Do the weights of two liquids not add when mixed?

I was given an interesting dilemma today. A co-worker saw me adding a liquid (Diisopropyl ethylamine AKA DIPEA) to a flask filled with another liquid (Tetrahydrofuran AKA THF). I needed to weigh out exactly 5 grams of DIPEA into the THF and so I zero'd the scale with the flask+THF on it, then proceeded to add the DIPEA until the scale said 5.000g. Since masses are additive I assumed this was fine.

My co-worker, however, stopped and told me that although masses of two liquids are additive, the combined weights would not be, and since the scale measures weight as opposed to mass I had apparently just added an incorrect amount of DIPEA. He explained the reasoning to me but I'm a chemist, not a physicist and certainly not skilled in fluid mechanics, so I would like someone to dumb it down for me a bit or tell me if I'm way off.

From what I understand, the scale measures weight which is a function of gravitational force. But gravitational force is a function of buoyant force (its less if the buoyant force is greater since the buoyant force pushed a liquid up). Finally, buoyant force is a function of density. This means that my THF (which had a density of .9 g/ml) had a greater buoyant force than my THF/DIPEA solution (DIPEA density is only .74 g/ml so the solution would be somewhere between .74 and .90). And this means that technically as I'm adding DIPEA, the added mass is not the only thing causing the weight to increase; but rather the decreased buoyant force is also causing that.

And so, when the scale finally read 5.000g, I had possibly only added 4.950 or maybe 4.990 etc (something less than 5.000). Is my reasoning correct? Any help is appreciated.

edited Dec 16 '18 at 15:51

knzhou

45.9k11123222

asked Dec 12 '18 at 14:41

Brian

380135

31

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From a practical standpoint, I'd worry much more about the fact that if you pour directly into the liquid, you won't be able to correct the mistake if you pour too much. Pouring 5.01 g would be a failure, having to scrape back 0.01 g from a separate weighing would be nothing.
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– Chieron
Dec 13 '18 at 8:08

2

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If adding the two liquids causes a reaction of either evaporation or condensation then your final weight is wrong and you added either too much or too little of liquid #2.
$endgroup$
– MonkeyZeus
Dec 13 '18 at 15:11

2

$begingroup$
And then there's the case of two fluids (not these two) combining endothermically to form a new compound. The resultant material will weigh more than the sum of the inputs by an amount (endothermic energy)/c^2 .
$endgroup$
– Carl Witthoft
Dec 13 '18 at 19:09

9

$begingroup$
You should be able to perform a simple experiment. Do it your way. Do it their way. Check if they differ in weight.
$endgroup$
– Schwern
Dec 13 '18 at 21:20

2

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Given you were measuring the Flask and Contents, the argument is specious. The scales will measure the combined mass, calibrated according to the common gravitational attraction (acceleration). The Yakk caveat does apply should either of the components, or the reaction products become gaseous as those gasses will became part of the general air pressure after a while (you can manage to weigh CO2.. )
$endgroup$
– Philip Oakley
Dec 13 '18 at 23:07

|
show 4 more comments

And so, when the scale finally read 5.000g, I had possibly only added 4.950 or maybe 4.990 etc (something less than 5.000). Is my reasoning correct? Any help is appreciated.

edited Dec 16 '18 at 15:51

knzhou

45.9k11123222

asked Dec 12 '18 at 14:41

Brian

380135

31

$begingroup$
From a practical standpoint, I'd worry much more about the fact that if you pour directly into the liquid, you won't be able to correct the mistake if you pour too much. Pouring 5.01 g would be a failure, having to scrape back 0.01 g from a separate weighing would be nothing.
$endgroup$
– Chieron
Dec 13 '18 at 8:08

2

$begingroup$
If adding the two liquids causes a reaction of either evaporation or condensation then your final weight is wrong and you added either too much or too little of liquid #2.
$endgroup$
– MonkeyZeus
Dec 13 '18 at 15:11

2

$begingroup$
And then there's the case of two fluids (not these two) combining endothermically to form a new compound. The resultant material will weigh more than the sum of the inputs by an amount (endothermic energy)/c^2 .
$endgroup$
– Carl Witthoft
Dec 13 '18 at 19:09

9

$begingroup$
You should be able to perform a simple experiment. Do it your way. Do it their way. Check if they differ in weight.
$endgroup$
– Schwern
Dec 13 '18 at 21:20

2

$begingroup$
Given you were measuring the Flask and Contents, the argument is specious. The scales will measure the combined mass, calibrated according to the common gravitational attraction (acceleration). The Yakk caveat does apply should either of the components, or the reaction products become gaseous as those gasses will became part of the general air pressure after a while (you can manage to weigh CO2.. )
$endgroup$
– Philip Oakley
Dec 13 '18 at 23:07

|
show 4 more comments

And so, when the scale finally read 5.000g, I had possibly only added 4.950 or maybe 4.990 etc (something less than 5.000). Is my reasoning correct? Any help is appreciated.

edited Dec 16 '18 at 15:51

knzhou

45.9k11123222

asked Dec 12 '18 at 14:41

Brian

380135

And so, when the scale finally read 5.000g, I had possibly only added 4.950 or maybe 4.990 etc (something less than 5.000). Is my reasoning correct? Any help is appreciated.

mass everyday-life density buoyancy weight

edited Dec 16 '18 at 15:51

knzhou

45.9k11123222

asked Dec 12 '18 at 14:41

Brian

380135

edited Dec 16 '18 at 15:51

knzhou

45.9k11123222

asked Dec 12 '18 at 14:41

Brian

380135

edited Dec 16 '18 at 15:51

knzhou

45.9k11123222

edited Dec 16 '18 at 15:51

knzhou

45.9k11123222

edited Dec 16 '18 at 15:51

knzhou

45.9k11123222

asked Dec 12 '18 at 14:41

Brian

380135

asked Dec 12 '18 at 14:41

Brian

380135

asked Dec 12 '18 at 14:41

Brian

380135

31

$begingroup$
From a practical standpoint, I'd worry much more about the fact that if you pour directly into the liquid, you won't be able to correct the mistake if you pour too much. Pouring 5.01 g would be a failure, having to scrape back 0.01 g from a separate weighing would be nothing.
$endgroup$
– Chieron
Dec 13 '18 at 8:08

2

$begingroup$
If adding the two liquids causes a reaction of either evaporation or condensation then your final weight is wrong and you added either too much or too little of liquid #2.
$endgroup$
– MonkeyZeus
Dec 13 '18 at 15:11

2

$begingroup$
And then there's the case of two fluids (not these two) combining endothermically to form a new compound. The resultant material will weigh more than the sum of the inputs by an amount (endothermic energy)/c^2 .
$endgroup$
– Carl Witthoft
Dec 13 '18 at 19:09

9

$begingroup$
You should be able to perform a simple experiment. Do it your way. Do it their way. Check if they differ in weight.
$endgroup$
– Schwern
Dec 13 '18 at 21:20

