1. Chemistry 11 Unit 11

2. I.Introduction: Organic chemistry is the chemistry of CARBON compounds. The name “organic” refers to how many of these compounds are derived from living things, “organisms”. However, many organic compounds are synthetically produced. Plastics Synthetic fibres (& natural) Dyes & drugs Petroleum products Flavouring agents & many others

3. There are over 8 million known organic compounds Organic compounds are composed of COVALENT bonds. Other elements often found in organic compounds include: H O N P S Cl Other halogens Etc.

4. II.Carbon Bonding: Carbon atoms have 4 valence electrons and can form 4 bonds with up to 4 other atoms. C + 4 H How many bonds can O & N form? O + 2 H N + 3 H

5. Draw electron dot diagrams for the following: C2H6C2H6 ethane C2H4C2H4 ethene C2H2C2H2 ethyne

6. III.Empirical, Molecular & Structural Formula Empirical formula shows the smallest whole number ratio of atoms in a molecule. Molecular formula shows the actual number of each atom in a molecule. Structural formula shows the relative positions of each atom in a molecule. ethaneetheneethyne EmpiricalCH 3 CH 2 CH MolecularC2H6C2H6 C2H4C2H4 C2H2C2H2 StructuralCH 3 CH 2 CHCH

7. IV.Formula Calculations: 1.A charcoal briquette is composed of 43.2g of carbon. When it is burned it combines with oxygen to form a compound with a mass of 159.0g. What is the empirical formula of the resulting compound? 2.A 10.00 g sample of a compound is composed of 8.00g of carbon and 2.00g of hydrogen. The molar mass of the compound is 30g. What is the empirical and molecular formula of the compound? mass of C = 43.2g mass of O = 159.0 - 43.2 = 115.8g moles of C = 43.2g 1mol 12g = 3.60mol moles of O = 115.8g 1mol 16g = 7.24mol Ratio of C to O = 1 : 2 Empirical Formula = CO 2 moles of C = 8.00g 1mol 12g = 0.667mol moles of H = 2.00g 1mol 1.0g = 2.0mol 1 3 Empirical Formula = CH 3 Molar mass of compound = 30g/mol Molar mass of empirical formula = 15g/mol 15 x 2 = 30  Empirical formula must be doubled. Molecular Formula = C 2 H 6