Modern Chemistry Chapter 7.ppt

chemical equations

1. Chemical Formulas &
Chemical Compounds
Dr Shikha Sharma

2. •A chemical formula
indicates the kind
and relative number
of atoms in a
chemical compound.
•C
8H
18(octane) has 8
carbon and 18
hydrogen atoms.

3. 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 Ca
2+
cationforms an ionic bond with an O
2-
anion, the
resulting compound will be neutral and the formula would
be CaO.
•However, if the Ca
2+
bonds with a F
-
anion, it would require
two F
-
ions to neutralize the Ca
2+
CaF
2

4. Calcium ( Ca
2+
) combines with oxygen ( O
2-
) CaO:
+-----------
Ca
2+
O
2-
+-----------
Calcium (Ca
2+
) combines with fluorine (F
1-
) CaF
2:
+----------- F
1-
Ca
2+
+----------- F
1-

5. 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”.

6. Binary Ionic Compounds
•binary compound-a compound composed of two
elements
•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 Al
3+
O
2-
Al
2O
3

7. Naming Binary Ionic Compounds
•nomenclature-a naming system
•Naming ionic compounds:
1)Write the name of the cation in the formula.
2)Write the name of the anion in the formula.
Al
2O
3aluminum oxide
Do practice problems #1 & 2 on page 223.

8. Problems-page 223
1)a-potassium (K
+
) & iodide (I
-
) 
KI
b-magnesium (Mg
2+
) & chloride (Cl
-
) 
MgCl
2
c-sodium (Na
+
) & sulfide (S
2-
) 
Na
2S
d-aluminum (Al
3+
) & sulfide (S
2-
) 
Al
2S
3
e-aluminum (Al
3+
) & nitride (N
3-
) 
AlN

9. #2a)AgCl
silver chloride
b)ZnO
zinc oxide
c)CaBr
2
calcium bromide
d)SrF
2
strontium fluoride
e)BaO
barium oxide
f)CaCl
2
calcium chloride

10. 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
chargeafter the name of the metal.
Cu
1+
copper (I)Fe
2+
iron (II)
Cu
2+
copper (II) Fe
3+
iron (III)

11. Using the Stock System
1)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 cationwith the charge
followed by the name of the anion.
CuClcopper (I) chloride
CuCl
2copper (II) chloride
Do practice problems #1 & 2 on page 225.

12. Practice-page 225
#1a)Cu
2+
& Br-
CuBr
2copper II bromide
b)Fe
2+
& O
2-

FeOiron II oxide
c)Pb
2+
& Cl
-

PbCl
2lead II chloride
d)Hg
2+
& S
2-

HgSmercury II sulfide
e)Sn
2+
& F
-

SnF
2tin II fluoride
f)Fe
3+
& O
2-

Fe
2O
3iron III oxide

13. Practice-page 225
#2a)CuO 
copper II oxide
b)CoF
3
cobalt III fluoride
c)SnI
4
tin IV iodide
d)FeS 
iron II sulfide

14. Polyatomic Ions
•polyatomic ion-a group of
covalently bonded atoms with
an ionic charge
•oxyanion-a negatively charged
polyatomic ion that contains
oxygen

15. 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(NO
3)
2
•Do practice problems #1 & 2 on page 227.

16. Practice Problems #1 page 227
1a-sodium iodide
NaI
b-calcium chloride
CaCl
2
c- potassium sulfide
K
2S
d-lithium nitrate
LiNO
3

17. e-copper (II) sulfate
CuSO
4
f- sodium carbonate
Na
2CO
3
g-calcium nitrite
Ca(NO
2)
2
h-potassium perchlorate
KClO
4

18. 2a-Ag
2O
silver oxide
b-Ca(OH)
2
calcium hydroxide
c-KClO
3
potassium chlorate
d-NH
4OH
ammonium hydroxide
e-Fe
2(CrO
4)
3
iron (III) chromate
f-KClO
potassium hypochlorite

