1. 1 The Mole 6.02 X 10 23

2. 2 What is “The Mole??” A counting unit (similar to a dozen) 6.02 X 10 23 (in scientific notation) = 602 billion trillion = 602,000,000,000,000,000,000,000 Amedeo Avogadro (1776 – 1856)This number is named in honor of Amedeo Avogadro (1776 – 1856)

3. 3 A mole is like a dozen. 1. How many paper clips in 1 dozen? a) 1b) 4c) 12 2. How many oranges in 2.0 dozen? a) 4b) 12c) 24 3. How many dozen contain 36 donuts? a) 3b) 12c) 36 Learning Check

4. 4 = 6.02 x 10 23 C atoms = 6.02 x 10 23 Na atoms A Mole of Particles A Mole of Particles Contains 6.02 x 10 23 particles 1 mole C 1 mole Na

5. 5 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!

6. 6 Avogadro’s number The number of particles in 1 mole is called Avogadro’s Number (6.02 x 10 23 ) This is the magic number that allows us to turn atomic masses (amu) into grams

7. 7 Atomic Masses Unit is the amu. –atomic mass unit –1 amu = 1.66 x 10 -24 g We define the masses of atoms and molecules in terms of atomic mass units. Amu’s can be found on the Periodic Table. –1 Carbon atom = 12.01 amu –1 Zinc atom = 65.39 amu

8. 8 Molecular masses Molecular mass of a molecule = atomic masses of all the atoms of that molecule –1 O 2 molecule = 2(16.00 amu) = 32.00 amu –1 MgCl 2 = 24.31 + 2(35.45) = 95.21 amu Atomic mass, molecular mass, and formula mass are ALL measured in amu.

9. 9 Practice Calculate the molecular mass (in amu) of the following compounds: Sodium chloride (NaCl): (22.99 amu + 35.45 amu =) 58.44 amu Sulfur dioxide (SO 2 ): (32.07 amu + 2 x 16.00 amu =) 64.07 amu Iron(III) hydroxide (Fe(OH) 3 ): (55.85 amu + 3 x (16.00 amu + 1.01 amu)) = 106.87 amu

10. 10 Molar Mass The molar mass is the mass in grams of one mole of a compound The molecular mass can be calculated from atomic masses: water = H 2 O = 2(1.01 amu) + 16.00 amu = 18.02 amu

11. 11 1 molecule has a molecular mass of 18.02 amu 1 mole of H 2 O will weigh 18.02 g, therefore the molar mass of H 2 O is 18.02 g

12. 12 1 mole of C atoms (= 12.01 amu) = 12.01 g 1 mole of Zn atoms (= 65.39 amu) = 65.39 g 1 mole of O 2 molecules (= 32.00 amu) = 32.00 g 1 mole of MgCl 2 formula unit (= 95.21 amu) = 95.21 g Notice:Notice: When you just have a SINGLE element, the molar mass = atomic mass

13. 13 IMPORTANT!! The masses of individual atoms or molecules are measured in AMU The mass of a mole of (6.02 x 10 23 ) atoms or (6.02 x 10 23 ) molecules are measured in GRAMS

14. 14 6.02 x 10 23 particle 1 mole or 1 mole 6.02 x 10 23 particle Note that a particle can be an atom or a molecule Avogadro’s Number as Conversion Factor

15. 15 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 2.Number of moles of S in 1.8 x 10 24 S atoms a) 1.0 mole S atoms b) 3.0 mole S atoms c) 1.1 x 10 48 mole S atoms Learning Check 0.500 mol x 6.02 x 10 23 atoms = 1 mol 1.8 x 10 24 atoms x 1 mol_____ = 6.02 x 10 23 atoms

16. 16 molar mass (g) of atom 1 mole or 1 mole molar mass (g) of atom Molar Mass as Conversion Factor

17. 17 Aluminum is often used for the structure of light-weight bicycle frames. How many grams of Al are in 3.00 moles of Al? 3.00 moles Al ? g Al Converting Moles and Grams

18. 18 1. Molar mass of Al1 mole Al = 27.0 g Al 2. Conversion factors for Al 27.0g Al or 1 mol Al 1 mol Al 27.0 g Al 3. Setup3.00 moles Al x 27.0 g Al 1 mole Al Answer = 81.0 g Al Aluminum is often used for the structure of light-weight bicycle frames. How many grams of Al are in 3.00 moles of Al? REMEMBER: Use the conversion factor that has the given unit ON THE BOTTOM!

