Modern Chemistry Chapter 7 Chemical Formulas & Chemical Compounds

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


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


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


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


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


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

Chapter 7 part 1 worksheetChapter 7 part 1 worksheet Write the formula for the following ionic compounds.Write the formula for the following ionic compounds.

1-1- magnesium phosphatemagnesium phosphate

MgMg33(PO(PO44))22

2-2- calcium hydroxidecalcium hydroxide

Ca(OH)Ca(OH)22

3-3- iron (II) nitrateiron (II) nitrate

Fe(NOFe(NO33))22

4-4- iron (III) sulfateiron (III) sulfate

FeFe22(SO(SO44))33

5-5- ammonium carbonateammonium carbonate

(NH(NH44))22COCO33

Write the name of the following ionic compounds.Write the name of the following ionic compounds.

6-6- FeSOFeSO44

iron (II) sulfateiron (II) sulfate

7-7- FePOFePO44

iron (III) phosphateiron (III) phosphate

8-8- KNOKNO33

potassium nitratepotassium nitrate

9-9- CuSOCuSO44

copper (II) sulfatecopper (II) sulfate

10-10- CuCu22SOSO44

copper (I) sulfatecopper (I) sulfate

Write the formula of the following molecular compounds.Write the formula of the following molecular compounds.

11-11- dinitrogen pentoxidedinitrogen pentoxide

NN22OO55

12-12- triphosphorus heptasulfidetriphosphorus heptasulfide

PP33SS77

13-13- silicon dioxidesilicon dioxide

SiOSiO22

14-14- carbon tetrachloridecarbon tetrachloride

CClCCl4415-15- disulfur trioxidedisulfur trioxide

SS22OO33

Write the name of the following molecular Write the name of the following molecular compounds using numerical prefixes.compounds using numerical prefixes.

16-16- HH22OO22

dihydrogen dioxidedihydrogen dioxide

17-17- PP22OO66

diphosphorus hexoxidediphosphorus hexoxide

18-18- SiSSiS22

silicon disulfidesilicon disulfide

19-19- NN44OO1010

tetranitrogen decoxidetetranitrogen decoxide

20-20- PIPI33

phosphorus triiodidephosphorus triiodide

Write the name of the following molecular Write the name of the following molecular compounds using the compounds using the Stock systemStock system..

21-21- HH22OO

hydrogen (I) oxidehydrogen (I) oxide

22-22- PP22OO55

phosphorus (V) oxidephosphorus (V) oxide

23-23- SiSSiS22

silicon (IV) sulfidesilicon (IV) sulfide

24-24- NN44OO1010

nitrogen (V) oxidenitrogen (V) oxide

25-25- PIPI33

phosphorus (III) iodidephosphorus (III) iodide

Determine the oxidation numbers assigned to each Determine the oxidation numbers assigned to each element in the following compounds or ions.element in the following compounds or ions.

26-26- NN22OO55

N = 5+N = 5+ O = 2-O = 2-

27-27- COCO22

C = 4+C = 4+ O = 2-O = 2-

28-28- SOSO33

S = 6+S = 6+ O = 2-O = 2-

29-29- POPO443-3-

P = 5+P = 5+ O = 2-O = 2-

30-30- NONO331-1-

N = 5+N = 5+ O = 2-O = 2-

Honors Ch 7 part 1Honors Ch 7 part 134 multiple choice:34 multiple choice:

chemical formulas represent ? (3)chemical formulas represent ? (3)ionic formulas from names (5)ionic formulas from names (5)ionic compound names from formulas (4)ionic compound names from formulas (4)molecular compound names from molecular compound names from

formulas (4)formulas (4)molecular formulas from names (4)molecular formulas from names (4)oxidation number assignment rules (4)oxidation number assignment rules (4)determining oxidation numbers (5)determining oxidation numbers (5)naming binary molecular compounds naming binary molecular compounds

using the stock system (5)using the stock system (5)

Honors Ch 7 part 1Honors Ch 7 part 1

1 short answer:1 short answer:What type of compound cannot be What type of compound cannot be

represented by a molecular formula? represented by a molecular formula? Explain.Explain.

