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Thread: Bonds Probability

  1. #1
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    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?
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  2. #2
    Junior Member Cervesa's Avatar
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    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 $\displaystyle \binom{4}{4} = 1$
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    Forum Admin topsquark's Avatar
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    Re: Bonds Probability

    Quote Originally Posted by Estermont View Post
    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
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    Re: Bonds Probability

    Quote Originally Posted by Cervesa View Post
    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?
    $\displaystyle \binom{7}{0} = 1$
    $\displaystyle \binom{7}{1} = 7$
    $\displaystyle \binom{7}{2} = 21$
    ...

    Quote Originally Posted by topsquark View Post
    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?
    Last edited by Estermont; Feb 22nd 2019 at 11:17 AM.
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  5. #5
    Forum Admin topsquark's Avatar
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    Re: Bonds Probability

    Quote Originally Posted by Estermont View Post
    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
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