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Chapter 11, The Mole Objectives: The student will be able to:

Chapter 11, The Mole Objectives: The student will be able to:. compare and contrast the mole concept and the concept of dozen; convert moles into numbers of particle and the reverse; relate the mass of an atom to the mass of a mole of atoms;

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Chapter 11, The Mole Objectives: The student will be able to:

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  1. Chapter 11, The Mole Objectives: The student will be able to: compare and contrast the mole concept and the concept of dozen; convert moles into numbers of particle and the reverse; relate the mass of an atom to the mass of a mole of atoms; apply conversion factor concepts to problems related to mass and moles; analyze a chemical formula for the molar ratios present; determine the molar mass of a compound; solve problems using numbers of particles, mass and moles; compare and contrast concepts of percent composition and chemical formula; compare and contrastempirical formula and molecular formula.

  2. One mole is 6.0221367X1023 particles of any substance. Only atomic scale particles should use the mole concept! These particles include: atoms, ions, molecules and formula units. 1 mole of Au atoms = 6.022X1023atoms of Au 1 mole of Na+1ions = 6.022X1023ions of Na 1 mole of CO2molecules = 6.022A. X1023molecules of CO2 1 mole of NaCl formula units = 6.022X1023formula units of NaCl Remember that the term “molecules” is used for covalent compounds and the term “formula units” is used for ionic compounds.

  3. 12 eggs 3.5 dozen eggs 1 dozen ( ) Using the mole concept is very much like using “dozen”. How many eggs are in 3.5 dozen eggs? Equivalence statement: 1 dozen eggs = 12 eggs ( ) ( ) = 42 eggs How many Mg atoms are in 3.5 moles of Mg? Equivalence statement: 1 mole Mg = 6.022X1023 Mg atoms ( ) 6.022X1023 Mg atoms 3.5 mole Mg = 2.1X1024 Mg atoms 1 mole Mg

  4. End class 1

  5. ( ) ( ) 3.423X1022 molecules Cl2 How many moles of Cl2 do you have if you have 3.423X1022 molecules of Cl2? 1 mole of Cl2 = 6.022X1023 molecules of Cl2 1 mole Cl2 6.022X1023 molecules Cl2 = 5.684X10-2 mol Cl2 or = 0.05684 mol Cl2 6.022X1023 is known as Avogadro’s Number Avogadro’s number can be used as a conversion factor just like those used in chapter 2.

  6. 1 mole of Cl26.022X1023 molecules of Cl2 = 1 mole of Cl2 1 mole of Cl2 6.022X1023molecules of Cl2 1 mole of Cl2 1 mole of Cl26.022X1023 molecules of Cl2 = 6.022X1023 molecules of Cl2 6.022X1023 molecules of Cl2 1 mole of Cl2 6.022X1023molecules of Cl2 All equivalence statements can be used to make two different conversion factors. 1 mole of Cl2 = 6.022X1023 molecules of Cl2 Dividing both sides of the equation by “1 mole of Cl2” would give: This is a conversion factor that would convert from moles of Cl2 to molecules of Cl2 Dividing both sides by 6.022X1023 molecules of Cl2 would give: This is a conversion factor that would convert from molecules of Cl2 to moles of Cl2

  7. Since a mole is such a large number of things, it really should only be used with very small particles. These particles are called representative particles. Representative Particles include: atoms, molecules, ions, and formula units. 1 mole of atoms = 6.022X1023atoms 1 mole of molecules = 6.022X1023molecules 1 mole of ions = 6.022X1023ions 1 mole of formula units = 6.022X1023formula units Since we usually know what type of representative particles are being used, we often do not write the words: atoms, molecules, ions, and formula units, but we should remember that they are implied!. 1 mole of Ne = 6.022X1023 Ne Element, so “atoms” Covalent, so “molecules” 1 mole of H2O = 6.022X1023 H2O Ion, so “ions” 1 mole of Ba+2 = 6.022X1023 Ba+2 1 mole of NaCl = 6.022X1023 NaCl Ionic, so “formula units”

  8. Complete Practice Problems 1-3 on page 311 and 4a,b on page 312.

  9. A large number of small things is very difficult to count by hand. Let’s say that 100 jelly beans has a mass of 235.32 g Imagine a dump truck full of jelly beans. How long might it take you to count them all by hand and how many mistakes would you make? A faster way to “count” them would be to measure the mass of 100 jelly beans and then measure the mass of all the jelly beans in the truck.