4 completion:4 completion:-name an ionic compound-name an ionic compound-name a polyatomic ion-name a polyatomic ion-determine oxidation numbers in a -determine oxidation numbers in a

polyatomic ion polyatomic ion and a compound and a compound

1 essay:1 essay:-eliminated (it will be on next test)-eliminated (it will be on next test)

Chemistry Ch 7 part 1 testChemistry Ch 7 part 1 test

25 multiple choice questions:25 multiple choice questions:chemical formulas & what they represent (2)chemical formulas & what they represent (2)determine ionic formula from name (4)determine ionic formula from name (4)determine ionic name from ionic formula (4)determine ionic name from ionic formula (4)determine molecular name from formula (4)determine molecular name from formula (4)determine molecular formula from name (4)determine molecular formula from name (4)rules for assigning oxidation numbers (3)rules for assigning oxidation numbers (3)determine oxidation numbers in compounds determine oxidation numbers in compounds

(4)(4)

Chemistry Chapter 7 part 1 Practice Chemistry Chapter 7 part 1 Practice TestTest

What do the letters and the subscripts in What do the letters and the subscripts in a chemical formula represent?a chemical formula represent?– The identities and the numbers of atoms of The identities and the numbers of atoms of

each element in a compound.each element in a compound. Name the following ionic compounds.Name the following ionic compounds. NaNa22S S sodium sulfidesodium sulfide

FeSOFeSO44 iron (II) sulfateiron (II) sulfate

FeFe33(PO(PO44))22 iron (II) phosphateiron (II) phosphate


What is the formula of the following What is the formula of the following ionic compounds?ionic compounds?

copper (I) phosphatecopper (I) phosphate CuCu33POPO44

copper (II) phosphatecopper (II) phosphate CuCu33(PO(PO44))22

magnesium nitridemagnesium nitrideMgMg33NN22

iron (III) sulfateiron (III) sulfate FeFe22(SO(SO44))33


Name the following molecular compounds.Name the following molecular compounds. NN22OO55 dinitrogen pentoxidedinitrogen pentoxide

PFPF33 phosphorus trifluoridephosphorus trifluoride

CBrCBr44 carbon tetrabromidecarbon tetrabromide What is the formula of the following What is the formula of the following

molecular compounds?molecular compounds? sulfur dichloridesulfur dichloride SClSCl22 diphosphorus pentoxidediphosphorus pentoxide PP22OO55

silicon disulfidesilicon disulfide SiSSiS22


What is the oxidation number of each What is the oxidation number of each element in the following molecular element in the following molecular compounds?compounds?

NN22OO55 N = N = 5+5+ O = 2-O = 2-

SOSO442-2- S =S = 6+6+ O = 2-O = 2-

HH33POPO44 H =H = 1+1+ P =P = 5+ 5+ O = O = 2-2-

Chemistry In ActionChemistry In Action

Read “Mass Spectrometry: Identifying Read “Mass Spectrometry: Identifying Molecules” on page 236.Molecules” on page 236.

Answer questions #1 & 2 at the end of the Answer questions #1 & 2 at the end of the reading.reading.

Modern ChemistryModern Chemistry

Chapter 7Chapter 7

Part 2Part 2

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 #1 page 238Practice #1 page 238a) Ha) H22SOSO44 2 H x 1.0 = 2.02 H x 1.0 = 2.0

1 S x 32.1 = 32.11 S x 32.1 = 32.14 O x 16.0 = 64.04 O x 16.0 = 64.0

2.0 + 32.1 + 64.0 = 98.1 amu2.0 + 32.1 + 64.0 = 98.1 amu

b) Ca(NOb) Ca(NO33))22 1 Ca x 40.1 = 1 Ca x 40.1 = 40.140.1

2 N x 14.0 = 28.02 N x 14.0 = 28.06 O x 16.0 = 96.06 O x 16.0 = 96.0

40.1 + 28.0 + 96.0 = 164.1 amu40.1 + 28.0 + 96.0 = 164.1 amu

c) = 95.0 amuc) = 95.0 amud) = 95.3 amud) = 95.3 amu

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

Review Quiz (10 pts)Review Quiz (10 pts)

Calculate the molar mass of each of Calculate the molar mass of each of the following compounds. Please the following compounds. Please show your workshow your work and use the and use the correct correct labellabel for each molar mass. for each molar mass.

1- 1- CaFCaF22

2-2- HH22OO

3-3- COCO22

4-4- PBrPBr33

5-5- AlAl22(SO(SO44))33

Molar Mass as a Conversion Molar Mass as a Conversion FactorFactor

# moles# moles

÷ molar mass÷ molar mass x molar mass x molar mass

# grams# grams #grams#grams

Do Practice problems #1 & 3 on page Do Practice problems #1 & 3 on page 242.242.

