Is the negative excess volume of ethanol/water the cause of my math error?

Question

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?

Answer 1

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!

Answer 2

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.

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

Answer 3

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.