19. Practice
Do the following formulas match the names given?
IF they do not match, provide the CORRECT name or
formula.
CuSO
4 copper I sulfate
Fe
2(SO
4)
3 iron III sulfate
FeSO
4 iron II sulfate
copper I nitrate CuNO
3
copper II nitrateCu
2NO
3

20. Practice
Do the following formulas match the names given?
CuSO
4copper I sulfate
NO [copperII]
Fe
2(SO
4)
3iron III sulfate
YES
FeSO
4iron II sulfate
YES
copper I nitrate CuNO
3
YES
copper II nitrateCu
2NO
3
NO [Cu(NO
3)
2]

21. Ionic Compound Nomenclature
Name the following compounds:
1)MgBr
2
2)CuO
3)Cu
2O
4)FeSO
4
5)Fe
2(SO
4)
3
6)CaSO
4
7)Cu
2SO
4
8)CuSO
4
9)FePO
4
10)Fe
3(PO
4)
2

22. Ionic Compound Nomenclature
Name the following compounds:
MgBr
2
magnesium bromide
CuO
copper II oxide
Cu
2O
copper I oxide
FeSO
4
iron II sulfate
Fe
2(SO
4)
3
iron III sulfate

23. CaSO
4
calcium sulfate
Cu
2SO
4
copper I sulfate
CuSO
4
copper II sulfate
FePO
4
iron III phosphate
Fe
3(PO
4)
2
iron II phosphate

24. Ionic Compound Nomenclature
Write the formulas of the following ionic compounds:
1)aluminum nitrate
2)aluminum nitride
3)magnesium phosphate
4)magnesium bromide
5)copper I sulfate
6)copper II sulfate
7)iron II nitrate
8)iron III fluoride
9)calcium hydroxide
10)calcium phosphate

25. aluminum nitrate
Al
3+
NO
3
1-
Al(NO
3)
3
aluminum nitride
Al
3+
N
3-
AlN
magnesium phosphate
Mg
2+
PO
4
3-
Mg
3(PO
4)
2

26. magnesium bromideMg
2+
Br
1-
MgBr
2
copper I sulfate Cu
1+
SO
4
2-
Cu
2SO
4
copper II sulfateCu
2+
SO
4
2-
CuSO
4
iron II nitrate Fe
2+
NO
3
1-
Fe(NO
3)
2

27. iron III fluorideFe
3+
F
1-
FeF
3
calcium hydroxideCa
2+
OH
1-
Ca(OH)
2
calcium phosphateCa
2+
PO
4
3-
Ca
3(PO
4)
2

28. 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 & CO
2or H
2O & H
2O
2

29. Naming of Binary Molecular Compounds From Formulas
1)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. P
2O
5diphosphoruspentoxide
1 = mono 6 = hexa
2 = di 7 = hepta
3 = tri 8 = octa
4 = tetra 9 = nona
5 = penta 10 = deca

30. Binary Molecular Compounds
P
4O
10tetra + phosphorus & dec+ oxide
tetraphosphorusdecoxide
CO carbon & mon+ oxide
carbon monoxide
CO
2carbon & di+ oxide
carbon dioxide

31. 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 ONLYif 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.

32. 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 CO
2
Do practice problems #1 & 2 on page 229.

33. Practice Problems #1 & 2 page 229
1-a-SO
3
sulfur trioxide
b-ICl
3
iodine trichloride
c- PBr
5
phosphorus pentabromide
2-a-carbon tetraiodide
CI
4
b-phosphorus trichloride
PCl
3
c- dinitrogentrioxide
N
2O
3

34. Molecular Compound Nomenclature
Name the following molecular compounds.
1)N
2O
5
2)SO
2
3)P
4O
10
4)CO
5)CO
2
6)SiO
2
7)H
2O
2
8)CF
4
9)PBr
3
10)SF
2