  10. ( ) ( ) How many jelly beans are in a dump truck if the mass of jelly beans in the truck is 4.532 X104 kg? Remember that 100 jelly beans = 235.32 g ( ) 1000 g jelly beans 100 jelly beans 4.532X104 kg jelly beans 1 kg jelly beans 235.32 g jelly beans = 1.926X107 jelly beans or = 19,260,000 jelly beans This method is called “counting by weighing” and it is how we count large numbers of atoms, molecules, ions, and formula units.

  11. When we count atoms by “weighing”, we need to know how much one mole of atoms “weighs”. molar mass is the mass in grams of 1 mole of a pure substance For elements, the molar mass is the same as the atomic mass expressed in grams. Remember that atomic mass is given on the periodic table. Example: the atomic mass of Carbon is 12.011 amu So 1 mole C = 12.011 g C If we have 12.011 g of carbon we have 1 mole of C which means we have 6.022X1023 atoms of C.

  12. ( ) ( ) How many grams C are in 0.382 mol C? 1 mole C = 12.011 g C ( ) 12.011 g C 0.382 mol C = 4.58 g C 1 mol C How many moles C are in 42.573 g C? 1 mole C = 12.011 g C ( ) 1 mole C 42.573 g C = 3.5445 mol C 12.011 g C atoms

  13. Complete Practice Problems 11 and 12 on page 316.

  14. ( ) } Avogadro’s Number ( ) ( ) 6.022X1023 atoms 1 mol Mass in g = number of atoms } atomic mass in g 1 mol Molar mass Avogadro’s number is the link between number of particles and moles. Molar mass is the link between mass in grams and moles. If we are given “mass” in grams, we can calculate the number of particles of a substance that are present.

  15. ( ) } Molar mass ( ) ( ) 1 mol number of atoms atomic mass in g = Mass in g } 1 mol 6.022X1023 atoms Avogadro’s Number The same type of problem can be worked in reverse as well. If we are given number of particles of a substance that are present, we can calculate the “mass” in grams.

  16. ( ( ) ) } } Avogadro’s Number Molar mass ( ( ) ) ( ) ( ) 1 mol 6.022X1023 atoms number of atoms atomic mass in g 1 mol = Mass in g Mass in g = number of atoms } } 1 mol atomic mass in g 6.022X1023 atoms 1 mol Avogadro’s Number Molar mass Notice the similarities and differences between these two related problems.

  17. ( ( ( ( ) ) ) ) How many grams Xe are 2.52X1025 atoms of Xe? 1 mole Xe = 131.3 g Xe 1 mole of Xe = 6.022X1023 atoms Xe ( ) 1 mol Xe 131.3 g Xe 2.52X1025 atoms Xe 1 mol Xe 6.022X1023 atoms Xe = 5,490 g Xe How many atoms Xe are contained in 42.5 g Xe? 1 mole Xe = 131.3 g Xe 1 mole of Xe = 6.022X1023 atoms Xe ( ) 1 mol Xe 6.022X1023 atoms Xe 42.5 g Xe 131.3 g Xe 1 mol Xe = 1.95X1023 atoms Xe

  18. Complete Practice Problems 13 and 14 on page 318.

  19. Moles and chemical formulas If we have 10 molecules of P2O5 How many O atoms are present? How many P atoms are present? We can create “molecule ratios” from the chemical formula to help us solve these problems. ( ) This is what we use to convert from molecules of P2O5 to atoms of P 2 atoms P 1 molecule P2O5 ( ) In a similar fashion, this is what we use to convert from molecules of P2O5 to atoms of O 5 atoms O 1 molecule P2O5 What if we have 0.2 moles of P2O5?

  20. What if we have 0.2 moles of P2O5? We could change the language. Now we can write “mole ratios” from the formula. ( ) This mole ratio converts from moles of P2O5 to moles of P atoms 2 moles P 1 mole P2O5 ( ) In a similar fashion, this mole ratio converts from moles of P2O5 to moles of O atoms 5 moles O 1 mole P2O5 Mole ratios of this type can be written for any formula!

