modern chemistry chapter 7 chemical formulas & chemical compounds modern chemistry chapter 7...
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Modern Chemistry Chapter 7Modern Chemistry Chapter 7Chemical Formulas & Chemical CompoundsChemical Formulas & Chemical Compounds
A chemical A chemical
formula indicates formula indicates the kind and the kind and relative number relative number of atoms in a of atoms in a chemical chemical compound.compound.
CC88HH1818 (octane) has (octane) has 8 carbon and 18 8 carbon and 18 hydrogen atoms.hydrogen atoms.
Forming Ionic CompoundsForming Ionic Compounds Compounds that have the elements held together by Compounds that have the elements held together by
ionic bonds are called ionic compounds.ionic bonds are called ionic compounds. For an ionic compound to exist, the algebraic sum of For an ionic compound to exist, the algebraic sum of
the positive and the negative charges of the ions the positive and the negative charges of the ions MUST = 0.MUST = 0.
For instance, when a calcium atom becomes an ion, it For instance, when a calcium atom becomes an ion, it has an overall 2+ charge which must be neutralized has an overall 2+ charge which must be neutralized by ion(s) that have a 2- charge.by ion(s) that have a 2- charge.
IF a CaIF a Ca2+ 2+ cation forms an ionic bond with an Ocation forms an ionic bond with an O2-2- anion, anion, the resulting compound will be neutral and the the resulting compound will be neutral and the formula would be CaO.formula would be CaO.
However, if the CaHowever, if the Ca2+ 2+ bonds with a Fbonds with a F- - anion, it would anion, it would require two Frequire two F-- ions to neutralize the Ca ions to neutralize the Ca2+ 2+ CaF CaF22
Calcium ( CaCalcium ( Ca2+2+ ) combines with oxygen ( O ) combines with oxygen ( O2- 2- ) ) CaO :CaO :
++ ---------- ---------- --
CaCa2+2+ O O2-2-
++ ---------- ---------- --
Calcium (CaCalcium (Ca2+ 2+ ) combines with fluorine (F) combines with fluorine (F1-1- ) ) CaFCaF22::
++ ---------- ---------- -- FF1-1-
CaCa2+2+
++ ---------- ---------- -- FF1-1-
Binary Ionic CompoundsBinary Ionic Compounds
monatomic ions-monatomic ions- ions ions formed from a single atomformed from a single atom
– IF the ion has a IF the ion has a positivepositive charge, use charge, use the name of the the name of the elementelement
– IF the ion has a IF the ion has a negativenegative charge, charge, replace the ending of replace the ending of the element name with the element name with “ide”.“ide”.
Binary Ionic CompoundsBinary Ionic Compounds binary compound-binary compound- a compound composed of two a compound composed of two
elementselements
Writing binary Writing binary ionic compound formulasionic compound formulas::1)1) Write the symbols for the ions side by side with Write the symbols for the ions side by side with
the cation being first.the cation being first.2)2) IF the charges of the two ions do not add to zero, IF the charges of the two ions do not add to zero,
cross over the charges by using the absolute value cross over the charges by using the absolute value of each ion’s charge as the subscript for the other of each ion’s charge as the subscript for the other ion so the algebraic sum of the ions equals zero.ion so the algebraic sum of the ions equals zero.
3)3) Check the subscripts and make sure they are in Check the subscripts and make sure they are in the smallest whole number ratio possible.the smallest whole number ratio possible.
e.g.e.g. aluminum oxide aluminum oxide AlAl3+3+OO2- 2- AlAl22OO33
Naming Binary Ionic Naming Binary Ionic CompoundsCompounds
nomenclature-nomenclature- a naming system a naming system
Naming ionic compounds:Naming ionic compounds:1)1) Write the name of the cation in the Write the name of the cation in the
formula.formula.2)2) Write the name of the anion in the formula.Write the name of the anion in the formula.
AlAl22OO33 aluminum oxide aluminum oxide
Do practice problems #1 & 2 on page 223.Do practice problems #1 & 2 on page 223.
