# Taking volume contraction into account when mixing water with ethanol

To calculate the amount of purified water and the amount of ethanol (96% Vol.) which is needed for a desired/variable amount of a mixture - given a desired/variable percentage of ethanol - I use the following formula:

$$W = \left(\frac{T * P}{ 100}\right) \times \frac{1}{0.96}$$

and

$$E = T - W$$

Where $$T$$ is the desired Total amount of product (in liters); $$P$$ is the desired Percentage of ethanol in the final product (in %Vol.); $$W$$ is the calculated amount of Water (in liters) and $$E$$ is the calculated amount of Ethanol (in liters containing 96 %Vol ethanol).

For example - using this formula - when I need $$\pu{40 liters}$$ product containing 50% vol. ethanol, the calculated amounts of water and ethanol for the mixture become $$19.1667 : 20.833$$.

The problem in the above calculation is that because of volume contraction the desired total amount of the final product will be less as given in the formula. To me, the problem became even more confusing after I learned that volume contraction depends on the mixing ratio of both components in a nonlinear way.

What changes should I made to the formula to take volume contraction into account?

• How did you derive above equation? Or it is given? Dec 7, 2019 at 17:54
• @MathewMahindaratne I tried to figure out this myself. I am not a chemistry guy. I use the industrial 96% ethanol as solvent for aroma's in a small lab setting. If measures scale up to 50-100 liters the volume contraction starts to play a significant role.
– iep
Dec 7, 2019 at 18:10
• This is almost a non-question. Separately adding the calculated volumes is actually fine as the total volume, even though it's contracted, does not preclude that they were in the volume ratio you wanted at the beginning. Analytical chemists will know all about this. To not mix to a known volume by adding one solvent to the other, but to measure them all separately and then mix together. If you need 40L and the calculated amount gives less than this, simply measure out the volume ratios and top up. Furthermore, any temperature change due to mixing is going to add more inaccuracy. Dec 7, 2019 at 19:59
• @Beerhunter It is a real question. I measure about 10% difference in calculated and measured values
– iep
Dec 7, 2019 at 20:03
• Well since there is a volume change and you - I assume - do this at atmoshperic pressure there will be pressure volume work involved. This process comes toghether with other changes like temperature. I guess that the pressure volume work should be avoided or at least be sutracted when determining the mixing enthalpy. Jul 23, 2020 at 18:35

Measuring out volumes can sometimes be easier than dealing with mass, but in this case, converting the calculations to mass-centric can make the process easier. (I'll use your symbols, but they will now be in mass units (kg). If you desire percentages of ethanol by volume, you still need to convert to mass units because of the volume contraction!))

First, calculate the total mass 𝑇 in kg of the volume of solution needed; divide the volume by the density of the specific ethanol-water solution. You can look up this density in a handbook or Google it. One simple table is at https://www.engineeringtoolbox.com/ethanol-water-mixture-density-d_2162.html, although many tables are available, even to tenths of a percent.

Second, calculate the mass of 96% ethanol needed: 𝐸 = (𝑇∗𝑃/100)×1/0.96. The ratio 𝐸/𝑇 will be higher than 𝑃, the % desired, because the ethanol is only 96%. Convert this to volume in liters, if you desire, by dividing by 0.80138, the density of 96% ethanol.

Third, calculate the amount of pure water needed: 𝑊 = 𝑇 − 𝐸. 𝑊/𝑇 will be less than 100 - 𝑃 because of the 4% water already in the ethanol. But the total mass will give you the volume required, and the % ethanol will be correct - BY MASS.

Commercial isopropanol solutions are frequently sold as 70% by volume; this calculates out to 64.7% by weight. Your procedure, if you desire to make variable % solutions BY VOLUME, will need to do a calculation before you even do the first step: convert the % ethanol by volume to % ethanol by mass.

