CHAPTER 3 - Lecture Notes

Chapter 3: Chemical Compounds; A. Types of Compounds and Their Formulas

Molecular View of Elements and Compounds

Tro, Chemistry: A Molecular Approach

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Chapter 3: Chemical Compounds; A. Types of Compounds and Their Formulas

Molecular Elements • Certain elements occur as 2 atom molecules – H2, N2, O2, F2, Cl2, Br2, I2

• Other elements occur as polyatomic molecules – P4, S8, Se8

Chemistry A Molecular Aprroach, Tro, 1st edition, 2008, Pearson Prentice Hall, p.94

Chapter 3: Chemical Compounds; A. Types of Compounds and Their Formulas A. Types of Compounds and Their Formulas

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Molecular Compounds – Made up of nonmetal atoms bound with covalent bonds (sharing electrons) – The smallest unit of a molecular compound is the molecule –

Example: one water molecule

the molecular compound water

CHEM101 F14

Chapter 2: Chemical Compounds

A. Types of Compounds and Their Formulas

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Formulas for Molecular Compounds –

Molecular formula – Compounds are represented by the symbols of the atoms found in the molecule. A subscript represents the number of atoms of each element. •

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H2O (water) contains two hydrogen atoms and one oxygen atom C12H22O11 (sucrose) contains 12 carbon atoms, 22 hydrogen atoms and 11 oxygen atoms No information about how atoms are connected to one another in a molecular formula

Chapter 3: Chemical Compounds; A. Types of Compounds and Their Formulas

A. Types of Compounds and Their Formulas

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Formulas for Molecular Compounds –

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Empirical formula – the simplest formula for a compounds; the relative numbers of atoms are shown (empirical formulas are not too informative) Structural formula – shows how atoms are connected and the types of bonds Condensed structural formula – shows how atoms are connected and is written on a single line Line-angle formula – lines represent bonds; carbon and hydrogen atoms are not shown (all other elements shown)

Chapter 3: Chemical Compounds; A. Types of Compounds and Their Formulas

A. Types of Compounds and Their Formulas

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Formulas for Acetic Acid

Condensed structural formula

CH3COOH or CH3CO2H

General Chemistry: Principles and Modern Applications, Petrucci, Harwood, Herring, Madura, 9th edition, 2007, Pearson Prentice Hall, p. 69

Chapter 3: Chemical Compounds; A. Types of Compounds and Their Formulas

A. Types of Compounds and Their Formulas

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Example of a line-angle formula H3C CH3 CH3

OH

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Structure of cholesterol

CH3 CH3

Chapter 3: Chemical Compounds; A. Types of Compounds and Their Formulas

A. Types of Compounds and Their Formulas

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Notes about organic compounds/structures – – –

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In organic compounds, carbon shares four electrons with other atoms and thus forms four bonds with other atoms Carbon can form large chains based on this bonding scheme A double line ( = ) between two atoms represents a double bond. This is a stronger type of covalent bond in which two atoms are sharing two electrons A triple line ( ≡ ) between two atoms represents a triple bond. This is the strongest type of covalent bond. Two atoms share three electrons in a triple bond.

Chapter 3: Chemical Compounds; A. Types of Compounds and Their Formulas

A. Types of Compounds and Their Formulas

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Ionic Compounds

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Made up of metal and nonmetal atoms bound with ionic attractions (electrons are transferred and opposite charges attract )

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Ionic compounds are crystalline network solids

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The smallest unit of an ionic compound is called the formula unit

Chapter 3: Chemical Compounds; A. Types of Compounds and Their Formulas

Example: NaCl

Electron transfer Electron

Cl

+

Na

Na

ClIonic interaction

Na

+

Cl-

Chapter 3: Chemical Compounds; A. Types of Compounds and Their Formulas

A. Types of Compounds and Their Formulas

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Formulas for Ionic Compounds – give the formula for the smallest electrically neutral collection of ions – Example: NaCl

General Chemistry: Principles and Modern Applications, Petrucci, Harwood, Herring, Madura, 9th edition, 2007, Pearson Prentice Hall, p. 71

Chapter 3: Chemical Compounds; A. Types of Compounds and Their Formulas A. Types of Compounds and Their Formulas

Example: What is the formula unit for the salt formed by magnesium and chlorine (Mg forms +2 ions, chlorine forms -1 ions)? •