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


6.25 mol of copper (II) nitrate = ? g6.25 mol of copper (II) nitrate = ? g

copper (II) nitrate = Cu(NOcopper (II) nitrate = Cu(NO33))22

Cu = 1 x 63.5 = 63.5Cu = 1 x 63.5 = 63.5N = 2 x 14.0 = 28.0N = 2 x 14.0 = 28.0O = 6 x 16.0 = O = 6 x 16.0 = 96.096.0

187.5 g/mol187.5 g/mol

6.25 mol x 187.5 g/mol = 1172 g 6.25 mol x 187.5 g/mol = 1172 g Cu(NOCu(NO33))22

mass-mole & mole-mass review mass-mole & mole-mass review quizquiz

1)1) How many moles of HHow many moles of H22O are there in 45.0 grams of O are there in 45.0 grams of HH22O? ( molar mass of HO? ( molar mass of H22O = 18.0 g/mol)O = 18.0 g/mol)

2)2) How many moles of COHow many moles of CO22 are there in 220.0 grams of are there in 220.0 grams of COCO22? (molar mass of CO? (molar mass of CO22 = 44.0 g/mol) = 44.0 g/mol)

3)3) How many grams of HHow many grams of H22O are in 5.5 moles of water?O are in 5.5 moles of water?

4)4) How many grams of COHow many grams of CO22 are in 0.05 moles of CO are in 0.05 moles of CO22 ? ?

5)5) How many grams of HHow many grams of H22COCO33 are in 1.75 moles of the are in 1.75 moles of the substance? (molar mass of Hsubstance? (molar mass of H22COCO33 = 62.0 g/mol) = 62.0 g/mol)

Honors Class- mass-mole & mole-mass review Honors Class- mass-mole & mole-mass review quizquiz

1)1) How many moles of HHow many moles of H22O are there in 45.0 grams of O are there in 45.0 grams of HH22O? O?

2)2) How many moles of COHow many moles of CO22 are there in 220.0 grams are there in 220.0 grams of COof CO22? ?

3)3) How many grams of HHow many grams of H22O are in 5.5 moles of water?O are in 5.5 moles of water?

4)4) How many grams of COHow many grams of CO22 are in 0.05 moles of CO are in 0.05 moles of CO22 ? ?

5)5) How many grams of HHow many grams of H22COCO33 are in 1.75 moles of the are in 1.75 moles of the substance? substance?

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.


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

1-b)1-b) Ba(NOBa(NO33))22 Ba = 1 x 137.3 = 137.3Ba = 1 x 137.3 = 137.3N = 2 x 14.0 = 28.0N = 2 x 14.0 = 28.0O = 6 x 16.0 = O = 6 x 16.0 = 96.0 96.0

261.3261.3

Ba = Ba = 137.3137.3 x 100 = 52.5% x 100 = 52.5% 261.3261.3

N = N = 28.0 28.0 x 100 = 10.7%x 100 = 10.7% 261.3261.3

O = O = 96.0 96.0 x 100 = 36.7% x 100 = 36.7% 261.3261.3


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


Mg(OH)Mg(OH)22 = 175 g = 175 g oxygen = oxygen = 54.87%54.87%

175 x 175 x 54.854.8 = 95.9 g oxygen = 95.9 g oxygen 100100

95.9 g95.9 g = 6.0 mol oxygen = 6.0 mol oxygen16.0 g/mol16.0 g/mol

Section Review #1 page 244Section Review #1 page 244

(NH(NH44))22COCO33

N = 2 x 14.0 = 28.0N = 2 x 14.0 = 28.0

H = 8 x 1.0 = 8.0H = 8 x 1.0 = 8.0

C = 1 x 12.0 = 12.0C = 1 x 12.0 = 12.0

O = 3 x 16.0 = O = 3 x 16.0 = 48.048.0

96.0 amu 96.0 amu

96.0 g/mol96.0 g/mol

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


%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

% composition by mass quiz% composition by mass quiz

1-1- Find the % composition by mass of Find the % composition by mass of each each element in the compound element in the compound HH33POPO44..

2-2- Find the % composition by mass of Find the % composition by mass of each each element in the compound element in the compound NN22OO55..

