

Percent Solution to Molarity
Name: Mark
Status: educator
Grade: other
Location: England
Country: N/A
Date: September 2007
Question:
It is a bit tragic after teaching for 20 years I now have to
ask this question. I am in a school without a laboratory assistant and I
have to dilute Nitric, Hydrochloric, Acetic and Sulfuric Acid. The problem
is they come as percentages, not molarities and I want to get them down to
2M, 1M and 0.1M etc concentrations. Is there a suitable site for me to find
out how to do this or is there an easy formula to use so that I can work it
out for each?
Replies:
Unfortunately, the history of chemistry is such that there are a daunting
number of concentration units. These developed because chemists historically
used units of mass, volume, and concentration that suited their particular
needs AT THE TIME  but "Times They are a' Changing" and we are stuck with
dozens of concentrationrelated units of "quantity".
An example of how technology has changed our use of concentration: Historically,
chemists, for generations (centuries?) have used burettes to do titrations.
These are typically 100 ml that can be 0.1 ml precision. So one might be
titrating a solution and obtain a reading: 45.3(?) +/ 0.05 ml. The (?) indicating
that some people with a magnifying glass assert they are able to interpolate
between two marks on the burette and obtain a meaningful reading  an
assertion I believe is unwarranted for two reasons: One problem is that
the burette is assumed to be exactly cylindrical. To the extent that is
not true affects the "accuracy" of the reading. Two, the meniscus shape
is dependent on the interfacial tension between the glass wall of the
burette and the solution. That is not something one has under control
however.
With the invention and widespread use and availability of electronic scales
(the term "balance" is not even appropriate any more, since nothing is
"balanced"), it is possible to weigh both sample and titrant to 0.01 gm,
and with more expensive scales to 0.001 gm, or even 0.0001 gm. Such
"weight" titrations totally eliminate volumetric measurements (except
for the buoyancy correction in the case of really careful measurements).
So mol / liter units are archaic. Units of mols / gm are the "natural"
units of concentration. Or possibly mole fraction in some cases. With
little sharpening of skills, titrations on a mass basis are available
even to first year students.
Pardon my lecture, but I think "the word" has to be spread about how weighing
technology has changed for the better, how we should do simple weighings.
Now to your question.
There are many web sites where you can refresh your grasp of "weight/weight"
and "weight/volume" measurements. One needs to know the molecular weight of
the solute to convert gmstomols. For example for salt (NaCl),
M.W. = 58.5 gm / mol, (I rounded the numbers to avoid clutter.) so
3.60 gm NaCl = 3.60 gm / 58.5 gm / mol = 0.061(5) mols NaCl. You can
dissolve this in a sufficient amount of water so that the TOTAL volume
is let's say 125.4 ml = 0.1254 liters. Then the concentration on a
(mol / liter) basis is 0.061(5) mol NaCl / 0.1254 liters = 0.490 mol
NaCl / liter.
There are numerous web sites with explanations and worked examples where
you can practice, e.g.:
http://pages.towson.edu/ladon/concas.html
http://www.chemprofessor.com/conc.htm
There are also several web sites that do the conversion arithmetic for
you. Two I stumbled on are"
http://www.unitconversion.org/
http://www.cleavebooks.co.uk/scol/index.htm
Finally, always, always use "dimensional analysis" when converting
various units. "Dimensional analysis" also is called "carrying the
units along". While students sometimes find it "boring". Well "boring"
it may be, but it prevents many computational errors when doing
conversions. You can review that subject at:
http://www.chemtutor.com/numbr.htm
http://www.physics.uoguelph.ca/tutorials/dimanaly/
http://chemistry.alanearhart.org/Tutorials/DimAnal/
I am not endorsing any of these web sites as being superior to others.
I just happened to find these.
Vince Calder
Mark,
First, you need to determine if the percentage is mass/volume, mass/mass
or volume/volume.
It is usually mass/vvolume and in which case, if you assume 100mL of solution,
then the mass in grams of dissolved solute in that 100mL is the percentage.
For example, 100mL of a 30% (mass/volume) solution has 30g of solute.
If the percentage is mass/mass, then 100g of the solution contains a mass
in grams equivalent to the percentage.
If the percentage is volume/volume, then 100mL of the solution contains a
volume in mL equivalent to the percentage and this will have to be converted
to mass by way of the density of the solute. For example, 100mL a 15% (v/v)
solution contains 15mL of solute and if the density of the solute is 0.95g/mL,
then dividing the mL of solute by the density gives the grams of that solute
in every 100mL of solution.
From there, you can figure out the molarities of the solution by using the
definition of a molarity which is: M = n/L, where n is moles of solute, and
L is liters of solution. By expanding the definition of n as = g/MM, where
g is grams and MM is molar mass, we get M = g/(MM x L). Thus: a 10% solution
of HCl must have a M of 10g/([1+35.4]g/mol x [100/1000]L).
To convert the primary solution into the target secondary solution, use the
equation McVc = MdVd, where Mc and Vc are the molarities and volumes of the
concentrated solution, and Md and Vd are the molarities and volumes of the
dilute solution. Thus, if you have a 5M primary solution and you want 100mL
of a 2M solution, you need: Vc = 2M x 100mL / 5M.
Greg (Roberto Gregorius)
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Update: June 2012

