Transcript Modern Chemistry Chapter 7 Chemical Formulas & Chemical

Modern Chemistry Chapter 7
Chemical Formulas & Chemical Compounds

A chemical formula
indicates the kind
and relative
number of atoms
in a chemical
compound.

C8H18 (octane) has
8 carbon and 18
hydrogen atoms.
Forming Ionic Compounds





Compounds that have the elements held together by
ionic bonds are called ionic compounds.
For an ionic compound to exist, the algebraic sum of
the positive and the negative charges of the ions
MUST = 0.
For instance, when a calcium atom becomes an ion, it
has an overall 2+ charge which must be neutralized
by ion(s) that have a 2- charge.
IF a Ca2+ cation forms an ionic bond with an O2- anion,
the resulting compound will be neutral and the
formula would be CaO.
However, if the Ca2+ bonds with a F- anion, it would
require two F- ions to neutralize the Ca2+  CaF2
Calcium ( Ca2+ ) combines with oxygen ( O2- )  CaO :
+ ----------
-
Ca2+
O2+ ----------
-
Calcium (Ca2+ ) combines with fluorine (F1- )  CaF2:
+ ----------
-
F1-
+ ----------
-
F1-
Ca2+
Binary Ionic Compounds
monatomic ions- ions
formed from a single atom
– IF the ion has a positive
charge, use the name of
the element
– IF the ion has a
negative charge, replace
the ending of the element
name with “ide”.
Binary Ionic Compounds

binary compound- a compound composed of two

Writing binary ionic compound formulas:
1) Write the symbols for the ions side by side with
the cation being first.
2) IF the charges of the two ions do not add to zero,
cross over the charges by using the absolute
value of each ion’s charge as the subscript for the
other ion so the algebraic sum of the ions equals
zero.
3) Check the subscripts and make sure they are in
the smallest whole number ratio possible.
e.g. aluminum oxide Al3+O2-  Al2O3
elements
Naming Binary Ionic Compounds

nomenclature- a naming system

Naming ionic compounds:
Write the name of the cation in the formula.
2) Write the name of the anion in the formula.
1)
Al2O3  aluminum oxide
Do practice problems #1 & 2 on page 223.
Problems- page 223
1)
ab-
cde-
potassium (K+) & iodide (I-) 
KI
magnesium (Mg2+) & chloride (Cl-) 
MgCl2
sodium (Na+) & sulfide (S2-) 
Na2S
aluminum (Al3+) & sulfide (S2-) 
Al2S3
aluminum (Al3+) & nitride (N3-) 
AlN
#2
a)
AgCl 
silver chloride
b)
ZnO 
zinc oxide
c)
CaBr2 
calcium bromide
d)
SrF2 
strontium fluoride
e)
BaO 
barium oxide
f)
CaCl2 
calcium chloride
Stock System of Nomenclature

Some metallic elements that form cations such as
chromium, cobalt, copper, iron, lead, manganese,
mercury, nickel, and tin can form cations of more than
one charge. (See ion chart)

For cations that can have multiple ionic charges, place
a Roman numeral in parentheses that is equal to the
ionic charge after the name of the metal.
Cu1+  copper (I)
Cu2+  copper (II)
Fe2+  iron (II)
Fe3+  iron (III)
Using the Stock System
Write the formula of the ionic compound.
2) Use the charge of the anion to determine the
charge of the cation.
3) Write the name of the cation with the charge
followed by the name of the anion.
1)
CuCl  copper (I) chloride
CuCl2  copper (II) chloride
Do practice problems #1 & 2 on page 225.
Practice- page 225
#1
a)
b)
c)
d)
e)
f)
Cu2+ & Br- 
CuBr2  copper II bromide
Fe 2+ & O2- 
FeO  iron II oxide
Pb 2+ & Cl- 
PbCl2  lead II chloride
Hg 2+ & S2- 
HgS  mercury II sulfide
Sn 2+ & F- 
SnF2  tin II fluoride
Fe 3+ & O2- 
Fe2O3  iron III oxide
Practice- page 225
#2 a)
CuO 
copper II oxide
b)
CoF3 
cobalt III fluoride
c)
SnI4 
tin IV iodide
d)
FeS 
iron II sulfide
Polyatomic Ions

polyatomic ion- a
group of covalently
bonded atoms with an
ionic charge

oxyanion- a
negatively charged
polyatomic ion that
contains oxygen
Ionic Compounds & Polyatomic Ions

