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Question:

A gaseous mixture with volume 16.8 dm3 contains two unsaturated neighboring hydrocarbons belonging to the same group. Density of the mixture was 14.4 times as hydrogen gas.

What could be the two hydrocarbons?

(a) $\ce{CH4}$, $\ce{C2H6}$

(b) $\ce{C2H2}$, $\ce{C3H4}$

(c) $\ce{C3H4}$, $\ce{C4H6}$

(d) $\ce{C3H6}$, $\ce{C4H9}$

(e) none of the above

I'm not really sure how to approach this question. The answer according to the mark scheme is b). What is the method to get this answer? Also, how can $\ce{C2H2}$ and $\ce{C3H4}$ be in the same group? One is an alkyne and the other a cycloalkene.

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  • $\begingroup$ Read what you have written, the quoted task does not match your comment. $\ce{C2H2 means CH#CH}$, $\ce{C3H4 means CH3-C#CH}$, $\ce{C4H6 means CH3-CH2-C#CH}$ $\endgroup$
    – Poutnik
    Commented Feb 1, 2023 at 13:19
  • $\begingroup$ Some mixtures are too dense, and some aren't unsaturated. $\endgroup$
    – Ivan Neretin
    Commented Feb 1, 2023 at 13:25
  • $\begingroup$ @IvanNeretin All is allowed in incorrect answers. :-) $\endgroup$
    – Poutnik
    Commented Feb 1, 2023 at 13:43
  • $\begingroup$ @Poutnik Fixed it. Oh right... I was thinking about cyclopropene for C3H4 $\endgroup$
    – Jane902
    Commented Feb 1, 2023 at 14:48

1 Answer 1

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The density of $\ce{H2}$ is such that $1$ mole, or $2$ g $\ce{H2}$ occupies a volume $22.4$ L at $0$°C at $1$ atm. Its density is $\pu{\frac{2 g}{22.4 L} = 0.0892 g/L}$. Doing the same calculation with $\ce{C2H2}$ and $\ce{C3H4}$ gives densities equal to $\ce{1.1607}$ g/L for $\ce{C2H2}$ and $\pu{1.7857}$ g/L for $\ce{C3H4}$

The density of the gas mixture is $14.4 · 0.0892 = 1.2857$ g/L. This value is between the densities of $\ce{C2H2}$ and $\ce{C3H4}$. It means that these two gases can be mixed to give the observed density. So the choice b) can be an acceptable answer.

And indeed a mixture $80$% $\ce{C2H2}$ + $20$% $\ce{C3H4}$ has the observed density.

The solution c) contains too heavy hydrocarbons. The solution d) contains a non-existant compound ($\ce{C4H9}$)

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    $\begingroup$ But the question states the compounds have to be unsaturated so that answer can't be valid. $\endgroup$
    – matt_black
    Commented Feb 1, 2023 at 20:14
  • $\begingroup$ @matt_black. You are right. I edit my answer to take this remark into account. $\endgroup$
    – Maurice
    Commented Feb 1, 2023 at 21:35

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