Lecture 8: Sensitive Drugs Requiring Titration and Heparin Infusion

Critical IV Medications

So far we have learned to calculate the flow rate and infusion and completion times for common IV fluids and for some medications that may be administered via IV infusion. The calculations we have made so far have primarily involved fluids and drugs like antibiotics for which the range of safety is fairly large. Obviously we would want these calculations to be as accurate as possible, but some variance from the ideal dose is fine as long as it is within the range of safety and effectiveness. So, when the range of safety and effectiveness is large, error introduced by rounding off to the nearest drop when using macrodrip tubing or by a manual IV setup varying in rate somewhat over time because it is not automatically adjusted is acceptable as long as this error is not too large.

In contrast, drugs used in critical care situations or which are used to control vital body functions (such as blood pressure, heart rate and kidney function, for example) have a much more narrow range of safety; so, for these drugs, we will need to use electronic infusion devices (such as volumetric pumps, or syringe pumps, for example), which can automatically control and monitor the flow rate. In rare cases where we might use a manual setup, we will always use microdrip tubing. Microdrip tubing is more accurate than macrodrip tubing because it has smaller drops, and therefore when we round off to the nearest drop, we are rounding off by a smaller amount and introducing a smaller amount of error.

Dosage Infusion Rate

Many of the drugs in this section have a quick action and short duration; in other words, they begin to work quickly, but don't last very long. Because of this, many of these drugs will often be ordered using a dosage infusion rate. Remember that dosage refers to the way that the strength of the drug is measured, i.e. the way the drug would be measured in a drug order or prescription. So dosage units that we have encountered so far are: mg, mcg, g, units, and mEq. Notice that mL are not a unit used to measure dosage; rather mL are used to measure volume.

A rate is something that is measured over time. So, dosage infusion rates, which measure the rate at which a drug is infusing, might have any of the following units: mg/min, mg/h, mcg/min, units/h. (Other units are possible, but the ones listed here are the most commonly used.) Dosage infusion rates are necessary because they tell us how much of the drug is infusing each minute, which is important because it is the amount of the drug and not the amount of fluid which determines the patient's response to the drug.

In contrast, flow rates, which we have already encountered, measure volume over time in gtt/min or mL/h; they measure the rate at which the liquid solution which contains the drug is infusing. We need the flow rate in order to actually administer the drug, because once a drug is mixed into a liquid, the only way we have to measure the drug is to use the concentration of the drug to determine the volume we need, and then we can measure out the volume of the drug which will give us the proper dosage.

For example,

Titration

Sometimes one of these more sensitive drugs are titrated; titration is a process in which a drug is ordered to be infused at a dosage infusion rate range, the nurse's job would then be to begin giving the patient the smallest dosage infusion rate in the range; then periodically the dosage would be raised until a particular desired response is obtained or until the maximum dosage infusion rate given by the range is reached. If the maximum dosage is reached before the desired response is obtained, then the nurse must obtain a new order from the doctor before adjusting the dosage further.

For example,

Calculating Flow Rate from Medication Orders of Critical and Sensitive Drugs

Just as in previous problems, whenever we receive an IV order, we will need to calculate the required flow rate before we can set up the IV. The only difference between these calculations in this section and those in the previous one are the way the drug is ordered. In this section we will have to use the dosage infusion rate ordered to determine the flow rate.

Also, because the drugs used in this section are very sensitive, often they are also ordered by weight, just like the problems we encounted a few lectures ago. So we may see a dosage infusion rate that depends upon weight; if this happens, we will need to use the patient's weight to eliminate the kg in the dosage infusion rate, and then we can proceed as in any other problem. For example, if a drug is ordered to infuse at 5 mcg/kg/min, we just need to multiply by the patient's weight in kg to cancel out the kg in the fraction. So, for example, if the patient weighs 100 kg, this would become 500 mcg/min, which is just a regular dosage infusion rate.

Let's look at some example problems:

First we will look at some simple examples. In these problems, the IV medication has been ordered by dosage infusion rate, and we must convert this to flow rate in mL/h so that we can set up the IV.

Example:

The order is for 5 mg of Levophed in 250 mL of D5S to infuse at 2 mcg/min.
Calculate the flow rate in mL/h to the nearest whole mL.

We want our answer to be in

, because all the problems in this chapter involve an electronic infusion device to control the flow rate of the IV, so the speed must be in mL/h, because this is the unit of measure accepted by electronic infusion devices.

_____

Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. Well, we don't have a rate in volume over time, but we have a rate in dosage over time: 2 mcg/min, and we know that we can get from mcg to mL by using the concentration of the drug: 5 mg of Levophed in 250 mL of D5S.

So this gives us:

_____

2 mcg

1 min

Now we want to get from mcg to mL. So, to cancel out mcg in the bottom of the fraction, we need to multiply by a fraction that has mcg in the top. Because we know that 1 mg = 1000 mcg, we have such a fraction:

1 mg

1000 mcg

. So, multiplying by this yields:

_____

2 mcg

1 min

1 mg

1000 mcg

Now we need to cancel out the mg, which appears in the top of the fraction, so we need to multiply by a fraction that has mg in the bottom. The concentration gives us such a fraction:

250 mL

5 mg

. Multiplying by this yields:

_____

2 mcg

1 min

1 mg

1000 mcg

250 mL

5 mg

Now we notice that we want our answer to be in mL/h, but the only units of time we can see in the bottom of our fractions on the right side of the equation is in min. So we need to multiply by a fraction which is equal to one and has min in the top so that we can cancel out the min in the bottom of the fraction on the right side of the equation. (Remember that it doesn't matter what order we multiply things in; because to multiply we will just multiply straight across, so for two units to cancel we don't need them to be right next to each other - we only need one of them to be somewhere in the top of a fraction and the other to be somewhere in the bottom.) We can convert this min to the unit h, which we want, because we know that 1 h = 60 min. So we can use the fraction:

60 min

1 h

, which is equal to one, and which has min in the top, to cancel out the min in the bottom.

So this yields:

_____

2 mcg

1 min

1 mg

1000 mcg

250 mL

5 mg

60 min

1 h

Now we need to cancel out all units that appear in both the top and the bottom of the equation:

_____

2 ~~mcg~~

1 ~~min~~

1 mg

1000 ~~mcg~~

250 mL

5 mg

60 ~~min~~

1 h

So this simplifies to:

_____

1000

250

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:

We can divide both 5 and 60 by 5. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

1000

250

5₁

60¹²

_____

1000

250

We can divide both 1000 and 12 by 4. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

~~1000~~₂₅₀

250

12³

_____

250

We can divide both 250 and by 2. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

2¹

~~250~~₁₂₅

250

_____

125

250

We can divide both 125 and 250 by 125. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

~~125~~₁

~~250~~²

_____

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:

_____

= 6

So our answer is 6

Example:

The order is for 1 g of Brevibloc in 2 L of NS to infuse at 310 mcg/min.
Calculate the flow rate in mL/h to the nearest whole mL.

We want our answer to be in

_____

Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. Well, we don't have a rate in volume over time, but we have a rate in dosage over time: 310 mcg/min, and we know that we can get from mcg to mL by using the concentration of the drug: 1 g of Brevibloc in 2 L of NS.

