1. What is a Mole? A mole is a specific amount of atoms. Atoms are very, very, very small. The fact that atoms are so small makes it hard to take measurements (such as mass, volume, density, etc.) of just one atom. So, instead of measuring individual atoms, we will group atoms into what we call a MOLE. 1 Mole = 602 billion trillion atoms or (6.02 x 10 23 atoms) Avogadro’s Number It’s the same as saying: 1 mile = 5,280 feet 1 cup = 8 ounces 1 foot = 12 inches

2. Moles don’t always have to be referring to atoms. They can also be referring to molecules, compounds, ions, or other particles. Just remember, 1 mole of anything is equal to 6.02 x 10 23. Question: What number of chloride ions are present if there are 2 moles of chloride ions? 1 mole = 6.02 x 10 23 ions 2 moles of chloride ions x 6.02 x 10 23 chloride ions = 1.204 x 10 24 chloride ions 1 mole of chloride ions

3. Question: Determine the formula and molar mass of sodium oxide. Formula: Na 2 O Formula Mass The mass of one compound of Na 2 O Molar Mass The mass of one mole of Na 2 O Na: 23 amu (atomic mass units) O: 16 amu 2 Na = 2 (23 amu) = 46 amu 1 O = 1 (16 amu) = +16 amu 62 amu Note: 1amu = 1.66 x 10 -24 grams 1.03 x 10 -22 g 1 mole of Na: 23 grams 1 mole of O: 16 grams 2 moles of Na = 2 (23 g) = 46 grams 1 mole of O = 1 (16 g) = 16 grams 62 grams *We can use our measuring equipment to achieve this value.

4. So, we realize now that when the atomic mass of an element is reported in AMU’s it is referring to 1 atom of that element. If the mass is reported in grams it is referring to 1 mole of atoms of that element. 4 Be 9.012 9.0 grams is the mass of 1 mole of Be atoms.

5. How many atoms of oxygen are in this room if this room contains 0.5 moles of oxygen atoms? Answer:.5 moles.5 moles O x 6.02 x 10 23 atoms of O = 3.01 x 10 23 atoms 1 of O

6. In order to perform a proper fireworks show, 3.5 x 10 24 formula units of CaC 2 must be used. Determine the number of moles of CaC 2 and then determine the mass of CaC 2 needed. 3.5 x 10 24 FU’s CaC 2 x 1 mole_CaC 2 = 6.02 x 10 23 FU’s CaC 2 5.81 moles CaC 2 5.81 moles CaC 2 x (40.0 g + 12.0 g + 12.0 g) 1 mole of CaC 2 = 371.8 grams CaC 2

7. Example: Fe CO CuCl Atom Molecule Formula Unit (Also called a molecular compound) (Also called an ionic compound)

8. Determine the molar mass of Copper (II) Chloride. Example: Step 1: Write the compound formula. Step 2: Using the periodic table, write down the molar mass of each element in the compound. (Round the mass to the nearest tenth) Step 3: Add together the molar masses of each element. CuCl 2 63.5 g35.5 gX 2 = 71 g 63.5 g + 71 g = 134.5 grams

9. Mass Moles Atoms/ Molecules/ Formula units ÷ molar mass ÷ 6.02 x 10 23 x molar mass x 6.02 x 10 23 This flow chart can be used to do any mole-mass, mass-mole, mole-molecule, molecule-mole, mass-molecule, or molecule- mass conversion. See the next few examples for details Gas Volume ÷ 22.4 x 22.4

10. Example: Determine the mass of 5.5 moles of Potassium Carbonate. Step 1: Write the formula for the compound. Step 2: Determine the molar mass of the compound. Step 3: Multiply the moles given by the molar mass. K 2 CO 3 (39.1gX 2)12.0g(16.0g X 3) + = 138.2 grams X 138.2g K 2 CO 3 5.5 moles K 2 CO 3 1 mole K 2 CO 3 = 760.1 grams K 2 CO 3

