A Few Pharmacology Problems

Solved by Dimensional Analysis

ChuckOates.com

I received an e-mail from a former Math for Health Careers student who was having difficulty with some pharmacology problems.  She asked if I could show her how to do them using dimension analysis, as we did in Math for Health Careers.  An exchange of several e-mail messages followed, and, to make a long story short, we arrived at the correct answers.  At her suggestion, I’m posting below an edited version of the questions and answers in hopes that it’ll be of some use to others who are comfortable with the dimensional analysis approach to problem solving.  Please e-mail me at Chuck@ChuckOates.com if you find errors in the solutions, or if you find the solutions useful, for that matter.  Thanks!

Chuck Oates

1)     There is a pre-filled syringe in your ambulance.  It contains 2 percent lidocaine.  It also contains 5 mL.  How many milligrams does it contain?  The answer is 100 mg.

(Revised explanation)
The definition of “% lidocaine” is 1 g lidocaine (or whatever solute) per 100 mL of solution.  Therefore, we can write

5 mL   x   2   x   1 g / 100 mL     ( in words, “amount of solution” x  2  x  “one percent soln.”)

=   (5 x 2) / 100    g

=   0.10 g   x   1000 mg / 1 g

=   100 mg

See http://en.wikipedia.org/wiki/Percentage_solution for the definition of percent solution.  See also the Mass-Volume Ratio heading in the article at http://en.wikipedia.org/wiki/Molarity.

Calling   1 g / 100 mL    a “percentage” (mass/volume percentage or weight/volume percentage, really) may make more sense if you consider that one milliliter of pure water “weighs” (has a mass of) one gram.  That means that 100 mL of pure water weighs 100 grams.

2)    The doctor orders 0.5 g of aminophylline to be placed in a 100 mL bag of D5W for an IV piggyback.  What is the per milliliter concentration?  The answer is 5 mg/mL.

To calculate the concentration of aminophylline in mg/mL, I'd start with the
information from the problem, namely:

0.5 g of aminophylline / 100 mL of D5W .  Then

0.5 g  / 100 mL   x   1000 mg / 1 g

=   500 mg / 100 mL

=   5 mg / mL

Let me know if that makes it clear (or clear as mud!).

3)    You are reading the label on a 250 mL IV bag in the ICU.  The label reads 400 mg dopamine added.  The bag now has 150 mL left in it.  What is the per milliliter concentration in the bag now?  Answer is 1600 mg/mL

I assume that the 400 mg of dopamine was added to a full bag of 250 mL.  If that's the case, the fact that there are 150 mL left is irrelevant.  The concentration will remain the same as the mixture flows out of the bag.  The concentration, then would be constant at

400 mg of dopamine / 250 mL of IV fluid

= 400 mg / 250 mL

= 1.6 mg / mL   x   1000 mcg / mg

= 1600 mcg / mL

I believe the answer given is off by 1000x.

4)    You are completing your report after delivering a patient to the emergency center, and you notice that the dopamine dose ordered by medical control is missing from your notes.  To avoid any problems, you decide to determine the ordered dose based on the information available.  The patient weighs 176 lb and the IV infusion is flowing through a microdrip administration set at 30 gtt/min.  The label you put on the 500 mL bag of normal saline reads "800 mg dopamine added."  What was the doctor's original dose per kilogram per minute order?

Problem number four is in some sense a "backward" version of those mcg/kg/min problems we worked in Module 3 of Math for Health Careers.  Let's see, we're trying to arrive at the physician's mg/kg/min order, based on the data supplied.  To make things a bit easier, let's first observe that

176 lb   x   1 kg / 2.2 lb   =   80 kg .

We can get part-way to the answer by figuring out how many total mg / kg as follows:

800 mg of dopamine  /  80 kg of pt. weight

=   800 mg  /  80 kg

=   10 mg / kg   (the total weight of dopamine per kg of pt. weight over all the time it took to infuse the whole 500 mL, whatever amount of time that was)

But how long did it take to infuse the 500 mL?  Well, we know how to do that.   It's the "240 miles to Amarillo at 60 miles/hr" problem, but we'll have to work a little harder, since the flow rate is give in gtt/min  (it would be really easy if it were in mL/min).  Let's see, a microdrip has 60 gtt / mL and we're running it at 30 gtt / min, so

500 mL   x   60 gtt / mL   x   1 min / 30 gtt

= 1000 minutes .

Okay, now we can get to mg / kg / min by combining these two results as follows:

10 mg/kg     /    1000 min

= 0.010 mg / kg / min        (which we'd probably rather see as mcg / kg / min, so let's invoke 1000 mcg / mg to get ...)

= 0.010 mg / kg / min   x   1000 mcg / mg

= 10 mcg / kg / min .

I hope that's right, and I'm pretty sure it is.  There are probably a thousand ways to work it, but looking at the units should lead you to one of the correct and, hopefully, easy ones.

I hope this helps!