According to the UK proof standard, a 100 proof spirit was defined as a spirit with gravity of $\frac{12}{13}$ that of water or $\pu{923 kg m-3}$, which is equivalent to $\mathrm{57.15\% \ ABV}$.

One liter of $\mathrm{57.15\% \ ABV}$ spirit contains $\pu{571.5 cm3}$ ethanol and $\pu{428.5 cm3}$ water. If the densities of ethanol and water at $\pu{20 ^\circ C}$ are $\pu{0.789 g cm-3}$ and $\pu{0.998 g cm-3}$, respectively, then the density of spirit should be $\pu{0.8785 g cm-3}$ or $\pu{878.5 kg m-3}$.

Where did I go wrong with that? How can a $\mathrm{57.15\% \ ABV}$ spirit have a density of $\pu{923 kg m-3}$?

  • 3
    $\begingroup$ See also wikipedia: Volumefraction : In chemistry, the volume fraction φi is defined as the volume of a constituent Vi divided by the volume of all constituents of the mixture V prior to mixing. $\endgroup$
    – Poutnik
    Feb 8 at 11:05
  • 10
    $\begingroup$ The flip side of Poutnik's comment is that mixing 100.00 ml of water and 100.00 ml of ethanol doesn't yield 200.00 ml of solution, but rather something less. $\endgroup$
    – MaxW
    Feb 8 at 20:12
  • 7
    $\begingroup$ You also may understand this contraction by thinking that H2O are tiny molecules that are able to fill the vacuums between the bulky and cumbersome ethanol molecule $\endgroup$
    – Maurice
    Feb 8 at 20:14
  • 1
    $\begingroup$ @MaxW About 196 mL. $\endgroup$
    – Poutnik
    Feb 10 at 11:41

There is nothing wrong with your calculations: but some liquids don't play nice

Your calculations are, likely, correct.

But liquids, unlike ideal gases, don't necessarily obey simple rules when mixed. In particular the two components, ethanol and water, strongly interact with each other when mixed and those interactions mean that the volumes of the pure liquids are not additive when they are mixed.

The effect is well known and a chart of it looks like this: https://commons.wikimedia.org/wiki/File:Excess_Volume_Mixture_of_Ethanol_and_Water.png

This is a fairly common behaviour for liquid mixtures where the components of the mixture interact. But the effect for ethanol/water is particularly strong and of practical significance both to chemists, industry and the tax authorities.

  • 4
    $\begingroup$ I'll note that water and ethanol are the typical choice when providing a "lost volume" classroom experiment. Often, such a demonstration is preceded by a similar experiment using ping pong balls, ball bearings, and sand. $\endgroup$
    – Brian
    Feb 9 at 2:52
  • 1
    $\begingroup$ This is very kind of you all! I know so much more than before I posted my question. The last afternoon I spent with understanding molar fractions and volume changes of water-ethanol mixtures yet I could not come up with an equation that would result 923 kg/m3. I can't tell if it's due to my poor arithmetic skills or getting the volume change incorrectly. If it was possible, could someone please show me the calculation done right? Thank you very much in advance! $\endgroup$
    – effmate
    Feb 9 at 10:02
  • 2
    $\begingroup$ @effmate - Because volumes don't add, you also can't interpolate to get the densities of mixtures. I'm sure that there is some curve fit formula for such an often used mixture like water and ethanol, but it is fit to experimental data. There is as of yet no good theoretical way to calculate the volume or densities of mixtures. $\endgroup$
    – MaxW
    Feb 9 at 10:51
  • $\begingroup$ @effmate this might help you if you're interested mathscinotes.com/2016/07/… $\endgroup$ Feb 9 at 10:53
  • $\begingroup$ What is the order of the maximum relative deviation? 10%? 1%? 0.1% Sub ppm? $\endgroup$ Feb 10 at 2:17

OP’s question 1: Where did I go wrong with that?

OP’s only mistake on the calculation here is the assumption of alcohol and water are additive. However, as matt_black pointed out in his answer that alcohol and water are not additive since the smaller water molecules can take up some of the space between the larger alcohol molecules, causing volume reduction (see the amount of volume reduction of various aqueous alcoholic mixtures displayed in the plot in matt_black’s answer).

OP’s question 2: How can a $57.15\% \ \mathrm{ABV}$ spirit have a density of $\pu{923 kg m−3}$?