Writing and naming strategies are the same for ionic
compounds with polyatomic ions. However, if more
than one polyatomic ion is needed in the formula, the
formula of the polyatomic ion is placed in parentheses
and a subscript is used outside the parenthesis to
show how many of the polyatomic ions are needed.
e.g.
iron (II) nitrate  Fe(NO3)2
 Do practice problems #1 & 2 on page 227.
Practice Problems #1 page 227
1
a-
sodium iodide
NaI
bcd-
calcium chloride
potassium sulfide
lithium nitrate
CaCl2
K2S
LiNO3
e-
copper (II) sulfate
CuSO4
f-
sodium carbonate
Na2CO3
g-
calcium nitrite
Ca(NO2)2
h-
potassium perchlorate
KClO4
2a-
Ag2O
silver oxide
b-
Ca(OH)2
calcium hydroxide
c-
KClO3
potassium chlorate
d-
NH4OH
ammonium hydroxide
e-
Fe2(CrO4)3
iron (III) chromate
f-
KClO
potassium hypochlorite
Practice
Do the following formulas match the names given?
IF they do not match, provide the CORRECT name
or formula.
CuSO4

copper I sulfate
Fe2(SO4)3

iron III sulfate
FeSO4

iron II sulfate
copper I nitrate 
CuNO3
copper II nitrate 
Cu2NO3
Practice
Do the following formulas match the names given?
CuSO4
copper I sulfate
NO [copperII]
Fe2(SO4)3 
iron III sulfate
YES
FeSO4

iron II sulfate
YES
copper I nitrate 
CuNO3
YES
copper II nitrate 
Cu2NO3
NO
[Cu(NO3)2]

Ionic Compound Nomenclature
Name the following compounds:
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
MgBr2
CuO
Cu2O
FeSO4
Fe2(SO4)3
CaSO4
Cu2SO4
CuSO4
FePO4
Fe3(PO4)2
Ionic Compound Nomenclature
Name the following compounds:
MgBr2
magnesium bromide
CuO
copper II oxide
Cu2O
copper I oxide
FeSO4
iron II sulfate
Fe2(SO4)3
iron III sulfate
CaSO4
calcium sulfate
Cu2SO4
copper I sulfate
CuSO4
copper II sulfate
FePO4
iron III phosphate
Fe3(PO4)2
iron II phosphate
Ionic Compound Nomenclature
Write the formulas of the following ionic compounds:
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
aluminum nitrate
aluminum nitride
magnesium phosphate
magnesium bromide
copper I sulfate
copper II sulfate
iron II nitrate
iron III fluoride
calcium hydroxide
calcium phosphate
aluminum nitrate
Al 3+ NO3 1Al(NO3)3
aluminum nitride
Al 3+ N 3AlN
magnesium phosphate
Mg 2+
PO4 3Mg3(PO4)2
magnesium bromide
MgBr2
Mg
2+
Br
copper I sulfate
Cu2SO4
Cu
1+
SO4 2-
copper II sulfate
CuSO4
Cu
2+
SO4
2-
iron II nitrate
Fe(NO3)2
Fe
2+
NO3
1-
1-
iron III fluoride
FeF3
Fe
3+
F
calcium hydroxide
Ca
Ca(OH)2
2+
OH
calcium phosphate
Ca
Ca3(PO4)2
2+
PO4
1-
1-
3-
Binary Molecular Compounds
For this course, molecular compounds consist
of two non-metals. For our purposes,
hydrogen will be considered a non-metal.
The ratio of the elements is NOT determined
by their individual ionic charges.
e.g. CO & CO2
or
H 2O & H 2 O 2
Naming of Binary Molecular Compounds From Formulas
Write the name of the first element in the formula.
2) Write the name of the second element using the
suffix “ide”.
3) Use numerical prefixes to show the number of
atoms of each element.
e.g. P2O5  diphosphorus pentoxide
1)
1
2
3
4
5
=
=
=
=
=
mono
di
tri
tetra
penta
6 = hexa
7 = hepta
8 = octa
9 = nona
10 = deca
Binary Molecular Compounds
P4O10  tetra + phosphorus & dec + oxide
tetraphosphorus decoxide
CO  carbon & mon + oxide
carbon monoxide
CO2  carbon & di + oxide
carbon dioxide
Formulas for Molecular Compounds
1)
The element with the smaller group number is
usually given first. If both elements are in the same
group, the element with the larger period number is
given first. This element is given a prefix ONLY if it
contributes more than one atom to the molecule of
the compound.
2)
The second element is named by combining a prefix
for the number of atoms of the element in the
compound, the root of the name of the element, and
the suffix “ide”.
3)
The “o” or the “a” at the end of a prefix is usually
dropped when the word following the prefix begins
with another vowel.
Writing Molecular Formulas
1)
Write the formula of the first element in the
compound name followed by the numerical
subscript that shows how many there are (if there
is no numerical prefix, there is one atom of the
element).
2)
Write the formula of the second element in the
compound name followed by a subscript that
shows how many atoms of the element are
designated by the numerical prefix in the name.
carbon dioxide  CO2
Do practice problems #1 & 2 on page 229.
Practice Problems #1 & 2 page 229
1- a-
SO3
sulfur trioxide
b-
ICl3
iodine trichloride
c2- a-
bc-
PBr5
phosphorus pentabromide
carbon tetraiodide
CI4
phosphorus trichloride
PCl3
dinitrogen trioxide
N2O3
Molecular Compound Nomenclature
Name the following molecular compounds.
1) N2O5
2) SO2
3) P4O10
4) CO
5) CO2
6) SiO2
7) H2O2
8) CF4
9) PBr3
10) SF2
Name the following molecular compounds.
1) N2O5
2) SO2
3) P4O10
4) CO
5) CO2
6) SiO2
7) H2O2
8) CF4
9) PBr3
10) SF2
dinitrogen pentoxide
sulfur dioxide
tetraphosphorus decoxide
carbon monoxide
carbon dioxide
silicon dioxide
dihydrogen dioxide
carbon tetrafluoride
phosphorus tribromide
sulfur difluoride
Molecular Compound Nomenclature
Write the formula for the following compounds.
1) carbon tetraiodide
2) trinitrogen heptoxide
3) triphosphorus hexasulfide
4) oxygen dichloride
5) disilicon triphosphide
6) tetranitrogen heptoxide
7) carbon disulfide
8) dihydrogen monosulfide
9) trihydrogen monophosphide
10)silicon disulfide
Molecular Compound Nomenclature
Write the formula for the following compounds.
1) carbon tetraiodide
CI4
2) trinitrogen heptoxide
N3O7
3) triphosphorus hexasulfide
P3S6
4) oxygen dichloride
OCl2
5) disilicon triphosphide
Si2P3
6) tetranitrogen heptoxide
N4O7
7) carbon disulfide
CS2
8) dihydrogen monosulfide
H2S
9) trihydrogen monophosphide
H3P
10)silicon disulfide
SiS2
Section Review Problem #2 page 231
2- a-
bcdefgh-
aluminum + bromine 
AlBr3
sodium + oxygen 
Na2O
magnesium + iodine 
MgI2
lead (II) + oxygen 
PbO
tin (II) + iodine 
SnI2
iron (III) + sulfur  Fe2S3
copper (II) + nitrate 
Cu(NO3)2
ammonium + sulfate 
(NH4)2SO4
Section Review Problem #3 page 231
3
a-
NaI 
sodium iodide
b-
MgS 
magnesium sulfide
c-
CaO 
calcium oxide
d-
K2S 
ef-
potassium sulfide
CuBr  copper (I) bromide
FeCl2  iron (II) chloride
Section Review Problem #4 (a-e) page 231
4
ab-
cde-
sodium hydroxide 
NaOH
lead (II) nitrate 
Pb(NO3)2
iron (II) sulfate 
FeSO4
diphosphorus trioxide 
P2O3
carbon diselenide 
CSe2
Oxidation Numbers

