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CHAPTER 2 ATOMS and MOLECULES

CHAPTER 2

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CHAPTER 2. ATOMS and MOLECULES. Periodic Table. Atomic Mass number below the element not whole numbers because the masses are averages of the masses of the different isotopes of the elements. Ions. Are charged species Result when elements gain electrons or lose electrons. - PowerPoint PPT Presentation

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Page 1: CHAPTER 2

CHAPTER 2

ATOMS and MOLECULES

Page 2: CHAPTER 2

Periodic Table

• Atomic Mass –number below the element–not whole numbers because

the masses are averages of the masses of the different isotopes of the elements

Page 3: CHAPTER 2

Ions

• Are charged species

• Result when elements gain electrons or lose electrons

Page 4: CHAPTER 2

2 Types of Ions

• Anions – (-) charged•Example: F-

• Cations – (+) charged•Example: Na+

Page 5: CHAPTER 2

Highly Important!

• Gain of electrons makes element (-) = anion

• Loss of electrons makes element (+) = cation

Page 6: CHAPTER 2

Charges• When elements combine, they have to be

in the form of IONS.• Cations and anions combine to form

compounds.• For a neutral compound, the sum of the

charges must be ZERO.• For a polyatomic ion, the sum of the

charges must equal the charge of the ION.

Page 7: CHAPTER 2

Isotopes

• Are atoms of a given element that differ in the number of neutrons and consequently in atomic mass.

Page 8: CHAPTER 2

Example

Isotopes % Abundance12C 98.89 %13C 1.11 %14C 11C

Page 9: CHAPTER 2

–For example, the mass of C = 12.01 a.m.u is the average of the masses of 12C, 13C and 14C.

Page 10: CHAPTER 2

Determination of Aver. Mass

• Ave. Mass = [(% Abund./100) (atomic mass)] + [(% Abund./100) (atomic mass)]

Page 11: CHAPTER 2

Take Note:• If there are more than 2

isotopes, then formula has to be re-adjusted

Page 12: CHAPTER 2

Sample Problem 1

• Assume that element Uus is synthesized and that it has the following stable isotopes:–284Uus (283.4 a.m.u.) 34.6 %–285Uus (284.7 a.m.u.) 21.2 %–288Uus (287.8 a.m.u.) 44.20 %

Page 13: CHAPTER 2

Solution

• Ave. Mass of Uus =• [284Uus] (283.4 a.m.u.)(0.346)• [285Uus] +(284.7 a.m.u.)(0.212)• [288Uus] +(287.8 a.m.u.)(0.4420)• = 97.92 + 60.36 + 127.21 • = 285.49 a.m.u (FINAL ANS.)

Page 14: CHAPTER 2

Oxidation Numbers

• Is the charge of the ions (elements in their ion form)

• Is a form of electron accounting

• Compounds have total charge of zero (positive charge equals negative charge)

Page 15: CHAPTER 2

Oxidation States

• Are the partial charges of the ions. Some ions have more than one oxidation states.

Page 16: CHAPTER 2

Oxidation States

• - generally depend upon the how the element follows the octet rule• Octet Rule – rule allowing

elements to follow the noble gas configuration

Page 17: CHAPTER 2

Nomenclature

• - naming of compounds

Page 18: CHAPTER 2

Periodic Table

• Rows (Left to Right) - periods

• Columns (top to bottom) - groups

Page 19: CHAPTER 2

Rule 1 – IONIC COMPOUNDS

• Metals w/ Fixed Oxidation States–Name metal or first element as is- Anion always ends in “–ide”

Page 20: CHAPTER 2

Terminal element or anion

• O - oxide P - phosphide• N - nitride Se - selenide• S - sulfide Cl - chloride• F - fluoride I - iodide• Br - bromideC - carbide

Page 21: CHAPTER 2

Note

• Only elements that come directly from the periodic table WILL end in –IDE.

• POLYATOMIC IONS will be named AS IS.

Page 22: CHAPTER 2

Name the following:

• CaO -• NaCl -•MgO - • CaS- • Na3N -

Page 23: CHAPTER 2

Answers:

• CaO - calcium oxide• NaCl - sodium chloride•MgO - magnesium oxide• CaS- calcium sulfide• Na3N - sodium nitride

Page 24: CHAPTER 2

Where do the subscripts come from?

• Answer: From the oxidation states of the ions.

• Remember: Ions are the species that combine.

• Target: Compounds! (No charges!)

Page 25: CHAPTER 2

Second Rule

• II. Ionic Compounds - Metals with no fixed oxidation states (Transition Metals) except for Ag, Zn and Al

• Metal(Roman #) + 1st syllable + ide

–Use Roman numerals after the metal to indicate oxidation state

Page 26: CHAPTER 2

Name the following:

• Copper (I) sulfide• Iron (II) oxide• Tin (II) iodide• Iron (III) nitride

Page 27: CHAPTER 2

Answers:

• Copper (I) sulfide Cu2S• Iron (II) oxide FeO• Tin (II) iodide SnI2

• Iron (III) nitride FeN

Page 28: CHAPTER 2

What about…….?

• Cesium hydroxide• Iron (III) acetate• Lithium phosphate• Aluminum Sulfite• Lead (II) sulfate• Silver nitrate

Page 29: CHAPTER 2

POLYATOMIC IONS

• Consist of more than 1 element.

• Have charges.

• Ex. SO4 2-, SO3 2-, PO4 3-,PO3 3-

Page 30: CHAPTER 2

Rule 3 – Covalent Compounds• III. For Non-metals (grps IV, V, VI VII), use

prefixes.Mono – 1 Hepta - 7Di - 2 Octa - 8Tri – 3 Nona - 9Tetra – 4 Deca - 10Penta – 5Hexa - 6

Page 31: CHAPTER 2

Rule 3 – Covalent Compounds (only have Non- Metals)

• Name 1st element as is. Use prefix, if necessary.

• Prefix + 1st element + prefix + 1st syllable of anion + ide

Page 32: CHAPTER 2

Name the following compounds

• CO2 - carbon dioxide

• N2O – dinitrogen oxide

• SO3 – sulfur trioxide

• N2O5 – dinitrogen pentoxide

• P2S5 – diphosphorus pentasulfide• CO – carbon monoxide

Page 33: CHAPTER 2

Naming Acids

• I. Acids without Oxygen–Use hydro + 1st syllable + “- ic acid”

• Example: HCl = hydrochloric acidHCN = hydrocyanic acid HBr = hydrobromic acid

Page 34: CHAPTER 2

II. Acids with oxygen• Polyatomic “ate” converts to “ic” + acid• Polyatomic “ite” converts to “ous” + acid

- H2SO3 sulfurous acid

–H2SO4 sulfuric acid

–HNO3 nitric acid

–HNO2 nitrous acid

–H3PO4 phosphoric acid

Page 35: CHAPTER 2

Trick!

• If anion ends in “ – ate”, acid ends in “ – ic”

• Example:• HClO4 perchlorate perchloric acid

• HClO3 chlorate chloric acid

Page 36: CHAPTER 2

Trick!

• If anion ends in “ – ite”, acid ends in “ – ous”

• Example:• HClO2 chlorite chlorous acid

• HClO hypochlorite hypochlorous acid

Page 37: CHAPTER 2

Name the following:

• HBrO4 (perbromate)

• HBrO3 (bromate)

• HBrO2 (bromite)• HBrO (hypobromite)

Page 38: CHAPTER 2

Pure Acids (not diluted)

• Simply use Rule 3. Use prefixes, if necessary.