A neutral collection of ions of Mg and Cl would need to include: ONE Mg2+ (total of +2 charge) and TWO Cl- (total of -2 charge)

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The formula is then MgCl2

Chapter 3: Chemical Compounds; A. Types of Compounds and Their Formulas

Practice Example: Classify the following substances as atomic elements, molecular elements, ionic compounds or molecular compounds • MgCl2 • Br2 • Li • LiF • Oxygen CLICKER QUESTION • CH3CH2OH • S8 • Ti • TiO2

Chapter 3: Chemical Compounds; B. The Mole and Chemical Compounds

B. The Mole and Chemical Compounds

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Because molecules are made up of atoms, the mass of a molecule can be calculated using the atomic masses of atoms from the periodic table – Molecular mass Example: What is the mass of one molecule of H2SO4? 2 H atoms = 2 x (1.0079 u) = 2.0158 u 1 S atom = 1 x (32.066 u) = 32.066 u 4 O atoms = 4 x (15.9994 u) = 63.9976 u 2.0158 u + 32.066 u+ 63.9976 u = 98.0794 u

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Note on significant figures: – Do not round when using atomic weights from the Periodic Table: Generally keep four significant digits after the decimal

Chapter 3: Chemical Compounds; B. The Mole and Chemical Compounds

B. The Mole and Chemical Compounds

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Formula mass - the mass of a formula unit of an ionic compound

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Example: One formula unit of NaCl weighs ? 1 Na atom = 22.9898 u 1 Cl atom = 35.4527 u 22.9898 + 35.4527 u = 58.4425 u

Chapter 3: Chemical Compounds; B. The Mole and Compounds

Practice examples: Calculate the molecular or formula weights of the following: • SO2

CLICKER QUESTION •

Ba(OH)2

Chapter 3: Chemical Compounds; B. The Mole and Compounds

B. The Mole and Chemical Compounds

The mole concept can be extended and applied to compounds Example: water, H2O • One molecule of water contains two hydrogen atoms and one oxygen atom • One molecule of water weighs 18.0152 u • One mole of water weighs 18.0152 g • One mole of water contains 6.022 x 1023 water molecules

Chapter 3: Chemical Compounds; C. Composition of Chemical Compounds

C. Composition of Chemical Compounds

Moles and chemical formulas The chemical formula can be used to relate moles of a compound to moles of the atoms making up the compound: Example: water, H2O • One mole of water contains two moles of hydrogen atoms and one mole of oxygen atoms • 6.022 x 1023 water molecules contains 12.044 x 1023 hydrogen atoms and 6.022 x 1023 oxygen atoms

Chapter 3: Chemical Compounds; C. Composition of Chemical Compounds

Practice examples: • How many moles of Carbon and Oxygen are in one mole of carbon dioxide, CO2?

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Calculate the number of molecules in 5.02 g of carbon dioxide

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Calculate the number of oxygen atoms in 5.02 g of carbon dioxide

Chapter 3: Chemical Compounds; C. Composition of Chemical Compounds

C. Composition of Chemical Compounds

Practice Example:

How many ions are present in 0.10 mg of MgCl2?

CLICKER QUESTION

Chapter 3: Chemical Compounds; C. Composition of Chemical Compounds

Percent Composition For a compound, the mass percentage of one element that makes up the compound is calculated as follows:

mass of element Mass percentage  100% mass of compound The mass percentage of hydrogen in water is: 2.0158 u Mass % of H  100% 18.0152 u = 11.1894 %

Chapter 3: Chemical Compounds; C. Composition of Chemical Compounds

Practice Example: What is the mass percent composition of acetic acid, CH3COOH?

Chapter 3: Calculations with Chemical Formulas and Equations; F. Determining Formulas

Determining formulas

Formulas from percent composition When the formula for a compound is unknown, the compound can be analyzed for the percentages of the elements that make up the compound

To determine a formula from percent composition data: STEP 1: Assume 100 g of compound is present and set the masses in g of each element to their percentages. STEP 2: Convert all masses to moles STEP 3: Write a formula based on the numbers of moles STEP 4: Convert subscripts to small whole numbers (divide each subscript by the smallest number) to get the empirical formula STEP 5: Use the experimentally-determined molecular mass to calculate the molecular formula

Chapter 3: Calculations with Chemical Formulas and Equations; F. Determining Formulas

