1. 11/20 Today you will need… The paper from the side table Take out the paper title “types of chemical bonds”. You have 10 minutes to finish this. Please do this ON YOUR OWN so you wont be confused when we start naming/writing formulas for these two types of compounds Turn in when finished

2. Chapter 7-Chemical Formulas & Chemical Compounds 7.1-Chemical Names & Formulas 7.2-Oxidation Numbers 7.3-Using Chemical Formulas 7.4-Determining Chemical Formulas

3. 7.1-Chemical Names & Formulas Pages 203-215

4. Significance of a Chemical Formula Molecular formula-tells the number of each type of atom that make up one molecule of the compound. C 8 H 18 There are 8 carbon atoms in one molecule of octane There are 18 hydrogen atoms in one molecule of octane

5. Significance of a Chemical Formula Ionic formula-represents one formula unit – The simplest whole number ratio of cations to anions in the compound CaBr 2 No subscript=implied “1” For every calcium ion in the compound, there are 2 bromide ions

6. Molecule vs. Ionic Compound

7. Monatomic Ions Ions formed from a single atom Gain or lose electrons to attain a noble gas configuration

8. 1+ 2+3+NA3-2-1- 0 Monatomic Ions: Common Charges Chart with charges of ions on page 205

9. Naming Monatomic Ions Cations-same as element’s name – Example- Li + is a lithium cation Anions-drop the ending of the element’s name and at –ide – Example- F- is a fluoride anion

10. Binary Ionic Compounds Compounds composed of 2 different elements that are ionically bonded Charges of ions must “cancel out” so the overall charge on the compound is 0. Use the “crisscross method” to write the formula.

11. Binary Ionic Compunds: Crisscross method Find the formula for a compound formed by the ions of aluminum and oxygen. 1. Write the symbols for the ions side by side, CATION FIRST. 1. Al 3+ O 2- 2. Cross over the charges by using the number only from each ions’ charge. Make this the subscript for the other ion. 1. Al 3+ O 2- 2. Al 2 O 3 3. Check the subscripts to make sure that they cannot be simplified. Al 2 O 3

12. Binary Ionic Compounds: Crisscross Method Write the formulas for the binary ionic compounds formed between the following elements: – Sodium and phosphorus – Gallium and nitrogen Na 3 P GaN

13. Naming Binary Ionics Combine the names of the cation and anion, cation first. – Example: Al 2 O 3 Cation: aluminum Anion: oxide Compound: Aluminum oxide Most elements form only one ion, so you usually don’t have to indicate the charge in the name.

14. Stock System Some elements can form more than one ion. – Example: Iron can form a +2 and +3 ion When you name these compounds, you have to indicate which ion is present. – If I asked you to write the formula for iron oxide without telling you if it was Fe +2 or Fe +3, then you could write 2 different formulas FeOFe 2 O 3 Use Roman numerals in parentheses in the name to indicate the charge on the ion. – FeO is Iron (II) oxide Fe 2 O 3 is Iron (III) oxide

15. Stock System ONLY use the Stock system for elements that can form more than one ion Almost always found in the d-block Always the cation. Never use Stock system on anions. See Table 7-1: Some Common Monatomic Ions on page 205

16. Naming Practice Write the formula and give the name for the compound formed by the ions: Cr 3+ and F - – CrF 3, Chromium (III) fluoride Na + and S 2- – Na 2 S, Sodium sulfide Sn 2+ and S 2- – SnS, Tin (II) sulfide Pb 2+ and Cl - – PbCl 2, Lead (II) chloride

17. Naming Practice Name the following compounds CuO – Copper (II) oxide LiBr – Lithium bromide CoF 3 – Cobalt (III) fluoride

18. Compounds Containing Polyatomics Chart of polyatomics on page 210. Compounds containing polyatomics are named the same way at binary ionics. 1. Name the cation 2. Name the anion Example: Name the compound AgNO 3 Cation: silver Anion: nitrate Compound: silver nitrate

19. Polyatomic Examples Write the formula for tin (IV) sulfate – Cation: Sn +4 – Anion: SO 4 2 - – Formula using crisscross method: Sn 2 (SO 4 ) 4 If you need more than one polyatomic ion, put it in parentheses to avoid confusion! – Wrong:Sn 2 SO 44 Right:Sn 2 (SO 4 ) 4

20. Polyatomic Examples Write formulas for the following ionic compounds: Sodium iodide – NaI Calcium nitrite – Ca(NO 3 ) 2 Copper (II) sulfate – CuSO 4

21. Polyatomic Examples Give the names for the following compounds Ag 2 O – Silver oxide KClO 3 – Potassium chlorate FeCrO 4 – Iron (II) chromate

22. Naming Binary Molecular Compounds Named using a prefix system to indicate how many atoms of each element are present. Table 7-3: Numerical Prefixes (pg. 212) 1Mono- 2Di- 3Tri- 4Tetra- 5Penta- 6Hexa- 7Hepta 8Octa- 9Nona- 10Deca-

23. Rules for Naming Binary Molecular Compounds 1. The less-electronegative element is given first. It is given a prefix only if it contributes more than one atom to a molecule. (Names will never start with mono-.) Order of nonmetals: C, P, N, H, S, I, Br, Cl, O, F 2. The second element is named by combining a prefix indicating the number of atoms, the root of the name of the element, and the ending –ide. You may drop an o or a at the end of the prefix if the name of the element starts with a vowel. Example: pentoxide, not pentaoxide.

