Great question, @Bonevader! @Roguelab @7000G @TrueScience …the denaturant does make a difference!
Using ebulliometry, the boiling point changes that occur for a solvent when solutes are dissolved in it, to tell concentration doesn’t work very well for a partial azeotrope. I’m sure you can guess why, knowing the definition or at least the concept of an azeotrope.
Using hydrometry, which IS literally a measurement of water concentration, most often in ethanol, is unlikely to work properly in heptane-denatured ethanol, specifically because of the biphasic (2 layer) liquid system and the differences in density between all 3 of them at a given temperature! Hydrometry uses the higher density of water than ethanol to determine their concentrations, relative to one another. Inside a transparent cylinder, filled about 2/3rds full of ethanol:water solution, floats the hydrometer; a precisely weight-calibrated, graduated, air-filled, glass ampoule-shaped bulb of neutral buoyancy (at room temperatures), which has a thin tube that protrudes to a measured height above the solution. Along this tube, atop the sealed floating hydrometer bulb, are horizontal graduation lines at which the meniscus of the liquid is read to determine the proof or ethanol percentage, and, by subtraction of the reading from the full scale (200 proof or 100%), the water percentage. The reading is also corrected for differences in density at temperatures above or below the normal room temperature of 20°C, by simultaneously reading a thermometer that is submerged in the same cylinder of liquid, and adding or subtracting points of the hydrometer scale to or from the hydrometer reading, as denoted by the temperature correction table, included with the hydrometer.
Now, knowing this sensitive density measurement is the basis for hydrometry, we must look at the density tables of each component. That is water, ethanol and the denaturant, n-heptane… or at least their temperature normalized densities (d in g/cm³ @ 20°C)
n-heptane = 0.6837 g/cm³
ethanol = 0.7894 g/cm³
water = 0.9982 g/cm³
Hmm… Well, as you can see, they each have very different densities, and since we do not actually know any of their concentrations already, figuring those concentrations out from density is like trying to solve a polynomial equation with 3 variable/constant pairs. If n-heptane had the same density as either ethanol or water, the hydrometer might be effective… but it isn’t. So now what? Well, we need some sort of combined density for these compounds with a known set of concentrations. Oh! Well we have the ternary azeotrope! As long as there is enough of each, we can distill some of the volume of this mixture at its precise azeotropic boiling point, right? Well… sort of. Let’s just go look it up to be more certain.
_________ b.p. . b.p.mix . % . density
ethanol w/ … 78.4 ------- … — . ---------
n-heptane … 98.5 . 70.9 … 51 . 0.729
or
water … … 100 … 78.1 … 5 . 0.804
and
… water … ethanol … n-heptane = Total, Up layer, Lo layer
%T 6.1 … T 33.0 … T 60.9 … b.p.mix 68.8
%U 0.2 … U 5.0 … U 94.8 … d.U 0.686
%L 15.0 … L 75.9 … L 9.1 … d.L 0.801
Ugh. So if we have the whole azeotrope in liquid form without layers (not really possible under normal conditions), then:
6.1% is water = 0.9982 g/cm³
33% is ethanol = 0.7894 g/cm³
60.9% is n-heptane = 0.6837 g/cm³ …right? Well, no. Water 5% and ethanol 95% have a special hydrogen bond induced density, and heptane 51% and ethanol 49% also have a special density. Therefore, the actual density must be ascertained from the Upper and Lower layers of the ternary heptane:ethanol:water bilayer average. Finally, a break in the fact that these 2 layers are split very nearly half and half! So the sum of 50% of each density is 0.7435, which is 50.5% of the sum of (ethanol+heptane) normal densities! Iow, it is almost equidistant (right in the middle) from heptane and ethanol’s normal densities. Confused? I am!
Basically, this means that if you keep the 2 layers agitated together, somehow, without altering the apparent density on the hydrometer with bubbles or vortexes, their average density composes about 95% of the actual ethanol:water density, as read by the hydrometer. So, as long as your liquid is on or pretty close to the hydrometer’s calibrated temperature (probably 20°C), you should be able to measure the density with a hydrometer, adjust for the actual liquid temperature using a thermometer and your little paper table that should have come with your hydrometer, then assume the water is about 1.7% higher than that adjusted reading.
Caveat: I am hella sleep-deprived, so feel free to tear my math apart!
Just a weird probably meaningless note I picked up in my research for this post…
By the way, there is some evidence to suggest that azeotropes are pressure dependent, and some are rather sensitive. Some sources say, for example, that the 95:5 ethanol:water azeotrope does not occur at reduced pressure! So vacuum distillation, alone, can allow ethanol to boil out of the 5% water in 190 proof! This is a rather promising theory, but sadly it is not entirely true when the 190 proof is mixed with other compounds, such as cannabis resin. However, it is still effective to “reproof” any 190 with excess water in it back to 190 proof by vacuum distillation.
