# Question #8bc4e

Jan 1, 2016

Indeed, the compound's molecular formula is ${\text{C"_16"H}}_{34}$.

#### Explanation:

All you have to do here is use the concept of mass conservation to figure out how many moles of carbon and how many moles of hydrogen were a part of the original compound.

As you know, cracking is simply a process used to split complex organic compounds into simpler molecules by breaking carbon - carbon bonds.

This means that after the cracking takes place, the number of atoms that were a part of compound $\text{Y}$ must now be a part of the three listed products.

Your strategy here will be to list the products that result from the cracking of your compound $\text{Y}$

• two moles of ethene, ${\text{C"_2"H}}_{4}$
• one mole of 1-butene, ${\text{C"_4"H}}_{8}$
• one mole of octane, ${\text{C"_8"H}}_{18}$

and figure out how many moles of carbon and of hydrogen were produced. You will have

$\text{For C: " overbrace(2 xx 2)^(color(red)("from ethene")) + overbrace(1 xx 4)^(color(blue)("from 1-butene")) + overbrace(1 xx 8)^(color(green)("from octane")) = "16 moles C}$

$\textcolor{w h i t e}{x}$

$\text{For H: " overbrace(2 xx 4)^(color(red)("from ethene")) + overbrace(1 xx 8)^(color(blue)("from 1-butene")) + overbrace(1 xx 18)^(color(green)("from octane")) = "34 moles H}$

So, one mole of compound $\text{Y}$ contained $16$ moles of carbon and $34$ moles of hydrogen. This of course means that one molecule of compound $\text{Y}$ will contain $16$ atoms of carbon and $34$ atoms of hydrogen.

Therefore, the compound's molecular formula will be

${\text{C"_16"H}}_{34} \to$ hexadecane 