1. 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.

2. 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

3. Calcium ( Ca2+ ) combines with oxygen ( O2- )  CaO : Ca2+ O2- Calcium (Ca2+ ) combines with fluorine (F1- )  CaF2: F1- Ca F1-

4. 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”.

5. Binary Ionic Compounds
binary compound- a compound composed of two elements
Writing binary ionic compound formulas:
Write the symbols for the ions side by side with the cation being first.
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.
Check the subscripts and make sure they are in the smallest whole number ratio possible.
e.g. aluminum oxide Al3+O  Al2O3

6. Naming Binary Ionic Compounds
nomenclature- a naming system
Naming ionic compounds:
Write the name of the cation in the formula.
Write the name of the anion in the formula.
Al2O3  aluminum oxide
Do practice problems #1 & 2 on page 223.

7. Problems- page 223 a- potassium (K+) & iodide (I-)  KI
b- magnesium (Mg2+) & chloride (Cl-) 
MgCl2
c- sodium (Na+) & sulfide (S2-) 
Na2S
d- aluminum (Al3+) & sulfide (S2-) 
Al2S3
e- aluminum (Al3+) & nitride (N3-) 
AlN

8. #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

9. 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) Fe2+  iron (II)
Cu2+  copper (II) Fe3+  iron (III)

10. Using the Stock System Write the formula of the ionic compound.
Use the charge of the anion to determine the charge of the cation.
Write the name of the cation with the charge followed by the name of the anion.
CuCl  copper (I) chloride
CuCl2  copper (II) chloride
Do practice problems #1 & 2 on page 225.

11. Practice- page 225 CuBr2  copper II bromide b) Fe 2+ & O2- 
#1 a) Cu2+ & Br- 
CuBr2  copper II bromide
b) Fe 2+ & O2- 
FeO  iron II oxide
c) Pb 2+ & Cl- 
PbCl2  lead II chloride
d) Hg 2+ & S2- 
HgS  mercury II sulfide
e) Sn 2+ & F- 
SnF2  tin II fluoride
f) Fe 3+ & O2- 
Fe2O3  iron III oxide

12. Practice- page 225
#2 a) CuO 
copper II oxide
b) CoF3 
cobalt III fluoride
c) SnI4 
tin IV iodide
d) FeS 
iron II sulfide

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

14. 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.

15. Practice Problems #1 page 227
a- sodium iodide
NaI
b- calcium chloride
CaCl2
c- potassium sulfide
K2S
d- lithium nitrate
LiNO3

16. e- copper (II) sulfate
CuSO4
f- sodium carbonate
Na2CO3
g- calcium nitrite
Ca(NO2)2
h- potassium perchlorate KClO4

17. 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

18. 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

19. Practice CuSO4  copper I sulfate NO [copperII]
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
copper I nitrate  CuNO3
copper II nitrate  Cu2NO3
NO [Cu(NO3)2]

20. Ionic Compound Nomenclature
Name the following compounds:
MgBr2
CuO
Cu2O
FeSO4
Fe2(SO4)3
CaSO4
Cu2SO4
CuSO4
FePO4
Fe3(PO4)2

21. 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

22. CaSO4
calcium sulfate
Cu2SO4
copper I sulfate
CuSO4
copper II sulfate
FePO4
iron III phosphate
Fe3(PO4)2
iron II phosphate

23. Ionic Compound Nomenclature
Write the formulas of the following ionic compounds:
aluminum nitrate
aluminum nitride
magnesium phosphate
magnesium bromide
copper I sulfate
copper II sulfate
iron II nitrate
iron III fluoride
calcium hydroxide
calcium phosphate

24. aluminum nitrate Al 3+ NO3 1- Al(NO3)3 aluminum nitride Al 3+ N 3- AlN magnesium phosphate Mg 2+ PO4 3- Mg3(PO4)2

