1. Chapter #6 Chemical Composition

2. Chapter Contents 6-1 Chemical Composition 6-2 Counting Units 6-3 Counting Atoms By the Gram 6-4 Counting Molecules by the Gram 6-5 Mole Conversions 6-6 Mass Percent Composition 6-8 Empirical Formula Calculation 6-9 Molecular Formula Calculation

3. 6-1 The Chemical Package The baker uses a package called the dozen. All dozen packages contain 12 objects. The stationary store uses a package called a ream, which contains 500 sheets of paper. So what is the chemistry package? About Packages

4. 6-2 The Chemical Package The baker uses a package called the dozen. All dozen packages contain 12 objects. The stationary store uses a package called a ream, which contains 500 sheets of paper. So what is the chemistry package? Well, it is called the mole (Latin for heap). About Packages Each of the above packages contain a number of objects that are convenient to work with, for that particular discipline.

5. The atomic weights listed on the periodic chart are the weights of a mole of atoms. For example a mole of hydrogen atoms weighs 1.00797 g and a mole of carbon atoms weighs 12.01 g. 6-3 The Mole A mole contains 6.022X10 23 particles, which is the number of carbon-12 atoms that will give a mass of 12.00 grams, which is a convenient number of atoms to work with in the chemistry laboratory.

6. 6-4 Moles of Objects Suppose we order a mole of marshmallows for a chemistry party. How much space here at Central would be required to store the marshmallows?

7. 6-4 Moles of Objects Suppose we order a mole of marshmallows for a chemistry party. How much space here at Central would be required to store the marshmallows? Would cover the entire 50 states 60 miles deep

8. 6-4 Moles of Objects Suppose we order a mole of marshmallows for a chemistry party. How much space here at Central would be required to store the marshmallows? Would cover the entire 50 states 60 miles deep How about a mole of computer paper instead of a ream of computer paper, how far would that stretch?

9. 6-4 Moles of Objects Suppose we order a mole of marshmallows for a chemistry party. How much space here at Central would be required to store the marshmallows? Would cover the entire 50 states 60 miles deep How about a mole of computer paper instead of a ream of computer paper, how far would that stretch? Way past the planet Pluto!

10. To calculate the molar mass of a compound we sum together the atomic weights of the atoms that make up the formula of the compound. This is called the formula weight (MW, M). Formula weights are the sum of atomic weights of atoms making up the formula. The following outlines how to find the formula weight of water symbol weight number H O 1.01 16.0 2 1 X X = =2.02 16.0 18.0 g/mole 6-5 Formula Weight Calculation

11. 6-7 Percent Composition Find the formula weight and the percent composition of glucose (C 6 H 12 O 6 ) symbol weightnumber H O C 16.0 1.01 12.0 6 12 6 x x x = = = 72.0 12.12 96.0 180.1 g/mole %C = %H = %O = 72.0 12.12 96.0 180.1 X 100 = 40.0 %C 6.73 %H 53.3 %O

12. 6-7 Percent Conversions Seawater contains approximately 3.5% NaCl by mass and has a density of 1.02 g/mL. What volume of seawater would be required to produce 1.0 g of NaCl when evaporated?

13. 6-7 Percent Conversions Seawater contains approximately 3.5% NaCl by mass and has a density of 1.02 g/mL. What volume of seawater would be required to produce 1.0 g of NaCl when evaporated? 1.02 g seawater mL seawater

14. 6-7 Percent Conversions Seawater contains approximately 3.5% NaCl by mass and has a density of 1.02 g/mL. What volume of seawater would be required to produce 1.0 g of NaCl when evaporated? 1.02 g seawater mL seawater 3.5 g NaCl 100 g seawater

15. 6-7 Percent Conversions Seawater contains approximately 3.5% NaCl by mass and has a density of 1.02 g/mL. What volume of seawater would be required to produce 1.0 g of NaCl when evaporated? 1.02 g seawater mL seawater 3.5 g NaCl 100 g seawater 1.00 g NaCl

