1. Chapter 7 Lesson Starter CCl 4 MgCl 2  What kind of information can you discern from the formulas?  Which of the compounds represented is molecular and which is ionic?  Guess the name of each of the above compounds based on the formulas written.  Chemical formulas form the basis of the language of chemistry and reveal much information about the substances they represent.

2. Chapter 7 Chemical formula- shows the number and kinds of atoms in the smallest representative unit of the substance. Molecular formula- chemical formula written for a molecule (covalent compounds) Ex. CO 2 O 2 or H 2 O 2 The formula for this compound would be C 5 H 12 O Formula unit- chemical formula written for an ionic compound. Is the lowest whole-number ratio of ions in the compound The formula for this compound would be NaCl

3. Chapter 7 Which of the following is molecular and which is ionic?  NaBr  CO 2 O3O3  NO  KCl  AlBr 3 Ionic Molecular Ionic Two or more nonmetals Low melting point Low boiling point Example CO 2 Metal and nonmetals High melting point High boiling point Solid at room temperature Electrically neutral Example NaCl

4. Chapter 7  A chemical formula indicates the relative number of atoms of each kind in a chemical compound.  For a molecular compound, the chemical formula reveals the number of atoms of each element contained in a single molecule of the compound.  example: octane — C 8 H 18 Significance of a Chemical Formula Molecular Formulas: The subscript after the C indicates that there are 8 carbon atoms in the molecule. The subscript after the H indicates that there are 18 hydrogen atoms in the molecule.

5. Chapter 7 Significance of a Chemical Formula Formula Units:  Note also that there is no subscript for sulfur: when there is no subscript next to an atom, the subscript is understood to be 1.

6. Chapter 7 Names and Formulas  We’re going to learn how to write names and formulas for  Binary ionic compounds  ionic compounds (stock system)  Polyatomic ions / compounds  Molecular compounds  Acids and Salts Take out 3 sheets of paper and fold them over to make a flip chart like the picture above. Label our flip chart “Chemical Names and Formulas” Chemical Names and Formulas

7. Chapter 7 Formulas for Binary Ionic Compounds  Binary ionic compound: ionic compound composed of two elements  The formula for a binary ionic compound can be written using the identities of the compound’s ions and balancing the total numbers of positive and negative charges to be equal. Example: magnesium bromide Ions combined: Mg 2+, Br –, Br – Chemical formula: MgBr 2 “Crossing Over” 1. Write the symbols for the ions 2. Cross over the charges using the absolute value of the charge as the subscript for the other ion 3. Reduce and drop the charges

8. Chapter 7 Examples Write the formula for the binary ionic compounds formed between the following elements 1. Zinc and iodine 2. Zinc and sulfur 3. Potassium and iodine 4. Magnesium and chlorine 5. Sodium and sulfur 6. Aluminum and sulfur 7. Aluminum and nitrogen

9. Chapter 7 Naming Binary Ionic Compounds  The nomenclature, or naming system, or binary ionic compounds involves combining the names of the compound’s positive and negative ions.  The name of the cation is first, followed by the anion, the anions ending changes to “ide”:  example: Al 2 O 3 — aluminum oxide For most simple ionic compounds, the ratio of the ions is not given in the compound’s name, because it is understood based on the relative charges of the compound’s ions.

10. Chapter 7 Examples Name the binary ionic compounds indicated by the following formulas 1. AgCl 2. ZnO 3. CaBr 2 4. SrF 2 5. BaO 6. CaCl 2 7. Cs 2 S

11. Chapter 7 Binary Ionic Compounds (Stock System)  Some elements such as iron, form two or more cations with different charges.  To distinguish the ions formed by such elements, scientists use the Stock system of nomenclature.  The system uses a Roman numeral to indicate an ion’s charge.  examples:Fe 2+ iron(II) Fe 3+ iron(III)  Use the charge and follow the “cross over” technique to write the formula

12. Chapter 7 Examples Write the formula and give the name for the compounds formed between the following ions: 1. Cu (II) and Br 2. Fe (II) and O 3. Fe (III) and O 4. Pb (II) and Cl 5. Hg (II) and S 6. Hg (IV) and S 7. Sn (II) and F

