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Chapter 7 – Chemical Formulas and Chemical Compounds

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  1. Chapter 7 – Chemical Formulas and Chemical Compounds Taken from Modern Chemistry written by Davis, Metcalfe, Williams & Castka

  2. HW – Notes on section 7.1 pgs 203-215 Section 7.1 – Chemical Names and Formulas • Students will be able to : • Explain the significance of a chemical formula • Determine the formula of an ionic compound formed between two given ions • Name an ionic compound given its formula • Using prefixes, name a binary molecular compound from its formula. • Write the formula of a binary molecular compound given its name. Objectives

  3. Section 7.1 – Chemical Names and Formulas The chemical formula indicates the relative number of atoms of each element in a chemical compound. Al2(SO4)3 C12H22O11 Significance of a chemical formula Elements’ subscripts indicate the number of atoms in the compound. Note how in this example parenthesis surround a polyatomic anion and the subscript refers to the entire unit

  4. By gaining or losing electrons many main-group elements form ions with stable configurations. Group 1 metals lose one e- to give 1+ cations. Group 2 metals lose two e- to give 2+ cations. Section 7.1 – Chemical Names and Formulas Ions formed from a single atom are known as monatomic ions Monatomic Ions

  5. Not all main-group elements readily form ions, C and Si form covalent bonds where they share electrons. Section 7.1 – Chemical Names and Formulas The nonmetals in groups 15, 16 & 17 gain e- to form anions. Monatomic Ions (continued)

  6. K+ Mg2+ Section 7.1 – Chemical Names and Formulas Elements which give up 1 or more e-and take a positive (+) chargeare called cations. Elements which gain 1 or more e-and take a negative (-) charge are called anions. Monatomic Ions (continued) F- N3-

  7. Mg2+ K+ Potassium cation Magnesium cation Section 7.1 – Chemical Names and Formulas Cation naming is simple, element name and the word cation. For anions you drop the end of the element name and add –ide to the root. Naming Monatomic Ions (continued) N3- F- Fluorine  fluoride Nitrogen  Nitride

  8. Compounds composed of two different elements are known as binary compounds Section 7.1 – Chemical Names and Formulas Cation goes first : Mg2+, Br-, Br- Binary Ionic Compounds Balance to become electrically neutral And you get MgBr2

  9. The naming system involves combining the names of the compound’s positive and negative ions Section 7.1 – Chemical Names and Formulas Al2O3 Aluminum cation & oxide Naming Binary Ionic Compounds And you get Aluminum Oxide

  10. The Stock system of nomenclature Some elements form more than one cation, (no elements form more than one monoatomic anion) each with a different charge – add Roman Numerals Section 7.1 – Chemical Names and Formulas Naming Binary Ionic Compounds Fe2+ FeO Iron(II) oxide Fe3+ Fe2O3 Iron(III) oxide

  11. Compounds Containing Polyatomic Ions Oxyanions each is a polyatomic ion that contains oxygen. Section 7.1 – Chemical Names and Formulas Most Common (-ate) ending NO3- Nitrate Naming Binary Ionic Compounds One Less O (-ite) ending NO2- Nitrite

  12. Less electronegative element given first, prefix only for multiples Second element named with prefix indicating # of atoms, with few exceptions ends with –ide (only 2 elements) The o or a at the end of the prefix is dropped when the word following begins with another vowel. • mono- • di- • tri- • tetra- • penta- • hexa- • hepta- • octa- • nona- • deca- Section 7.1 – Chemical Names and Formulas Naming Binary Molecular Compounds

  13. Example pg 212 P4O10 Section 7.1 – Chemical Names and Formulas Naming Binary Molecular Compounds Prefix need if more than one Less-electronegative element Prefix indicating number Root name +ide tetraphosphorusdecoxide

  14. Similar to naming molecular compounds Section 7.1 – Chemical Names and Formulas Si3N4 SiO2 SiC Covalent-Network Compounds Trisilicon tetranitride Silicon Carbide Silicon dioxide

  15. Acids are a specific class of compound which usually refer to a solution of water of one of these special compounds. Section 7.1 – Chemical Names and Formulas H2SO4 HCl Acids and Salts Hydrochloric acid Sulfuric acid

  16. An ionic compound composed of a cation and the anion from an acid is often referred to as a salt. Section 7.1 – Chemical Names and Formulas NaCl Common Table Salt Acids and Salts- (continued)

  17. PRACTICE – pg 207 Q 1 & 2 all Practice Naming Section 7.1 – Chemical Names and Formulas PRACTICE – pg 209 Q 1 & 2 all PRACTICE – pg 211 Q 1 & 2 all Practices PRACTICE – pg 213 Q 1 & 2 all SECTION REVIEW– pg 215 Q 2,3 & 4 all