16. 6-7 Percent Conversions Seawater contains approximately 3.5% NaCl by mass and has a density of 1.02 g/mL. What volume of seawater would be required to produce 1.0 g of NaCl when evaporated? 1.02 g seawater mL seawater 3.5 g NaCl 100 g seawater 1.00 g NaCl = 28 mL

17. A mole of glucose (C 6 H 12 O 6 ) contains 6.022 X 10 23 molecules of glucose. And 6 X 6.022 X 10 23 atoms of C. Since a mole is 6.022 X 10 23 particles then a mole of glucose must contain 6 moles of C atoms. How many moles of hydrogen atoms are contained in a mole of glucose? In 5 moles of H 2 SO 4 how many moles of oxygen atoms is there? 6-8 Mole Concepts

18. A mole of glucose (C 6 H 12 O 6 ) contains 6.022 X 10 23 molecules of glucose. And 6 X 6.022 X 10 23 atoms of C. Since a mole is 6.022 X 10 23 particles then a mole of glucose must contain 6 moles of C atoms. How many moles of hydrogen atoms are contained in a mole of glucose? 12 Moles of hydrogen. How many moles of oxygen and hydrogen are in one mole of H 2 O contains: 6-8 Mole Concepts

19. A mole of glucose (C 6 H 12 O 6 ) contains 6.022 X 10 23 molecules of glucose. And 6 X 6.022 X 10 23 atoms of C. Since a mole is 6.022 X 10 23 particles then a mole of glucose must contain 6 moles of C atoms. How many moles of hydrogen atoms are contained in a mole of glucose? 12 Moles of hydrogen. How many moles of oxygen and hydrogen are in one mole of H 2 O contains: One mole of oxygen atoms Two moles of hydrogen atoms 6-8 Mole Concepts

20. A mole of glucose (C 6 H 12 O 6 ) contains 6.022 X 10 23 molecules of glucose. And 6 X 6.022 X 10 23 atoms of C. Since a mole is 6.022 X 10 23 particles then a mole of glucose must contain 6 moles of C atoms. How many moles of hydrogen atoms are contained in a mole of glucose? 12 Moles of hydrogen. How many moles of oxygen and hydrogen are in one mole of H 2 O contains: One mole of oxygen atoms Two moles of hydrogen atoms In 5 moles of H 2 SO 4 how many moles of oxygen atoms are there? 6-8 Mole Concepts

21. A mole of glucose (C 6 H 12 O 6 ) contains 6.022 X 10 23 molecules of glucose. And 6 X 6.022 X 10 23 atoms of C. Since a mole is 6.022 X 10 23 particles then a mole of glucose must contain 6 moles of C atoms. How many moles of hydrogen atoms are contained in a mole of glucose? 12 Moles of hydrogen. How many moles of oxygen and hydrogen are in one mole of H 2 O contains: One mole of oxygen atoms Two moles of hydrogen atoms In 5 moles of H 2 SO 4 how many moles of oxygen atoms is there? 20 moles of O atoms. 6-8 Mole Concepts

22. In 50.0g of H 2 SO 4 how many moles of sulfuric acid are there? 6-8 Mole Conversions 50.0g of H 2 SO 4

23. In 50.0g of H 2 SO 4 how many moles of sulfuric acid are there? 6-8 Mole Conversions 50.0g of H 2 SO 4 = 98.0g of H 2 SO 4 mole H 2 SO 4

24. In 50.0g of H 2 SO 4 how many moles of sulfuric acid are there? 6-8 Mole Conversions 50.0g of H 2 SO 4 = 0.510 mole H 2 SO 4 98.0g of H 2 SO 4 mole H 2 SO 4

25. In 50.0g of H 2 SO 4 how many moles of oxygen atoms are there? 6-8 Mole Conversions

26. In 50.0g of H 2 SO 4 how many moles of oxygen atoms are there? 6-8 Mole Conversions 50.0g of H 2 SO 4 =

27. In 50.0g of H 2 SO 4 how many moles of oxygen atoms are there? 6-8 Mole Conversions 50.0g of H 2 SO 4 = 98.0g of H 2 SO 4 mole H 2 SO 4

