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determination of alcohol values at 20 °C, which is the required reporting temperature for many countries, though 15.56 °C (or 60 °F) is still sometimes used in the U.S. For dealing with other temperatures the reader is referred to Chen.3 As further aid to the reader a table tabulating all chemical terms is attached as an Appendix at the end of this article.

TA B L E 1

illustration of the WATER PARTIAL MOLAR VOLUME of ethanol present. Note, the molar mass of e CALCULATION PROCEDURES definitions). The 4 column refers to the num th

Partial molar table of of ethanol present. Note, the molar mass of ethanol is 46.068 (see appended 99.98% (w/w) ethanol stock willofhave (100-99 ofV(total) ethanol present. Note,g/mol then /n molar mass ethanol is 4 ABW density th n1 n2 x1 V/n volume-water 1 2 1 number of moles of water in the composition. Thus definitions). The 4 column refers to the th molar mass ofrefers watertoisthe 18.015 g/mol definitions).The The 4 column number of (see mola 99.98 0.7893 2.1703 0.0011 of 126.695 0.9995 58.3773 0.0005 13.8235 ethanol present. Note, thewill molar of ethanol is 46 99.98% (w/w) ethanol stock will have 99.98% (100-99.98) g/ ethanol 18.015 (g/mol) = 0.0011 moles of water. (w/w) stock havemass (100-99.98) g/ com 18.0 V(total) refers to the total volume of this th definitions). Theof4 58.7201 column refers to the number of mole The99.02 molar mass (see also appended table of definitions). 0.7923of water 2.1494is 18.015 0.0544g/mol 126.215 0.9753 0.0253 14.2183 The molar mass water is is 18.015 g/mol (see also appen 0.7893 g/mL. It simply calculated as 100 g/0 99.98% (w/w) ethanol stock will have (100-99.98) g/ 18.0 97.99 refers 0.7954 0.1116of this 125.723 0.9502 when 59.1061 0.0525 14.3097density is V(total) to the2.1271 total volume composition theismass is 100g V(total) refers total volume of thisand composition wh The molar mass ofthe water 18.015 g/mol (see Theto mole fraction (x1) column afteralso the appen V(tota 0.7893 g/mL. It is simply calculated as 100 g/0.7893 (g/mL) = 126.695 mL for 99.98 % (w/w). 96.99 0.7984 2.1054 0.1671 125.251 g/mL. 0.9265 59.4911 calculated 0.0794 0.7893 It is simply as14.6435 100 g/0.7893 (g/m V(total) refers toare theinterested total volume of this composition whm If we in calculating the partial The95.99 mole fraction ) column 0.2226 after the V(total) column is calculated as n 1/(n 1+n 2) 0.8013 (x12.0837 124.797 0.9035 59.8933 0.1068 14.7351 ) column after the V(total) column is The mole fraction (x 1 0.7893 g/mL. It is simplyeverything calculatedhas as 100 (g/m illustration), to beg/0.7893 normalized If we are interested in calculating the partial molar volumeinofcalculating (the 2ndpartial component in ourV component inwater our illustration). This includes If wemole are interested molar volum Finding Partial The fraction (x1) column afterthe the V(total) column is st illustration), everything normalized byillustration). the number of moles alcohol (the 1 columns of Table 1.beVof(total)/n in our This includes , and nby /n Molar Volumes for Ethanol and Water has to be component illustration), everything has to normalized 1 2 1the num Ifcomponent inofcalculating theincludes partial molar shown inwe theare next two columns Table , and n2/n1.1 shown in the next twovolum component in our illustration). This includes V interested (total)/n 1illustration). in our This Vcan (total)/n In order to calculate partial molar volumes for ethanol and water, 1, The partial molar volume of water then be Theillustration), partial molar volume of waterhas canto then be calculated as:by the everything be normalized num columns of Table 1. percentages a set of density data with various corresponding weight columns of Table 1. component in our illustration). of alcohol (ABW) is needed. Fortunately, Table Vb of the legal   Vtotal This  includes V (total)/n1, The partial molar volume of water can then be calculated as:  of water The partial volume metrology tables known as the OIML tables (www.oiml.org/en/files/   can then be calculated columns of molar Table 1.  V    n1   pdf_r/r022-e75.pdf/view) consists of such a set of data. Figure      V table compared The partial molar 1 shows a plot of data selected from that    of  nwater   can then be calculated OIML    nV2 total volume  total 2 n1      with data obtained at the UW-Green  n      n1   using an  V Bay  (UWGB)   campus  V   1   1  n1      Equation Anton Paar density meter. The OIML data are consistentwith that  Vtotal     Equa    n2  n at theUWGB    n2    n2  n data obtained from the author’s laboratory 1   n2  campus.  