Question

Question #8bc4e

Answer 1

Indeed, the compound's molecular formula is `"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 `"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 `"Y"`

• two moles of ethene, `"C"_2"H"_4`
• one mole of 1-butene, `"C"_4"H"_8`
• one mole of octane, `"C"_8"H"_18`

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

`"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"`

`color(white)(x)`

`"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 `"Y"` contained `16` moles of carbon and `34` moles of hydrogen. This of course means that one molecule of compound `"Y"` will contain `16` atoms of carbon and `34` atoms of hydrogen.

Therefore, the compound's molecular formula will be

`"C"_16"H"_34 ->` hexadecane

![https://en.wikipedia.org/wiki/Hexadecane]()