Problems- page 223Problems- page 223
1)1) a-a- potassium (Kpotassium (K++) & iodide (I) & iodide (I--) ) KIKI
b-b- magnesium (Mgmagnesium (Mg2+2+) & chloride (Cl) & chloride (Cl--) ) MgClMgCl22
c-c- sodium (Nasodium (Na++) & sulfide (S) & sulfide (S2-2-) )
NaNa22SSd-d- aluminum (Alaluminum (Al3+3+) & sulfide (S) & sulfide (S2-2-) )
AlAl22SS33
e-e- aluminum (Alaluminum (Al3+3+) & nitride (N) & nitride (N3-3-) ) AlNAlN
#2#2 a)a) AgCl AgCl
silver chloridesilver chloride
b)b) ZnO ZnO
zinc oxidezinc oxide
c)c) CaBrCaBr22
calcium bromidecalcium bromide
d)d) SrFSrF22
strontium fluoridestrontium fluoride
e)e) BaO BaO
barium oxidebarium oxide
f)f) CaClCaCl22
calcium chloridecalcium chloride
Stock System of Stock System of NomenclatureNomenclature
Some metallic elements that form cations such as Some metallic elements that form cations such as chromium, cobalt, copper, iron, lead, manganese, chromium, cobalt, copper, iron, lead, manganese, mercury, nickel, and tin can form cations of more mercury, nickel, and tin can form cations of more than one charge. (See ion chart)than one charge. (See ion chart)
For cations that can have For cations that can have multiple ionic chargesmultiple ionic charges, , place a place a Roman numeral Roman numeral in parentheses that is in parentheses that is equal to the ionic chargeequal to the ionic charge after the name of the after the name of the metal.metal.
CuCu1+1+ copper (I) copper (I) FeFe2+2+ iron (II) iron (II)
CuCu2+2+ copper (II) copper (II) FeFe3+3+ iron (III) iron (III)
Using the Stock SystemUsing the Stock System
1)1) Write the formula of the ionic compound.Write the formula of the ionic compound.
2)2) Use the charge of the anion to determine the Use the charge of the anion to determine the charge of the cation.charge of the cation.
3)3) Write the name of the cation with the charge Write the name of the cation with the charge followed by the name of the anion.followed by the name of the anion.
CuCl CuCl copper (I) chloride copper (I) chloride
CuClCuCl22 copper (II) chloride copper (II) chloride
Do practice problems #1 & 2 on page 225.Do practice problems #1 & 2 on page 225.
Practice- page 225Practice- page 225
#1#1 a)a) CuCu2+2+ & Br- & Br-
CuBrCuBr22 copper II bromide copper II bromide
b)b) Fe Fe 2+2+ & O & O2-2-
FeO FeO iron II oxide iron II oxide
c)c) Pb Pb 2+2+ & Cl & Cl--
PbClPbCl22 lead II chloride lead II chloride
d)d) Hg Hg 2+2+ & S & S2-2-
HgS HgS mercury II sulfide mercury II sulfide
e)e) Sn Sn 2+2+ & F & F--
SnFSnF22 tin II fluoride tin II fluoride
f)f) Fe Fe 3+3+ & O & O2-2-
FeFe22OO33 iron III oxide iron III oxide
Practice- page 225Practice- page 225
#2#2 a)a) CuO CuO
copper II oxidecopper II oxide
b)b) CoFCoF33
cobalt III fluoridecobalt III fluoride
c)c) SnISnI44
tin IV iodidetin IV iodide
d)d) FeS FeS
iron II sulfideiron II sulfide
Polyatomic IonsPolyatomic Ions
polyatomic ion- polyatomic ion- a group of a group of covalently bonded covalently bonded atoms with an ionic atoms with an ionic chargecharge
oxyanion-oxyanion- a a negatively charged negatively charged polyatomic ion that polyatomic ion that contains oxygencontains oxygen
Ionic Compounds & Polyatomic IonsIonic Compounds & Polyatomic Ions
Writing and naming strategies are the same for Writing and naming strategies are the same for ionic compounds with polyatomic ions. ionic compounds with polyatomic ions. However, if more than one polyatomic ion is However, if more than one polyatomic ion is needed in the formula, needed in the formula, the formula of the the formula of the polyatomic ion is placed in parentheses and a polyatomic ion is placed in parentheses and a subscript is used outside the parenthesis to subscript is used outside the parenthesis to show how many of the polyatomic ions are show how many of the polyatomic ions are needed.needed.
e.g.e.g. iron (II) nitrate iron (II) nitrate Fe(NO Fe(NO33))22
Do practice problems #1 & 2 on page 227.Do practice problems #1 & 2 on page 227.
