# Ethanol+Water Lab Questions

So we did a lab where we mixed 10 ml of 95%ethanol with 10 ml of water and the volume of the mixture we measured was 19.60 ml. Even though we know the volume of the mixture is 19.60 our teacher wants us to calculate what the volume of the mixture should have really been and compare it to what we got.

Now the first thing we need is the densities of 95% ethanol,water and the mixture
which ethanol we know it's .804 according to a chart, water I think is just 1?, and the mixture I'm having a little bit of trouble. I know we used this equation c1v1=c2v2 and c2 is what were looking for so please correct me if I'm wrong I put this (.95)(10?)=(c2)(19.60) and I got C2=.48 and the density of this is .918.

Now I have the find the mass of each 10 ml liquid so would that just be mass= D x V
so for ethanol it would be 10*.95 and water 10*1 and then you add both mass so I got 19.5 and now
the last step is the find the volume using the sum of mass so would it be V=M*D which I got 19.5 and now the density I multiply would it be .918? If so I get this 17.90 does this sound about right or should my calculations have added up to 20 ml I know it's a lot but atleast I tried lol I just need someone to double check please

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Bystander
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So we did a lab where we mixed 10 ml of 95%ethanol with 10 ml of water and the volume of the mixture we measured was 19.60 ml. Even though we know the volume of the mixture is 19.60 our teacher wants us to calculate what the volume of the mixture should have really been and compare it to what we got.

Check your notes, please. You were probably asked to calculate the ideal solution volume you expect from mixing the two original volumes, and to compare that to the measured volume.

Now the first thing we need is the densities of 95% ethanol,water and the mixture which ethanol we know it's .804 according to a chart, water I think is just 1?
Close enough.

, and the mixture I'm having a little bit of trouble. I know we used this equation c1v1=c2v2

This is the equation you use for an ideal solution, good.

and c2 is what were looking for

You are interested in V2, not C2.

so please correct me if I'm wrong I put this (.95)(10?)=(c2)(19.60) and I got C2=.48
Now we get to the ambiguities of usage of the concept of percentage in chemistry (and a whole lot of other fields); it is unspeakably common to specify solutions, mixtures, assays, analytical results, impurities, as xx.x% without stating a basis, mass %, volume %, mole %. The basis is left to the intuition of the user, implicit in context or custom of the particular subfield (sugar refining, moonshining, what have you).

If you choose to understand 95% as a volume percentage, "(.95)(10) = what?" And, to get this moving, it's the volume of ethanol. How much water is in the first solution? How much water is in the final solution? And, how much ethanol?

and the density of this is .918.
Now I have the find the mass of each 10 ml liquid so would that just be mass= D x V so for ethanol it would be 10*.95 and water 10*1 and then you add both mass so I got 19.5 and now the last step is the find the volume using the sum of mass so would it be V=M*D which I got 19.5 and now the density I multiply would it be .918? If so I get this 17.90 does this sound about right or should my calculations have added up to 20 ml I know it's a lot but atleast I tried lol I just need someone to double check please
The red highlighting is where you've started mixing volume and mass percentage calculations, and we'll sort that out if you're interested. For the moment, just do the volume calculations for water and ethanol in the two solutions separately, and for the mixture of the two solutions.

• Ihate Mymajor
Check your notes, please. You were probably asked to calculate the ideal solution volume you expect from mixing the two original volumes, and to compare that to the measured volume.

Yes I was suppose to calculate the ideal solution volume.

Close enough.

This is the equation you use for an ideal solution, good.

You are interested in V2, not C2.

Here I'm a little confused he wrote the equation in the white board and I feel like he underlined C2 either way I'll do it both ways just to be sure. What would my C2 be then?

Now we get to the ambiguities of usage of the concept of percentage in chemistry (and a whole lot of other fields); it is unspeakably common to specify solutions, mixtures, assays, analytical results, impurities, as xx.x% without stating a basis, mass %, volume %, mole %. The basis is left to the intuition of the user, implicit in context or custom of the particular subfield (sugar refining, moonshining, what have you).

