Unit 5 Bonding and Nomenclature

Unit 5 Bonding and Nomenclature

lithium nitrate lead (II) sulfide lithium nitride barium sulfide lithium nitrite Chemistry sulfur dioxide Unit 4 Part 2: Bonding and Inorganic Nomenclature NaClO3 NO2 Fe(ClO3)2 N2O4 Fe(ClO3)3 N2O5

PowerPoint Presentation by Mr. John Bergmann Chemical Bonding Ionic Bonds: atoms give up or gain e and are attracted to each other by coulombic attraction Na loses e Na1+ Na1+ + Cl1 Cl gains e NaCl ionic compounds = salts K1+ + NO31 KNO3 where NO31 is a polyatomic ion: a charged group of atoms that stay together

Cl1 Properties of Salts 1. very hard each ion is bonded to several oppositely -charged ions 2. high melting points many bonds must be broken 3. brittle with sufficient force, like atoms are brought next to each other and repel calcite Covalent Bonds atoms share e to get a full valence shell C (4 v.e) 1s2 2s2 2p2 1s2 2s2 2p5 (7 v.e) F both need 8 v.e for a full outer shell (octet rule) a model of a covalent molecule Lewis structure: that shows all of the valence e

1. Two shared e make a single covalent bond, four make a double bond, etc. 2. unshared pairs: pairs of unbonded valence e 3. Each atom needs a full outer shell, i.e., 8 e . Exception: H needs 2 e carbon tetrafluoride (CF4) o x x o o C o x x x x F x x x x F

x x x x x x x x x x F x x x x F x x x x o

o o C o H H x o H x H x o H x o x

o H Hx C x o Hx x x F x o x x x x x x x Co F x F x x o x x x x methane (CH4)

x x x x x x x x x x x x F nitrogen triiodide (NI3) o x x o o N o

o x x x x o o I N o x x o x x x x o x x x I

x x x x x I x x x x x x x x Ix x x x Ix x x x x

x x x x x I carbon dioxide (CO2) o o x x o C o x x xx Ox x

xx Ox x x x x x Ox x o o o C o x xx Ox

x x xx xx O =C=O xx xx covalent compounds = molecular compounds -- have lower melting points than do ionic compounds (consist of two nonmetal elements) butter Metallic Bonds In metals, valence shells of atoms overlap, so v.e are free to travel between atoms through material. In insulators (like wood), the v.e are attached to particular atoms.

Not so in metals. Properties of Metals ductile malleable conduct heat and electricity All due to free-moving v.e. Other Types of Bonds dipole-dipole forces hydrogen bonds London dispersion forces ion-dipole forces boiling H2O DNA These are much weaker than ionic, covalent, or metallic bonds, but very important in determining states of matter, boiling and melting points, and

molecular shape (among other things). Writing Formulas of Ionic Compounds chemical formula: has neutral charge; shows types of atoms and how many of each To write an ionic compounds formula, we need: 1. the two types of ions (i.e., pink and blue) 2. the charge on each ion Na1+ and F1 NaF Ba2+ and O2 BaO Na1+ and

O2 Na2O Ba2+ and F1 BaF2 charge on cation / anion criss-cross rule: becomes subscript of anion / cation ** Warning: Reduce to lowest terms. Al3+ and O2 Ba2+ and S2 In3+ and Br1 Al 2 O 3 Ba 2S 2 In 1 Br 3

Al2O3 BaS InBr3 Writing Formulas w/Polyatomic Ions Parentheses are required only when you need more than one bunch of a particular polyatomic ion. Ba2+ and SO42 BaSO4 Mg2+ and NO21 Mg(NO2)2 NH41+ and

ClO31 NH4ClO3 Sn4+ and SO42 Sn(SO4)2 Fe3+ and Cr2O72 Fe2(Cr2O7)3 N3 (NH4)3N NH41+ and Inorganic Nomenclature

potassium nitrate KNO3 copper (II) sulfate Cu2SO4 dinitrogen monoxide N2O sodium hydroxide NaOH Ionic Compounds (cation/anion combos) Single-Charge Cations with Elemental Anions i.e., pulled off the Table anions The single-charge cations are: groups 1, 2, 13, and Ag1+, Cd2+, and Zn2+ Na A. To name, given the formula: Ba 1. Use name of cation. 2. Use name of anion (it has the ending ide).

