Presentation on theme: "Chapter 7 – Chemical Formulas and Chemical Compounds"— Presentation transcript:
1Chapter 7 – Chemical Formulas and Chemical Compounds Taken from Modern Chemistry written by Davis, Metcalfe, Williams & Castka
2Section 7.1 – Chemical Names and Formulas HW – Notes on section 7.1 pgsObjectivesStudents will be able to :Explain the significance of a chemical formulaDetermine the formula of an ionic compound formed between two given ionsName an ionic compound given its formulaUsing prefixes, name a binary molecular compound from its formula.Write the formula of a binary molecular compound given its name.
3Section 7.1 – Chemical Names and Formulas Significance of a chemical formulaThe chemical formula indicates the relative number of atoms of each element in a chemical compound.Al2(SO4)3C12H22O11Note how in this example parenthesis surround a polyatomic anion and the subscript refers to the entire unitElements’ subscripts indicate the number of atoms in the compound.
4Section 7.1 – Chemical Names and Formulas By gaining or losing electrons many main-group elements form ions with stable configurations.Monatomic IonsGroup 1 metals lose one e- to give 1+ cations.Group 2 metals lose two e- to give 2+ cations.Ions formed from a single atom are known as monatomic ions
5Section 7.1 – Chemical Names and Formulas Not all main-group elements readily form ions, C and Si form covalent bonds where they share electrons.Monatomic Ions(continued)The nonmetals in groups 15, 16 & 17 gain e- to form anions.
6Section 7.1 – Chemical Names and Formulas Monatomic Ions(continued)K+Mg2+Elements which give up 1 or more e- and take a positive (+) charge are called cations.Elements which gain 1 or more e- and take a negative (-) charge are called anions.F-N3-
7Section 7.1 – Chemical Names and Formulas Mg2+K+Naming Monatomic Ions(continued)PotassiumcationMagnesiumcationCation 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.N3-F-Fluorine fluorideNitrogen Nitride
8Section 7.1 – Chemical Names and Formulas Binary Ionic CompoundsCompounds composed of two different elements are known as binary compoundsCation goes first : Mg2+, Br-, Br-Balance to become electrically neutralAnd you getMgBr2
9Section 7.1 – Chemical Names and Formulas Naming Binary Ionic CompoundsThe naming system involves combining the names of the compound’s positive and negative ionsAl2O3Aluminum cation & oxideAnd you getAluminum Oxide
10Section 7.1 – Chemical Names and Formulas Naming Binary Ionic CompoundsThe Stock system of nomenclatureSome elements form more than one cation, (no elements form more than one monoatomic anion) each with a different charge – add Roman NumeralsFe2+FeOIron(II) oxideFe3+Fe2O3Iron(III) oxide
11Section 7.1 – Chemical Names and Formulas Naming Binary Ionic CompoundsCompounds Containing Polyatomic IonsOxyanions each is a polyatomic ion that contains oxygen.MostCommon(-ate) endingNO3-NitrateOneLess O(-ite) endingNO2-Nitrite
12Section 7.1 – Chemical Names and Formulas Naming Binary Molecular CompoundsLess electronegative element given first, prefix only for multiplesSecond 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-
13Section 7.1 – Chemical Names and Formulas Naming Binary Molecular CompoundsExample pg 212P4O10Prefix need if more than oneLess-electronegative elementPrefix indicating numberRoot name +idetetraphosphorus decoxide
14Section 7.1 – Chemical Names and Formulas Covalent-Network CompoundsSimilar to naming molecular compoundsSiCSiliconCarbideSiO2SilicondioxideSi3N4Trisilicontetranitride
15Section 7.1 – Chemical Names and Formulas Acids and SaltsAcids are a specific class of compound which usually refer to a solution of water of one of these special compounds.HClHydrochloricacidH2SO4Sulfuricacid
16Section 7.1 – Chemical Names and Formulas Acids and Salts - (continued)An ionic compound composed of a cation and the anion from an acid is often referred to as a salt.NaClCommonTable Salt
24Section 7.1 – Chemical Names and Formulas - POGIL Naming AcidsKEY
25Section 7.2 - Oxidation Numbers ObjectivesHW – Notes on section 7.2 pgsStudents 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.
26Section 7.2 - Oxidation Numbers 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.Some are arbitrary, but they are useful in naming compounds, in writing formulas and in balancing equations.
27Section 7.2 - Oxidation Numbers Assigning Oxidation Numbers – the rulesThe following are guidelines...Atoms of pure elements have an oxidation number of zero.NaO2P4S8All zero.
28Section 7.2 - Oxidation Numbers Assigning Oxidation Numbers – the rulesThe 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.
29Section 7.2 - Oxidation Numbers Assigning Oxidation Numbers – the rulesFluorine has an oxidation number of -1 as it is the most electronegative element.
30Section 7.2 - Oxidation Numbers Assigning Oxidation Numbers – the rulesOxygen has an oxidation number of -2 in almost all compounds. Exceptions peroxides is -1, compounds with halogens +2Hydrogen 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.
