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Q: Interesting question ( Answered,   5 Comments )
Subject: Interesting question
Category: Science
Asked by: mdouglas04-ga
List Price: $40.00
Posted: 19 Dec 2004 12:10 PST
Expires: 18 Jan 2005 12:10 PST
Question ID: 444736
I'm trying to determine how much soluble content there is in a
product. Please note the product itself is in fine powder format and
is in-soluble in water, but there's a very small percentage
(approximately 0.2%) of soluble powder-format additive in it.

I was told that you can take a sample and weigh it, wash it , and
weigh it again to compare the weight difference. The difference will
be the soluble additive.

I understand the theory but am having a hard time doing it. Can you
recommend a efficient/easy/scientific way of doing this? Or maybe
recommend a lab that does this kind of work? Thank you.

Request for Question Clarification by answerguru-ga on 19 Dec 2004 13:49 PST
Hello there,

This is a very simple process, in fact, I recall doing much the same
type of thing in numerous high school chemistry classes. The process
you are describing is called "crystallization" and is an effective and
inexpensive way of determining the amount of soluble additive in the

You do need some equipment to do this with significant accuracy and
precision. The most important thing is having the right equipment:

A digital scale (preferably something that is sensitive to a
thousandth of a gram, depending on the level of precision you need).

A heat lamp to crystallize the solution.

A petrie dish to hold your sample.

I'm not certain if this is sufficient enough for you - please let me
know if you need anything else. If you like, I can write up detailed
steps for performing this experiment assuming you have the equipment
listed above.


Clarification of Question by mdouglas04-ga on 19 Dec 2004 18:51 PST

When you use a heat lamp to crystallize the solution, I assume you are
trying to dry the powder?

I don't have any of the equipments you described. Can you recommend a
website/store from where I can buy these equipments. I visited Staples
and the smallest scale they have is Postal meter, which is not

Yes, I would like to have the detailed steps. Thank you.

Request for Question Clarification by answerguru-ga on 20 Dec 2004 00:38 PST
Hi again,

I did a little searching for the equipment I mentioned, but to be
honest I couldn't give you a recommendation. My only point on that
topic is that you're probably not going to find what you need in a
regular store like Staples - you would probably need to get the
equipment from a store that sells lab supplies in your area.

Since this is outside of your original question, I suggest posting
another question to find out more about lab suppliers online or in
your area.

That being said, I'm still willing to provide you with the steps for
the experiment if you want them. I didn't want to answer without first
clarifying that I would not be able to provide supplier information.


Clarification of Question by mdouglas04-ga on 20 Dec 2004 08:19 PST
That is fine. Please provide detailed information on conducting the
experiment. Thanks!
Subject: Re: Interesting question
Answered By: answerguru-ga on 21 Dec 2004 00:45 PST
Hello mdouglas04,

As requested, I have provided the procedure required to conduct the
experiment described in your question.

Experiment: Calculating the Mass of a Soluble Additive

Purpose: To determine the percentage of soluble additive contained
within a power which itself is in-water soluble.

Hypothesis: The powder product contains approximately 0.2% soluble additive.

Key Equipment:

Petrie Dish
Graduated Cylinder
Heat Lamp


1. Determine and record the mass (in grams) of the empty petrie dish
when it is clean and dry.

2. Repeat step (1) for the empty graduated cylinder.

3. Take a small amount of powder (half a teaspoon or less), place it
in the graduated cylinder.

4. Record the combined mass of the cylinder and powder, and subtract
the intial mass of the cylinder to obtain the intial mass of the

5. Slowly add small amounts of water (preferably filtered) to the
graduated cylinder while continuously swirling the cylinder to help
dissolve the powder entirely.

6. Repeat (5) until the powder has completely dissolved and no
particles are visible in the mixture.

7. Record the volume (at least 0.1 mL precision) of the solution

8. Carefully transfer some of the solution to the petrie dish. This
should certainly be less than a third of the amount required to fill
the dish. If multiple attempts are required, ensure the volume of the
cylinder is recorded each time (this ensures all 'lost' solution is
accounted for). Record the volume of the solution added to the petrie

9. Place the petrie dish under the heat lamp and leave it until no
liquid remains in the petrie dish. The final content on the dish
should resemble crystals (which can then be ground back into the
original powder if necessary but are the same mass)

10. Record the mass of the petrie dish with crystallized powder.

11. Subtract the original mass of the petrie dish to obtain the mass
of the crystals.

12. Calculate the percentage of the total solution used:
Total % of solution used = Volume of solution used / Total solution volume

13. Use the percentage from (12) to calculate the amount of powder
actually used for crystallization:

Mass of powder used = (Total powder mass)* (Total % of solution used)

This calculation is justified by the fact that a solution contains a
uniform concentration of solute by volume.