19. 19 ÷ molar mass ÷ Avogadro’s number Grams Moles Particles x molar mass x Avogadro’s number Everything must go through Moles!!! Calculations

20. 20 Atoms/Molecules and Grams How many atoms of Cu are present in 35.4 g of Cu? = 3.36 X 10 23 atoms Cu

21. 21 Learning Check! How many atoms of K are present in 78.4 g of K? = 1.21 x 10 24 atoms K

22. 22 Practice How many grams are 4.25 x 10 23 molecules of CH 4 ? How many moles are there in 27.2 g of potassium oxide?

23. 23 Molar Volume At STP, 1 mol (6.02 x 10 23 particles) of ANY gas occupies a volume of 22.4 L. –The quantity 22.4 L is called the molar volume of a gas. –STP = standard temperature and pressure 0 o C and 1 atm What volume will 0.375 mol of O 2 gas occupy at STP?

24. 24

25. 25 Mass Percent Measures the mass (by percent) of a single atom in a compound. Mass Percent = Element mass x 100 % Total mass Ex. – A 35.0 g sample contains 11.2 g of potassium. What is the mass percent of potassium?

26. 26 Percent Composition Percentage of each element in a compound –By mass Can be determined from: -the formula of the compound or -the experimental mass analysis of the compound The percentages may not always total to 100% due to rounding

27. 27

28. 28 Examples 1. A 23.9 g sample contains 14.3 g C, 3.2 g H, and 6.4 g O. What is the % composition? 2. What is the % composition of calcium chloride?

29. 29 Formulas Empirical formula: formula in its simplest form; gives the lowest ratio of the atoms in a compound. Molecular formula: actual formula of a compound.

30. 30 Examples C 2 H 6 : Empirical formula = CH 3 N 2 O 4 : Empirical formula = NO 2 C 6 H 12 O 6 : Empirical formula = CH 2 O

31. 31 Example An unknown compound contains 0.426 grams of carbon and 0.107 grams of hydrogen. Determine the empirical formula.

32. 32 Calculations 1. Determine the amount in grams of all the elements in the compound 2. Convert those grams into moles for each element 3. Divide the number of moles by the smallest and round up or down 4. Turn those rounded numbers into whole numbers when necessary

33. 33 C: 0.426 g / 12.01 g/mol = 0.0355 mol H: 0.107 g / 1.008 g/mol = 0.106 mol C: 0.0355 / 0.0355 = 1.00; round to 1 H: 0.106 / 0.0355 = 2.99; round to 3 Empirical formula: CH 3

34. 34 Percentages Convert percentages in grams assuming 100 % = 100 grams Treat as grams problems

35. 35 Practice A compound is made out of 30.4 % of nitrogen and 69.6 % of oxygen. Determine its empirical formula.

36. 36 N: 30.4 % = 30.4 g; 30.4 g / 14.01 g/mol = 2.17 mol O: 69.6 % = 69.6 g; 69.6 g / 16.00 g/mol = 4.35 mol N: 2.17 / 2.17 = 1.00 O: 4.35 / 2.17 = 2.00 Empirical formula: NO 2

37. 37 Formulas Empirical formula: formula in its simplest form. Molecular formula: actual formula of a compound.

38. 38 Examples C 2 H 6 : Empirical formula = CH 3 N 2 O 4 : Empirical formula = NO 2 C 6 H 12 O 6 : Empirical formula = CH 2 O Molecular formula = (empirical formula) n where n is whole number C 2 H 6 = (CH 3 ) 2 N 2 O 4 = (NO 2 ) 2 C 6 H 12 O 6 = (CH 2 O) 6

39. 39 Example problem The empirical formula of a compound is C 2 H 5 and its molar mass is 58.12 g/mol. What is the molecular formula of this compound?

40. 40 Calculations 1. Convert empirical formula to empirical formula molar mass by using the molar mass 2. Determine n by dividing the actual molar mass by the empirical formula molar mass 3. Molecular formula = n x empirical formula

41. 41 C 2 H 5 : empirical molar mass = 2 x 12.01 + 5 x 1.008 = 29.06 g/mol n = actual molar mass / empirical molar mass = 58.12 g/mol / 29.06 g/mol = 2 Molecular formula = 2 x empirical formula = C 4 H 10

42. 42 Percentages When percentages are given, first calculate the empirical formula. Use this empirical formula to calculate the molecular formula.

43. 43 Practice A compound is made out of 30.4 % of nitrogen and 69.6 % of oxygen. Determine its empirical formula. The molar mass of the compound is: 92.02 g/mol Determine the actual formula.

44. 44 Molar mass of empirical formula: (1 x 14.01 g/mol + 2 x 16.00 g/mol) = 46.01 g/mol n = 92.02 g/mol / 46.01 g/mol = 2 The actual formula is: (NO 2 ) x 2 = N 2 O 4