So this gives us:

_____

310 mcg

1 min

1 mg

1000 mcg

. So, multiplying by this yields:

_____

310 mcg

1 min

1 mg

1000 mcg

But we notice that our concentration is 1 g of Brevibloc in 2 L of NS, which contains g, but our set of fractions still contains mg instead of g. We can fix this; all we need to do is multiply by a fraction that has mg on the bottom to cancel out the mg which appears on the top. Because we know that 1 g = 1000 mg, we have such a fraction:

1 g

1000 mg

. So, multiplying by this yields:

_____

310 mcg

1 min

1 mg

1000 mcg

1 g

1000 mg

Now we need to cancel out the g, which appears in the top of the fraction, so we need to multiply by a fraction that has g in the bottom. The concentration gives us such a fraction:

2 L

1 g

. Multiplying by this yields:

_____

310 mcg

1 min

1 mg

1000 mcg

1 g

1000 mg

2 L

1 g

We need to multiply by a fraction which is equal to one and has L in the bottom so that we can cancel out the L in the top of the fraction on the right of the equation.

So, we first need to multiply by a fraction that is equal to one and which cancels out L and gets us to mL. Since we know that 1L=1000mL, the fraction

1000 mL

1 L

will be equal to one, and since it has L in the bottom, will also cancel out the L in the top of the fraction in the right side of the equation we already have:

_____

310 mcg

1 min

1 mg

1000 mcg

1 g

1000 mg

2 L

1 g

1000 mL

1 L

60 min

1 h

, which is equal to one, and which has min in the top, to cancel out the min in the bottom.

So this yields:

_____

310 mcg

1 min

1 mg

1000 mcg

1 g

1000 mg

2 L

1 g

1000 mL

1 L

60 min

1 h

Now we need to cancel out all units that appear in both the top and the bottom of the equation:

_____

310 ~~mcg~~

1 ~~min~~

1 mg

1000 ~~mcg~~

1 g

1000 mg

2 L

1 g

1000 mL

1 L

60 ~~min~~

1 h

So this simplifies to:

_____

310

1000

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:

We can divide both 1 and 60 by 1. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

310

1000

1₁

1000

60⁶⁰

_____

310

1000

We can divide both 1000 and 1000 by 1000. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

310

~~1000~~₁

1000

~~1000~~¹

_____

310

1000

We can divide both 1000 and 60 by 20. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

310

~~1000~~₅₀

60³

_____

310

We can divide both 50 and 310 by 10. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

~~310~~³¹

50₅

_____

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:

_____

186

= 37.2

= 37

So our answer is 37

Example:

The order is for 2 g of procanimide in 250 mL of D5W to infuse at 4 mg/min.
Calculate the flow rate in mL/h to the nearest whole mL.

We want our answer to be in

_____

Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. Well, we don't have a rate in volume over time, but we have a rate in dosage over time: 4 mg/min, and we know that we can get from mg to mL by using the concentration of the drug: 2 g of procanimide in 250 mL of D5W.

So this gives us:

_____

4 mg

1 min

Now we want to get from mg to mL. So, to cancel out mg in the bottom of the fraction, we need to multiply by a fraction that has mg in the top. Because we know that 1 g = 1000 mg, we have such a fraction:

1 g

1000 mg

. So, multiplying by this yields:

_____

4 mg

1 min

1 g

1000 mg

Now we need to cancel out the g, which appears in the top of the fraction, so we need to multiply by a fraction that has g in the bottom. The concentration gives us such a fraction:

250 mL

2 g

. Multiplying by this yields:

_____

4 mg

1 min

1 g

1000 mg

250 mL

2 g

60 min

1 h

, which is equal to one, and which has min in the top, to cancel out the min in the bottom.

So this yields:

_____

4 mg

1 min

1 g

1000 mg

250 mL

2 g

60 min

1 h

Now we need to cancel out all units that appear in both the top and the bottom of the equation:

_____

4 mg

1 ~~min~~

1 g

1000 mg

250 mL

2 g

60 ~~min~~

1 h

So this simplifies to:

_____

1000

250

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:

We can divide both 2 and 60 by 2. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

1000

250

2₁

60³⁰

_____

1000

250

We can divide both 1000 and 30 by 10. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

~~1000~~₁₀₀

250

30³

_____

100

250

We can divide both 100 and by 4. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

4¹

~~100~~₂₅

250

_____

250

We can divide both 25 and 250 by 25. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

25₁

~~250~~¹⁰

_____

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:

_____

= 30

So our answer is 30

Now we will look at examples that have one extra step. In these problems, we will be given a dosage infusion rate that depends upon the patient's weight. So these problems will be exactly the same as the previous ones, except now we will need to convert the dosage infusion rate by weight to just a dosage infusion rate by cancelling out the kg.

Example:

The order is for 300 mg of a medication in 75 mL of NS to infuse at 46 mcg/kg/min. The patient weighs 72.7 kg.
Calculate the flow rate in mL/h to the nearest whole mL.

We want our answer to be in

_____

Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. Well, we don't have a rate in volume over time, but we have a rate in dosage over time: 46 mcg/kg/min, and we know that we can get from mcg to mL by using the concentration of the drug: 300 mg of a medication in 75 mL of NS.

Because this rate is in terms of mcg/kg/min, we want to go ahead and convert it to mcg/min by mulitplying by 72.7 kg:

_____

mcg

min

46	mcg
	kg

min

72.7 kg

Because we can put kg over 1 without changing it, this becomes:

_____

mcg

min

46	mcg
	kg

min

72.7	kg
	1

And cancelling kg, which appear in both the top and the bottom of the numerator yields:

_____

mcg

min

46	mcg
	kg

min

72.7	kg
	1

46 mcg

min

72.7

Multiplying yields:

_____

mcg

min

46 mcg

min

72.7

3344.2 mcg

min

= 3344.2

mcg

min

So this gives us:

_____

3344.2 mcg

1 min

1 mg

1000 mcg

. So, multiplying by this yields:

_____

3344.2 mcg

1 min

1 mg

1000 mcg

Now we need to cancel out the mg, which appears in the top of the fraction, so we need to multiply by a fraction that has mg in the bottom. The concentration gives us such a fraction:

75 mL

300 mg

. Multiplying by this yields:

_____

3344.2 mcg

1 min

1 mg

1000 mcg

75 mL

300 mg

60 min

1 h

, which is equal to one, and which has min in the top, to cancel out the min in the bottom.

So this yields:

_____

3344.2 mcg

1 min

1 mg

1000 mcg

75 mL

300 mg

60 min

1 h

Now we need to cancel out all units that appear in both the top and the bottom of the equation:

_____

3344.2 ~~mcg~~

1 ~~min~~

1 mg

1000 ~~mcg~~

75 mL

300 mg

60 ~~min~~

1 h

So this simplifies to:

_____

3344.2

1000

300

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:

We can divide both 300 and 60 by 60. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

3344.2

1000

~~300~~₅

60¹

_____

3344.2

1000

We can divide both 1000 and 75 by 25. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

3344.2

~~1000~~₄₀

75³

_____

3344.2

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:

_____

10032.6

200

= 50.163

= 50

So our answer is 50

Example:

The order is for 30 mg of dobutamine in 100 mL of D5W to infuse at 2.5 mcg/kg/min. The patient weighs 71.8 kg.
Calculate the flow rate in mL/h to the nearest whole mL.

We want our answer to be in

_____

Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. Well, we don't have a rate in volume over time, but we have a rate in dosage over time: 2.5 mcg/kg/min, and we know that we can get from mcg to mL by using the concentration of the drug: 30 mg of dobutamine in 100 mL of D5W.