If you choose to understand 95% as a volume percentage, "(.95)(10) = what?" And, to get this moving, it's the volume of ethanol. How much water is in the first solution? How much water is in the final solution? And, how much ethanol?

The red highlighting is where you've started mixing volume and mass percentage calculations, and we'll sort that out if you're interested. For the moment, just do the volume calculations for water and ethanol in the two solutions separately, and for the mixture of the two solutions.
Ok I think I see how I did my volume calculations wrong. For ethanol it should have been 10ml* .804(because this is the density) and that should give me the volume? or am I still on the wrong path haha. I think my volume calculation for water was right (10 * 1(density of water). So now my total mass is 18.04. Now I'm confused on how to find the volume of 18.04 grams because we should need volume ( which I assume the c1v1=c2v2 calculation is for but I'm not sure what to put as c2 to get v2) and the density which I'm now lost on where we get that is it the density of C2? sort of how the density of c1 was .804 because of 95% ethanol.? Thanks

Chestermiller
Mentor
Check your notes, please. You were probably asked to calculate the ideal solution volume you expect from mixing the two original volumes, and to compare that to the measured volume.
If it's an ideal solution, then shouldn't the final volume be 20 ml? Of course, another way of answering this is to find the initial volumes of water and ethanol in the 95% solution, and then applying the ideal mixing rule (no volume change on mixing) to the combination of the two solutions. Of course, you get two different answers doing it this way. I'm guessing that the 20 ml answer is what the teacher was looking for.

Chet

Borek
Mentor
Now I'm confused on how to find the volume of 18.04 grams
Mass is additive, so you can use the information given to calculate percent concentration. Then, consult your density tables.

Bystander
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Ok I think I see how I did my volume calculations wrong. For ethanol it should have been 10ml* .804(because this is the density) and that should give me the volume?

If density is defined as mass per unit volume, or mass divided by unit volume, and you multiply density by volume, what will you get as a result?

Let's stick with the volume calculation for starters: you have measured volumes of 95% ethanol and of water that are both 10 ml; if the ethanol solution is 95% by volume, how much water by volume and how much ethanol by volume are there?

or am I still on the wrong path haha. I think my volume calculation for water was right (10 * 1(density of water). So now my total mass is 18.04. Now I'm confused on how to find the volume of 18.04 grams because we should need volume ( which I assume the c1v1=c2v2 calculation is for but I'm not sure what to put as c2 to get v2) and the density which I'm now lost on where we get that is it the density of C2? sort of how the density of c1 was .804 because of 95% ethanol.? Thanks
"C" stands for concentration, "V" for volume, and the subscript "2" indicates that the values are for the diluted solution resulting from the addition of water.

For now just focus on understanding what is in the 10 ml of 95% ethanol. We'll walk through this ONE step at a time.

Chestermiller
Mentor
If density is defined as mass per unit volume, or mass divided by unit volume, and you multiply density by volume, what will you get as a result?

Let's stick with the volume calculation for starters: you have measured volumes of 95% ethanol and of water that are both 10 ml; if the ethanol solution is 95% by volume, how much water by volume and how much ethanol by volume are there?

"C" stands for concentration, "V" for volume, and the subscript "2" indicates that the values are for the diluted solution resulting from the addition of water.

For now just focus on understanding what is in the 10 ml of 95% ethanol. We'll walk through this ONE step at a time.
Is the original solution 95% by volume or 95% by mass? According to this reference
http://www.pharmcoaaper.com/pages/Quality_Systems/work_instructions/quality_work_instructions/calculating_weight_percent_for_190_proof.pdf
it's by mass.

Chet

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If it's an ideal solution, then shouldn't the final volume be 20 ml? Of course, another way of answering this is to find the initial volumes of water and ethanol in the 95% solution, and then applying the ideal mixing rule (no volume change on mixing) to the combination of the two solutions. Of course, you get two different answers doing it this way. I'm guessing that the 20 ml answer is what the teacher was looking for.

Chet
Yes I also think he wants us to calculate 20 ml and compare it to the 19.60 ml we measured the problem is my calculations just aren't adding up to 20ml.