Practice Problems #1 page 227Practice Problems #1 page 227
11 a-a- sodium iodidesodium iodideNaINaI
b-b- calcium chloridecalcium chloride
CaClCaCl22c-c- potassium sulfidepotassium sulfide
KK22SSd-d- lithium nitratelithium nitrate
LiNOLiNO33
e-e- copper (II) sulfatecopper (II) sulfate
CuSOCuSO44
f-f- sodium carbonatesodium carbonate
NaNa22COCO33
g-g- calcium nitritecalcium nitrite
Ca(NOCa(NO22))22
h-h- potassium perchloratepotassium perchlorateKClOKClO44
2a-2a- AgAg22OO
silver oxidesilver oxide
b-b- Ca(OH)Ca(OH)22
calcium hydroxidecalcium hydroxide
c-c- KClOKClO33
potassium chloratepotassium chlorate
d-d- NHNH44OHOH
ammonium hydroxideammonium hydroxide
e-e- FeFe22(CrO(CrO44))33
iron (III) chromateiron (III) chromate
f-f- KClOKClO
potassium hypochloritepotassium hypochlorite
PracticePracticeDo the following formulas match the names Do the following formulas match the names
given?given?IF they do not match, provide the CORRECT IF they do not match, provide the CORRECT
name or formula.name or formula.
CuSOCuSO44 copper I sulfatecopper I sulfate
FeFe22(SO(SO44))33 iron III sulfateiron III sulfate
FeSOFeSO44 iron II sulfateiron II sulfate
copper I nitrate copper I nitrate CuNOCuNO33
copper II nitratecopper II nitrate CuCu22NONO33
PracticePracticeDo the following formulas match the names given?Do the following formulas match the names given?
CuSOCuSO44 copper I sulfatecopper I sulfateNO [copperII]NO [copperII]
FeFe22(SO(SO44))33 iron III sulfateiron III sulfateYESYES
FeSOFeSO44 iron II sulfateiron II sulfateYESYES
copper I nitrate copper I nitrate CuNOCuNO33
YESYES
copper II nitratecopper II nitrate CuCu22NONO33
NO [Cu(NONO [Cu(NO33))22]]
Ionic Compound NomenclatureIonic Compound NomenclatureName the following compounds:Name the following compounds:
1)1) MgBrMgBr22
2)2) CuOCuO3)3) CuCu22OO4)4) FeSOFeSO44
5)5) FeFe22(SO(SO44))33
6)6) CaSOCaSO44
7)7) CuCu22SOSO44
8)8) CuSOCuSO44
9)9) FePOFePO44
10)10) FeFe33(PO(PO44))22
Ionic Compound NomenclatureIonic Compound Nomenclature
Name the following compounds:Name the following compounds:
MgBrMgBr22
magnesium bromidemagnesium bromideCuOCuO
copper II oxidecopper II oxideCuCu22OO
copper I oxidecopper I oxideFeSOFeSO44
iron II sulfateiron II sulfateFeFe22(SO(SO44))33
iron III sulfateiron III sulfate
CaSOCaSO44
calcium sulfatecalcium sulfateCuCu22SOSO44
copper I sulfatecopper I sulfateCuSOCuSO44
copper II sulfatecopper II sulfateFePOFePO44
iron III phosphateiron III phosphateFeFe33(PO(PO44))22
iron II phosphateiron II phosphate
Ionic Compound NomenclatureIonic Compound Nomenclature
Write the formulas of the following ionic compounds:Write the formulas of the following ionic compounds:
1)1) aluminum nitratealuminum nitrate2)2) aluminum nitridealuminum nitride3)3) magnesium phosphatemagnesium phosphate4)4) magnesium bromidemagnesium bromide5)5) copper I sulfatecopper I sulfate6)6) copper II sulfatecopper II sulfate7)7) iron II nitrateiron II nitrate8)8) iron III fluorideiron III fluoride9)9) calcium hydroxidecalcium hydroxide10)10) calcium phosphatecalcium phosphate
aluminum nitratealuminum nitrate
Al Al 3+3+ NONO33 1-1-
Al(NOAl(NO33))33
aluminum nitridealuminum nitride
Al Al 3+3+ N N 3-3-
AlNAlN
magnesium phosphatemagnesium phosphate
Mg Mg 2+2+ POPO44 3-3-
MgMg33(PO(PO44))22
magnesium bromidemagnesium bromide Mg Mg 2+2+ Br Br 1-1-
MgBrMgBr22
copper I sulfatecopper I sulfate Cu Cu 1+1+ SOSO4 4 2-2-
CuCu22SOSO44
copper II sulfatecopper II sulfate Cu Cu 2+2+ SOSO44 2-2-
CuSOCuSO44
iron II nitrateiron II nitrate Fe Fe 2+2+ NONO33 1-1-
Fe(NOFe(NO33))22
iron III fluorideiron III fluoride Fe Fe 3+3+ F F 1-1-
FeFFeF33
calcium hydroxidecalcium hydroxide Ca Ca 2+2+ OH OH 1-1-
Ca(OH)Ca(OH)22
calcium phosphatecalcium phosphate Ca Ca 2+2+ POPO44 3-3-
CaCa33(PO(PO44))22
Binary Molecular Binary Molecular CompoundsCompounds
For this course, molecular compounds For this course, molecular compounds consist of two non-metals. For our consist of two non-metals. For our purposes, hydrogen will be considered a purposes, hydrogen will be considered a non-metal. non-metal.