Because this rate is in terms of mcg/kg/min, we want to go ahead and convert it to mcg/min by mulitplying by 71.8 kg:

_____

mcg

min

2.5	mcg
	kg

min

71.8 kg

Because we can put kg over 1 without changing it, this becomes:

_____

mcg

min

2.5	mcg
	kg

min

71.8	kg
	1

And cancelling kg, which appear in both the top and the bottom of the numerator yields:

_____

mcg

min

2.5	mcg
	kg

min

71.8	kg
	1

2.5 mcg

min

71.8

Multiplying yields:

_____

mcg

min

2.5 mcg

min

71.8

179.5 mcg

min

= 179.5

mcg

min

So this gives us:

_____

179.5 mcg

1 min

1 mg

1000 mcg

. So, multiplying by this yields:

_____

179.5 mcg

1 min

1 mg

1000 mcg

Now we need to cancel out the mg, which appears in the top of the fraction, so we need to multiply by a fraction that has mg in the bottom. The concentration gives us such a fraction:

100 mL

30 mg

. Multiplying by this yields:

_____

179.5 mcg

1 min

1 mg

1000 mcg

100 mL

30 mg

60 min

1 h

, which is equal to one, and which has min in the top, to cancel out the min in the bottom.

So this yields:

_____

179.5 mcg

1 min

1 mg

1000 mcg

100 mL

30 mg

60 min

1 h

Now we need to cancel out all units that appear in both the top and the bottom of the equation:

_____

179.5 ~~mcg~~

1 ~~min~~

1 mg

1000 ~~mcg~~

100 mL

30 mg

60 ~~min~~

1 h

So this simplifies to:

_____

179.5

1000

100

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:

We can divide both 30 and 60 by 30. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

179.5

1000

100

30₁

60²

_____

179.5

1000

100

We can divide both 1000 and 2 by 2. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

179.5

~~1000~~₅₀₀

100

2¹

_____

179.5

500

100

We can divide both 500 and 100 by 100. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

179.5

~~500~~₅

~~100~~¹

_____

179.5

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:

_____

179.5

= 35.9

= 36

So our answer is 36

Example:

The order is for 400 mg of dopamine in 250 mL of D5W to infuse at 4 mcg/kg/min. The patient weighs 65 kg.
Calculate the flow rate in mL/h to the nearest whole mL.

We want our answer to be in

_____

Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. Well, we don't have a rate in volume over time, but we have a rate in dosage over time: 4 mcg/kg/min, and we know that we can get from mcg to mL by using the concentration of the drug: 400 mg of dopamine in 250 mL of D5W.

Because this rate is in terms of mcg/kg/min, we want to go ahead and convert it to mcg/min by mulitplying by 65 kg:

_____

mcg

min

4	mcg
	kg

min

65 kg

Because we can put kg over 1 without changing it, this becomes:

_____

mcg

min

4	mcg
	kg

min

65	kg
	1

And cancelling kg, which appear in both the top and the bottom of the numerator yields:

_____

mcg

min

4	mcg
	kg

min

65	kg
	1

4 mcg

min

Multiplying yields:

_____

mcg

min

4 mcg

min

260 mcg

min

= 260

mcg

min

So this gives us:

_____

260 mcg

1 min

1 mg

1000 mcg

. So, multiplying by this yields:

_____

260 mcg

1 min

1 mg

1000 mcg

Now we need to cancel out the mg, which appears in the top of the fraction, so we need to multiply by a fraction that has mg in the bottom. The concentration gives us such a fraction:

250 mL

400 mg

. Multiplying by this yields:

_____

260 mcg

1 min

1 mg

1000 mcg

250 mL

400 mg

60 min

1 h

, which is equal to one, and which has min in the top, to cancel out the min in the bottom.

So this yields:

_____

260 mcg

1 min

1 mg

1000 mcg

250 mL

400 mg

60 min

1 h

Now we need to cancel out all units that appear in both the top and the bottom of the equation:

_____

260 ~~mcg~~

1 ~~min~~

1 mg

1000 ~~mcg~~

250 mL

400 mg

60 ~~min~~

1 h

So this simplifies to:

_____

260

1000

250

400

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:

We can divide both 400 and 60 by 20. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

260

1000

250

~~400~~₂₀

60³

_____

260

1000

250

We can divide both 1000 and by 20. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

~~260~~¹³

~~1000~~₅₀

250

_____

250

We can divide both 50 and 250 by 50. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

50₁

~~250~~⁵

_____

We can divide both 20 and 5 by 5. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

5¹

20₄

_____

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:

_____

= 9.75

= 10

So our answer is 10

Now we will look at some simple examples going the other direction. In these examples, we know the flow rate and we want to convert the flow rate to a dosage infusion rate so that we can find out what dosage of the drug the patient is getting over time. We might have to do this if we want to check the safety of the drug order, or if we want to check to see if an IV set up by another nurse infusing at this rate is actually infusing at the dosage rate which has been ordered.

Example:

The order is for 2 g of lidocaine in 500 mL of D5W to infuse at 30 mL/h.
Calculate the dosage rate in mg/min to the nearest tenth.

We want our answer to be in

min

_____

min

Now since we are looking for a rate, which is dosage over time, we need to find something to put on the right side of the equation that will be a rate which expresses dosage over time. Well, we don't have a rate in dosage over time, but we have a rate in volume over time: 30 gtt/min. We know that all problems in this section use microdrip tubing, so we know that our flow rate in gtt/min is equal to our flow rate in ml/h (since the 60 gtt/mL will cancel out the 60 min/h in our equation). So we can rewrite 30 gtt/min as:30 mL/h, and we know that we can get from mL to mg by using the concentration of the drug: 2 g of lidocaine in 500 mL of D5W.

So this gives us:

_____

min

30 mL

1 h

Now we want to cancel out the mL units which appear in the top of the fraction at right and convert it to some kind of dosage units, because we are trying to get to mg units. We can use the concentration to do this: 2 g of lidocaine in 500 mL of D5W. So now we multiply by

2 g

500 mL

_____

min

30 mL

1 h

2 g

500 mL

Now we want to get from g to mg. So, to cancel out g in the top of the fraction, we need to multiply by a fraction that has g in the bottom. Because we know that 1000 mg = 1 g, we have such a fraction:

1000 mg

1 g

. So, multiplying by this yields:

_____

min

30 mL

1 h

2 g

500 mL

1000 mg

1 g

Now we notice that we want our answer to be in mg/min, but the only units of time we can see in the bottom of our fractions on the right side of the equation is in h. So we need to multiply by a fraction which is equal to one and has h in the top so that we can cancel out the h in the bottom of the fraction on the right side of the equation. (Remember that it doesn't matter what order we multiply things in; because to multiply we will just multiply straight across, so for two units to cancel we don't need them to be right next to each other - we only need one of them to be somewhere in the top of a fraction and the other to be somewhere in the bottom.) We can convert this h to the unit min, which we want, because we know that 1 h = 60 min. So we can use the fraction:

1 h

60 min

, which is equal to one, and which has h in the top, to cancel out the h in the bottom.

So this yields:

_____

min

30 mL

1 h

2 g

500 mL

1000 mg

1 g

1 h

60 min

Now we need to cancel out all units that appear in both the top and the bottom of the equation:

_____

min

30 mL

1 h

2 g

500 mL

1000 mg

1 g

1 h

60 min

So this simplifies to:

_____

min

500

1000

min

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:

We can divide both 2 and 500 by 2. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

min

250

1000

min

We can divide both 1 and 60 by 1. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

min

250

1000

min

We can divide both 30 and 60 by 30. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

min

250

1000

min

We can divide both 1000 and 2 by 2. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

min

250

500

min

We can divide both 500 and 250 by 250. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

min

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:

_____

min

= 2

min

= 2

min

So our answer is 2

min

Example:

The order is for 330 mg of a medication in 250 mL of NS to infuse at 28 mL/h.
Calculate the dosage rate in mcg/min to the nearest tenth.

