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

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Presentation on theme: "Chapter 7 – Chemical Formulas and Chemical Compounds Taken from Modern Chemistry written by Davis, Metcalfe, Williams & Castka."— Presentation transcript:

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

2 Section 7.1 – Chemical Names and Formulas Objectives 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. HW – Notes on section 7.1 pgs

3 Section 7.1 – Chemical Names and Formulas Significance of a chemical formula The chemical formula indicates the relative number of atoms of each element in a chemical compound. C 12 H 22 O 11 Elements’ subscripts indicate the number of atoms in the compound. Al 2 (SO 4 ) 3 Note how in this example parenthesis surround a polyatomic anion and the subscript refers to the entire unit

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

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

6 Section 7.1 – Chemical Names and Formulas Monatomic Ions (continued) 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.

7 Section 7.1 – Chemical Names and Formulas Naming Monatomic Ions (continued) 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. Potassium cation Magnesium cation Fluorine  fluorideNitrogen  Nitride

8 Section 7.1 – Chemical Names and Formulas Binary Ionic Compounds Compounds composed of two different elements are known as binary compounds Cation goes first : Mg 2+, Br -, Br - Balance to become electrically neutral And you get

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

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

11 Section 7.1 – Chemical Names and Formulas Naming Binary Ionic Compounds Oxyanions each is a polyatomic ion that contains oxygen. Nitrite Nitrate Compounds Containing Polyatomic Ions

12 Section 7.1 – Chemical Names and Formulas Naming Binary Molecular Compounds 1mono- 2di- 3tri- 4tetra- 5penta- 6hexa- 7hepta- 8octa- 9nona- 10deca- 1.Less electronegative element given first, prefix only for multiples 2.Second element named with prefix indicating # of atoms, with few exceptions ends with –ide (only 2 elements) 3.The o or a at the end of the prefix is dropped when the word following begins with another vowel.

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

14 Section 7.1 – Chemical Names and Formulas Covalent-Network Compounds Similar to naming molecular compounds

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

16 Section 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.

17 Section 7.1 – Chemical Names and Formulas Practices PRACTICE – pg 207 Q 1 & 2 all PRACTICE – pg 209 Q 1 & 2 all PRACTICE – pg 211 Q 1 & 2 all PRACTICE – pg 213 Q 1 & 2 all Practice Naming

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

19 Section 7.1 – Chemical Names and Formulas - POGIL KEY

20 Section 7.1 – Chemical Names and Formulas - POGIL KEY

21 Section 7.1 – Chemical Names and Formulas - POGIL KEY

22 Section 7.1 – Chemical Names and Formulas - POGIL KEY

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

24 Section 7.1 – Chemical Names and Formulas - POGIL KEY

25 Section Oxidation Numbers Objectives 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. HW – Notes on section 7.2 pgs

26 Section Oxidation Numbers 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.

27 Section Oxidation Numbers Assigning Oxidation Numbers – the rules The following are guidelines... 1.Atoms of pure elements have an oxidation number of zero. All zero.

28 Section Oxidation Numbers Assigning Oxidation Numbers – the rules 2.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 Oxidation Numbers Assigning Oxidation Numbers – the rules 3.Fluorine has an oxidation number of -1 as it is the most electronegative element.

30 Section Oxidation Numbers Assigning Oxidation Numbers – the rules 4.Oxygen has an oxidation number of -2 in almost all compounds. Exceptions peroxides is -1, compounds with halogens +2 5.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.

31 Section Oxidation Numbers Assigning Oxidation Numbers – the rules 6.The algebraic sum of the oxidation numbers of all atoms in a neutral compound = zero. 7.The algebraic sum of the oxidation numbers of all atoms in a polyatomic ion = the charge of the ion. 8.Rules 1-7 apply to covalently bonded atoms, oxidation numbers can also be assigned to atoms in ionic compounds.

