Download - Dalton’s Atomic Theory
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Dalton’s Atomic Theory
• Elements - made up of atoms• Same elements, same atoms.• Different elements, different
atoms.• Chemical reactions involve
bonding of atoms
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Law of Definite Composition
• A compound always contains the same proportion of elements by mass
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Law of Multiple Proportions
• Compounds form from specific combinations of atoms
• H2O vs H2O2
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Chemical Bonds
• Holds compounds together
• Need to be broken for chemical and physical changes to occur
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The Atom
• Made up of:–Protons – (+) charged
–Electrons – (-) charged
–neutrons
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Periodic Table
• Alkaline Metals – Grps. I & II• Transition Metals• Non-metals• Halogens – Group VII• Noble Gases –Group VIII - little
chemical activity
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Periodic Table
• Atomic Mass - # at bottom•how much element weighs
• Atomic Number - # on top•gives # protons = # electrons
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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
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Ions
• Are charged species
• Result when elements gain electrons or lose electrons
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2 Types of Ions
• Anions – (-) charged•Example: F-
• Cations – (+) charged•Example: Na+
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Highly Important!
• Gain of electrons makes element (-) = anion
• Loss of electrons makes element (+) = cation
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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.
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Examples
• In CO2, the charge of C is + 4
• In CO, the charge of C is +2.
• In KMnO4, since the charge of K is +1, O is -2 so -2 x 4 = -8, Mn must be +7.
• In (PO4)3-, the charge of O is -2, so -2 x 4= -8, then P must have a charge of +5, so the sum when the charges are added will be -3.
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Isotopes
• Are atoms of a given element that differ in the number of neutrons and consequently in atomic mass.
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Example
Isotopes % Abundance12C 98.89 %13C 1.11 %14C 11C
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–For example, the mass of C = 12.01 a.m.u is the average of the masses of 12C, 13C and 14C.
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Determination of Aver. Mass
• Ave. Mass = [(% Abund./100) (atomic
mass)] + [(% Abund./100) (atomic mass)]
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Take Note:• If there are more than 2
isotopes, then formula has to be re-adjusted
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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 %
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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.)
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Periodic Table
• Mendeleev – arranged elements in the (.) table
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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
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–For example, the mass of C = 12.01 a.m.u is the average of the masses of 12C, 13C and 14C.
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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)
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Oxidation States
• Are the partial charges of the ions. Some ions have more than one oxidation states.
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Oxidation States
• - generally depend upon the how the element follows the octet rule
• Octet Rule – rule allowing elements to follow the noble gas configuration
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Nomenclature
• - naming of compounds
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Periodic Table
• Rows (Left to Right) - periods
• Columns (top to bottom) - groups
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Rule 1 – IONIC COMPOUNDS
• Metals w/ Fixed Oxidation States
–Name metal or first element as is
- Anion always ends in “–ide”
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Terminal element or anion
• O - oxide P - phosphide
• N - nitride Se - selenide
• S - sulfide Cl - chloride
• F - fluoride I - iodide
• Br - bromide C - carbide
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Note
• Only elements that come directly from the periodic table WILL end in –IDE.
• POLYATOMIC IONS will be named AS IS.
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Name the following:
• CaO -• NaCl -• MgO - • CaS - • Na3N -
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Answers:
• CaO - calcium oxide• NaCl - sodium chloride• MgO - magnesium oxide• CaS - calcium sulfide• Na3N - sodium nitride
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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!)
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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
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Name the following:
• Copper (I) sulfide
• Iron (II) oxide
• Tin (II) iodide
• Iron (III) nitride
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Answers:
• Copper (I) sulfide Cu2S
• Iron (II) oxide FeO
• Tin (II) iodide SnI2
• Iron (III) nitride FeN
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What about…….?
• Cesium hydroxide
• Iron (III) acetate
• Lithium phosphate
• Aluminum Sulfite
• Lead (II) sulfate
• Silver nitrate
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POLYATOMIC IONS
• Consist of more than 1 element.
• Have charges.
• Ex. SO4 2-, SO3 2-, PO4 3-,PO3 3-
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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
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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
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Name the following compounds
• CO2 - carbon dioxide
• N2O – dinitrogen oxide
• SO3 – sulfur trioxide
• N2O5 – dinitrogen pentoxide
• P2S5 – diphosphorus pentasulfide
• CO – carbon monoxide
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Naming Acids
• I. Acids without Oxygen
–Use hydro + 1st syllable + “- ic acid”
• Example: HCl = hydrochloric acidHCN = hydrocyanic acid
HBr = hydrobromic acid
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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
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Trick!
• If anion ends in “ – ate”, acid ends in “ – ic”
• Example:
• HClO4 perchlorate perchloric acid
• HClO3 chlorate chloric acid
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Trick!
• If anion ends in “ – ite”, acid ends in “ – ous”
• Example:• HClO2 chlorite chlorous acid
• HClO hypochlorite hypochlorous acid
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Name the following:
• HBrO4 (perbromate)
• HBrO3 (bromate)
• HBrO2 (bromite)
• HBrO (hypobromite)
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Fundamental laws
• Law of Conservation of Mass
•Mass is neither created or destroyed
•Conversion from one form to another
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Determination of Aver. Mass
• Ave. Mass = [(% Abund./100) (atomic
mass)] + [(% Abund./100) (atomic mass)]
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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 %
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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.)
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Chemical Formula
• Gives the combining whole number ratios of the elements in a compound
• C6H12O6
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Structural Formula
• Gives the spatial arrangement of atoms in the compound
• Structural formula for H2O is H – O – H
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Empirical Formula
• Only gives the types of elements in the compound and the ratio of the elements in the formula
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Empirical Formula
• Does not tell exactly how many of the elements are in the compound
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Molecular Formula
• Gives you the exact elemental composition of the compound
• Formula of the compound as it would actually exist.
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EF vs. MF
Sucrose or table sugar:
Molecular Formula = C6H12O6
Empirical Formula = CH2O
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Sample Problem
• The compound adrenaline contains % C = 56.79 % H = 6.56 % O = 28.37 % N = 8.28 by mass. Find the empirical formula.
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Empirical Formula
• EF Determination when % Masses are given
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Steps to Solve for EF
• Step 1: Sum up all given percentages. If total equals 100%, go to step 2. If total does not equal 100, the missing % is due to one of the component elements.
• Step 2: Convert Mass % to grams.
• Step 3: Calculate moles using mole = gram/molar mass
•
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Empirical Formula
• Step 4. To get simplest ratios, divide the moles calculated by the smallest calculated mole. You must have a ration of 1 for at least one of the element. (Follow rule for rounding).
• Step 5. You now have the ratios or subscripts for the EF.
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Molecular Formula Detn.
Step 1. Obtain empirical formula mass by adding atomic masses of all elements in empirical formula
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Molecular Formula Detn.
Step 2. Get ratio by applying the formula below:
Molecular Formula = given molar mass
Empirical formula mass
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Molecular Formula Detn.
Step 3.
Multiply empirical formula subscripts by obtained ratio
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Sample Problem
• Caffeine, a stimulant found in coffee, contains 49.5 % C, 5.15% H, 28.9 % N, and 16.5 % O by mass. The molar mass of the compound is 195 g/mol. Determine the empirical and molecular formula of caffeine.
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Sample Problem
• Ibuprofen, a headache remedy, contains 75.69 % C, 8.80% H, and 15.51 % O by mass. The molar mass of the compound is 206 g/mol. Determine the empirical and molecular formula of ibuprofen.