How to dilute 91% alcohol to 70%?

[HRG]

TL;DR

dilute alcohol to 70%

We have some 91% alcohol and I want to dilute it to 70% alcohol to disinfect for covid-19.

Example:
Have 6.7 ounces of 91% alcohol.
How many ounces of water to add to get 70% alcohol?

What is the math to do that using algebra so I can calculate it with various amounts of 91% alcohol?

 

[TheDS1337]

So, you have 0.91 * 6.7 ounces = 6.097‬ ounces of pure alcohol.

You want to have a solution with 70% alcohol, so, 6.097 = 0.7 * x, therefore the, x = 8.71‬

So you should add 8.71‬ - 6.7 = 2.01‬ ounces of water to get there.

That's atleast what the maths says, I don't know chemistry so if there are some chemical properties or something, you're better of getting a respond for a chemist.

 

[jbriggs444]

That's atleast what the maths says, I don't know chemistry so if there are some chemical properties or something, you're better of getting a respond for a chemist.

There are some implicit assumptions here -- that it is percentage by volume and that volumes add when solutions are created from pure substances.

But it ought to be close enough. The densities are not that far apart, the volumes add pretty well and I doubt that the delta between 75% IPA and 65% IPA is a deal breaker.

Edit:

The "by volume" bit...

You can have a solution that is "70% by volume" or a solution that is "70% by mass". Or perhaps even "70% by number of molecules". The three all mean different things.

70% alcohol by volume would mean that if you had one liter of solution and somehow extracted the pure alcohol, you would have 700 mL of pure alcohol as a result.

70% alcohol by mass would mean that if you had one kilogram of solution and somehow extracted the pure alcohol, you would have 700 grams of pure alcohol as a result.

70% alcohol by molar fraction would mean that if you had one mole of solution (6.02 x 10²³ molecules) and somehow extracted the pure alcohol, you would have 0.7 moles of pure alcohol as a result.

The mass density of alcohol is different from the mass density of alcohol/water solution. So you would not expect the percent by volume to match the percent by mass. [They are close -- like 0.79 g/ml for pure alcohol versus 0.88 g/ml for the solution. Thank you, Google]

The molar density (atoms per liter) for alcohol and for alcohol/water solution differ. So the percent by volume and percent by molar fraction will not match.

Similar for mass density versus molar density. The molecular weights of water and ethanol are very different. So percent by mass and percent by molar fraction will not match.

If you look on your bottle of isopropyl alcohol ... takes a trip upstairs to the bathroom ... you will find a label like "50% by volume". That is a clue.The "volumes add" bit...

If you have 0.7 liters of pure alcohol and 0.3 liters of pure water, there is no guarantee that the result will be 1.0 liters of alcohol/water solution. I expect that it will be darned close, but it likely will not be exact.

 

[HRG]

So, you have 0.91 * 6.7 ounces = 6.097‬ ounces of pure alcohol.
You want to have a solution with 70% alcohol, so, 6.097 = 0.7 * x, therefore the, x = 8.71‬
So you should add 8.71‬ - 6.7 = 2.01‬ ounces of water to get there.
That's atleast what the maths says, I don't know chemistry so if there are some chemical properties or something, you're better of getting a respond for a chemist.

Hey DS1337,

Your first step is brilliant to multiply 0.91 * 6.7 oz = 6.097 oz of pure alcohol. That's the step I didn't think of.

Now I figure with 100% alcohol, mixing 7 parts of pure alcohol to 3 parts of water = 70% alcohol. (hope that's right)

So 6.097 oz / 7 = 0.871 oz per part.
3 * 0.871 = 2.613 oz of water.

But your calculation is to add 2.01 oz of water.

What am I missing?

 

[TheDS1337]

Hey DS1337,

Your first step is brilliant to multiply 0.91 * 6.7 oz = 6.097 oz of pure alcohol. That's the step I didn't think of.

Now I figure with 100% alcohol, mixing 7 parts of pure alcohol to 3 parts of water = 70% alcohol. (hope that's right)

So 6.097 oz / 7 = 0.871 oz per part.
3 * 0.871 = 2.613 oz of water.

But your calculation is to add 2.01 oz of water.

What am I missing?

Remember, the extra 2.613 - 2.01 oz was already included in that 9% of the solution

 

[HRG]

Remember, the extra 2.613 - 2.01 oz was already included in that 9% of the solution

Brilliant again! Thanks so much for helping me with this problem to combat covid-19.
HRG

 

[TheDS1337]

Brilliant again! Thanks so much for helping me with this problem to combat covid-19.
HRG

No problem, althought you might carefully read jbriggs444's comment if you haven't already.

 

[HRG]

No problem, althought you might carefully read jbriggs444's comment if you haven't already.

Frankly jbriggs444's comments go right over my head.
I want to mark your post as the answer but don't see a way to do that. But thanks again.