31Section 7.2 - Oxidation Numbers Assigning Oxidation Numbers – the rulesThe algebraic sum of the oxidation numbers of all atoms in a neutral compound = zero.The algebraic sum of the oxidation numbers of all atoms in a polyatomic ion = the charge of the ion.Rules 1-7 apply to covalently bonded atoms, oxidation numbers can also be assigned to atoms in ionic compounds.
32Section 7.2 - Oxidation Numbers Using Oxidation Numbers for Formulas and NamesMany non-metals have more than one oxidation state. Recall the use of Roman numerals to denote charges.FormulaPrefix systemStock SystemPCl3phosphorus trichloridephosphorus(III) chlorideNOnitrogenmonoxidenitrogen(II)oxidePbO2leaddioxidelead (IV) oxide
33Practices Section 7.2 Ba(NO3)2 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 = +2Which elements have specific rules?Oxygen has a rule in most compoundsOxidation # for O = -2Which element does not have a specific rule?N does not have a specific rule.Use rule 8 to find the oxidation # of NLet N = Oxidation # for nitrogen(# Ba) (Oxid. # of Ba) + (# N) (Oxid. # N) + (# of O) (Oxid. # of O) = 01(+2) + 2(N) + 6(-2) = 0N = +5Ba(NO3)2
34Practices Section 7.2 NF3 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 = -1Which element does not have a specific rule?N does not have a specific rule.Use rule 8 to find the oxidation # of NLet N = oxidation # of N(# N) (Oxid. # N) + (# F) (Oxid. # F) = 01(N) + 3(-1) = 0N = +3
35Practices Section 7.2 (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 = -2Which 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) = 0The two ions present are NH4+ and SO42-.N + 4(+1) = +1 so N = -3S + 4(-2) = -2 so S = +6
38Section 7.3 - Using Chemical Formulas ObjectivesHW – Notes on section 7.3 pgsStudents 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.
39Section 7.3 - Using Chemical Formulas Formula MassesThe 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.Book exampleAverage atomic mass of H : 1.01 amuAverage atomic mass of O : amuH2O (2 x 1.01 amu) + 16 amu = amu
40Section 7.3 - Using Chemical Formulas Molar MassesA compound’s molar mass is numerically equal to its formula mass.Book exampleFormula mass of H2O = amuWhich is also the molar mass of water g/mol.
41Section 7.3 – Using Chemical Formulas Molar Mass as a Conversion Factor – Remember our old friends. . .There are 3 mole equalities. They are:1 mol = 6.02 x 1023 particles1 mol = g-formula-mass (periodic table)1 mol = 22.4 L for a gas at STP*These become. . .[ ] or [ ]1 mol6.02 x 1023 particles[ ] or [ ]1 molg-formula-mass (periodic table)[----] OR [----]1 mol22.4 L
42Section 7.3 - Using Chemical Formulas Percentage CompositionThe percentage by mass of each element in a compound is known as the percentage composition of the compound.Mass of X in sample of compoundX 100 %=Mass % X in compoundMass of sample of compound
43Section 7.3 - Using Chemical Formulas Percentage Composition - examplePercent Composition Example:Calculate the percent composition of Mg(NO3)21 Mg = 1 x 24 = 242 N = 2 x 14 = 286 O = 6 x 16 = 96148g/moleDouble check -do they total 100%?% Mg = 24/148 x 100 = 16.2%% N = 28/148 x 100 = 18.9%% O = 96/148 x 100 = 64.0%
44Section 7.3 - Using Chemical Formulas Percentage Composition - practiceQuiz BreakKey
45Section 7.4 - Determining Chemical Formulas ObjectivesHW – Notes on section 7.4 pgsStudents 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
46Section 7.4 - Determining Chemical Formulas Calculation of Empirical FormulasAn 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.
47Section 7.4 - Determining Chemical Formulas Calculation of Empirical Formulas - exampleLet’s Determine the empirical formula for a compound which is 54.09% Ca, 43.18% O, and 2.73% H1) Divide each percent by that element's atomic weight.1.352/1.352 = 12.699/1.352 = 22.73/1.352 = 2Ca = 54.09/40 = 1.352O = 43.18/16 = 2.699H = 2.73/1 = 2.732) To get the answers to whole numbers, divide through by the smallest one.This gives us CaO2H2better yet Ca(OH)2
48Section 7.4 - Determining Chemical Formulas Calculations of Molecular FormulasAn empirical formula may or may not be a correct molecular formula.Book example , diborane1’s empirical formula is BH3, any multiple of that equals the same ratio – B2H6,B3H9,B4H12 etcIt 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.
49Section 7.4 - Determining Chemical Formulas Calculations of Molecular Formulas – (continued)The relationship between an empirical formula and a molecular formula is seen below:X(empirical formula) = molecular formulaX is a whole-number multiple indicating the factor that you need to multiply the empirical formula by to get the molecular formula.
50Section 7.4 - Determining Chemical Formulas Calculations of Molecular Formulas – (continued)X = molecular formulaempirical formulaFormula mass of diborane = amuEmpirical mass of diborane amuThe molecular formula of diborane is therefore B2H6(BH3)2 = B2H6
51Section 7.4 - Determining Chemical Formulas PRACTICEEmpirical Formula PracticeMolecular Formula Practice