oxidation numbers (oxidation states)-
assigned to the atoms composing a molecular
compound or polyatomic ion that indicate the
general distribution of electrons among the
bonded atoms in the compound or ion
Assigning Oxidation Numbers
1)
2)
3)
4)
The atoms in a pure element are assigned
an oxidation number of zero.
The more electronegative (second) element
in a binary molecular compound is assigned
the number equal to the negative charge it
would have if it were an anion.
Fluorine always has an oxidation number of
-1 because it is the most electronegative
element.
Oxygen has an oxidation number of -2 in
almost all compounds.
5) Hydrogen has an oxidation number of
+1 in compounds where it is listed first
and -1 when it is listed last in the
compound formula.
6) The algebraic sum of all oxidation
numbers in a neutral compound is
equal to zero.
7) The algebraic sum of the oxidation
numbers of the atoms in a polyatomic
ion equal the ion’s charge.
8) Oxidation numbers can also be assigned
to atoms in an ionic compound.
Using Oxidation Numbers
 Do
practice
problem #1 on
page 234.
Practice #1 pg 234
a) HCl
H = 1+
b) CF4
C = 4+
c) PCl3
P = 3+
d) SO2
S = 4+
e) HNO3
H = 1+
f) KH
K = 1+
g) P4O10
P= 5+
h) HClO3
H = 1+
i) N2O5
N = 5+
j) GeCl2
Ge = 2+
Cl = 1F = 1Cl = 1O = 2N = 5+
H = 1O = 2Cl = 5+
O = 2Cl = 1-
O = 2-
O = 2-
Oxidation Number problems
What would be the oxidation number of each
element in the following compounds &
polyatomic ions?
H2 O
H=
O=
H2SO4
H=
S=
N2O5
N=
O=
SO42-
S=
O=
PO43-
P=
O=
O=
What would be the oxidation number of each element in
the following compounds & polyatomic ions?
H2O
H2SO4
N2O5
SO42PO43-
H = 1+
O = 2-
H = 1+
S = 6+
N = 5+
O = 2-
S = 6+
O = 2-
P = 5+
O = 2-
O = 2-
Oxidation Numbers & the Stock System

We can use oxidation numbers assigned to the
less electronegative (first) element to name
binary molecular compounds by using the
oxidation number as if it were a cation.
PCl3 
phosphorus trichloride 
phosphorus (III) chloride
Do section review problems #1-2 on page 235.