 

[jbriggs444]

If a post gives you the answer you requested and does so in a way that satisfies you then clicking the Like button is appropriate. That is as close to "this was the answer to my question" as we get around here.

And thank you for the feedback on my answer. Even though it means that I could have done better. I've added verbiage. Maybe that will add clarity.

 

[HRG]

OK, I've reduced the formula down to this:

R = oRiginal oz of 91% alcohol
W = oz of water to add to get 70% alcohol

W = (.91 R / .7) - R

My algebra at my senior age is too rusty to simplify that equation further. Can a math guru simplify it more or is that as simple as it gets? If you can simplify it more, please list the steps so I can refresh my memory.

Thanks,
HRG

 

[jbriggs444]

After thinking about this a bit, I realized that with a little bit of density information, we have what we need to make the calculation by @TheDS1337 more exact.

We are talking percent by volume. So we can ignore the fancy stuff about mass fraction, molar fraction and what not.

We start with 6.7 fluid ounces of 91% IPA.

The density of 90% IPA is 0.8192. The density of 100% IPA is 0.786. If we do a linear interpolation, that is an estimated

9×0.8192+1×0.78610=0.816 specific gravity for 91% IPA9×0.8192+1×0.78610=0.816 specific gravity for 91% IPA​

That means that we have

6.7×0.816=5.47 ounce mass of 91% IPA6.7×0.816=5.47 ounce mass of 91% IPA​

@TheDS1337 calculates that we need a resulting volume of 8.71 fluid ounces to attain the desired 70% concentration by volume. Volume of solvent plus volume of solute may not equal volume of solution. But mass of solute plus mass of solvent will be equal to the mass of the resulting solution. And we can figure out how much mass we need to end up with.

The density of 70% IPA is 0.88. That's

0.88×8.71=7.66 ounce mass of 70% IPA0.88×8.71=7.66 ounce mass of 70% IPA​

So we need

7.66−5.47=2.19 ounce mass of pure water to be added7.66−5.47=2.19 ounce mass of pure water to be added​

The specific gravity of water is 1, so that's 2.19 fluid ounces.

[The 2.19 is on the ragged edge of too many significant figures -- three sig figs in a result based on a two sig fig density of 0.88 g/ml for 70% IPA. Don't trust that 9 very much]

 

[TheDS1337]

OK, I've reduced the formula down to this:

R = oRiginal oz of 91% alcohol
W = oz of water to add to get 70% alcohol

W = (.91 R / .7) - R

My algebra at my senior age is too rusty to simplify that equation further. Can a math guru simplify it more or is that as simple as it gets? If you can simplify it more, please list the steps so I can refresh my memory.

Thanks,
HRG

W = (.91 R / .7) - R = (.91 R - .7 R) / .7 = .21 R / .7 = 0.3 R

After thinking about this a bit, I realized that with a little bit of density information, we have what we need to make the calculation by @TheDS1337 more exact.

We are talking percent by volume. So we can ignore the fancy stuff about mass fraction, molar fraction and what not.

We start with 6.7 fluid ounces of 91% IPA.

The density of 90% IPA is 0.8192. The density of 100% IPA is 0.786. If we do a linear interpolation, that is an estimated$$\frac{9 \times 0.8192 + 1 \times 0.786}{10}= 0.816 \text{ specific gravity for 91% IPA}$$That means that we have $$6.7 \times 0.816 = 5.47 \text{ ounce mass of 91% IPA}$$
@TheDS1337 calculates that we need a resulting volume of 8.71 fluid ounces to attain the desired 70% concentration by volume. Volume of solvent plus volume of solute may not equal volume of solution. But mass of solute plus mass of solvent will be equal to the mass of the resulting solution. And we can figure out how much mass we need to end up with.

The density of 70% IPA is 0.88. That's$$0.88 \times 8.71 = 7.66 \text{ ounce mass of 70% IPA}$$So we need$$7.66-5.47 = 2.19 \text{ ounce mass of pure water to be added}$$The specific gravity of water is 1, so that's 2.19 fluid ounces.

Indeed, my calculations puts into consideration the molecules of water and ethanol are both treated equally, although it's not the case because ethanol and water density is not the same. with ethanol being 789 kg/m³ and water being 997 kg/m³

In other words, because water is heavier, it'll take more water to fill up the volume gap

 

[HRG]

W = (.91 R / .7) - R = (.91 R - .7 R) / .7 = .21 R / .7 = 0.3 R

Thanks.
Just figured out another way to simplify:

W = (.91R / .7) - R
W = (1.3R) - R
W = 1.3R - 1R
W = .3R

W = .3 * 6.7 oz
W = 2.01 oz of water to add.

Your answer (.3R) was verification that my simplification method is valid. Thanks.

 

[hutchphd]

And this is not rocket science...better too much alcohol than too little...of course one does need to procure it.