1. UNIT 9: THE MOLE VOCABULARY: representative particle mole Avogadro’s number molar mass percent composition empirical formula molecular formula hydrateanhydrousdilutionmolarity

2. 10.1 Measuring Matter Matter can be measured in 3 ways: Counting particles Mass Volume

3. The Mole We need a chemical unit that equals the number of atoms or molecules which has a mass in grams which will be numerically equal to the relative mass. The basic unit is the MOLE. It is the SI base unit used to measure the amount of a substance.

4. 1 mole = 6.02  10 23 particles The number is called Avogadro’s number. How large is a mole? 602,213,670,000,000,000,000,000 A mole of pennies would be $602,213,670,000,000,000,000,000 six hundred and two sextillion, two hundred and thirteen quintillion, six hundred and seventy quadrillion dollars six sextillion, twenty-two quintillion, one hundred thirty-six quadrillion, seven hundred trillion dollars 6.02 x 1023 particles of matter

5. 6.02 x 10 23 = Avogadro’s number The number of representative particles in one mole of a pure substance Representative particles = Atom / Ions Molecule (covalent compounds) Formula unit (ionic compounds)

6. ONE MOLE OF Fe = 6.02 x 10 23 atoms H 2 O = 6.02 x 10 23 molecules CaCl 2 = 6.02 x 10 23 formula units

7. MOLESREPRESENTATIVE PARTICLE Conversion factor = 6.02 x 10 23 particles 1 mole How many atoms are in 2.00 mol of argon?

8. How many molecules are in 7.25 mol of carbon dioxide? How many formula units are in 3.15 mol of sodium oxide?

9. Representative particles moles Use the inverse of Avogadro’s number as the conversion factor.

10. Representative particles moles How many moles are in 1.62 x 10 23 atoms of argon? How many moles are in 4.35 x 10 24 molecules of dinitrogen trifluoride?

11. Molar Mass The molar mass of any element is numerically equivalent to its atomic mass and has the units g/mol. Mass of one mole of a substance g/mol Mass of atoms is found on the periodic table H atom = 1.01 g He atom = 4.00 g Mg atom = 24.3 g O atom = 16.0 g K = Mass of compounds must be calculated H 2 0 CO 2

12. The molar mass of a compound equals the molar mass of each element, multiplied by the moles of that element in the chemical formula, added together. “ADD ‘M UP”

13. H 2 0 = H2 x 1.00= 2.00 gO1 x 16.0= 16.0 g 18.0 g/mol CO 2 = C + 2(O) = 12.0 + 2(16.0) = 44.0 g/mol

14. Calculating Molar Mass What is the molar mass of barium acetate? What is the molar mass of ammonium sulfate?

15. Mass Moles How many moles are in 12.5 g of carbon dioxide? How many moles are in 83.2 g of H 2 SO 4 ? How many moles are in 12.5 g of carbon dioxide? How many moles are in 83.2 g of H 2 SO 4 ?

16. MolesMass What is the mass in grams of 1.5 mol of sodium chloride? What is the mass in grams of 0.774 mol of oxygen? What is the mass in grams of 1.5 mol of sodium chloride? What is the mass in grams of 0.774 mol of oxygen?

17. Molar Volume STP Standard temperature and pressure T = 0 o C P = 1 atmosphere (atm) STP Standard temperature and pressure T = 0 o C P = 1 atmosphere (atm)

18. At STP 1 mole of gas occupies a volume of 22.4L 22.4L = molar volume of a gas 22.4L of a gas contains 6.02 x 10 23 particles of that gas. At STP 1 mole of gas occupies a volume of 22.4L 22.4L = molar volume of a gas 22.4L of a gas contains 6.02 x 10 23 particles of that gas.

