Chemistry - Taking volume contraction into account when mixing water with ethanol

Solution 1:

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 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, although many tables are available, even to tenths of a percent.

Second, calculate the mass of 96% ethanol needed: = (∗/100)×1/0.96. The ratio / will be higher than , 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: = − . / will be less than 100 - 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.

Solution 2:

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:


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 $(2)$,

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

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