**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!