35. Name the following molecular compounds.
1)N
2O
5 dinitrogen pentoxide
2)SO
2 sulfur dioxide
3)P
4O
10 tetraphosphorus decoxide
4)CO carbon monoxide
5)CO
2 carbon dioxide
6)SiO
2 silicon dioxide
7)H
2O
2 dihydrogen dioxide
8)CF
4 carbon tetrafluoride
9)PBr
3 phosphorus tribromide
10)SF
2 sulfur difluoride

36. 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

37. Molecular Compound Nomenclature
Write the formula for the following compounds.
1)carbon tetraiodide CI
4
2)trinitrogen heptoxide N
3O
7
3)triphosphorus hexasulfideP
3S
6
4)oxygen dichloride OCl
2
5)disilicon triphosphide Si
2P
3
6)tetranitrogen heptoxide N
4O
7
7)carbon disulfide CS
2
8)dihydrogen monosulfide H
2S
9)trihydrogen monophosphide H
3P
10)silicon disulfide SiS
2

38. Section Review Problem #2 page 231
2-a-aluminum + bromine 
AlBr
3
b-sodium + oxygen 
Na
2O
c- magnesium + iodine 
MgI
2
d-lead (II) + oxygen 
PbO
e-tin (II) + iodine  SnI
2
f- iron (III) + sulfur Fe
2S
3
g-copper (II) + nitrate  Cu(NO
3)
2
h-ammonium + sulfate  (NH
4)
2SO
4

39. Section Review Problem #3 page 231
3a-NaI 
sodium iodide
b-MgS 
magnesium sulfide
c-CaO 
calcium oxide
d-K
2S 
potassium sulfide
e-CuBr  copper (I) bromide
f- FeCl
2 iron (II) chloride

40. Section Review Problem #4 (a-e) page 231
4a-sodium hydroxide 
NaOH
b-lead (II) nitrate 
Pb(NO
3)
2
c-iron (II) sulfate 
FeSO
4
d-diphosphorus trioxide 
P
2O
3
e-carbon diselenide 
CSe
2

41. 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

42. Assigning Oxidation Numbers
1)The atoms in a pure elementare assigned an oxidation number of zero.
2)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.
3)Fluorine always has an oxidation number of -1 because it is the most
electronegative element.
4)Oxygen has an oxidation number of -2 in almost all compounds.

43. 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 compoundis
equal to zero.
7)The algebraic sum of the oxidation numbers of the atoms in a
polyatomic ionequal the ion’s charge.
8)Oxidation numbers can also be assigned to atoms in an ionic
compound.

44. Periodic Table with Oxidation Number

45. Using Oxidation Numbers
•Do practice
problem #1 on
page 234.

46. Practice #1 pg 234
a)HCl H = 1+Cl= 1-
b)CF
4 C = 4+F = 1-
c)PCl
3 P = 3+Cl= 1-
d) SO
2 S = 4+O = 2-
e)HNO
3 H = 1+N = 5+O = 2-
f)KH K = 1+H = 1-
g)P
4O
10 P= 5+O = 2-
h)HClO
3H = 1+Cl= 5+O = 2-
i)N
2O
5 N = 5+O = 2-
j)GeCl
2 Ge= 2+Cl= 1-

47. Oxidation Number problems
What would be the oxidation number of each element
in the following compounds & polyatomic ions?
H
2O H = O =
H
2SO
4 H = S = O =
N
2O
5 N = O =
SO
4
2-
S = O =
PO
4
3-
P = O =

48. What would be the oxidation number of each element in the following
compounds & polyatomic ions?
H
2O
H = 1+O = 2-
H
2SO
4
H = 1+S = 6+O = 2-
N
2O
5
N = 5+O = 2-
SO
4
2-
S = 6+O = 2-
PO
4
3-
P = 5+O = 2-

49. 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.
PCl
3phosphorus trichloride 
phosphorus (III) chloride
Do section review problems #1-2 on page 235.