NaF sodium fluoride BaO barium oxide Na2O sodium oxide BaF2 barium fluoride Zn Ca Ag B. To write formula, given the name: 1. Write symbols for the two types of ions. 2. Balance charges to write formula. silver sulfide

Ag1+ S2 Ag2S zinc phosphide Zn2+ P3 Zn3P2 calcium iodide Ca2+ I1 CaI2 Multiple-Charge Cations with Elemental Anions i.e., pulled off the Table anions The multiple-charge cations are: Pb, Sn, and the transition elements

(but of course! not Ag, Cd, or Zn) A. To name, given the formula: 1. Figure out charge on cation. 2. Write name of cation. 3. Write Roman numerals in ( ) to show cations charge. 4. Write name of anion. Fe Cu Stock System of nomenclature FeO iron Fe?2+ oxide O2 Fe2O3 iron Fe?3+ oxide Fe?3+ O2 O2 O2 iron (III) oxide

CuBr 1 copper Cu?1+ Brbromide copper (I) bromide CuBr2 1 copper Cu?2+ Brbromide Br1 copper (II) bromide iron (II) oxide B. To find the formula, given the name: 1. Write symbols for the two types of ions. 2. Balance charges to write formula. Co Sn cobalt (III) chloride Co3+

Cl1 CoCl3 tin (IV) oxide Sn4+ O2 SnO2 tin (II) oxide Sn2+ O2 SnO Compounds Containing Polyatomic Ions Insert name of ion where it should go in the compounds name. Write formulas: iron (III) nitrite

Fe3+ NO21 Fe(NO2)3 ammonium phosphide NH41+ P3 (NH4)3P ammonium chlorate NH41+ ClO31 NH4ClO3 zinc phosphate Zn2+ PO43 Zn3(PO4)2 lead (II) permanganate Pb2+ MnO41 Pb(MnO4)2 Write names: (NH4)2SO2 ammonium hyposulfite

AgBrO2 silver bromite (NH4)3N ammonium nitride Mn(CrO4)2 Mn?4+CrO42 manganese (IV) chromate CrO42 Cr2(SO2)3 Cr?3+ SO32 chromium (III) hyposulfite Cr?3+ SO32 SO32 Covalent Compounds -- contain two types of nonmetals ** Key: FORGET CHARGES! Use Greek prefixes to indicate how What to do: many atoms of each element, but dont use mono on first element. 1 mono 2 di 3 tri

6 hexa 7 hepta 8 octa 4 tetra 5 penta 9 nona 10 dec EXAMPLES: carbon dioxide CO2 CO dinitrogen trioxide carbon monoxide N2O3 N2O5 dinitrogen pentoxide carbon tetrachloride

CCl4 NI3 nitrogen triiodide Dihydrogen Monoxide: A Tale of Danger and Irresponsibility -- major component of acid rain -- found in all cancer cells -- inhalation can be deadly -- excessive ingestion results in acute physical symptoms: e.g., frequent urination, bloated sensation, profuse sweating -- often an industrial byproduct of chemical reactions; dumped wholesale into rivers and lakes Traditional System of Nomenclature (i.e., NOT the Stock System) used historically (and still some today) to name compounds w/multiple-charge cations Fe

Cu Sn Au Pb 1. Use Latin root of cation. To use: 2. Use -ic ending for higher charge; -ous ending for lower charge. 3. Then say name of anion, as usual. ous gold, Au aurAu3+ Au1+ lead, Pb plumbPb4+ Pb2+ tin, Sn stannSn4+ Sn2+ copper, Cu cuprCu2+ Cu1+ 3+ Write formulas: names: iron, Fe ferr- Write Fe

