1. INTRODUCTORY CHEMISTRY INTRODUCTORY CHEMISTRY Concepts and Critical Thinking Seventh Edition by Charles H. Corwin Chapter 8 Lecture Chapter 8 The Mole Concept by Christopher G. Hamaker Illinois State University © 2014 Pearson Education, Inc.

2. Chapter 8© 2014 Pearson Education, Inc. Avogadro’s Number Avogadro’s number (symbol N) is the number of atoms in 12.01 grams of carbon. Its numerical value is 6.02 x 10 23. Therefore, a 12.01 g sample of carbon contains 6.02 x 10 23 carbon atoms.

3. Chapter 8© 2014 Pearson Education, Inc. Analogies for Avogadro’s Number The volume occupied by one mole of softballs would be about the size of Earth. One mole of Olympic shot put balls has about the same mass as that of Earth. One mole of hydrogen atoms laid side by side would circle Earth about 1 million times.

4. Chapter 8© 2014 Pearson Education, Inc. Mole Calculations I 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 x 10 23 particles. 1 mol = Avogadro’s number = 6.02 x 10 23 units We can use the mole relationship to convert between the number of particles and the mass of a substance.

5. Chapter 8© 2014 Pearson Education, Inc. Mole Calculations I, Continued How many sodium atoms are in 0.240 mol Na? 1.We want atoms of Na. 2.We have 0.240 mol Na. 3.1 mole Na = 6.02 x 10 23 atoms Na. = 1.44 x 10 23 atoms Na 0.240 mol Na x 1 mol Na 6.02 x 10 23 atoms Na

6. Chapter 8© 2014 Pearson Education, Inc. Mole Calculations I, Continued How many moles of aluminum are in 3.42 x 10 21 atoms Al? 1.We want moles Al. 2.We have 3.42 x 10 21 atoms Al. 3.1 mol Al = 6.02 x 10 23 atoms Al. = 5.68 x 10 – 3 mol Al3.42 x 10 21 atoms Al x 1 mol Al 6.02 x 10 23 atoms Al

7. Chapter 8© 2014 Pearson Education, Inc. Mole Calculations II We will be using the unit analysis method again. Recall the following steps:

8. Chapter 8© 2014 Pearson Education, Inc. Molar Mass The atomic mass of any substance expressed in grams is the molar mass (MM) of that substance. The atomic mass of carbon is 12.01 amu. Therefore, the molar mass of carbon is 12.01 g/mol. Since nitrogen occurs naturally as a diatomic, N 2, the molar mass of nitrogen gas is two times 14.01 g or 28.02 g/mol.

9. Chapter 8© 2014 Pearson Education, Inc. One Mole of Several Substances C 12 H 22 O 11 H2OH2O mercury sulfur NaCl copper lead K 2 Cr 2 O 7

10. Chapter 8© 2014 Pearson Education, Inc. Calculating Molar Mass The molar mass of a substance is the sum of the molar masses of each element. What is the molar mass of copper(II) nitrite, Cu(NO 2 ) 2 ? The sum of the atomic masses is as follows: 63.55 + 2(14.01 + 16.00 + 16.00) = 63.55 + 2(46.01) = 155.57 amu The molar mass for Cu(NO 2 ) 2 is 155.57 g/mol.

11. Chapter 8© 2014 Pearson Education, Inc. Mole Calculations II 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 x 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.

12. Chapter 8© 2014 Pearson Education, Inc. Mole–Mass Calculation What is the mass of 1.33 moles of titanium, Ti? 1.We want grams. 2.We have 1.33 moles of titanium. 3.Use the molar mass of Ti: 1 mol Ti = 47.88 g Ti. = 63.7 g Ti 1.33 mole Ti x 47.88 g Ti 1 mole Ti

13. Chapter 8© 2014 Pearson Education, Inc. Atoms–Mass Calculation What is the mass of 2.55 x 10 23 atoms of lead? 1.We want grams. 2.We have atoms of lead. 3.Use Avogadro’s number and the molar mass of Pb. = 87.9 g Pb 2.55 × 10 23 atoms Pb x 1 mol Pb 6.02×10 23 atoms Pb 207.2 g Pb 1 mole Pb x

14. Chapter 8© 2014 Pearson Education, Inc. Mass–Molecule Calculation How many F 2 molecules are present in 2.25 g of fluorine gas? 1.We want molecules F 2. 2.We have grams F 2. 3.Use Avogadro’s number and the molar mass of F 2. 3.56 x 10 22 molecules O 2 2.25 g F 2 x 1 mol F 2 38.00 g F 2 6.02 x 10 23 molecules F 2 1 mol F 2 x

15. Chapter 8© 2014 Pearson Education, Inc. Mass of an Atom or Molecule 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 as follows: 1.06 x 10 –22 g/molecule 64.07 g SO 2 1 mol SO 2 6.02 x 10 23 molecules SO 2 1 mol SO 2 x

16. Chapter 8© 2014 Pearson Education, Inc. Molar Volume At standard temperature and pressure, 1 mol of any gas occupies 22.4 L. The volume occupied by 1 mol of gas (22.4 L) is called the molar volume. Standard temperature and pressure are 0  C and 1 atm.

