How do I find the final concentration after an additional dilution?

I took poor notes and didn't record what the final concentration of an amylase solution I made last year. I recorded that I took $\rm 0.2\ g/10\ ml$ of the amylase solution, and then took $\rm 1\ ml$ of that and diluted to $\pu{35 ml}$.


How would I solve for the final concentration? Would I use


$$
\begin{align}
c\_1V\_1 &= c\_2V\_2\tag{1}\\
(\pu{0.2 g}/\pu{10 ml})(\pu{1 ml}) &= c\_2(\pu{35 ml})\tag{2}\\
c\_2 &= \frac{\pu{0.02 g}}{\pu{35 ml}}?\tag{3}
\end{align}
$$


Therefore, I would take $\pu{0.02 g}$ of amalyase powder and add up to $\pu{35 ml}$ with distilled $\ce{H2O}$ to achieve the final concentration that I would have previously had after dilution?




Check your math: It looks like you forgot to divide by $ 10\ \mathrm{ml} $ in finding $ c\_2 $. 


Also: the process you are describing in the first paragraph will not produced the same concentration that you started with. If you want $ 35\ \mathrm{ml} $ at the same concentration ($ 0.02\ \mathrm{g/ml} $) by taking some fraction of your original solution and diluting it, you will need to add more solute.


e.g.
$$ \rm(0.02\ g/ml) \cdot (35\ ml) = 0.07\ g $$
$$ \rm(0.07\ g + \mathit{m}) / 35\ ml = (0.02\ g/ml) $$
$$ m \rm = (0.02\ g/ml) \cdot (35\ ml) - 0.07\ g $$


The first expression is the concentration of your fraction (which doesn't change from the original solution, assuming a homogenous solution).


The second expression is equating the concentration that you want (conc. of the original solution, for example, with the concentration of your diluted fraction with $ m\ \mathrm{g} $ of solute added (using the mass we just found).


Then I rearranged this to express the mass of solute needed to obtain this concentration. 


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