Dalton’s Atomic Theory

Dalton’s Atomic Theory

• Elements - made up of atoms• Same elements, same atoms.• Different elements, different

atoms.• Chemical reactions involve

bonding of atoms

Law of Definite Composition

• A compound always contains the same proportion of elements by mass

Law of Multiple Proportions

• Compounds form from specific combinations of atoms

• H2O vs H2O2

Chemical Bonds

• Holds compounds together

• Need to be broken for chemical and physical changes to occur

The Atom

• Made up of:–Protons – (+) charged

–Electrons – (-) charged

–neutrons

Periodic Table

• Alkaline Metals – Grps. I & II• Transition Metals• Non-metals• Halogens – Group VII• Noble Gases –Group VIII - little

chemical activity

Periodic Table

• Atomic Mass - # at bottom•how much element weighs

• Atomic Number - # on top•gives # protons = # electrons

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

2 Types of Ions

• Anions – (-) charged•Example: F-

• Cations – (+) charged•Example: Na+

Highly Important!

• Gain of electrons makes element (-) = anion

• Loss of electrons makes element (+) = cation

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.

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.

Isotopes

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

Example

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

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

Determination of Aver. Mass

• Ave. Mass = [(% Abund./100) (atomic

mass)] + [(% Abund./100) (atomic mass)]

Take Note:• If there are more than 2

isotopes, then formula has to be re-adjusted

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 %

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.)

Periodic Table

• Mendeleev – arranged elements in the (.) table

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

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

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)

Oxidation States

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

Oxidation States

• - generally depend upon the how the element follows the octet rule

• Octet Rule – rule allowing elements to follow the noble gas configuration

Nomenclature

• - naming of compounds

Periodic Table

• Rows (Left to Right) - periods

• Columns (top to bottom) - groups

Rule 1 – IONIC COMPOUNDS

• Metals w/ Fixed Oxidation States

–Name metal or first element as is

- Anion always ends in “–ide”

Terminal element or anion

• O - oxide P - phosphide

• N - nitride Se - selenide

• S - sulfide Cl - chloride

• F - fluoride I - iodide

• Br - bromide C - carbide

Note

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

• POLYATOMIC IONS will be named AS IS.

Name the following:

• CaO -• NaCl -• MgO - • CaS - • Na3N -

Answers:

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

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!)

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

Name the following:

• Copper (I) sulfide

• Iron (II) oxide

• Tin (II) iodide

• Iron (III) nitride

Answers:

• Copper (I) sulfide Cu2S

• Iron (II) oxide FeO

• Tin (II) iodide SnI2

• Iron (III) nitride FeN

What about…….?

• Cesium hydroxide

• Iron (III) acetate

• Lithium phosphate

• Aluminum Sulfite

• Lead (II) sulfate

• Silver nitrate

POLYATOMIC IONS

• Consist of more than 1 element.

• Have charges.

• Ex. SO4 2-, SO3 2-, PO4 3-,PO3 3-

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

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

Name the following compounds

• CO2 - carbon dioxide

• N2O – dinitrogen oxide

• SO3 – sulfur trioxide

• N2O5 – dinitrogen pentoxide

• P2S5 – diphosphorus pentasulfide

• CO – carbon monoxide

Naming Acids

• I. Acids without Oxygen

–Use hydro + 1st syllable + “- ic acid”

• Example: HCl = hydrochloric acidHCN = hydrocyanic acid

HBr = hydrobromic acid

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

Trick!

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

• Example:

• HClO4 perchlorate perchloric acid

• HClO3 chlorate chloric acid

Trick!

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

• Example:• HClO2 chlorite chlorous acid

• HClO hypochlorite hypochlorous acid

Name the following:

• HBrO4 (perbromate)

• HBrO3 (bromate)

• HBrO2 (bromite)

• HBrO (hypobromite)

Fundamental laws

• Law of Conservation of Mass

•Mass is neither created or destroyed

•Conversion from one form to another

Determination of Aver. Mass

• Ave. Mass = [(% Abund./100) (atomic

mass)] + [(% Abund./100) (atomic mass)]

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 %

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.)

Chemical Formula

• Gives the combining whole number ratios of the elements in a compound

• C6H12O6

Structural Formula

• Gives the spatial arrangement of atoms in the compound

• Structural formula for H2O is H – O – H

Empirical Formula

• Only gives the types of elements in the compound and the ratio of the elements in the formula

Empirical Formula

• Does not tell exactly how many of the elements are in the compound

Molecular Formula

• Gives you the exact elemental composition of the compound

• Formula of the compound as it would actually exist.

EF vs. MF

Sucrose or table sugar:

Molecular Formula = C6H12O6

Empirical Formula = CH2O

Sample Problem

• The compound adrenaline contains % C = 56.79 % H = 6.56 % O = 28.37 % N = 8.28 by mass. Find the empirical formula.

Empirical Formula

• EF Determination when % Masses are given

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

•

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.

Molecular Formula Detn.

Step 1. Obtain empirical formula mass by adding atomic masses of all elements in empirical formula

Molecular Formula Detn.

Step 2. Get ratio by applying the formula below:

Molecular Formula = given molar mass

Empirical formula mass

Molecular Formula Detn.

Step 3.

Multiply empirical formula subscripts by obtained ratio

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.

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.