1. Christopher G. Hamaker, Illinois State University, Normal IL © 2008, Prentice Hall Chapter 9 The Mole Concept INTRODUCTORY CHEMISTRY INTRODUCTORY CHEMISTRY Concepts & Connections Fifth Edition by Charles H. Corwin

2. Chapter 9 2

3. 3 Avogadro’s number (symbol N) is the number of atoms in 12.01 grams of carbon. Its numerical value is 6.02 × 10 23. Therefore, a 12.01 g sample of carbon contains 6.02 × 10 23 carbon atoms. Avogadro’s Number

4. Chapter 9 4 The mole (mol) is a unit of measure for an amount of a chemical substance. A mole is Avogadro’s number of particles, which is 6.02 × 10 23 particles. 1 mol = Avogadro’s number = 6.02 × 10 23 units We can use the mole relationship to convert between the number of particles and the mass of a substance. The Mole

5. Chapter 9 5 The volume occupied by one mole of softballs would be about the size of the Earth. One mole of Olympic shotput balls has about the same mass as the Earth. Analogies for Avogadro’s Number

6. Chapter 9 6 One Mole of Several Substances C 12 H 22 O 11 H2OH2O mercury sulfur NaCl copper lead K 2 Cr 2 O 7

7. Chapter 9 7 We will be using the Unit Analysis Method again. Recall: –First, we write down the unit requested. –Second, we write down the given value. –Third, we apply unit factor(s) to convert the given units to the desired units. Mole Calculations

8. Chapter 9 8 How many sodium atoms are in 0.120 mol Na? –Step 1: we want atoms of Na –Step 2: we have 0.120 mol Na –Step 3: 1 mole Na = 6.02 × 10 23 atoms Na = 7.22 × 10 22 atoms Na 0.120 mol Na × 1 mol Na 6.02 × 10 23 atoms Na Mole Calculations I

9. Practice Exercise (from Corwin) Calculate the number of molecules present in 0.0763 moles of sodium chloride. Calculate the number of particles present in 0.15 moles of Cadmium. Chapter 9 9

10. 10 How many moles of potassium are in 1.25 × 10 21 atoms K? –Step 1: we want moles K –Step 2: we have 1.25 × 10 21 atoms K –Step 3: 1 mole K = 6.02 × 10 23 atoms K = 2.08 × 10 -3 mol K1.25 × 10 21 atoms K × 1 mol K 6.02 × 10 23 atoms K Mole Calculations I

11. Calculate the number of moles of potassium iodide in 5.34 x 10 25 formula units of potassium iodide. Chapter 9 11

12. Chapter 9 12 The atomic mass of any substance expressed in grams is the molar mass (MM) of that substance. The atomic mass of iron is 55.85 amu. Therefore, the molar mass of iron is 55.85 g/mol. Since oxygen occurs naturally as a diatomic, O 2, the molar mass of oxygen gas is 2 times 16.00 g or 32.00 g/mol. Molar Mass

13. Chapter 9 13 The molar mass of a substance is the sum of the molar masses of each element. What is the molar mass of magnesium nitrate, Mg(NO 3 ) 2 ? The sum of the atomic masses is: 24.31 + 2(14.01 + 16.00 + 16.00 + 16.00) = 24.31 + 2(62.01) = 148.33 amu The molar mass for Mg(NO 3 ) 2 is 148.33 g/mol. Calculating Molar Mass

14. Molar Mass Problems Calculate the molar mass for each of the following substances: Silver metal Ammonia gas Magnesium nitrate Manganese metal Sulfur hexafluoride Strontium acetate Chapter 9 14

15. Chapter 9 15 Now we will use the molar mass of a compound to convert between grams of a substance and moles or particles of a substance. 6.02 × 10 23 particles = 1 mol = molar mass If we want to convert particles to mass, we must first convert particles to moles and then we can convert moles to mass. Mole Calculations II

16. Chapter 9 16 What is the mass of 1.33 moles of titanium, Ti? We want grams; we have 1.33 moles of titanium. Use the molar mass of Ti: 1 mol Ti = 47.88 g Ti = 63.7 g Ti 1.33 mole Ti × 47.88 g Ti 1 mole Ti Mole-Mole Calculations

17. Molar Mass Calculations What is the mass of 0.424 moles of Lead? Chapter 9 17

18. Chapter 9 18 What is the mass of 2.55 × 10 23 atoms of lead? We want grams; we have atoms of lead. Use Avogadro’s number and the molar mass of Pb. = 87.9 g Pb 2.55 × 10 23 atoms Pb × 1 mol Pb 6.02×10 23 atoms Pb 207.2 g Pb 1 mole Pb × Mole Calculations III