  18. Section 7.1 – Chemical Names and Formulas Quiz Break Quiz Key

  19. Section 7.1 – Chemical Names and Formulas - POGIL Ions How are ions made from neutral atoms? KEY

  20. Section 7.1 – Chemical Names and Formulas - POGIL Naming Ionic Compounds What are the structural units that make up Ionic compounds and how are they named? KEY

  21. Section 7.1 – Chemical Names and Formulas - POGIL Polyatomic Ions Can a group of atoms have a charge? KEY

  22. Section 7.1 – Chemical Names and Formulas - POGIL Naming Molecular Compounds How are the chemical formula and name of a molecular compound related? KEY

  23. Section 7.1 – V2 – Ions and Compounds Quiz Break Quiz Key

  24. Section 7.1 – Chemical Names and Formulas - POGIL Naming Acids KEY

  25. HW – Notes on section 7.2 pgs 216-219 • Students will be able to : • List the rules for assigning oxidation numbers. • Give the oxidation number for each element in the formula of a chemical compound. • Name the binary molecular compounds using oxidation numbers and the Stock sytem. Section 7.2 - Oxidation Numbers Objectives

  26. To indicate the general distribution of electrons among bonded atoms in molecular compounds , oxidation numbers (or states) are assigned to the atoms that compose the same. Section 7.2 - Oxidation Numbers Oxidation Numbers Some are arbitrary, but they are useful in naming compounds, in writing formulas and in balancing equations.

  27. The following are guidelines... Section 7.2 - Oxidation Numbers Atoms of pure elements have an oxidation number of zero. Na O2 P4 S8 Assigning Oxidation Numbers– the rules All zero.

  28. Section 7.2 - Oxidation Numbers Assigning Oxidation Numbers– the rules The more-electronegative element in a binary compound is assigned the number equal to the negative charge it would have as an anion. The less-electronegative is assigned the number equal to the positive charge it would have as a cation.

  29. Section 7.2 - Oxidation Numbers Assigning Oxidation Numbers– the rules Fluorine has an oxidation number of -1 as it is the most electronegative element.

  30. Oxygen has an oxidation number of -2 in almost all compounds. Exceptions peroxides is -1, compounds with halogens +2 Section 7.2 - Oxidation Numbers Hydrogen has an oxidation number of +1 in all compounds containing elements that are more- electronegative; it has an oxidation number of -1 in compounds with metals. Assigning Oxidation Numbers– the rules

  31. The algebraic sum of the oxidation numbers of all atoms in a neutral compound = zero. Section 7.2 - Oxidation Numbers The algebraic sum of the oxidation numbers of all atoms in a polyatomic ion = the charge of the ion. Assigning Oxidation Numbers– the rules Rules 1-7 apply to covalently bonded atoms, oxidation numbers can also be assigned to atoms in ionic compounds.

  32. Many non-metals have more than one oxidation state. Recall the use of Roman numerals to denote charges. Section 7.2 - Oxidation Numbers Using Oxidation Numbers for Formulas and Names

  33. Is the substance elemental? • No, three elements are present. • Is the substance ionic? • Yes, metal + non-metal. • Are there any monoatomic ions? • Yes, barium ion is monoatomic. • Barium ion = Ba2+ • Oxidation # for Ba = +2 • Which elements have specific rules? • Oxygen has a rule....-2 in most compounds • Oxidation # for O = -2 • Which element does not have a specific rule? • N does not have a specific rule. • Use rule 8 to find the oxidation # of N • Let N = Oxidation # for nitrogen • (# Ba) (Oxid. # of Ba) + (# N) (Oxid. # N) + (# of O) (Oxid. # of O) = 0 • 1(+2) + 2(N) + 6(-2) = 0 • N = +5 Ba(NO3)2 Section 7.2 Practices

  34. NF3 Section 7.2 • Is the substance elemental? • No, two elements are present. • Is the substance ionic? • No, two non-metals. • Are there any monoatomic ions? • Since it is molecular, there are no ions present. • Which elements have specific rules? • F = -1 • Which element does not have a specific rule? • N does not have a specific rule. • Use rule 8 to find the oxidation # of N • Let N = oxidation # of N • (# N) (Oxid. # N) + (# F) (Oxid. # F) = 0 • 1(N) + 3(-1) = 0 • N = +3 Practices

