In this chapter of IBDP Chemistry, we will have an overview of organic chemistry and use alkanes as an example.

Organic chemistry is the study of the compounds of Carbon. The simplest compound of Carbon is a hydrocarbon, which contains Hydrogen and Carbon only.

Alkanes are an example of a homologous series, which is a family of compounds in which all:

· obey a general formula

· differ from one to the next by a CH₂ unit

· have similar chemical properties and gradually changing physical properties

The empirical formula is the simplest whole number ratio of the atoms in a compound.

Eg. 1. Find the empirical formula of a compound which is 82.76% Carbon and 17.24% Hydrogen.

Carbon: 82.76/12 = 6.896

Hydrogen: 17.24/1 = 17.24

6.896/6.896 = 1 17.24/6.896 = 2.5 (then multiply by 2)

= C₂H₅

relative molecular mass = 29.

2. Given that it’s Mr is 58, find its molecular formula.

58/29 = 2 (C₂H₅) x2 = C₄H₁₀

Refer to IBDP Chemistry, the compound has two isomers – compounds with the same molecular formula but a different structure. Draw displayed formulae of the isomers.

Butane

This is a straight chain.

Methylpropane

This is a branched chain

The melting points and boiling points of alkanes increase with increasing chain length. The liquid state alkanes at room temperature become increasingly more dense.

C₁ – C₄ Gases at room temperature

C₅ – C₁₉ Liquids at room temperature

C₂₀ + Solids at room temperature

When it comes to IBDP Chemistry, All hydrocarbons are non-polar, the molecules are held together by Van der Waal forces. The longer the chain, the more Van der Waal forces there are between molecules, so more energy is needed to cause them to reach their boiling points and melting points.

When liquid molecules pack together tightly, there are more molecules in a smaller space, therefore the density increases.

Alkanes are unreactive because they are non-polar. However, they do burn.

Complete combustion produces CO₂ and H₂O.

· CH₄ + 2O₂ --> CO₂ + 2H₂O

· C₃H₈ + 5O₂ --> 3CO₂ + 4H₂O

· C₂H₆ + ⁷⁄₂ O₂ -->2CO₂ + 3H₂O

· C₈H₁₈ + ²⁵⁄₂ O₂ --> 8CO₂ + 9H₂O

Incomplete combustion can produce Carbon Monoxide (CO) and Carbon (soot) (C)

· C₆H₁₄ + ¹³⁄₂ O₂ --> 6CO + 7H₂O

· C₄H₁₀ + ⁵⁄₂ O₂--> 4C + 5H₂O

This is the end of this topic.

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