25. magnesium bromide Mg 2+ Br 1- MgBr2 copper I sulfate Cu 1+ SO4 2- Cu2SO4 copper II sulfate Cu 2+ SO4 2- CuSO4 iron II nitrate Fe 2+ NO3 1- Fe(NO3)2

26. iron III fluoride Fe 3+ F 1- FeF3 calcium hydroxide Ca 2+ OH 1- Ca(OH)2 calcium phosphate Ca 2+ PO4 3- Ca3(PO4)2

27. 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 H2O & H2O2

28. Naming of Binary Molecular Compounds From Formulas
Write the name of the first element in the formula.
Write the name of the second element using the suffix “ide”.
Use numerical prefixes to show the number of atoms of each element.
e.g. P2O5  diphosphorus pentoxide
1 = mono 6 = hexa
2 = di 7 = hepta
3 = tri 8 = octa
4 = tetra 9 = nona
5 = penta = deca

29. Binary Molecular Compounds
P4O10  tetra + phosphorus & dec + oxide
tetraphosphorus decoxide
CO  carbon & mon + oxide
carbon monoxide
CO2  carbon & di + oxide
carbon dioxide

30. Formulas for Molecular Compounds
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.
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”.
The “o” or the “a” at the end of a prefix is usually dropped when the word following the prefix begins with another vowel.

31. Writing Molecular Formulas
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).
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.

32. Practice Problems #1 & 2 page 229
1- a- SO3
sulfur trioxide
b- ICl3
iodine trichloride
c- PBr5
phosphorus pentabromide
2- a- carbon tetraiodide
CI4
b- phosphorus trichloride
PCl3
c- dinitrogen trioxide
N2O3

33. Molecular Compound Nomenclature
Name the following molecular compounds.
N2O5
SO2
P4O10 CO
CO2
SiO2
H2O2
CF4
PBr3
SF2

34. Name the following molecular compounds.
N2O5 dinitrogen pentoxide
SO2 sulfur dioxide
P4O10 tetraphosphorus decoxide
CO carbon monoxide
CO2 carbon dioxide
SiO2 silicon dioxide
H2O2 dihydrogen dioxide
CF4 carbon tetrafluoride
PBr3 phosphorus tribromide
SF2 sulfur difluoride

35. Molecular Compound Nomenclature
Write the formula for the following compounds.
carbon tetraiodide
trinitrogen heptoxide
triphosphorus hexasulfide
oxygen dichloride
disilicon triphosphide
tetranitrogen heptoxide
carbon disulfide
dihydrogen monosulfide
trihydrogen monophosphide
silicon disulfide

36. Molecular Compound Nomenclature
Write the formula for the following compounds.
carbon tetraiodide CI4
trinitrogen heptoxide N3O7
triphosphorus hexasulfide P3S6
oxygen dichloride OCl2
disilicon triphosphide Si2P3
tetranitrogen heptoxide N4O7
carbon disulfide CS2
dihydrogen monosulfide H2S
trihydrogen monophosphide H3P
silicon disulfide SiS2

37. Section Review Problem #2 page 231
2- a- aluminum + bromine 
AlBr3
b- sodium + oxygen 
Na2O
c- magnesium + iodine 
MgI2
d- lead (II) + oxygen 
PbO
e- tin (II) + iodine  SnI2
f- iron (III) + sulfur  Fe2S3
g- copper (II) + nitrate  Cu(NO3)2
h- ammonium + sulfate  (NH4)2SO4

38. Section Review Problem #3 page 231
a- NaI 
sodium iodide
b- MgS 
magnesium sulfide
c- CaO 
calcium oxide
d- K2S 
potassium sulfide
e- CuBr  copper (I) bromide
f- FeCl2  iron (II) chloride

39. Section Review Problem #4 (a-e) page 231
a- sodium hydroxide 
NaOH
b- lead (II) nitrate 
Pb(NO3)2
c- iron (II) sulfate 
FeSO4
d- diphosphorus trioxide 
P2O3
e- carbon diselenide 
CSe2

40. 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

41. Assigning Oxidation Numbers
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.

42. 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.