Determining formulas

Formulas from percent composition Example: A compound that contains only nitrogen and oxygen is 30.4% nitrogen by mass; the molecular mass of the compound is 92 u. STEP 1: Assume 100 g of compound is present: Mass (N) = 30.4 g Mass (O) = 69.6 g STEP 2: Convert all masses to moles 1mol N = 2.17 moles Moles (N) = 30.4 g x 14.0067 g Moles (O) = 69.6 g x

1mol O 15.9994 g

= 4.35 moles

STEP 3: Write a formula based on the numbers of moles N2.17O4.35

Chapter 3: Calculations with Chemical Formulas and Equations; F. Determining Formulas

Determining formulas

Formulas from percent composition Example: continued N2.17O4.35 STEP 4: Convert subscripts to small whole numbers (divide each subscript by the smallest number) to get the empirical formula

NO2 STEP 5: Use the experimentally-determined molecular mass to calculate the molecular formula empirical formula molecular mass: 14.0067 + 2(15.9994) = 46.0055 g molecular formula molecular mass: 92 g/mol Correct formula must be N2O4

Chapter 3: Calculations with Chemical Formulas and Equations; F. Determining Formulas

Determining formulas Practice Example: A compound with a molecular mass of 98.0 u contains 3.09 % Hydrogen by mass; 31.6% Phosphorus by mass and 65.3% Oxygen by mass. Determine its molecular formula.

Chapter 3: Calculations with Chemical Formulas and Equations; F. Determining Formulas

Determining formulas

Combustion Analysis The empirical formulas for compounds that are easily burned can be found via combustion analysis: In a combustion reaction, molecules are burned in a stream of oxygen gas. If the molecule contains H and C, the products are always carbon dioxide and water.

Chapter 3: Calculations with Chemical Formulas and Equations; F. Determining Formulas

Determining formulas Combustion Analysis An apparatus for combustion analysis: The masses of the products H2O and CO2 are recorded; the mass of the original sample is known

General Chemistry: Principles and Modern Applications, Petrucci, Harwood, Herring, Madura, 9th edition, 2007, Pearson Prentice Hall, p. 80

Chapter 3: Calculations with Chemical Formulas and Equations; F. Determining Formulas

Determining formulas

Combustion Analysis

To determine an empirical formula via combustion analysis: STEP 1: Determine the mass of carbon in the original sample. (Find the number of moles of C in the CO2 product and convert to grams of C) STEP 2: Determine the mass of hydrogen in the original sample. (Find the number of moles of H in the H2O product and convert to grams of H) STEP 3: Determine the mass of oxygen in the original sample. (Find the difference in the original mass and the known C and H masses). STEP 4: Determine a formula for the compound from the number of moles of C, H and O (C and H already known, calculate O from mass) STEP 5: Find the empirical formula by converting subscripts to smallest whole numbers

Chapter 3: Calculations with Chemical Formulas and Equations; F. Determining Formulas

Determining formulas

Combustion Analysis Example: What is the empirical formula for Vitamin C? Combustion analysis on a 0.2000 g sample yields 0.2998 g CO2 and 0.0819 g H2O. STEP 1: Determine the mass of carbon in the original sample. (Find the number of moles of C in the CO2 product and convert to grams of C) 0.2998 g CO2 x 1 mol CO 2 x 1 mol C x 12.011 g C 44.010 g CO 2 1 mol CO 2 1 mol C

= 0.08182 g C

STEP 2: Determine the mass of hydrogen in the original sample. (Find the number of moles of H in the H2O product and convert to grams of H) 0.0819 g H2O x

1 mol H 2 O 2 mol H 1.00794 g H x x = 0.009164 g H 18.0153 g H 2 O 1 mol H 2O 1 mol H

Chapter 3: Calculations with Chemical Formulas and Equations; F. Determining Formulas

Determining formulas

Combustion Analysis Example continued: (0.08182 g C

0.009164 g H) STEP 3: Determine the mass of oxygen in the original sample. (Find the difference in the original mass and the known C and H masses). Mass O = sample mass – mass C – mass H = 0.2000 g – 0.08182g – 0.009164 g = 0.1090 g O

STEP 4: Determine a formula for the compound from the number of moles of C, H and O (C and H already known, calculate O from mass)

1 mol C = 0.006812 moles C 12.011 g C 0.009164 g H x 1 mol H = 0.009092 moles H 1.00794 g H 1 mol O 0.1090 g O x = 0.006813 moles O 15.9994 g O 0.08182 g C x

