# Molecular sequencer

Investigate the molecular masses in this sequence of molecules and deduce which molecule has been analysed in the mass spectrometer.

## Problem

Methanal, ethanal, propanal and butanal form the first four compounds in a sequence of alkyl aldehydes with molecular formulae

CH$_2$O, C$_2$H$_4$O, C$_3$H$_6$O, C$_4$H$_8$O

What would be the molecular formula for pentanal and hexanal?

What would be the general formula for the $n$th molecule in the sequence? What masses could it have, taking into account these common isotopes of C, H and O:

$^{16}$O $99.76\%$, $^{17}$O $0.04\%$, $^{18}$O $0.2\%$

$^{12}$C $98.9\%$, $^{13}$C $1.1\%$

$^1$H $99.99\%$, $^2$H $0.01\%$

The relative abundance of the lightest form of one particular type of alkyl aldehyde is almost exactly $8$ times that of its next lightest form. Can you work out its molecular formula?

## Getting Started

For the last part of the question, using a spreadsheeting program
to solve numerically is a good idea!

## Student Solutions

Pentanal is $C_5H_{10}O$

Hexanal is $C_6H_{12}O$

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:

$$\begin{matrix} (0.989)^n (0.9999)^{2n} (0.9976) \\ = 8[(0.989)^n (0.9999)^{2n} (0.0004) \\ + (0.989)^{n}\ ^{2n}C_1 (0.9999)^{2n-1} (0.0001)(0.9976) \\ + ^nC_1(0.989)^{n-1}(0.011)(0.9999)^{2n}(0.9976)] \end{matrix}$$

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.

## Teachers' Resources

It is not envisaged that this problem would be used as a class problem. It is more appropriate for an enthusiastic student or small group of students looking for a challenge to work on independently.