11. Example: Determine the number of moles in 34.5 grams of Potassium Carbonate. Step 1: Write the formula for the compound. Step 2: Determine the molar mass of the compound. Step 3: Divide the mass given by the molar mass. K 2 CO 3 138.2 grams 34.5 grams K 2 CO 3 X 1mole K 2 CO 3 138.2 grams K 2 CO 3 =.25 moles K 2 CO 3

12. Example: Determine the number of molecules of CO 2 in 2.5 moles of CO 2. Step 1: Multiply the number of moles by 6.02 x 10 23. 2.5 moles CO 2 X 6.02 x 10 23 molecules CO 2 1.0 mole CO 2 = 1.51 x 10 24 molecules CO 2

13. Example: Determine the number of moles of Carbon dioxide if you have 3.55 x 10 23 molecules of Carbon dioxide. Step 1: Divide the number of molecules by 6.02 x 10 23. 3.55 x 10 23 molecules CO 2 X 1 mole CO 2 6.02 x 10 23 molecules CO 2 =.59 moles CO 2

14. Example: Determine the number of molecules of carbon monoxide in 75.0 grams of carbon monoxide. Step 1: Write the formula for the compound. Step 2: Determine the molar mass of the compound. Step 3: Divide the mass given by the molar mass of the compound. Step 4: Multiply moles by 6.02 x 10 23. CO 12.0 g16.0g+= 28.0g 75.0g CO x 1 mole CO 28.0 grams CO = 2.7 moles CO X 6.02 molecules CO 1 mole CO = 1.6 x 10 24 molecules CO

15. Practice Problem: Determine the mass of 2.5 x 10 23 formula units of Calcium Oxide. Show Work: 2.5 x 10 23 formula units of CaO X 1 mole CaO 6.02 x 10 23 formula units of CaO =.42 moles CaO.42 moles CaO X 56.1 grams CaO 1 mole CaO = 23.6 grams CaO

16. Percent Composition of a compound is a statement of the relative mass each contributes to the mass of the compound as a whole. Example: The percent composition of water is: 89% Oxygen 11% Hydrogen How did we get these values? Answer: H2OH2O 2.0 grams16.0 grams+ =18.0 grams Oxygen: 16g/18g =.89 or 89% Hydrogen: 2g/18g =.11 or 11%

17. An empirical formula of a compound is the simplest ratio of the elements in that compound. For example, the compound hydrogen peroxide has a molecular formula of H 2 O 2. Thus, the empirical formula for hydrogen peroxide is HO. Example Problem: Determine the empirical formula of a compound that contains.90 grams of calcium and 1.6 grams of chlorine. Step 1: Convert grams of each element into moles. Step 2: Divide each number of moles by the lowest value. Step 3: Use final values for the subscripts in the formula..90 grams Ca 1.6 grams Cl X 1 mole Ca X 1 mole Cl 40.1 grams Ca 35.5 grams Cl =.02244 moles Ca =.04507 moles Cl.02244 = 1 mole Ca = 2 moles Cl CaCl 2

18. A molecular formula shows the actual number of atoms of each element in a molecule, as well as the ratio of atoms. Example: Determine the molecular formula of a compound that has a percentage composition of 40.9% carbon, 4.58% hydrogen, and 54.8% oxygen. Its molar mass is 176.1 grams. Step 1: Assume the mass of the compound is 100 grams. This would make 4.58% become 4.58 grams. Step 2: Convert grams of each element into moles. Step 3: Divide each mole value by the lowest value. Step 4: Determine the empirical formula. Step 5: Calculate the mass of the empirical formula. Step 6: Divide the molar mass of the compound by the mass of its empirical formula. Multiply all subscripts by the answer you get from step 6.

19. Example Problem (Continued) 40.9g C 4.58g H 54.8g O X 1 mole C X 1 mole H X 1 mole O 12.0 g C 1.0 g H 16.0 g O = 3.4 moles C = 4.58 moles H = 3.425 moles O 3.4 moles = 1 mole C = 1.35 mole H = 1 mole O X 3 = 3 = 4 = 3 Empirical Formula: C 3 H 4 O 3 Mass = 88 grams Molar Mass = 176.1 g = 2 Molecular Formula: C 6 H 8 O 6 (Ascorbic Acid or Vitamin C)