19. Moles Volume What is the volume of 1.50 mol of nitrogen at STP? What is the volume of 3.75 mol of helium at STP What is the volume of 1.50 mol of nitrogen at STP? What is the volume of 3.75 mol of helium at STP

20. Volume Moles How many moles are in 75.3 L of water vapor at STP? How many moles are in 250 L of nitrogen dioxide at STP? How many moles are in 75.3 L of water vapor at STP? How many moles are in 250 L of nitrogen dioxide at STP?

21. A Little Review: H atom = 1.01 amu 1 mole of H atom = 1.01 gram = 6.02  10 23 H atoms molecule for covalent compounds O 2 molecule = 32.0 amu O 2 molecule = 32.0 amu 1 mole of O 2 = 32.0 g = contains 6.02  10 23 O 2 molecules H 2 O molecule = 18.0 amu 1 mole of H 2 O = 18.0 grams = 6.02  10 23 H 2 O molecules formula unit for ionic compounds formula unit NaCl = 23.0 35.5 35.5 58.5 amu 58.5 amu 1 mole of NaCl = 58.5 g = 6.02  10 23 NaCl formula units

22. Converting Between Units Must ALWAYS go to moles first

23. Moles ParticlesVolumeMassParticlesVolumeMass

24. Given Unit → Mole → Wanted Unit Conversions Conversion factors must be used. Example 1. How many atoms are there in 3.60 liters of Ne? ? atoms = 3.60 literss 3.60 L Ne x 1 mole x 6.02 x 10 23 atoms = 22.4 L 1 mole 22.4 L 1 mole

25. Example 2. What is the mass of 5.0 x 10 22 atoms of potassium? ? g = 5.00 moles K ? g = 5.00 moles K 5.00 atoms K x 1 mole_________ x 39.1 g = g 5.00 atoms K x 1 mole_________ x 39.1 g = g 6.02 x 10 23 atoms 1 mole 6.02 x 10 23 atoms 1 mole Example 3. How many molecules are present in 77.5 g of carbon dioxide? ? molecules 77.5 g CO2 ? molecules 77.5 g CO2 77.5 g CO2 x 1 mole x 6.02 x 10 23 molecules = molecule 44.0 g 1 mole 44.0 g 1 mole add 12.0 carbon 32.0 oxygen 32.0 oxygen 44.0 g

26. Chemical Formulas and the Mole Chemical formulas indicate the numbers and types of atoms contained in one unit of the compound.Chemical formulas indicate the numbers and types of atoms contained in one unit of the compound. One mole of CCl 2 F 2 contains one mole of C atoms, two moles of Cl atoms, and two moles of F atoms.One mole of CCl 2 F 2 contains one mole of C atoms, two moles of Cl atoms, and two moles of F atoms. One mole of CCl 2 F 2 contains one carbon atom, two Chlorine atoms, and two Fluorine atoms.One mole of CCl 2 F 2 contains one carbon atom, two Chlorine atoms, and two Fluorine atoms. iron(III)sulfate is Fe2(SO 4 ) 3 2 moles of Fe and 3 moles of SO 4 1 mole Fe 2 (SO 4 ) 3 1 mole Fe 2 (SO 4 ) 3 1 mole Fe 2 (SO 4 ) 3 1 mole Fe 2 (SO 4 ) 3

27. Example 4. How many moles of Fe 3+ ions and SO 4 2- ions are present in 4.68 moles of iron(III)sulfate? Fe2(SO 4 ) 3 ? moles Fe 3+ = 4.68 moles Fe 2 (SO 4 ) 3 x 2 moles of Fe 1 mole Fe 2 (SO 4 ) 3 1 mole Fe 2 (SO 4 ) 3 9.36 moles of Fe 3+ ? moles SO 4 2- = 4.68 moles Fe 2 (SO 4 ) 3 x 3 moles of SO 4 2- 1 mole Fe 2 (SO 4 ) 3 1 mole Fe 2 (SO 4 ) 3 14.0 moles SO 4 2- 14.0 moles SO 4 2-