HONORS- % composition by mass HONORS- % composition by mass quizquiz

1-1- Find the % composition by mass of Find the % composition by mass of each each element in the compound element in the compound hydrogen hydrogen phosphate.phosphate.

2-2- Find the % composition by mass of Find the % composition by mass of each each element in the compound element in the compound dinitrogen dinitrogen pentoxide.pentoxide.

Determining Chemical Determining Chemical FormulasFormulas

empirical formula- empirical formula- consists of the consists of the symbols for the elements combined in symbols for the elements combined in a compound, with subscripts showing a compound, with subscripts showing the smallest whole number mole ratio the smallest whole number mole ratio of the different atoms in the compoundof the different atoms in the compound

CHCH33 = empirical formula (does not = empirical formula (does not exist)exist)

CC22HH66 = molecular formula (ethene) = molecular formula (ethene)

Empirical FormulasEmpirical Formulas The formulas of ionic The formulas of ionic

compounds are compounds are empirical formulasempirical formulas by the definition of by the definition of ionic formulas.ionic formulas.

The formulas of The formulas of molecular molecular compounds compounds may or may or may notmay not be the same be the same as its empirical as its empirical formula.formula.

Calculating an Empirical Calculating an Empirical FormulaFormula

1)1) If the elements are in % composition by If the elements are in % composition by mass form, covert the percentages to mass form, covert the percentages to grams.grams.

2)2) Convert the masses of each element to Convert the masses of each element to moles by dividing the mass of the element moles by dividing the mass of the element by its molar mass.by its molar mass.

3)3) Select the element with the smallest Select the element with the smallest number of moles and divide the number number of moles and divide the number of moles of each element by that number of moles of each element by that number which will give you a 1:---:--- ratio.which will give you a 1:---:--- ratio.

4)4) IF the ratio is very close to a whole IF the ratio is very close to a whole number ratio, apply the numbers to each number ratio, apply the numbers to each element. If one of the number is not close element. If one of the number is not close to a whole number, use a multiplier to to a whole number, use a multiplier to convert the ratio to a whole number ratio.convert the ratio to a whole number ratio.

1-1- If the elements are in % If the elements are in % composition by mass form, covert composition by mass form, covert the percentages to grams.the percentages to grams.

e.g.e.g. C = 40.0% C = 40.0% 40.0 g 40.0 gH = 6.67% H = 6.67% 6.67 g 6.67 gO = 53.3% O = 53.3% 53.3 53.3

gg

2-2- Convert the masses of each Convert the masses of each element to moles by dividing the element to moles by dividing the mass of the element by its molar mass of the element by its molar mass.mass.

e.g.e.g. C = 40.0/12 = 3.33 molC = 40.0/12 = 3.33 molH = 6.67/1 = 6.67 molH = 6.67/1 = 6.67 molO = 53.3/16 = 3.33 molO = 53.3/16 = 3.33 mol

3-3- Select the element with the Select the element with the smallest number of moles and smallest number of moles and divide the number of moles of each divide the number of moles of each element by that number which will element by that number which will give you a 1:---:--- ratio.give you a 1:---:--- ratio.

e.g.e.g. C = 3.33/3.33 = 1C = 3.33/3.33 = 1H = 6.67/3.33 = 2H = 6.67/3.33 = 2O = 3.33/3.33 = 1O = 3.33/3.33 = 1

4-4- IF the ratio is very close to a whole IF the ratio is very close to a whole number ratio, apply the numbers to number ratio, apply the numbers to each element. If one of the number each element. If one of the number is not close to a whole number, use is not close to a whole number, use a multiplier to convert the ratio to a a multiplier to convert the ratio to a whole number ratio.whole number ratio.

e.g.e.g. 1:2:1 ratio 1:2:1 ratio CH CH22OO

Calculating an Empirical FormulaCalculating an Empirical Formula Sample Problem L page 246.Sample Problem L page 246. 32.38% Na, 22.65% S, & 44.99% O.32.38% Na, 22.65% S, & 44.99% O.1-1- convert to 32.38 g Na, 22.65 g S, & 44.99 g Oconvert to 32.38 g Na, 22.65 g S, & 44.99 g O 2-2- 32.38 32.38 ÷ 22.99 = 1.408 mol Na÷ 22.99 = 1.408 mol Na