13. Chapter 7 Binary Ionic Compounds (Stock System)  Naming ionic compounds using the stock system nomenclature: 1. Reverse cross the subscripts 2. Double check that the nonmetals charge is accurate 3. Add a positive for the metal and negative for the nonmetal Example: Cr 1 F 3  Cr 3 F 1   Cr 3+ F 1-

14. Chapter 7 Examples Name the binary ionic compounds indicated by the following formulas using stock system nomenclature 1. SnI 4 2. CoF 3 3. CuO 4. CoI 2 5. HgI 2 6. FeS 7. PbS 2

15. Chapter 7 The name of the ion with the greater number of oxygen atoms ends in -ate. = nitrate The name of the ion with the smaller number of oxygen atoms ends in -ite. = nitrite Naming Binary Ionic Compounds Polyatomic Ions 

16. Chapter 7 Naming Binary Ionic Compounds Containing Polyatomic Ions,  Some elements can form more than two types of oxyanions.  example: chlorine can form In this case, an anion that has one fewer oxygen atom than the -ite anion has is given the prefix hypo-. An anion that has one more oxygen atom than the -ate anion has is given the prefix per-. hypochlorite chlorite chlorateperchlorate

17. Chapter 7 Polyatomic Ions

18. Chapter 7 Understanding Formulas for Polyatomic Ionic Compounds

19. Chapter 7 Examples Name the binary ionic compounds containing polyatomic ions as indicated by the following formulas 1. Ca(OH) 2 2. KClO 3 3. NH 4 OH 4. Fe 2 (CrO 4 ) 3 5. KClO 6. Fe(ClO) 2 7. Ba(OH)S 2

20. Chapter 7 Formulas for Binary Ionic Compounds Containing Polyatomic Ions 

21. Chapter 7 Examples Write the formulas for the binary ionic compounds containing polyatomic ions 1. Lithium nitrate 2. Copper (II) sulfate 3. Sodium carbonate 4. Calcium nitrite 5. Potassium perchlorate 6. Barium hydroxide 7. Iron (II) hypochlorate

22. Chapter 7 Naming Binary Molecular Compounds  Unlike ionic compounds, molecular compounds are composed of individual covalently bonded units, or molecules.  As with ionic compounds, there is also a Stock system for naming molecular compounds.  The old system of naming molecular compounds is based on the use of prefixes.  examples:CCl 4 — carbon tetrachloride (tetra- = 4) CO — carbon monoxide (mon- = 1) CO 2 — carbon dioxide (di- = 2)

23. Chapter 7 Prefixes for Naming Covalent Compounds

24. Chapter 7 Examples Write the names for the binary ionic compounds containing polyatomic ions 1. Carbon tetraiodide 2. Phosphorus trichloride 3. Dinitrogen trioxide 4. Carbon dioxide 5. Dinitrogen pentoxide 6. Sulfur hexafluoride 7. Iodine trichloride

25. Chapter 7 Formulas for Binary Molecular Compounds  Use the stock system for naming molecular compounds  The ending of the name of the last atom is changed to “ide”  Note that if the first element is only one atom you do not need to write the “mono”  examples:carbon tetrachloride — CCl 4 carbon monoxide — CO carbon dioxide — CO 2

26. Chapter 7 Examples Write the names for the binary covalent compounds 1. SO 3 2. ICl 5 3. PBr 5 4. XeF 4 5. N 2 O 5 6. SF 6 7. N 2 O 3

27. Chapter 7 Acids and Salts  An acid is a certain type of molecular compound. Most acids used in the laboratory are either binary acids or oxyacids.  Binary acids are acids that consist of two elements, usually hydrogen and a halogen.  Oxyacids are acids that contain hydrogen, oxygen, and a third element (usually a nonmetal).  In the laboratory, the term acid usually refers to a solution in water of an acid compound rather than the acid itself.  example: hydrochloric acid refers to a water solution of the molecular compound hydrogen chloride, HCl

28. Chapter 7 Acids and Salts  Many polyatomic ions are produced by the loss of hydrogen ions from oxyacids.  examples: An ionic compound composed of a cation and the anion from an acid is often referred to as a salt.  Table salt, NaCl, contains the anion from hydrochloric acid, HCl.  Calcium sulfate, CaSO 4, is a salt containing the anion from sulfuric acid, H 2 SO 4. sulfuric acidH 2 SO 4  sulfate nitric acidHNO 3  nitrate phosphoric acidH 3 PO 4  phosphate