28. H atom = 1.01 amu 1 mole of H atom = 1.01 gram = 6.02  10 23 H atoms O 2 molecule = 32.0 amu O 2 molecule = 32.0 amu 1 mole of O 2 = 32.0 g = contains 6.02  10 23 O 2 molecules H 2 O molecule = 18.0 amu 1 mole of H 2 O = 18.0 grams = 6.02  10 23 H 2 O molecules formula unit for ionic compounds formula unit NaCl = 23.0 35.5 35.5 58.5 amu 58.5 amu 1 mole of NaCl = 58.5 g = 6.02  10 23 NaCl formula units

29. 10.2 PERCENT COMPOSITION The percent by mass of any element in a compound can be found by dividing the mass of the element by the mass of the compound and multiplying by 100.The percent by mass of any element in a compound can be found by dividing the mass of the element by the mass of the compound and multiplying by 100.  Find the total molar mass of each element in the compound.  Find the molar mass of the entire compound.  Divide the total molar mass of each element by the molar mass of the compound then multiply by 100  Check that all your percentages add up to 100

30. % composition of C 3 F 6 Total mass of C = total mass of F = Mass of compound C 3 F 6 = % of C = % of F = Check your work!

31. What is the percent composition of H2SO4?

32. 10.3 EMPIRICAL FORMULA AND MOLECULAR FORMULA The empirical formula for a compound is the smallest whole-number mole ratio of the elements. Molecular formula – gives the actual number of each kind of atom in a molecule Molecular formula of hydrogen peroxide = H 2 O 2 Molecular formula is always a whole- number multiple of the empirical formula. Empirical formula of hydrogen peroxide = HO

33. You can calculate the empirical formula from percent by mass by assuming you have 100.00 g of the compound. Then, convert the mass of each element to moles. The empirical formula may or may not be the same as the molecular formula. The molecular formula specifies the actual number of atoms of each element in one molecule or formula unit of the substance.

34. Calculating Empirical Formulas 1) Change % to g (assume 100 g of compound, so 30% = 30 g) 2) Convert each element from g to moles 3) Divide each mole amount by the smallest number from step 2 4) Change to a whole number = subscript in empirical formula

35. Example: 48.64% C, 8.16% H, 43.20% O 48.64g C x 1 mol C 12.01 g C 12.01 g C 8.16g H x 1 mol H 1.008 g H 1.008 g H 43.20g O x 1 mol O 16.00 g O 16.00 g O

36. 4.050 mol C = 1.5 mol C 2.700 8.10 mol H = 3 mol H 2.700 2.700 mol O = 1 mol O 2.700

37. The simplest mole ratio is The simplest mole ratio is (1.5 mol C), (3 mol H), (1 mol O) (1.5 mol C), (3 mol H), (1 mol O) Multiply each number in the ratio-in this case 2-that yields a ratio of whole numbers. 1.5 mol C x 2 = 3 mol C 3 mol H x 2 = 6 mol H 1 mol O x 2 = 2 mol O The simplest whole number ratio is 3 atoms C, 6 atoms H, 2 atoms O. Thus, the formula is C 3 H 6 O 2

38.  Example: 30.5% n and 69.5% O g of N = g of O = Convert to moles Divide Formula =

39. Example: 43.7 % P and 56.3 % O

40. MOLECULAR FORMULAS To find the molecular formula you must know -Empirical formula -Molar mass of compound Can be the same as the empirical formula but often is not

41.  Calculating molecular formula Calculate the mass of the empirical formula Divide molar mass of compound by mass of empirical formula Multiply each subscript in the empirical formula by the answer from step 2

42. Example: empirical = NO 2, molar mass = 92 g/mol  Mass of NO 2 =   Molecular formula =  Mass of NO 2 =   Molecular formula =

43. An unknown compound contains 58.5% carbon, 9.8% hydrogen, and 31.4% oxygen. Its molar mass is 102 g/mol. What are its empirical and molecular formulas? An unknown compound contains 58.5% carbon, 9.8% hydrogen, and 31.4% oxygen. Its molar mass is 102 g/mol. What are its empirical and molecular formulas?