Problems page 235
#1aHF
bc-
H = 1+
F = 1-
C = 4+
I = 1-
H = 1+
O = 2-
CI4
H2O
d-
PI3
efgh-
P = 3+
I = 1CS2
C = 4+
S = 2This is a rare case when O = 1-.
H2CO3 H = 1+ C = 4+ O = 2NO21- N = 3+
O = 2-

Problems page 235
#2a- CI4 
carbon (IV) iodide
b- SO3 
sulfur (VI) oxide
c- As2S3 
arsenic (III) sulfide
d- NCl3 
nitrogen (III) chloride
Oxidation Numbers & the Stock System

Using oxidation numbers & the stock system, what
would be the names of the following binary molecular
compounds? (fill in the blank with the roman numeral)
N2O5
 nitrogen __ oxide
SiO2
 silicon __ oxide
CF4
 carbon __ fluoride
PI3
 phosphorus __ iodide
SiBr4
 silicon __ bromide

Using oxidation numbers & the stock system, what would be
the names of the following binary molecular compounds?
N2O5

SiO2

CF4

PI3

nitrogen V oxide
silicon IV oxide
carbon IV fluoride
SiBr4 
phosphorus III iodide
silicon IV bromide
Chapter 7 part 1 worksheet

Write the formula for the following ionic compounds.
1-magnesium phosphate
Mg3(PO4)2
2-calcium hydroxide
Ca(OH)2
3-iron (II) nitrate
Fe(NO3)2
4-iron (III) sulfate
Fe2(SO4)3
5-ammonium carbonate
(NH4)2CO3

Write the name of the following ionic compounds.
6-
FeSO4
iron (II) sulfate
7-
FePO4
iron (III) phosphate
8-
KNO3
potassium nitrate
9-
CuSO4
copper (II) sulfate
10- Cu2SO4
copper (I) sulfate

Write the formula of the following molecular compounds.
11- dinitrogen pentoxide
N2O5
12- triphosphorus heptasulfide
P3S7
13- silicon dioxide
SiO2
14- carbon tetrachloride
CCl4
15- disulfur trioxide
S2O3

Write the name of the following molecular
compounds using numerical prefixes.
16-
H2O2
dihydrogen dioxide
17-
P2O6
diphosphorus hexoxide
18-
SiS2
silicon disulfide
19-
N4O10
tetranitrogen decoxide
20-
PI3
phosphorus triiodide
Write the name of the following molecular
compounds using the Stock system.
21- H2O
hydrogen (I) oxide
22- P2O5
phosphorus (V) oxide
23- SiS2
silicon (IV) sulfide
24- N4O10
nitrogen (V) oxide
25- PI3
phosphorus (III) iodide
Determine the oxidation numbers assigned to each
element in the following compounds or ions.
2627282930-
N2O5
N = 5+
O = 2-
C = 4+
O = 2-
S = 6+
O = 2-
P = 5+
O = 2-
N = 5+
O = 2-
CO2
SO3
PO43NO31-
Honors Ch 7 part 1
34 multiple choice:
chemical formulas represent ? (3)
ionic formulas from names (5)
ionic compound names from formulas (4)
molecular compound names from formulas (4)
molecular formulas from names (4)
oxidation number assignment rules (4)
determining oxidation numbers (5)
naming binary molecular compounds using
the stock system (5)
Honors Ch 7 part 1
1 short answer:
What type of compound cannot be
represented by a molecular formula?
Explain.
4 completion:
-name an ionic compound
-name a polyatomic ion
-determine oxidation numbers in a polyatomic ion
and a compound
1 essay:
-eliminated (it will be on next test)
Chemistry Ch 7 part 1 test
25 multiple choice questions:
chemical formulas & what they represent (2)
determine ionic formula from name (4)
determine ionic name from ionic formula (4)
determine molecular name from formula (4)
determine molecular formula from name (4)
rules for assigning oxidation numbers (3)
determine oxidation numbers in compounds (4)
Chemistry Chapter 7 part 1 Practice Test

What do the letters and the subscripts in a
chemical formula represent?
– The identities and the numbers of atoms of
each element in a compound.

Name the following ionic compounds.

Na2S
sodium sulfide

FeSO4
iron (II) sulfate

Fe3(PO4)2
iron (II) phosphate

Chemistry Chapter 7 part 1 Practice Test
What is the formula of the following ionic
compounds?

copper (I) phosphate
Cu3PO4

copper (II) phosphate
Cu3(PO4)2

magnesium nitride
Mg3N2

iron (III) sulfate
Fe2(SO4)3

Chemistry Chapter 7 part 1 Practice Test
Name the following molecular compounds.