24. Naming Binary Molecular Compounds Example Name the following compound: P 4 O 10. P 4 O 10 Prefix needed if less electronegative element contributes more than one atom Name of less-electronegative element + Prefix indicating number of atoms contributed by more- electronegative element Root name of more-electronegative element + -ide. + tetraphosphorus decoxide

25. Naming Binary Molecular Compounds Practice Name the following compounds: As 2 O 5 – Diarsenic pentoxide XeF 4 – Xenon tetrafluoride CCl 4 – Carbon tetrachloride

26. Naming Binary Molecular Compounds Practice Write formulas for the following compounds: Carbon diselenide – CSe 2 Sulfur pentachloride – SCl 5 Dihydrogen monoxide –H2O–H2O

27. 7.2-Oxidation Numbers Pages 216-219

28. What are oxidation numbers? In ionic compounds, charges on ions reflect the electron distribution – Negative ions have a greater electron density in the compound. Oxidation numbers or oxidation states reflect to distribution of electrons among the bonded atoms in a molecular compound – Do not have an actual physical meaning – Useful in naming compounds, writing formulas, and balancing chemical equations.

29. How do you assign oxidation numbers?? Rules are on page 591. You will need to know them! (rules on page 216 are not complete!) 1. Atoms in a pure compound have an oxidation number of zero. Example: atoms in Na, O 2, P 4, and S 8 all have oxidation numbers of 0.

30. 2. More electronegative-element in a binary molecular compound is assigned the oxidation number equal to the negative charge it would have as an anion. Example: The oxidation number of O in NO is -2. Oxidation Number Rules

31. 3. Fluorine always has an oxidation number of -1 in all compounds because it is the most electronegative element. Example: The oxidation number of F in LiF is -1.

32. 4. Oxygen has an oxidation number of -2 unless it is combined with F, when it is +2, or it is in a peroxide, such as H 2 O 2, when it is -1. Examples: In H 2 SO 4, O -2 In H 2 O 2 (a peroxide due to O 2 2- ), O -1 In OF 2, O +2 Oxidation Number Rules

33. 5. The oxidation state of hydrogen in most compounds is +1 unless it is combined with a metal, in which case it is -1. Examples: In KH, H’s oxidation number is -1 because K is a metal. In HF, H’s oxidation number is +1. Oxidation Number Rules

34. 6. In compounds, Group 1 and 2 elements and aluminum have oxidation numbers of +1, +2, and +3, respectively. – Example: The oxidation number of Ca in CaCO 3 is +2. Oxidation Number Rules

35. 7. The algebraic sum of the oxidation numbers of all atoms in a neutral compound is equal to zero. Example: In H 2 SO 4 H 2(+1)=2 O 4(-2)=-8 Subtotal: 2-8=-6 Therefore, S must be +6 to give a sum of zero. Oxidation Number Rules

36. 8. The sum of the oxidation numbers of all atoms in an polyatomic ion is equal to the charge of the ion. Example: In SO 4 2- : O=4(-2)=-8 Therefore, S must be +6 to give a sum of -2 (equal to the charge of the ion) Oxidation Number Rules

37. 9. Oxidation numbers can also be assigned to atoms in ionic compounds. A monatomic ion has an oxidation number equal to the charge on the ion. Na + has an oxidation number of +1 N -3 has an oxidation number of -3 Oxidation Number Rules

38. Examples Cl 2 PI 3 NO 2 - As 2 S 3 ClO 2 - 0 -1, +3 +3, -2

39. Using oxidation numbers: Stock System Oxidation numbers can be used in naming molecular compounds as an alternative to the prefix system. Example: lead dioxide – PbO 2 – +4, -2 – Lead(IV) oxide

40. Practice Phosphorus trichloride – PCl 3 +3, -1 – Phosphorus(III) chloride Dimolybdenum trioxide – Mo 2 O 3 +3, -2 – Molybdenum(III) oxide

41. 7.3-Using Chemical Formulas Pages 221-228

42. Formula Masses The average atomic mass of an atom is given on the periodic table. Like atoms, compounds have average masses. Formula mass-the sum of the average atomic masses of all the atoms represented in it formula. – Round masses from the periodic table to the hundredths position.