2

$begingroup$
Given you were measuring the Flask and Contents, the argument is specious. The scales will measure the combined mass, calibrated according to the common gravitational attraction (acceleration). The Yakk caveat does apply should either of the components, or the reaction products become gaseous as those gasses will became part of the general air pressure after a while (you can manage to weigh CO2.. )
$endgroup$
– Philip Oakley
Dec 13 '18 at 23:07

|
show 4 more comments

31

$begingroup$
From a practical standpoint, I'd worry much more about the fact that if you pour directly into the liquid, you won't be able to correct the mistake if you pour too much. Pouring 5.01 g would be a failure, having to scrape back 0.01 g from a separate weighing would be nothing.
$endgroup$
– Chieron
Dec 13 '18 at 8:08

2

$begingroup$
If adding the two liquids causes a reaction of either evaporation or condensation then your final weight is wrong and you added either too much or too little of liquid #2.
$endgroup$
– MonkeyZeus
Dec 13 '18 at 15:11

2

$begingroup$
And then there's the case of two fluids (not these two) combining endothermically to form a new compound. The resultant material will weigh more than the sum of the inputs by an amount (endothermic energy)/c^2 .
$endgroup$
– Carl Witthoft
Dec 13 '18 at 19:09

9

$begingroup$
You should be able to perform a simple experiment. Do it your way. Do it their way. Check if they differ in weight.
$endgroup$
– Schwern
Dec 13 '18 at 21:20

2

$begingroup$
Given you were measuring the Flask and Contents, the argument is specious. The scales will measure the combined mass, calibrated according to the common gravitational attraction (acceleration). The Yakk caveat does apply should either of the components, or the reaction products become gaseous as those gasses will became part of the general air pressure after a while (you can manage to weigh CO2.. )
$endgroup$
– Philip Oakley
Dec 13 '18 at 23:07

From a practical standpoint, I'd worry much more about the fact that if you pour directly into the liquid, you won't be able to correct the mistake if you pour too much. Pouring 5.01 g would be a failure, having to scrape back 0.01 g from a separate weighing would be nothing.

– Chieron
Dec 13 '18 at 8:08

If adding the two liquids causes a reaction of either evaporation or condensation then your final weight is wrong and you added either too much or too little of liquid #2.

– MonkeyZeus
Dec 13 '18 at 15:11

And then there's the case of two fluids (not these two) combining endothermically to form a new compound. The resultant material will weigh more than the sum of the inputs by an amount (endothermic energy)/c^2 .

– Carl Witthoft
Dec 13 '18 at 19:09

You should be able to perform a simple experiment. Do it your way. Do it their way. Check if they differ in weight.

– Schwern
Dec 13 '18 at 21:20

Given you were measuring the Flask and Contents, the argument is specious. The scales will measure the combined mass, calibrated according to the common gravitational attraction (acceleration). The Yakk caveat does apply should either of the components, or the reaction products become gaseous as those gasses will became part of the general air pressure after a while (you can manage to weigh CO2.. )

– Philip Oakley
Dec 13 '18 at 23:07

|
show 4 more comments

5 Answers
5

active

oldest

votes

132

Of course, by common sense, if you put together two objects with masses $m_1$ and $m_2$, and nothing comes out, then you end up with mass $m_1 + m_2$.

Weights are a little more complicated because of buoyant forces. All objects on Earth continuously experience a buoyant force from the volume of the air they displace. This doesn't matter as long as volume is conserved: if you stack two solid blocks their weights add because the total buoyant force is the same as before. But when you mix two liquids the total buoyant force can change, because the volume of the mixed liquid might not be equal to the sum of the individual volumes.

To estimate this effect, let's say (generously) that mixing two liquids might result in a change of total volume of $10%$. The density of air is about $0.1%$ that of a typical liquid. So the error of this effect will be, at most, around $0.01%$, which is irrelevant. Thus we can conclude that, rather than trying to help, your coworker just wanted to feel smug for a second.

answered Dec 12 '18 at 14:53

knzhou

45.9k11123222

44

$begingroup$
"at most, around 0.01%, which is irrelevant. Thus we can conclude that, rather than trying to help, your coworker just wanted to feel smug for a second." This assumes that 0.01% insignificant for their purposes in his lab.
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– Shufflepants
Dec 12 '18 at 16:32

69

$begingroup$
It's more accurate than the method of adding the liquids by hand. It's even more accurate than the scale (s)he's using. Therefore, 0.01% is irrelevant. Also, I believe this answer is, as stated, quite generous with the changes of volume. 10% is a lot, and probably more than you'd ever find outside of specific combinations for purposes of demonstrating otherwise.
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– Jacco van Dorp
Dec 12 '18 at 16:33

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@Dancrumb It's not a matter of compressibility, but potentially a matter of miscibility (solutions are typically more dense than bare solutes) and potentially a matter of chemical reaction resulting in a substance of wildly different density.
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– Beanluc
Dec 12 '18 at 18:30

54

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@Dancrumb As a rough analogy, imagine a beaker full of spheres with radius 10 mm, and another with spheres with radius 1mm. When you add them together, the volume will be subadditive because the small spheres can fit in the holes between the large ones. And it's not necessarily a physical process: you might have two molecules combining to form a molecule that is smaller than the sum of the volumes of the original molectules.
$endgroup$
– Acccumulation
Dec 12 '18 at 18:32

26

$begingroup$
If you scale is sensitive enough to "feel" the effect of the buoyant force from the air then even when you measure a single liquid the weight you read is not the "real" weight.
$endgroup$
– nasu
Dec 12 '18 at 22:18

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show 11 more comments

@knzhou supplied a good answer. I’m going to offer a couple of other interpretations.

The first has nothing to do with the fact that you’re mixing liquids—it’s just that there are difficulties is determining mass precisely by measuring weight. As already pointed out, there is the buoyancy of the air—that produces a mass error of about $-0.0013$ g/l at sea level, or about $-0.14$% in the case of a substance with a density of $0.9$ g/l. Then, the gravitational acceleration varies by location by up to half a percent, largest near the poles and smallest near the equator (also smaller at high elevation). So if your scale measures force (uses springs or load cells rather than counterbalancing mass) and is calibrated for use in Paris, you’ll get an additional ~$ -0.1$% error if you weigh your sample in Mexico City. Of course, these errors would be the same if you weighed each liquid separately as if you weigh the mixture.

Second, I don’t know any chemistry, but if your two liquids react and produce gas which escapes from the top of the container, that is obviously lost mass. If this is not the case, then your buddy is pretty much full of crap, because the only other effect I can imagine which would produce a difference based on weight pre- or post-mixing is the one already alluded to:

In oceanography it’s called cabbeling. Two substances mixed together don’t necessarily have exactly the density you’d compute from a weighted average of the densities of each. For liquids I’m familiar with, the discrepancy is far less than $10$%: more like $1$% at most, which, when combined with air buoyancy makes for an error of 0.001%. You might as well worry about the error caused by the tidal force from the moon.

answered Dec 12 '18 at 16:55

Ben51

4,011827

1

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Your second point is the one which might have by far the most impact. The weight could be $0N$ after the reaction.
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– Eric Duminil
Dec 13 '18 at 11:45

2

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Liquids also spontaneously evaporate; depending on the conditions, this can be rather dramatic. In any case, the important thing to take from this is that you should follow the procedure. If you make changes to the procedure, you need to understand the possible effects of that change. It is ridiculous to assume that changing the procedure will have no effect. A chemist should be well acquainted with e.g. the difference between adding acid to water and water to acid - even though the process is "additive", there's marked practical differences to the two approaches :)
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– Luaan
Dec 13 '18 at 13:59

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The section about calibration is rather silly in this context and I think it detracts from the point.
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– pipe
Dec 13 '18 at 15:18

2

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"You might as well worry about the error caused by the tidal force from the moon" - No. That force is a perturbation of the order of $1$ part in $10^7$, considerably smaller than the effects you are talking about. Air buoyancy effects (which depend on atmospheric pressure, and there therefore weather-dependent even at a fixed location on earth) ARE routinely included when weighing to an accuracy of $0.01%$ or better.
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– alephzero
Dec 14 '18 at 18:52

1

$begingroup$
@alephzero Of course it is important that the equipment must be correctly calibrated but its also important that they aren't pouring the liquid into a beaker with a hole in the bottom, a sponge or a pair of pants. The question is about mixing two liquids and their weights changing, the answer should address that and not standard lab practice.
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– Lio Elbammalf
Dec 15 '18 at 22:32

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show 3 more comments

Based on your question I assume the precision is very important in your lab. If so, then yes, your colleague is right. At least to some degree.