19. Molar Mass Calculations contd What is the mass in grams of 2.01 x 10 22 atoms of sulfur? What is the mass of 7.75 x 10 22 formula units (molecules) of lead (II) sulfide? Chapter 9 19

20. Chapter 9 20 How many O 2 molecules are present in 0.470 g of oxygen gas? We want molecules O 2 ; we have grams O 2. Use Avogadro’s number and the molar mass of O 2 : 8.84 × 10 21 molecules O 2 0.470 g O 2 × 1 mol O 2 32.00 g O 2 6.02×10 23 molecules O 2 1 mole O 2 × Mole Calculations III

21. Chapter 9 21 What is the mass of a single molecule of sulfur dioxide? The molar mass of SO 2 is 64.07 g/mol. We want mass/molecule SO 2, we have the molar mass of sulfur dioxide. Use Avogadro’s number and the molar mass of SO 2 : 1.06 × 10 -22 g/molecule 64.07 g SO 2 1 mol SO 2 6.02×10 23 molecules SO 2 1 mole SO 2 × Mass of a Single Molecule

22. Chapter 9 22 At standard temperature and pressure, 1 mole of any gas occupies 22.4 L. The volume occupied by 1 mole of gas (22.4 L) is called the molar volume. Standard temperature and pressure are 0  C and 1 atm. Molar Volume

23. Chapter 9 23 We now have a new unit factor equation: 1 mole gas = 6.02 × 10 23 molecules gas = 22.4 L gas Molar Volume of Gases

24. Chapter 9 24 One Mole of a Gas at STP The box below has a volume of 22.4 L, the volume occupied by 1 mole of a gas at STP.

25. Chapter 9 25 The density of gases is much less than that of liquids. We can calculate the density of any gas at STP easily. The formula for gas density at STP is: = density, g/L molar mass in grams molar volume in liters Gas Density

26. Chapter 9 26 What is the density of ammonia gas, NH 3, at STP? First we need the molar mass for ammonia: –14.01 + 3(1.01) = 17.04 g/mol The molar volume NH 3 at STP is 22.4 L/mol. Density is mass/volume: = 0.761 g/L 17.04 g/mol 22.4 L/mol Calculating Gas Density

27. Chapter 9 27 We can also use molar volume to calculate the molar mass of an unknown gas. 1.96 g of an unknown gas occupies 1.00 L at STP. What is the molar mass? We want g/mol; we have g/L. 1.96 g 1.00 L 22.4 L 1 mole ×= 43.9 g/mol Molar Mass of a Gas

28. Chapter 9 28 We now have three interpretations for the mole: –1 mol = 6.02 × 10 23 particles –1 mol = molar mass –1 mol = 22.4 L at STP for a gas This gives us 3 unit factors to use to convert between moles, particles, mass, and volume. Mole Unit Factors

29. Chapter 9 29 A sample of methane, CH 4, occupies 4.50 L at STP. How many moles of methane are present? We want moles; we have volume. Use molar volume of a gas: 1 mol = 22.4 L. 4.50 L CH 4 ×= 0.201 mol CH 4 1 mol CH 4 22.4 L CH 4 Mole-Volume Calculation

30. Chapter 9 30 What is the mass of 3.36 L of ozone gas, O 3, at STP? We want mass O 3 ; we have 3.36 L O 3. Convert volume to moles, then moles to mass: = 7.20 g O 3 3.36 L O 3 ×× 22.4 L O 3 1 mol O 3 48.00 g O 3 1 mol O 3 Mass-Volume Calculation

31. Chapter 9 31 How many molecules of hydrogen gas, H 2, occupy 0.500 L at STP? We want molecules H 2 ; we have 0.500 L H 2. Convert volume to moles, and then moles to molecules: 0.500 L H 2 × 1 mol H 2 22.4 L H 2 6.02×10 23 molecules H 2 1 mole H 2 × = 1.34 × 10 22 molecules H 2 Molecule-Volume Calculation

32. Chapter 9 32 The percent composition of a compound lists the mass percent of each element. For example, the percent composition of water, H 2 O is: –11% hydrogen and 89% oxygen All water contains 11% hydrogen and 89% oxygen by mass. Percent Composition

