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Measuring Without a Scale

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Crashing

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This is an idea i came up with that is theoretically more accurate than using an electronic scale:

SOLUBILITIES MUST BE KNOWN FOR THIS!

So i recently got a bag of Aluminum but my scale broke.....;)

The apartment I crash at currently is kept at 24*C, so for this sake lets assume my water solution after measured is 25*C in a stable environment.

So if 36.4g Al2(SO4)3 (aluminum sulfate) is soluble in 100g H20 at 20*C, and 40.4g into 100g H20 at 30*C, to find it's solubility in 100g H20 at 25*C:

40.4g - 36.4g = 4g / 2g = 2g

40.4g - 2g = 38.4g + 2g = 40.4g

So at 25*C, it takes exactly 100g H20 to dissolve exactly 38.4g Al2(SO4)3.


The above section is NOT TRUE FOR ALL SUBSTANCES, read below ^

Let's continue assuming that at 25*C, it takes exactly 100g H20 to dissolve exactly 38.4g Al2(SO4)3, as the solubility for aluminum is linear at these temps.


But, I've only got a small unknown sample amount of Al2(SO4)3 and a 1mL syringe of H20.

So, lets place the powder into a dish and add water slowly until complete saturation.

"hey dude, check it out man, it totally JUST went clear after exactly 20 units (.2mL) ! stellar!<3"

So, given the temp is static, and that 100g = 100.43mL -

38.4g/100.43mL = Y / .2mL

.2(38.4) = 100.43(Y)

7.86 = 100.43(Y)

7.86 / 100.43 = .078g = I have 78mg of Al2(SO4)3 in my bag.

Please review and check my work and comment.
 
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To clarify, I have 78mg in a 20cc solution. There was 78mg in the bag to start with. Now you can use dilution to apply aluminum to a substrate, for instance.
 
Now I may be wrong, someone please correct me if I am, but my understanding of the matter is thus:

You can't figure out the solubility of a compound at a given temperature like that. In some cases you may get close, but it won't be nearly exact. Solubility does not increase/decrease linearly with temperature with just about any compound I'm aware of. You're looking at more of a log scale with some compounds, but that isn't accurate either. You're not going to find an equation that will accurately tell you the solubility of an unknown organic compound in a given amount of solvent at a given temperature. The rate of change differs radically from compound to compound.

The solubility of a given compound is determined by taking a given amount of solvent CONTROLLED (as in much more controlled than you will find anywhere outside of a lab) to a certain temperature, and adding said compound (already equalized to the same temperature as the solvent) until the solvent is saturated. Any undissolved solute is filtered away, the solvent is evaporated, and the solute which remains is weighed. Then you do the math and determine W grams of X is soluble in Y milliliters of Z.

Unfortunately your method isn't really an option unless you have the ability to control the conditions and use the known solubility as a reference, as in controlling the temperature of the experiment so it is at the value given in the known solubility and adding solvent dropwise (with plenty of time and gentle stirring in between each drop) until the solute goes completely into solution, and then derive your answer based on that.
 
You're absolutely right. So, therefore the mean temperature would not be accurate or calculable, so one would have to use the exact solubility:temperature. I shouldn't have assumed that all substances have a linear temp:solubility relationship.

Sodium sulfate is a good example. It is more soluble at 30 degrees than at 40, but much less soluble at 20 degrees than 30.

One would have to be in a temperature, pressure and humidity controlled environment with a very patient method of administering solution.

Also, this calculation assumes absolute purity of substance, which is another ordeal.

Thanks for catching that and replying.
 
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