43. Formula Mass Example Find the formula mass of potassium chlorate, KClO 3.

44. Molar Mass Molar mass-the mass of one mole of a compound A compound’s molar mass is numerically equal to is formula mass – Just as one atom of oxygen has a mass of 16.00 amu and one mole of oxygen (6.022 X 10 23 atoms) has a mass of 16 grams.

45. Molar Mass Example Find the molar mass of barium nitrate, Ba(NO 3 ) 2. Don’t forget to distribute the “2” to each atom in the nitrate ion!

46. Molar mass practice Find the molar mass of each of the following compounds: – Al 2 S 3 150.17 g/mol – NaNO 3 85.00 g/mol – Ba(OH) 2 171.35 g/mol

47. Molar Mass as a Conversion Factor Molar mass can be used to relate the amount of a compound in moles to a mass in grams.

48. Molar Mass Conversion Example 1 What is the mass in grams of 2.50 mol of oxygen gas, O 2 ? – Given: 2.50 mol O 2 – Unknown: ? g O 2

49. Molar Mass Conversion Example 2 How many moles of sulfur dioxide are in 3.82 g of the substance? – Given: 3.82 g SO 2 – Unknown: ? mol SO 2

50. Percentage Composition Percentage by mass of each element in a compound Percentages should always total very close to 100! Always check!

51. Percentage Composition Example 1 Find the percentage composition of sodium nitrate. Formula of sodium nitrate: NaNO 3 Molar mass of NaNO 3 =85.00 g/mol Total100%

52. Percentage Composition Example 2 Find the percentage composition of silver sulfate. Formula of silver sulfate: Ag 2 SO 4 Molar mass of Ag 2 SO 4 =311.81 g/mol Total100.01%

53. Percentage Composition Example 3 How many grams of copper are in a 38.0 g sample of copper(I) sulfide? Formula of copper(I) sulfide: Cu 2 S Molar mass of Cu 2 S=159.17 g/mol

54. February 2 nd Pick up an “Empirical Formula” worksheet from side lab table. Turn in word search from yesterday if you haven’t done so already and then have a seat.

55. 7.4-Determining Chemical Formulas Pages 229-233

56. Why % composition? When a new compound is made or discovered, it is analyzed to determine its % composition. From this info, its empirical formula can be determined.

57. Empirical Formula Simplified formula for a compound – For an ionic compound, empirical formula is the compound’s formula unit because we always simplify the formula  Calcium sulfide: Ca 2 S 2  CaS – For a molecular compound, may not represent actual number of atoms present in a molecule  Molecular formula: B 2 H 6  Empirical formula: BH 3

58. Empirical Formula C2H6C2H6 CH 3 reduce subscripts Smallest whole number ratio of atoms in a compound

59. How do you calculate empirical formula? 1. Find mass (or %) of each element. 2. Convert grams to moles of each element. 3. Divide moles by the smallest # to find subscripts. 4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.

60. Empirical Formula Example 1 A compound consist of 75% carbon and 25% hydrogen by mass. Determine the empirical formula for this compound 75 g 1 mol 12.01 g = 6.24 mol N 25 g 1 mol 1.01 g = 24.75 mol O 6.24 mol = 1 C = 4 H

61. Empirical Formula Example 2 Find the empirical formula for a sample of 25.9% N and 74.1% O. 25.9 g 1 mol 14.01 g = 1.85 mol N 74.1 g 1 mol 16.00 g = 4.63 mol O 1.85 mol = 1 N = 2.5 O

62. Empirical Formula Example N 1 O 2.5 Need to make the subscripts whole numbers  multiply by 2 N2O5N2O5

63. Molecular Formula “True Formula” - the actual number of atoms in a compound CH 3 C2H6C2H6 empirical formula molecular formula ?

64. How do you calculate molecular formula? 1. Find the empirical formula. 2. Find the empirical formula mass. 3. Divide the molecular mass by the empirical mass. 4. Multiply each subscript by the answer from step 3.

65. Molecular Formula Example The empirical formula for ethylene is CH 2. Find the molecular formula if the molecular mass is 28.1 g/mol. 28.1 g/mol 14.03 g/mol = 2.00 empirical mass (CH 2 ) = 14.03 g/mol (CH 2 ) 2  C 2 H 4

66. General pathway to solve problems Mass of each element Percent composition of compound Empirical formula for compound Molecular formula for compound

67. Molecular Formula Example A sample of a compound with a molar mass of 34.00 g/mol is found to consist of 0.44 g H and 6.92 g O. Find its molecular formula. – 5.98 % H, 94.02 % O – Empirical formula: HO – Molecular formula: (HO) 2 =H 2 O 2