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I have a binary mixture, say ethanol at $90\%wt$ ethanol, $10\%wt$ water. According to Perry's Chemical Engineers' Handbook 9th Ed., that mixture has a density of $0.81797g/cc$ (at $20^{\circ}C$). So my $1L$ of mixture should have a mass of $817.97g$.

Now, I want to create this mixture. I know the total mass and the mass %'s, so I just multiply them out: we find that we need $736.17g$ ethanol and $81.80g$ water. Ok, so far so good. However, say all of my balances are broken, so I only have graduated cylinders to work with. Easy, just convert those masses back to volumes using their respective pure substance densities: $0.78934g/cc$ for ethanol and $0.99823g/cc$ for water.

However, doing this tells me that I should have $932.6mL$ of ethanol and $81.9mL$ of water, which adds up to a total volume of $1014.6mL$ of binary mixture, so somehow I magicked $14.6mL$ of liquid out of thin air using this conversion algorithm.

Is this due to the negative excess volume of the ethanol/water mixture upon mixing? That's my current guess, but for the life of me I'm having a hard time figuring out the math here. It seems like the change in volume should be much smaller: $10\%wt$ water in ethanol is about $10\%mol$ and there's about $4.5mol$ of water in the mixture, so based on this dataset it seems like there should only be a difference of about $3mL$. Even with all the rounding I just did, it seems like this can't be the only error source.

UPDATE: That last step was where my error was. For some reason, I calculated the excess volume based on only the number of moles of water, not the total moles of the mixture. When using the total volume of the mixture (as one should), and a new set of tables (since the old ones can no longer be accessed 4 years later), the total volume effect comes out to -14.4mL, which is nearly exactly what my calculations predicted. Thanks to Anger Density for finally narrowing down the error, and do check out Ananta's answer for a thorough breakdown of this kind of calculation.

Is there something wrong with the logic of my algorithm here? Can anyone point me to the reason I'm seeing this numerical error whenever I run this calculation?

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  • $\begingroup$ Hmmmm. I see 0.81942 in a different reference for 90 wt.% ethanol. $\endgroup$
    – Jon Custer
    Jan 7, 2019 at 23:23
  • $\begingroup$ @JonCuster Ahh, is it hard copy or could you link it? $\endgroup$ Jan 7, 2019 at 23:25
  • $\begingroup$ wissen.science-and-fun.de/chemistry/chemistry/density-tables/… for online, and 0.8194 in my old CRC... $\endgroup$
    – Jon Custer
    Jan 7, 2019 at 23:27
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    $\begingroup$ @JonCuster Interesting! I usually trust both CRC and Perry's quite a lot, I wonder why the disparity. I'm at the gym, I'll look into it when I get home, or maybe when I get back to work tomorrow $\endgroup$ Jan 7, 2019 at 23:33
  • $\begingroup$ My CRC has 0.8180 as the 'relative density at 20C, kg/l' while 0.8194 is the 'specific gravity at 20C'. $\endgroup$
    – Jon Custer
    Jan 7, 2019 at 23:43

3 Answers 3

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I think your calculations are exactly right. Because of the negative excess volume of ethanol-water solutions, you will mix a larger total volume of separate ethanol and water to make your 90% weight solution of ethanol.

In fact, the excess volume is "built in" to the density measurement listed in Perry's. Let's calculate the required excess volume to reproduce this density. Using your already calculated values, I get:

$n_{total}=n_{ethanol}+n_{water}=16.0+4.5=20.5$ moles total solution

$x_{ethanol}=\frac{n_{ethanol}}{n_{total}}=\frac{16.0}{20.5}=0.78$

$v_{excess}=\frac{\Delta v}{n_{total}}=\frac{\pu{-14.6mL}}{\pu{20.5mol}}=\pu{-0.71mL/mol}$

Comparing this calculated value to the closest one I can find in the tables you referenced, it looks like they list -0.678 ml/mol for an ethanol mole fraction of 0.7805. Seeing as the DDBST data are for 25 degrees Celsius and yours are at 20, I would call the agreement excellent!

