Another math/chemistry question

A drug called Levamisole used to come as a 13.5% solution, but now is only available in granular form (presumedly 100% Levamisole?). How much of the granular form do I need to mix into a 1 liter bottle of sterile water to obtain my desired 13.5% dilution?

I remember this kind of calculation being fairly simplistic, but now I can’t figure out out how to start. A straight up 13.5% (135g/L) seems like too much. Do I really need to convert stuff to moles?


That depends on whether the original 13.5% was by weight, by volume, or by molarity.

did some more googling and have discovered that I need to get more information from the person who posed this question to me.

Reserve the right to resume this thread when I have more info.

I would generally assume it was by weight. I would add 135 g to a liter. % by mole would an unusual unit, and % by volume for a solid would be pretty unusual as well.

Since it’s a solid, it would be by weight over volume (for water this is almost equivalent to weight on weight). If it was a liquid, volume over volume. And you don’t need to make a whole liter, just 13.5g to 100mL works.

It isn’t really a chemistry question but a pharma one: I had some serious problems in a college lab because my labmate was pharma, so I remember it well. I’ve confirmed time and again that the custom of doing weight-or-volume over volume isn’t a “regional” peculilarity but a professional one.
% by mole has its own symbol.


As for the volume, 1L was actually on the small side. We’re (me and the Vet I’m shadowing) are looking at running this medication through the water lines for a poultry house. If we go ahead with the plan, we’ll be mixing it hundreds of gallons at a time :slight_smile:

If you added 135 grams to the water, wouldn’t it overfill the container?

If we assume it’s a liquid the same density as the water, the 13.5% is by volume, and when it mixes with water we get a 1+1 = 2 situation (rubbing alcohol and water, for example, won’t equal 2 cups if you mix a cup of each, since the molecules can “fit” between each other more tightly than they do by themselves).

We’ll have two equations with two unknowns, with the drug (L) and the water (w)

L / w = .135 [to get the 13.5%]
L + w = 1000 [because our total volume at the end needs to be 1L, or 1000ml]

Solving gives L = 119 (grams and w = 881 mL

This would be useful if you wanted to make exactly 1L of solution.

However, the granules are probably going to dissolve (so its molecules “fit” between the water molecules), so the point of saturation would be useful at the working temperature, as would the solubility*.

*I have no idea if that’s the right word or how it is measured.

ETA: Should probably preview, but Nava nailed it since you’re not trying to make a specific amount.

It wouldn’t be by molarity, as molarity is not expressed in terms of percentages.

For a solid dissolved in an aqeous solution, it is usually a weight/weight percent (w/w %). This unit can easily be converted to parts-per-million (ppm), which is concentration expressed in terms of mass.

It could also be weight/volume percent (w/v %).

However, in practice, for solutes dissolved in an aqueous solution, there is usually little difference between weight/weight percentage and weight/volume percentage. (The difference is due to the fact that the density of solution expressed in terms of mg/L is only approximately 1 mg/L, which under dilute conditions is a good approximation. You run into problems with this assumption with more concentrated solutions, though.)

Volume percentages are usually only used for liquid solutes and are generally called out specifically as volume/volume percent (v/v %).

You add the solid, then fill it to a liter. Yes, that means that you don’t add a full liter of water. You could get technical and compensate for the densities of the variuos components, but for most things the density is pretty darn near 1 and it won’t make much difference. If you tell a chemist 13% by volume, they will assume you mean 130 grams added to a liter then filled with water. If you want something more precise, you would go with molarity or even molality. I’ve never heard of % by mole.

So you’re saying the solubility* term I made up is pretty close to 1? i.e. almost all the solute “fits” between the water?

So anyway, for this example, there actually is a difference between w/w% and w/v%, because the solution is fairly concentrated.

I would expect this to properly be a w/w%, which is almost certainly what you had when you had a 13.5% solution in the past.

To prepare this, you would add the following to 1 L of water (assumed to have a mass of 1000 g):

13.5 w/w% = (x g of solute)/(x g + 1000 g)

Solving for x gives x = 156 g of solute (Levamisole) should be added to 1 L of water to prepare the solution.

This gives you 156 g of solute in 1156 g of solution, which is the correct w/w%.

This would generally work for dilute solutions, but does not work for this problem because the solution is relatively concentrated. See my previous post.

For most common solutions, the solute will not add significantly to the final volume, so you can just make it by adding 13.5g of solute to 100ml of diluent. However, when it will, you’ll generally see the recipe read something like this:

Add 13.5 g solute to 80ml diluent. Mix to dissolve. Q.S. to 100ml.

Where Q.S. stands for Quantum sufficit, or as much as suffices.

This is not correct, either. The link in the OP indicates that the solute is a solid, so it would not be a volume percentage (v/v%).

If you added 119 g of solid Levamisole to 881 g of water, you would end up with 119 g of Levamisole in 1000 g of solution, which is only an 11.9 % solution (w/w).

I’m saying that this is the way these solutions are made and everybody makes them the same way so the units are always consistent. I’ve never seen anybody take the time to calculate out the fine detail of these solutions to make sure you had precisely X% by weight or volume. For anything that requires that type of accuracy, I would use a different unit like molarity or molality.

And yes, I am saying that even if they did calculate it out precisely, the result would most likely be within 10% of the way you mix them in the first place, because the density of most chemicals is pretty close to 1 and the “fit between the molecules” effect is not huge. There will always be exceptions of course. Somebody working with gold nanoparticles may have a different take on it.

To be exactly sure that you are doing this properly, Pullet, I would find your old container (or catalog entry) of 13.5% Levamisole solution, and see if the container says “13.5% (w/w)” or “13.5% (w/v)”.

As you can see, it makes a difference in this case.

For the former, you would add 156 g of solid Levamisole to 1 L of water.

For the latter, you would add 135 g of solid Levamisole to 1 L of water.

Thanks, Robby. Finding out some info about the old container is what I left the thread to do earlier. Unfortunately, it has either been out of production for a while (which is saying something the way poultry guys will hold on to stuff), or it was some other kind of formulation where Levamisole was just one of several active ingredients. Hopefully, will have more news soon.

Also, could you come teach chemistry at my school? Maybe I could remember some of this then.

P.S. Got to use the other formula you helped me with (CiVi=CfVf) in real life the other day! Woot!

Actually, Santo Rugger, this would be (approximately) correct if it was a w/v%.

This may be correct, but you are assuming that it is 13.5% (w/v).

You’re welcome!

(Can you tell I’d rather be teaching chemistry than working on the sewer design on my desk?)