29. Chapter 7 Acid Formulas  Binary acids: Add hydro and change the anions ending to “ic”  Example: HF  hydroflouric acid  Oxyacids: For the polyatomic ion with the greater amount of oxygen's change the anions ending from “ate” to “ic”. For the one with fewer oxygens change the anions ending from “ite” to “ous”  Examples  HNO 3  nitric acid  HNO 2  nitrous acid

30. Chapter 7 Examples Write the names for the binary covalent compounds 1. HCl 2. HBr 3. HI 4. H 3 PO 4 5. H 2 SO 3 6. H 2 SO 4 7. CH 3 COOH

31. Chapter 7 Section 2

32. Chapter 7 Oxidation Numbers  In an ionic compound: the charges of the ions reflect the electron distribution of the compound.  oxidation numbers are assigned to the atoms in the compound or ion.  In order to indicate the general distribution of electrons among the bonded atoms in a compound or a polyatomic ion,  Unlike ionic charges, oxidation numbers do not have an exact physical meaning: rather, they serve as useful “bookkeeping” devices to help keep track of electrons.

33. Chapter 7 Assigning Oxidation Numbers 1. The atoms in a pure element have an oxidation number of zero. examples: Na, O 2, P 4, and, S 8, have oxidation # = 0 2. Fluorine has an oxidation number of –1 in all of its compounds because it is the most electronegative element.  examples: HF the F = -1 3. Oxygen usually has an oxidation number of –2.  Exceptions:  In peroxides, such as H 2 O 2, O = –1.  With fluorine, such as OF 2, O = +2.

34. Chapter 7 Assigning Oxidation Numbers 4. Hydrogen has an oxidation number of +1; it has an oxidation number of –1 when with metals.  Examples:  HF  H = +1 5. The more-electronegative element (right, top) is assigned a negative number equal to its charge examples: NaCl: Cl = -1

35. Chapter 7 Assigning Oxidation Numbers 6. The algebraic sum of the oxidation numbers of all atoms in an neutral compound is equal to zero. 7. The algebraic sum of the oxidation numbers of all atoms in a polyatomic ion is equal to the charge of the ion.

36. Chapter 7 Assigning Oxidation Numbers Sample Problem: Assign oxidation numbers to each atom in the following compounds or ions: a. UF 6 b. H 2 SO 4 c. ClO 3 - d. CI 4 e. Na 2 O 2 f. NO 2 -

37. Chapter 7 Using Oxidation Numbers for Formulas and Names  As shown in the table in the next slide, many nonmetals can have more than one oxidation number.  These numbers can sometimes be used in the same manner as ionic charges to determine formulas.  example: What is the formula of a binary compound formed between sulfur and oxygen? From the common +4 and +6 oxidation states of sulfur, you could predict that sulfur might form SO 2 or SO 3. Both are known compounds.

38. Chapter 7 Common Oxidation States of Nonmetals

39. Chapter 7 Using Oxidation Numbers for Formulas and Names Using oxidation numbers, the Stock system, introduced in the previous section for naming ionic compounds, can be used as an alternative to the prefix system for naming binary molecular compounds. Prefix systemStock system PCl 3 phosphorus trichloridephosphorus(III) chloride PCl 5 phosphorus pentachloridephosphorus(V) chloride N2ON2Odinitrogen monoxidenitrogen(I) oxide NOnitrogen monoxidenitrogen(II) oxide Mo 2 O 3 dimolybdenum trioxidemolybdenum(III) oxide

40. Chapter 7 Section 3

41. Chapter 7  A chemical formula indicates:  the elements present in a compound  the relative number of atoms or ions of each element present in a compound  Chemical formulas also allow chemists to calculate a number of other characteristic values for a compound:  formula mass  molar mass  percentage composition Section 3 Using Chemical Formulas

42. Chapter 7 Formula Masses  The formula mass is the sum of the average atomic masses of all atoms represented in its formula.  example:formula mass of water, H 2 O average atomic mass of H: 1.01 amu average atomic mass of O: 16.00 amu Section 3 Using Chemical Formulas average mass of H 2 O molecule: 18.02 amu