33. Chapter 9 33 There are a few steps to calculating the percent composition of a compound. Let’s practice using H 2 O. –Assume you have 1 mole of the compound. –One mole of H 2 O contains 2 mol of hydrogen and 1 mol of oxygen. –2(1.01 g H) + 1(16.00 g O) = molar mass H 2 O –2.02 g H + 16.00 g O = 18.02 g H 2 O Calculating Percent Composition

34. Chapter 9 34 Next, find the percent composition of water by comparing the masses of hydrogen and oxygen in water to the molar mass of water: 2.02 g H 18.02 g H 2 O × 100% = 11.2% H 16.00 g O 18.02 g H 2 O × 100% = 88.79% O Calculating Percent Composition

35. Chapter 9 35 TNT (trinitrotoluene) is a white crystalline substance that explodes at 240 °C. Calculate the percent composition of TNT, C 7 H 5 (NO 2 ) 3. 7(12.01 g C) + 5(1.01 g H) + 3 (14.01 g N + 32.00 g O) = g C 7 H 5 (NO 2 ) 3 84.07 g C + 5.05 g H + 42.03 g N + 96.00 g O = 227.15 g C 7 H 5 (NO 2 ) 3. Percent Composition Problem

36. Chapter 9 36 84.07 g C 227.15 g TNT × 100% = 37.01% C Percent Composition of TNT 1.01 g H 227.15 g TNT × 100% = 2.22% H 42.03 g N 227.15 g TNT × 100% = 18.50% N 96.00 g O 227.15 g TNT × 100% = 42.26% O

37. Chapter 9 37 The empirical formula of a compound is the simplest whole number ratio of ions in a formula unit or atoms of each element in a molecule. The molecular formula of benzene is C 6 H 6. –The empirical formula of benzene is CH. The molecular formula of octane is C 8 H 18. –The empirical formula of octane is C 4 H 9. Empirical Formulas

38. Chapter 9 38 We can calculate the empirical formula of a compound from its composition data. We can determine the mole ratio of each element from the mass to determine the formula of radium oxide, Ra ? O ?. A 1.640 g sample of radium metal was heated to produce 1.755 g of radium oxide. What is the empirical formula? We have 1.640 g Ra and 1.755-1.640 = 0.115 g O. Calculating Empirical Formulas

39. Chapter 9 39 The molar mass of radium is 226.03 g/mol, and the molar mass of oxygen is 16.00 g/mol. We get Ra 0.00726 O 0.00719. Simplify the mole ratio by dividing by the smallest number. We get Ra 1.01 O 1.00 = RaO is the empirical formula. Calculating Empirical Formulas 1 mol Ra 226.03 g Ra 1.640 g Ra ×= 0.00726 mol Ra 1 mol O 16.00 g O 0.115 g O ×= 0.00719 mol O

40. Chapter 9 40 We can also use percent composition data to calculate empirical formulas. Assume that you have 100 grams of sample. Acetylene is 92.2% carbon and 7.83% hydrogen. What is the empirical formula? If we assume 100 grams of sample, we have 92.2 g carbon and 7.83 g hydrogen. Empirical Formulas from Percent Composition

41. Chapter 9 41 Calculate the moles of each element: 1 mol C 12.01 g C 92.2 g C ×= 7.68 mol C 1 mol H 1.01 g H 7.83 g H ×= 7.75 mol H The ratio of elements in acetylene is C 7.68 H 7.75. Divide by the smallest number to get the formula: 7.68 C = C 1.00 H 1.01 = CH 7.75 7.68 H Empirical Formula for Acetylene

42. Chapter 9 42 The empirical formula for acetylene is CH. This represents the ratio of C to H atoms on acetylene. The actual molecular formula is some multiple of the empirical formula, (CH) n. Acetylene has a molar mass of 26 g/mol. Find n to find the molecular formula: = CH (CH) n 26 g/mol 13 g/mol n = 2 and the molecular formula is C 2 H 2. Molecular Formulas

43. Chapter 9 43 Avogadro’s number is 6.02 × 10 23, and is 1 mole of any substance. The molar mass of a substance is the sum of the atomic masses of each element in the formula. At STP, 1 mole of any gas occupies 22.4 L. Chapter Summary

44. Chapter 9 44 We can use the following flow chart for mole calculations: Chapter Summary, continued

45. Chapter 9 45 The percent composition of a substance is the mass percent of each element in that substance. The empirical formula of a substance is the simplest whole number ratio of the elements in the formula. The molecular formula is a multiple of the empirical formula. Chapter Summary, continued