50. •Problems page 235
#1a-HF
H = 1+ F = 1-
b-CI
4
C = 4+ I = 1-
c-H
2O
H = 1+ O = 2-
d-PI
3
P = 3+ I = 1-
e-CS
2 C = 4+ S = 2-
f-This is a rare case when O = 1-.
g-H
2CO
3H = 1+ C = 4+ O = 2-
h-NO
2
1-
N = 3+ O = 2-

51. •Problems page 235
#2a-CI
4
carbon (IV) iodide
b-SO
3
sulfur (VI) oxide
c-As
2S
3
arsenic (III) sulfide
d-NCl
3
nitrogen (III) chloride

52. 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)
N
2O
5nitrogen __ oxide
SiO
2silicon __ oxide
CF
4 carbon __ fluoride
PI
3 phosphorus __ iodide
SiBr
4silicon __ bromide

53. •Using oxidation numbers & the stock system, what would be the names of the
following binary molecular compounds?
N
2O
5 
nitrogen V oxide
SiO
2 
silicon IV oxide
CF
4
carbon IV fluoride
PI
3
phosphorus III iodide
SiBr
4 
silicon IV bromide

54. Chapter 7 part 1 worksheet
•Write the formula for the following ionic compounds.
1-magnesium phosphate
Mg
3(PO
4)
2
2-calcium hydroxide
Ca(OH)
2
3-iron (II) nitrate
Fe(NO
3)
2
4-iron (III) sulfate
Fe
2(SO
4)
3
5-ammonium carbonate
(NH
4)
2CO
3

55. •Write the name of the following ionic compounds.
6-FeSO
4
iron (II) sulfate
7-FePO
4
iron (III) phosphate
8-KNO
3
potassium nitrate
9-CuSO
4
copper (II) sulfate
10-Cu
2SO
4
copper (I) sulfate

56. •Write the formula of the following molecular compounds.
11-dinitrogen pentoxide
N
2O
5
12-triphosphorus heptasulfide
P
3S
7
13-silicon dioxide
SiO
2
14-carbon tetrachloride
CCl
4
15-disulfur trioxide
S
2O
3

57. •Write the name of the following molecular compounds using numerical
prefixes.
16-H
2O
2
dihydrogendioxide
17-P
2O
6
diphosphorushexoxide
18-SiS
2
silicon disulfide
19-N
4O
10
tetranitrogendecoxide
20-PI
3
phosphorus triiodide

58. Write the name of the following molecular compounds
using the Stock system.
21-H
2O
hydrogen (I) oxide
22-P
2O
5
phosphorus (V) oxide
23-SiS
2
silicon (IV) sulfide
24-N
4O
10
nitrogen (V) oxide
25-PI
3
phosphorus (III) iodide

59. Determine the oxidation numbers assigned to each
element in the following compounds or ions.
26-N
2O
5
N = 5+O = 2-
27-CO
2
C = 4+O = 2-
28-SO
3
S = 6+O = 2-
29-PO
4
3-
P = 5+O = 2-
30-NO
3
1-
N = 5+ O = 2-

60. 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)

61. 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)

62. 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)

63. 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.
• Na
2S sodium sulfide
• FeSO
4 iron (II) sulfate
• Fe
3(PO
4)
2 iron (II) phosphate

64. Chemistry Chapter 7 part 1 Practice Test
•What is the formula of the following ionic compounds?
• copper (I) phosphate Cu
3PO
4
• copper (II) phosphate Cu
3(PO
4)
2
• magnesium nitride Mg
3N
2
• iron (III) sulfate Fe
2(SO
4)
3

65. Chemistry Chapter 7 part 1 Practice Test
•Name the following molecular compounds.
• N
2O
5 dinitrogen pentoxide
• PF
3 phosphorus trifluoride
• CBr
4 carbon tetrabromide
•What is the formula of the following molecular compounds?
• sulfur dichloride SCl
2
• diphosphorus pentoxideP
2O
5
• silicon disulfide SiS
2

66. Chemistry Chapter 7 part 1 Practice Test
•What is the oxidation number of each element in the following
molecular compounds?
• N
2O
5 N = 5+O = 2-
• SO
4
2-
S =6+O = 2-
• H
3PO
4H =1+P =5+O = 2-

67. Chemistry In Action
•Read “Mass Spectrometry: Identifying Molecules” on page 236.
•Answer questions #1 & 2 at the end of the reading.