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

44. Practice #1 pg 234
a) HCl H = 1+ Cl = 1-
b) CF4 C = 4+ F = 1-
c) PCl3 P = 3+ Cl = 1-
d) SO2 S = 4+ O = 2-
e) HNO3 H = 1+ N = 5+ O = 2-
f) KH K = 1+ H = 1-
g) P4O10 P= 5+ O = 2-
h) HClO3 H = 1+ Cl = 5+ O = 2-
i) N2O5 N = 5+ O = 2-
j) GeCl2 Ge = 2+ Cl = 1-

45. Oxidation Number problems
What would be the oxidation number of each element in the following compounds & polyatomic ions?
H2O H = O =
H2SO4 H = S = O =
N2O5 N = O =
SO42- S = O =
PO43- P = O =

46. H = 1+ O = 2- H2SO4 H = 1+ S = 6+ O = 2- N2O5 N = 5+ O = 2- SO42-
What would be the oxidation number of each element in the following compounds & polyatomic ions?
H2O
H = 1+ O = 2-
H2SO4
H = 1+ S = 6+ O = 2-
N2O5
N = 5+ O = 2-
SO42-
S = 6+ O = 2-
PO43-
P = 5+ O = 2-

47. 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.

48. Problems page 235
#1a- HF
H = 1+ F = 1-
b- CI4
C = 4+ I = 1-
c- H2O
H = 1+ O = 2-
d- PI3
P = 3+ I = 1-
e- CS2 C = 4+ S = 2-
f- This is a rare case when O = 1-.
g- H2CO3 H = 1+ C = 4+ O = 2-
h- NO21- N = 3+ O = 2-

49. Problems page 235
#2a- CI4 
carbon (IV) iodide
b- SO3 
sulfur (VI) oxide
c- As2S3 
arsenic (III) sulfide
d- NCl3 
nitrogen (III) chloride

50. 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

51. nitrogen V oxide SiO2  silicon IV oxide CF4  carbon IV fluoride
Using oxidation numbers & the stock system, what would be the names of the following binary molecular compounds?
N2O 
nitrogen V oxide
SiO 
silicon IV oxide
CF 
carbon IV fluoride
PI 
phosphorus III iodide
SiBr4 
silicon IV bromide

52. 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.

53. Practice #2 page 239
a) Al2S3 
2 Al x 27.0 = 54.0
3 S x 32.1 = 96.3
= 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
= 85.0 g/mol
c) Ba(OH)2  1 Ba x = 137.3
2 O x 16.0 = 32.0
2 H x 1.0 = 2.0
= g/mol

54. Problem #1 page 242 a) 6.60 g (NH4)2SO4 N = 2 x 14.0 = 28.0
H = 8 x = 8.0
S = 1 x 32.1 = 32.1
O = 4 x 16.0 = 64.0
129.1
6.60/129.1 = 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

55. 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%

56. eg H2O = H = 2 x 1.0 =
O = 1 x 16.0 = 16.0
18.0 g/mol
%H in H2O = x 100 = 11.1%
18.0
% O in H2O = x 100 = 88.9%

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

58. Problem #1 page 244
a) PbCl2 Pb = 1 x = 207.2
Cl = 2 x = 71.0
278.2
Pb = x 100 = %
Cl = x 100 = %

59. 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 = x 100 = 43.8%

60. Section Review #3
mass of 3.25 mol Fe2(SO4)3 ?
Fe = 2 x 55.8 = 111.6
S = 3 x 32.1 =
O = 12 x 16.0 = 192.0
399.9 g/mol
3.25 mol x g/mol = g

61. Section Review #5
% composition of each element of (NH4)2CO3
N = 2 x 14.0 = 28.0
H = 8 x = 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 = x 100 = 8.3%
%C = 12.0 x 100 = 12.5%
%O = 48.0 x 100 = 50.0%

62. 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.

63. 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