22.65 ÷ 32.07 = 0.7063 mol S22.65 ÷ 32.07 = 0.7063 mol S44.99 ÷ 16.00 = 2.812 mol O44.99 ÷ 16.00 = 2.812 mol O

3-3- 1.408 ÷ 0.7063 = 1.993 mol Na 1.408 ÷ 0.7063 = 1.993 mol Na 2 2

0.7063 ÷ 0.7063 = 1 mol S0.7063 ÷ 0.7063 = 1 mol S2.812 ÷ 0.7063 = 3.981 mol O 2.812 ÷ 0.7063 = 3.981 mol O 4 4

4-4- Rounding Rounding 2:1:4 2:1:4 Na Na22SOSO44


Review sample problem M on page Review sample problem M on page 247.247.

Do practice problems #1, 2, & 3 on Do practice problems #1, 2, & 3 on page 247.page 247.

Practice problem #1 page 247Practice problem #1 page 247

63.52% iron (Fe)63.52% iron (Fe) 36.48% sulfur (S)36.48% sulfur (S)

Convert % to grams:Convert % to grams: Fe = 63.52g S = 36.48gFe = 63.52g S = 36.48g

Divide each element by its molar mass:Divide each element by its molar mass:Fe = 63.52/55.8 = 1.14 molFe = 63.52/55.8 = 1.14 molS = 36.48/32.1 = 1.14 molS = 36.48/32.1 = 1.14 mol

Divide each number of moles by the smallest Divide each number of moles by the smallest number:number:

Fe = 1.14/1.14 = 1Fe = 1.14/1.14 = 1 S = 1.14/1.14 = 1S = 1.14/1.14 = 1

Ratio = 1:1 so Ratio = 1:1 so FeSFeS is the empirical formula is the empirical formula


K = 26.56%K = 26.56% Cr = 35.41% Cr = 35.41% O = 38.03% O = 38.03%

K = 26.56/39.1 = 0.679 molK = 26.56/39.1 = 0.679 molCr = 35.41/52.0 = 0.681 molCr = 35.41/52.0 = 0.681 molO = 38.03/16.0 = 2.38 molO = 38.03/16.0 = 2.38 mol

K = 0.679/0.679 = 1K = 0.679/0.679 = 1Cr = 0.681/0.679 = 1.003Cr = 0.681/0.679 = 1.003O = 2.38/0.679 = 3.51O = 2.38/0.679 = 3.51

1:1:3.5 ratio1:1:3.5 ratioDouble the ratio to get whole numbers Double the ratio to get whole numbers 2:2:7 2:2:7

Empirical formula is Empirical formula is KK22CrCr22OO77

Practice problem #3 page 247.Practice problem #3 page 247.

20.0 g calcium & bromine20.0 g calcium & bromine

4.00 g Ca so 16.00 g Br4.00 g Ca so 16.00 g Br

Already in grams so divide by molar mass:Already in grams so divide by molar mass:4.00/ 40.1 = .0997 mol Ca4.00/ 40.1 = .0997 mol Ca16.00/79.9 = .2003 mol Br16.00/79.9 = .2003 mol Br

Ca = .0997/.0997 = 1Ca = .0997/.0997 = 1Br = .2003/.0997 = 2.009 Br = .2003/.0997 = 2.009 2 2

Empirical formula is Empirical formula is CaBrCaBr22

Ch 7 part 2 quiz #4Ch 7 part 2 quiz #4Empirical FormulasEmpirical Formulas

1-1- A compound is 27.3% carbon and A compound is 27.3% carbon and 72.7% 72.7% oxygen by mass. What is the oxygen by mass. What is the empirical empirical formula of the compound?formula of the compound?

2-2- A compound is 11.1% hydrogen A compound is 11.1% hydrogen and and

88.9% oxygen. What is its empirical 88.9% oxygen. What is its empirical formula?formula?

Calculating a Molecular Calculating a Molecular FormulaFormula

molecular formula-molecular formula- the actual formula of the actual formula of a a molecularmolecular compound (it may or may not compound (it may or may not be the same as the empirical formula of the be the same as the empirical formula of the compound)compound)

1)1) The molar mass of a compound is The molar mass of a compound is determined by analytical means & is given.determined by analytical means & is given.

2)2) Calculate the formula mass of the empirical Calculate the formula mass of the empirical formula. Divide the molar mass of the formula. Divide the molar mass of the compound by its empirical mass.compound by its empirical mass.