28. In 50.0g of H 2 SO 4 how many moles of oxygen atoms are there? 6-8 Mole Conversions 50.0g of H 2 SO 4 = mole H 2 SO 4 4mole O 98.0g of H 2 SO 4 mole H 2 SO 4

29. In 50.0g of H 2 SO 4 how many moles of oxygen atoms are there? 6-8 Mole Conversions 50.0g of H 2 SO 4 = mole H 2 SO 4 4mole O 2.04 mole O 98.0g of H 2 SO 4 mole H 2 SO 4

30. In 5 moles of H 2 SO 4 how many atoms of oxygen are present? 6-8 Mole Conversions

31. In 5 moles of H 2 SO 4 how many atoms of oxygen are present? 6-8 Mole Conversions 5 moles H 2 SO 4 =

32. In 5 moles of H 2 SO 4 how many atoms of oxygen are present? 6-8 Mole Conversions 5 moles H 2 SO 4 mole H 2 SO 4 4 mole O

33. In 5 moles of H 2 SO 4 how many atoms of oxygen are present? 6-8 Mole Conversions 5 moles H 2 SO 4 mole H 2 SO 4 4 mole O 6.02 x 10 23 atoms O mole O = 1.20 x 10 25 atoms

34. 6-8 Empirical Formulas Empirical formula is the smallest whole number ratio between atoms and can be calculated from the percent composition. Molecular formulas happen to be the exact number of atoms making up a molecule, and may or may no be the simplest whole number ratio. Molecular formulas are whole number multiples of the empirical formula.

35. 6-8 Empirical Formula Steps 1.Assume 100 g of compound. 2.Convert percent to a mass number. 3.Convert the mass to moles. 4.Divide each mole number by the smallest mole number. 5.Rounding: a.If the decimal is ≤ 0.1, then drop the decimals b.If the decimal is ≥0.9, then round up. c.All other decimal need to be multiplied by a whole number until roundable.

36. 6-8 Empirical Formula Example A compound is composed of 75.0% C and 25.0% H. Find its empirical formula. Step #1 Assume 100 g of compound 75.0 g C 25.0 g H

37. 6-8 Empirical Formula Example A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #2 Convert grams to moles. 75.0 g C 25.0 g H 12.01 g Mole C 1.008 g H Mole H = 6.225 mole C = 24.802 mole H

38. 6-8 Empirical Formula Example A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #3 Divide each mole number by the smallest. 75.0 g C 25.0 g H 12.01 g Mole C 1.008 g H Mole H = 6.225 mole C = 24.802 mole H 6.225 24.802 = 1.00= 3.98

39. 6-8 Empirical Formula Example A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #4 Rounding; Decimal ≤ 0.1, drop decimals 75.0 g C 25.0 g H 12.01 g Mole C 1.008 g H Mole H = 6.225 mole C = 24.802 mole H 6.225 24.802 = 1.00= 3.98

40. 6-8 Empirical Formula Example A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #4 Rounding; Decimal ≤ 0.1, drop decimals 75.0 g C 25.0 g H 12.01 g Mole C 1.008 g H Mole H = 6.225 mole C = 24.802 mole H 6.225 24.802 = 1 C= 3.98

41. 6-8 Empirical Formula Example A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #4 Rounding; Decimal ≥ 0.9, round up 75.0 g C 25.0 g H 12.01 g Mole C 1.008 g H Mole H = 6.225 mole C = 24.802 mole H 6.225 24.802 = 1 C= 3.98

42. 6-8 Empirical Formula Example A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #4 Rounding; Decimal ≥ 0.9, round up 75.0 g C 25.0 g H 12.01 g Mole C 1.008 g H Mole H = 6.225 mole C = 24.802 mole H 6.225 24.802 = 1 C= 3.98