  21. Mole ratios from formulas can be used as conversion factors! ( ) ( ) 2 mol K+1 1 mol SO4-2 1 mole K2SO4 1 mole K2SO4 ( ) ( ) 2 mol K+1 0.381 mole K2SO4 1 mole K2SO4 Assume we have 1.0 mole of K2SO4 How many moles of K+1 ions are present? How many moles of SO4-2 ions are present? If we have 0.381 mole of K2SO4, how many moles of K+1 ions do we have? = 0.762 mol K+1

  22. ( ( ( ( ) ) ) ) ( ) ( ) 5 mol C 7 mol H 5 mol C 3 mol Cl 7 mol H 3 mol Cl 1 mole C5H7Cl3 1 mole C5H7Cl3 1 mole C5H7Cl3 1 mole C5H7Cl3 1 mole C5H7Cl3 1 mole C5H7Cl3 = 2.5 mol C atoms ( ( ( ) ) ) = 3.5 mol H atoms 0.50 mole C5H7Cl3 0.50 mole C5H7Cl3 0.50 mole C5H7Cl3 = 1.5 mol Cl atoms If we have 0.50 mole of C5H7Cl3 How many moles of C atoms are present? How many moles of H atoms are present? How many moles of Cl atoms are present?

  23. Complete Practice Problems 20 - 24 on page 321.

  24. The atomic mass from the periodic table is the mass in grams of one mole of the element. When we have a compound, we can find the molar mass of the compound by adding up the atomic masses for each atom present in the compound. For example, H2O one mole of H2O has 2 moles of H and one mole of O in it. Therefore the molar mass of H2O is: (2)*(1.008 g/mol) + (1)*(16.00 g/mol) = 18.02 g/mol 18.02 g/mol is the molar mass of water

  25. What would be the molar mass of C5H7Cl3? 1 mole of C5H7Cl3 contains 5 mol C, 7 mol H and 3 mol Cl 1 mol C = 12.01 g C 12.01 g X 5 = 60.05 g 1.008 g X 7 = 7.056 g 35.45 g X 3 = 106.4 g 60.05 g + 7.056 g + 106.4 g 1 mol H = 1.008 g H 1 mol Cl = 35.45 g Cl 173.5 g So, 1 mole of C5H7Cl3 = 173.5 g C5H7Cl3 173.5 g is the molar mass of C5H7Cl3 This is also written as 173.5 g/mol Molar Mass is used as a conversion factor in many problems!

  26. ( ) How many moles of C5H7Cl3 do we have if we have 37.32 g of C5H7Cl3? Molar mass definition: 1 mole of C5H7Cl3 = 173.5 g C5H7Cl3 ( ) 1 mole C5H7Cl3 = 0.2152 mol C5H7Cl3 37.32 g C5H7Cl3 } 173.5 g C5H7Cl3 Molar mass (shown in conversion factor form converting from grams to moles)

  27. ( ) ( ) How many grams of C5H7Cl3 do we have if we have 0.725 mol of C5H7Cl3? Molar mass definition: 1 mole of C5H7Cl3 = 173.5 g C5H7Cl3 173.5 g C5H7Cl3 = 126 g C5H7Cl3 0.725 mol C5H7Cl3 } 1 mole C5H7Cl3 Molar mass (shown in conversion factor form converting from moles to grams)

  28. ( ) ( ( ) ) Look at the similarities and differences in how the same molar mass can be used. ( ) 1 mole C5H7Cl3 = 0.2152 mol C5H7Cl3 37.32 g C5H7Cl3 } 173.5 g C5H7Cl3 Molar mass (shown in conversion factor form converting from grams to moles) 173.5 g C5H7Cl3 = 126 g C5H7Cl3 0.725 mol C5H7Cl3 } 1 mole C5H7Cl3 Molar mass (shown in conversion factor form converting from moles to grams)

  29. Complete Practice Problems 25 - 30 on pages 322-324.

  30. ( ) ( ) 1 mole Ag2CrO4 25.8 g Ag2CrO4 ( ) 331.8 g Ag2CrO4 2 mole Ag+ 1 mole Ag2CrO4 ( ) 6.022X1023 ions 1 mole Ag+ Practice problem 31 on page 326. A 25.8 g sample of Ag2CrO4 is used in this problem. a) How many silver ions are in the sample? The thought process is: mass of sample to moles of sample tomoles of silver ionsto number of silver ions. In order to do this we need three equivalence statements: 1 mole of Ag2CrO4 = 331.8 g of Ag2CrO4this was calculated from masses in the periodic table 2 mole of Ag = 1mole of Ag2CrO4 1 mole of Ag+ = 6.022X1023 ions Ag+ = 9.37X1022 Ag+ ions