We want our answer to be in

mcg

min

_____

mcg

min

Now since we are looking for a rate, which is dosage over time, we need to find something to put on the right side of the equation that will be a rate which expresses dosage over time. Well, we don't have a rate in dosage over time, but we have a rate in volume over time: 28 gtt/min. We know that all problems in this section use microdrip tubing, so we know that our flow rate in gtt/min is equal to our flow rate in ml/h (since the 60 gtt/mL will cancel out the 60 min/h in our equation). So we can rewrite 28 gtt/min as:28 mL/h, and we know that we can get from mL to mcg by using the concentration of the drug: 330 mg of a medication in 250 mL of NS.

So this gives us:

_____

mcg

min

28 mL

1 h

Now we want to cancel out the mL units which appear in the top of the fraction at right and convert it to some kind of dosage units, because we are trying to get to mcg units. We can use the concentration to do this: 330 mg of a medication in 250 mL of NS. So now we multiply by

330 mg

250 mL

_____

mcg

min

28 mL

1 h

330 mg

250 mL

Now we want to get from mg to mcg. So, to cancel out mg in the top of the fraction, we need to multiply by a fraction that has mg in the bottom. Because we know that 1000 mcg = 1 mg, we have such a fraction:

1000 mcg

1 mg

. So, multiplying by this yields:

_____

mcg

min

28 mL

1 h

330 mg

250 mL

1000 mcg

1 mg

Now we notice that we want our answer to be in mcg/min, but the only units of time we can see in the bottom of our fractions on the right side of the equation is in h. So we need to multiply by a fraction which is equal to one and has h in the top so that we can cancel out the h in the bottom of the fraction on the right side of the equation. (Remember that it doesn't matter what order we multiply things in; because to multiply we will just multiply straight across, so for two units to cancel we don't need them to be right next to each other - we only need one of them to be somewhere in the top of a fraction and the other to be somewhere in the bottom.) We can convert this h to the unit min, which we want, because we know that 1 h = 60 min. So we can use the fraction:

1 h

60 min

, which is equal to one, and which has h in the top, to cancel out the h in the bottom.

So this yields:

_____

mcg

min

28 mL

1 h

330 mg

250 mL

1000 mcg

1 mg

1 h

60 min

Now we need to cancel out all units that appear in both the top and the bottom of the equation:

_____

mcg

min

28 mL

1 h

330 mg

250 mL

1000 mcg

1 mg

1 h

60 min

So this simplifies to:

_____

mcg

min

330

250

1000

mcg

min

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:

We can divide both 330 and 250 by 10. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

mcg

min

1000

mcg

min

We can divide both 33 and 60 by 3. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

mcg

min

1000

mcg

min

We can divide both 28 and 20 by 4. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

mcg

min

1000

mcg

min

We can divide both 1000 and 5 by 5. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

mcg

min

200

mcg

min

We can divide both 200 and 25 by 25. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

mcg

min

mcg

min

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:

_____

mcg

min

616

mcg

min

= 616

mcg

min

= 616

mcg

min

So our answer is 616

mcg

min

Example:

The order is for 2.5 g of a medication in 350 mL of D5W to infuse at 36 mL/h.
Calculate the dosage rate in mg/h to the nearest tenth.

We want our answer to be in

_____

Now since we are looking for a rate, which is dosage over time, we need to find something to put on the right side of the equation that will be a rate which expresses dosage over time. Well, we don't have a rate in dosage over time, but we have a rate in volume over time: 36 gtt/min. We know that all problems in this section use microdrip tubing, so we know that our flow rate in gtt/min is equal to our flow rate in ml/h (since the 60 gtt/mL will cancel out the 60 min/h in our equation). So we can rewrite 36 gtt/min as:36 mL/h, and we know that we can get from mL to mg by using the concentration of the drug: 2.5 g of a medication in 350 mL of D5W.

So this gives us:

_____

36 mL

1 h

Now we want to cancel out the mL units which appear in the top of the fraction at right and convert it to some kind of dosage units, because we are trying to get to mg units. We can use the concentration to do this: 2.5 g of a medication in 350 mL of D5W. So now we multiply by

2.5 g

350 mL

_____

36 mL

1 h

2.5 g

350 mL

1000 mg

1 g

. So, multiplying by this yields:

_____

36 mL

1 h

2.5 g

350 mL

1000 mg

1 g

Now we need to cancel out all units that appear in both the top and the bottom of the equation:

_____

36 mL

1 h

2.5 g

350 mL

1000 mg

1 g

So this simplifies to:

_____

2.5

350

1000

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:

We can divide both 2.5 and 350 by 1. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

2.5

350

1000

We can divide both 1000 and 350 by 50. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

_____

2.5

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:

_____

1800

= 257.14285714286

= 257.1

So our answer is 257.1

Now we look at some examples of drug orders for titrated drugs. Remember that these are drugs ordered using a dosage range, and the nurse must begin the IV at the lowest rate and then periodically increase the rate until the desired response is obtained. First, we would want to convert the dosage infusion rate range to flow rate range so that we can figure out what to enter in the electronic infusion device as the rate of infusion. This gives us a minimum and maximum flow rate, so we can set up the IV to infuse at the minimum flow rate and increase the flow rate periodically until we reach the maximum flow rate, or until the desired response is obtained, whichever comes first. Then, when the desired response is obtained, we can see what the flow rate of the IV is at that moment; we will want to record what the rate was when the patient stabilized, but we won't want to put it down in mL/h becuase this doesn't really indicate how much medication they were receiving over time; rather, it only indicates the amount of fluid they were receiving over time. So, we would want to convert this flow rate in mL/h back to a dosage infusion rate, which we can then record on the patient's chart.

So, all of these problems will have two steps:

Converting the ordered dosage infusion rate range to a flow rate range.
Converting the flow rate of the stabilizing dosage to the equivalent dosage infusion rate.

Clearly if we do this correctly, the stabilizing dosage infusion rate should fall between the minimum and maximum ordered dosage infusion rates.

Example:

The order is for 2 g of lidocaine in 500 mL of NS to infuse at 2-4 mg/min.
Calculate the flow rate range in mL/h to the nearest whole mL. The patient stabilizes at 45 mL/h. Calculate the dosage infusion rate at which the patient stabilized in mg/min.

We want our answer to be in

_____

We have a range for the dosage infusion rate: 2-4 mg/min So this means that we need our answer to be a range as well; in other words, we need a minimum and a maximum flow rate for our answer.

We will find the minimum first, and then we will use the same method to find the maximum.

Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. Well, we don't have a rate in volume over time, but we have a rate in dosage over time: 2 mg/min for the minimum, and 4 mg/min for the maximum, and we know that we can get from mg to mL by using the concentration of the drug: 2 g of lidocaine in 500 mL of NS.

So this gives us:

Min: _____

2 mg

1 min

1 g

1000 mg

. So, multiplying by this yields:

Min: _____

2 mg

1 min

1 g

1000 mg

Now we need to cancel out the g, which appears in the top of the fraction, so we need to multiply by a fraction that has g in the bottom. The concentration gives us such a fraction:

500 mL

2 g

. Multiplying by this yields:

Min: _____

2 mg

1 min

1 g

1000 mg

500 mL

2 g

60 min

1 h

, which is equal to one, and which has min in the top, to cancel out the min in the bottom.

So this yields:

Min: _____

2 mg

1 min

1 g

1000 mg

500 mL

2 g

60 min

1 h

Now we need to cancel out all units that appear in both the top and the bottom of the equation:

Min: _____

2 mg

1 ~~min~~

1 g

1000 mg

500 mL

2 g

60 ~~min~~

1 h

So this simplifies to:

Min: _____

1000

500

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:

We can divide both 2 and 60 by 2. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

Min: _____

1000

500

2₁

60³⁰

Min: _____

1000

500

We can divide both 1000 and 30 by 10. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

Min: _____

~~1000~~₁₀₀

500

30³

Min: _____

100

500

We can divide both 100 and by 2. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

Min: _____

2¹

~~100~~₅₀

500

Min: _____

500

We can divide both 50 and 500 by 50. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:

Min: _____

50₁

~~500~~¹⁰

Min: _____

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:

Min: _____

= 30

So our minimum is 30

We want our answer to be in

_____

We have a range for the dosage infusion rate: 2-4 mg/min So this means that we need our answer to be a range as well; in other words, we need a minimum and a maximum flow rate for our answer.

