Getting 70% Isopropyl Alcohol down to exactly 50% based on a 32oz bottle

February 6, 2013 8:10 AM

For one of the steps to a science project, I need to take a 32oz bottle of drugstore 70% isopropyl alcohol and reduces down to 50% by adding distilled water. Can anyone tell me how many ounces of water I would need to add to the 32oz bottle of isopropyl to make it 50% alcohol? The mathematical formula would be helpful if it's easy enough to calculate in the future. Thank you.
posted by orehek to Science & Nature (19 answers total)

You have 32oz of 70% alcohol, so it contains 32*0.7 = 22.4 oz of alcohol.

You want it to be 50% alcohol, so you need the total volume of liquid to be twice that, or 44.8 oz.

Since you already have 32oz of liquid, you need 12.8 more oz of distilled water.
posted by Elementary Penguin at 8:12 AM on February 6, 2013 [2 favorites]

12.8 oz.

There's 22.4 oz of alcohol in the bottle already (.7 * 32). This doesn't change. The only thing you change is the 32 (the total amount of liquid).

.7 * 32 = 22.4
.5 * x = 22.4

Solve for x: 22.4/.5 = 44.8.

You need 44.8 of liquid, 44.8 - 32 = 12.8 oz of water you have to add.
posted by theodolite at 8:15 AM on February 6, 2013 [4 favorites]

Here's the mathematical formula: 32 * (0.7) + ? * (0) = (32 + ?) * 0.5

You want to know how much "?" to add. So with a little algebra, we get:
(32 * 0.7 ) - (32 * 0.5) = ? * 0.5
32 * 0.2 * 2 = ?
12.8 = ?

In the future, if you have different size bottles of alcohol, or different strengths, or different targets, just change the number in the first formula accordingly.
posted by willbaude at 8:16 AM on February 6, 2013 [1 favorite]

Best answer: You can also use this Wolfram Alpha link if you don't want to do the math yourself. I can't get it to give the answer in imperial units, though. Sorry.
posted by Elementary Penguin at 8:21 AM on February 6, 2013

Presumably you want the answer in U.S. customary units rather than imperial units, which are quite different in the case of fluid ounces.
posted by grouse at 8:24 AM on February 6, 2013 [2 favorites]

Best answer: T = Total volume of starting liquid
A = Percentage of the starting liquid that is alcohol
E = Extra water you need to add to make the resulting mixture 50% alcohol

E = (A x T) - (T - (A x T))

So using your amounts:
T = 32
A = 70% (or 0.7 in decimal)

E = (0.7 x 32) - (32 - (0.7 x 32))
E = 12.8oz
posted by EndsOfInvention at 8:49 AM on February 6, 2013 [1 favorite]

Note that adding X oz of water to Y oz of 70% alcohol/water will give you less than X + Y oz due to miscibility (at least for ethyl alcohol. Not sure about isopropyl).
posted by ShooBoo at 9:18 AM on February 6, 2013 [3 favorites]

Response by poster: Thank you everyone for all of your help. The formula was very helpful. But how about this notion of miscibility? Will that factor throw my final concentration off a tiny bit?
posted by orehek at 9:40 AM on February 6, 2013

MeTa
posted by grouse at 9:42 AM on February 6, 2013 [1 favorite]

ShooBoo is correct. Rather than measuring water and adding that much, you should add water until the total volume is 44.8 oz.

I don't know how big the difference would be but if you find out by measuring how much water you add to get it up to 44.8 oz, please report back!
posted by aubilenon at 10:19 AM on February 6, 2013 [1 favorite]

Fill another container with 15 ounces of water. Add enough water from that container to the alcohol to get to 44.8 ounces. Measure the water left in the water container to find out how much you added from the original 15 ounces.
posted by maudlin at 10:37 AM on February 6, 2013 [1 favorite]

if you have an n% solution and want a m% solution.

Take m units of your original solution. Now add water (0%) until you have n total units. Every time you will have m% in the final.

I'd avise you to double check the math first, but the general idea is

(n%*m volume)/n volume = m% (units are a bit awkward, but they fall our correctly)

It's a really useful trick when doing stuff like this.
posted by aspo at 12:29 PM on February 6, 2013

Best answer: Beyond the basic nominal mixing, are you interested in mixtures by volume or by mole/mass?

There's a fair nonlinear volume change in the isopropanol-water system, on the order of a few percent. See, for example, here. There's almost a 5% difference in density between the 70/30 and 50/50 isopropanol/water mixtures. There can be a few percent difference, therefore between a solution that 30% and 50% solutions in water by mass or by volume. This is enough to matter for some things, not enough to worry about for others.
posted by bonehead at 1:28 PM on February 6, 2013 [1 favorite]

Note that isopropanol is typically sold as % by volume, which can make this fairly tricky to do the math.
posted by bonehead at 1:39 PM on February 6, 2013

Ask the pharmacist for the 99% isopropyl alcohol, it's not on the open shelves, it's where they dispense pills.
posted by alicesshoe at 6:13 PM on February 6, 2013

Beyond the basic nominal mixing, are you interested in mixtures by volume or by mole/mass?

Yes, this is the first question that needs to be asked.

Assuming it's volume, the second question needs to be how that 70% was measured in the first place. Do they mix 70 gallons of pure isopropanol and 30 gallons of water? Or do they put 70 gallons into a vat and then keep adding water until they get 100 gallons? Knowing this, and replicating the method they used, will correct for the miscibility issue.
posted by gjc at 6:13 PM on February 6, 2013

Ask the pharmacist for the 99% isopropyl alcohol

Why would that help?
posted by grouse at 6:14 PM on February 6, 2013

Thought it would be easier to drop the alcohol to 50%, by adding 50% water, less a drop or so...

In any case, I think it's better to start with something that is nearer to 100% pure...not something that gas been cut so much to start with.
posted by alicesshoe at 6:45 PM on February 6, 2013

Response by poster: Thank you Bonehead for that interesting article. As you eluded to, a few percent difference give or take will not make a difference for my application. I have an antique wet specimen which was previously preserved in toxic formalin and I'm switching it over to 50% isopropyl which is significantly safer. Thanks again everyone. Much appreciated.
posted by orehek at 7:09 PM on February 6, 2013