1. SECTION 4.4 CONTINUED Molecular formula (True formula)

2.  For covalent compounds the empirical formula may or may not be the true molecular formula.  The molecular formula gives the actual number of atoms of each type of element found in one molecule of the compound

3. Empirical FormulaMolecular FormulaMolar Mass CH none13 C 2 H 2 (ethyne or acetylene) 2 X 13 = 26 C 6 H 6 (benzene)6 X 16 = 78 C 8 H 8 (styrene)8 X 13 = 104 CH 2 O CH 2 O (methanol)30 C 2 H 4 O 2 (ethanoic acid) 2 X 30 = 60 C 6 H 12 O 6 (glucose) 6 X 30 = 180

4.  If the molar mass found using the empirical formula is the same as the actual molar mass of the compound then the empirical formula is the molecular formula.  If they are not the same then the actual molar mass will be some whole number multiple of the empirical mass and the molecular formula will be the same whole number multiple of the empirical formula.

5. We will follow the following steps to find the molecular formula. 1. Write down the empirical formula and calculate the empirical mass. 2. Set up the following ratio. actual molar mass empirical mass 3. Multiply the subscripts in the empirical formula by the whole number from this ratio.

6. The empirical formula for a compound of sulfur and chlorine is SCl. The known molar mass of this compound 135 g/mol. Determine the molecular formula of the compound.

7. A compound contains 40% carbon, 6.70% hydrogen, and 53.3% oxygen by mass. The mass of 1 mole of the compound is 90.1 g. Determine the molecular formula of the compound ElementMassMolar Mass Number of moles RatioEmpirical formula

9.  A hydrated salt is a salt with water molecules attached to it. i.e. CuSO 4. 5H 2 O  An anhydrous salt is a salt with no water molecules attached to it. i.e. CuSO 4  Hydrated salt  anhydrous salt + water

10. When a 5.742 g of hydrated magnesium sulfate is heated until all the water has been released, 2.801 g of anhydrous magnesium sulfate remains. Determine the formula of hydrated magnesium.

11. MassMolar mass Number of moles RatioFormula

12. Carbon-Hydrogen analysis  A hydrocarbon is any compound that contains only hydrogen and carbon.  When a hydrocarbon undergoes complete combustion the only products will be carbon dioxide and water.  Hydrocarbon + oxygen  carbon dioxide + water

13. A sample of an unknown compound containing only C and H is burned in a C-H analyzer: 3.94 g of water and 9.62 g of CO 2 are produced. The measured molar mass is 84.0 g/mol. Determine the molecular formula of the compound C ? H ? + O 2  H 2 O + CO 2 3.94 g 9.62 g

16. ElementNumber of moles RatioFormula

18. Assignment  Lab prep is to be completed for Tuesday  The following questions must be completed for the day after the lab.  Page 159 practice problem  Page 162 practice problem  Page 16#5 1,3,4,6,7,8,10,11  Quiz on empirical and molecular formula on Wed. April 27