N2O5
dinitrogen pentoxide

PF3
phosphorus trifluoride

CBr4
carbon tetrabromide
 What is the formula of the following
molecular compounds?

sulfur dichloride
SCl2

diphosphorus pentoxide
P2O5

silicon disulfide
SiS2

Chemistry Chapter 7 part 1 Practice Test
What is the oxidation number of each
element in the following molecular
compounds?

N2O5
N = 5+ O = 2
SO42S = 6+ O = 2
H3PO4
H = 1+ P = 5+ O = 2
Chemistry In Action

Read “Mass Spectrometry: Identifying
Molecules” on page 236.

Answer questions #1 & 2 at the end of the
reading.
Modern Chemistry

Chapter 7

Part 2
Using Chemical Formulas

formula mass- the sum of the average
atomic masses of all atoms represented in its
formula
– Do practice #1 on page 238

molar mass- the mass of one mole of an
element or a compound (equal to the
formula mass expressed in grams)
– Do practice problems #1 & 2 on page 239.

Practice #1 page 238
a) H2SO4 
2 H x 1.0 = 2.0
1 S x 32.1 = 32.1
4 O x 16.0 = 64.0
2.0 + 32.1 + 64.0 = 98.1 amu
b) Ca(NO3)2 
1 Ca x 40.1 = 40.1
2 N x 14.0 = 28.0
6 O x 16.0 = 96.0
40.1 + 28.0 + 96.0 = 164.1 amu
c) = 95.0 amu
d) = 95.3 amu

Practice #2 page 239
a) Al2S3 
2 Al x 27.0 = 54.0
3 S x 32.1 = 96.3
54.0 + 96.3 = 150.3 g/mol
b) NaNO3 
1 Na x 23.0 = 23.0
1 N x 14.0 = 14.0
3 O x 16.0 = 48.0
23.0 + 14.0 + 48.0 = 85.0 g/mol
c) Ba(OH)2 
1 Ba x 137.3 = 137.3
2 O x 16.0 = 32.0
2 H x 1.0 = 2.0
137.3 + 32.0 + 2.0 = 171.3 g/mol
Review Quiz (10 pts)

Calculate the molar mass of each of the
following compounds. Please show your
work and use the correct label for each
molar mass.
12345-
CaF2
H2O
CO2
PBr3
Al2(SO4)3
Molar Mass as a Conversion Factor
# moles
÷ molar mass
# grams
x molar mass
#grams
Do Practice problems #1 & 3 on page 242.

Problem #1 page 242
a) 6.60 g (NH4)2SO4
N = 2 x 14.0 = 28.0
H = 8 x 1.0 = 8.0
S = 1 x 32.1 = 32.1
O = 4 x 16.0 = 64.0
129.1
6.60/129.1 = 0.051 mol (NH4)2SO4
b) 4.5 kg = 4500 g Ca(OH)2
Ca = 1 x 40.1 = 40.1
O = 2 x 16.0 = 32.0
H = 2 x 1.0 = 2.0
74.1
4500/74.1 = 60.7 mol Ca(OH)2

Problem #3 page 242
6.25 mol of copper (II) nitrate = ? g
copper (II) nitrate = Cu(NO3)2
Cu = 1 x 63.5 = 63.5
N = 2 x 14.0 = 28.0
O = 6 x 16.0 = 96.0
187.5 g/mol
6.25 mol x 187.5 g/mol = 1172 g Cu(NO3)2
mass-mole & mole-mass review quiz
1)
How many moles of H2O are there in 45.0 grams of
H2O? ( molar mass of H2O = 18.0 g/mol)
2)
How many moles of CO2 are there in 220.0 grams of
CO2? (molar mass of CO2 = 44.0 g/mol)
3)
How many grams of H2O are in 5.5 moles of water?
4)
How many grams of CO2 are in 0.05 moles of CO2 ?
5)
How many grams of H2CO3 are in 1.75 moles of the
substance? (molar mass of H2CO3 = 62.0 g/mol)
Honors Class-
mass-mole & mole-mass review quiz
1)
How many moles of H2O are there in 45.0 grams of
H2O?
2)
How many moles of CO2 are there in 220.0 grams of
CO2?
3)
How many grams of H2O are in 5.5 moles of water?
4)
How many grams of CO2 are in 0.05 moles of CO2 ?
5)
How many grams of H2CO3 are in 1.75 moles of the
substance?
Percentage Composition

percentage composition- the percentage of
the total mass of each element in a compound
mass of element in 1 mole x 100%
molar mass of compound
eg. CO2
mass C = 1 x 12.0 = 12.0
mass O = 2 x 16.0 = 32.0
molar mass of CO2 = 44.0 g/mol
%C = 12.0/44.0 (100) = 27.3%
%O = 32.0/44.0 (100) = 72.7%
eg H2O =
%H in H2O =
% O in H2O =
H = 2 x 1.0 = 2.0
O = 1 x 16.0 = 16.0
18.0 g/mol
2.0 x 100 = 11.1%
18.0
16.0 x 100 = 88.9%
18.0
% composition by mass practice
Do Practice problems
#1-3 on page 244.
Do Section Review
problems #1, 3, &
5 on page 244.