  31. 215.8 g + 52.00 g + 64.00 g ( ) ( ) 1 mole Ag2CrO4 331.8 g Ag2CrO4 1 mole Ag2CrO4 6.022X1023 Ag2CrO4 Practice Problem 33b is like 33a. Part of 33c needed to be completed in order to do 33a or b. 33 c) What is the mass in grams of 1 formula unit of Ag2CrO4? We get this by dividing the mass of one mole of Ag2CrO4 by the number of formula units in 1 mole of Ag2CrO4 (Avogadro's number). Ag2CrO4 has what molar mass? Ag 107.9 g/mol X 2 mol = 215.8 g Cr 52.00 g/mol X 1 mol = 52.00 g O 16.00 g/mol X 4 mol = 64.00 g 331.8 g in 1 mol Ag2CrO4 this is used in 33a and 33b = 5.510X10-22 g/formula unit of Ag2CrO4

  32. Complete Practice Problems 32 - 35 on page 326.

  33. ( ) 2.016 g H X 100 = 11.19 % H % H = 18.02 g H2O ( ) 16.00 g O X 100 = 88.79 % O % O = 18.02 g H2O ( ) Mass of element in 1 mole of compound X 100 = % by mass of the element Mass of 1 mole of compound Percent Composition: H2O 2H 2X(1.008 g) = 2.016 g H O 1X(16.00 g) = 16.00 g O 1 mole H2O = 18.02 g

  34. ( ) 0.09835 g Na X 100 = 39.34% Na % Na = 0.25000 g sample ( ) 0.15163 g Cl X 100 = 60.652% Cl % Cl = 0.25000 g sample Percent Composition: The mass percent of each element present in a compound is called the percent composition. Percent composition can be determined from the formula of a compound as in the example for water on the previous “slide” or it can be measured by analytical chemists in a laboratory as in the example below. A chemist analyzes 0.25000 g of a salt and obtains the following results: the sample contained 0.09835 g Na and 0.15163 g Cl What is the percent composition of the salt?

  35. Complete Practice Problems 42-45 on page 331.

  36. H C 6 3 3 3 Empirical Formula: the formula of a compound that uses the smallest whole number mole ratios of each element present in the compound. Molecular Formula: the formula of a compound that uses the actual mole ratios of each element present in the compound. Compare and Contrast these two ideas Propene: one molecule of propene has 3 carbon atoms and 6 hydrogen atoms, so the molecular formula is: C3H6 but the smallest whole number ratio of carbon to hydrogen is 2 hydrogen atoms for every 1 carbon atom so the empirical formula is CH2 = CH2

  37. If the percent composition is known for a compound, it can be used to calculate the empirical formula. Step 1: Assume you have 100.00 g of the compound. Step 2: Multiply the mass percents by 100.00 g to find the mass of each element present in 100.00 g of the compound. Step 3: Convert the mass of each element into moles of each element. Step 4: Calculate mole ratios (divide the larger number of moles by the smallest number of moles whenever possible). Step 5: Write the formula using the mole ratios calculated. If a mole ratio is not a whole number, multiply each mole ratio by whatever number will convert them into whole numbers.

  38. ( ) (17.74 g C) 1 mol C = 1.477 mol C 12.01 g C ( ) (3.722 g H) 1 mol H = 3.692 mol H 1.008 g H ( ) (78.54 g Cl) 1 mol Cl = 2.216 mol Cl 35.45 g Cl A compound known to contain C, H and Cl is analyzed and the percent composition is determined to be: 17.74% C, 3.722 % H, and 78.54% Cl. What is the empirical formula of the compound? (100.00g)(17.74% C) = 17.74 g C Steps 1 and 2 (100.00g)(3.722% H) = 3.722 g H (100.00g)(78.54% Cl) = 78.54 g Cl Step 3 Step 5 CH2.5Cl1.5 Step 4 1.477/1.477 = 1.000 mol C/1 mol C 3.692/1.477 = 2.500 mol H/1 mol C C2H5Cl3 2.216/1.477 = 1.500 mol Cl/1 mol C

  39. If the molar mass and empirical formula of a compound are known, then the molecular formula can be determined. If the empirical formula is C2H5Cl3 and the molar mass is 406.2 g/mol, what is the molecular formula for the compound? C2H5Cl3 2C 2X(12.01 g) = 24.02 g C 5H 5X(1.008 g) = 5.040 g H 3Cl 3X(35.45 g) = 106.4 g Cl 135.4 g in 1 mol of C2H5Cl3 406.2 g Therefore: C(2*3)H(5*3)Cl(3*3) = C6H15Cl9 = 3.000 135.4 g

  40. Complete Practice Problems 51-53 on page 335. Attempt Practice Problems 54-57 on page 337.

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