3 P3 P 4+ ? ? 4+ ? 4+ Pb3P4 Pb Pb Pb cuprous Fe2+ sulfide P3 P3 plumbic phosphide Cu2S Cu1+ S2 Pb3P2 Pb?2+ Pb?2+ Pb?2+ P3 P3 auric nitride plumbous Au3+ N3 AuN 1 Cl phosphide SnCl 1 Sn?4+ Cl1 Cl1 ferrous fluoride 4

Fe2+ F1 FeF2 Cl stannic chloride Empirical Formula and Molecular Formula shows the true number and type of atoms in a mcule lowest-terms formula Compound Molecular Formula Empirical Formula glucose C6H12O6

CH2O propane C3H8 C3H8 butane C4H10 C2H5 naphthalene C10H8 C5H4 sucrose C12H22O11 C12H22O11 octane

C8H18 C4H9 Empirical Formulas Empirical Formulas How can you calculate the empirical formula of a compound? Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. . Empirical Formulas The empirical formula of a compound gives the lowest whole-number ratio of the atoms or moles of the elements in a compound. An empirical formula may or may not be the same as a molecular formula. For example, the lowest ratio of hydrogen to oxygen in hydrogen peroxide is 1:1. Thus, the empirical formula of hydrogen peroxide is HO. The molecular formula, H2O2, has twice the number of atoms as the empirical formula. Notice that the ratio of hydrogen to oxygen is still the same, 1:1. Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved.

. Empirical Formulas For carbon dioxide, the empirical and molecular formulas are the same CO2. Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. . Empirical Formulas The figure below shows two compounds of carbon and hydrogen having the same empirical formula (CH) but different molecular formulas. Ethyne (C2H2), also called acetylene, is a gas used in welders torches. Styrene (C8H8) is used in making polystyrene. Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved.

. Empirical Formulas The percent composition of a compound can be used to calculate the empirical formula of that compound. The percent composition tells the ratio of masses of the elements in a compound. The ratio of masses can be changed to ratio of moles by using conversion factors based on the molar mass of each element. The mole ratio is then reduced to the lowest wholenumber ratio to obtain the empirical formula of the compound. Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. . Sample Problem Problem 10.12 10.12 Determining the Empirical Formula of a Compound A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula of the compound?

Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. . Sample Problem Problem 10.12 10.12 1 Analyze List the knowns and the unknown. The percent composition gives the ratio of the mass of nitrogen atoms to the mass of oxygen atoms in the compound. Change the ratio of masses to a ratio of moles and reduce this ratio to the lowest whole-number ratio. KNOWNS percent by mass of N = 25.9% N percent by mass of O = 74.1% O UNKNOWN empirical formula = N?O? Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. . Sample Problem Problem 10.12

10.12 2 Calculate Solve for the unknown. Convert the percent by mass of each element to moles. 1 mol 25.9 g N = 1.85 mol N N 14.0 g N 1 mol 74.1 g O = 4.63 mol O O 16.0 g O The mole ratio of N to O is N1.85O4.63. Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. . Percent means parts per 100, so 100.0 g of the compound contains 25.9 g N and 74.1 g O.

Sample Problem Problem 10.12 10.12 2 Calculate Solve for the unknown. Divide each molar quantity by the smaller number of moles to get 1 mol for the element with the smaller number of moles. 1.85 mol = 1 mol N N 1.85 4.63 mol O 1.85 = 2.50 mol O The mole ratio of N to O is N1O2.5. Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. . Sample Problem Problem 10.12 10.12 2 Calculate Solve for the unknown.