Bystander
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Is the original solution 95% by volume or 95% by mass? According to this reference
http://www.pharmcoaaper.com/pages/Quality_Systems/work_instructions/quality_work_instructions/calculating_weight_percent_for_190_proof.pdf
it's by mass.

Chet
Yes, indeed, and that's been bothering me. The OP doesn't include details pointing to terribly elegant dilatometry (temperature, pipette and pycnometer calibrations), which suggests that this lab was intended more as a demonstration of non-ideal solution behavior (0.5 ml in 20 is pretty hard to miss), than as a test of actual lab technique (19.6 is ~ 0.1 ml off, depending on T).

Hence, the attempt to just walk through a couple different approaches to a "predicted" volume.

Yes, indeed, and that's been bothering me. The OP doesn't include details pointing to terribly elegant dilatometry (temperature, pipette and pycnometer calibrations), which suggests that this lab was intended more as a demonstration of non-ideal solution behavior (0.5 ml in 20 is pretty hard to miss), than as a test of actual lab technique (19.6 is ~ 0.1 ml off, depending on T).

Hence, the attempt to just walk through a couple different approaches to a "predicted" volume.
Yeah I'm sorry guys this wasn't the main focus our our lab just a small section needed. Our main question for this section was why 10ml of ethanol+10ml of water added together doesn't add up to 20ml. I just needed these calculations for a table. I'm not sure if the 95% is mass or volume I just know we used ethanol that was at 95% not sure if that helps.

Bystander
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Yes I also think he wants us to calculate 20 ml and compare it to the 19.60 ml we measured the problem is my calculations just aren't adding up to 20ml.
Your concern is understandable --- "10 + 10 = 20" is just too simple to feel comfortable --- "What color was George Washington's white horse?" The answer is just sitting there staring at you, and it is not obvious that the question is at all intended as written.

Chestermiller
Mentor
Yes, indeed, and that's been bothering me. The OP doesn't include details pointing to terribly elegant dilatometry (temperature, pipette and pycnometer calibrations), which suggests that this lab was intended more as a demonstration of non-ideal solution behavior (0.5 ml in 20 is pretty hard to miss), than as a test of actual lab technique (19.6 is ~ 0.1 ml off, depending on T).

Hence, the attempt to just walk through a couple different approaches to a "predicted" volume.
So, if I understand correctly, you are trying to predict the final volume (without assuming ideal solution behavior) so that you can compare it with the 19.6? Is this correct?

Chet

If density is defined as mass per unit volume, or mass divided by unit volume, and you multiply density by volume, what will you get as a result?

Let's stick with the volume calculation for starters: you have measured volumes of 95% ethanol and of water that are both 10 ml; if the ethanol solution is 95% by volume, how much water by volume and how much ethanol by volume are there?

"C" stands for concentration, "V" for volume, and the subscript "2" indicates that the values are for the diluted solution resulting from the addition of water.

For now just focus on understanding what is in the 10 ml of 95% ethanol. We'll walk through this ONE step at a time.

If density is defined as mass per unit volume, or mass divided by unit volume, and you multiply density by volume, what will you get as a result?

Let's stick with the volume calculation for starters: you have measured volumes of 95% ethanol and of water that are both 10 ml; if the ethanol solution is 95% by volume, how much water by volume and how much ethanol by volume are there?

"C" stands for concentration, "V" for volume, and the subscript "2" indicates that the values are for the diluted solution resulting from the addition of water.

For now just focus on understanding what is in the 10 ml of 95% ethanol. We'll walk through this ONE step at a time.
hey I'm sorry I meant mass not volume when I wrote this

"For ethanol it should have been 10ml* .804(because this is the density) and that should give me the volume?"
ok so my mass calculated would be 8.04 grams for ethanol

So, if I understand correctly, you are trying to predict the final volume (without assuming ideal solution behavior) so that you can compare it with the 19.6? Is this correct?

Chet
Yes that is correct

Borek
Mentor
95% v/v is equivalent of 93.7% w/w.

10 mL of 95% w/w ethanol plus 10 mL of water (which is just a 0% w/w ethanol solution) yields 19.4 mL

Bystander
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Okay, now we're all getting to the same page.

