1. CHEMISTRY CHAPTER 7 SECTIONS 3-4 SECTION 3. USING CHEMICAL FORMULAS Introduction A chemical formula indicates: the elements present in a compound the relative number of atoms or ions of each element present in a compound

2. Chemical formulas also allow chemists to calculate a number of other characteristic values for a compound: formula mass molar mass percentage composition

3. Formula Masses The formula mass of any molecule, formula unit, or ion is the sum of the average atomic masses of all atoms represented in its formula. The molar mass is the same number, but expressed in grams instead of AMU. It is the mass of one mole of the compound.

4. Example 1 (molecular compound): formula mass of water, H 2 O average atomic mass of H: 1.01 amu ave. atomic mass of O: 16.00 amu average mass of H 2 O molecule: 18.02 amu

5. Average mass of H 2 O molecule = 18.02 amu Molar mass of H 2 O = 18.02 g/mol Example 2 (ionic compound) calcium nitrate = Ca(NO 3 ) 2 A formula unit has 1 Ca 2+ ion and 2 NO 3 - ions

6. Formula mass: 1 Ca atom x 40.08 amu/atom = 40.08 amu 2 N atoms x 14.01 amu/atom = 28.02 amu 6 O atoms x 16.00 amu/atom = 96.00 amu total 164.10 amu What is the molar mass of Ca(NO 3 ) 2 ?

7. Formula mass: 1 Ca atom x 40.08 amu/atom = 40.08 amu 2 N atoms x 14.01 amu/atom = 28.02 amu 6 O atoms x 16.00 amu/atom = 96.00 amu total 164.10 amu What is the molar mass of Ca(NO 3 ) 2 ? 164.10 g/mol

9. 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. This uses the method of dimensional analysis (factor-label method) just as we did for elements.

10. Mole-Mass Calculations

11. Example 1: grams to moles How many moles of CO 2 in 325 g? First you need to calculate the molar mass of CO 2 : C: 1 x 12.0 = 12.0 O: 2 x 16.0 = 32.0 44.0 g/mol

12. Then convert g to mol:

13. Example 2: moles to grams What is the mass of 0.15 mol MgSO 4 ? First you need to calculate the molar mass of MgSO 4 : Mg 1 x 24.3 = 24.3 S 1 x 32.1 = 32.1 O 4 x 16.0 = 64.0 120.4 g/mol

14. Then convert mol to g:

15. Molar Mass as a Conversion Factor - 75159

16. Recall that Avogadro’s number ( = ) is the number of particles (atoms, molecules, or formula units) in 1 mole of a substance. So we can convert between grams or moles of a compound and numbers of molecules or formula units.

17. Recall that Avogadro’s number ( = 6.02 x 10 23 ) is the number of particles (atoms, molecules, or formula units) in 1 mole of a substance. So we can convert between grams or moles of a compound and numbers of molecules or formula units.

18. Converting Between Amount in Moles and Number of Particles

19. Example 1: How many formula units and atoms are there in 25.0 g MgCl 2 ? Calculate molar mass: Mg 1 x 24.3 = 24.3 Cl 2 x 35.3 = 71.0 95.3 g/mol

20. Then convert g to formula units: = 1.58 x 10 23 atoms of Mg and 3.16 x 10 23 atoms of Cl

21. Example 2: What is the mass of 8.0 x 10 30 molecules of CO 2 ? Calculate molar mass: C 1 x 12.0 = 12.0 O 2 x 16.0 = 32.0 44.0 g/mol Then convert molecules to g:

22. Percentage Composition ( = Percent by Mass) = the percent of the mass of a compound contributed by each element. Use the following equation:

23. The mass percentage of an element in a compound is the same regardless of the sample’s size. If you know the molar mass (or calculate it from the formula), it is convenient to use one mole as the sample size. The sum of the percents for all elements in the compound must add up to 100%.

24. Percentage Composition of Iron Oxides

25. Example 1: Calculate the percentage composition of a sample that is 1.67 g Ce and 4.54 g I. Total mass = 1.67 g + 4.54 g = 6.21 g

26. Example 2: Calculate the percent by mass of each element in Pb(NO 3 ) 4. First calculate molar mass: Pb 1 x 207.2 = 207.2 N O

27. Example 2: Calculate the percent by mass of each element in Pb(NO 3 ) 4. First calculate molar mass: Pb 1 x 207.2 = 207.2 N 4 x 14.0 = 56.0 O 12 x 16.0 = 192.0 455.2 g/mol

28. We can then consider the sample size to be 1 mol, and use the fraction that each element contributed to the total: Check: 45.5%+12.3%+42.2% = 100%

29. 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. Determine the formula of a hydrated crystal.

30. An empirical formula gives the relative numbers of atoms in a compound, using the smallest whole number ratios. For an ionic compound, it is usually the same as the formula unit.

