The way I would perform the dimensional analysis you give in your question would be slightly different. That is mainly due to the fact that I don't consider chemical names or formulae as units, but as designators.
As an aside, I think you were starting from a question like the following:
How much $\ce{C7H4O}$ would you have to weigh in for it to fully react
with $150~\mathrm{g}$ of $\ce{KNO3}$ if the reaction is described by
the following equation? $$ \ce{C7H4O + 6KNO3 -> Fumes^ + Tar} $$
The result is $m_{\ce{C7H4O}}$, given is $m_{\ce{KNO3}} = 150~\mathrm{g}$.
The amount of $\ce{C7H4O}$ used per $\ce{KNO3}$ is
$$|\nu_{\ce{C7H4O}} / \nu_{\ce{KNO3}}| = \frac{1}{6}$$
As such, the result is composed as follows:
$$\begin{align}
n_{\ce{C7H4O}} &= \frac{1}{6} n_{\ce{KNO3}} \\
\tag{$n=m/M$}\frac{m_{\ce{C7H4O}}}{M_{\ce{C7H4O}}} &= \frac{1}{6} \frac{m_{\ce{KNO3}}}{M_{\ce{KNO3}}} \\ \hline
m_{\ce{C7H4O}} &= \frac{1}{6} \frac{m_{\ce{KNO3}}}{M_{\ce{KNO3}}} M_{\ce{C7H4O}}
\end{align}$$
If we now apply dimension analysis to what we think is the result, we can very quickly see how the two molar masses cancel each others units out and only the mass of $\ce{KNO3}$ remains to give the right hand side of the equation a dimension of $\mathrm{g}$. As the only symbol left on the left hand side is a $m$ we know that it needs a dimension $\mathrm{g}$ and as such we can happily plug in the numbers into our calculator for the result.
Frankly, I am so used to "my" method that I had to look at your example quite closely to see what was actually going on. To me that method just seems unwieldy, as you clutter up equation space with chemical formulae.
That being said; find the method that works the best for you (meaning it is robust enough to also solve your chemical engineering, biology and other problems using dimensional analysis) and stick to it.
