I. Oxidation Numbers II. Nomenclature III. The Mole

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I. Oxidation Numbers

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◦ A positive or negative whole number assigned to an element in a molecule or ion on the basis of a set of formal rules; to some degree it reflects the positive or negative character of an atom

1. Keep track of electrons 2. Tell if electrons gained, lost, or unequally shared 3. Allows us to predict formulas of chemical

compound

Not to be confused with oxidation, a process in which a substance loses one or more electrons….

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1. Electronegativity ◦ Higher EN = negative oxidation number

◦ Lower EN = positive oxidation number

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Remember always refer back to: ◦ Electronegativity

◦ Electrostatic forces

◦ Bonding Characteristics

◦ There will be exceptions to the following rules…

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1. Oxidation # of free atoms, pure elements, and polyatomic elements is ZERO ◦ Both atoms have equal EN, no transfer or shift of

electrons

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2. Oxidation # of monatomic ion is equal to the charge of the ion

If an atom loses 2e-, the other atom(s) must gain those 2e-. Electrons DO NOT just float around….

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3. The sum of all oxidation numbers in a compound must be ZERO.

Compounds are not electrically charged!

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4. Alkali metals always have a +1 oxidation # when not free

Hydrogen is not an alkali metal although it is in group 1

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5. Alkaline Earth Metals always have a +2 oxidation number. They form +2 ions when they bond.

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6a. Certain elements have the same oxidation # in almost all their compounds.

Halogens have oxidation number -1 when bond to metals

Halogen with higher EN than other bonded nonmental is assigned the negative number

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6b. Hydrogen assigned a +1 oxidation # in most compounds

BUT…..

Hydrogen + metal = metallic hydrides ◦ Hydrogen has a -1 oxidation number

◦ Hydrogen more EN than any other metal

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6c. Oxygen has an -2 oxidation number in most compounds

Oxygen is bonded to highly EN elements does not have -2 oxidation number

Oxygen is VERY EN and pulls e- from most other elements ◦ Exception: Perioxide ion O2

2- where O has -1 O.N.

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7. Oxidation number of all atoms in a polyatomic ion add up to the charge of the ion

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Follow rules in order ◦ If rules contradict each other, the rule listed first

should be followed

Write an algebraic equation to solve for unknown oxidation numbers in compounds

Ionic Compounds ◦ “Criss-Cross” method

◦ Use charge of one ion as the subscript for the other

◦ Simplify ratios of atoms

◦ Exception: Peroxide ion O22- example: Na2O2 , not

NaO

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II. Nomenclature

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IUPAC developed a systematic way to name compounds.

Names reveal the composition and qualities of certain substances

Indicate the types of bonds and intermolecular attractions

Covalent Compounds

Binary Ionic Compounds

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Composed of ONLY nonmetals

Two word names Use Greek Prefix

System Least EN element

first, then more EN element

Ending of last element is changed to -ide.

Number Prefix

1 Mono*

2 Di

3 Tri

4 Tetra

5 Penta

6 Hexa

7 Hepta

8 Octa

9 Nona

10 deca

*omit mono- for first atom

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1. Antimony tribromide

2. Hexaboron silicide

3. Chlorine dioxide

4. Iodide pentafluoride

5. P4S5

6. Si2Br6

7. CH4

8. NF3

SbBr3

B6Si

ClO2

IF5

Tetraphorsphorus pentasulfide

Disilicon hexabromide

Methane

Nitrogen trifluoride

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Compounds formed between metals and nonmetals! Ionic compounds DO NOT use Greek Prefix System Named according to the two elements or polyatomic

ions Positive ions use the same name as their parent

atoms (ex. sodium atoms form sodium ions). Named FIRST. Negative ions have an –ide ending. Named SECOND.

Polyatomic ions made up of 2+

types of atoms with covalent bonds! They act like a

single unit.

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Oxyanions: anions composed of oxygen and one other element/ polyatomic ion (table p. 174).

2 Forms of oxyanion: ◦ More oxygens = name ends in “___-ate”

◦ Less oxygens = name ends in “____-ite”

3+ Forms of oxyanion: ◦ Most oxygens = name “per-________-ate”

◦ Least oxygens = name “hypo-______-ite”

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Polyatomic compounds: compounds that contain polyatomic ions (see greengreen handout)

Same as binary compounds, but: ◦ Name the cation first

◦ Name the anion second

◦ Replace “-ide” ending with polyatomic ion name

◦ If 2 polyatomic ions, use polyatomic ion names

◦ Do not use Greek prefix system

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1. MgO 2. K2S 3. Na2SO4

4. Ba(ClO3)2

5. NH4Cl 6. K2Cr2O7

7. CaSO4

8. Zn3(PO4)2

Magnesium oxide

Potassium sulfide

Sodium sulfate

Barium chlorate

Ammonium chloride

Potassium dichromate

Calcium sulfate

Zinc phosphate

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Stock System (Roman Numeral System): ◦ If an element can have more than 1 oxidation state,

a roman numeral is placed in parenthesis after the element’s name.