Problem #1 page 244
a) PbCl2
Pb = 1 x 207.2 = 207.2
Cl = 2 x 35.5 = 71.0
278.2
Pb = 207.2 x 100 = 74.5%
278.2
Cl = 71.0 x 100 = 25.5%
278.2
1-b) Ba(NO3)2
Ba = 1 x 137.3 = 137.3
N = 2 x 14.0 = 28.0
O = 6 x 16.0 = 96.0
261.3
Ba = 137.3 x 100 = 52.5%
261.3
N=
28.0 x 100 = 10.7%
261.3
O = 96.0 x 100 = 36.7%
261.3

Problem #2 page 244

ZnSO4·7H2O
Zn = 1 x 65.4 = 65.4
S = 1 x 32.1 = 32.1
O = 4 x 16.0 = 64.0
H2O = 7 x 18.0 = 126.0
287.5
%H2O = 126.0 x 100 = 43.8%
287.5

Problem #3 page 244
Mg(OH)2 = 175 g
oxygen = 54.87%
175 x 54.8 = 95.9 g oxygen
100
95.9 g
= 6.0 mol oxygen
16.0 g/mol

Section Review #1
(NH4)2CO3
page 244
N = 2 x 14.0 = 28.0
H = 8 x 1.0 = 8.0
C = 1 x 12.0 = 12.0
O = 3 x 16.0 = 48.0
96.0 amu
96.0 g/mol

Section Review #3
mass of 3.25 mol Fe2(SO4)3 ?
Fe = 2 x 55.8 = 111.6
S = 3 x 32.1 = 96.3
O = 12 x 16.0 = 192.0
399.9 g/mol
3.25 mol x 399.9 g/mol = 1299.7 g

Section Review #5
% composition of each element of (NH4)2CO3
N = 2 x 14.0 = 28.0
H = 8 x 1.0 = 8.0
C = 1 x 12.0 = 12.0
O = 3 x 16.0 = 48.0
96.0 g/mol
%N = 28.0 x 100 = 29.2%
96.0
%H = 8.0 x 100 = 8.3%
96.0
%C = 12.0 x 100 = 12.5%
96.0
%O = 48.0 x 100 = 50.0%
96.0
% composition by mass quiz
1-
Find the % composition by mass of each
element in the compound H3PO4.
2-
Find the % composition by mass of each
element in the compound N2O5.
HONORS- % composition by mass quiz
1-
Find the % composition by mass of each
element in the compound hydrogen
phosphate.
2-
Find the % composition by mass of each
element in the compound dinitrogen
pentoxide.
Determining Chemical Formulas

empirical formula- consists of the
symbols for the elements combined in a
compound, with subscripts showing the
smallest whole number mole ratio of the
different atoms in the compound
CH3 = empirical formula (does not exist)
C2H6 = molecular formula (ethene)
Empirical Formulas

The formulas of ionic
compounds are
empirical formulas
by the definition of ionic
formulas.

The formulas of
molecular compounds
may or may not be the
same as its empirical
formula.
Calculating an Empirical Formula
If the elements are in % composition by mass
form, covert the percentages to grams.
2) Convert the masses of each element to moles
by dividing the mass of the element by its
molar mass.
3) Select the element with the smallest number of
moles and divide the number of moles of each
element by that number which will give you a
1:---:--- ratio.
4) IF the ratio is very close to a whole number
ratio, apply the numbers to each element. If
one of the number is not close to a whole
number, use a multiplier to convert the ratio to
a whole number ratio.
1)
1- If the elements are in % composition by
mass form, covert the percentages to
grams.
e.g.
C = 40.0%  40.0 g
H = 6.67%  6.67 g
O = 53.3%  53.3 g
2- Convert the masses of each element to
moles by dividing the mass of the
element by its molar mass.
e.g.
C = 40.0/12 = 3.33 mol
H = 6.67/1 = 6.67 mol
O = 53.3/16 = 3.33 mol
3- Select the element with the smallest
number of moles and divide the number
of moles of each element by that
number which will give you a 1:---:--ratio.
e.g.
C = 3.33/3.33 = 1
H = 6.67/3.33 = 2
O = 3.33/3.33 = 1
4- IF the ratio is very close to a whole
number ratio, apply the numbers to each
element. If one of the number is not
close to a whole number, use a multiplier
to convert the ratio to a whole number
ratio.
e.g.
1:2:1 ratio  CH2O
Calculating an Empirical Formula
Sample Problem L page 246.
 32.38% Na, 22.65% S, & 44.99% O.
1- convert to 32.38 g Na, 22.65 g S, & 44.99 g O

2- 32.38 ÷ 22.99 = 1.408 mol Na
22.65 ÷ 32.07 = 0.7063 mol S
44.99 ÷ 16.00 = 2.812 mol O
3- 1.408 ÷ 0.7063 = 1.993 mol Na  2
0.7063 ÷ 0.7063 = 1 mol S
2.812 ÷ 0.7063 = 3.981 mol O  4
4- Rounding  2:1:4  Na2SO4
Calculating an Empirical Formula

Review sample problem M on page 247.