Now that we have found the minimum, we will find the maximum.

Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. Well, we don't have a rate in volume over time, but we have a rate in dosage over time: 2 mg/min for the minimum, and 4 mg/min for the maximum, and we know that we can get from mg to mL by using the concentration of the drug: 2 g of lidocaine in 500 mL of NS.

So this gives us:

Max: _____

=

4 mg

1 min

Now we want to get from mg to mL. So, to cancel out mg in the bottom of the fraction, we need to multiply by a fraction that has mg in the top. Because we know that 1 g = 1000 mg, we have such a fraction:

1 g

1000 mg

. So, multiplying by this yields:
Max: _____

mL

h

=

4 mg

1 min

×

1 g

1000 mg

Now we need to cancel out the g, which appears in the top of the fraction, so we need to multiply by a fraction that has g in the bottom. The concentration gives us such a fraction:

500 mL

2 g

. Multiplying by this yields:
Max: _____

mL

h

=

4 mg

1 min

×

1 g

1000 mg

×

500 mL

2 g

Now we notice that we want our answer to be in mL/h, but the only units of time we can see in the bottom of our fractions on the right side of the equation is in min. So we need to multiply by a fraction which is equal to one and has min in the top so that we can cancel out the min in the bottom of the fraction on the right side of the equation. (Remember that it doesn't matter what order we multiply things in; because to multiply we will just multiply straight across, so for two units to cancel we don't need them to be right next to each other - we only need one of them to be somewhere in the top of a fraction and the other to be somewhere in the bottom.) We can convert this min to the unit h, which we want, because we know that 1 h = 60 min. So we can use the fraction:

60 min

1 h

, which is equal to one, and which has min in the top, to cancel out the min in the bottom.

So this yields:
Max: _____

mL

h

=

4 mg

1 min

×

1 g

1000 mg

×

500 mL

2 g

×

60 min

1 h

Now we need to cancel out all units that appear in both the top and the bottom of the equation:
Max: _____

mL

h

=

4 mg

1 ~~min~~

×

1 g

1000 mg

×

500 mL

2 g

×

60 ~~min~~

1 h

So this simplifies to:
Max: _____

mL

h

=

4

1

×

1

1000

×

500

2

×

60

1

mL

h

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 2 and 60 by 2. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

4

1

×

1

1000

×

500

2₁

×

60³⁰

1

mL

h

Max: _____

mL

h

=

4

1

×

1

1000

×

500

1

×

30

1

mL

h

We can divide both 1000 and 30 by 10. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

4

1

×

1

~~1000~~₁₀₀

×

500

1

×

1

1

×

30³

1

mL

h

Max: _____

mL

h

=

4

1

×

1

100

×

500

1

×

1

1

×

3

1

mL

h

We can divide both 100 and by 4. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

4¹

1

×

1

~~100~~₂₅

×

500

1

×

1

1

×

3

1

mL

h

Max: _____

mL

h

=

1

1

×

1

25

×

500

1

×

1

1

×

3

1

mL

h

We can divide both 25 and 500 by 25. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

1

1

×

1

25₁

×

~~500~~²⁰

1

×

1

1

×

3

1

mL

h

Max: _____

mL

h

=

1

1

×

1

1

×

20

1

×

1

1

×

3

1

mL

h

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:
Max: _____

mL

h

=

60

1

mL

h

= 60

mL

h

= 60

mL

h

So our maximum is 60

mL

h

.
So our answer is 30-60

mL

h

.
Now that we've calculated the ordered flow rate range, we want to calculate the dosage rate for the stabilizing dosage. Since the stabilizing dosage flow rate given in the problem was 45 mL/h, what we actually need to do is to solve the following problem:
The order is for 2 g of lidocaine in 500 mL of NS and the patient stabilized at 45 mL/h.
Calculate the stabilizing dosage rate in mg/min to the nearest tenth.
We want our answer to be in

mg

min

.

_____

mg

min

=
Now since we are looking for a rate, which is dosage over time, we need to find something to put on the right side of the equation that will be a rate which expresses dosage over time. Well, we don't have a rate in dosage over time, but we have a rate in volume over time: 45 gtt/min. We know that all problems in this section use microdrip tubing, so we know that our flow rate in gtt/min is equal to our flow rate in ml/h (since the 60 gtt/mL will cancel out the 60 min/h in our equation). So we can rewrite 45 gtt/min as:45 mL/h, and we know that we can get from mL to mg by using the concentration of the drug: 2 g of lidocaine in 500 mL of NS.
So this gives us:
_____

mg

min

=

45 mL

1 h

Now we want to cancel out the mL units which appear in the top of the fraction at right and convert it to some kind of dosage units, because we are trying to get to mg units. We can use the concentration to do this: 2 g of lidocaine in 500 mL of NS. So now we multiply by

2 g

500 mL

:
_____

mg

min

=

45 mL

1 h

×

2 g

500 mL

Now we want to get from g to mg. So, to cancel out g in the top of the fraction, we need to multiply by a fraction that has g in the bottom. Because we know that 1000 mg = 1 g, we have such a fraction:

1000 mg

1 g

. So, multiplying by this yields:
_____

mg

min

=

45 mL

1 h

×

2 g

500 mL

×

1000 mg

1 g

Now we notice that we want our answer to be in mg/min, but the only units of time we can see in the bottom of our fractions on the right side of the equation is in h. So we need to multiply by a fraction which is equal to one and has h in the top so that we can cancel out the h in the bottom of the fraction on the right side of the equation. (Remember that it doesn't matter what order we multiply things in; because to multiply we will just multiply straight across, so for two units to cancel we don't need them to be right next to each other - we only need one of them to be somewhere in the top of a fraction and the other to be somewhere in the bottom.) We can convert this h to the unit min, which we want, because we know that 1 h = 60 min. So we can use the fraction:

1 h

60 min

, which is equal to one, and which has h in the top, to cancel out the h in the bottom.

So this yields:
_____

mg

min

=

45 mL

1 h

×

2 g

500 mL

×

1000 mg

1 g

×

1 h

60 min

Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____

mg

min

=

45 mL

1 h

×

2 g

500 mL

×

1000 mg

1 g

×

1 h

60 min

So this simplifies to:
_____

mg

min

=

45

1

×

2

500

×

1000

1

×

1

60

mg

min

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 2 and 500 by 2. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mg

min

=

45

1

×

1

250

×

1000

1

×

1

1

mg

min

We can divide both 1 and 60 by 1. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mg

min

=

45

1

×

1

250

×

1000

1

×

1

60

mg

min

We can divide both 45 and 60 by 15. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mg

min

=

3

1

×

1

250

×

1000

1

×

1

4

mg

min

We can divide both 1000 and 4 by 4. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mg

min

=

3

1

×

1

1

×

1

250

×

250

1

×

1

1

mg

min

We can divide both 250 and 250 by 250. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mg

min

=

3

1

×

1

1

×

1

1

×

1

1

×

1

1

mg

min

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:
_____

mg

min

=

3

1

mg

min

= 3

mg

min

= 3

mg

min

So our stabilizing dosage rate is 3

mg

min

.
Example:
The order is for 75 mg of a medication in 100 mL of NS to infuse at 130-641 mcg/min.
Calculate the flow rate range in mL/h to the nearest whole mL. The patient stabilizes at 24 mL/h. Calculate the dosage infusion rate at which the patient stabilized in mg/h.
We want our answer to be in

mL

h

, because all the problems in this chapter involve an electronic infusion device to control the flow rate of the IV, so the speed must be in mL/h, because this is the unit of measure accepted by electronic infusion devices.