31. For a molecular compound, the molecular formula gives the actual number of atoms in a molecule. It may be the same as the empirical formula, or it may be a whole number multiple of it.

32. Examples: molecular formula empirical formula CO 2 Pb 2 O 4 Hg 2 I 2 glucose C 6 H 12 O 6

33. Examples: molecular formula empirical formula CO 2 CO 2 Pb 2 O 4 PbO 2 Hg 2 I 2 HgI glucose C 6 H 12 O 6 CH 2 O

34. Calculation of Empirical Formulas 1. Convert grams to moles for each element (if using percentage composition, assume a total mass of 100 g). 2. Divide by the smallest number of moles to get subscripts. Change to whole numbers if necessary. 3. Write the formula using the subscripts.

35. Example 1: What is the empirical formula of a compound containing 28.3 g Ca and 14.6 g P?

36. Divide by smaller number (0.471 mol): Change to whole number ratio: 1.5:1 = 3:2. Formula: Ca 3 P 2 (Name =)

37. Divide by smaller number (0.471 mol): Change to whole number ratio: 1.5:1 = 3:2. Formula: Ca 3 P 2 (Name = calcium phosphide )

38. Example 2: A compound has a percent composition of 40.0%C, 6.71% H, and 53.3% O. What is its empirical formula? For a total mass of 100 g, there are 40.0 g C, 6.71 g H, and 53.3 g O.

39. Divide by smallest number (3.33 mol): Change to whole number ratio: 1:2:1. Empirical formula: CH 2 O

40. 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.

41. The relationship between a compound’s empirical formula and its molecular formula can be written as follows: n(empirical formula) = molecular formula The formula masses have a similar relationship: n(empirical formula mass) = molecular formula mass

42. Comparing Empirical and Molecular Formulas

43. To determine the molecular formula of a compound, you must know the compound’s molecular formula mass (molar mass) [usually given in a problem]. First determine the empirical formula and the empirical formula mass.

44. Then determine n from the ratio: Finally, multiply the empirical formula by n to get the molecular formula.

45. Example 1: A sample contains 0.076 mol C and 0.0765 mol H, and has a molecular mass of 78 AMU. What are the empirical and molecular formulas? First determine the empirical formula: Empirical formula =

46. Example 1: A sample contains 0.076 mol C and 0.0765 mol H, and has a molecular mass of 78 AMU. What are the empirical and molecular formulas? First determine the empirical formula: Empirical formula = CH

47. Next determine the empirical formula mass (work in AMU since molecular mass was given in AMU): C: 1 x 12.0 = 12.0 H: 1 x 1.0 = 1.0 13.0 AMU Molecular formula = n(empirical formula) =

48. Next determine the empirical formula mass (work in AMU since molecular mass was given in AMU): C: 1 x 12.0 = 12.0 H: 1 x 1.0 = 1.0 13.0 AMU Molecular formula = n(empirical formula) = 6(CH) = C 6 H 6

49. Example 2: (p. 249, #4): 4.04 g of N combine with 11.46 g of O to produce a compound with a formula mass of 108.0 amu. What is the molecular formula?

50. Empirical formula: Whole number ratio: 5:2. Empirical formula =

51. Empirical formula: Whole number ratio: 5:2. Empirical formula = N 2 O 5

52. Next determine the empirical formula mass: N: 2 x 14.0 = 28.0 O: 5 x 16.0 = 80.0 108.0 AMU Molecular formula = n(empirical formula) = N 2 O 5

53. Hydrated Crystals Some compounds crystallizing from water solution incorporate water molecules into the crystal. These hydrates contain a specific ratio of water to compound: (formula for compound)xH 2 O Example: CuSO 4 2H 2 O

54. Formulas for hydrates are determined by determining the mass of the hydrated compound, heating to remove all water, and then determining the mass of the anhydrous (= without water) compound. From this the moles of water per mole of anhydrous compound can be calculated.

55. Example: a 10.407 g sample of hydrated barium iodide (BaI 2 ) is heated to drive off water. The dry sample has a mass of 9.520 g. What is the formula for the hydrate? First determine the mass of water: hydrated (= BaI 2 + H 2 O) 10.407 g anhydrous (= BaI 2 ) -9.520 g water alone 0.887 g

56. Then determine moles of each compound and their ratio: BaI 2 : Ba 1 x 137.3 = 137.3 I 2 x 126.9 = 253.8 391.1 g/mol

57. H 2 O: H 2 x 1.0 = 2.0 O 1 x 16.0 = 16.0 18.0 g/mol Then calculate the ratio: Formula =

58. H 2 O: H 2 x 1.0 = 2.0 O 1 x 16.0 = 16.0 18.0 g/mol Then calculate the ratio: Formula = BaI 2 2H 2 O