Multiply each part of the ratio by the smallest whole number that will convert both subscripts to whole numbers. 1 mol N 2 = 2 mol N 2.5 mol O 2 = 5 mol O The empirical formula is N2O5. Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. . Sample Problem Problem 10.12 10.12 3 Evaluate Does the result make sense? The subscripts are whole numbers, and the percent composition of this empirical formula equals the percents given in the original problem. Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. . Molecular Formulas Molecular Formulas

How does the molecular formula of a compound compare with the empirical formula? Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. . Interpret Data Data Ethyne and benzene have the same empirical formulaCH. Comparison of Empirical and Molecular Formulas Formula (name) Classification of formula Molar mass (g/mol) CH Empirical 13

C2H2 (ethyne) Molecular 26 (2 13) C6H6 (benzene) Molecular 78 (6 13) CH2O (methanol) Empirical and molecular 30 C2H4O2 (ethanoic acid) Molecular 60 (2 30) C6H12O6 (glucose) Molecular

180 (6 30) Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. . Interpret Data Data Methanal, ethanoic acid, and glucose have the same empirical formula CH2O. Comparison of Empirical and Molecular Formulas Formula (name) Classification of formula Molar mass (g/mol) CH Empirical 13 C2H2 (ethyne)

Molecular 26 (2 13) C6H6 (benzene) Molecular 78 (6 13) CH2O (methanal) Empirical and molecular 30 C2H4O2 (ethanoic acid) Molecular 60 (2 30) C6H12O6 (glucose) Molecular 180 (6 30)

Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. . Interpret Data Data Notice that the molar masses of the compounds in these two groups are simple whole-number multiples of the molar masses of the empirical formulas, CH and CH2O. Comparison of Empirical and Molecular Formulas Formula (name) Classification of formula Molar mass (g/mol) CH Empirical 13 C2H2 (ethyne)

Molecular 26 (2 13) C6H6 (benzene) Molecular 78 (6 13) CH2O (methanal) Empirical and molecular 30 C2H4O2 (ethanoic acid) Molecular 60 (2 30) C6H12O6 (glucose) Molecular 180 (6 30)

Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. . Molecular Formulas Methanal (formaldehyde), ethanoic acid (acetic acid), and glucose have the same empirical formula CH2O. Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. . Molecular Formulas The molecular formula of a compound is either the same as its experimentally determined empirical formula, or it is a simple whole-number multiple of its empirical formula. Once you have determined the empirical formula of a compound, you can determine its molecular formula, if you know the compounds molar mass.

Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. . Molecular Formulas You can calculate the empirical formula mass (efm) of a compound from its empirical formula. This is simply the molar mass of the empirical formula. Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. . Molecular Formulas You can calculate the empirical formula mass (efm) of a compound from its empirical formula. Then you can divide the experimentally determined molar mass by the empirical formula mass. This quotient gives the number of empirical formula units in a molecule of the compound and is the multiplier to convert the empirical formula to the molecular formula. Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved.

. Sample Problem Problem 10.13 10.13 Finding the Molecular Formula of a Compound Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH4N. Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. . Sample Problem Problem 10.13 10.13 1 Analyze List the knowns and the unknown. Divide the molar mass by the empirical formula mass to obtain a whole number. Multiply the empirical formula subscripts by this value to get the molecular formula. KNOWNS empirical formula = CH4N

molar mass = 60.0 g/mol UNKNOWN molecular formula = C?H? N? Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. . Sample Problem Problem 10.13 10.13 2 Calculate Solve for the unknown. First calculate the empirical formula mass. efm of CH4N = 12.0 g/mol + 4(1.0 g/mol) + 14.0 g/mol = 30.0 g/mol Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. .

Sample Problem Problem 10.13 10.13 2 Calculate Solve for the unknown. Divide the molar mass by the empirical formula mass. molar mass efm 60.0 g/mol = 2 = 30.0 g/mol Multiply the formula subscripts by this value. (CH4N) 2 = C2H8N2 Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. . Sample Problem Problem 10.13 10.13 3 Evaluate Does the result make sense?

The molecular formula has the molar mass of the compound. Copyright Pearson Education, Inc., or its affiliates. All Rights Reserved. .

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