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  • $\begingroup$ Interesting! This makes it seem like I might have made a mistake not in my original calculation, but in my assessment of how large the excess volume effect should be. I estimated 3mL when I originally posted this question a few years ago (vs the ~14mL actual), but I may have messed up a conversion between mass and moles or some similar thing. I'm not in a situation to work through the math right now, but I'll try to remember to look at this during the workday tomorrow and check the math! $\endgroup$ May 14, 2023 at 23:48
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    $\begingroup$ I just redid all the math and you're totally right. Working it through the way I did, there are indeed about 20.5 total moles in the solution, and with an excess volume of mixing of -0.7mL/mol (which I found here since I can no longer access the tables I linked years ago) this comes out to -14.4mL excess volume, nearly exactly what I found! Accepted. $\endgroup$ May 15, 2023 at 16:35
  • $\begingroup$ As additional followup: I can't reconstruct exactly what I did back when I first posted this, since I can't get at the tables. However, based on the values from the new set of tables I linked in the last comment, it looks like I originally (for some reason) calculated excess volume based off only the 4.5 moles of water. I have no idea why I did that, but it gives approximately -3mL excess volume, which is the number that caused my initial confusion. I will update the original question to highlight where my error was. $\endgroup$ May 15, 2023 at 16:38
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First Things First

Ethanol and water do form a solution with a negative excess volume, like you correctly said. This is pertinent to answering your question. For further reading, refer to this Wikipedia page on Excess Properties.

Evaluating Your Answer Step by Step

Step 1: Calculation of Mass of Solution.

Since density of the solution is given. This calculation is accurate.

$$ m_\text{solution} = \rho\times V = \pu{0.81797 g cm^3} \times \pu{1000 cm^3} = \pu{817.97 g} $$

Step 2: Calculating Weights of Ethanol and Water

Since weight ratios are given, this calculation also seems straightforward.

$$ m_\text{ethanol} = \pu{0.9} \times m_\text{solution} = \pu{0.9\times 817.97 g = 7.4\times 10^{3} g}\\ m_\text{water} = \pu{0.1} \times m_\text{solution} = \pu{0.1\times 817.97 g = 8.2\times 10^{1} g} $$

Step 3: Calculating Volumes of Ethanol and Water

This is where your problem lies. The thing is, when ethanol and water are mixed together, these don't mix ideally. That is, $V_\text{ethanol}+V_\text{water} \neq V_\text{solution}$. Actually,

$$ V_\text{ethanol}+V_\text{water} > V_\text{solution} $$

Thus, you will need more volumes of ethanol and water than required by calculations using mass.

Remember: mass and volume analysis often contradict due this deviation. This is why some scientists, especially physicists, prefer molalilty as opposed to molarity.

The Fix

Start with volume, we are required to make $\pu{1000 cm^3}$ of ethanol-water mixture. Assume $V_\text{ethanol}$ and $V_\text{water}$ are required to make this solution. Now, We know that

$$ \begin{align} &V_\text{ethanol}+V_\text{water} > V_\text{solution}\\ \implies &=\dfrac{m_\text{ethanol}}{\rho_\text{ethanol}}+\dfrac{m_\text{water}}{\rho_\text{water}} > \dfrac{m_\text{solution}}{\rho_\text{solution}} \end{align} $$

Here you can use the exact values.

$$ \dfrac{m_\text{ethanol}}{\pu{0.78934 g cm^{-3}}} + \dfrac{m_\text{water}}{\pu{0.99823 g cm^{-3}}} > \dfrac{\pu{817.97 g}}{\pu{0.81797 g cm^{-3}}} $$

This is an equation of the type

$$ \dfrac{x}{\pu{0.78934 g cm^{-3}}} + \dfrac{y}{\pu{0.99823 g cm^{-3}}} > \dfrac{\pu{817.97 g}}{\pu{0.81797 g cm^{-3}}} \tag{1} $$

The Problem with the Fix

Unfortunately, based on theoretical analysis, this is as far as I can get. I can help you with a strategy to find the exact masses. Since it is given that

$$ \dfrac{m_\text{ethanol}}{m_\text{water}} = 9 \tag{2} $$

Try different values of $m_\text{ethanol}$ and $m_\text{ethanol}$ following Eqs. (1) and (2). Unfortunately, without a quantitative relation between $m_\text{ethanol}$ and $m_\text{water}$ in Eq. (1), this is as far as we can get. An (informed) trial and error seems like your best option here.

