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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: yngvai on October 18, 2005, 09:30:05 AM

Title: Calculating alcoholic content of a mixed drink?
Post by: yngvai on October 18, 2005, 09:30:05 AM

Trying to figure out the correct way to calculate the alcoholic content of a mixed drink.

I am only using two liquids, one alcoholic and the other water (zero alcohol)

25.4 oz of 190 proof/95% alcohol
64 oz. of 0 proof/0% alcohol

I added the two volumes together, 89.4 oz.

I calculated the percentage of alc. liquid to total liquid =  25.4/89.4 = 28.4%.

I multiplied the above 28.4% and the proof of the alcohol = .284 x 190 = 53.96 proof or 26.98% alcohol.

So I've concluded that the resulting liquid is about 54 proof.

Have I calculated correctly?  If not, what step(s) do I need to adjust?

TIA,
Steve

Title: Re:Calculating alcoholic content of a mixed drink?
Post by: Borek on October 18, 2005, 10:10:03 AM

Quote from: yngvai on October 18, 2005, 09:30:05 AM

26.98% alcohol.

Seems OK.

Somehow I don't like the way you did the calculations. I feel it is better to first calculate pure ethanol mass in the 95% solution, then to divide it by the final mass - the final result will be the same, but you are not loosing contact with reality  ;)

Title: Re:Calculating alcoholic content of a mixed drink?
Post by: yngvai on October 18, 2005, 10:21:31 AM

Thanks, Borek!  And the drinkers of my limoncello thank you!