The ratio of the elements is NOT determined The ratio of the elements is NOT determined by their individual ionic charges.by their individual ionic charges.
e.g. CO & COe.g. CO & CO22 or H or H22O & HO & H22OO22
Naming of Binary Molecular Compounds From Naming of Binary Molecular Compounds From FormulasFormulas
1)1) Write the Write the name of the first element name of the first element in the in the formula.formula.
2)2) Write the Write the name of the second element using the name of the second element using the suffix “ide”suffix “ide”..
3)3) Use numerical prefixes Use numerical prefixes to show the number of to show the number of atoms of each element.atoms of each element.
e.g. Pe.g. P22OO55 diphosphorus pentoxide diphosphorus pentoxide
1 = mono1 = mono 6 = hexa6 = hexa2 = di2 = di 7 = hepta7 = hepta3 = tri3 = tri 8 = octa8 = octa4 = tetra4 = tetra 9 = nona9 = nona5 = penta5 = penta 10 = deca 10 = deca
Binary Molecular CompoundsBinary Molecular Compounds
PP44OO1010 tetra + phosphorus & dec + oxide tetra + phosphorus & dec + oxide
tetraphosphorus decoxidetetraphosphorus decoxide
CO CO carbon & mon + oxide carbon & mon + oxide
carbon monoxidecarbon monoxide
COCO22 carbon & di + oxide carbon & di + oxide
carbon dioxidecarbon dioxide
Formulas for Molecular CompoundsFormulas for Molecular Compounds
1)1) The element with the smaller group number is The element with the smaller group number is usually given first. If both elements are in the usually given first. If both elements are in the same group, the element with the larger period same group, the element with the larger period number is given first. This element is given a number is given first. This element is given a prefix prefix ONLYONLY if it contributes more than one if it contributes more than one atom to the molecule of the compound.atom to the molecule of the compound.
2)2) The second element is named by combining a The second element is named by combining a prefix for the number of atoms of the element prefix for the number of atoms of the element in the compound, the root of the name of the in the compound, the root of the name of the element, and the suffix “ide”.element, and the suffix “ide”.
3)3) The “o” or the “a” at the end of a prefix is The “o” or the “a” at the end of a prefix is usually dropped when the word following the usually dropped when the word following the prefix begins with another vowel.prefix begins with another vowel.
Writing Molecular FormulasWriting Molecular Formulas
1)1) Write the formula of the first element in the Write the formula of the first element in the compound name followed by the numerical compound name followed by the numerical subscript that shows how many there are (if subscript that shows how many there are (if there is no numerical prefix, there is one atom there is no numerical prefix, there is one atom of the element).of the element).
2)2) Write the formula of the second element in the Write the formula of the second element in the compound name followed by a subscript that compound name followed by a subscript that shows how many atoms of the element are shows how many atoms of the element are designated by the numerical prefix in the name.designated by the numerical prefix in the name.
carbon dioxide carbon dioxide CO CO22
Do practice problems #1 & 2 on page 229.Do practice problems #1 & 2 on page 229.
Practice Problems #1 & 2 page 229Practice Problems #1 & 2 page 2291-1- a-a- SOSO33
sulfur trioxidesulfur trioxide
b-b- IClICl33
iodine trichlorideiodine trichloride
c-c- PBrPBr55
phosphorus pentabromidephosphorus pentabromide
2-2- a-a- carbon tetraiodidecarbon tetraiodide
CICI44
b-b- phosphorus trichloridephosphorus trichloride
PClPCl33c-c- dinitrogen trioxidedinitrogen trioxide
NN22OO33
Molecular Compound Molecular Compound NomenclatureNomenclature
Name the following molecular compounds.Name the following molecular compounds.1)1) NN22OO55
2)2) SOSO22
3)3) PP44OO1010
4)4) COCO
5)5) COCO22
6)6) SiOSiO22
7)7) HH22OO22
8)8) CFCF44
9)9) PBrPBr33
10)10) SFSF22
Name the following molecular Name the following molecular compounds.compounds.