43. Chapter 7 Formula Masses  The mass of a water molecule can be referred to as a molecular mass.  The mass of one formula unit of an ionic compound, such as NaCl, is not a molecular mass.  The mass of any unit represented by a chemical formula (H 2 O, NaCl) can be referred to as the formula mass. Section 3 Using Chemical Formulas

44. Chapter 7 Formula Masses, continued Sample Problem F Find the formula mass of potassium chlorate, KClO 3. Section 3 Using Chemical Formulas

45. Chapter 7 Formula Masses, continued Sample Problem F Solution The mass of a formula unit of KClO 3 is found by adding the masses of one K atom, one Cl atom, and three O atoms. Atomic masses can be found in the periodic table in the back of your book. In your calculations, round each atomic mass to two decimal places. Section 3 Using Chemical Formulas

46. Chapter 7 Formula Masses, continued Sample Problem F Solution, continued Chapter 7 Section 3 Using Chemical Formulas formula mass of KClO 3 = 122.55 amu

47. Chapter 7 Molar Masses  The molar mass of a substance is equal to the mass in grams of one mole, or approximately 6.022 × 10 23 particles, of the substance.  example: the molar mass of pure calcium, Ca, is 40.08 g/mol because one mole of calcium atoms has a mass of 40.08 g.  The molar mass of a compound is calculated by adding the masses of the elements present in a mole of the molecules or formula units that make up the compound. Section 3 Using Chemical Formulas

48. Chapter 7 Molar Masses, continued  One mole of water molecules contains exactly two moles of H atoms and one mole of O atoms. The molar mass of water is calculated as follows. Section 3 Using Chemical Formulas molar mass of H 2 O molecule: 18.02 g/mol A compound’s molar mass is numerically equal to its formula mass.

49. Chapter 7 Calculating Molar Masses for Ionic Compounds Chapter 7 Section 3 Using Chemical Formulas

50. Chapter 7 Molar Masses, continued Sample Problem G What is the molar mass of barium nitrate, Ba(NO 3 ) 2 ? Section 3 Using Chemical Formulas

51. Chapter 7 Molar Masses, continued Sample Problem G Solution One mole of barium nitrate, contains one mole of Ba, two moles of N (1 × 2), and six moles of O (3 × 2). Chapter 7 Section 3 Using Chemical Formulas molar mass of Ba(NO 3 ) 2 = 261.35 g/mol

52. Chapter 7 Molar Mass as a Conversion Factor  The molar mass of a compound can be used as a conversion factor to relate an amount in moles to a mass in grams for a given substance.  To convert moles to grams, multiply the amount in moles by the molar mass: Amount in moles × molar mass (g/mol) = mass in grams Section 3 Using Chemical Formulas

53. Chapter 7 Mole-Mass Calculations Chapter 7 Section 3 Using Chemical Formulas

54. Chapter 7 Molar Mass as a Conversion Factor, continued Sample Problem H What is the mass in grams of 2.50 mol of oxygen gas? Section 3 Using Chemical Formulas

55. Chapter 7 Molar Mass as a Conversion Factor, continued Sample Problem H Solution Given: 2.50 mol O 2 Unknown: mass of O 2 in grams Solution: moles O 2 grams O 2 amount of O 2 (mol) × molar mass of O 2 (g/mol) = mass of O 2 (g) Section 3 Using Chemical Formulas

56. Chapter 7 Molar Mass as a Conversion Factor, continued Sample Problem H Solution, continued Calculate the molar mass of O 2. Section 3 Using Chemical Formulas Use the molar mass of O 2 to convert moles to mass.

