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Element #8 : Oxygen, Isotopes. 16 8 O 8 Protons 8 Neutrons 99.759% 15.99491462 amu 17 8 O 8 Protons 9 Neutrons 0.037% 16.9997341 amu 18 8 O 8 Protons 10 Neutrons - PowerPoint PPT Presentation
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Element #8 : Oxygen, Isotopes
• 168O 8 Protons 8 Neutrons
99.759% 15.99491462 amu
• 178O 8 Protons 9 Neutrons
0.037% 16.9997341 amu
• 188O 8 Protons 10 Neutrons
0.204 % 17.999160 amu
Fig.2.A
Fig.2.B
Calculating the “Average” Atomic Mass of an Element
24Mg (78.7%) 23.98504 amu 25Mg (10.2%) 24.98584 amu
26Mg (11.1%) 25.98636 amu
Total =
With Significant Digits = amu
Problem: Calculate the average atomic mass of Magnesium! Magnesium Has three stable isotopes, 24Mg ( 78.7%); 25Mg (10.2%); 26Mg (11.1%).
Calculating the “Average” Atomic Mass of an Element
24Mg (78.7%) 23.98504 amu x 0.787 = 18.876226 amu 25Mg (10.2%) 24.98584 amu x 0.102 = 2.548556 amu26Mg (11.1%) 25.98636 amu x 0.111 = 2.884486 amu
24.309268 amu
With Significant Digits = 24.3 amu
Problem: Calculate the average atomic mass of Magnesium! Magnesium Has three stable isotopes, 24Mg ( 78.7%); 25Mg (10.2%); 26Mg (11.1%).
Calculate the Average Atomic Mass of Zirconium, Element #40
Zirconium has five stable isotopes: 90Zr, 91Zr, 92Zr, 94Zr, 96Zr.
Isotope (% abd.) Mass (amu) (%) Fractional Mass
90Zr (51.45%) 89.904703 amu X 0.5145 = 46.2560 amu91Zr (11.27%) 90.905642 amu X 0.1127 = 10.2451 amu92Zr (17.17%) 91.905037 amu X 0.1717 = 15.7801 amu94Zr (17.33%) 93.906314 amu X 0.1733 = 16.2740 amu96Zr (2.78%) 95.908274 amu X 0.0278 = 2.6663 amu
91.2215 amu
With Significant Digits = 91.22 amu
Problem: Estimate the abundance of the two Bromine isotopes, given that the average mass of Br is 79.904 amu. Since exact masses of isotopes not give, estimate from: mass in amu = #p+ + #n:
79Br = 79 g/mol and 81Br = 81 g/mol (approximately).Plan: Let the abundance of 79Br = X and of 81Br = Y and X + Y = 1.0Solution:
Problem: Estimate the abundance of the two Bromine isotopes, given that the average mass of Br is 79.904 amu. Since exact masses of isotopes not give, estimate from: mass in amu = #p+ + #n:
79Br = 79 g/mol and 81Br = 81 g/mol (approximately).Plan: Let the abundance of 79Br = X and of 81Br = Y and X + Y = 1.0Solution: X(79) + Y(81) = 79.904
X + Y = 1.00 therefore X = 1.00 - Y (1.00 - Y)(79) + Y(81) = 79.904
79 - 79 Y + 81Y = 79.904
2 Y = 0.904 = 1 w/ sig. figs. so Y = 0.5
X = 1.00 - Y = 1.00 - 0.5 = 0.5
%X = % 79Br = 0.5 x 100% = 50% (Actual: 50.67% = 79Br)%Y = % 81Br = 0.5 x 100% = 50% (Actual: 49.33% = 81Br)
Modern Reassessment of the Atomic Theory1. All matter is composed of atoms. Although atoms are composed of smaller particles (electrons, protons, and neutrons), the atom is the smallest body that retains the unique identity of the element.
2. Atoms of one element cannot be converted into atoms of another element in a chemical reaction. Elements can only be converted into other elements in Nuclear reactions in which protons are changed.
3. All atoms of an element have the same number of protons and electrons, which determines the chemical behavior of the element. Isotopes of an element differ in the number of neutrons, and thus in mass number, but not in chemical behavior (much). A sample of the element is treated as though its atoms have an average mass.
4. Compounds are formed by the chemical combination of two or more elements in specific ratios, as originally stated by Dalton.