C0.006812H0.009092O0.006813

Chapter 3: Calculations with Chemical Formulas and Equations; F. Determining Formulas

Determining formulas

Combustion Analysis Example continued: C0.006812H0.009092O0.006813 STEP 5: Find the empirical formula by converting subscripts to smallest whole numbers Divide by 0.006812 first: CH1.33O For smallest whole numbers, multiply each number by 3: C3H4O3 (actual formula for vitamin C6H8O6)

Chapter 3: Calculations with Chemical Formulas and Equations; F. Determining Formulas

Practice Example A 0.1888-g sample of a hydrocarbon produces 0.6260 g CO2 and 0.1602 g H2O in combustion analysis. Its molecular mass is found to be 106 amu. Determine the empirical and molecular formulas for this hydrocarbon.

Chapter 3: Chemical Compounds; D. Oxidation States and Chemical Compounds D. Oxidation States (AKA Oxidation Numbers) and Chemical Compounds

Rules for assigning oxidation states: (follow these rules in order) 1. An atom or molecule in its elemental state has an oxidation state of 0. Examples: Na(0), S(0), H2(0), Br2(0), O2(0)

2. The sum of oxidation states in: a. a neutral compound is 0. b. a polyatomic ion is equal to the charge on the polyatomic ion. Example: NO3- (O.S. of O + O.S. of N = -1)

3. In a compound, group 1 metals have an oxidation state of +1 and group 2 metals have an oxidation state of +2 Examples: Na+(+1), Ca2+(+2)

4. In a compound, fluorine has an oxidation state of -1. 5. In a compound, hydrogen usually has an oxidation state of +1. 6. In a compound, oxygen usually has an oxidation state of -2. 7. In binary compounds with metals group 17 elements have an oxidation state of -1; group 16 of -2; and group 15 of -3. Examples: Cl-(-1), S2-(-2)

Chapter 3: Chemical Compounds; D. Oxidation States and Chemical Compounds

Practice examples: Assign oxidation numbers to the atoms in the following compounds: CuF2 Fe2O3 C (s) H2O NaH Fe (s) Fe 2+ H2SO4

CLICKER QUESTION

Chapter 3: Chemical Compounds; E. Naming Compounds

E. Naming Compounds: Binary Ionic Compounds

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The formula for a binary ionic compound represents the simplest combining ratio of the ions in the compound –

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NaCl contains one sodium ion for every chloride ion

The simplest combining ratio of the ions is determined by the number of electrons that are transferred between the metal and nonmetal atoms –

The formula for the ionic compound that is formed between magnesium ions and bromide ions is MgBr2 Electron transfer

Br

Br

Mg Ionic Interactions

Mg Br-

2+

Br-

Chapter 3: Chemical Compounds; E. Naming Compounds

E. Naming Compounds: Binary Ionic Compounds

Formulas: • Net charge of the formula must equal zero • Write the elemental symbol for the cation first and the elemental symbol for the anion second Names: • Write the name of the cation first and the name of the anion second • Leave out the word “ion”

Chapter 3: Chemical Compounds; E. Naming Compounds

General Chemistry: Principles and Modern Applications, Petrucci, Harwood, Herring, Madura, 9th edition, 2007, Pearson Prentice Hall, p. 85

Chapter 3: Chemical Compounds; E. Naming Compounds E. Naming Compounds: Binary Ionic Compounds

Examples: What is the name and formula of the ionic compound made up of the: • sodium ion and the chloride ion – – –

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Na+ and ClNeed one of each ion for a net charge of zero NaCl, sodium chloride

Aluminum ion and the iodide ion – – –

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Al3+ and INeed one aluminum ion and three iodide ions for a net zero charge AlI3, aluminum iodide

Iron (III) ion and the oxide ion – – –

Fe3+ and O2Need two iron and three oxygen for a net zero charge Fe2O3, iron (III) oxide (or ferric oxide)

Chapter 3: Chemical Compounds; E. Naming Compounds E. Naming Compounds: Binary Ionic Compounds

Practice examples: • What is the name and formula of the ionic compound made up of the: – –

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calcium ion and the sulfide ion Nickel (II) ion and the chloride ion