There are main three possibilities why this may happen. From your description the reason is purely physical, so option 1 on my list and I will explain it in more details. The other two are added just for completeness of the answer.

Why adding weights doesn't produce the sum of weights?

Consider these options.

If the substances mix (i.e. a result is a solution of the two substances) the volume of resultant solution is smaller than volumes of ingredients. The masses remain the same though. I'll explain this phenomenon in a bit. The reason is purely physical.

If the substances react and some of the resulting substances leaves the solution (usually it will be a gas that evaporates) the mass will reduce by the mass of removed substance. Usually the volume will also be smaller since on one hand some of the substance is removed and the resultant solution with whatever remains should follow the same principle as in 1. But the newly formed particles could in theory prevent that and cause the volume to actually increase. I can't think of any examples (or even if there exists some). The reason is mixed chemical and physical.

If the substances react and as a result you get a soluble substance you end up with a solution of a new substance. In most (if not all) cases you have less particles to pack so your volume will drop with the mass remaining the same.If additionally the solvents mix you end up with added effect as in one so the difference can be even higher. This is a mix of chemical and physical reasons.

So what happens?

Your colleague refers specifically to point 1 on my list so I'll focus on it only.

In general the main reason is that the particles in solutions might have different size (and sometimes to a degree also shape, but usually it's less important factor) and as a result they can "pack" more densely so in the same volume you can fit more particles and as a result - more mass. The most extreme case is that the substances that are solvents themselves mix well with a significant volume reduction and the solved substance have just a tiny impact on the overall density.

As a result you end up with a mass being sum of masses but volume noticeably smaller than sum of volumes. And here enters the buoyancy on stage. The difference in volume causes a difference in weight. It's not big but if you need a huuuge precision it might make you fail meeting it.

So how big an impact can be?

In general the bigger the difference in particles size the stronger effect you'll notice.

I know nothing about the substances you're referring to but let's use something well known to most people. And something you can easily test yourself (and then consume the by-products to cool down your head a bit after thinking it all over ;-) ). There are two popular good solvents that mix rather easily - water and ethanol. I'll make a perfect case scenario where we mix pure water with pure ethanol. If you want to make an experiment you'll actually use some kind of solutions where those are solvents but the results will be noticeable for sure.

First let's calculate how much volume are we going to loose. Based on this water-ethanol mixtures density we'll calculate the difference for 40% alcohol solution (popularly known as vodka).

Pure water has a density of $0.998202 text{ g}/text{cm}^3$

Pure ethanol has a density of $0.79074 text{ g}/text{cm}^3$

40% ethanol in water solution (calculated by weight) has a density of $0.93684 text{ g}/text{cm}^3$

Those densities are based normalised atmosphere so weight based.

So let's take (by weight) $60 text{ g}$ water and $40 text{ g}$ ethanol.
The volumes are respectively $60/0.998202 = 60.108074text{ cm}^3$ of water and $40/0.79074 = 50.585527text{ cm}^3$ ethanol.

A common sense would suggest we are going to have $100 text{ g}$ of solution with a volume of $60.108074 + 50.585527 = 110.693601text{ cm}^3$. Then we realise the density is going to be different, so $100 text{ g}$ of solution that measures $100/0.93684 = 106.741813text{ cm}^3$. But neither of those is correct.

To have accurate results we need to go through masses.

The density of air is $0.001204 text{ g}/text{cm}^3$

The experienced weight loss for water is $60.108074 cdot 0.001204 = 0.07237text{ g}$ and the mass of water is $60.07237text{ g}$

The experienced weight loss for ethanol is $50.585527 cdot 0.001204 = 0.060905text{ g}$ and the mass of ethanol is $40.060905text{ g}$

The mass of resultant solution is $60.07237 + 40.060905 = 100.137275text{ g}$

To calculate accurately the weight let's use proportion.

If we had $100text{ g}$ of $40%$ ethanol its volume would be $106.741813text{ cm}^3$ calculated earlier.

The experienced weight loss from buoyancy would be $106.741813 cdot 0.001204 = 0.128517 text{ g}$

The mass of it would be $100.128517text{ g}$

So the weigh of solution, which mass is $100.137275 text{ g}$ is going to be
$100/100.128517cdot100.137275 = 100.008747 text{ g}$

i.e. higher by $0.008747$ (or $0.008747%$) than expected.

If we applied this result to your case, for $5text{ g}$ you'd end up with $5.000437 text{ g}$.

The difference seems to be less than your precision. If it's acceptable or not - you need to decide yourself. Also your case might get more extreme.

Oh, just for reference - the volume of our solution would be $106.741813/100.128517cdot100.137275 = 106.751149 text{ cm}^3$

It's still significantly less than the sum of volumes, which as we've calculated was $110.693601 text{ cm}^3$.

What can you do to avoid that

Assuming the difference is significant enough to bother you, you have two options.

Mix the solution first and then measure the requested amount

Experimentally check the actual change of weight and calculate amounts to use. Note though that you risk a rounding errors this way.

Test volume loss yourself!

If you want to check it yourself take neutral spirit and water (I think apple juice will do and will add colour to better see the difference). Pour water into glass tube. Then carefully pour spirit so that it doesn't mix with water (tilt the glass tube and slowly pour it on the tube's side). Mark the current volume. Now stir or shake until both liquids mix well and wait until it settles.

The resultant mixture is safe to consume ;-)

You may also check this video.

answered Dec 13 '18 at 11:12

Ister

4516

2

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Nice answer! I was even able to comprehend the calculation methods because you used numbers instead of just letters. Thank you
$endgroup$
– undefined
Dec 13 '18 at 13:16

$begingroup$
Perfect answer - the example at the end really sold it for me.
$endgroup$
– Lio Elbammalf
Dec 15 '18 at 22:38

add a comment |

I just adhere to the scenario you have described.

Just put a container filled with water on a scale. Weight reads W1.
Put a floating object of mass w. Reads is now W1 + w.

Repeat with a fluid having different density on which the above object is still floating. If the weight of vessel and fluid is now W2 ask your colleague about the reading he/she expects after dropping in the floater of weigh w.......

edited Dec 14 '18 at 12:55

answered Dec 14 '18 at 12:47

Alchimista

636410

add a comment |

-1

I'll offer another consideration since you've mentioned buoyancy, density which is not only a function of mass but volume as well.