3)3) ““Multiply” the empirical formula by this Multiply” the empirical formula by this factor.factor.


1)1) empirical formula = Pempirical formula = P22OO55

2)2) molecular mass is 283.89molecular mass is 283.89

3)3) empirical mass is 141.94empirical mass is 141.94

4)4) Dividing the molecular mass by Dividing the molecular mass by the empirical mass gives a the empirical mass gives a multiplication factor of : multiplication factor of :

283.89 283.89 ÷ 141.94 ÷ 141.94 = 2.0001 = 2.0001 2 2

5)5) 2 x (P2 x (P22OO55) ) P P44OO1010

Chapter 7 ProblemsChapter 7 Problems

Do practice Do practice problems #1 & problems #1 & 2 on page 249.2 on page 249.

Do section Do section review review problems #1-4 problems #1-4 on on page 249.page 249.


empirical formula = CHempirical formula = CH

formula mass = 78.110 amuformula mass = 78.110 amu

empirical mass = ? = 12.0 + 1.0 = 13.0 empirical mass = ? = 12.0 + 1.0 = 13.0 amuamu

molecular mass / empirical mass = molecular mass / empirical mass = 78.110/13.0 = 6.008 78.110/13.0 = 6.008 multiplication multiplication factor of 6factor of 6

molecular formula = CH x 6 molecular formula = CH x 6 C C66HH66


formula mass = 34.00 amuformula mass = 34.00 amu0.44 g H & 6.92 g O0.44 g H & 6.92 g O11stst find empirical formula: find empirical formula:

H = 0.44/1.0 = 0.44H = 0.44/1.0 = 0.44O = 6.92/16.0 = 0.43O = 6.92/16.0 = 0.43

0.44/0.43 0.44/0.43 1 H & 0.43/0.43 1 H & 0.43/0.43 1 O 1 Oempirical formula = HOempirical formula = HOempirical mass = 17.0empirical mass = 17.0

formula mass / empirical mass = formula mass / empirical mass = 34.00/17.0 = 234.00/17.0 = 2

HO x 2 HO x 2 H H22OO22

Section review problem #1 page 249.Section review problem #1 page 249.

36.48% Na36.48% Na 25.41% S25.41% S 38.11% O 38.11% O

36.48/23.0 = 1.58 mol Na36.48/23.0 = 1.58 mol Na

25.41/32.1 = 0.792 mol S25.41/32.1 = 0.792 mol S

38.11/16.0 = 2.38 mol O38.11/16.0 = 2.38 mol O

1.58/0.792 = 1.995 1.58/0.792 = 1.995 2 2

0.792/0.792 = 1 0.792/0.792 = 1 1 1

2.38/0.792 = 3.005 2.38/0.792 = 3.005 3 3

2:1:3 2:1:3 Na Na22SOSO33

Section review problem #2 page 249.Section review problem #2 page 249.

53.70% Fe53.70% Fe 46.30% S46.30% S

53.70/55.8 = 0.962 mol Fe53.70/55.8 = 0.962 mol Fe46.30/32.1 = 1.44 mol S46.30/32.1 = 1.44 mol S

0.962/0.962 = 1 Fe0.962/0.962 = 1 Fe1.44/0.962 = 1.50 S1.44/0.962 = 1.50 S

1:1.5 doubled 1:1.5 doubled 2:3 2:3 Fe Fe22SS33

Section review problem #3 page 249Section review problem #3 page 249

1.04 g K1.04 g K 0.70 g Cr0.70 g Cr 0.86 g O0.86 g O

1.04/39.1 = .0266 mol K1.04/39.1 = .0266 mol K0.70/52.0 = .0135 mol Cr0.70/52.0 = .0135 mol Cr0.86/16.0 = .0538 mol O0.86/16.0 = .0538 mol O

.0266/.0135 = 1.97 .0266/.0135 = 1.97 2 2

.0135/.0135 = 1.0135/.0135 = 1

.0538/.0135 = 3.99 .0538/.0135 = 3.99 4 4

Empirical formula = KEmpirical formula = K22CrOCrO44

Section Review problem #4 page 249Section Review problem #4 page 249

4.04 g N 11.46 g O4.04 g N 11.46 g O f.m. = 108.0 amu f.m. = 108.0 amu

4.04/14.0 = .289 mol N4.04/14.0 = .289 mol N11.46/16.0 =.716 mol O11.46/16.0 =.716 mol O

.289/.289 = 1.289/.289 = 1 .716/.289 = 2.45.716/.289 = 2.45double ratio double ratio 2:5 2:5 N N22OO55

e.f.m. = 108e.f.m. = 108f.m./e.f.m. = 108/108 = 1f.m./e.f.m. = 108/108 = 1

empirical formula is same as molecular formula empirical formula is same as molecular formula NN22OO55

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


1)1) If the elements are in % composition by If the elements are in % composition by mass form, convert the percentages to mass form, convert the percentages to grams.grams.

