For the purpose of this example lets say I have two ingredients:
The goal is to mix them together to create a mix that is exactly $50\%$ water and $50\%$ chocolate.
How can I calculate exactly how much of the 2 ingredients to combine to create 500ml without waste?
If you take $s$ liters (or quarts, etc.) number 1, and $(1-s)$ liters of #2, you'll end up with $1$ liter of mix. That liter will contain $$ (1-s) 0.8 = 0.8 - 0.8s $$ liters of chocolate; you want that to be $0.5$ liters. So set \begin{align} 0.8 - 0.8s &= 0.5 \\ -0.8s &= -0.3 \\ s &= \frac{-0.3}{-0.8} = \frac{3}{8} \end{align} So: take 3/8 pure water, and 5/8 chocolate mix.
If you have, say, 7 liters of the 80% mix (what I've called "#2"), you'll need $\frac{3}{5} \times 7$ of the water to make your 50-50 mix.
In short: let's say the amount of chocolate mix that you have (measured in cups, gallons, liters, whatever) is $A$. You'll need to add $0.6 \times A$ of water to it. You'll end up with $1.6 \times A$ of mix.
For instance, if you have 9 cups of the 80% mix, you'll need to add $0.6 \times 9 = 5.4 cups$ of water to it to get to a 50% mix; when you're done, you'll have the original 9 cups of mix plus 5.4 cups of water to get 14.4 cups of mix, whic is $1.6 \times 9$.
(I got $s$ and $1-s$ screwed up in my earlier response, which I've now fixed up.)
Final Edit
Now that the problem's been clarified -- you want to get some amount $C$ of final 50-50 mix -- the solution is this:
quantity of pure water: (3/8) * C
quantity of 80% choc: (5/8) * C.
In your example where $C = 500$ml, you get
water: (3/8) * 500 = 187.5 ml
mix: (5/8) * 500 = 312.5 ml
Suppose, that you have $x$ liters of the mixture. Then you have $0,2x$ water. Now you add $y$ liters of pure water. The sum of these two liquids has to have the average water proportion. It is $0.5\cdot (x+y)$
Thus the equation is $0.2x+y=0.5x+0.5y$
Solve this equation for y. This gives you the amount of pure water, which you have to add-in relation to x.
There are two related problems of this sort that come up, and it is worth distinguishing them carefully. They involve solving the same equations but with different givens.
Problem 1:
You have $V_2$ liters of mixture 2, which has concentration $0.8$, and you want to dilute it with $V_1$ liters of mixture 1, which has concentration $0$, in such a way that the result has concentration $0.5$. So here $V_2$ is given, $V_1$ is unknown, and you want
$$\frac{0.8V_2}{V_1+V_2}=0.5.$$ Solving this you find $0.5V_1=0.3V_2$ so $V_1=0.6V_2$.
Problem 2:
You have as much of mixtures 1 and 2 as is required, and you want a total volume of $V$ which has concentration $0.5$. So here $V$ is given and $V_1,V_2$ are both unknown. Then you want to solve the two equations
$$V_1+V_2=V \\ \frac{0.8V_2}{V_1+V_2}=0.5.$$
I will solve this case in detail. Using the first equation, replace $V_1+V_2$ by $V$ in the second equation:
$$\frac{0.8V_2}{V}=0.5.$$
Multiply both sides by $V$:
$$0.8V_2=0.5V.$$
Divide both sides by $0.8$:
$$V_2=\frac{5}{8} V.$$
Now substitute into the first equation to get $V_1=V-\frac{5}{8} V = \frac{3}{8} V.$
One thing that simplifies the situation in practice is that with these types of equations, $V_1,V_2$ will always be directly proportional to $V$, if all the other numbers are held constant. That means you can solve the problem for $V=1$ and then multiply by $V$. This is done in John Hughes' answer.
