Why does a 57.15% ABV spirit (ethanol+water) have a density of 923 kg per cubic meter?

Question

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 $\pu{923 kg m-3}$?

Answer 1

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:

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.

Answer 2

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):

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):

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:

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