In non-relativistic physics mass, as determined by a weight scale in a consistent gravitational field is always conserved. So add two masses of reactants, you should expect to get the sum of the masses in the product assuming the reaction did not result in any gases that were lost from the solution.

But in a relativistic sense, energy is equivalent to mass. So if your chemical reaction is exothermic and the heat dissipates away from the solution you would find that you actually lost mass. But the equivalent mass would be so small (by a factor of $1/c^2$) you probably could not measure it. I don't think that's affecting your result.

Although conservation of mass is an absolute, conservation of volume is not. It's possible to mix volumes of two reactants, get the same mass, but the product volume may be either smaller or larger than the volume of the reactants. Why? Two possible reasons. As a chemical reaction you are forming a new molecule that may pack together spatially better than the two reactants. So since mass is the same, density would increase in that case.

Is this possibly what you might be experiencing?

answered Dec 12 '18 at 21:12

docscience

8,64821850

$begingroup$
I'm not sure "relativistic" is the correct term here. E = mc^2 is a relationship not based on special relativity. And mass gain/loss in an endo/exothermic reaction can be measured (albeit very carefully)
$endgroup$
– Carl Witthoft
Dec 13 '18 at 20:05

add a comment |

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

132

Of course, by common sense, if you put together two objects with masses $m_1$ and $m_2$, and nothing comes out, then you end up with mass $m_1 + m_2$.

answered Dec 12 '18 at 14:53

knzhou

45.9k11123222

44

$begingroup$
"at most, around 0.01%, which is irrelevant. Thus we can conclude that, rather than trying to help, your coworker just wanted to feel smug for a second." This assumes that 0.01% insignificant for their purposes in his lab.
$endgroup$
– Shufflepants
Dec 12 '18 at 16:32

69

$begingroup$
It's more accurate than the method of adding the liquids by hand. It's even more accurate than the scale (s)he's using. Therefore, 0.01% is irrelevant. Also, I believe this answer is, as stated, quite generous with the changes of volume. 10% is a lot, and probably more than you'd ever find outside of specific combinations for purposes of demonstrating otherwise.
$endgroup$
– Jacco van Dorp
Dec 12 '18 at 16:33

15

$begingroup$
@Dancrumb It's not a matter of compressibility, but potentially a matter of miscibility (solutions are typically more dense than bare solutes) and potentially a matter of chemical reaction resulting in a substance of wildly different density.
$endgroup$
– Beanluc
Dec 12 '18 at 18:30

54

$begingroup$
@Dancrumb As a rough analogy, imagine a beaker full of spheres with radius 10 mm, and another with spheres with radius 1mm. When you add them together, the volume will be subadditive because the small spheres can fit in the holes between the large ones. And it's not necessarily a physical process: you might have two molecules combining to form a molecule that is smaller than the sum of the volumes of the original molectules.
$endgroup$
– Acccumulation
Dec 12 '18 at 18:32

26

$begingroup$
If you scale is sensitive enough to "feel" the effect of the buoyant force from the air then even when you measure a single liquid the weight you read is not the "real" weight.
$endgroup$
– nasu
Dec 12 '18 at 22:18

|
show 11 more comments

132

Of course, by common sense, if you put together two objects with masses $m_1$ and $m_2$, and nothing comes out, then you end up with mass $m_1 + m_2$.

answered Dec 12 '18 at 14:53

knzhou

45.9k11123222

44

$begingroup$
"at most, around 0.01%, which is irrelevant. Thus we can conclude that, rather than trying to help, your coworker just wanted to feel smug for a second." This assumes that 0.01% insignificant for their purposes in his lab.
$endgroup$
– Shufflepants
Dec 12 '18 at 16:32

69

$begingroup$
It's more accurate than the method of adding the liquids by hand. It's even more accurate than the scale (s)he's using. Therefore, 0.01% is irrelevant. Also, I believe this answer is, as stated, quite generous with the changes of volume. 10% is a lot, and probably more than you'd ever find outside of specific combinations for purposes of demonstrating otherwise.
$endgroup$
– Jacco van Dorp
Dec 12 '18 at 16:33

15

$begingroup$
@Dancrumb It's not a matter of compressibility, but potentially a matter of miscibility (solutions are typically more dense than bare solutes) and potentially a matter of chemical reaction resulting in a substance of wildly different density.
$endgroup$
– Beanluc
Dec 12 '18 at 18:30

54

$begingroup$
@Dancrumb As a rough analogy, imagine a beaker full of spheres with radius 10 mm, and another with spheres with radius 1mm. When you add them together, the volume will be subadditive because the small spheres can fit in the holes between the large ones. And it's not necessarily a physical process: you might have two molecules combining to form a molecule that is smaller than the sum of the volumes of the original molectules.
$endgroup$
– Acccumulation
Dec 12 '18 at 18:32

26

$begingroup$
If you scale is sensitive enough to "feel" the effect of the buoyant force from the air then even when you measure a single liquid the weight you read is not the "real" weight.
$endgroup$
– nasu
Dec 12 '18 at 22:18

|
show 11 more comments

132

Of course, by common sense, if you put together two objects with masses $m_1$ and $m_2$, and nothing comes out, then you end up with mass $m_1 + m_2$.

answered Dec 12 '18 at 14:53

knzhou

45.9k11123222

Of course, by common sense, if you put together two objects with masses $m_1$ and $m_2$, and nothing comes out, then you end up with mass $m_1 + m_2$.

answered Dec 12 '18 at 14:53

knzhou

45.9k11123222

answered Dec 12 '18 at 14:53

knzhou

45.9k11123222

answered Dec 12 '18 at 14:53

knzhou

45.9k11123222

answered Dec 12 '18 at 14:53

knzhou

45.9k11123222

44

$begingroup$
"at most, around 0.01%, which is irrelevant. Thus we can conclude that, rather than trying to help, your coworker just wanted to feel smug for a second." This assumes that 0.01% insignificant for their purposes in his lab.
$endgroup$
– Shufflepants
Dec 12 '18 at 16:32

69

$begingroup$
It's more accurate than the method of adding the liquids by hand. It's even more accurate than the scale (s)he's using. Therefore, 0.01% is irrelevant. Also, I believe this answer is, as stated, quite generous with the changes of volume. 10% is a lot, and probably more than you'd ever find outside of specific combinations for purposes of demonstrating otherwise.
$endgroup$
– Jacco van Dorp
Dec 12 '18 at 16:33

15

$begingroup$
@Dancrumb It's not a matter of compressibility, but potentially a matter of miscibility (solutions are typically more dense than bare solutes) and potentially a matter of chemical reaction resulting in a substance of wildly different density.
$endgroup$
– Beanluc
Dec 12 '18 at 18:30

54

$begingroup$
@Dancrumb As a rough analogy, imagine a beaker full of spheres with radius 10 mm, and another with spheres with radius 1mm. When you add them together, the volume will be subadditive because the small spheres can fit in the holes between the large ones. And it's not necessarily a physical process: you might have two molecules combining to form a molecule that is smaller than the sum of the volumes of the original molectules.
$endgroup$
– Acccumulation
Dec 12 '18 at 18:32