14. Compare the mass from (13) to the mass found in (11).


If the two calculated masses are equal, the the powder does not
contain any additive. However, since our hypothesis states that we
expect there to be a discrepancy, we can calculate the amount of
soluble additive by simply calculating the % difference between the
two masses:

% additive = (Powder mass - Crystal mass)/(Crystal mass)

I hope this provides you with a clear understanding of how this type
of experiment should be conducted. If any of the information above is
unclear, please let me know with a clarification request.



Request for Answer Clarification by mdouglas04-ga on 21 Dec 2004 13:30 PST
Hello answerguru,

Thank you for answering my question.

I'm not clear on step 14, where you compare the powder mass from step
13 with crystal mass from step 11. They could never be equal, right?

In any case, here is my understand of your proposal:

Let's say I use 100 grams of powder and mixed with distilled water to
create a solution, and let's say the volume of the solution (water +
powder) is 1 liter.

Then I transfer 500 ml (1/2 liter) of solution to the petrie dish, use
heat lamp to evaporate the water, and in the end I get 0.1 gram of
crystals, this is the result of step 11.

Now since I used 50% of solution, based on the step 12 and 13, the
amount of powder actually used for crystallization is: 100 grams x 50%
= 50 grams.

So now I have 50 grams of original powder versus 0.1 gram of crystals,
the % of additive should be  (50 - 0.1)/50

Please confirm this is the correct understanding.

Lastly, in regards to step 7 where you record the volume of the
solution, should you be recording the volume of the water you added?
The volume of step 7 is really the volume of water added PLUS the

Thanks again.

Clarification of Answer by answerguru-ga on 21 Dec 2004 23:16 PST
Hi again,

In step 14, the comparison of the two calculated masses can only be
equal when there is no additive. In your case, since you know an
additive exists, there will certainly be a difference.

Your outlined understanding is correct - you have got the hang of it now :)

Your clarification regarding step 7 is correct, but difficult to
obtain unless you know the level of solubility. Recall that you are
adding small amounts of water until no solid material remains. You
could do a test run to find out approximately how much water is
required for a given amount of powder and then repeat by using a
precise amount of water (recording the volume) and then adding a
predetermined mass of powder.

Thanks for using Google Answers!

Subject: Re: Interesting question
From: zn833-ga on 20 Dec 2004 10:44 PST
A digital scale like what you are going to require, as well as the
petri dishes, can be bought from lab supply companies like Fisher
( and VWR ( but from either of these
stores it will be rather expensive.  If there is a lab in the area,
and you are just going to need to make a couple measurements, I
suggest you just ask if you can borrow theirs.  They will probably let

Also, when doing this test, it is important never to touch the dishes
that you are weighing in with bare hands once you have found the
weight of the dish.  Your hands will leave oils that will skew your
measurements if they are very small.
Subject: Re: Interesting question
From: bluck-ga on 20 Dec 2004 12:04 PST
You've named the equipment incorrectly.  In a science laboratory the
name for an analytical scale used to measure these small quantities
with accuracy is 'balance.'  If mdouglas04 goes searching the internet
for a 'scale' rather than a 'balance' he/she will spend time looking
at needless results.  Please be accurate in nomenclature.  However,
zn833 did provide a link to balances!

Zn833 commented that you should ask a 'lab in your area ... if you can
borrow theirs.'  Our school balances (college) cost a minimum of $5000
(USD), require special tables to sit on, and paid technicians to
readjust when out-of-calibration.  The probability that a commercial
lab will allow you near their balances, which are required to be
calibrated on schedule according to government regulations, is zero. 
However, a nearby highschool chemistry teacher might allow it.
Subject: Re: Interesting question
From: zn833-ga on 21 Dec 2004 08:16 PST
Thanks for the clarification on nomenclature Bluck.  I should have
known better.  Also, a high school will probably let you use their
balance, but I would guess that if there is a university in the area,
a research lab there would consent as well, although they may want to
make sure you use it correctly.

I just want to add a few things to the protocol.  answerguru says that
filtered water is better, when in fact it is crucial.  If you don't
use distilled water you will end up crystalizing minerals that are
contained in the water with you powder.  Also, and I am not sure on
this, but I would think that the percent lost would be (powder mass -
crystal mass)/POWDER mass.  Although at ~0.2% the difference that
makes could be minimal.
Subject: Re: Interesting question
From: nanoalchemist-ga on 23 Dec 2004 08:55 PST
An analytical balance would be crucial. If you do go to a university
lab, the bonus is that they may be so particular about the equipment
that for such a simple experiment they may just do it for you. =)
Subject: Re: Interesting question
From: dops-ga on 06 Jan 2005 14:15 PST
answerguru-ga has provided a reasonable outline for you to follow.
However, steps #5 and #6 "Slowly add small amounts of water
(preferably filtered) to the
graduated cylinder while continuously swirling the cylinder to help
dissolve the powder entirely.

6. Repeat (5) until the powder has completely dissolved and no
particles are visible in the mixture." are not so helpful is you
expect 98% insoluble material. Also the percent additive = mass of
crystal/the mass of the powder used as calculated in #13

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