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'Heavy Hydrocarbons' printed from https://nrich.maths.org/
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.
$^{12}\text{CH}_4$ = 16 gmol$^{-1}$
$^{12}\text{CH}_3\text{D}$, $^{13}\text{CH}_4$ = 17
gmol$^{-1}$
$^{12}\text{CH}_2\text{D}_2$, $^{13}\text{CH}_3\text{D}$ = 18
gmol$^{-1}$
$^{12}\text{CHD}_3$, $^{13}\text{CH}_2\text{D}_2$ = 19
gmol$^{-1}$
$^{12}\text{CD}_4$, $^{13}\text{CHD}_3$ = 20 gmol$^{-1}$
$^{13}\text{CD}_4$ = 21 gmol$^{-1}$
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:
16 gmol$^{-1}$
[$^{12}\text{CH}_4$]
$\textbf P(16 \text{gmol}^{-1}) = 0.989 \times (0.99985)^4 = 0.988$
(3sf)
17 gmol$^{-1}$
[$^{12}\text{CH}_3\text{D}$, $^{13}\text{CH}_4$]
$\textbf P(17 \text{gmol}^{-1}) =
(0.989\times^4\textbf{C}_1(0.00015)\times (0.99985)^3
) + (0.011 \times (0.99985)^4) = 0.0116$ (3sf)
18 gmol$^{-1}$
[$^{12}\text{CH}_2\text{D}_2$,
$^{13}\text{CH}_3\text{D}$]
$\textbf P(18 \text{gmol}^{-1}) =
(0.989\times^4\textbf{C}_2(0.00015)^2\times (0.99985)^2 ) +
(0.011\times^4\textbf{C}_1(0.00015)\times (0.99985)^3) =
6.73\times10^{-6}$ (3sf)
19 gmol$^{-1}$
[$^{12}\text{CHD}_3$, $^{13}\text{CH}_2\text{D}_2$]
$\textbf P(19 \text{gmol}^{-1}) =
(0.989\times^4\textbf{C}_3(0.00015)^3\times 0.99985 ) +
(0.011\times^4\textbf{C}_2(0.00015)^2\times (0.99985)^2) =
1.50\times10^{-9}$ (3sf)
20 gmol$^{-1}$
[$^{12}\text{CD}_4$, $^{13}\text{CHD}_3$]
$\textbf P(20 \text{gmol}^{-1}) =
(0.989\times^4\textbf{C}_4(0.00015)^4) +
(0.011\times^4\textbf{C}_3(0.00015)^3\times 0.99985) =
1.49\times10^{-13}$ (3sf)
21 gmol$^{-1}$
[$^{13}\text{CD}_4$]
$\textbf P(21 \text{gmol}^{-1}) =
(0.011\times^4\textbf{C}_4(0.00015)^4) = 5.57\times10^{-18}$
(3sf)
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.
$^{12}\text{C}_2\text{H}_6$ = 30 gmol$^{-1}$
$^{12}\text{CD}_2\text{H}_5$,
$^{12}\text{C}^{13}\text{C}\text{H}_6$ = 31 gmol$^{-1}$
$^{12}\text{C}_2\text{H}_4\text{D}_2$,
$^{12}\text{C}^{13}\text{CD}\text{H}_5$,
$^{13}\text{C}_2\text{H}_6$ = 32 gmol$^{-1}$
$\textbf P(30 \text{gmol}^{-1}) = (0.989)^2 \times (0.99985)^6 =
0.977$ (3sf)
$\textbf P(31 \text{gmol}^{-1}) = ((0.989)^2 \times
^6\textbf{C}_10.00015 \times (0.99985)^ 5) +
((0.989)\times^2\textbf{C}_1(0.011)\times(0.99985)^ 6) = 0.0226$
(3sf)
$\textbf P(32 \text{gmol}^{-1}) = ((0.989)^2 \times
^6\textbf{C}_2(0.00015)^2 \times (0.99985)^ 4) +
((0.989)\times^2\textbf{C}_1(0.011)\times^6\textbf{C}_1(0.00015)\times(0.99985)^
5)$
$+ (^2\textbf{C}_2(0.011)^2 \times (0.99985)^6) = 1.41 \times
10^{-4} \text{ (3sf)}$
The three most likely possibilities for the molecular mass of
propane are 44, 45 and 46 gmol$^{-1}$ in order of likelihood.
$^{12}\text{C}_3\text{H}_8$ = 44 gmol$^{-1}$
$^{12}\text{C}_2{}^{13}\text{C}\text{H}_8$,
$^{12}\text{C}_3\text{D}\text{H}_7 = 45\text {gmol}^{-1}$
$^{12}\text{C}^{13}\text{C}_2\text{H}_8$,
$^{12}\text{C}_2^{13}\text{C}\text{D}\text{H}_7,
^{12}\text{C}_3\text{D}_2\text{H}_ 6 = 46\text {gmol}^{-1}$
$\textbf P(44 \text{gmol}^{-1}) = (0.989)^3 \times (0.99985)^8 =
0.966$ (3sf)
$\textbf P(45 \text{gmol}^{-1}) = (^3\textbf{C}_1 \times 0.011
\times (0.989)^2 \times (0.99985)^ 8) +
((0.989)^3\times^8\textbf{C}_1(0.00015)\times(0.99985)^ 7) =
0.0334$ (3sf)
$\textbf P(46 \text{gmol}^{-1}) = (^3\textbf{C}_2 \times (0.011)^2
\times 0.989 \times (0.99985)^ 8) + (^3\textbf{C}_1 \times 0.011
\times (0.989)^2 \times^8\textbf{C}_1(0.00015)\times(0.99985)^
7)$
$+ ((0.989)^3\times^8\textbf{C}_2(0.00015)^2\times(0.99985)^ 6) =
3.98 \times 10^{-4}$ (3sf)
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:
$\textbf{P}(^{13}\text{C}_4\text{D}_{10}) = 0.011^4 \times
0.00015^{10} = 8.44 \times 10^{-47}$ (3sf)
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.
$\therefore\text{E}(^{13}\text{C}_4\text{D}_{10}) = 2.51 \times
10^{22} \times 8.44 \times 10^{-47} = 2.12 \times 10^{-24}$
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:
$\textbf{P} = 0.989^n \times ^{2n + 2}\textbf{C}_1(0.99985)^{2n +
1}\times 0.00015$
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.
$0.989^n \times ^{2n + 2}\textbf{C}_1 \times (0.99985)^{2n +
1}\times 0.00015$ > $0.989^n \times (0.99985)^{2n + 2}$
$0.00015 (2n + 2)$ > 0.99985
2n + 2 > 6665.666667
n > 3331.833333
n = 3332
Is this long chain alkane likely
to exist in a real sample?
The probability of encountering an isotopologue containing
$^{13}\text{C}$ is given by:
$\textbf{P} = ^n\textbf{C}_1\times 0.011 \times 0.989^{n-1} \times
(0.99985)^{2n + 2}$
This probability must be greater than the likelihood of finding the
alkane molecule consisting entirely of $^{12}\text{C}$ and H.
So:
$^n\textbf{C}_1\times 0.011 \times (0.989)^{n-1} \times
(0.99985)^{2n + 2}$ > $0.989^n \times (0.99985)^{2n + 2}$
$ \text{n} \times (0.989)^{n-1} \times 0.011$ >
$(0.989)^{n}$
$0.011\text{n}$ > 0.989
$\text{n}$ > 89.90909091
n = 90
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.