I am not too certain of what you are asking. If you are asking why do we not look at a molecule and then just identify the number of independent modes of vibration the answer is that it would be too hard. Real oscillations are a linear superposition of these normal modes, even undergrad spectra are often complicated. Only very simple diatomic give nice Raman and IR spectra, or very symmetrical molecules.
In general we can expand a potential function about an equilibrium configuration and retain second order terms. The first term is the absolute value so will be set to zero. The second term vanishes by assumption that we are at an equilibrium point.
\begin{equation}
U(\boldsymbol q)=U(q^i_0)+\sum _i\frac{\partial U}{\partial q^i}\bigg|_{0}\eta _i+\sum_{i,j}\frac 12 \bigg(\frac{\partial ^2U}{\partial q^i\partial q^j}\bigg)\bigg|_0\eta _i\eta _j+\sum _{i,j,k}\frac 1 6\bigg(\frac{\partial ^3U}{\partial q^i\partial q^j\partial q^k}\bigg)\bigg|_0\eta _i\eta _j\eta _k+\dots
\end{equation}
Where $\eta_i = q^i-q^i_0$ and $\dot \eta _i=\dot q^i-\dot q^i_0=\dot q^i$ are variations to the equilibrium configuration. This analysis leads to the following Lagrangian,
\begin{equation}
\mathscr L=\frac 12\sum _{i,j}(M_{ij}\dot \eta _i\dot \eta _j-V_{ij}\eta_i\eta_j)
\end{equation}
And a series of coupled harmonic oscillator ordinary differential equations.
\begin{equation}
\sum _{j}(M_{ij}\ddot \eta_j+V_{ij}\eta_j)=0
\end{equation}
A special type of solution is a normal mode,
\begin{equation}
\eta _j=a_j\cos (\omega t-\varphi)
\end{equation}
The general solution to the small oscillations problem is then a linear superposition of normal modes,
\begin{equation}
\eta _r=\sum _r\boldsymbol a_r\cos (\omega _rt-\varphi _r)
\end{equation}
So far we have not counted modes per say. Although mechanically the aim is to compute $\eta$ in chemistry we normally only bother with the frequency of the mode. Performing this analysis with molecules is a little tricky except for linear symmetric molecules (the symmetry simplifies the problem).
If we consider a linear symmetric triatomic molecule like CO2 for instance then we expect $3(3)-5=4$. Performing the above analysis will give two modes ($\omega_1$ is discounted as it corresponds to translation in the internuclear axis so was already counted).
\begin{equation}
\omega _1=0,\ \ \ \ \ \ \ \omega_2=\sqrt{\frac{k}{m_1}},\ \ \ \ \ \ \omega _3=\sqrt{\frac{km_T}{m_2m_1}}
\end{equation}
Where $m_T=2m_1+m_2$ (total mass). (follow the procedure in Goldstein and interpretation of their oscillations). You may ask why this does't agree with the number of modes we expect by the simple counting above?
We can split the total number of vibrations into transverse and longitudinal modes. For an $N$ atom system, we look down the internuclear axis, one is free translation leaving $N-1$ longitudinal oscillations. Therefore the total number of transverse oscillations is $2N-4$. Which solves that problem.
So what about the transverse modes? These are generally complicated and depend on the phase of the oscillations that we choose. This area can lead to ro-vibrational coupling and orbital angular momentum. The quantisation of this is important to quantum chemical spectroscopy.
So you can see that in general real modes are complicated and their interpretation and form depend on phase too. Could we use modes as a basis for other modes? Yes, that is what we do. Could we use modes of different molecules as a basis to predict other molecules modes? Yes ... PROVIDED they are of the same symmetry. In this way we can use the same modes of CO2 as for Carbon disulphide CS2, or silicon tetrachlorides SCl4 modes for methane CH4.
Loss of symmetry changes the problem. How "okay" this is depends really. It is really the symmetry of the molecule that dictates this. This is seen when computing the modes of a non-linear triatomic molecule, they are not too far removed from the linear case as shown here but the number of them is not the same and they are slightly different. Therefore using vibrations of one molecule as a basis for another is flawed. The frequencies we get are not the same!
Does this answer your question?
EDIT TO ANSWER
These two degenerate modes of the CO2 are the complicated transverse modes I was referring to in my answer. If they are degenerate then they have the same roots of the characteristic equation and hence the same expression of the frequencies. However they are complex because of phase and rotational effects and they need NOT be the same. See Goldstein and Lissajous figures which should answer your question :)
EDIT TO EDIT
Perhaps it would be more intuitive to say that the absence of rotational angular momentum about the internuclear axis is a result of there being no degenerate modes down this axis, or perhaps that there is a $C_{\infty}$ symmetry axis? Or perhaps the fact that there is rotational angular momentum about the transverse axes means there is ro-vibrational coupling, due to the degeneracies? Asking are there any degeneracies along the internuclear axis of polyatomic molecules is a harder question to answer. Certainly we would attempt to construct the analysis from a point where this was not the case.
The standard free energy of formation of carbon dioxide from graphite and oxygen is $394\, \rm kJ\, mol^{-1}$ and so in the formation of one molecule of carbon dioxide $6.5 \times 10^{-19} \,\rm J$ of energy are liberated.
$6.5 \times 10^{-19} \,\rm J$ of energy is $4.1 \, \rm eV$ or has the mass equivalent of $4.4 \times 10^{-9}\,\rm u$.
So you will see that even your very accurate and precise value for the mass of an oxygen atom $M(^{16}\rm O)=15.994915 \, u$ is not good enough to find the energy of formation of carbon dioxide.
This is a good example of the difference between Chemistry and Nuclear Physics in that Chemistry deals with energy changes of order $\rm eV$ whereas Nuclear Physics deals with energies of order $\rm MeV$.
Best Answer
I answered a question much like this in my Chemistry finals, and that was a several page essay. You'll excuse me if this answer is a rather shorter!
The definitive way to measure molecular size is X-Ray crystallography. This gives you the structure of the crystal including the positions of all the atoms, so you automatically get the molecule size. This method works for any material that you can crystallise even including huge molecules like DNA and proteins.
In the gas phase the size of molecules can be easily measured for simple molecules using rotational spectroscopy. This gives you the moments of inertia about various axes and from these you can calculate the bond lengths. The method doesn't work for molecules that are too big and complex or (like succinic acid) that aren't easily vaporised, however the bond lengths are pretty constant across most molecules. So once you have bond lengths for simple molecules you can put these into molecules like succinic acid and calculate the size of the molecule that way.