prefix-system name
3
Tri-
N2 O
Dinitrogen monoxide
4
Tetra-
NO
Nitrogen monoxide
5
Penta-
NO2
Nitrogen dioxide
6
Hexa-
N2O3
Dinitrogen trioxide
7
Hepta-
N2O4
Dinitrogen tetroxide
8
Octo-
N2O5
Dinitrogen pentoxide
9
Nona-
10
Deca18
Rules for naming binary molecular
compounds
1. The first element has a prefix if there is more
than 1 atom of that element in the formula.
2. Second element has a prefix and ends in –
ide. (Some exceptions.)
3. o or a at end of a prefix is usually dropped
when another vowel comes next: monoxide
or pentoxide
19
Practice with binary molecular
compounds
Naming:
SO3
Write formula for:
Carbon tetraiodide
ICl3
Phosphorus trichloride
PBr5
Dinitrogen trioxide
20
Acids and Salts
• Binary acids consist of 2 elements
(usually H and a halogen)
• Oxyacids contain H, O, and
usually a nonmetal. Learn more
in Ch14.
• Salts are composed of a cation
and the anion from an acid.
Acid formulas and name
HF hydrofluoric acid
HNO2 nitrous acid
HClO hypochlorous
HCl hydrochloric
HNO3 nitric
HClO2 chlorous
HBr hydrobromic
H2SO3 sulfurous
HClO3 chloric
HI hydriodic
H2SO4 sulfuric
HClO4 perchloric
H3PO4 phosphoric
CH3COOH acetic
HC2H302
H2CO3 carbonic acid
21
Practice
Write formulas for these:
Sodium and oxygen
Cu2+ and NO3NH4+ and SO42Lead(II) nitrate
Iron(II) sulfate
Diphosphorus trioxide
Carbon diselenide
Acetic acid
Name these:
NaI
MgS
CaO
K2S
CuNO3
Pb(NO3)2
22
Oxidation Numbers or States
Used to indicate the general
distribution of electrons
among bonded atoms in a
molecular compound or
polyatomic ion.
1. Pure elements are 0.
2. More electronegative
element has Ox. # equal to
its negative charge as an ion.
(Less e.n. has + Ox. # equal
to its positive charge as an
ion.)
3. Fluorine is always -1.
4. Oxygen is usually -2. In H2O2 it
is -1)
5. H is usually +1 except in
compounds w/metals it is -1.
6. Algebraic sum of ox. # of all
atoms in a neutral compound
is 0.
7. Algebraic sum of ox. # of all
atoms in a polyatomic ion =
charge of the ion.
8. Although rules 1-7 apply to
covalently bonded atoms, ox.
# can also be assigned to ionic
compound.
23
When asked to assign oxidation numbers, you work
backward using the rules on the previous slide.
HCl
Or you can criss cross:
Cl is group 17 so it is -1
PCl3
H is group 1 and comes
Put +3 above P
first so it is +1
Put -1 above Cl (Grp 17)
CF4
HNO3
F is always -1
O is -2 (-6 total so H
to make + added to - = 0
and N must total +6)
C must be +4
H is +1
So N must be +5
24
Practice
Assign Ox. #
HF
CI4
H20
Na2O2
H2CO3
NO2 -
Name
CI4
SO3
As2S3
NCl3
25
I. Formula Masses (called Gram
Molecular Weight or Gram
Formula Weight on p6 of notes)
• The formula mass of any molecule, formula
unit, or ion is the sum of the average atomic
masses of all atoms represented in its formula.
• YOU MUST HAVE A PC (periodic chart) and a
calculator.
26
Find the formula mass (GFW) of H2SO4
(in your handout these are called
gram molecular or formula weights - see p.6)
2 H atoms 1.01 amu = 2.02 amu
1 H atom
1 S atom 32.06 amu = 32.06 amu
1 S atom
4 0 atoms 16.00 amu = 64.00 amu
1 0 atom
Formula mass of H2SO4 is 98.08 amu
27
Find the formula mass of Ca(NO3)2
1 Ca atom 40.08 amu = 40.08 amu
Ca atom
2 N atoms 14.01 amu = 28.02 amu
N atom
6 0 atoms 16.00 amu = 96.00 amu
O atom
Formula mass of Ca(NO3)2 is 164.10 amu
28
Your turn…
PO43-
MgCl2
1 P x 30.97 = 30.97
4 O x 16.00 = 64.00
1 Mg x 24.31 = 24.31
2 Cl x 35.45 = 70.90
Ans. 94.97
Ans. 95.21
29
A compound’s Molar Mass is
numerically equal to its formula mass.
Using a PC, sum the masses of the elements present in a
mole of what is present in the compound.