57. Chapter 7 Converting Between Amount in Moles and Number of Particles Chapter 7 Section 3 Using Chemical Formulas

58. Chapter 7 Molar Mass as a Conversion Factor, continued Sample Problem I Ibuprofen, C 13 H 18 O 2, is the active ingredient in many nonprescription pain relievers. Its molar mass is 206.31 g/mol. a. If the tablets in a bottle contain a total of 33 g of ibuprofen, how many moles of ibuprofen are in the bottle? b. How many molecules of ibuprofen are in the bottle? c. What is the total mass in grams of carbon in 33 g of ibuprofen? Section 3 Using Chemical Formulas

59. Chapter 7 Molar Mass as a Conversion Factor, continued Sample Problem I Solution Given: 33 g of C 13 H 18 O 2 molar mass 206.31 g/mol Unknown:a. moles C 13 H 18 O 2 b. molecules C 13 H 18 O 2 c. total mass of C Solution:a. grams moles Section 3 Using Chemical Formulas

60. Chapter 7 Molar Mass as a Conversion Factor, continued Sample Problem I Solution, continued b. molesmolecules Section 3 Using Chemical Formulas c. moles C 13 H 18 O 2 moles C grams C

61. Chapter 7 Molar Mass as a Conversion Factor, continued Sample Problem I Solution, continued Section 3 Using Chemical Formulas a. b. c.

62. Chapter 7 Percentage Composition  It is often useful to know the percentage by mass of a particular element in a chemical compound.  To find the mass percentage of an element in a compound, the following equation can be used. Section 3 Using Chemical Formulas The mass percentage of an element in a compound is the same regardless of the sample’s size.

63. Chapter 7 Percentage Composition, continued  The percentage of an element in a compound can be calculated by determining how many grams of the element are present in one mole of the compound. Section 3 Using Chemical Formulas The percentage by mass of each element in a compound is known as the percentage composition of the compound.

64. Chapter 7 Percentage Composition of Iron Oxides Chapter 7 Section 3 Using Chemical Formulas

65. Chapter 7 Click below to watch the Visual Concept. Visual Concept Section 3 Using Chemical Formulas Percentage Composition

66. Chapter 7 Percentage Composition Calculations Chapter 7 Section 3 Using Chemical Formulas

67. Chapter 7 Percentage Composition, continued Sample Problem J Find the percentage composition of copper(I) sulfide, Cu 2 S. Section 3 Using Chemical Formulas

68. Chapter 7 Percentage Composition, continued Sample Problem J Solution Given: formula, Cu 2 S Unknown: percentage composition of Cu 2 S Solution: formula molar mass mass percentage of each element Section 3 Using Chemical Formulas

69. Chapter 7 Percentage Composition, continued Sample Problem J Solution, continued Section 3 Using Chemical Formulas Molar mass of Cu 2 S = 159.2 g

70. Chapter 7 Percentage Composition, continued Sample Problem J Solution, continued Section 3 Using Chemical Formulas

71. Chapter 7 Preview Lesson Starter Objectives Calculation of Empirical Formulas Calculation of Molecular Formulas Section 4 Determining Chemical Formulas

72. Chapter 7 Section 4 Determining Chemical Formulas Lesson Starter  Compare and contrast models of the molecules NO 2 and N 2 O 4.  The numbers of atoms in the molecules differ, but the ratio of N atoms to O atoms for each molecule is the same.

73. Chapter 7 Section 4 Determining Chemical Formulas Objectives  Define empirical formula, and explain how the term applies to ionic and molecular compounds.  Determine an empirical formula from either a percentage or a mass composition.  Explain the relationship between the empirical formula and the molecular formula of a given compound.  Determine a molecular formula from an empirical formula.

74. Chapter 7 Empirical and Actual Formulas Chapter 7 Section 4 Determining Chemical Formulas

75. Chapter 7 Section 4 Determining Chemical Formulas  An empirical formula consists of the symbols for the elements combined in a compound, with subscripts showing the smallest whole-number mole ratio of the different atoms in the compound.  For an ionic compound, the formula unit is usually the compound’s empirical formula.  For a molecular compound, however, the empirical formula does not necessarily indicate the actual numbers of atoms present in each molecule.  example: the empirical formula of the gas diborane is BH 3, but the molecular formula is B 2 H 6.

76. Chapter 7 Section 4 Determining Chemical Formulas Calculation of Empirical Formulas  To determine a compound’s empirical formula from its percentage composition, begin by converting percentage composition to a mass composition. Assume that you have a 100.0 g sample of the compound. Then calculate the amount of each element in the sample. example: diborane The percentage composition is 78.1% B and 21.9% H. Therefore, 100.0 g of diborane contains 78.1 g of B and 21.9 g of H.

