[Tinctures, Control of Alcohol Level] © Bjørn Larsen & D. Foley, 1997. We can't place enough emphasis on the point that mathematics is a language. If we can translate the question into an algebraic expression, or model, we often get quickly to the answer. For example: Forty percent alcohol by volume => Forty percent of the volume of the fluid is alcohol. 40%Et-OH / vol => (0.40) * (FluidVolume)mL = (Alcohol)mL. (a)If we want 25%Et-OH/vol and have 40% we must dilute the tincture. Twenty-five percent of the final volume is to be alcohol. Find the new volume. (0.25) * Vol[final]mL = (Alcohol)mL. so we have: (0.25) * Vol[final]mL = (Alcohol)mL, Vol[final]mL = (Alcohol)mL / (0.25), and from above we can substitute equivalent expressions, Vol[final]mL = {(0.40)*FluidVol[initial]} / (0.25), Vol[final]mL = (0.40)/(0.25) * FluidVol[initial]. How much water to add? Add enough water to get to Vol[final]. nb:Water and alcohol do not mix 1:1. (b)If we want 55%Et-OH/vol and have 40% we must reinforce the tincture with more alcohol. This is easy with pure alcohol, using the above definition of `percent alcohol by volume'. It gets more involved using vodka, etc. as water is added along with the alcohol. I call this the "clown shoes effect". As we approach our goal our attempts to get there push it yet further away. Clown shoes type problems are often solved using the calculus. We shall use algebra only. We want the Et-OH%/vol[final] to be 55%. We have two primary volumes of interest: (Et-OH)[final] & (FluidVolume)[final]. Fifty-five percent(P) of the final volume of the fluid is alcohol. (I): (P) * (FluidVolume) = (Et-OH). Or: (II): (Et-OH)[final] / (FluidVolume)[final] = (P)[final]. (III): FluidVolume[1] + FluidVolume[2] = FluidVolume[final]. (IV): (Et-OH[1]) + (Et-OH[2]) = (Et-OH[final]). And, by substituting equivalent expressions into (II) above, we get: (V): [(Et-OH[initial]) + (E-OH[final])] / (FluidVolume[initial] + FluidVolume[final]) = (P[final]). (VI): ( P[1] * FluidVolume[1] + P[2] * FluidVolume[2] ) / (FluidVolume[1] + FluidVolume[2]) = (P[final]). Which I'd prefer to think of as: (VI): (ab + cd) / (b + d) = X. Find d: (VII): (ab + cd) = X(b + d), (VIII): ab + cd = Xb + Xd, (IX): (ab - Xb) = (Xd - cd), (X): b(a - X) = d(X - c), (XI): d = b(a - X) / (X - c). Or, substituting back to a more useful form: (XI): FluidVolume[2] = FluidVolume[1] * (Percent[1] - Percent[final]) / (Percent[final] - Percent[2]). (XI): V[2] = V[1] * (P[1]-P[final]) / (P[final]-p[2]). (XII): V[2] = V[1] * (P[final]-P[1]) / (P[2]-P[final]). V[2] is the amount of the stronger alcoholic solution to add to the original volume of tincture(V[1]) to get the desired concentration of alcohol. If the final volume does not add up to the sum (V[1]+V[2]) then top up with distilled water. Now try it with 100mL of our 25% tincture and 90% Polish vodka. How much vodka to add to the tincture to get:(a) a 55% tincture? (b) a 30% tincture? Summary: To reduce the strength of an alcohol tincture: V[final] = V[initial] * P[initial] / P[final]. nb: Dilute the tincture up to V[final] with pure water. To increase the alcoholic strength of a tincture: V[2] = V[1] * (P[final]-P[1]) / (P[2]-P[final]). nb: The reinforcing alcohol V[2] is higher in proof than fluid V[1] the tincture. Add V[2] to V[1]. If the final volume does not add up to the sum (V[1]+V[2]) then top up with distilled water. As I mentioned, the tincture fortification model is built upon an approximation. Here is a calculation for the error: Let P[final] = 1, let V[1] + V[2] = 1, (the predicted final volume) & let V[actual] = 2/3, (the measured tincture volume) then P[actual] = P[final]/(1-E), = P[final] * 1/(2/3), = 3/2. So: P[actual]= P[final](final volume predicted by model)/(actual final volume). (III): FluidVolume[1] + FluidVolume[2] = FluidVolume[final]. This assumption above may work as an estimate...but below is a more accurate model. (III.b) ( WaterVol + AlcVol) + (HydratedAlcohol/WaterAlcoholate)Volume = FluidVolume[final] Each mole of (HydratedAlcohol/WaterAlcoholate) removes a mole each of water and alcohol from the solution. What is the distribution of the total water and alcohol in these groups? These ratios are a mystery to me, so I write the following: To calculate the error(E) in terms of the estimated P[final]: E = (V[1] + V[2] - actual volume) / (V[1] + V[2]). This is the percent deviation of the measured volume from the calculated quantity. E, the error [percent deviation], is the RATIO of the difference [absolute deviation] of the theoretical [modeled] quantity from the measured or actual quantity. This error tends to over-estimate the size of the denominator in the percent equation. Our final tincture volume, V[1]+V[2], is (E*100) percent less than estimated. The corrected strength is: P[actual] = P[final]/(1-E), (+ or -) the uncertainty due to the limits of measurement accuracy. n.b. (1-E) = (actual tincture volume) / (V[1] + V[2]). (1-E) = 1 - {(V[1] + V[2] - actual volume) / (V[1] + V[2])}, = 1 - (V[1] + V[2])/ (V[1] + V[2]) + (actual volume) / (V[1] + V[2]), = 1 - 1 + (actual volume) / (V[1] + V[2]), = (actual volume) / (V[1] + V[2]), = (actual final tincture volume) / V[final from the model]. One can calculate the actual, corrected alcohol strength. It seems more convenient to adjust the volume of the mixed fluids. Top up to estimated volume; add more water. Check to see that the correction of the percentage alcohol is significant. Is the variation within the limits of measurement accuracy or not? Perhaps this approach works for you? € Bjørn & Donna 1997. This document may be copied. Please remit $1.00 per copy to Island Enviromint. Additional copies may be ordered for $2.00 each. 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