1)1) NN22OO55 dinitrogen pentoxidedinitrogen pentoxide
2)2) SOSO22 sulfur dioxidesulfur dioxide
3)3) PP44OO1010 tetraphosphorus decoxidetetraphosphorus decoxide
4)4) COCO carbon monoxidecarbon monoxide
5)5) COCO22 carbon dioxidecarbon dioxide
6)6) SiOSiO22 silicon dioxidesilicon dioxide
7)7) HH22OO22 dihydrogen dioxidedihydrogen dioxide
8)8) CFCF44 carbon tetrafluoridecarbon tetrafluoride
9)9) PBrPBr33 phosphorus tribromidephosphorus tribromide
10)10) SFSF22 sulfur difluoridesulfur difluoride
Molecular Compound Molecular Compound NomenclatureNomenclature
Write the formula for the following Write the formula for the following compounds.compounds.
1)1) carbon tetraiodidecarbon tetraiodide
2)2) trinitrogen heptoxidetrinitrogen heptoxide
3)3) triphosphorus hexasulfidetriphosphorus hexasulfide
4)4) oxygen dichlorideoxygen dichloride
5)5) disilicon triphosphidedisilicon triphosphide
6)6) tetranitrogen heptoxidetetranitrogen heptoxide
7)7) carbon disulfidecarbon disulfide
8)8) dihydrogen monosulfidedihydrogen monosulfide
9)9) trihydrogen monophosphidetrihydrogen monophosphide
10)10)silicon disulfide silicon disulfide
Molecular Compound Molecular Compound NomenclatureNomenclature
Write the formula for the following compoundsWrite the formula for the following compounds..1)1) carbon tetraiodidecarbon tetraiodide CICI44
2)2) trinitrogen heptoxidetrinitrogen heptoxide NN33OO77
3)3) triphosphorus hexasulfidetriphosphorus hexasulfide PP33SS66
4)4) oxygen dichlorideoxygen dichloride OClOCl225)5) disilicon triphosphidedisilicon triphosphide SiSi22PP33
6)6) tetranitrogen heptoxidetetranitrogen heptoxide NN44OO77
7)7) carbon disulfidecarbon disulfide CSCS22
8)8) dihydrogen monosulfidedihydrogen monosulfide HH22SS
9)9) trihydrogen monophosphidetrihydrogen monophosphide HH33PP
10)10)silicon disulfide silicon disulfide SiSSiS22
Section Review Problem #2 page 231Section Review Problem #2 page 231
2-2- a-a- aluminum + bromine aluminum + bromine AlBrAlBr33
b-b- sodium + oxygen sodium + oxygen
NaNa22OOc-c- magnesium + iodine magnesium + iodine
MgIMgI22
d-d- lead (II) + oxygen lead (II) + oxygen PbOPbO
e-e- tin (II) + iodine tin (II) + iodine SnISnI22
f-f- iron (III) + sulfur iron (III) + sulfur FeFe22SS33
g-g- copper (II) + nitrate copper (II) + nitrate Cu(NOCu(NO33))22
h-h- ammonium + sulfate ammonium + sulfate (NH(NH44))22SOSO44
Section Review Problem #3 page 231Section Review Problem #3 page 231
33 a-a- NaI NaI sodium iodidesodium iodide
b-b- MgS MgS magnesium sulfidemagnesium sulfide
c-c- CaO CaO calcium oxidecalcium oxide
d-d- KK22S S
potassium sulfidepotassium sulfide
e-e- CuBr CuBr copper (I) bromidecopper (I) bromide
f-f- FeClFeCl22 iron (II) chlorideiron (II) chloride
Section Review Problem #4 (a-e) Section Review Problem #4 (a-e) page 231page 231
44 a-a- sodium hydroxide sodium hydroxide NaOHNaOH
b-b- lead (II) nitrate lead (II) nitrate Pb(NOPb(NO33))22
c-c- iron (II) sulfate iron (II) sulfate FeSOFeSO44
d-d- diphosphorus trioxide diphosphorus trioxide PP22OO33
e-e- carbon diselenide carbon diselenide CSeCSe22
Oxidation NumbersOxidation Numbers
oxidation numbers (oxidation states)-oxidation numbers (oxidation states)- assigned to the atoms composing a molecular assigned to the atoms composing a molecular compound or polyatomic ion that indicate the compound or polyatomic ion that indicate the general distribution of electrons among the general distribution of electrons among the bonded atoms in the compound or ionbonded atoms in the compound or ion
Assigning Oxidation NumbersAssigning Oxidation Numbers
1)1) The atoms in a The atoms in a pure elementpure element are are assigned an oxidation number of zero.assigned an oxidation number of zero.