44. 10.4 Hydrates A hydrate is a compound that has a specific number of water molecules bound to its atoms.A hydrate is a compound that has a specific number of water molecules bound to its atoms. The number of water molecules associated with each formula unit of the compound is written following a dot.The number of water molecules associated with each formula unit of the compound is written following a dot. Sodium carbonate decahydrate = Na 2 CO 3 10H 2 O

45. PrefixMolecules H 2 OFormulaName Mono-1 (NH 4 ) 2 C 2 O 4. H 2 O Di-2 CaCl 2. 2H 2 O Tri-3 NaC 2 H 3 O 2. 3H 2 O Tetra-4 FePO 4.. 4H 2 O Penta-5 CuSO 4.. 5H 2 O Hexa-6 CoCl 2. 6H 2 O Hepta-7 MgSO 4.. 7H 2 O Octa-8 Ba(OH) 2.. 8H 2 O Nona-9. 9H 2 O Deca10 Na 2 CO 3.. 10H 2 O

46. Analyzing a Hydrate When heated, water molecules are released from a hydrate leaving an anhydrous compound.When heated, water molecules are released from a hydrate leaving an anhydrous compound. To determine the formula of a hydrate, find the number of moles of water associated with 1 mole of hydrate.To determine the formula of a hydrate, find the number of moles of water associated with 1 mole of hydrate.

47. Analyzing a Hydrate Weigh hydrate.Weigh hydrate. Heat to drive off the water.Heat to drive off the water. Weigh the anhydrous compound.Weigh the anhydrous compound. Subtract and convert the difference to moles.Subtract and convert the difference to moles. The ratio of moles of water to moles of anhydrous compound is the coefficient for water in the hydrate.The ratio of moles of water to moles of anhydrous compound is the coefficient for water in the hydrate.

48. Use of Hydrates Anhydrous forms of hydrates are often used to absorb water, particularly during shipment of electronic and optical equipment.Anhydrous forms of hydrates are often used to absorb water, particularly during shipment of electronic and optical equipment. In chemistry labs, anhydrous forms of hydrates are used to remove moisture from the air and keep other substances dry.In chemistry labs, anhydrous forms of hydrates are used to remove moisture from the air and keep other substances dry.

49. 10.5 MOLARITY One way to represent concentration (how much solute is present in a given amount of solution). Molarity (M) = the moles of solute present in a liter of solution 6 moles HCl moles of solute 6 moles HCl 0.500 L of solution L of solution 1 L solution = 12 M = 6 M (6 molar)

50. Dilution and Volumes of Solutions Required for Reactions Dilution is the process of making a solution less concentrated by adding some of the solution to more solvent. In all dilutions, the number of moles of solute remains the same. After dilution, the new molarity is simply the number of moles of solute present divided by the total volume of solution.

51. The number of moles of solute in a given volume of a solution of known concentration can always be determined by multiplying the volume times the molarity. liters x moles solute = moles solute liter liter Volume x Molarity = amount of solute

52. Problems using Molarity 1) What is the Molarity of a solution which contains 29.3 g NaCl in 250 mL of solution? ?M = moles of solute = 29.3 g NaCl x 1 mole = 2.00 M L of solution.250 L solution 58.5 g L of solution.250 L solution 58.5 g 2) How many moles of KNO 3 are contained in 300 mL in 0.500 M solution? ? moles = 0.300 L x 0.500 moles = 0.150 mole 1 L 1 L 3) How many grams of KNO 3 are needed to prepare 500 mL of a 0.200 M solution? ? g = 0.500 L x 0.200 mole x 101.1 g = 10.1 grams KNO 3 1 L 1 mole 1 L 1 mole

53. What is the concentration of a solution with a volume of 2.5 liters containing 660 grams of calcium phosphate? 0.85 M How many grams of potassium carbonate are needed to make 200 mL of a 2.5 M solution? 69.1 g