Mixing alcohol

Posted by J. Robert Michael PhD on Fri 20 March 2020 Updated on Fri 20 March 2020

The problem

I was recently asked about how much rubbing alcohol to mix together with non-alcoholic solutions to create a mixture and what the final concentration would be. I'd like to note that the below information is not FDA, CDC, ABC, or XYZ approved. These are the thoughts of a previous math teacher and should be considered solely at your own discretion and at your own risk. Let me say that again, use at your own risk!

Let's phrase this as a standard math problem. You want to create A cups of a% alcohol and have an alcoholic solution (S1) of concentration b% and want to mix it with another solution (S2) of 0% alcohol. How much of each solution to mix?

The solution

Let's assign some variables. Let's call the final amount of the alcoholic solution (S1) to be B and the amount of non-alcoholic solution (S2) to be C. Then we can state our first equation describing the total amount of liquid:

A = B + C

Now, we also need to describe how much alcohol is in the final solution. We know that our final mixture will have A cups and be comprised of a% alcohol so the total amount of alcohol will be a * A. Now let's consider the things we are mixing together. Generally speaking this will be the amount of alcohol in each thing we are mixing, so in this case it will be b * B + 0 * C since the S1 solution is b% mixture and the S2 solution has no alcohol at all. This gives us our second equation defining the amount of alcohol:

a*A = b*B

Now, remember that in this situation, we know how much solution we want to finally create (A) and we know what percentage of alcohol we want the final mixture to be (a) and we also know that percentage of our S1 alcoholic solution is b. So we can solve for B (the amount of the alcoholic solution S1) using the second equation:

B = (a*A)/b

Now that A and B are known, we can figure out how much of the non-alcoholic solution (S2) we should have:

C = A*(1 - a/b)

Example 1 (solving for a)

Let's start off with a more straightforward example. Rather than figure out how much of S1 and S2 we need, let's say we know that information and want to figure out how much the final alcoholic content would be.

Suppose we are making 2 cups of a solution by combining 1.75 cups of a 95% solution with 1/4 cups of a non-alcoholic solution. In this example, A = 2, B = 1.75, and C = 0.25. We also know the concentration of the 'input' alcoholic solution is 95% so b = 0.95.

As a sanity check, our first equation checks out. In this example, A = B + C just means that 2 = 1.75 + .25. Now let's look at the second equation. The total amount of alcohol in the final solution will be b*B + 0*C which is just 0.95 * 1.75 or ~1.66 cups of alcohol. Finally, since a*A = b*B we know that a*2 = 0.95 * 1.75 so the final alcoholic concentration would be a = 0.831 or about 83.1%.

Example 2 (solving for B, C)

Let's modify this slightly and say that we really only want to create 2 cups of a 75% mixture. How much of solution S1 and S2 do we mix?

Again, we know that A = 2 but this time we don't know B or C. However we do still know that b = 0.95 and in this case we know that the final concentration should be a = 0.75. This time we can jump straight to the equation which compares the total amount of alcohol and say that a*A = b*B implies that 0.75 * 2 = 0.95 * B which means that B = (2*0.75)/0.95 = 1.579. Now we can easily solve for C since A = B + C so that 2 = 1.579 + C gives us C.


This is a pretty classic example of systems of equations in the real world and is often used for 'word problems' in undergraduate Algebra classes. However, I'll state again that if anyone is mixing chemicals at home then you should use at your own risk any equations or methods described above.

I'm just a guy writing a blog while his daughter is napping and- Wait ... Is that ... Yeah, Ellie's up - I gotta run!

tags: math, covid19