Al2S3
2 mol Al 26.98 g Al = 53.96 g Al
mol Al
3 mol S 32.06 g S = 96.18 g S
mol S
Molar mass of Al2S3 is = the sum or 150.14 g
in one mole
Make sure to show your work and cancel units!!
30
III. Using molar mass to convert
Problem: How many moles are in 4.5 kg Ca(OH)2?
Find molar mass:
1 mol Ca 40.08 g Ca = 40.08 g Ca
1 mol Ca
2 mol 0 16.00 g O = 32.00 g O
1 mol O
2 mol H
1.01 g H
= 2.02 g H
Now add…
1 mol H
Therefore there are
74.10 g /mol
31
Next step, convert mass to moles
4.5 kg Ca(OH)2 ____g ? (make a change so units cancel)
4.5 kg 1000g = 4500 g (4.5 x 1000)
1kg
4500 g Ca(OH)2 1 mol Ca(OH)2 = 60.73 mol Ca(OH)2
74.10 g Ca(OH)2
(4500/74.50)
Bottom line: in 4.5 kg, there are 60.73 mol of Ca(OH)2
32
Now, how about this? How many
formula units are in Ca(OH)2?
Remember that there are 6.02 x 1023 in a mole?
Convert moles to units:
60.73 mol Ca(OH)2 6.02 x 1023 units of Ca(OH)2
1 mol Ca(OH)2
(60.73 x 6.02x1023)
Answer: 3.66 x 1025 units of Ca(OH)2
[Put in Calculator as: 60.73 x 6.02 EE 23 ]
33
IV. Percent Composition
Pb3(PO4)4
CaC03
Pb: 3 x 207 = 621
Ca: 1 x 40 = 40
P: 4 x 31 = 124
C: 1 x 12 = 12
O: 16x12 = 256
O: 3 x 16 = 48
GFW = 1001g
GFW = 100g
Pb = 621/1001 x 100 = 62.04%
Ca= 40/100 x 100 = 40%
P = 124/1001 x 100 = 12.39%
C = 12/100 x 100 = 12%
O = 256/1001 x 100 = 25.57%
O = 48/100 x 100 = 48%
34
V. % Hydration in a compound – see
your handout for more info/samples on
p.6-9
CuSO4-5H2O
H20 = 5x18 = 90
Cu = 1x63.5 = 63.5
S = 1x32 = 32
O = 4x16 = 64
GMW/GFW = 249.5g
H20 = 90/249.5 x 100 =
36.07%
Na2SO4-10H20
H20 = 10x18 = 180
Na = 2x23 = 46
S = 1x32 = 32
O = 4x16 = 64
GMW/GFW = 322g
H20 = 180/322 x 100 =
55.90%
35
VI. Empirical Formula – simplest ratio
of subscripts. Turn to p8 of notes
Find the empirical formula of a compound composed of
7.22g of Nickel, 2.53g of Phosphorus, and 5.25g of
Oxygen.
Ni: 7.22 = 0.123 P: 2.53 = 0.082 O = 5.25 = 0.328
58.7
31
16
0.082 – smallest molar concentration
Ni: 0.123 = 1.5 P: 0.082 = 1
0.082
0.082
0 = 0.328 = 4
0.082
36
Ni: 7.22 = 0.123 P: 2.53 = 0.082 O = 5.25 = 0.328
58.7
31
16
Ni: 0.123 = 1.5 P: 0.082 = 1
0 = 0.328 = 4
0.082
0.082
0.082
NOTE: When one of the values is 0.5, all values
must be doubled.
Ni = 1.5 x 2 = 3 P = 1 x 2 = 2 O = 4 x 2 = 8
Therefore, the empirical formula is Ni3P2O8, but we
have 3 elements so the last 2 must be a
polyatomic ion (PO4-3), so the answer is:
Ni3(PO4)2 or Nickel (II) Phosphate
37
VII. Empirical Formula from %
Composition (p.8&9 of notes)
Determine the empirical formula of a compound
composed of 22.1% Al, 25.4% P, and 52.5% O.
Al: 22.1/27 =0.82 P:25.4/31 =0.82 O: 52.5/16 =3.28
Al: 0.82/0.82 =1 P: 0.82/0.82 =1 O: 3.28/0.82=4
The empirical formula is: AlPO4
38
VIII. True molecular formula of a
compound (p.9 of notes).
Determine the true molecular formula of a
compound composed of 75.0% Carbon and
25.0% Hydrogen having a molecular mass of
96g.
First: C = 75/12 = 6.25
H = 25/1 = 25
C = 6.25/6.25 = 1
H = 25/6.25 = 4
Empirical Formula is CH4 (subscripts are reduced)
We are not done yet…
39
Grand finale for CH4…
2nd : C = 1x12 = 12
H = 4x1 = 4
GMW = 16
3rd: 96/16 = 6
Last: True molecular formula is C6H24
40