26

$begingroup$
If you scale is sensitive enough to "feel" the effect of the buoyant force from the air then even when you measure a single liquid the weight you read is not the "real" weight.
$endgroup$
– nasu
Dec 12 '18 at 22:18

|
show 11 more comments

44

$begingroup$
"at most, around 0.01%, which is irrelevant. Thus we can conclude that, rather than trying to help, your coworker just wanted to feel smug for a second." This assumes that 0.01% insignificant for their purposes in his lab.
$endgroup$
– Shufflepants
Dec 12 '18 at 16:32

69

$begingroup$
It's more accurate than the method of adding the liquids by hand. It's even more accurate than the scale (s)he's using. Therefore, 0.01% is irrelevant. Also, I believe this answer is, as stated, quite generous with the changes of volume. 10% is a lot, and probably more than you'd ever find outside of specific combinations for purposes of demonstrating otherwise.
$endgroup$
– Jacco van Dorp
Dec 12 '18 at 16:33

15

$begingroup$
@Dancrumb It's not a matter of compressibility, but potentially a matter of miscibility (solutions are typically more dense than bare solutes) and potentially a matter of chemical reaction resulting in a substance of wildly different density.
$endgroup$
– Beanluc
Dec 12 '18 at 18:30

54

$begingroup$
@Dancrumb As a rough analogy, imagine a beaker full of spheres with radius 10 mm, and another with spheres with radius 1mm. When you add them together, the volume will be subadditive because the small spheres can fit in the holes between the large ones. And it's not necessarily a physical process: you might have two molecules combining to form a molecule that is smaller than the sum of the volumes of the original molectules.
$endgroup$
– Acccumulation
Dec 12 '18 at 18:32

26

$begingroup$
If you scale is sensitive enough to "feel" the effect of the buoyant force from the air then even when you measure a single liquid the weight you read is not the "real" weight.
$endgroup$
– nasu
Dec 12 '18 at 22:18

"at most, around 0.01%, which is irrelevant. Thus we can conclude that, rather than trying to help, your coworker just wanted to feel smug for a second." This assumes that 0.01% insignificant for their purposes in his lab.

– Shufflepants
Dec 12 '18 at 16:32

It's more accurate than the method of adding the liquids by hand. It's even more accurate than the scale (s)he's using. Therefore, 0.01% is irrelevant. Also, I believe this answer is, as stated, quite generous with the changes of volume. 10% is a lot, and probably more than you'd ever find outside of specific combinations for purposes of demonstrating otherwise.

– Jacco van Dorp
Dec 12 '18 at 16:33

@Dancrumb It's not a matter of compressibility, but potentially a matter of miscibility (solutions are typically more dense than bare solutes) and potentially a matter of chemical reaction resulting in a substance of wildly different density.

– Beanluc
Dec 12 '18 at 18:30

@Dancrumb As a rough analogy, imagine a beaker full of spheres with radius 10 mm, and another with spheres with radius 1mm. When you add them together, the volume will be subadditive because the small spheres can fit in the holes between the large ones. And it's not necessarily a physical process: you might have two molecules combining to form a molecule that is smaller than the sum of the volumes of the original molectules.

– Acccumulation
Dec 12 '18 at 18:32

If you scale is sensitive enough to "feel" the effect of the buoyant force from the air then even when you measure a single liquid the weight you read is not the "real" weight.

– nasu
Dec 12 '18 at 22:18

|
show 11 more comments

@knzhou supplied a good answer. I’m going to offer a couple of other interpretations.

answered Dec 12 '18 at 16:55

Ben51

4,011827

1

$begingroup$
Your second point is the one which might have by far the most impact. The weight could be $0N$ after the reaction.
$endgroup$
– Eric Duminil
Dec 13 '18 at 11:45

2

$begingroup$
Liquids also spontaneously evaporate; depending on the conditions, this can be rather dramatic. In any case, the important thing to take from this is that you should follow the procedure. If you make changes to the procedure, you need to understand the possible effects of that change. It is ridiculous to assume that changing the procedure will have no effect. A chemist should be well acquainted with e.g. the difference between adding acid to water and water to acid - even though the process is "additive", there's marked practical differences to the two approaches :)
$endgroup$
– Luaan
Dec 13 '18 at 13:59

1

$begingroup$
The section about calibration is rather silly in this context and I think it detracts from the point.
$endgroup$
– pipe
Dec 13 '18 at 15:18

2

$begingroup$
"You might as well worry about the error caused by the tidal force from the moon" - No. That force is a perturbation of the order of $1$ part in $10^7$, considerably smaller than the effects you are talking about. Air buoyancy effects (which depend on atmospheric pressure, and there therefore weather-dependent even at a fixed location on earth) ARE routinely included when weighing to an accuracy of $0.01%$ or better.
$endgroup$
– alephzero
Dec 14 '18 at 18:52

1

$begingroup$
@alephzero Of course it is important that the equipment must be correctly calibrated but its also important that they aren't pouring the liquid into a beaker with a hole in the bottom, a sponge or a pair of pants. The question is about mixing two liquids and their weights changing, the answer should address that and not standard lab practice.
$endgroup$
– Lio Elbammalf
Dec 15 '18 at 22:32

|
show 3 more comments

@knzhou supplied a good answer. I’m going to offer a couple of other interpretations.

answered Dec 12 '18 at 16:55

Ben51

4,011827

1

$begingroup$
Your second point is the one which might have by far the most impact. The weight could be $0N$ after the reaction.
$endgroup$
– Eric Duminil
Dec 13 '18 at 11:45

2

$begingroup$
Liquids also spontaneously evaporate; depending on the conditions, this can be rather dramatic. In any case, the important thing to take from this is that you should follow the procedure. If you make changes to the procedure, you need to understand the possible effects of that change. It is ridiculous to assume that changing the procedure will have no effect. A chemist should be well acquainted with e.g. the difference between adding acid to water and water to acid - even though the process is "additive", there's marked practical differences to the two approaches :)
$endgroup$
– Luaan
Dec 13 '18 at 13:59

1

$begingroup$
The section about calibration is rather silly in this context and I think it detracts from the point.
$endgroup$
– pipe
Dec 13 '18 at 15:18

2

$begingroup$
"You might as well worry about the error caused by the tidal force from the moon" - No. That force is a perturbation of the order of $1$ part in $10^7$, considerably smaller than the effects you are talking about. Air buoyancy effects (which depend on atmospheric pressure, and there therefore weather-dependent even at a fixed location on earth) ARE routinely included when weighing to an accuracy of $0.01%$ or better.
$endgroup$
– alephzero
Dec 14 '18 at 18:52

1

$begingroup$
@alephzero Of course it is important that the equipment must be correctly calibrated but its also important that they aren't pouring the liquid into a beaker with a hole in the bottom, a sponge or a pair of pants. The question is about mixing two liquids and their weights changing, the answer should address that and not standard lab practice.
$endgroup$
– Lio Elbammalf
Dec 15 '18 at 22:32

|
show 3 more comments

@knzhou supplied a good answer. I’m going to offer a couple of other interpretations.

answered Dec 12 '18 at 16:55

Ben51

4,011827

@knzhou supplied a good answer. I’m going to offer a couple of other interpretations.