Fixing the Problem with the Fix

Start from the ideal case (the dotted line in the graph) and follow Eq. (2) (dark solid line in the graph), that is Eq. (2), which in terms of volume given below.

$$\dfrac{m_\text{ethanol}}{m_\text{water}} = 9 \text{ or } \dfrac{\pu{0.78934} \times V_\text{ethanol}}{\pu{0.99823} \times V_\text{water}} = 9 $$

This means you need to follow the dark solid line (eq2) in the green-shaded region.

Geogebra graph for experimental volumes in case of positive deviation

until you get the desired result. That is,

$$ V_\text{solution} = \pu{1000 cm^{3}} $$

Note: Since other answers have verified densities, I am assuming these are correct

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    $\begingroup$ And so: "Is the negative excess volume of ethanol/water the cause of the OP's math error?" $\endgroup$ May 13, 2023 at 0:03
  • $\begingroup$ @ToddMinehardt not only are OP's calculations somewhat inaccurate (since OP reasoned it with "negative volume of mixing"), but all the answers are also wrong suggesting you need a higher than required volume of ethanol and water because of "negative volume of mixing," because ethanol and water show "positive deviation from ideal mixing behavior." Remember, we do need to prepare a $\pu{1000 cm^{3}}$ of solution using only graduated cylinders and the answer I posted provides a well-reasoned method to do just that. $\endgroup$
    – ananta
    May 13, 2023 at 2:42
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    $\begingroup$ This answer is not correct. The statements "Ethanol and water do form a solution with a negative excess volume..." and "you will need less volumes of ethanol and water than required by calculations using mass" are incompatible. $\endgroup$ May 13, 2023 at 12:11
  • $\begingroup$ @AngerDensity thank you for pointing out the discrepancy. I have made the correction. $\endgroup$
    – ananta
    May 13, 2023 at 13:12
  • $\begingroup$ I appreciate the effort you put into this answer, and to revise it to make sure it is accurate. I don't think it answers my exact original question, since I was aware of the concept of excess volume but the effect seemed to be of the wrong magnitude. However I think that for anyone finding this Q&A in the future it's very valuable. I saw you comment elsewhere that you may extend it to the general case, and I hope you do! Perhaps one of us could also find a new table of values for excess volume in the EtOH/H2O system, the one I linked 4 years ago seems to have expired. $\endgroup$ May 14, 2023 at 23:57
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When you mix ethanol and water they take up less space due to the way the molecules fit together, so you will have what is called a negative volume of mixing. In this case, the 932.6 mL of ethanol and the 81.9 mL of water will give you a 100 mL solution of 90% ethanol.

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    $\begingroup$ Presumably you mean a 1 liter solution (1000mL)? $\endgroup$
    – Jon Custer
    Nov 2, 2020 at 17:30
  • $\begingroup$ Ethanol and water show positive deviation from ideal mixing behaviour. That is, $\mathrm{V_\text{actual}}>\mathrm{V_\text{ethanol}}+\mathrm{V_\text{water}}$. $\endgroup$
    – ananta
    May 12, 2023 at 18:54
  • $\begingroup$ @Ananta Exactly the opposite, V_act < V_eth + V_water. You may confuse it with positive deviation from Raoult law, where mixture has higher then ideal vapour pressure. $\endgroup$
    – Poutnik
    May 13, 2023 at 6:03
  • $\begingroup$ @Poutnik Oh! I see it now. Think I will be deleting this answer then. Still, does the analysis stand correct for solutions with positive excess volumes? $\endgroup$
    – ananta
    May 13, 2023 at 6:06
  • $\begingroup$ @ananta I think the answer could be easily tweaked to address the general case of nonzero excess mixing volume, both signs simultaneously. $\endgroup$
    – Poutnik
    May 13, 2023 at 6:20

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