◦ Transition metals

Common Name ◦ Use suffix at the end of the first element (metal)

◦ Smaller oxidation number “-ous”

◦ Larger oxidation number “-ic”

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1. Hg2I2

2. CuBr

3. FeCl2

4. Co2(C2O4)3

5. SnO

6. SnO2

7. PbSO4

8. Pb(SO4)2

Mercury (I) iodide

Copper (I) bromide

Iron (II) chloride

Cobalt (III) oxalate

Tin (II) oxide

Tin (IV) oxide

Lead (II) sulfate

Lead (IV) sulfate

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Hydrates are compounds that have water molecules in their crystalline structure. ◦ Hold water = “water of hydration”

◦ Formulas written followed by a “dot” and number of water molecules.

◦ Compounds name + greek prefix + hydrate

Ex. Na2CO3 · 7H2O

sodium carbonate heptahydrate

◦ Anhydrates = compounds with NO water in their structures…

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Binary acids form when binary compounds dissolve in water

“hydro”- ___________ acid

◦ HCl: hydrogen chloride hydrochloric acid

◦ HBr: hydrogen bromide hydrobromic acid

◦ H2S: hydrogen sulfide hydrosulfuric acid

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Ternary acids contain three elements, generally containing a polyatomic ion, or combination of hydrogen, oxygen, and a nonmetal.

◦ Rule1: Addition of 1 oxygen to the acid: Per__________-ic acid.

◦ Rule 2: Subtraction of 1 oxygen from the acid:

__________-ous acid. ◦ Rule 3: Subtraction of 2 oxygens from the acid:

hypo________-ous acid.

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III. The Mole

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It is a counting number (like a dozen)

Used to count really, really small things…

Avagadro’s Number (NA)

1 MOLE = 6.022x1023 units (4 sig figs)

A mole is a amount!!!!!

That is why it is used to count very small things, like atoms and molecules…

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1 mole of hockey pucks would equal the mass of the moon!

1 mole of basketballs would fill a bag the size of the earth!

1 mole of pennies would cover the Earth ¼ mile deep!

1 mole of sand would fill all the Great Lakes 10 times!

1 mole of popcorn kernels would cover the United States 9 miles deep!

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1 Mole of ANYTHING is equal to

6.022 x 106.022 x 102323 items

Remember……!Remember……!

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We can use the concept of the mole to solve for problems like:

◦ How many copper atoms are in one penny?

◦ What is the mass of a single atom?

◦ What is the mass of 0.500 moles of helium atoms?

◦ How many atoms are in 33mg of gold?

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Because a mole is so large, measurements aren’t counted, they are weighed.

Molar mass is the mass (grams) of 1 mole (NA) of particles of an element or compound

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The same number! Different units! Look at the periodic table Scientists chose Avagadro’s # (NA) to: ◦ relate atomic mass units to the larger, more

practical unit of GRAMS. ◦ Represent the # of particles in a mole so that the

atomic mass of an element and mass of a mole of the element have the same numeric value, just different units! (Hydrogen experiment)

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Carbon 12.01 g/mol

Aluminum 26.98 g/mol

Zinc 65.39 g/mol

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Sodium bicarbonate ◦ NaHCO3

◦ 22.99 + 1.01 + 12.01 + 3(16.00) = 84.01 g/mol

Use atomic mass from periodic table

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Examples:

H2O 2(1.01) + 16.00 = 18.02 g/mol

NaCl 22.99 + 35.45 = 58.44 g/mol

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P.184 in textbook flow chart

Mass (g) ↔ Moles ↔ Number of units (atoms, etc..)

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How many molecules or atoms are in a certain amount of a substance?

How many grams are there in a mole of a substance?

How many moles are there in ??? grams of a substance?

What is the percent composition of a substance? (how much do each of the different types of atoms weigh in the compound?)

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Use dimensional analysis to SOLVE problems… ◦ Sample Problems:

P. 184

P. 185

P. 186

P. 187

P. 189

P. 190

P. 192

P. 193

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Example: How many moles are in 22 grams of copper metal?

In all problems like this, you need to go through four steps to find a solution.