2)2) The more electronegative (second) The more electronegative (second) element in a binary molecular element in a binary molecular compound is assigned the number compound is assigned the number equal to the negative charge it would equal to the negative charge it would have if it were an anion.have if it were an anion.
3)3) Fluorine always has an oxidation Fluorine always has an oxidation number of -1 because it is the most number of -1 because it is the most electronegative element.electronegative element.
4)4) Oxygen has an oxidation number of -2 Oxygen has an oxidation number of -2 in almost all compounds.in almost all compounds.
5)5) Hydrogen has an oxidation number of Hydrogen has an oxidation number of +1 in compounds where it is listed first +1 in compounds where it is listed first and -1 when it is listed last in the and -1 when it is listed last in the compound formula.compound formula.
6)6) The algebraic sum of all oxidation The algebraic sum of all oxidation numbers in a numbers in a neutral compoundneutral compound is is equal to zero.equal to zero.
7)7) The algebraic sum of the oxidation The algebraic sum of the oxidation numbers of the atoms in a numbers of the atoms in a polyatomic polyatomic ionion equal the ion’s charge. equal the ion’s charge.
8)8) Oxidation numbers can also be assigned Oxidation numbers can also be assigned to atoms in an ionic compound.to atoms in an ionic compound.
Using Oxidation NumbersUsing Oxidation Numbers
Do practice Do practice problem #1 on problem #1 on page 234.page 234.
Practice #1 pg 234Practice #1 pg 234
a)a) HClHCl H = 1+H = 1+ Cl = 1-Cl = 1-
b)b) CFCF44 C = 4+C = 4+ F = 1-F = 1-
c)c) PClPCl33 P = 3+P = 3+ Cl = 1-Cl = 1-
d) d) SOSO22 S = 4+S = 4+ O = 2-O = 2-
e)e) HNOHNO33 H = 1+H = 1+ N = 5+N = 5+ O = 2-O = 2-
f)f) KHKH K = 1+K = 1+ H = 1-H = 1-
g)g) PP44OO1010 P= 5+P= 5+ O = 2-O = 2-
h)h) HClOHClO33 H = 1+H = 1+ Cl = 5+Cl = 5+ O = 2-O = 2-
i)i) NN22OO55 N = 5+N = 5+ O = 2-O = 2-
j)j) GeClGeCl22 Ge = 2+Ge = 2+ Cl = 1-Cl = 1-
Oxidation Number problemsOxidation Number problemsWhat would be the oxidation number of each What would be the oxidation number of each
element in the following compounds & element in the following compounds & polyatomic ions?polyatomic ions?
HH22OO H = H = O = O =
HH22SOSO44 H = H = S = S = O =O =
NN22OO55 N = N = O =O =
SOSO442-2- S =S = O =O =
POPO443-3- P =P = O =O =
What would be the oxidation number of each element What would be the oxidation number of each element in the following compounds & polyatomic ions?in the following compounds & polyatomic ions?
HH22OOH = 1+H = 1+ O = 2- O = 2-
HH22SOSO44
H = 1+H = 1+ S = 6+S = 6+ O = 2-O = 2-
NN22OO55
N = 5+N = 5+ O = 2-O = 2-
SOSO442-2-
S = 6+S = 6+ O = 2-O = 2-
POPO443-3-
P = 5+P = 5+ O = 2-O = 2-
Oxidation Numbers & the Stock Oxidation Numbers & the Stock SystemSystem
We can use oxidation numbers assigned to We can use oxidation numbers assigned to the less electronegative (first) element to the less electronegative (first) element to name binary molecular compounds by using name binary molecular compounds by using the oxidation number as if it were a cation.the oxidation number as if it were a cation.
PClPCl33 phosphorus trichloride phosphorus trichloride
phosphorus (III) chloridephosphorus (III) chloride
Do section review problems #1-2 on page 235.Do section review problems #1-2 on page 235.