  35. (NH4)2SO4 • Is the substance elemental? • No, four elements are present. • Is the substance ionic? • Yes, even though there are no metals present, the ammonium ion is a common polyatomic cation. • Are there any monoatomic ions? • No, the cation and anion are both polyatomic. • Which elements have specific rules? • H = +1 because it is attached to a non-metal (N) • O = -2 • Which elements do not have a specific rule? • Neither N nor S has a specific rule. • You must break the compound into the individual ions that are present and then use rule 9 to find the oxidation numbers of N and S. Notice that if you try to use rule 8, you end up with one equation with two unknowns: 2N + 8(+1) + 1S + 4(-2) = 0 • The two ions present are NH4+ and SO42-. • N + 4(+1) = +1 so N = -3 • S + 4(-2) = -2 so S = +6 Section 7.2 Practices

  36. PRACTICE – pg 218 Q 1 all Section 7.2 SECTION REVIEW– pg 219 Q 1 & 2 all Practices Practice Sheet Key

  37. Key Section 7.2 Quiz

  38. HW – Notes on section 7.3 pgs 221-228 • Students will be able to : • Calculate the formula mass or molar mass of any given compound. • Use molar mass to convert between mass in grams and amount in moles of a chemical compound. • Calculate the # of molecules, formula units, or ions in a given molar amount of a chemical compound. • Calculate the % composition of a given chemical compound. Section 7.3 - Using Chemical Formulas Objectives

  39. The formula mass of any molecule, formula unit, or ion is the sum of the average atomic masses of all the atoms represented in the formula. Section 7.3 - Using Chemical Formulas Book example Average atomic mass of H : 1.01 amu Average atomic mass of O : 16.00 amu H2O  (2 x 1.01 amu) + 16 amu = 18.02 amu Formula Masses

  40. A compound’s molar mass is numerically equal to its formula mass. Section 7.3 - Using Chemical Formulas Book example Formula mass of H2O = 18.02 amu Which is also the molar mass of water 18.02 g/mol. Molar Masses

  41. There are 3 mole equalities. They are: 1 mol = 6.02 x 1023 particles 1 mol = g-formula-mass (periodic table) 1 mol = 22.4 L for a gas at STP* Section 7.3 – Using Chemical Formulas These become. . . [----] OR [----] [-------------] or [-------------] 1 mol [-------------] or [-------------] 1 mol 6.02 x 1023 particles 22.4 L 1 mol g-formula-mass (periodic table) 22.4 L 1 mol 1 mol 6.02 x 1023 particles g-formula-mass (periodic table) 1 mol Molar Mass as a Conversion Factor – Remember our old friends. . .

  42. The percentage by mass of each element in a compound is known as the percentage composition of the compound. Section 7.3 - Using Chemical Formulas Percentage Composition

  43. Percent Composition Example: Calculate the percent composition of Mg(NO3)2 Section 7.3 - Using Chemical Formulas Double check - do they total 100%? Percentage Composition - example

  44. Quiz Break Section 7.3 - Using Chemical Formulas Key Percentage Composition - practice

  45. HW – Notes on section 7.4 pgs 229-233 • Students will be able to : • Define empirical formula, and explain how the terms 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 Section 7.4 - Determining Chemical Formulas Objectives

  46. An empirical formula consists of the symbols for the elements combined in a compound, with subscripts showing the smallest whole-number mole ratios of the different atoms in the compound. Section 7.4 - Determining Chemical Formulas Calculation of Empirical Formulas

  47. Let’s Determine the empirical formula for a compound which is 54.09% Ca, 43.18% O, and 2.73% H Section 7.4 - Determining Chemical Formulas 1) Divide each percent by that element's atomic weight. Calculation of Empirical Formulas - example 2) To get the answers to whole numbers, divide through by the smallest one. This gives us  CaO2H2 better yet  Ca(OH)2

  48. An empirical formula may or may not be a correct molecular formula. Section 7.4 - Determining Chemical Formulas Book example , diborane1’s empirical formula is BH3, any multiple of that equals the same ratio – B2H6,B3H9,B4H12 etc Calculations of Molecular Formulas It is a colorless gas at room temperature with a repulsively sweet odor. Diborane mixes well with air, easily forming explosive mixtures. Diborane will ignite spontaneously in moist air at room temperature.

  49. The relationship between an empirical formula and a molecular formula is seen below: Section 7.4 - Determining Chemical Formulas X(empirical formula) = molecular formula X is a whole-number multiple indicating the factor that you need to multiply the empirical formula by to get the molecular formula. Calculations of Molecular Formulas – (continued)

  50. X = molecular formula empirical formula Section 7.4 - Determining Chemical Formulas Formula mass of diborane = 27.67 amu Empirical mass of diborane 13.84 amu Calculations of Molecular Formulas – (continued) The molecular formula of diborane is therefore B2H6 (BH3)2 = B2H6