Adding water to a solution adds no solute. That sentence is the whole of $C_1 V_1 = C_2 V_2$, read aloud C-one V-one equals C-two V-two, and if you ever forget the algebra you can rebuild it from the sentence in one line: the moles before $(C_1 V_1)$ are the moles after $(C_2 V_2)$, because none of them went anywhere.

Say the subscript convention out loud once and it will never trouble you again: subscript 1 is before the dilution — the stock, the strong side, the drum you draw from. Subscript 2 is after — the batch, the weak side, the tank you are making. So $C_1$ is the stock concentration and $V_1$ the volume of stock drawn; $C_2$ is the finished strength and $V_2$ the total final volume of the batch, stock and water together. The draw sits inside $V_2$, never on top of it — which is why a batch is made up to a mark and not at one.

The equation is scale-free in its volumes. Both sides in millilitres, both in litres, both in cubic metres: it does not care, because the volumes divide each other and the unit cancels. What it will not survive is mixed volumes across the equals sign, and that is the classic thousandfold slip — a litre of stock destined for a 500 mL flask. The same freedom applies to the concentrations: mol/L both sides, or % w/w both sides, or mg/L both sides. One currency, chosen once.

Rearranged for the draw, $V_1 = \dfrac{C_2 V_2}{C_1}$ — the moles the batch needs, divided by how densely the stock carries them. Rearranged for the strength you end up with, $C_2 = \dfrac{C_1 V_1}{V_2}$.

One field habit to finish. Run the number before opening a valve, and check it against what is actually on the pad. A part-used drum that looks full enough is not a calculation, and finding out at two in the morning, with a half-made batch, is how a night shift becomes a story.