Chemistry 11 Unit 11

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

• 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.

II. Carbon Bonding:

• Carbon atoms have 4 valence electrons and can form 4 bonds with up to 4 other atoms.

C + 4 H C

H

H H

H• How many bonds can O & N form?

O + 2 H O H

H

N + 3 H H N H

H

Draw electron dot diagrams for the following:

C2H6

ethane

C2H4

ethene

C2H2

ethyne

H C C

H

H H

H

H C CH

H H

HC CH H

C CH HC CH

H

H

H

H

H C CH

H H

H

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.

ethane ethene ethyne

Empirical CH3 CH2 CH

Molecular C2H6 C2H4 C2H2

Structural CH3CH3 CH2CH2 CHCH

H C C

H

H H

H

H C CH

H H

H

C CH H

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 = 43.2gmass 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 : 2Empirical Formula = CO2

moles of C = 8.00g 1mol 12g

= 0.667mol

moles of H = 2.00g 1mol 1.0g

= 2.0mol

1

3Empirical Formula = CH3

Molar mass of compound = 30g/molMolar mass of empirical formula = 15g/mol

15 x 2 = 30 Empirical formula must be doubled.

Molecular Formula = C2H6