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Step 1: Figure out how many parts in your calculation you will have by using this diagram

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Step 2: Make a T-chart, and put whatever information the problem gave you in the top left. After that, put the units of whatever you were given in the bottom right of the T, and the units of what you want to find in the top right.

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Step 3: Put the conversion factors into the T-chart in front of the units on the right.

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Step 4: Cancel out the units from the top left and bottom right, then find the answer by multiplying all the stuff on the top together and dividing it by the stuff on the bottom.

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Pau!

Continue by adding another section in the

T-chart… repeat steps…

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…and there you go.

Molarity is the amount of a substance dissolved in one liter of solution.

Molarity (M) = moles/ liters of solution

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Chemical compounds contain two or more atoms chemically combined to behave as one unit.

Masses of compound units can be found by adding the masses of the atoms contained in them.

Formula unit = a single unit of a compound ◦ NaCl = one unit of sodium chloride

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The mass of a mole of a substance: ◦ Gram-atomic mass = mass of a mole of atoms

◦ Gram-molecular mass = mass of a mole of molecules

◦ Gram-formula mass= mass of a mole of formula units in an ionic compound

All have the units g/mol

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Formulas tell us the proportions of atoms in a molecule or ionic compound

We can relate # of atoms # of moles ◦ Example: gram-molecular mass of NH3?

◦ 1 mole of nitrogen = (1) 14.01 = 14.01 g/mol

◦ 3 moles of hydrogen = (3) 1.008 = 3.024 g/mol

◦ 14.01 + 3.024 = 17.03 g/mol

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Example: gram-formula mass of Al2(SO4)3?

Each formula unit contains two Al, three S, and 12 O. A mole of Al2(SO4)3 consists of 2 moles of Al atoms, 3 moles of S atoms, and 12 moles of O atoms.

Find the molar mass….. 342.1 g/mol Al2(SO4)3

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Structural Formulas – ◦ Show the types of atoms

◦ Exact composition of each molecule

◦ Arrangement of chemical bonds

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Molecular Formula – “TRUE FORMULA” ◦ Shows the types of atoms

◦ Exact composition of ATOMS in each molecule

◦ Does not show shape, location of bonds, or bond type

H2O = water

C2H4 = ethene

Cl2 = chlorine

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Empirical Formulas – ◦ Tell what elements are present in simple ratios

◦ Used for both ionic compounds and molecules

◦ Careful when writing empirical formulas for molecules…

May be the actual molecular composition OR

May only show the simplest ratio of atoms in the molecule

H2O = water Empirical formula = H20

C2H4 = ethene Empirical formula = CH2

Cl2 = chlorine Empirical formula = Cl

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Empirical Formulas – REDUCE SUBSCRIPTS Example:

C2H6 CH3

◦ 1. FIND MASS (OR %) OF EACH ELEMENT

◦ 2. Find moles of each element

◦ 3. Divide moles by the smallest # to find subscripts

◦ 4. When necessary, multiply subscripts by 2,3, or 4 to get whole #’s

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Find the empirical formula for a sample of 25.9% N and 74.1% O.

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Empirical formula:

N1O2.5

Need to make subscripts whole numbers x2

N2O5

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Mole ratio in the EF mass ratio

Example: ◦ Water formula – H20 2 moles hydrogen for every 1

mole of oxygen

◦ Express in mass – 1 mole of water contains 2.016g of hydrogen atoms and 16.00g of oxygen atoms

◦ Convert moles to mass (grams)

2 mole H (1.008g/1 mole H) = 2.016g H

1 mole O (16.00g/ 1 mole O) = 16.00g O

Total mass of water = 2.016g + 16.00g = 18.02g

Now find % composition…..

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Molecular Formula: 1. Find the empirical formula

2. Find the empirical formula mass

3. Divide the molecular mass by the empirical mass

4. Multiply each subscript by the answer from step 3

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The empirical formula for ethene is CH2. Find the molecular formula is the molecular mass is 28.1 g/mol.

Empirical mass = 14.03 g/mol

(28.1 g/mol)/ (14.03 g/mol) = 2.00

(CH2)2 C2H4

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Percent composition = the mass composition of a compound

All other formulas describe # of atoms

%Comp. describes masses of atoms

Which atoms make up the most mass in a compound or molecule?

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Percent = “per hundred”

General setup: ◦ Part/whole x 100%

Example: ◦ Lab analysis of 30.00g Al2(SO4)3

4.731g Al (4.731/30.00) x 100% = 15.77% Al

8.433g S (8.433/30.00) x 100% = 28.11% S

16.836g O (16.836/ 30.00) x 100% = 56.12% O

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Sample Problem p. 192

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