2)2) Convert the masses of each element to Convert the masses of each element to moles by dividing the mass of the element moles by dividing the mass of the element by its molar mass.by its molar mass.

3)3) Select the element with the smallest Select the element with the smallest number of moles and divide the number number of moles and divide the number of moles of each element by that number of moles of each element by that number which will give you a 1:---:--- ratio.which will give you a 1:---:--- ratio.

4)4) IF the ratio is very close to a whole IF the ratio is very close to a whole number ratio, apply the numbers to each number ratio, apply the numbers to each element. If one of the number is not close element. If one of the number is not close to a whole number, use a multiplier to to a whole number, use a multiplier to convert the ratio to a whole number ratio.convert the ratio to a whole number ratio.


molecular formula-molecular formula- the actual formula of the actual formula of a a molecularmolecular compound (it may or may not compound (it may or may not be the same as the empirical formula of the be the same as the empirical formula of the compound)compound)

1)1) The molar mass of a compound is The molar mass of a compound is determined by analytical means & is given.determined by analytical means & is given.

2)2) Calculate the formula mass of the empirical Calculate the formula mass of the empirical formula. Divide the molar mass of the formula. Divide the molar mass of the compound by its empirical mass.compound by its empirical mass.

3)3) ““Multiply” the empirical formula by this Multiply” the empirical formula by this factor.factor.

Chapter 7 part 2 quiz #5Chapter 7 part 2 quiz #5Calculating molecular formulasCalculating molecular formulas

1-1- A molecular compound has an A molecular compound has an empirical formula of CHempirical formula of CH33. Its molecular . Its molecular

formula mass is 30 amu. What is formula mass is 30 amu. What is the the molecular formula of this molecular formula of this compound?compound?

HONORS- Chapter 7 part 2 quiz #5HONORS- Chapter 7 part 2 quiz #5Calculating molecular formulasCalculating molecular formulas

1-1- A molecular compound is 80% A molecular compound is 80% carbon carbon and 20% hydrogen. Its and 20% hydrogen. Its molecular molecular formula mass is 30 amu. formula mass is 30 amu. What is the What is the molecular formula molecular formula of of this compound?this compound?

Final Practice- chapter 7 part 2Final Practice- chapter 7 part 21-1- Determine the molar mass of the Determine the molar mass of the

compound compound NaNa33POPO4 4 ..

2-2- How many moles of COHow many moles of CO22 are in 198 g ? are in 198 g ?

3-3- What is the mass of 2.25 moles of HWhat is the mass of 2.25 moles of H22O ?O ?

4-4- What is the % composition of each element What is the % composition of each element of of the compound Pthe compound P44OO1010 ? ?

5-5- What is the empirical formulas of a What is the empirical formulas of a compound that compound that is 25.9% N and 74.1% O ? is 25.9% N and 74.1% O ? What is it molecular What is it molecular formula if its molecular formula if its molecular mass is 216 ?mass is 216 ?

Final Practice- chapter 7 part 2Final Practice- chapter 7 part 2

1-1- Determine the molar mass of the Determine the molar mass of the compound compound NaNa33POPO4 4 ..

Na = 3 x 23.0 = 69.0Na = 3 x 23.0 = 69.0

P = 1 x 31.0 = 31.0P = 1 x 31.0 = 31.0

O = 4 x 16.0 = O = 4 x 16.0 = 64.064.0

164.0 g/mol164.0 g/mol

2-2- How many moles of COHow many moles of CO22 are in 198 g ? are in 198 g ?

C = 1 x 12.0 = 12.0C = 1 x 12.0 = 12.0

O = 2 x 16.0 = O = 2 x 16.0 = 32.032.0


198/44.0 = 4.5 mol CO198/44.0 = 4.5 mol CO22

3-3- What is the mass of 2.25 moles of HWhat is the mass of 2.25 moles of H22O ?O ?