answered Dec 12 '18 at 16:55

Ben51

4,011827

answered Dec 12 '18 at 16:55

Ben51

4,011827

answered Dec 12 '18 at 16:55

Ben51

4,011827

answered Dec 12 '18 at 16:55

Ben51

4,011827

1

$begingroup$
Your second point is the one which might have by far the most impact. The weight could be $0N$ after the reaction.
$endgroup$
– Eric Duminil
Dec 13 '18 at 11:45

2

$begingroup$
Liquids also spontaneously evaporate; depending on the conditions, this can be rather dramatic. In any case, the important thing to take from this is that you should follow the procedure. If you make changes to the procedure, you need to understand the possible effects of that change. It is ridiculous to assume that changing the procedure will have no effect. A chemist should be well acquainted with e.g. the difference between adding acid to water and water to acid - even though the process is "additive", there's marked practical differences to the two approaches :)
$endgroup$
– Luaan
Dec 13 '18 at 13:59

1

$begingroup$
The section about calibration is rather silly in this context and I think it detracts from the point.
$endgroup$
– pipe
Dec 13 '18 at 15:18

2

$begingroup$
"You might as well worry about the error caused by the tidal force from the moon" - No. That force is a perturbation of the order of $1$ part in $10^7$, considerably smaller than the effects you are talking about. Air buoyancy effects (which depend on atmospheric pressure, and there therefore weather-dependent even at a fixed location on earth) ARE routinely included when weighing to an accuracy of $0.01%$ or better.
$endgroup$
– alephzero
Dec 14 '18 at 18:52

1

$begingroup$
@alephzero Of course it is important that the equipment must be correctly calibrated but its also important that they aren't pouring the liquid into a beaker with a hole in the bottom, a sponge or a pair of pants. The question is about mixing two liquids and their weights changing, the answer should address that and not standard lab practice.
$endgroup$
– Lio Elbammalf
Dec 15 '18 at 22:32

|
show 3 more comments

1

$begingroup$
Your second point is the one which might have by far the most impact. The weight could be $0N$ after the reaction.
$endgroup$
– Eric Duminil
Dec 13 '18 at 11:45

2

$begingroup$
Liquids also spontaneously evaporate; depending on the conditions, this can be rather dramatic. In any case, the important thing to take from this is that you should follow the procedure. If you make changes to the procedure, you need to understand the possible effects of that change. It is ridiculous to assume that changing the procedure will have no effect. A chemist should be well acquainted with e.g. the difference between adding acid to water and water to acid - even though the process is "additive", there's marked practical differences to the two approaches :)
$endgroup$
– Luaan
Dec 13 '18 at 13:59

1

$begingroup$
The section about calibration is rather silly in this context and I think it detracts from the point.
$endgroup$
– pipe
Dec 13 '18 at 15:18

2

$begingroup$
"You might as well worry about the error caused by the tidal force from the moon" - No. That force is a perturbation of the order of $1$ part in $10^7$, considerably smaller than the effects you are talking about. Air buoyancy effects (which depend on atmospheric pressure, and there therefore weather-dependent even at a fixed location on earth) ARE routinely included when weighing to an accuracy of $0.01%$ or better.
$endgroup$
– alephzero
Dec 14 '18 at 18:52

1

$begingroup$
@alephzero Of course it is important that the equipment must be correctly calibrated but its also important that they aren't pouring the liquid into a beaker with a hole in the bottom, a sponge or a pair of pants. The question is about mixing two liquids and their weights changing, the answer should address that and not standard lab practice.
$endgroup$
– Lio Elbammalf
Dec 15 '18 at 22:32

Your second point is the one which might have by far the most impact. The weight could be $0N$ after the reaction.

– Eric Duminil
Dec 13 '18 at 11:45

Liquids also spontaneously evaporate; depending on the conditions, this can be rather dramatic. In any case, the important thing to take from this is that you should follow the procedure. If you make changes to the procedure, you need to understand the possible effects of that change. It is ridiculous to assume that changing the procedure will have no effect. A chemist should be well acquainted with e.g. the difference between adding acid to water and water to acid - even though the process is "additive", there's marked practical differences to the two approaches :)

– Luaan
Dec 13 '18 at 13:59

The section about calibration is rather silly in this context and I think it detracts from the point.

– pipe
Dec 13 '18 at 15:18

"You might as well worry about the error caused by the tidal force from the moon" - No. That force is a perturbation of the order of $1$ part in $10^7$, considerably smaller than the effects you are talking about. Air buoyancy effects (which depend on atmospheric pressure, and there therefore weather-dependent even at a fixed location on earth) ARE routinely included when weighing to an accuracy of $0.01%$ or better.

– alephzero
Dec 14 '18 at 18:52

@alephzero Of course it is important that the equipment must be correctly calibrated but its also important that they aren't pouring the liquid into a beaker with a hole in the bottom, a sponge or a pair of pants. The question is about mixing two liquids and their weights changing, the answer should address that and not standard lab practice.

– Lio Elbammalf
Dec 15 '18 at 22:32

|
show 3 more comments

Based on your question I assume the precision is very important in your lab. If so, then yes, your colleague is right. At least to some degree.

Why adding weights doesn't produce the sum of weights?

Consider these options.

If the substances mix (i.e. a result is a solution of the two substances) the volume of resultant solution is smaller than volumes of ingredients. The masses remain the same though. I'll explain this phenomenon in a bit. The reason is purely physical.

If the substances react and some of the resulting substances leaves the solution (usually it will be a gas that evaporates) the mass will reduce by the mass of removed substance. Usually the volume will also be smaller since on one hand some of the substance is removed and the resultant solution with whatever remains should follow the same principle as in 1. But the newly formed particles could in theory prevent that and cause the volume to actually increase. I can't think of any examples (or even if there exists some). The reason is mixed chemical and physical.

If the substances react and as a result you get a soluble substance you end up with a solution of a new substance. In most (if not all) cases you have less particles to pack so your volume will drop with the mass remaining the same.If additionally the solvents mix you end up with added effect as in one so the difference can be even higher. This is a mix of chemical and physical reasons.

So what happens?

Your colleague refers specifically to point 1 on my list so I'll focus on it only.

So how big an impact can be?

In general the bigger the difference in particles size the stronger effect you'll notice.

First let's calculate how much volume are we going to loose. Based on this water-ethanol mixtures density we'll calculate the difference for 40% alcohol solution (popularly known as vodka).

Those densities are based normalised atmosphere so weight based.

So let's take (by weight) $60 text{ g}$ water and $40 text{ g}$ ethanol.
The volumes are respectively $60/0.998202 = 60.108074text{ cm}^3$ of water and $40/0.79074 = 50.585527text{ cm}^3$ ethanol.

To have accurate results we need to go through masses.

To calculate accurately the weight let's use proportion.

So the weigh of solution, which mass is $100.137275 text{ g}$ is going to be
$100/100.128517cdot100.137275 = 100.008747 text{ g}$

i.e. higher by $0.008747$ (or $0.008747%$) than expected.

If we applied this result to your case, for $5text{ g}$ you'd end up with $5.000437 text{ g}$.

The difference seems to be less than your precision. If it's acceptable or not - you need to decide yourself. Also your case might get more extreme.

Oh, just for reference - the volume of our solution would be $106.741813/100.128517cdot100.137275 = 106.751149 text{ cm}^3$

It's still significantly less than the sum of volumes, which as we've calculated was $110.693601 text{ cm}^3$.