68. Modern Chemistry
•Chapter 7
•Part 2

69. 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.

70. •Practice #1 page 238
a) H
2SO
4 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(NO
3)
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

71. •Practice #2 page 239
a) Al
2S
3
2 Al x 27.0 = 54.0
3 S x 32.1 = 96.3
54.0 + 96.3 = 150.3 g/mol
b) NaNO
3
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 Bax 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

72. Review Quiz (10 pts)
•Calculate the molar mass of each of the following compounds. Please
show your workand use the correct labelfor each molar mass.
1-CaF
2
2-H
2O
3-CO
2
4-PBr
3
5-Al
2(SO
4)
3

73. Molar Mass as a Conversion Factor
# moles
÷molar mass x molar mass
# grams #grams
Do Practice problems #1 & 3 on page 242.

74. •Problem #1 page 242
a) 6.60 g (NH
4)
2SO
4N = 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 (NH
4)
2SO
4
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

75. •Problem #3 page 242
6.25 mol of copper (II) nitrate = ? g
copper (II) nitrate = Cu(NO
3)
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(NO
3)
2

76. mass-mole & mole-mass review quiz
1)How many moles of H
2O are there in 45.0 grams of H
2O?
( molar mass of H
2O = 18.0 g/mol)
2)How many moles of CO
2are there in 220.0 grams of CO
2?
(molar mass of CO
2= 44.0 g/mol)
3)How many grams of H
2O are in 5.5 moles of water?
4)How many grams of CO
2are in 0.05 moles of CO
2?
5)How many grams of H
2CO
3are in 1.75 moles of the
substance? (molar mass of H
2CO
3= 62.0 g/mol)

77. Honors Class-mass-mole & mole-mass review quiz
1)How many moles of H
2O are there in 45.0 grams of H
2O?
2)How many moles of CO
2are there in 220.0 grams of CO
2?
3)How many grams of H
2O are in 5.5 moles of water?
4)How many grams of CO
2are in 0.05 moles of CO
2?
5)How many grams of H
2CO
3are in 1.75 moles of the
substance?

78. Percentage Composition
•percentage composition-the percentage of the total mass of each element in a
compound
mass of element in 1 molex 100%
molar mass of compound
eg. CO
2mass C = 1 x 12.0 = 12.0
mass O = 2 x 16.0 = 32.0
molar mass of CO
2= 44.0 g/mol
%C = 12.0/44.0 (100) = 27.3%
%O = 32.0/44.0 (100) = 72.7%

79. eg H
2O = H = 2 x 1.0 = 2.0
O = 1 x 16.0 = 16.0
18.0 g/mol
%H in H
2O = 2.0 x 100 = 11.1%
18.0
% O in H
2O = 16.0x 100 = 88.9%
18.0

80. % composition by mass practice
Do Practice problems #1-3 on page
244.
Do Section Review problems #1, 3,
& 5 on page 244.

81. •Problem #1 page 244
a) PbCl
2 Pb = 1 x 207.2 = 207.2
Cl = 2 x 35.5 = 71.0
278.2
Pb = 207.2x 100 = 74.5%
278.2
Cl = 71.0 x 100 = 25.5%
278.2

82. 1-b)Ba(NO
3)
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= 137.3x 100 = 52.5%
261.3
N = 28.0 x 100 = 10.7%
261.3
O = 96.0 x 100 = 36.7%
261.3

83. •Problem #2 page 244
•ZnSO
4·7H
2O Zn = 1 x 65.4 = 65.4
S = 1 x 32.1 = 32.1
O = 4 x 16.0 = 64.0
H
2O = 7 x 18.0 =126.0
287.5
%H
2O = 126.0x 100 = 43.8%
287.5

84. •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

85. •Section Review #1 page 244
(NH
4)
2CO
3
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

86. •Section Review #3
mass of 3.25 mol Fe
2(SO
4)
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

87. •Section Review #5
% composition of each element of (NH
4)
2CO
3
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 = 28.0x 100 = 29.2%
96.0
%H = 8.0x 100 = 8.3%
96.0
%C = 12.0 x 100 = 12.5%
96.0
%O = 48.0x 100 = 50.0%
96.0

88. % composition by mass quiz
1-Find the % composition by mass of each element in the compound
H
3PO
4.
2-Find the % composition by mass of each element in the compound
N
2O
5.