1 n    V  1   EQUATION   the  with   VEqua Table 1 presents a set of data values dealing      n11   1   partial total    nhow      Considering the first two data rows: Considering the first two data rows: n    n  molar volumes of water are calculated.  2  n1   2   n1   The first two columns in Table 1 are taken from the OIML table:  Vtotal    n1    and  VtotalmL   (58the  .7201 two 58.3773 mL  0 .3428 Considering the first twoor data the first column refers to alcohol weight percentage (w/w) ABW, rows: Considering   (58.7201  first data)rows:   n n  1  n1  0.0248 mo the second column is the corresponding density in g/mL.  V ) mol  2   (0.02531  0.00051  The third column refers to the number of moles of ethanol in the  two data rows:  total   (58.7201  n1first Considering the n1   the mixture is 100 g. Thus,  n of composition assuming the total mass  calculated  . Thus  n2mol composition 99.98% (w/w) is asthe  (0.02531  0.00051 ) moland  0.0248  2of 99.98 the partial molar volume ofmol water for    ( 0 . 02531 0 . 00051 ) mol  0 . 0248  . Thus the a 99.98 % (w/w) solution consists g of ethanol in 100   n  1  nn1  g of the ethanol-water mixture and this will have 99.98 g/46.068   Vtotal   composition 99.98% (w/w)mass is calculated  0(w/w) .00051 mol as2   (0.02531    0.0248 composition 99.98% as mol . Thus the (g/mol) or 2.1703 moles of ethanol present. Note, the molar  is) calculated n n   0 . 3428 V mL    1   Thus the partial molar volume of water for the composition 99.98 of ethanol is 46.068 g/mol (see appended table of definitions).      1     13 . 82     V % (w/w) is calculated  99.98%  isn2calculated composition as The fourth column refers to the number ofmoles of water mol asnV2 total   in the  total  n1 (w/w)      0.0248  composition. Thus a 99.98 %(w/w) ethanol stock will have (100  n1   0.3428 V    mL V mL   n1   2 0.3428 n1     moles  molar  13.82   of water.  Equation  99.98) g/18.015 (g/mol) = 0.0011 The     V   13.82  Equa  total   0 . 0248 n mol    n     n2   2(see  n1 also 2 0.0248 n2  n mol  appended  table of mass of water is 18.015 g/mol      n  partial 0.3428 mLof water for th V  1 This  n   molar volume definitions).      isn1the   13 . 82 Equa   1   1    mol  n2  n1   n2   0.0248 V(total) refers to the total volume of this composition when the to calculate the parti  n are Readers  encouraged mass is 100 g and density is 0.7893 g/mL.partial It is simply calculated This is the molar volume of water the first with 99.98% (w/w) alcohol. Thisfor is the partial molar 1   volume of water for the first com  composition as 100 g/0.7893 (g/mL) = 126.695 mL for 99.98 % (w/w). The final plots of the partial molar volumes of EQUATION 2ethanol in Readers to calculate the partialare molar volumesto ofcalculate a similar way. The mole fraction (x1) column afterare theencouraged V(total) column is Readers encouraged the partial molar vo This is the partial molar of for the first com shown inwater Figure-2. Plotting the fraction (x1) isvolume calculated as n1/(n1+n2). This is the partial molar volume of water for the first composition The final the partial molar volumes of both water and ethanol asvolumes a the function thewate mole has thethe advantage in that the partial calculated value The final plots of partial molar of of both If we are interested in calculating theplots partialofmolar volume of Readers are encouraged to calculate molar vo with 99.98 % (w/w) alcohol. in Figure-2. Plotting the partial molar volumes against the mole fraction (x1) is shown due to the fact that different labs might have water (the second componentfraction in our illustration), everything has Readers are encouraged to calculate the partial molar volumes of fraction (x1) is shown in Figure-2. Plotting the partial mol The final plots of the volumes both to be normalized by the number of moles of alcohol (the the first calculated has the advantage in that values are ofmolar lab data and isofimportant ethanol with 99.98 % independent (w/w) alcohol. composition data. has the advantage in partial that the calculated values are water indep is shown inofFigure-2. the partial mola fraction (x1)fact due to the fact that different labs might have different ways presenting the have weight due to the that different labsPlotting might different w has the advantage in that the calculated values are indep composition data. WWW.ARTISANSPIRITMAG.COM   81 composition data. due to the fact that different labs might have different w

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Artisan Spirit: Spring 2017  

The magazine for craft distillers and their fans.

Artisan Spirit: Spring 2017  

The magazine for craft distillers and their fans.