2. The Mole

3. 1 dozen = 1 century= 1 pair = 1 mole = 12 100 2 6.02 x 10 23 There are exactly 12 grams of carbon-12 in one mole of carbon-12.

4. 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 × 10 23. Therefore, a 12.01 g sample of carbon contains 6.02 × 10 23 carbon atoms.

5. The Mole The mole (mol) is a unit of measure for an amount of a chemical substance. A mole is Avogadro’s number of particles, that 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.

6. How Big Is a Mole? The volume occupied by one mole of softballs would be about the size of the Earth. One mole of Olympic shot put balls has about the same mass as the Earth.

7. The Mole 1 dozen cookies = 12 cookies 1 mole of cookies = 6.02 X 10 23 cookies 1 dozen cars = 12 cars 1 mole of cars = 6.02 X 10 23 cars 1 dozen Al atoms = 12 Al atoms 1 mole of Al atoms = 6.02 X 10 23 atoms Note that the NUMBER is always the same, but the MASS is very different! Mole is abbreviated mol (gee, that’s a lot quicker to write, huh?)

8. = 6.02 x 10 23 C atoms = 6.02 x 10 23 H 2 O molecules = 6.02 x 10 23 NaCl “molecules” (technically, ionics are compounds not molecules so they are called formula units) 6.02 x 10 23 Na + ions and 6.02 x 10 23 Cl – ions A Mole of Particles A Mole of Particles Contains 6.02 x 10 23 particles 1 mole C 1 mole H 2 O 1 mole NaCl

9. 1. Number of atoms in 0.500 mole of Al a) 500 Al atoms b) 6.02 x 10 23 Al atoms c) 3.01 x 10 23 Al atoms Learning Check

10. Mole Relationships Mole Atoms or molecules Atoms or molecules Liters Grams 6.02 x 10 23 AtomicMassAtomicMass 22. 4 L 22.4 L

11. Molar Mass 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.

12. The Mass of 1 mole (in grams) Equal to the numerical value of the average atomic mass (get from periodic table) 1 mole of C atoms= 12.0 g 1 mole of Mg atoms =24.3 g 1 mole of Cu atoms =63.5 g Molar Mass

13. Other Names Related to Molar Mass Molecular Mass/Molecular Weight: If you have a single molecule, mass is measured in amu’s instead of grams. But, the molecular mass/weight is the same numerical value as 1 mole of molecules. Only the units are different. (This is the beauty of Avogadro’s Number!) Formula Mass/Formula Weight: Same goes for compounds. But again, the numerical value is the same. Only the units are different. THE POINT: You may hear all of these terms which mean the SAME NUMBER… just different units

14. Find the molar mass (usually we round to the tenths place) Learning Check! A.1 mole of Br atoms B.1 mole of Sn atoms =79.9 g/mole = 118.7 g/mole

15. Calculating Molecular Mass Calculate the formula mass of carbon dioxide, CO 2. 12.01 g + 2(16.00 g) = 44.01 g  One mole of CO 2 (6.02 x 10 23 molecules) has a mass of 44.01 grams

16. 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 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.

17. A.Molar Mass of K 2 O = ? Grams/mole B. Molar Mass of antacid Al(OH) 3 = ? Grams/mole Learning Check!

18. Prozac, C 17 H 18 F 3 NO, is a widely used antidepressant that inhibits the uptake of serotonin by the brain. Find its molar mass. Learning Check

19. Avogadro’s Law Equal volumes of gases at the same temperature and pressure have equal numbers of particles. At STP, it is a mole or 6.02 x 10 23 atoms or molecules depending on the gas. He or Ne would be atoms while CO 2 or F 2 would be molecules

20. 1) Volume: 1 mole of CO 2 at STP = ____________22.4 L 2) Number of molecules of CO 2 : 1 mole = ________ 6.02 x 10 23 molecules 3) Mass: 1 mole of CO 2 = _______________ 12.0 + 16.0 + 16.0 = 44 g/mol There are 3 things we know about a mole. For example: Using the gas, carbon dioxide (CO 2 ) we know its volume, number of molecules and mass. 1) Volume: 1 mole of N 2 at STP = _______________ 2) Number of molecules of N 2 : 1 mole = __________ 3) Mass: 1 mole of N 2 = ___________ What changes between these 2 gases? ____________ 22.4 L 6.02 x 10 23 molecules 14 + 14 = 28 g/mol mass only

21. 1. How many moles are in 25 grams of CO 2 2. How many grams would 42 moles of CO 2 mass? Using the mole formula (Found in Table T) Number of moles = given mass gram-formula mass

22. 3. How many moles are in 58 grams of N 2 ? 4. How many grams would 72 moles of N 2 mass?