Problems page 235Problems page 235
#1a-#1a- HFHF
H = 1+H = 1+ F = 1-F = 1-
b-b- CICI44
C = 4+C = 4+ I = 1-I = 1-
c-c- HH22OO
H = 1+H = 1+ O = 2-O = 2-
d-d- PIPI33
P = 3+P = 3+ I = 1-I = 1-
e-e- CSCS22 C = 4+C = 4+ S = 2-S = 2-
f-f- This is a rare case when O = 1-.This is a rare case when O = 1-.
g-g- HH22COCO33 H = 1+ C = 4+ O = 2- H = 1+ C = 4+ O = 2-
h-h- NONO221-1- N = 3+N = 3+ O = 2-O = 2-
Problems page 235Problems page 235
#2a-#2a- CICI44
carbon (IV) iodidecarbon (IV) iodide
b-b- SOSO33
sulfur (VI) oxidesulfur (VI) oxide
c-c- AsAs22SS33
arsenic (III) sulfidearsenic (III) sulfide
d-d- NClNCl33
nitrogen (III) chloridenitrogen (III) chloride
Oxidation Numbers & the Stock Oxidation Numbers & the Stock SystemSystem
Using oxidation numbers & the stock system, what Using oxidation numbers & the stock system, what would be the names of the following binary would be the names of the following binary molecular compounds? (fill in the blank with the molecular compounds? (fill in the blank with the roman numeral)roman numeral)
NN22OO55 nitrogen __ oxide nitrogen __ oxide
SiOSiO22 silicon __ oxide silicon __ oxide
CFCF44 carbon __ fluoride carbon __ fluoride
PIPI33 phosphorus __ iodide phosphorus __ iodide
SiBrSiBr44 silicon __ bromide silicon __ bromide
Using oxidation numbers & the stock system, what Using oxidation numbers & the stock system, what would be the names of the following binary molecular would be the names of the following binary molecular compounds?compounds?
NN22OO5 5 nitrogen V oxidenitrogen V oxide
SiOSiO2 2 silicon IV oxidesilicon IV oxide
CFCF44 carbon IV fluoridecarbon IV fluoride
PIPI33 phosphorus III iodidephosphorus III iodide
SiBrSiBr4 4 silicon IV bromidesilicon IV bromide
Using Chemical FormulasUsing Chemical Formulas
formula mass-formula mass- the sum of the average the sum of the average atomic masses of all atoms represented in atomic masses of all atoms represented in its formulaits formula
– Do practice #1 on page 238Do practice #1 on page 238
molar mass- molar mass- the mass of one mole of an the mass of one mole of an element or a compound (equal to the element or a compound (equal to the formula mass expressed in grams)formula mass expressed in grams)
– Do practice problems #1 & 2 on page 239.Do practice problems #1 & 2 on page 239.
Practice #2 page 239Practice #2 page 239a) Ala) Al22SS33
2 Al x 27.0 = 54.02 Al x 27.0 = 54.03 S x 32.1 = 96.33 S x 32.1 = 96.3
54.0 + 96.3 = 150.3 g/mol54.0 + 96.3 = 150.3 g/mol
b) NaNOb) NaNO33 1 Na x 23.0 = 23.01 Na x 23.0 = 23.01 N x 14.0 = 14.01 N x 14.0 = 14.03 O x 16.0 = 48.03 O x 16.0 = 48.0
23.0 + 14.0 + 48.0 = 85.0 g/mol23.0 + 14.0 + 48.0 = 85.0 g/mol
c) Ba(OH)c) Ba(OH)22 1 Ba x 137.3 = 1 Ba x 137.3 = 137.3137.3
2 O x 16.0 = 32.02 O x 16.0 = 32.02 H x 1.0 = 2.02 H x 1.0 = 2.0
137.3 + 32.0 + 2.0 = 171.3 g/mol137.3 + 32.0 + 2.0 = 171.3 g/mol
Problem #1 page 242Problem #1 page 242
a) 6.60 g (NHa) 6.60 g (NH44))22SOSO44 N = 2 x 14.0 = 28.0N = 2 x 14.0 = 28.0H = 8 x 1.0 = 8.0H = 8 x 1.0 = 8.0S = 1 x 32.1 = 32.1S = 1 x 32.1 = 32.1O = 4 x 16.0 = O = 4 x 16.0 = 64.064.0 129.1129.1
6.60/129.1 = 0.051 mol (NH6.60/129.1 = 0.051 mol (NH44))22SOSO44
b) 4.5 kg = 4500 g Ca(OH)b) 4.5 kg = 4500 g Ca(OH)22
Ca = 1 x 40.1 = 40.1Ca = 1 x 40.1 = 40.1O = 2 x 16.0 = 32.0O = 2 x 16.0 = 32.0H = 2 x 1.0 = H = 2 x 1.0 = 2.02.0 74.174.1
4500/74.1 = 60.7 mol Ca(OH)4500/74.1 = 60.7 mol Ca(OH)22
Percentage CompositionPercentage Composition percentage composition-percentage composition- the percentage of the percentage of
the total mass of each element in a compoundthe total mass of each element in a compound
mass of element in 1 molemass of element in 1 mole x 100% x 100%molar mass of compoundmolar mass of compound
eg. COeg. CO22 mass C = 1 x 12.0 = 12.0mass C = 1 x 12.0 = 12.0mass O = 2 x 16.0 = mass O = 2 x 16.0 = 32.032.0 molar mass of COmolar mass of CO22 = 44.0 g/mol = 44.0 g/mol
%C = 12.0/44.0 (100) = 27.3%%C = 12.0/44.0 (100) = 27.3%%O = 32.0/44.0 (100) = 72.7%%O = 32.0/44.0 (100) = 72.7%
eg Heg H22O = O = H = 2 x 1.0 = 2.0 H = 2 x 1.0 = 2.0 O = 1 x 16.0 = 16.0O = 1 x 16.0 = 16.0 18.0 g/mol18.0 g/mol
%H in H%H in H22O = O = 2.0 2.0 x 100 = 11.1%x 100 = 11.1% 18.018.0
% O in H% O in H22O = O = 16.016.0 x 100 = 88.9% x 100 = 88.9% 18.018.0
% composition by mass % composition by mass practicepractice
Do Practice Do Practice problems #1-3 problems #1-3 on page 244.on page 244.