17. Chapter 8© 2014 Pearson Education, Inc. Molar Volume of Gases We now have a new unit factor equation: 1 mol gas = 6.02 x 10 23 molecules gas = 22.4 L gas

18. Chapter 8© 2014 Pearson Education, Inc. One Mole of a Gas at STP The box below has a volume of 22.4 L, which is the volume occupied by 1 mol of a gas at STP.

19. Chapter 8© 2014 Pearson Education, Inc. Gas Density The density of gases is much less than that of liquids. We can easily calculate the density of any gas at STP. The formula for gas density at STP is as follows: = density, g/L molar mass in grams molar volume in liters

20. Chapter 8© 2014 Pearson Education, Inc. Calculating Gas Density What is the density of methane gas, CH 4, at STP? First we need the molar mass for ammonia. 12.01 + 4(1.01) = 16.05 g/mol The molar volume CH 4 at STP is 22.4 L/mol. Density is mass/volume. = 0.717 g/L 16.05 g/mol 22.4 L/mol

21. Chapter 8© 2014 Pearson Education, Inc. Molar Mass of a Gas We can also use molar volume to calculate the molar mass of an unknown gas. An unknown gas has 4.29 g and occupies 1.50 L at STP. What is the molar mass? We want g/mol; we have g/L. 4.29 g 1.50 L 22.4 L 1 mol x = 64.1 g/mol

22. Chapter 8© 2014 Pearson Education, Inc. Mole Calculations III We now have three interpretations for the mole: 1.1 mol = 6.02 x 10 23 particles 2.1 mol = molar mass 3.1 mol = 22.4 L at STP This gives us three unit factors to use to convert among moles, particles, mass, and volume.

23. Chapter 8© 2014 Pearson Education, Inc. Calculating Molar Volume 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 x = 0.201 mol CH 4 1 mol CH 4 22.4 L CH 4

24. Chapter 8© 2014 Pearson Education, Inc. Calculating Mass Volume 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 x x 22.4 L O 3 1 mol O 3 48.00 g O 3 1 mol O 3

25. Chapter 8© 2014 Pearson Education, Inc. Calculating Molecule Volume How many molecules of argon gas, Ar, occupy 0.430 L at STP? We want molecules Ar; we have 0.430 L Ar. Convert volume to moles, and then moles to molecules. 0.430 L Ar x 1 mol Ar 22.4 L Ar 6.02 x 10 23 molecules Ar 1 mol Ar x = 1.16 x 10 22 molecules Ar

26. Chapter 8© 2014 Pearson Education, Inc. Percent Composition 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.

27. Chapter 8© 2014 Pearson Education, Inc. Calculating Percent Composition There are a few steps to calculating the percent composition of a compound. Let’s practice using H 2 O. 1.Assume you have 1 mol of the compound. 2.One mole of H 2 O contains 2 mol of hydrogen and 1 mol of oxygen. Therefore, 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

28. Chapter 8© 2014 Pearson Education, Inc. More Calculating Percent Composition 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 x 100% = 11.2% H 16.00 g O 18.02 g H 2 O x 100% = 88.79% O

29. Chapter 8© 2014 Pearson Education, Inc. Percent Composition Problem 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

30. Chapter 8© 2014 Pearson Education, Inc. Percent Composition of TNT 84.07 g C 227.15 g TNT x 100% = 37.01% C 5.05 g H 227.15 g TNT x 100% = 2.22% H 42.03 g N 227.15 g TNT x 100% = 18.50% N 96.00 g O 227.15 g TNT x 100% = 42.26% O

31. Chapter 8© 2014 Pearson Education, Inc. Empirical Formula 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.

32. Chapter 8© 2014 Pearson Education, Inc. Calculating Empirical Formulas 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 empirical 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.

33. Chapter 8© 2014 Pearson Education, Inc. Calculating Empirical Formulas, Continued – 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. 1 mol Ra 226.03 g Ra 1.640 g Ra x = 0.00726 mol Ra 1 mol O 16.00 g O 0.115 g O x = 0.00719 mol O

34. Chapter 8© 2014 Pearson Education, Inc. Empirical Formulas from Mass Composition 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.

35. Chapter 8© 2014 Pearson Education, Inc. Empirical Formula for Acetylene Calculate the moles of each element. 1 mol C 12.01 g C 92.2 g C x = 7.68 mol C 1 mol H 1.01 g H 7.83 g H x = 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 following formula: 7.68 C = C 1.00 H 1.01 = CH 7.75 7.68 H

36. Chapter 8© 2014 Pearson Education, Inc. Molecular Formula 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.

37. Chapter 8© 2014 Pearson Education, Inc. Critical Thinking: Avogadro’s Number In 1911, Ernest Rutherford estimated the value of Avogadro’s number as 6.11 x 10 23 using alpha particles from radium. The most recent measurements of Avogadro’s number were made using X-ray diffraction of silicon crystals. Currently, the most accurate value for Avogadro’s number is 6.0221415 x 10 23.

38. Chapter 8© 2014 Pearson Education, Inc. Chapter Summary Avogadro’s number is 6.02 x 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.

39. Chapter 8© 2014 Pearson Education, Inc. Chapter Summary, Continued We can convert between the number of particles and moles of a substance using Avogadro’s number (1 mol = 6.02 x 10 23 particles). We can convert between mass of a substance and moles of a substance using the molar mass. We can convert between the volume of a gas at STP and moles of a gas using molar volume at STP (1 mol = 22.4 L).

40. Chapter 8© 2014 Pearson Education, Inc. Chapter Summary, Continued 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.