In petroleum refining, catalytic cracking breaks down long-chain hydrocarbons into smaller… — JAMB UTME

In petroleum refining, catalytic cracking breaks down long-chain hydrocarbons into smaller, economically valuable fractions. If one mole of dodecane ( $C_{12}H_{26}$ ) undergoes thermal-catalytic cracking to yield one mole of hexane ( $C_6H_{14}$ ) and two moles of an alkene product $X$ , what is the molecular formula of $X$ ?

$C_3H_6$ Answer
B
$C_3H_8$
C
$C_6H_{12}$
D
$C_6H_{14}$

Answer

The molecular formula of alkene

X

C_3H_6

(propene).

In catalytic cracking, atomic mass is conserved. Subtracting the atoms of hexane (

C_6H_{14}

) from dodecane (

C_{12}H_{26}

) leaves 6 carbon atoms and 12 hydrogen atoms. Since 2 moles of product

X

are formed (

2X = C_6H_{12}

), dividing by 2 yields

C_3H_6

, which is propene, a valid alkene.

Step-by-Step Solution

Write the balanced stoichiometric equation for the cracking reaction.

C_{12}H_{26} \rightarrow C_6H_{14} + 2X

Cracking conserves the total number of carbon and hydrogen atoms between reactants and products.

Determine the remaining number of carbon and hydrogen atoms allocated to

2X

Carbons:

12 - 6 = 6

, Hydrogens:

26 - 14 = 12

. Total fragment formula is

C_6H_{12}

Subtract the atoms present in one mole of hexane from dodecane.

Divide the carbon and hydrogen count by 2 to find the formula of 1 mole of product

X

Carbon count for

X = \frac{6}{2} = 3

, Hydrogen count for

X = \frac{12}{2} = 6

. Formula of

X = C_3H_6

Since two moles of alkene

X

are produced, each mole contains half the remaining atoms, fitting the general formula for alkenes (

C_nH_{2n}

Key Concept

Catalytic Cracking Stoichiometry of Alkanes

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