Do practice problems #1, 2, & 3 on page
247.

Practice problem #1 page 247
63.52% iron (Fe)
36.48% sulfur (S)
Convert % to grams: Fe = 63.52g
S = 36.48g
Divide each element by its molar mass:
Fe = 63.52/55.8 = 1.14 mol
S = 36.48/32.1 = 1.14 mol
Divide each number of moles by the smallest number:
Fe = 1.14/1.14 = 1
S = 1.14/1.14 = 1
Ratio = 1:1 so FeS is the empirical formula

Practice problem #2 page 247
K = 26.56%
Cr = 35.41%
O = 38.03%
K = 26.56/39.1 = 0.679 mol
Cr = 35.41/52.0 = 0.681 mol
O = 38.03/16.0 = 2.38 mol
K = 0.679/0.679 = 1
Cr = 0.681/0.679 = 1.003
O = 2.38/0.679 = 3.51
1:1:3.5 ratio
Double the ratio to get whole numbers  2:2:7
Empirical formula is K2Cr2O7

Practice problem #3 page 247.
20.0 g calcium & bromine
4.00 g Ca so 16.00 g Br
Already in grams so divide by molar mass:
4.00/ 40.1 = .0997 mol Ca
16.00/79.9 = .2003 mol Br
Ca = .0997/.0997 = 1
Br = .2003/.0997 = 2.009  2
Empirical formula is CaBr2
Ch 7 part 2 quiz #4
Empirical Formulas
1-
A compound is 27.3% carbon and 72.7%
oxygen by mass. What is the empirical
formula of the compound?
2-
A compound is 11.1% hydrogen and
88.9% oxygen. What is its empirical formula?
Calculating a Molecular Formula

molecular formula- the actual formula of a
molecular compound (it may or may not be the
same as the empirical formula of the compound)
The molar mass of a compound is determined by
analytical means & is given.
2) Calculate the formula mass of the empirical
formula. Divide the molar mass of the compound
by its empirical mass.
3) “Multiply” the empirical formula by this factor.
1)
Calculating a Molecular Formula
1)
2)
3)
4)
5)
empirical formula = P2O5
molecular mass is 283.89
empirical mass is 141.94
Dividing the molecular mass by the
empirical mass gives a multiplication
factor of :
283.89 ÷ 141.94 = 2.0001  2
2 x (P2O5)  P4O10
Chapter 7 Problems
 Do
practice
problems #1 & 2
on page 249.
 Do
section
review problems
#1-4 on
page 249.

Practice problem #1 page 249
empirical formula = CH
formula mass = 78.110 amu
empirical mass = ? = 12.0 + 1.0 = 13.0 amu
molecular mass / empirical mass = 78.110/13.0
= 6.008  multiplication factor of 6
molecular formula = CH x 6  C6H6

Practice problem #2 page 249
formula mass = 34.00 amu
0.44 g H & 6.92 g O
1st find empirical formula:
H = 0.44/1.0 = 0.44
O = 6.92/16.0 = 0.43
0.44/0.43  1 H & 0.43/0.43  1 O
empirical formula = HO
empirical mass = 17.0
formula mass / empirical mass = 34.00/17.0 = 2
HO x 2  H2O2

Section review problem #1 page 249.
36.48% Na 25.41% S 38.11% O
36.48/23.0 = 1.58 mol Na
25.41/32.1 = 0.792 mol S
38.11/16.0 = 2.38 mol O
1.58/0.792 = 1.995  2
0.792/0.792 = 1  1
2.38/0.792 = 3.005  3
2:1:3  Na2SO3

Section review problem #2 page 249.
53.70% Fe
46.30% S
53.70/55.8 = 0.962 mol Fe
46.30/32.1 = 1.44 mol S
0.962/0.962 = 1 Fe
1.44/0.962 = 1.50 S
1:1.5 doubled  2:3  Fe2S3
Section review problem #3 page 249
1.04 g K 0.70 g Cr
1.04/39.1 = .0266 mol K
0.70/52.0 = .0135 mol Cr
0.86/16.0 = .0538 mol O
.0266/.0135 = 1.97  2
.0135/.0135 = 1
.0538/.0135 = 3.99  4
Empirical formula = K2CrO4
0.86 g O