_____

mL

h

=
We have a range for the dosage infusion rate: 130-641 mcg/min So this means that we need our answer to be a range as well; in other words, we need a minimum and a maximum flow rate for our answer.
We will find the minimum first, and then we will use the same method to find the maximum.
Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. Well, we don't have a rate in volume over time, but we have a rate in dosage over time: 130 mcg/min for the minimum, and 641 mcg/min for the maximum, and we know that we can get from mcg to mL by using the concentration of the drug: 75 mg of a medication in 100 mL of NS.
So this gives us:
Min: _____

mL

h

=

130 mcg

1 min

Now we want to get from mcg to mL. So, to cancel out mcg in the bottom of the fraction, we need to multiply by a fraction that has mcg in the top. Because we know that 1 mg = 1000 mcg, we have such a fraction:

1 mg

1000 mcg

. So, multiplying by this yields:
Min: _____

mL

h

=

130 mcg

1 min

×

1 mg

1000 mcg

Now we need to cancel out the mg, which appears in the top of the fraction, so we need to multiply by a fraction that has mg in the bottom. The concentration gives us such a fraction:

100 mL

75 mg

. Multiplying by this yields:
Min: _____

mL

h

=

130 mcg

1 min

×

1 mg

1000 mcg

×

100 mL

75 mg

Now we notice that we want our answer to be in mL/h, but the only units of time we can see in the bottom of our fractions on the right side of the equation is in min. So we need to multiply by a fraction which is equal to one and has min in the top so that we can cancel out the min in the bottom of the fraction on the right side of the equation. (Remember that it doesn't matter what order we multiply things in; because to multiply we will just multiply straight across, so for two units to cancel we don't need them to be right next to each other - we only need one of them to be somewhere in the top of a fraction and the other to be somewhere in the bottom.) We can convert this min to the unit h, which we want, because we know that 1 h = 60 min. So we can use the fraction:

60 min

1 h

, which is equal to one, and which has min in the top, to cancel out the min in the bottom.

So this yields:
Min: _____

mL

h

=

130 mcg

1 min

×

1 mg

1000 mcg

×

100 mL

75 mg

×

60 min

1 h

Now we need to cancel out all units that appear in both the top and the bottom of the equation:
Min: _____

mL

h

=

130 ~~mcg~~

1 ~~min~~

×

1 mg

1000 ~~mcg~~

×

100 mL

75 mg

×

60 ~~min~~

1 h

So this simplifies to:
Min: _____

mL

h

=

130

1

×

1

1000

×

100

75

×

60

1

mL

h

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 75 and 60 by 15. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Min: _____

mL

h

=

130

1

×

1

1000

×

100

75₅

×

60⁴

1

mL

h

Min: _____

mL

h

=

130

1

×

1

1000

×

100

5

×

4

1

mL

h

We can divide both 1000 and 4 by 4. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Min: _____

mL

h

=

130

1

×

1

~~1000~~₂₅₀

×

100

5

×

1

1

×

4¹

1

mL

h

Min: _____

mL

h

=

130

1

×

1

250

×

100

5

×

1

1

×

1

1

mL

h

We can divide both 250 and by 10. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Min: _____

mL

h

=

~~130~~¹³

1

×

1

~~250~~₂₅

×

100

5

×

1

1

×

1

1

mL

h

Min: _____

mL

h

=

13

1

×

1

25

×

100

5

×

1

1

×

1

1

mL

h

We can divide both 25 and 100 by 25. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Min: _____

mL

h

=

13

1

×

1

25₁

×

~~100~~⁴

5

×

1

1

×

1

1

mL

h

Min: _____

mL

h

=

13

1

×

1

1

×

4

5

×

1

1

×

1

1

mL

h

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:
Min: _____

mL

h

=

52

5

mL

h

= 10.4

mL

h

= 10

mL

h

So our minimum is 10

mL

h

.
We want our answer to be in

mL

h

, because all the problems in this chapter involve an electronic infusion device to control the flow rate of the IV, so the speed must be in mL/h, because this is the unit of measure accepted by electronic infusion devices.

_____

mL

h

=
We have a range for the dosage infusion rate: 130-641 mcg/min So this means that we need our answer to be a range as well; in other words, we need a minimum and a maximum flow rate for our answer.
Now that we have found the minimum, we will find the maximum.
Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. Well, we don't have a rate in volume over time, but we have a rate in dosage over time: 130 mcg/min for the minimum, and 641 mcg/min for the maximum, and we know that we can get from mcg to mL by using the concentration of the drug: 75 mg of a medication in 100 mL of NS.
So this gives us:
Max: _____

mL

h

=

641 mcg

1 min

Now we want to get from mcg to mL. So, to cancel out mcg in the bottom of the fraction, we need to multiply by a fraction that has mcg in the top. Because we know that 1 mg = 1000 mcg, we have such a fraction:

1 mg

1000 mcg

. So, multiplying by this yields:
Max: _____

mL

h

=

641 mcg

1 min

×

1 mg

1000 mcg

Now we need to cancel out the mg, which appears in the top of the fraction, so we need to multiply by a fraction that has mg in the bottom. The concentration gives us such a fraction:

100 mL

75 mg

. Multiplying by this yields:
Max: _____

mL

h

=

641 mcg

1 min

×

1 mg

1000 mcg

×

100 mL

75 mg

Now we notice that we want our answer to be in mL/h, but the only units of time we can see in the bottom of our fractions on the right side of the equation is in min. So we need to multiply by a fraction which is equal to one and has min in the top so that we can cancel out the min in the bottom of the fraction on the right side of the equation. (Remember that it doesn't matter what order we multiply things in; because to multiply we will just multiply straight across, so for two units to cancel we don't need them to be right next to each other - we only need one of them to be somewhere in the top of a fraction and the other to be somewhere in the bottom.) We can convert this min to the unit h, which we want, because we know that 1 h = 60 min. So we can use the fraction:

60 min

1 h

, which is equal to one, and which has min in the top, to cancel out the min in the bottom.

So this yields:
Max: _____

mL

h

=

641 mcg

1 min

×

1 mg

1000 mcg

×

100 mL

75 mg

×

60 min

1 h

Now we need to cancel out all units that appear in both the top and the bottom of the equation:
Max: _____

mL

h

=

641 ~~mcg~~

1 ~~min~~

×

1 mg

1000 ~~mcg~~

×

100 mL

75 mg

×

60 ~~min~~

1 h

So this simplifies to:
Max: _____

mL

h

=

641

1

×

1

1000

×

100

75

×

60

1

mL

h

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 75 and 60 by 15. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

641

1

×

1

1000

×

100

75₅

×

60⁴

1

mL

h

Max: _____

mL

h

=

641

1

×

1

1000

×

100

5

×

4

1

mL

h

We can divide both 1000 and 4 by 4. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

641

1

×

1

~~1000~~₂₅₀

×

100

5

×

1

1

×

4¹

1

mL

h

Max: _____

mL

h

=

641

1

×

1

250

×

100

5

×

1

1

×

1

1

mL

h

We can divide both 250 and 100 by 50. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

641

1

×

1

~~250~~₅

×

~~100~~²

5

×

1

1

×

1

1

mL

h

Max: _____

mL

h

=

641

1

×

1

5

×

2

5

×

1

1

×

1

1

mL

h

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:
Max: _____

mL

h

=

1282

25

mL

h

= 51.28

mL

h

= 51

mL

h

So our maximum is 51

mL

h

.
So our answer is 10-51

mL

h

.
Now that we've calculated the ordered flow rate range, we want to calculate the dosage rate for the stabilizing dosage. Since the stabilizing dosage flow rate given in the problem was 24 mL/h, what we actually need to do is to solve the following problem:
The order is for 75 mg of a medication in 100 mL of NS and the patient stabilized at 24 mL/h.
Calculate the stabilizing dosage rate in mg/h to the nearest tenth.
We want our answer to be in

mg

h

.

