# Thread: Bonds Probability

1. ## Bonds Probability

When methane (CH4) is placed in a mass spectrometer, Johnny knows that all, some or none of the bonds may break. What are the number of different ways this can happen, taking into account all rotations and reflections separately? (note: multiple breaks are possible) What would the results be for ethane (C2H6)?

I'm struggling to understand what the "rotations and reflections" mean. I checked out whether methane had any isomers, but no, it didn't. Neither did ethane...

Anyway for methane I got 5, but my friend got 16 from considering each H as distinct from each other...

CH4
CH3 H
CH2 H H
CH H H H
CH2 CH H
CH2 C H H

With a similar method I got 11.

C2H6
CH3 CH3
CH2 CH3 H
CH2 CH2 H H
CH CH3 H H
CH CH2 H H H
CH CH H H H H
CH3 C H H H
CH2 C H H H H
C CH H H H H H
C C H H H H H H

Can anyone check if this is right?

2. ## Re: Bonds Probability

let 1, 2, 3, and 4 represent the individual bonds

no bonds broken ... one possible

one bond broken ... (1) (2) (3) (4) ... four possible

two bonds broken ... (1,2) (1,3) (1,4) (2,3) (2,4) (3,4) ... six possible

three bonds broken ... (1,2,3) (1,2,4) (1,3,4) (2,3,4) ... four possible

four bonds broken ... one possible

sixteen possibilities?

are you familiar with combinations and just maybe the notation $\displaystyle \binom{n}{k} = \dfrac{n!}{k!(n-k)!}$ ?

fyi, $\displaystyle \binom{4}{0} = 1$, $\displaystyle \binom{4}{1} = 4$, $\displaystyle \binom{4}{2} = 6$, $\displaystyle \binom{4}{3} = 4$, and $\displaystyle \binom{4}{4} = 1$

3. ## Re: Bonds Probability Originally Posted by Estermont I'm struggling to understand what the "rotations and reflections" mean. I checked out whether methane had any isomers, but no, it didn't. Neither did ethane...
Just a quick note. CH4 is shaped like a tetrahedron. So there are rotation and reflection symmetries. It has nothing to do with isomers.

-Dan

4. ## Re: Bonds Probability Originally Posted by Cervesa are you familiar with combinations and just maybe the notation $\displaystyle \binom{n}{k} = \dfrac{n!}{k!(n-k)!}$ ?
$\displaystyle \binom{4}{0} = 1$, $\displaystyle \binom{4}{1} = 4$, $\displaystyle \binom{4}{2} = 6$, $\displaystyle \binom{4}{3} = 4$, and $\displaystyle \binom{4}{4} = 1$
Yes, I do know them, just that I'm not used to using them.

Would the second part be something like this then?
$\displaystyle \binom{7}{0} = 1$
$\displaystyle \binom{7}{1} = 7$
$\displaystyle \binom{7}{2} = 21$
... Originally Posted by topsquark Just a quick note. CH4 is shaped like a tetrahedron. So there are rotation and reflection symmetries. It has nothing to do with isomers.

-Dan
I already knew that. But how does it affect the answer?

5. ## Re: Bonds Probability Originally Posted by Estermont I already knew that. But how does it affect the answer?
It doesn't. But you were asking about where the symmetries were coming from.

-Dan