Definitions• ELEMENT - A substance that cannot be separated into
simpler substances by chemical means
• COMPOUND - A substance composed of atoms of two or more elements chemically united in fixed proportions
• PERIODIC TABLE - “MENDELEEV TABLE” - A tabular arrangement of the elements, vertical groups or families of elements based upon their chemical properties - actually combining ratios with oxygen
Fig2.16
Fig.2.17
Groups in the Periodic Table
Main Group Elements (Vertical Groups) Group IA - Group IIA - Group IIIA - Group IVA - Group VA - Group VIA - Group VIIA - Group VIIIA -Other Groups ( Vertical and Horizontal Groups)Group IB - 8B -Period 6 Group -Period 7 Group -
Groups in the Periodic Table
Main Group Elements (Vertical Groups) Group IA - Alkali Metals Group IIA - Alkaline Earth Metals Group IIIA - Boron Family Group IVA - Carbon Family Group VA - Nitrogen Family Group VIA - Oxygen Family (Calcogens) Group VIIA - Halogens Group VIIIA - Noble GasesOther Groups ( Vertical and Horizontal Groups)Group IB - 8B - Transition MetalsPeriod 6 Group - Lanthanides (Rare Earth Elements)Period 7 Group - Actinides
OS
SeTePo
NPAsSbBi
CSiGeSnPb
BAlGaInTl
ZnCuCdHg
AgAu
NiPdPt
CoRhIr
FeRuOs
MnTcRe
CrMoW
VNbTa
TiZrHf
ScY
LaAc
The Periodic Table of the Elements
The Alkali Metals
The Alkaline Earth Metals
Ce Pr Nd PmSmEu Gd Tb Dy Ho Er TmYb LuTh Pa Np PuAmCmBk Cf Es FmMd No LrU
HLiNaK
RbCsFr
BeMgCaSrBaRa Rf Sg
The Halogens
The Noble Gases
HeNeArKrXeRn
FClBrI
AtDu Bo HaMe
The Periodic Table of the ElementsHLi BeNaMgK Ca ScRbCsFr
SrBaRa
Ti V CrMn FeY
LaAc
Co Ni Cu ZnZrHf
NbTa
Rf
MoW
TcRe
RuOs
RhIr
PdPt
AgAu
CdHg
FHeNeArCl
Br KrXeRn
IAt
Ce Pr Nd PmTh
SmEu Gd Tb Dy Ho Er Tm Yb LuPa U Np PuAmCm Bk Cf Es FmMd No Lr
Boron family
BAlGaInTl
Carbon Family
CSi
GeSnPb
Nitrogen family
NP
AsSbBi
Oxygen Family
OS
SeTePo
Du Sg Bo Ha Me
OS
SeTePo
NPAsSbBi
CSiGeSnPb
BAlGaInTl
The Periodic Table of the Elements
Lanthanides: The
Rare Earth ElementsThe Actinides
FClBrI
At
H HeNeArKrXeRn
LiNaKRbCsFr
BeMgCaSrBaRa
Ce
The Transition Metals
Pr Nd PmSmEu Gd Tb DyHo Er TmYb LuTh Pa U Np PuAmCmBk Cf Es FmMd No Lr
Sc Ti V CrMnYLa
Fe Co Ni Cu ZnZr NbMo Tc Ru Rh PdAg CdHf Ta W Re Os Ir Pt Au Hg
Ac Rf Sg HaDu Bo Me
Fig.2.18
Fig.2.19
Fig.2.20
Predicting the Ion an Element will form in Chemical Reactions
Problem: What monoatomic ions will each of the elements form?(a) Barium(z=56) (b) Sulfur(z=16) (c) Titanium(z =22) (d) Fluorine(z=9)Plan: We use the “z” value to find the element in the periodic table and which is the nearest noble gas. Elements that lie after a noble gas will loose electrons, and those before a noble gas will gain electrons.Solution: (a) Ba
(b) S
(c) Ti
(d) F
Predicting the Ion an Element will form in Chemical Reactions
Problem: What monoatomic ions will each of the elements form?(a) Barium(z=56) (b) Sulfur(z=16) (c) Titanium(z =22) (d) Fluorine(z=9)Plan: We use the “z” value to find the element in the periodic table and which is the nearest noble gas. Elements that lie after a noble gas will loose electrons, and those before a noble gas will gain electrons.Solution: (a) Ba+2, Barium is an alkaline earth element, Group 2A, and is expected to loose two electrons to attain the same number of electrons as the noble gas Xenon! (b) S -2, Sulfur is in the Oxygen family, Group 6A, and is expected to gain two electrons to attain the same number of electrons as the noble gas Argon! (c) Ti+4, Titanium is in Group 4B, and is expected to loose 4 electrons to attain the same number of electrons as the noble gas Argon! (d) F -, Fluorine is in a halogen, Group 7A, and is expected to gain one electron, to attain the same number of electrons as the noble gas Neon!
HeNeArKrXeRn
The Periodic Table of the Elements
CrMn Fe Co NiMoW
TcRe
RuOs
RhIr
PdPt
Most Probable Oxidation State
+1
+2
+3 +4
+3 +_4 - 3 - 2 - 1
0
HLiNaKRb
CsFr
Sc
Y
BeMgCaSr
Ba
RaLaAc
BAlGa
InTl
Ti
RfHf
Zr
CSi
Ge
SnPb
FClBrI
At
OSSe
TePo
NP
As
SbBi
Zn
CdHg
+ 2+1
CuAgAu
+5
VNb
Ta
CeTh
Pr Nd PmSmEu Gd Tb Dy Ho Er TmYb LuPa U Np Pu AmCmBk Cf Es FmMd No Lr
+3
+3
Du Sg Bo Ha Me
(same concept as Fig. 2.20)
Chemical Compounds and BondsChemical Bonds - The electrostatic forces that hold the atoms of elements together in the compound.