_____

mg

h

=
Now since we are looking for a rate, which is dosage over time, we need to find something to put on the right side of the equation that will be a rate which expresses dosage over time. Well, we don't have a rate in dosage over time, but we have a rate in volume over time: 24 gtt/min. We know that all problems in this section use microdrip tubing, so we know that our flow rate in gtt/min is equal to our flow rate in ml/h (since the 60 gtt/mL will cancel out the 60 min/h in our equation). So we can rewrite 24 gtt/min as:24 mL/h, and we know that we can get from mL to mg by using the concentration of the drug: 75 mg of a medication in 100 mL of NS.
So this gives us:
_____

mg

h

=

24 mL

1 h

Now we want to cancel out the mL units which appear in the top of the fraction at right and convert it to some kind of dosage units, because we are trying to get to mg units. We can use the concentration to do this: 75 mg of a medication in 100 mL of NS. So now we multiply by

75 mg

100 mL

:
_____

mg

h

=

24 mL

1 h

×

75 mg

100 mL

Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____

mg

h

=

24 mL

1 h

×

75 mg

100 mL

So this simplifies to:
_____

mg

h

=

24

1

×

75

100

mg

h

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 75 and 100 by 25. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mg

h

=

24

1

×

3

4

mg

h

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:
_____

mg

h

=

72

4

mg

h

= 18

mg

h

= 18

mg

h

So our stabilizing dosage rate is 18

mg

h

.
Example:
The order is for 50 mg of nitroglycerin in 250 mL of D5W to infuse at 10-80 mcg/min.
Calculate the flow rate range in mL/h to the nearest whole mL. The patient stabilizes at 9 mL/h. Calculate the dosage infusion rate at which the patient stabilized in mcg/min.
We want our answer to be in

mL

h

, because all the problems in this chapter involve an electronic infusion device to control the flow rate of the IV, so the speed must be in mL/h, because this is the unit of measure accepted by electronic infusion devices.

_____

mL

h

=
We have a range for the dosage infusion rate: 10-80 mcg/min So this means that we need our answer to be a range as well; in other words, we need a minimum and a maximum flow rate for our answer.
We will find the minimum first, and then we will use the same method to find the maximum.
Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. Well, we don't have a rate in volume over time, but we have a rate in dosage over time: 10 mcg/min for the minimum, and 80 mcg/min for the maximum, and we know that we can get from mcg to mL by using the concentration of the drug: 50 mg of nitroglycerin in 250 mL of D5W.
So this gives us:
Min: _____

mL

h

=

10 mcg

1 min

Now we want to get from mcg to mL. So, to cancel out mcg in the bottom of the fraction, we need to multiply by a fraction that has mcg in the top. Because we know that 1 mg = 1000 mcg, we have such a fraction:

1 mg

1000 mcg

. So, multiplying by this yields:
Min: _____

mL

h

=

10 mcg

1 min

×

1 mg

1000 mcg

Now we need to cancel out the mg, which appears in the top of the fraction, so we need to multiply by a fraction that has mg in the bottom. The concentration gives us such a fraction:

250 mL

50 mg

. Multiplying by this yields:
Min: _____

mL

h

=

10 mcg

1 min

×

1 mg

1000 mcg

×

250 mL

50 mg

Now we notice that we want our answer to be in mL/h, but the only units of time we can see in the bottom of our fractions on the right side of the equation is in min. So we need to multiply by a fraction which is equal to one and has min in the top so that we can cancel out the min in the bottom of the fraction on the right side of the equation. (Remember that it doesn't matter what order we multiply things in; because to multiply we will just multiply straight across, so for two units to cancel we don't need them to be right next to each other - we only need one of them to be somewhere in the top of a fraction and the other to be somewhere in the bottom.) We can convert this min to the unit h, which we want, because we know that 1 h = 60 min. So we can use the fraction:

60 min

1 h

, which is equal to one, and which has min in the top, to cancel out the min in the bottom.

So this yields:
Min: _____

mL

h

=

10 mcg

1 min

×

1 mg

1000 mcg

×

250 mL

50 mg

×

60 min

1 h

Now we need to cancel out all units that appear in both the top and the bottom of the equation:
Min: _____

mL

h

=

10 ~~mcg~~

1 ~~min~~

×

1 mg

1000 ~~mcg~~

×

250 mL

50 mg

×

60 ~~min~~

1 h

So this simplifies to:
Min: _____

mL

h

=

10

1

×

1

1000

×

250

50

×

60

1

mL

h

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 50 and 60 by 10. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Min: _____

mL

h

=

10

1

×

1

1000

×

250

50₅

×

60⁶

1

mL

h

Min: _____

mL

h

=

10

1

×

1

1000

×

250

5

×

6

1

mL

h

We can divide both 1000 and 6 by 2. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Min: _____

mL

h

=

10

1

×

1

~~1000~~₅₀₀

×

250

5

×

1

1

×

6³

1

mL

h

Min: _____

mL

h

=

10

1

×

1

500

×

250

5

×

1

1

×

3

1

mL

h

We can divide both 500 and by 10. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Min: _____

mL

h

=

10¹

1

×

1

~~500~~₅₀

×

250

5

×

1

1

×

3

1

mL

h

Min: _____

mL

h

=

1

1

×

1

50

×

250

5

×

1

1

×

3

1

mL

h

We can divide both 50 and 250 by 50. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Min: _____

mL

h

=

1

1

×

1

50₁

×

~~250~~⁵

5

×

1

1

×

3

1

mL

h

Min: _____

mL

h

=

1

1

×

1

1

×

5

5

×

1

1

×

3

1

mL

h

We can divide both 5 and 5 by 5. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Min: _____

mL

h

=

1

1

×

1

1

×

5¹

5₁

×

3

1

mL

h

Min: _____

mL

h

=

1

1

×

1

1

×

1

1

×

3

1

mL

h

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:
Min: _____

mL

h

=

3

1

mL

h

= 3

mL

h

= 3

mL

h

So our minimum is 3

mL

h

.
We want our answer to be in

mL

h

, because all the problems in this chapter involve an electronic infusion device to control the flow rate of the IV, so the speed must be in mL/h, because this is the unit of measure accepted by electronic infusion devices.