H = 2 x 1.0 = 2.0H = 2 x 1.0 = 2.0

O = 1 x 16.0 = O = 1 x 16.0 = 16.016.0


2.25 mol x 18.0 g/mol = 40.5 g H2.25 mol x 18.0 g/mol = 40.5 g H22OO

4-4- What is the % composition of each element What is the % composition of each element of of the compound Pthe compound P44OO1010 ? ?

P = 4 x 31.0 = 124.0P = 4 x 31.0 = 124.0

O = 10 x 16.0 = O = 10 x 16.0 = 160.0160.0

284.0 g/mol284.0 g/mol

P = 124/284(100) = 43.7%P = 124/284(100) = 43.7%

O = 160/284 (100) = 56.3%O = 160/284 (100) = 56.3%

5-5- What is the empirical formulas of a What is the empirical formulas of a compound that compound that is 25.9% N and 74.1% O ? is 25.9% N and 74.1% O ? What is it molecular What is it molecular formula if its molecular formula if its molecular mass is 216 ?mass is 216 ?

N = 25.9/14.0 = 1.85N = 25.9/14.0 = 1.85

O = 74.1/16.0 = 4.63O = 74.1/16.0 = 4.63

N = 1.85/1.85 = 1N = 1.85/1.85 = 1

O = 4.63/1.85 = 2.5O = 4.63/1.85 = 2.5

1:2.5 doubled 1:2.5 doubled 2:5 so empirical formula = N 2:5 so empirical formula = N22OO55

e.f.m. = (2 x 14) + (5 x 16) = 108e.f.m. = (2 x 14) + (5 x 16) = 108

216 (mfm)/108 (efm) = 2216 (mfm)/108 (efm) = 2 2 x 2:5 2 x 2:5 4:10 4:10 N N44OO1010

Honors Chemistry Chapter 7 part 2 Honors Chemistry Chapter 7 part 2 testtest

38 multiple choice:38 multiple choice:Definition of formula mass & molar massDefinition of formula mass & molar massCalculate formula mass of a compound (3)Calculate formula mass of a compound (3)Convert from mass to moles or moles to mass Convert from mass to moles or moles to mass

when given the amount & molar mass of a when given the amount & molar mass of a substance (7)substance (7)

Calculate % composition by mass (6)Calculate % composition by mass (6)Definition & what an empirical formula representsDefinition & what an empirical formula representsCalculate empirical formulas (7)Calculate empirical formulas (7)Know how to determine molecular formula from Know how to determine molecular formula from

empirical formula and determine what the empirical formula and determine what the empirical fromula of a molecular formula would beempirical fromula of a molecular formula would be

Calculate molecular formula when given empirical Calculate molecular formula when given empirical formula & formula mass (7)formula & formula mass (7)

Essay Question:Essay Question:

____ & ____ are examples of the ____ & ____ are examples of the empirical and the molecular empirical and the molecular

formula of formula of a compound, a compound, respectively. Explain the respectively. Explain the relationship between these two types relationship between these two types

of formulas.of formulas.

Chemistry Chapter 7 part 2 test Chemistry Chapter 7 part 2 test reviewreview

24 multiple choice questions24 multiple choice questions Definition of molar mass and formula massDefinition of molar mass and formula mass Calculate a formula massCalculate a formula mass Interpret a molar massInterpret a molar mass Convert from mass to moles or moles to mass when Convert from mass to moles or moles to mass when

given the molar mass of a compound (6)given the molar mass of a compound (6) Calculate % composition by mass (3)Calculate % composition by mass (3) Definition of empirical formula and what it represents Definition of empirical formula and what it represents

(4)(4) Calculate the empirical formula of compounds (3)Calculate the empirical formula of compounds (3) What is needed to determine the molecular formula What is needed to determine the molecular formula

from an empirical formulafrom an empirical formula Determine the molecular formula of a compound from Determine the molecular formula of a compound from

its formula mass and the empirical formula (3) its formula mass and the empirical formula (3)

Essay Question:Essay Question:

____ & ____ are examples of the ____ & ____ are examples of the empirical and the molecular empirical and the molecular

formula of formula of a compound, a compound, respectively. Explain the respectively. Explain the relationship between these two types relationship between these two types

of formulas.of formulas.

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Modern Chemistry Chapter 7 Chemical Formulas & Chemical Compounds

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