Note that in actuality the masses of the different isotopologues of
$\text{CH}_4$ are slightly different. These differences may be
noted by a very sensitive mass spectrometer. Take for
example:
RMM $^{12}\text{CH}_3\text{D}$ = 12 + 3(1.007825) + 2.014102 =
17.037577 gmol$^{-1}$
RMM $^{13}\text{CH}_4$ = 13.00335 + 4(1.007825) = 17.03465
gmol$^{-1}$
where the calculation is limited by the degree of accuracy of the
given data.
However in this question it is acceptable to take values to the
nearest gmol$^{-1}$, giving roughly equal molecular masses for
certain isotopes.
The three most common species encountered would be
$^{12}\text{CH}_4$, $^{13}\text{CH}_4$ and
$^{12}\text{CH}_3\text{D}$ in order of likelihood. This can bee
seen intuitively as the probability of encountering a
$^{13}\text{C}$ more likely than encountering a single
$^{2}\text{H}$, and for a small molecule such as methane, it is far
more likely to obtain $^{12}\text{CH}_4$ than either of the other
possibilities. The three most likely molecular masses as 16, 17 and
18 gmol$^{-1}$ as the introduction of more $^{13}\text{C}$ and
$^{2}\text{H}$ to a small molecule reduces its likelihood more than
any combinatorial effects can compensate for.
The actual probabilities of encountering each of these molecular
masses of methane are:
Following a similar principle to that above, the three most likely
possibilities for the molecular masses of ethane are 30, 31 and 32
gmol$^{-1}$ in order of likelihood.
A molecule of butane with molecular mass 72 is the isotopologue
$^{13}\text{C}_4\text{D}_{10}$. The probability of any butane
molecule being this isotopologue is:
24dm$^3$ of butane corresponds to roughly 1 mole of butane
molecules. Thus, as 1 dm$^3$ is equivalent to a litre, the number
of moles in the required volume is $\frac{1}{24}$.
The number of molecules in the sample is given by multiplying the
number of moles, by the number of molecules in a mole (the Avogadro's constant):
Number of molecules $= \frac{N_A}{24} = 2.51 \times 10^{22}$
So the likelihood can be found by multiplying the probability for
one molecule by the total number of molecules in the sample.
This question requires a little more algebraic appreciation of the
calculations thusfar.
The probability of encountering a generic alkane
$^{12}\text{C}_n{}^1\text{H}_{2n + 2}$ is given by:
$\textbf{P} = 0.989^n \times (0.99985)^{2n + 2}$
The probability of encountering an isotopologue containing
deuterium $^{12}\text{C}_n^{\ 1}\text{H}_{2n + 1}\text{D}$ is given
by:
The probability of encountering this isotopologue must be greater
than the likelihood of finding the butane molecule consisting
entirely of $^{12}\text{C}$ and H.
Consider the likelihood of the
existence of this molecule.
As an extension, does the
probability of the existence of such molecules change if the
molecule is produced via a method which involves polymerisation?
Try to construct an algebraic test.