89. 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.

90. 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
CH
3= empirical formula (does not exist)
C
2H
6= molecular formula (ethene)

91. Empirical Formulas
•The formulas of ionic
compounds are empirical
formulasby the definition
of ionic formulas.
•The formulas of molecular
compounds may or may
notbe the same as its
empirical formula.

92. Calculating an Empirical Formula
1)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.

93. 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

94. 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

95. 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

96. 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 CH
2O

97. 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 Na
2SO
4

98. Calculating an Empirical Formula
•Review sample problem M on page 247.
•Do practice problems #1, 2, & 3 on page 247.

99. •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 = 1S = 1.14/1.14 = 1
Ratio = 1:1 so FeSis the empirical formula

100. •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 K
2Cr
2O
7

101. •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 CaBr
2

102. 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?

103. 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)
1)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.

104. Calculating a Molecular Formula
1)empirical formula = P
2O
5
2)molecular mass is 283.89
3)empirical mass is 141.94
4)Dividing the molecular mass by the
empirical mass gives a multiplication
factor of : 283.89 ÷
141.94 = 2.0001 2
5)2 x (P
2O
5) P
4O
10

105. Chapter 7 Problems
•Do practice
problems #1 & 2 on
page 249.
•Do section review
problems #1-4 on
page 249.

106. •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 C
6H
6

107. •Practice problem #2 page 249
formula mass = 34.00 amu
0.44 g H & 6.92 g O
1
st
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 H
2O
2

108. •Section review problem #1 page 249.
36.48% Na25.41% 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 Na
2SO
3

109. •Section review problem #2 page 249.
53.70% Fe46.30% 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 Fe
2S
3

110. Section review problem #3 page 249
1.04 g K0.70 g Cr0.86 g O
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 = K
2CrO
4

111. •Section Review problem #4 page 249
4.04 g N 11.46 g Of.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 N
2O
5
e.f.m. = 108
f.m./e.f.m. = 108/108 = 1
empirical formula is same as molecular formula N
2O
5

112. 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.

113. 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

114. Calculating an Empirical Formula
1)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.

115. 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)
1)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.

116. Chapter 7 part 2 quiz #5
Calculating molecular formulas
1-A molecular compound has an empirical formula of CH
3. Its molecular
formula mass is 30 amu. What is the molecular formula of this
compound?

117. 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?

118. Final Practice-chapter 7 part 2
1- Determine the molar mass of the compound
Na
3PO
4 .
2- How many moles of CO
2are in 198 g ?
3- What is the mass of 2.25 moles of H
2O ?
4- What is the % composition of each element of
the compound P
4O
10?
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 ?

119. Final Practice-chapter 7 part 2
1- Determine the molar mass of the compound
Na
3PO
4 .
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 CO
2are 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 CO
2

120. 3- What is the mass of 2.25 moles of H
2O ?
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 H
2O
4- What is the % composition of each element of
the compound P
4O
10?
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%

121. 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 = N
2O
5
e.f.m. = (2 x 14) + (5 x 16) = 108
216 (mfm)/108 (efm) = 22 x 2:5 4:10 N
4O
10

122. 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)

123. •Essay Question:
____ & ____ are examples of the empirical and the molecular
formula of a compound, respectively. Explain the relationship
between these two types of formulas.

124. 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)

125. •Essay Question:
____ & ____ are examples of the empirical and the molecular
formula of a compound, respectively. Explain the relationship
between these two types of formulas.