43. 6-8 Empirical Formula Example A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #4 Rounding; Decimal ≥ 0.9, round up 75.0 g C 25.0 g H 12.01 g Mole C 1.008 g H Mole H = 6.225 mole C = 24.802 mole H 6.225 24.802 = 1 C= 4 H

44. 6-8 Empirical Formula Example A compound is composed of 75.0% C and 15.0% H. Find its empirical formula. Step #4 Rounding; Decimal ≥ 0.9, round up 75.0 g C 25.0 g H 12.01 g Mole C 1.008 g H Mole H = 6.225 mole C = 24.802 mole H 6.225 24.802 = 1 C= 4 H Empirical Formula = CH 4

45. 6-9 Molecular Formulas Empirical formula, is the smallest ratio between atoms in a molecular or formula unit. Molecular formula, is the exact number of atoms in a molecule; a whole number multiple of an empirical formula

46. 6-9 Possible Molecular Formulas Assume an empirical formula of C 3 H 5 O Empirical formula IntegerMolecular Formula C3H5OC3H5O 1234512345 C3H5OC3H5O C3H5OC3H5O C3H5OC3H5O C3H5OC3H5O C3H5OC3H5O

47. 6-9 Possible Molecular Formulas Assume an empirical formula of C 3 H 5 O Empirical formula IntegerMolecular Formula C3H5OC3H5O 1234512345 C3H5OC3H5O C3H5OC3H5O C3H5OC3H5O C3H5OC3H5O C3H5OC3H5O C 6 H 10 O 2

48. 6-9 Possible Molecular Formulas Assume an empirical formula of C 3 H 5 O Empirical formula IntegerMolecular Formula C3H5OC3H5O 1234512345 C3H5OC3H5O C3H5OC3H5O C3H5OC3H5O C3H5OC3H5O C3H5OC3H5O C 6 H 10 O 2 C 9 H 15 O 3

49. 6-9 Possible Molecular Formulas Assume an empirical formula of C 3 H 5 O Empirical formula IntegerMolecular Formula C3H5OC3H5O 1234512345 C3H5OC3H5O C3H5OC3H5O C3H5OC3H5O C3H5OC3H5O C3H5OC3H5O C 6 H 10 O 2 C 9 H 15 O 3 C 12 H 20 O 4 C 15 H 25 O 5

50. 6-9 Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole

51. 6-9 Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole Step #1 Assume 100g of compound 83.6 g C 16.3 g H

52. 6-9 Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole Step #2 Convert grams to moles 83.6 g C 16.3 g H 12.01 g C mole 1.008 g H

53. 6-9 Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole Step #2 Convert grams to moles 83.6 g C 16.3 g H 12.01 g C mole 1.008 g H = 6.961 mole = 16.17 mole

54. 6-9 Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole Step #3 Divide each mole number by the smallest. 83.6 g C 16.3 g H 12.01 g C mole 1.008 g H = 6.961 mole = 16.17 mole

55. 6-9 Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole Step #3 Divide each mole number by the smallest. 83.6 g C 16.3 g H 12.01 g C mole 1.008 g H = 6.961 mole = 16.17 mole 6.961 = 1.00 = 2.32

56. 6-9 Sample Problem Calculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole Step #4 Round if---Not Roundable 83.6 g C 16.3 g H 12.01 g C mole 1.008 g H = 6.961 mole = 16.17 mole 6.961 = 1.00 = 2.32 Step #4, Multiply by an integer until roundable 1.00 X 3 = 3 2.32 X 3 = 7 Empirical formula C 3 H 7

57. 6-9 Molecular Formula Integer Divide empirical weight into molecular weight 3x12 + 7x1 =43 43 86 2 Now multiply the empirical formula by 2

58. 6-9 Molecular Formula Integer Divide empirical weight into molecular weight 3x12 + 7x1 =43 43 86 2 Now multiply the empirical formula by 2 Molecular Formula is C 6 H 14

59. The End