23. 1. How many moles are in 25 grams of CO 2 2. How many grams would 42 moles of CO 2 mass? 3. How many moles are in 58 grams of N 2 ? 4. How many grams would 72 moles of N 2 mass? X = 25 g 44 g.56818…. =.57 mol 42= X 44 g 1848 g = 1800 g X = 58 g 28 g 2.071428…. = 2.1 mol 72 = X 28 2016 = 2.0 x 10 3 g

24. Mole Calculations I 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

25. Mole Calculations I 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

26. 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 × 10 23 particles = 1 mol = molar mass If we want to convert particles to mass, we must first convert particles to moles and than we can convert moles to mass.

27. Mass-Mole Calculations 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

28. Mole Calculations II 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 = 87.8 g Pb 2.55 × 10 23 atoms Pb × 1 mol Pb 6.02×10 23 atoms Pb 207.2 g Pb 1 mole Pb ×

29. Mole Calculations II 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 ×

30. Gas Density 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

31. Calculating Gas Density 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

32. Molar Mass of a Gas We can also use molar volume to calculate the molar mass of an unknown gas. 1.96 g of an unknown gas occupies 1.00L 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

33. Mole Unit Factors 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.

34. Mole-Volume Calculation 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

35. Mass-Volume Calculation 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

36. Molecule-Volume Calculation 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

37. Law of Definite Composition The law of definite composition states that “Compounds always contain the same elements in a constant proportion by mass”. Sodium chloride is always 39.3% sodium and 60.7% chlorine by mass, no matter what its source. Water is always 11.2% hydrogen and 88.8% oxygen by mass.

38. Law of Definite Composition A drop of water, a glass of water, and a lake of water all contain hydrogen and oxygen in the same percent by mass.

39. Chemical Formulas A particle composed of two or more nonmetal atoms is a molecule. A chemical formula expresses the number and types of atoms in a molecule. The chemical formula of sulfuric acid is H 2 SO 4.

40. Writing Chemical Formulas The number of each type of atom in a molecule is indicated with a subscript in a chemical formula. If there is only one atom of a certain type, no ‘1’ us used. A molecule of the vitamin niacin has 6 carbon atoms, 6 hydrogen atoms, 2 nitrogen atoms, and 1 oxygen atom. What is the chemical formula? C6H6N2OC6H6N2O

41. Interpreting Chemical Formulas Some chemical formulas use parenthesis to clarify atomic composition. Antifreeze has chemical formula C 2 H 4 (OH) 2. There are 2 carbon atoms, 4 hydrogen atoms, and 2 OH units, giving a total of 6 hydrogen atoms and 2 oxygen atoms. Antifreeze has a total of 10 atoms.

42. 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.

43. Calculating Percent Composition There are a few steps to calculating the percent composition of a compound. Lets 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

44. 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 × 100% = 11.2% H 16.00 g O 18.02 g H 2 O × 100% = 88.79% O

45. 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.

46. Percent Composition of TNT 84.07 g C 227.15 g TNT × 100% = 37.01% C 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

47. Empirical Formulas 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.

48. 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 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.

49. Calculating Empirical Formulas The molar mass of radium is 226.03 g/mol and the molar mass of oxygen is 16.00 g/mol. 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 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.

50. Empirical Formulas from Percent Composition We can also use percent composition data to calculate empirical formulas. Assume that you have 100 grams of sample. Benzene 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.

51. Empirical Formulas from Percent Composition 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 benzene 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

52. Molecular Formulas The empirical formula for benzene is CH. This represents the ratio of C to H atoms of benzene. The actual molecular formula is some multiple of the empirical formula, (CH) n. Benzene has a molar mass of 78 g/mol. Find n to find the molecular formula. = CH (CH) n 78 g/mol 13 g/mol n = 6 and the molecular formula is C 6 H 6.

53. Conclusions Avogadro’s number is 6.02 × 10 23 and is one 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.

54. Conclusions Continued We can use the following flow chart for mole calculations:

55. Conclusions 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.