8.04 g of 95% ethanol solution; multiply by .95 to get mass of ethanol; divide that by mass of 95% solution plus 10g water and get percentage by mass for final solution; take that to the tables and get a density; divide density by the total mass and get a final volume. That's tabulated non-ideal solution behavior.

Bystander
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95% v/v is equivalent of 93.7% w/w.

10 mL of 95% w/w ethanol plus 10 mL of water (which is just a 0% w/w ethanol solution) yields 19.4 mL
19.48, but we're both guessing at T.

95% v/v is equivalent of 93.7% w/w.

10 mL of 95% w/w ethanol plus 10 mL of water (which is just a 0% w/w ethanol solution) yields 19.4 mL

95% v/v is equivalent of 93.7% w/w.

10 mL of 95% w/w ethanol plus 10 mL of water (which is just a 0% w/w ethanol solution) yields 19.4 mL

Thank you so much for the video I appreciate it so if my calculations come out correctly I should get 19.40 and then I can compare it to the 19.60 I actually measued

Borek
Mentor
19.48, but we're both guessing at T.
Sure. Actually I don't even remember at the moment what is the temperature in tables used in CASC. And it is a midnight here and I am too lazy to check.

Borek
Mentor
Thank you so much for the video I appreciate it so if my calculations come out correctly I should get 19.40 and then I can compare it to the 19.60 I actually measued
I think you should first say "assuming volumes are additive yields 20 mL", then say what is the volume calculated after taking density tables into account (and you were already told how to get the correct result), and finally compare that with the volume measured.

Okay, now we're all getting to the same page.

8.04 g of 95% ethanol solution; multiply by .95 to get mass of ethanol; divide that by mass of 95% solution plus 10g water and get percentage by mass for final solution; take that to the tables and get a density; divide density by the total mass and get a final volume. That's tabulated non-ideal solution behavior.
Ok so here's how I set it up 8.04*.95/? +10 ok one last question though mass of the 95% solution I must divide is the total mass right so 18.04grams? If so I got this .42% the mass of this is .931 ok now for the final volume .931/18.04 and I got .051 aww like I was so close to the answer.

Ok so here's how I set it up 8.04*.95/? +10 ok one last question though mass of the 95% solution I must divide is the total mass right so 18.04grams? If so I got this .42% the mass of this is .931 ok now for the final volume .931/18.04 and I got .051 aww like I felt so close to the answer.

Borek
Mentor
8.04*.95/? +10
I am not sure what you mean by that. I guess you aim at the total mass of the final solution. If so, that's not correct. You forgot about mass of water present in the ethanol solution.

Chestermiller
Mentor
Pure ethanol = 0.78934 gm/ml
Pure water = 0.99823 gm/ml
95% by weight ethanol = 0.80424 gm/ml

Take as a basis 100 gm of solution.
Grams water = 5
Grams ethanol = 95
Volume of solution = 100/0.80424 = 124.34 ml
Grams water in 10 ml of this solution = (5)(10)/124.34=0.40212 gm
Grams ethanol in 10 ml of this solution = (95)(10)/124.34 =7.6403 gm

You are going to be mixing this with 10 ml of pure water.
Mass of 10 ml of pure water = (10)(0.99823)=9.9823

In the final solution, you have 9.9823 + 0.40212 = 10.384 gm water
7.6403 gm ethanol
Total mass of solution = 7.6403+10.384=18.024 gm
Mass percent ethanol = (7.6403)(100)/(7.6403+ 10.384)=42.4 %
Density of a 42.4% solution = 0.930
Volume of solution = 18.024/0.930 = 19.4 ml

Chet

Ok so here's how I set it up 8.04*.95/? +10 ok one last question though mass of the 95% solution I must divide is the total mass right so 18.04grams? If so I got this .42% the mass of this is .931 ok now for the final volume .931/18.04 and I got .051 aww I was so close to the answer.
I am not sure what you mean by that. I guess you aim at the total mass of the final solution. If so, that's not correct. You forgot about mass of water present in the ethanol solution.
ok I fixed it and I see I was dividing .931/18.05 instead of 18.05/.931 which would get me 19.40 thanks