Section Review problem #4 page 249
4.04 g N
11.46 g O
f.m. = 108.0 amu
4.04/14.0 = .289 mol N
11.46/16.0 =.716 mol O
.289/.289 = 1
.716/.289 = 2.45
double ratio  2:5  N2O5
e.f.m. = 108
f.m./e.f.m. = 108/108 = 1
empirical formula is same as molecular formula N2O5
To find molar mass: add the masses of the
elements in the formula of the compound.
To find number of grams (mass): multiply
# of moles times the molar mass of the
compound.
To find the number of moles: divide the
number of grams by the molar mass of
the compound.
To calculate % composition by mass:
1- find the molar mass of a compound
2- divide the mass of each element by the
molar mass of the compound
3- multiply by 100 to convert each ratio to a
percent
Calculating an Empirical Formula
If the elements are in % composition by mass
form, convert the percentages to grams.
2) Convert the masses of each element to moles
by dividing the mass of the element by its
molar mass.
3) Select the element with the smallest number of
moles and divide the number of moles of each
element by that number which will give you a
1:---:--- ratio.
4) IF the ratio is very close to a whole number
ratio, apply the numbers to each element. If
one of the number is not close to a whole
number, use a multiplier to convert the ratio to
a whole number ratio.
1)
Calculating a Molecular Formula

molecular formula- the actual formula of a
molecular compound (it may or may not be the
same as the empirical formula of the compound)
The molar mass of a compound is determined by
analytical means & is given.
2) Calculate the formula mass of the empirical
formula. Divide the molar mass of the compound
by its empirical mass.
3) “Multiply” the empirical formula by this factor.
1)
Chapter 7 part 2 quiz #5
Calculating molecular formulas
1-
A molecular compound has an
empirical formula of CH3. Its molecular
formula mass is 30 amu. What is the
molecular formula of this compound?
HONORS- Chapter 7 part 2 quiz #5
Calculating molecular formulas
1-
A molecular compound is 80% carbon
and 20% hydrogen. Its molecular
formula mass is 30 amu. What is the
molecular formula of this compound?
Final Practice- chapter 7 part 2
1-
Determine the molar mass of the compound
Na3PO4 .
2-
How many moles of CO2 are in 198 g ?
3-
What is the mass of 2.25 moles of H2O ?
4-
What is the % composition of each element
the compound P4O10 ?
5-
What is the empirical formulas of a compound that
is 25.9% N and 74.1% O ? What is it molecular
formula if its molecular mass is 216 ?
of
Final Practice- chapter 7 part 2
1-
Determine the molar mass of the compound
Na3PO4 .
Na = 3 x 23.0 = 69.0
P = 1 x 31.0 = 31.0
O = 4 x 16.0 = 64.0
164.0 g/mol
2-
How many moles of CO2 are in 198 g ?
C = 1 x 12.0 = 12.0
O = 2 x 16.0 = 32.0
44.0 g/mol
198/44.0 = 4.5 mol CO2
3-
What is the mass of 2.25 moles of H2O ?
H = 2 x 1.0 = 2.0
O = 1 x 16.0 = 16.0
18.0 g/mol
2.25 mol x 18.0 g/mol = 40.5 g H2O
4-
What is the % composition of each element
the compound P4O10 ?
P = 4 x 31.0 = 124.0
O = 10 x 16.0 = 160.0
284.0 g/mol
P = 124/284(100) = 43.7%
O = 160/284 (100) = 56.3%
of
5-
What is the empirical formulas of a compound that
is 25.9% N and 74.1% O ? What is it molecular
formula if its molecular mass is 216 ?
N = 25.9/14.0 = 1.85
O = 74.1/16.0 = 4.63
N = 1.85/1.85 = 1
O = 4.63/1.85 = 2.5
1:2.5 doubled  2:5 so empirical formula = N2O5
e.f.m. = (2 x 14) + (5 x 16) = 108
216 (mfm)/108 (efm) = 2
2 x 2:5  4:10  N4O10
Honors Chemistry Chapter 7 part 2 test

38 multiple choice:
Definition of formula mass & molar mass
Calculate formula mass of a compound (3)
Convert from mass to moles or moles to mass when
given the amount & molar mass of a substance (7)
Calculate % composition by mass (6)
Definition & what an empirical formula represents
Calculate empirical formulas (7)
Know how to determine molecular formula from empirical
formula and determine what the empirical fromula of a
molecular formula would be
Calculate molecular formula when given empirical formula
& formula mass (7)

Essay Question:
____ & ____ are examples of the
empirical and the molecular formula of
a compound, respectively. Explain the
relationship between these two types
of formulas.
Chemistry Chapter 7 part 2 test review

24 multiple choice questions
 Definition of molar mass and formula mass
 Calculate a formula mass
 Interpret a molar mass
 Convert from mass to moles or moles to mass when given
the molar mass of a compound (6)
 Calculate % composition by mass (3)
 Definition of empirical formula and what it represents (4)
 Calculate the empirical formula of compounds (3)
 What is needed to determine the molecular formula from an
empirical formula
 Determine the molecular formula of a compound from its
formula mass and the empirical formula (3)

Essay Question:
____ & ____ are examples of the
empirical and the molecular formula of
a compound, respectively. Explain the
relationship between these two types
of formulas.