What can you do to avoid that

Assuming the difference is significant enough to bother you, you have two options.

Mix the solution first and then measure the requested amount

Experimentally check the actual change of weight and calculate amounts to use. Note though that you risk a rounding errors this way.

Test volume loss yourself!

The resultant mixture is safe to consume ;-)

You may also check this video.

answered Dec 13 '18 at 11:12

Ister

4516

2

$begingroup$
Nice answer! I was even able to comprehend the calculation methods because you used numbers instead of just letters. Thank you
$endgroup$
– undefined
Dec 13 '18 at 13:16

$begingroup$
Perfect answer - the example at the end really sold it for me.
$endgroup$
– Lio Elbammalf
Dec 15 '18 at 22:38

add a comment |

Based on your question I assume the precision is very important in your lab. If so, then yes, your colleague is right. At least to some degree.

Why adding weights doesn't produce the sum of weights?

Consider these options.

If the substances mix (i.e. a result is a solution of the two substances) the volume of resultant solution is smaller than volumes of ingredients. The masses remain the same though. I'll explain this phenomenon in a bit. The reason is purely physical.

If the substances react and some of the resulting substances leaves the solution (usually it will be a gas that evaporates) the mass will reduce by the mass of removed substance. Usually the volume will also be smaller since on one hand some of the substance is removed and the resultant solution with whatever remains should follow the same principle as in 1. But the newly formed particles could in theory prevent that and cause the volume to actually increase. I can't think of any examples (or even if there exists some). The reason is mixed chemical and physical.

If the substances react and as a result you get a soluble substance you end up with a solution of a new substance. In most (if not all) cases you have less particles to pack so your volume will drop with the mass remaining the same.If additionally the solvents mix you end up with added effect as in one so the difference can be even higher. This is a mix of chemical and physical reasons.

So what happens?

Your colleague refers specifically to point 1 on my list so I'll focus on it only.

So how big an impact can be?

In general the bigger the difference in particles size the stronger effect you'll notice.

First let's calculate how much volume are we going to loose. Based on this water-ethanol mixtures density we'll calculate the difference for 40% alcohol solution (popularly known as vodka).

Those densities are based normalised atmosphere so weight based.

So let's take (by weight) $60 text{ g}$ water and $40 text{ g}$ ethanol.
The volumes are respectively $60/0.998202 = 60.108074text{ cm}^3$ of water and $40/0.79074 = 50.585527text{ cm}^3$ ethanol.

To have accurate results we need to go through masses.

To calculate accurately the weight let's use proportion.

So the weigh of solution, which mass is $100.137275 text{ g}$ is going to be
$100/100.128517cdot100.137275 = 100.008747 text{ g}$

i.e. higher by $0.008747$ (or $0.008747%$) than expected.

If we applied this result to your case, for $5text{ g}$ you'd end up with $5.000437 text{ g}$.

The difference seems to be less than your precision. If it's acceptable or not - you need to decide yourself. Also your case might get more extreme.

Oh, just for reference - the volume of our solution would be $106.741813/100.128517cdot100.137275 = 106.751149 text{ cm}^3$

It's still significantly less than the sum of volumes, which as we've calculated was $110.693601 text{ cm}^3$.

What can you do to avoid that

Assuming the difference is significant enough to bother you, you have two options.

Mix the solution first and then measure the requested amount

Experimentally check the actual change of weight and calculate amounts to use. Note though that you risk a rounding errors this way.

Test volume loss yourself!

The resultant mixture is safe to consume ;-)

You may also check this video.

answered Dec 13 '18 at 11:12

Ister

4516

2

$begingroup$
Nice answer! I was even able to comprehend the calculation methods because you used numbers instead of just letters. Thank you
$endgroup$
– undefined
Dec 13 '18 at 13:16

$begingroup$
Perfect answer - the example at the end really sold it for me.
$endgroup$
– Lio Elbammalf
Dec 15 '18 at 22:38

add a comment |

Based on your question I assume the precision is very important in your lab. If so, then yes, your colleague is right. At least to some degree.

Why adding weights doesn't produce the sum of weights?

Consider these options.

If the substances mix (i.e. a result is a solution of the two substances) the volume of resultant solution is smaller than volumes of ingredients. The masses remain the same though. I'll explain this phenomenon in a bit. The reason is purely physical.

If the substances react and some of the resulting substances leaves the solution (usually it will be a gas that evaporates) the mass will reduce by the mass of removed substance. Usually the volume will also be smaller since on one hand some of the substance is removed and the resultant solution with whatever remains should follow the same principle as in 1. But the newly formed particles could in theory prevent that and cause the volume to actually increase. I can't think of any examples (or even if there exists some). The reason is mixed chemical and physical.

If the substances react and as a result you get a soluble substance you end up with a solution of a new substance. In most (if not all) cases you have less particles to pack so your volume will drop with the mass remaining the same.If additionally the solvents mix you end up with added effect as in one so the difference can be even higher. This is a mix of chemical and physical reasons.

So what happens?

Your colleague refers specifically to point 1 on my list so I'll focus on it only.

So how big an impact can be?

In general the bigger the difference in particles size the stronger effect you'll notice.

First let's calculate how much volume are we going to loose. Based on this water-ethanol mixtures density we'll calculate the difference for 40% alcohol solution (popularly known as vodka).

Those densities are based normalised atmosphere so weight based.

So let's take (by weight) $60 text{ g}$ water and $40 text{ g}$ ethanol.
The volumes are respectively $60/0.998202 = 60.108074text{ cm}^3$ of water and $40/0.79074 = 50.585527text{ cm}^3$ ethanol.

To have accurate results we need to go through masses.

To calculate accurately the weight let's use proportion.

So the weigh of solution, which mass is $100.137275 text{ g}$ is going to be
$100/100.128517cdot100.137275 = 100.008747 text{ g}$

i.e. higher by $0.008747$ (or $0.008747%$) than expected.

If we applied this result to your case, for $5text{ g}$ you'd end up with $5.000437 text{ g}$.

The difference seems to be less than your precision. If it's acceptable or not - you need to decide yourself. Also your case might get more extreme.

Oh, just for reference - the volume of our solution would be $106.741813/100.128517cdot100.137275 = 106.751149 text{ cm}^3$

It's still significantly less than the sum of volumes, which as we've calculated was $110.693601 text{ cm}^3$.

What can you do to avoid that

Assuming the difference is significant enough to bother you, you have two options.

Mix the solution first and then measure the requested amount

Experimentally check the actual change of weight and calculate amounts to use. Note though that you risk a rounding errors this way.

Test volume loss yourself!

The resultant mixture is safe to consume ;-)

You may also check this video.

answered Dec 13 '18 at 11:12

Ister

4516

Based on your question I assume the precision is very important in your lab. If so, then yes, your colleague is right. At least to some degree.

Why adding weights doesn't produce the sum of weights?

Consider these options.

If the substances mix (i.e. a result is a solution of the two substances) the volume of resultant solution is smaller than volumes of ingredients. The masses remain the same though. I'll explain this phenomenon in a bit. The reason is purely physical.

