By looking at the sequence of molecular formulae, it can be seen
that as the carbon chain increases by one, the number of hydrogens
increases by 2, whereas the number of oxygens remains at one.
Therefore, the nth
molecule is given by $C_nH_{2n}O$.
Note that in actuality the masses of the different isotopologues
(i.e. molecules which differ only in their isotopic composition) of
say $\text{CH}_4$ 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 lightest isotopologue is $^{12}C_n\ ^1H_{2n}\ ^{16}O$ which has
a molecular mass of $14n +16$.
The heaviest isotopologue is $^{13}C_n\ ^2H_{2n}\ ^{18}O$ which has
a molecular mass of $17n +18$.
Additionally, all intermediate masses are also possible.
To work out the final part of this problem involves recognising
that the lightest general form of an aldehyde is $^{12}C_n\
^1H_{2n}\ ^{16}O$, whereas the next lightest is given by $^{12}C_n\
^1H_{2n}\ ^{17}O$ and $^{12}C_{n-1}\ ^{13}C\ ^1H_{2n}\ ^{16}O$ and
$^{12}C_n\ ^1H_{2n-1}\ ^2H\ ^{16}O$.
Therefore, the probability of the lightest isotopologue must be
equal to eight times the sum of the probabilities of the other
three forms given:
This equation is best solved using a spreadsheeting program, and
inputting different values of $n$ until both sides of the equation
are roughly equal. This reveals that $\mathbf{n=11}$
satisfies the equation.