According to this Wikipedia article, the United Kingdom has used the Alcohol by Volume (abbreviated as ABV) standard to measure alcohol content, as prescribed by the European Union (EU) since January 01, 1980:

ABV is a standard measure of how much alcohol (ethanol) is contained in a given volume of an alcoholic beverage (expressed as a volume percent). It is defined as the number of milliliters ($\pu{mL}$) of pure ethanol present in $\pu{100 mL}$ of solution at $\pu{20 ^\circ C}$ ($\pu{68 ^\circ F}$). This can be done in two ways. For example:

One of the two ways to make $50\% \ \mathrm{ABV}$ from pure alcohol: One would take 50 parts of alcohol and dilute it to 100 parts of solution with water while mixing the solution (thus, the amount of water added would be a little above 50 parts).

The second way to make $50\% \ \mathrm{ABV}$ is by volume fraction: Here, one would take 50 parts alcohol and 50 parts water measured separately, and then mix these two 50 parts together thoroughly. The resulting final volume would not be 100 parts, but a little less (between 96 and 97 parts).

These two mixtures would have different densities since the water content in each is different. I assume, the way EU suggest is by the second method. The following calculations confirm it.

You can calculate the densities of series of solutions of $\mathrm{ABV}$ by using experimental density values published for $\mathrm{ABW}$ here at $\pu{20 ^\circ C}$ ($\mathrm{ABW}$: Alcohol by Weight).

$$ \begin{array}{|c|c|c|c|} \hline\ \mathrm{ABW} & \text{density} & V_\ce{EtOH} & V_\ce{H2O} & V_\text{Total} & \mathrm{ABV} \\ \hline 0 & 0.998 & 0 & 100.200 & 100.200 & 0 \\ 10 & 0.982 & 12.675 & 90.180 & 102.855 & 12.32\\ 20 & 0.969 & 25.349 & 80.160 & 105.509 & 24.03\\ 30 & 0.954 & 38.023 & 70.140 & 108.163 & 35.15\\ 40 & 0.935 & 50.697 & 60.120 & 110.817 & 45.75\\ 50 & 0.914 & 63.372 & 50.100 & 113.472 & 55.85\\ 60 & 0.891 & 76.046 & 40.080 & 116.126 & 65.49\\ 70 & 0.868 & 88.720 & 30.060 & 118.780 & 74.69\\ 80 & 0.843 & 101.394 & 20.040 & 121.434 & 83.50\\ 90 & 0.818 & 114.068 & 10.020 & 124.088 & 91.91\\ 100 & 0.789 & 126.743 & 0 & 126.743 & 100\\ \hline \end{array} $$

Here, $V_\ce{EtOH}$ is volume of alcohol in the $\mathrm{ABV}$ solution and $V_\ce{H2O}$ is volume of water in the same solution (e.g., in $10\% \ \mathrm{ABW}$ at $\pu{20 ^\circ C}$, volume of $\ce{EtOH}$ is $\frac{\pu{10 g}}{\pu{0.789 g mL-1}} = \pu{12.675 mL}$). Consequently, $V_\text{Total}$ is $V_\ce{EtOH} + V_\ce{H2O}$, the total volume of alcohol and water in the $\pu{100 g}$ of particular $\mathrm{ABW}$ solution without considering the volume lost.

When you plot the $\mathrm{ABV}$ versus density you get a polynomial curve with equation $y = -1.0 \times 10^{-5}x^2 - 0.0007 x + 0.9956$ with $R^2 = 0.9996$ (an excellent agreement):

Plot of ABV Vs Density

Using the plot and the equation, now you can calculate the density of any $\mathrm{ABV}$ solution at $\pu{20 ^\circ C}$. For example, let's calculate density for $\mathrm{57.15\% \ ABV}$:

$$y = -1.0 \times 10^{-5}x^2 - 0.0007 x + 0.9956 \\ = -1.0 \times 10^{-5} \times (57.15)^2 - 0.0007 \times 57.15 + 0.9956 = 0.923$$

This is an excellent agreement with the definition.

Also, the beauty of the data analysis in above table is you can find the $\% \ \mathrm{ABW}$ of any given $\% \ \mathrm{ABW}$. For example, if you plot the $\% \ \mathrm{ABV}$ versus $\% \ \mathrm{ABW}$, you get a polynomial curve with equation $y = 0.0023 x^2 + 0.7617 x + 0.1878$ with $R^2 = 1$ (an excellent agreement):

Plot of %ABV versus %ABW

Using this plot and the corresponding equation, now you can calculate the $\% \ \mathrm{ABW}$ of any given solution if you know its $\% \ \mathrm{ABV}$ at $\pu{20 ^\circ C}$. For fun, let's calculate $\% \ \mathrm{ABW}$ for $\mathrm{57.15\% \ ABV}$ in OP's question:

$$y = 0.0023 x^2 + 0.7617 x + 0.1878 \\ = 0.0023 \times (57.15)^2 + 0.7617 \times 57.15 + 0.1878 = 51.23\%$$


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