• The ethanol-water-mixture-density table is a great resource. It clearly shows the significance of my "problem". How did you find the the density of 96%?
– iep
Dec 7, 2019 at 18:22
• I appreciate you took effort to demonstrate the calculations by mass. However, we have no tooling for precise measuring of masses > 50 kg. Measuring huge volumes is way easier in the current setting. Nevertheless, you made me thinking of "correcting" my formula with a "volume contraction ratio", probably derived from table values. Would such an approach work? I mean: say we have 96% ethanol at hand, and we want 100 liters of 70%, 60% or 35%, could we "just" multiply/divide with a value derived from the density table?
– iep
Dec 7, 2019 at 18:32
• @iep unless you're using calibrated flow meters and if for example you were using a graduated vessel, your error in measuring the volume probably knocks out any concerns around the percentage volume loss due to contraction. Dec 7, 2019 at 19:56
• @iep can't you use the ratios you derived above but use 11.5% more volume of each? You'll have more than 40L of 50-50 water-ethanol. Use 40L (since you can measure accurately) and keep the rest for the next batch Dec 7, 2019 at 21:04
• Start with finding the volume required, but convert to mass, then calculate the amounts needed, then convert back to volume - all using the density table. For very common % solutions required, make a reference table for, say, 100 liters, then adjust as needed for the needed quantity. For density of 96% ethanol, I used handymath.com/cgi-bin/ethanolwater3.cgi?submit=Entry. (I didn't want to overwhelm you with all the decimals.) Don't over-calculate to more decimals than you can measure in your product. Dec 8, 2019 at 14:36

Based on the comment to an answer elsewhere by OP, I'd try to solve this problem, purely based on volumes. However, these calculations has ignored the volume contraction of 96% ethanol may have showed initially (that should be minimal since it is only ~4% of water by volume in there).

I'd say, your equation, regardless of how OP has derived it, is erroneous. I also like to change the variables OP has used to followings:

• $$V_{Tot}$$ is the desired Total amount of desired ethanol-water solution (in liters);
• $$P_\%$$ is the desired Percentage of ethanol in the final solution (in $$\%(v/v)$$;
• $$V_{W}$$ is the calculated amount of Water (in liters); and
• $$V_{E}$$ is the calculated amount of $$96\%(v/v)$$ ethanol (in liters).

Accordingly, the needed volume of pure ethanol $$= V_{Tot} \times \frac{P_\%}{100}=0.96V_{E}$$. Thus, $$V_{E} = \frac{V_{Tot} \times P_\%}{100 \times 0.96}= \frac{V_{Tot} \times P_\%}{96} \tag{1}$$

Note that this is actually OP's first equation, but it is for $$V_{E}$$ ($$E$$ in OP's notation) instead of $$V_{W}$$ ($$W$$ in OP's notation).

Now we can derive the equation for $$V_{W}$$. Actual $$V_{W}$$ is:

$$V_{W}=V_{Tot}\left(\frac{100-P_\%}{100}\right)$$

Yet, we cannot use $$V_{W}=V_{Tot}-V_{E}$$, because some water is coming from $$V_{E}$$. That amount of water is $$0.04V_{E} = 0.04 \times \frac{V_{Tot} \times P_\%}{96}$$. Thus, we can manipulate this equation as follows:

$$V_{W}=(V_{Tot}-V_{E})-0.04 \times \frac{V_{Tot} \times P_\%}{96}= V_{Tot}\left(\frac{100-P_\%}{100}\right)-0.04 \times \frac{V_{Tot} \times P_\%}{96}\\=\frac{96-P_\%}{96}\times V_{Tot}$$

$$\therefore \; V_{W}=\frac{96-P_\%}{96}\times V_{Tot} \tag{2}$$

Now we apply these two equation to OP's example of making 50% solution:

If $$P_\% = 50$$ and $$V_{Tot} = \pu{50 L}$$, from equation $$(1)$$ and $$(2)$$,

$$V_{E} = \frac{V_{Tot} \times P_\%}{96}= \frac{50 \times 50}{96} = \pu{26.042 L}$$

$$V_{W}=\frac{96-P_\%}{96}\times V_{Tot} = \frac{96-50}{96}\times 50 = \pu{23.958 L}$$

Thus, theoretical total (disregarding contraction) is $$\pu{50 L}$$. Acutally, practical volume must be a little off but your percentage by volume is much close to 50% (only volume contraction did not account is initial 96% ethanol).

Late edition to fulfill OP's request:

If $$P_\% = 35$$ and $$V_{Tot} = \pu{100 L}$$, from equation $$(1)$$ and $$(2)$$,

$$V_{E} = \frac{V_{Tot} \times P_\%}{96}= \frac{100 \times 35}{96} = \pu{36.458 L}$$

$$V_{W}=\frac{96-P_\%}{96}\times V_{Tot} = \frac{96-35}{96}\times 100 = \pu{63.542 L}$$

• Do you mean notation rather than notification? Dec 7, 2019 at 20:17
• @airhuff : Yes, that's what I meant and corrected accordingly. I appreciate your careful reading. :-) Dec 7, 2019 at 20:57