32 Section Oxidation Numbers Using Oxidation Numbers for Formulas and Names Many non-metals have more than one oxidation state. Recall the use of Roman numerals to denote charges. FormulaPrefix systemStock System PCl 3 phosphorus trichloride phosphorus(III) chloride NO nitrogen monoxide nitrogen(II) oxide PbO 2 lead dioxide lead (IV) oxide

33 Section 7.2 Practices Ba(NO 3 ) 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 = +2 Which elements have specific rules? Oxygen has a rule 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

34 Section 7.2 Practices NF 3 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

35 Section 7.2 Practices (NH 4 ) 2 SO 4 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 NH 4 + and SO N + 4(+1) = +1 so N = -3 S + 4(-2) = -2 so S = +6

36 Section 7.2 Practices PRACTICE – pg 218 Q 1 all Practice Sheet Key

37 Section 7.2 Quiz Key

38 Section Using Chemical Formulas Objectives 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. HW – Notes on section 7.3 pgs

39 Section Using Chemical Formulas Formula Masses formula mass 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. Book example Average atomic mass of H : 1.01 amu Average atomic mass of O : amu H 2 O  (2 x 1.01 amu) + 16 amu = amu

40 Section Using Chemical Formulas Molar Masses A compound’s molar mass is numerically equal to its formula mass. Book example Formula mass of H 2 O = amu Which is also the molar mass of water g/mol.

41 Section 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 particles 1 mol = g-formula-mass (periodic table) 1 mol = 22.4 L for a gas at STP* These become... [ ] or [ ] 1 mol 6.02 x particles 1 mol 6.02 x particles [ ] or [ ] 1 mol g-formula-mass (periodic table) 1 mol g-formula-mass (periodic table) [----] OR [----] 1 mol 22.4 L1 mol 22.4 L

42 Section Using Chemical Formulas Percentage Composition The percentage by mass of each element in a compound is known as the percentage composition of the compound. Mass of X in sample of compound X 100 %=Mass % X in compound Mass of sample of compound

43 Section Using Chemical Formulas Percentage Composition - example 1 Mg = 1 x 24 = 24 2 N = 2 x 14 = 28 6 O = 6 x 16 = g/mole % Mg = 24/148 x 100 = 16.2% % N = 28/148 x 100 = 18.9% % O = 96/148 x 100 = 64.0% Percent Composition Example: Calculate the percent composition of Mg(NO 3 ) 2 Double check - do they total 100%?

44 Section Using Chemical Formulas Quiz Break Key Percentage Composition - practicepractice

45 Section Determining Chemical Formulas Objectives 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 HW – Notes on section 7.4 pgs

46 Section Determining Chemical Formulas Calculation of Empirical Formulas 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.

47 Section Determining Chemical Formulas Calculation of Empirical Formulas - example Let’s Determine the empirical formula for a compound which is 54.09% Ca, 43.18% O, and 2.73% H 1) Divide each percent by that element's atomic weight. Ca = 54.09/40 = O = 43.18/16 = H = 2.73/1 = ) To get the answers to whole numbers, divide through by the smallest one /1.352 = /1.352 = /1.352 = 2 This gives us  CaO 2 H 2 better yet  Ca(OH) 2

48 Section Determining Chemical Formulas Calculations of Molecular Formulas An empirical formula may or may not be a correct molecular formula. Book example, diborane 1 ’s empirical formula is BH 3, any multiple of that equals the same ratio – B 2 H 6, B 3 H 9, B 4 H 12 etc 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 Section 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 formula X is a whole-number multiple indicating the factor that you need to multiply the empirical formula by to get the molecular formula.

50 Section Determining Chemical Formulas Calculations of Molecular Formulas – (continued) X = molecular formula empirical formula Formula mass of diborane = amu Empirical mass of diborane amu The molecular formula of diborane is therefore B 2 H 6 (BH 3 ) 2 = B 2 H 6

51 Section Determining Chemical Formulas PRACTICE Empirical Formula Practice Molecular Formula Practice


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