If the substances react and some of the resulting substances leaves the solution (usually it will be a gas that evaporates) the mass will reduce by the mass of removed substance. Usually the volume will also be smaller since on one hand some of the substance is removed and the resultant solution with whatever remains should follow the same principle as in 1. But the newly formed particles could in theory prevent that and cause the volume to actually increase. I can't think of any examples (or even if there exists some). The reason is mixed chemical and physical.

If the substances react and as a result you get a soluble substance you end up with a solution of a new substance. In most (if not all) cases you have less particles to pack so your volume will drop with the mass remaining the same.If additionally the solvents mix you end up with added effect as in one so the difference can be even higher. This is a mix of chemical and physical reasons.

So what happens?

Your colleague refers specifically to point 1 on my list so I'll focus on it only.

So how big an impact can be?

In general the bigger the difference in particles size the stronger effect you'll notice.

First let's calculate how much volume are we going to loose. Based on this water-ethanol mixtures density we'll calculate the difference for 40% alcohol solution (popularly known as vodka).

Those densities are based normalised atmosphere so weight based.

So let's take (by weight) $60 text{ g}$ water and $40 text{ g}$ ethanol.
The volumes are respectively $60/0.998202 = 60.108074text{ cm}^3$ of water and $40/0.79074 = 50.585527text{ cm}^3$ ethanol.

To have accurate results we need to go through masses.

To calculate accurately the weight let's use proportion.

So the weigh of solution, which mass is $100.137275 text{ g}$ is going to be
$100/100.128517cdot100.137275 = 100.008747 text{ g}$

i.e. higher by $0.008747$ (or $0.008747%$) than expected.

If we applied this result to your case, for $5text{ g}$ you'd end up with $5.000437 text{ g}$.

The difference seems to be less than your precision. If it's acceptable or not - you need to decide yourself. Also your case might get more extreme.

Oh, just for reference - the volume of our solution would be $106.741813/100.128517cdot100.137275 = 106.751149 text{ cm}^3$

It's still significantly less than the sum of volumes, which as we've calculated was $110.693601 text{ cm}^3$.

What can you do to avoid that

Assuming the difference is significant enough to bother you, you have two options.

Mix the solution first and then measure the requested amount

Experimentally check the actual change of weight and calculate amounts to use. Note though that you risk a rounding errors this way.

Test volume loss yourself!

The resultant mixture is safe to consume ;-)

You may also check this video.

answered Dec 13 '18 at 11:12

Ister

4516

answered Dec 13 '18 at 11:12

Ister

4516

answered Dec 13 '18 at 11:12

Ister

4516

answered Dec 13 '18 at 11:12

Ister

4516

2

$begingroup$
Nice answer! I was even able to comprehend the calculation methods because you used numbers instead of just letters. Thank you
$endgroup$
– undefined
Dec 13 '18 at 13:16

$begingroup$
Perfect answer - the example at the end really sold it for me.
$endgroup$
– Lio Elbammalf
Dec 15 '18 at 22:38

add a comment |

2

$begingroup$
Nice answer! I was even able to comprehend the calculation methods because you used numbers instead of just letters. Thank you
$endgroup$
– undefined
Dec 13 '18 at 13:16

$begingroup$
Perfect answer - the example at the end really sold it for me.
$endgroup$
– Lio Elbammalf
Dec 15 '18 at 22:38

Nice answer! I was even able to comprehend the calculation methods because you used numbers instead of just letters. Thank you

– undefined
Dec 13 '18 at 13:16

Perfect answer - the example at the end really sold it for me.

– Lio Elbammalf
Dec 15 '18 at 22:38

add a comment |

I just adhere to the scenario you have described.

Just put a container filled with water on a scale. Weight reads W1.
Put a floating object of mass w. Reads is now W1 + w.

edited Dec 14 '18 at 12:55

answered Dec 14 '18 at 12:47

Alchimista

636410

add a comment |

I just adhere to the scenario you have described.

Just put a container filled with water on a scale. Weight reads W1.
Put a floating object of mass w. Reads is now W1 + w.

edited Dec 14 '18 at 12:55

answered Dec 14 '18 at 12:47

Alchimista

636410

add a comment |

I just adhere to the scenario you have described.

Just put a container filled with water on a scale. Weight reads W1.
Put a floating object of mass w. Reads is now W1 + w.

edited Dec 14 '18 at 12:55

answered Dec 14 '18 at 12:47

Alchimista

636410

I just adhere to the scenario you have described.

Just put a container filled with water on a scale. Weight reads W1.
Put a floating object of mass w. Reads is now W1 + w.

edited Dec 14 '18 at 12:55

answered Dec 14 '18 at 12:47

Alchimista

636410

edited Dec 14 '18 at 12:55

answered Dec 14 '18 at 12:47

Alchimista

636410

answered Dec 14 '18 at 12:47

Alchimista

636410

answered Dec 14 '18 at 12:47

Alchimista

636410

add a comment |

-1

I'll offer another consideration since you've mentioned buoyancy, density which is not only a function of mass but volume as well.

Is this possibly what you might be experiencing?

answered Dec 12 '18 at 21:12

docscience

8,64821850

$begingroup$
I'm not sure "relativistic" is the correct term here. E = mc^2 is a relationship not based on special relativity. And mass gain/loss in an endo/exothermic reaction can be measured (albeit very carefully)
$endgroup$
– Carl Witthoft
Dec 13 '18 at 20:05

add a comment |

-1

I'll offer another consideration since you've mentioned buoyancy, density which is not only a function of mass but volume as well.

Is this possibly what you might be experiencing?

answered Dec 12 '18 at 21:12

docscience

8,64821850

$begingroup$
I'm not sure "relativistic" is the correct term here. E = mc^2 is a relationship not based on special relativity. And mass gain/loss in an endo/exothermic reaction can be measured (albeit very carefully)
$endgroup$
– Carl Witthoft
Dec 13 '18 at 20:05

add a comment |

-1

I'll offer another consideration since you've mentioned buoyancy, density which is not only a function of mass but volume as well.

Is this possibly what you might be experiencing?

answered Dec 12 '18 at 21:12

docscience

8,64821850

I'll offer another consideration since you've mentioned buoyancy, density which is not only a function of mass but volume as well.

Is this possibly what you might be experiencing?

answered Dec 12 '18 at 21:12

docscience

8,64821850

answered Dec 12 '18 at 21:12

docscience

8,64821850

answered Dec 12 '18 at 21:12

docscience

8,64821850

answered Dec 12 '18 at 21:12

docscience

8,64821850

$begingroup$
I'm not sure "relativistic" is the correct term here. E = mc^2 is a relationship not based on special relativity. And mass gain/loss in an endo/exothermic reaction can be measured (albeit very carefully)
$endgroup$
– Carl Witthoft
Dec 13 '18 at 20:05

add a comment |

$begingroup$
I'm not sure "relativistic" is the correct term here. E = mc^2 is a relationship not based on special relativity. And mass gain/loss in an endo/exothermic reaction can be measured (albeit very carefully)
$endgroup$
– Carl Witthoft
Dec 13 '18 at 20:05

I'm not sure "relativistic" is the correct term here. E = mc^2 is a relationship not based on special relativity. And mass gain/loss in an endo/exothermic reaction can be measured (albeit very carefully)

– Carl Witthoft
Dec 13 '18 at 20:05

add a comment |

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