Question #8bc4e
1 Answer
Indeed, the compound's molecular formula is
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
Your strategy here will be to list the products that result from the cracking of your compound
- 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"#
#"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
Therefore, the compound's molecular formula will be
#"C"_16"H"_34 -># hexadecane