Do Section Review Do Section Review problems #1, 3, problems #1, 3, & 5 on page 244.& 5 on page 244.
Problem #1 page 244Problem #1 page 244
a) PbCla) PbCl22 Pb = 1 x 207.2 = 207.2Pb = 1 x 207.2 = 207.2Cl = 2 x 35.5 = Cl = 2 x 35.5 = 71.071.0 278.2278.2
Pb = Pb = 207.2207.2 x 100 = 74.5% x 100 = 74.5% 278.2278.2
Cl = Cl = 71.0 71.0 x 100 = 25.5% x 100 = 25.5% 278.2278.2
Problem #2 page 244Problem #2 page 244
ZnSOZnSO44·7H·7H22OO Zn = 1 x 65.4 = 65.4Zn = 1 x 65.4 = 65.4S = 1 x 32.1 = 32.1S = 1 x 32.1 = 32.1O = 4 x 16.0 = 64.0O = 4 x 16.0 = 64.0
HH22O = 7 x 18.0 =O = 7 x 18.0 = 126.0126.0
287.5287.5
%H%H22O = O = 126.0126.0 x 100 = 43.8% x 100 = 43.8% 287.5287.5
Section Review #3Section Review #3
mass of 3.25 mol Femass of 3.25 mol Fe22(SO(SO44))33 ? ?
Fe = 2 x 55.8 = 111.6Fe = 2 x 55.8 = 111.6S = 3 x 32.1 = 96.3S = 3 x 32.1 = 96.3O = 12 x 16.0 = O = 12 x 16.0 = 192.0192.0
399.9 g/mol399.9 g/mol
3.25 mol x 399.9 g/mol = 1299.7 3.25 mol x 399.9 g/mol = 1299.7 gg
Section Review #5Section Review #5% composition of each element of % composition of each element of
(NH(NH44))22COCO33
N = 2 x 14.0 = 28.0N = 2 x 14.0 = 28.0H = 8 x 1.0 = 8.0H = 8 x 1.0 = 8.0C = 1 x 12.0 = 12.0C = 1 x 12.0 = 12.0O = 3 x 16.0 = O = 3 x 16.0 = 48.048.0
96.0 g/mol96.0 g/mol
%N = %N = 28.028.0 x 100 = 29.2% x 100 = 29.2%96.096.0
%H = %H = 8.0 8.0 x 100 = 8.3% x 100 = 8.3% 96.096.0%C = %C = 12.0 12.0 x 100 = 12.5%x 100 = 12.5% 96.096.0%O = %O = 48.048.0 x 100 = 50.0% x 100 = 50.0% 96.096.0
To find molar mass: add the masses of To find molar mass: add the masses of the elements in the formula of the the elements in the formula of the compound.compound.
To find number of grams (mass): To find number of grams (mass): multiply # of moles times the molar multiply # of moles times the molar mass of the compound.mass of the compound.
To find the number of moles: divide To find the number of moles: divide the number of grams by the molar the number of grams by the molar mass of the compound.mass of the compound.
To calculate % composition by mass: To calculate % composition by mass:
1- find the molar mass of a compound1- find the molar mass of a compound
2- divide the mass of each element by 2- divide the mass of each element by thethe
molar mass of the compoundmolar mass of the compound
3- multiply by 100 to convert each 3- multiply by 100 to convert each ratio to a ratio to a
percentpercent