77. Chapter 7 Section 4 Determining Chemical Formulas Calculation of Empirical Formulas, continued  Next, the mass composition of each element is converted to a composition in moles by dividing by the appropriate molar mass. These values give a mole ratio of 7.22 mol B to 21.7 mol H.

78. Chapter 7 Section 4 Determining Chemical Formulas Calculation of Empirical Formulas, continued  To find the smallest whole number ratio, divide each number of moles by the smallest number in the existing ratio. Because of rounding or experimental error, a compound’s mole ratio sometimes consists of numbers close to whole numbers instead of exact whole numbers. In this case, the differences from whole numbers may be ignored and the nearest whole number taken.

79. Chapter 7 Section 4 Determining Chemical Formulas Calculation of Empirical Formulas, continued Sample Problem L Quantitative analysis shows that a compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula of this compound.

80. Chapter 7 Section 4 Determining Chemical Formulas Calculation of Empirical Formulas, continued Sample Problem L Solution Given: percentage composition: 32.38% Na, 22.65% S, and 44.99% O Unknown: empirical formula Solution: percentage composition mass composition composition in moles smallest whole- number mole ratio of atoms

81. Chapter 7 Section 4 Determining Chemical Formulas Calculation of Empirical Formulas, continued Sample Problem L Solution, continued

82. Chapter 7 Calculation of Empirical Formulas, continued Sample Problem L Solution, continued Smallest whole-number mole ratio of atoms: The compound contains atoms in the ratio 1.408 mol Na:0.7063 mol S:2.812 mol O. Section 4 Determining Chemical Formulas Rounding yields a mole ratio of 2 mol Na:1 mol S:4 mol O. The empirical formula of the compound is Na 2 SO 4.

83. Chapter 7 Section 4 Determining Chemical Formulas Calculation of Molecular Formulas  The empirical formula contains the smallest possible whole numbers that describe the atomic ratio.  The molecular formula is the actual formula of a molecular compound.  An empirical formula may or may not be a correct molecular formula.  The relationship between a compound’s empirical formula and its molecular formula can be written as follows. x(empirical formula) = molecular formula

84. Chapter 7 Section 4 Determining Chemical Formulas Calculation of Molecular Formulas, continued  The formula masses have a similar relationship. x(empirical formula mass) = molecular formula mass  To determine the molecular formula of a compound, you must know the compound’s formula mass.  Dividing the experimental formula mass by the empirical formula mass gives the value of x.  A compound’s molecular formula mass is numerically equal to its molar mass, so a compound’s molecular formula can also be found given the compound’s empirical formula and its molar mass.

85. Chapter 7 Comparing Empirical and Molecular Formulas Chapter 7 Section 4 Determining Chemical Formulas

86. Chapter 7 Click below to watch the Visual Concept. Visual Concept Section 4 Determining Chemical Formulas Comparing Molecular and Empirical Formulas

87. Chapter 7 Calculation of Molecular Formulas, continued Sample Problem N In Sample Problem M, the empirical formula of a compound of phosphorus and oxygen was found to be P 2 O 5. Experimentation shows that the molar mass of this compound is 283.89 g/mol. What is the compound’s molecular formula? Section 4 Determining Chemical Formulas

88. Chapter 7 Calculation of Molecular Formulas, continued Sample Problem N Solution Given: empirical formula Unknown: molecular formula Solution: x(empirical formula) = molecular formula Section 4 Determining Chemical Formulas

89. Chapter 7 Calculation of Molecular Formulas, continued Sample Problem N Solution, continued Molecular formula mass is numerically equal to molar mass. molecular molar mass = 283.89 g/mol molecular formula mass = 283.89 amu empirical formula mass mass of phosphorus atom = 30.97 amu mass of oxygen atom = 16.00 amu empirical formula mass of P 2 O 5 = 2 × 30.97 amu + 5 × 16.00 amu = 141.94 amu Section 4 Determining Chemical Formulas

90. Chapter 7 Calculation of Molecular Formulas, continued Sample Problem N Solution, continued Dividing the experimental formula mass by the empirical formula mass gives the value of x. Section 4 Determining Chemical Formulas The compound’s molecular formula is therefore P 4 O 10. 2 × (P 2 O 5 ) = P 4 O 10

91. End of Chapter 7 Show