ethanol and water in this case are then: The volume using Equation 5, now shows Vtotal = 236.774 mL, closer to what Spedding, et V1 final 54.592 mL/mol ; V mL/mol 2 17.777 al., had calculated. The real lesson learned from this example is that after mixing 100 mL of 95% V1using 54.Equation 592 mL/mol ; V 2shows 17.777 mL/mol The volume 5, now Vtotal = 236.774 to alcohol what Spedding, et ABV final and 144 mL of water, the final volume is not 244 mL.APPENDIX ThemL, finalcloser correct content will CHEMICAL TERMS al., had calculated. Thedepending real learned thiscalculated example isvolume that after mixing 100 mL of 95% Example 4:ABV Mixing 100lesson mL ofupon 95 %which (v/v)from be 40.00-40.12% ‘true’ is used. The final volume using Equation 5, now shows Vtotal = 236.774 mL, closer to what Spedding, et AND DEFINITIONS with mL of 40final % (v/v) ethanolis not 244 mL. The final correct alcohol content will ABV andethanol 144 mL of100 water, the volume al., had calculated. The real lesson learnedGram from this example is thatmass. after mixing 100 mL of 95% or gram molecular The mass in grams numerically equal to Example 4: Mixing 100 mL of 95% (v/v) ethanol with 100 mLformula of 40% (v/v) ethanol be This 40.00-40.12% ABV depending upon ‘true’ calculated volume is used. example is quite useful when we need towhich know the final weight of a substance or the correct sum of all thealcohol atomic masses in its molecular ABV and 144 mL of water, the final volumetheismolecular not 244 mL. The final content will volume of mixing of two ethanol solutions with different ABV formula. An atom of each element has a characteristic mass and in like manner each This example isbe quite useful when we need tothat know the final volume of(v/v) mixing of twoformula 40.00-40.12% ABV depending upon which ‘true’ calculated volume isethanol used. Example 4: professionals Mixing 100 mLtechnicians of 95% (v/v) ethanol with 100 mL of molecule of a 40% compound has aethanol characteristic mass. For example, the gramvalues. Most and are aware the molecular weight of water (H O) is 15.9994 (rounded 16.0) (atomic mass of oxygen) solutions with different ABV values. Mostproducts professionals and technicians are2 aware that theto final final volume of mixing 100 mL each of two ethanol of + (2 × 1.00794) (atomic mass of hydrogen), or(v/v) 18.02 g.ethanol Example 4: Mixing 100need mL ofto95% (v/v) ethanol with 100 mL ofof 40% This example is quite useful when we know the final volume of mixing two ethanol different of ABV will not100 be 200 mL. But, the question what volume mixing mL each of two ethanolis,products of different ABV will not be 200 mL. Molartechnicians mass. The molar mass is a physical property solutions with different ABV values. Most professionals and areMaware that thedefined finalas the mass of a given is the final volume? But, the question is, what is the final volume? This example is quite useful when we needsubstance to know theelement finalorvolume of mixing ofbytwo ethanol (chemical chemical compound) divided the amount of that For a of 95 mixing % ABV,100 the mL density 0.8114 g/mL, and the volume eachis of two ethanol products of different not be 200formL. substance. The baseABV systemwill international SI unit molar mass is kg/mol. However, solutions with different ABV values. Mostmolar professionals technicians areanaware the final masses areABW almostand always expressed example,that the molar mass of corresponding ABW isis,92.41 a final 40 %g/mL, ABV, the density For athe 95% ABV, the density 0.8114 and the corresponding is 92.41 %.in g/mol. For aAs40% But, question what%.isisFor the volume? water: M(H O) is 18.02 g/mol. mixing 100ABW mLiseach of two of different ABV will not be 200 mL. 2 is 0.94805 g/mLvolume , and the of corresponding 33.305 %. ethanol products ABV, the density is 0.94805 g/mL, and the correspondingMolar ABW is 33.305 %. Those values are The molar volume, symbol Vm, is the volume occupied by one mole the question is,towhat is the final