Name the following compounds: – – – –

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MgF2 CoCl3 Ag2S MnO2

Give formulas for the following compounds: – – – –

Copper (III) chloride Lithium fluoride Iron (II) oxide Barium chloride

Chapter 3: Chemical Compounds; E. Naming Compounds E. Naming Compounds: Molecular Compounds Formulas of molecular compounds • Write the element with the positive oxidation state first Naming molecular compounds • Name the first element in the formula, using a prefix if needed • Name the second element in the formula using –ide ending and a prefix if needed Prefixes: Number Prefix 1 2 3 4 5 6 7 8 9 10

monoditritetrapentahexaheptaoctanonadeca-

Chapter 3: Chemical Compounds; E. Naming Compounds E. Naming Compounds: Molecular Compounds

Examples: N2O5 BBr3 SO3 SF6

dinitrogen pentoxide boron tribromide sulfur trioxide sulfur hexafluoride

Chapter 3: Chemical Compounds; E. Naming Compounds E. Naming Compounds: Binary Acids

Binary acids are compounds of hydrogen and a non-metal in aqueous solution. We name them differently than pure binary molecular compounds because they behave like acids in water:

HF (aq) HCl(aq) HBr (aq) HI (aq) H2S(aq)

hydrofluoric acid hydrochloric acid hydrobromic acid hydroiodic acid hydrosulfuric acid

Chapter 3: Chemical Compounds; E. Naming Compounds

Practice Examples: Name the following compounds:

HF (aq) PCl3 HCl (g)

CO2 N2O3

Chapter 3: Chemical Compounds; E. Naming Compounds

E. Naming Compounds: Polyatomic ions

Polyatomic ions • A polyatomic ion is a group of covalently-bonded atoms with a net charge • Names usually end in –ite or –ate • Polyatomic ions containing oxygen are called oxoanions – – –

They are named according to the increasing number of oxygen atoms and increasing oxidation state of the nonmetal: Hypo____ite _____ite ____ate per____ate Eg. ClOClO2ClO3ClO4Hypochlorite Chlorite Chlorate Perchlorate (1 oxygen;Cl+1)

(2 oxygen;Cl+3) (3 oxygen;Cl+5) (4 oxygen;Cl+7)

Chapter 3: Chemical Compounds; E. Naming Compounds

E. Naming Compounds: Polyatomic ions

Polyatomic ions

General Chemistry: Principles and Modern Applications, Petrucci, Harwood, Herring, Madura, 9th edition, 2007, Pearson Prentice Hall, p. 89

Chapter 3: Chemical Compounds; E. Naming Compounds E. Naming Compounds: Polyatomic ions

Naming ionic compounds that contain polyatomic ions • Memorize names of polyatomic ions (use rules about oxoanions for help) • Naming rules are the same as they were for binary ionic compounds – – –

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Name cation first and anion second Net charge of compound must equal zero Use parentheses for polyatomic ions when more than one ion is needed in the formula

Examples: – – – –

KCN LiNO3 Fe(OH)3 (NH4)2SO4

potassium cyanide lithium nitrate iron (III) hydroxide ammonium sulfate

Chapter 3: Chemical Compounds; E. Naming Compounds

E. Naming Compounds: Oxoacids Naming oxoacids • Oxoacids are acids formed by hydrogen and oxoanions • Naming rules – – –

Do not include the word hydrogen Change oxoanion name by dropping “-ate” or “-ite” ending Change to “-ous” or “-ic” ending

ate  ic ite  ous

Chapter 3: Chemical Compounds; E. Naming Compounds

E. Naming Compounds: Oxoacids Naming oxoacids

General Chemistry: Principles and Modern Applications, Petrucci, Harwood, Herring, Madura, 9th edition, 2007, Pearson Prentice Hall, p. 90

Chapter 3: Chemical Compounds; E. Naming Compounds

E. Naming Compounds: Hydrates Hydrates Hydrates - ionic compounds that hold water molecules in the 3dimensional solid structure of the compound Hydrates carry a certain number of water molecules per formula unit of the ionic compound The number of water molecules per formula unit of the compound is denoted after the compound formula: eg. CoCl2•6H2O Naming Hydrates: Name the ionic compound first Add the prefix for the number of water molecules + “hydrate” eg. CoCl2•6H2O is cobalt (II) chloride hexahydrate

Chapter 3: Chemical Compounds; E. Naming Compounds

Practice examples: Name the following compounds MnO2 KMnO4 P2O5 CuO NO HClO3 CuSO4•5H2O

CLICKER QUESTION