Ionic Compounds - Electrons are transferred from one atom to another to form Ionic Cpds.
Covalent Compounds - Electrons are shared between atoms of different elements to form Covalent Cpds.
“Cation” - An atom that has lost electron(s) to form “ + ” ions.May be 1 or more e-s. Common with metal elements.“Anion” - An atom which has gained electron(s), to form “ - ”
ions. Common w/ nonmetal elements.Later we’ll learn that group of atoms can also be anion or cation
Mono-atomic (monatomic) ions form binary ionic compounds.
Fig.2.20
Fig 2.22 (P 65)
Thepolyatomicion
Chemical Formulas
Empirical Formula - Shows the relative number of atoms of each element in the compound. It is the simplest formula, and is derived from masses of the elements.
Molecular Formula - Shows the actual number of atoms of each element in the molecule of the compound.
Structural Formula - Shows the actual number of atoms, and the bonds between them ; that is, the arrangement of atoms in the molecule.
Empirical and Molecular Formulas Name Molecular Empiricalwater H2O
hydrogen H2O2 peroxide
ethane C2H6
sulfur S8
acetic acid CH3COOH
Empirical and Molecular Formulas Name Molecular Empiricalwater H2O H2O
hydrogen H2O2 HOperoxide
ethane C2H6 CH3
sulfur S8 S
acetic acid CH3COOH COH2
Fig.2.23
Start trying to learn those in bold. Best done by looking at name w/ use.
Give the Name and Chemical Formulas of the Compounds formed from the following pairs of Elements
a) Sodium and Oxygen Na2O Sodium Oxide
b) Zinc and Chlorine c) Calcium and Fluorine
d) Strontium and Nitrogen
e) Hydrogen and Iodine
f) Scandium and Sulfur
Give the Name and Chemical Formulas of the Compounds formed from the following pairs of Elements
a) Sodium and Oxygen Na2O Sodium Oxide
b) Zinc and Chlorine ZnCl2 Zinc Chloride c) Calcium and Fluorine CaF2 Calcium Fluoride
d) Strontium and Nitrogen Sr3N2 Strontium Nitride
e) Hydrogen and Iodine HI Hydrogen Iodide
f) Scandium and Sulfur Sc2S3 Scandium Sulfide
Start learning these boldface ones.
Determining Names and Formulas of Ionic Compounds of Elements That Form More Than One Ion.
Give the systematic names for the formulas or the formulas for the names of the following compounds.
a) Iron III Sulfide - Fe is +3, and S is -2 therefore the compound is: Fe2S3
b) CoF2 -
c) Stannic Oxide -
d) NiCl3 -
Determining Names and Formulas of Ionic Compounds of Elements That Form More Than One Ion.
Give the systematic names for the formulas or the formulas for the names of the following compounds.
a) Iron III Sulfide - Fe is +3, and S is -2 therefore the compound is: Fe2S3
b) CoF2 - the anion is Fluoride (F -1) and there are two F -1, the cation is Cobalt and it must be Co+2 therefore the compound is: Cobalt (II) Fluoridec) Stannic Oxide - Stannic is the common name for Tin (IV), Sn+4, the Oxide ion is O-2, therefore the formula of the compound is: SnO2
d) NiCl3 - The anion is chloride (Cl-1), there are three anions, so the Nickel cation is Ni+3, therefore the name of the compound is: Nickel (III) Chloride
Rules for Families of OxoanionsFamilies with Two Oxoanions
The ion with more O atoms takes the nonmetal root and the suffix “-ate”.
The ion with fewer O atoms takes the nonmetal root and the suffix “-ite”.
Families with Four Oxoanions (usually a Halogen)The ion with most O atoms has the prefix “per-”, the nonmetal root and the suffix “-ate”.
The ion with one less O atom has just the suffix “-ate”.
The ion with two less O atoms has the just the suffix “-ite”.
The ion with three less O atoms has the prefix “hypo-” and thesuffix “-ite”.
Fig.2.24
NAMING OXOANIONS - EXAMPLES
Prefixes Root Suffixes Chlorine Bromine Iodine
per “ ” ate perchlorate perbromate periodate [ ClO4
-] [ BrO4-] [ IO4
-] “ ” ate chlorate bromate iodate [ ClO3
-] [BrO3-] [ IO3
-]
“ ” ite chlorite bromite iodite [ ClO2
-] [ BrO2-] [ IO2
-]
hypo “ ” ite hypochlorite hypobromite hypoiodite [ ClO -] [ BrO -] [ IO -]
No.
of O
ato
ms