volume? volume.ABW Those values areBut, calculated according the procedures of corresponding For a 95% ABV, the density is 0.8114 g/mL, and the is 92.41 %. For a 40%at a given temperature and 1 presented of a in substance or chemical compound) in this calculated according to the procedures of Spedding, et al., their(chemical articleelement Spedding, et al., in their article1 presented in this magazine. pressure. is equal to the molar mass (M) divided by the ABV, the density is 0.94805 g/mL,we andcan the corresponding ABWxItis 33.305 %. Those values aremass density (ρ). It has the magazine. WithFor this information, and awe95% then1density is andinternational the ABW 92.41 %.although For ait 40% 1, n2,system 1: corresponding With this information, can ABV, calculate , n2, andcalculate x1:0.8114ng/mL, (SI) unit of cubic meters per is mole (m3/mol), is more 1 presented in this calculated according to the procedures of Spedding, et al., in their article 3 to use the units ofABW cubic centimeters per mole /mol) values for liquids and ABV, the density is 0.94805 g/mL, and thepractical corresponding is 33.305 %. (cm Those aresolids. 81.14With 0.9241 94.805 0.33305 23 magazine. thisinformation, we can2.calculate n2,Mole. andAxchemical 1: 1 presented mole is a in chemical unit, defined to be 6.022 x 10 n1 3procedures 130 moln1,Ethanol in molecules, this calculated according tothe of Spedding, et al., theirmass article atoms, or some other unit. The mass of a mole is the gram formula mass of a substance. 46.068 , n2of, and x12:O) has 6.022 x 1023 molecules and weighs 18.02 magazine..805 With this information, we can calculate As an example,n11mole water (H 81 [Ethanol] 81..14 1400..9241 075994 94 .80500..33305 66695 2.3130 mol Ethanol n (rounded down to 18.0) grams. 1 mole of ethanol (C2H5OH) has 6.022 x 1023 molecules 1 46.08 grams. n2 3.8517 mol Water 46 . 068 and weighs .14 0.9241 94.805 0.33305 1881 .015 n1 94.805 0.66695 2.3 130fraction. mol Aunit Ethanol 81n.14 0.0759 Mole of concentration. The mole fraction or molar fraction, symbolized 46.068 nx 2 3 . 8517 mol Water 1 (x ) is defined as the amount of a constituent (expressed in moles), ni, divided by the total 0 . 3752 i 1 1881 .015 amount of all constituents in the mixture (also expressed in moles), ntot. The sum of all the .14 0.0759 94.805 0.66695 n1 n2 mole fraction is a ratio of moles to moles, molar n 2 [Water] 3.mole 8517 molmustWater fractions equal 1. Whereas n1 18.015 x1 0.3752 concentration is a quotient of moles to volume. The mole fraction is one way of expressing The information of x1 now partial molar volumes of ethanol and n1 n2 the composition of a mixture with a dimensionless quantity; mass fraction (percentage by n1 allows for the calculation of the x 0 . 3752 weight, wt%) and volume fraction (percentage by volume, vol%) are others. water according to 1 Equations (3) & (4). The partial molar volumes are: n2 allows for the calculation of the The information of xn1 1now partialweight. molar volumes ethanol Molecular Molecular weight isof a measure of theand sum of the atomic weights of
atoms in a molecule. The according information to of Equations x1 now allows the The calculation water (3)for & (4). partialofmolarthe volumes are: Molecular weight is often used interchangeably with molecular The information of x now allows for the calculation ofbutthe molar of ethanol and mass in chemistry, there partial is a difference betweenvolumes the two. Molecular mass is a measure 1 the partial molar volumes of ethanol and water according to of mass and molecular weight is a measure of force acting on the molecular mass. We do according to Equations (3) & (4). The volumes Equations 3 and water 4. The partial molar volumes are: notpartial cover moremolar explicit details here. are:
Partial molar volume. This is broadly understood as the contribution that a component of a mixture makes to the overall volume of the solution. However, depending on which solutions are being mixed, there is rather more to this issue than seems to The calculated volume using Equation 8 becomes 198.264 at first glance. This relates binary water and ethanol solutions, for The calculated volume using Equation 8 becomes 198.264exist mL, a contraction of remarkably 1.736 mLto or mL, a contraction of 1.736 mL or 0.868 %. The calculated example, which are non-additive volume mixtures; the resulting volume upon mixing and 0.868%. The calculated alcohol content for this mixture and final volume can be seen to be temperature equilibration is not simply a sum of their respective component volumes. alcohol content for this mixture and final volume can be seen 68.09% ABV. When one mole of water is added to a large volume of water at a set temperature, the to be 68.09 % ABV. volume increases by a specific amount, typically around 18cm3. The molar volume of pure water at 25 °C, for example, would thus be reported as 18cm3 mol-1. However, addition of Conclusion 1 mole of water to a large volume of pure ethanol results in an increase in volume of only 14cm3. The thatthe the increase This article has provided an outline of the procedures involved in reason finding partialis different molar is that the volume occupied by a given This article has provided an outline of the procedures involved number of water molecules depends upon the identity of the surrounding molecules. The volumes of ethanol and water and the subsequent derivation of two useful regression 3 value 14cm is said to be the partial molar volume of water in ethanol in this case (25°C). in finding the partial molar volumes of ethanol and water and equations. Illustrations of the applications of the partial molar volumes for use in the alcoholX in a mixture is the change in volume In general, the partial molar volume of a substance the subsequent derivation of two useful regression equations. per mole applications of X added to the add mixture. Brewers and distillers work at 20 °C or at 16.65 °C beverage industry were presented with 4 key examples. These another Illustrations of the applications of the partial molar volumes for (60 °F), and this paper is directed to looking at the partial molar volumes of water and dimension to apply correct procedures for ethanol products prove useful ethanol at 20 °C which in order toshould allow them to perform correct alcohol dilutions on products use in the alcohol beverage industry dilution were presented with four and know the true final alcohol concentration of their products. for the brewer and distiller. key examples. These applications add another dimension to
V 1 57.178 mL
V 2 17.138 mL
apply correct dilution procedures for ethanol products, which References should prove useful for the brewer and distiller.
1. Spedding, G., Weygandt, A. and Linske, M. (2016).REFERENCES Alcohol dilution practices for distillers. Dr. Franklin Chen received Ph.D. in 19772016; from Princeton Artisanhis Spirit – Spring 65-70. University. He had Colgate-Palmolive (19771. Spedding, G., Weygandt, A.&and Linske, M.(2010), (2016). Alcohol Dilution Practices for 2. worked Engel, with T., and Reid, P., in “Thermodynamics, Statistical Thermodynamics Kinetics” Distillers. Artisan Spirit – Spring 2016; 65-70. 1980), Johnson and2Johnson (1980-1984), and Kimberlynd Edition, Pearson, Upper Saddle Rivers, NJ., page 216 Clark (1984-2002). He retired from Kimberly-Clark in 2002 2. Engel, T., and Reid, P., in “Thermodynamics, Statistical Thermodynamics & Kinetics” 3. Chen, F. and joined the faculty at UW-Green Bay. He was promoted (2010), 2nd Edition, Pearson, Upper Saddle Rivers, NJ., page 216. http://www2.stetson.edu?~wgrubbs/datadriven/fchen/bartender/partialmolarvolum to Associate Professor in 2008. His research interests focus on 3. Chen, F. http://www2.stetson.edu?~wgrubbs/datadriven/fchen/bartender/ echem.html Applied Research and Computation Research. He holds more partialmolarvolumechem.html than 30 patents and has more than 30 publications to his credit.
Proposed Editorial comments
The magazine for craft distillers and their fans.