1. THE MOLE

2. Atomic and molecular mass Masses of atoms, molecules, and formula units are given in amu (atomic mass units). Example: Sodium chloride: (22.99 amu + 35.45 amu =) 58.44 amu Sulfur dioxide: (32.07 amu + 2 x 16.00 amu =) 64.07 amu

3. The Mole The mole is a counting unit. 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

4. 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 1 mole of H 2 O will weigh 18.02 g, therefore the molar mass of H 2 O is 18.02 g

5. Keep in mind The masses of individual atoms or molecules are measured in AMU The mass of a mole of atoms or molecules are measured in GRAMS

6. Molar Volume At STP, 1 mol (6.02 x 10 23 rep. part.) 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?

8. Mass percent Measures the mass (by percent) of a single atom in a compound. Mass Percent = Element mass x 100 % Total mass

9. 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

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

11. 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 integer C 2 H 6 = (CH 3 ) 2 N 2 O 4 = (NO 2 ) 2 C 6 H 12 O 6 = (CH 2 O) 6

12. 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

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

14. 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

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

16. Calculations molecular formulas 1. Convert empirical formula to empirical formula 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

17. 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?

18. 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

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