_____

mL

h

=
We have a range for the dosage infusion rate: 10-80 mcg/min So this means that we need our answer to be a range as well; in other words, we need a minimum and a maximum flow rate for our answer.
Now that we have found the minimum, we will find the maximum.
Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. Well, we don't have a rate in volume over time, but we have a rate in dosage over time: 10 mcg/min for the minimum, and 80 mcg/min for the maximum, and we know that we can get from mcg to mL by using the concentration of the drug: 50 mg of nitroglycerin in 250 mL of D5W.
So this gives us:
Max: _____

mL

h

=

80 mcg

1 min

Now we want to get from mcg to mL. So, to cancel out mcg in the bottom of the fraction, we need to multiply by a fraction that has mcg in the top. Because we know that 1 mg = 1000 mcg, we have such a fraction:

1 mg

1000 mcg

. So, multiplying by this yields:
Max: _____

mL

h

=

80 mcg

1 min

×

1 mg

1000 mcg

Now we need to cancel out the mg, which appears in the top of the fraction, so we need to multiply by a fraction that has mg in the bottom. The concentration gives us such a fraction:

250 mL

50 mg

. Multiplying by this yields:
Max: _____

mL

h

=

80 mcg

1 min

×

1 mg

1000 mcg

×

250 mL

50 mg

Now we notice that we want our answer to be in mL/h, but the only units of time we can see in the bottom of our fractions on the right side of the equation is in min. So we need to multiply by a fraction which is equal to one and has min in the top so that we can cancel out the min in the bottom of the fraction on the right side of the equation. (Remember that it doesn't matter what order we multiply things in; because to multiply we will just multiply straight across, so for two units to cancel we don't need them to be right next to each other - we only need one of them to be somewhere in the top of a fraction and the other to be somewhere in the bottom.) We can convert this min to the unit h, which we want, because we know that 1 h = 60 min. So we can use the fraction:

60 min

1 h

, which is equal to one, and which has min in the top, to cancel out the min in the bottom.

So this yields:
Max: _____

mL

h

=

80 mcg

1 min

×

1 mg

1000 mcg

×

250 mL

50 mg

×

60 min

1 h

Now we need to cancel out all units that appear in both the top and the bottom of the equation:
Max: _____

mL

h

=

80 ~~mcg~~

1 ~~min~~

×

1 mg

1000 ~~mcg~~

×

250 mL

50 mg

×

60 ~~min~~

1 h

So this simplifies to:
Max: _____

mL

h

=

80

1

×

1

1000

×

250

50

×

60

1

mL

h

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 50 and 60 by 10. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

80

1

×

1

1000

×

250

50₅

×

60⁶

1

mL

h

Max: _____

mL

h

=

80

1

×

1

1000

×

250

5

×

6

1

mL

h

We can divide both 1000 and 6 by 2. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

80

1

×

1

~~1000~~₅₀₀

×

250

5

×

1

1

×

6³

1

mL

h

Max: _____

mL

h

=

80

1

×

1

500

×

250

5

×

1

1

×

3

1

mL

h

We can divide both 500 and by 20. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

80⁴

1

×

1

~~500~~₂₅

×

250

5

×

1

1

×

3

1

mL

h

Max: _____

mL

h

=

4

1

×

1

25

×

250

5

×

1

1

×

3

1

mL

h

We can divide both 25 and 250 by 25. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

4

1

×

1

25₁

×

~~250~~¹⁰

5

×

1

1

×

3

1

mL

h

Max: _____

mL

h

=

4

1

×

1

1

×

10

5

×

1

1

×

3

1

mL

h

We can divide both 5 and 10 by 5. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

4

1

×

1

1

×

10²

5₁

×

3

1

mL

h

Max: _____

mL

h

=

4

1

×

1

1

×

2

1

×

3

1

mL

h

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:
Max: _____

mL

h

=

24

1

mL

h

= 24

mL

h

= 24

mL

h

So our maximum is 24

mL

h

.
So our answer is 3-24

mL

h

.
Now that we've calculated the ordered flow rate range, we want to calculate the dosage rate for the stabilizing dosage. Since the stabilizing dosage flow rate given in the problem was 9 mL/h, what we actually need to do is to solve the following problem:
The order is for 50 mg of nitroglycerin in 250 mL of D5W and the patient stabilized at 9 mL/h.
Calculate the stabilizing dosage rate in mcg/min to the nearest tenth.
We want our answer to be in

mcg

min

.

_____

mcg

min

=
Now since we are looking for a rate, which is dosage over time, we need to find something to put on the right side of the equation that will be a rate which expresses dosage over time. Well, we don't have a rate in dosage over time, but we have a rate in volume over time: 9 gtt/min. We know that all problems in this section use microdrip tubing, so we know that our flow rate in gtt/min is equal to our flow rate in ml/h (since the 60 gtt/mL will cancel out the 60 min/h in our equation). So we can rewrite 9 gtt/min as:9 mL/h, and we know that we can get from mL to mcg by using the concentration of the drug: 50 mg of nitroglycerin in 250 mL of D5W.
So this gives us:
_____

mcg

min

=

9 mL

1 h

Now we want to cancel out the mL units which appear in the top of the fraction at right and convert it to some kind of dosage units, because we are trying to get to mcg units. We can use the concentration to do this: 50 mg of nitroglycerin in 250 mL of D5W. So now we multiply by

50 mg

250 mL

:
_____

mcg

min

=

9 mL

1 h

×

50 mg

250 mL

Now we want to get from mg to mcg. So, to cancel out mg in the top of the fraction, we need to multiply by a fraction that has mg in the bottom. Because we know that 1000 mcg = 1 mg, we have such a fraction:

1000 mcg

1 mg

. So, multiplying by this yields:
_____

mcg

min

=

9 mL

1 h

×

50 mg

250 mL

×

1000 mcg

1 mg

Now we notice that we want our answer to be in mcg/min, but the only units of time we can see in the bottom of our fractions on the right side of the equation is in h. So we need to multiply by a fraction which is equal to one and has h in the top so that we can cancel out the h in the bottom of the fraction on the right side of the equation. (Remember that it doesn't matter what order we multiply things in; because to multiply we will just multiply straight across, so for two units to cancel we don't need them to be right next to each other - we only need one of them to be somewhere in the top of a fraction and the other to be somewhere in the bottom.) We can convert this h to the unit min, which we want, because we know that 1 h = 60 min. So we can use the fraction:

1 h

60 min

, which is equal to one, and which has h in the top, to cancel out the h in the bottom.

So this yields:
_____

mcg

min

=

9 mL

1 h

×

50 mg

250 mL

×

1000 mcg

1 mg

×

1 h

60 min

Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____

mcg

min

=

9 mL

1 h

×

50 mg

250 mL

×

1000 mcg

1 mg

×

1 h

60 min

So this simplifies to:
_____

mcg

min

=

9

1

×

50

250

×

1000

1

×

1

60

mcg

min

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 50 and 250 by 50. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mcg

min

=

9

1

×

1

5

×

1000

1

×

1

1

mcg

min

We can divide both 1 and 60 by 1. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mcg

min

=

9

1

×

1

5

×

1000

1

×

1

60

mcg

min

We can divide both 9 and 60 by 3. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mcg

min

=

3

1

×

1

5

×

1000

1

×

1

20

mcg

min

We can divide both 1000 and 20 by 20. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mcg

min

=

3

1

×

1

1

×

1

5

×

50

1

×

1

1

mcg

min

We can divide both 50 and 5 by 5. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mcg

min

=

3

1

×

1

1

×

1

1

×

10

1

×

1

1

mcg

min

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:
_____

mcg

min

=

30

1

mcg

min

= 30

mcg

min

= 30

mcg

min

So our stabilizing dosage rate is 30

mcg

min

.
Example:
The order is for 5 mg of Primacor in 500 mL of D5W to infuse at 0.375-0.75 mcg/kg/min. The patient weighs 92 kg.
Calculate the flow rate range in mL/h to the nearest whole mL. The patient stabilizes at 250 mL/h. Calculate the dosage infusion rate at which the patient stabilized in mcg/min.
We want our answer to be in

mL

h

, because all the problems in this chapter involve an electronic infusion device to control the flow rate of the IV, so the speed must be in mL/h, because this is the unit of measure accepted by electronic infusion devices.

_____

mL

h

=
We have a range for the dosage infusion rate: 0.375-0.75 mcg/kg/min So this means that we need our answer to be a range as well; in other words, we need a minimum and a maximum flow rate for our answer.
We will find the minimum first, and then we will use the same method to find the maximum.
Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. First we will consider the minimum. We don't have a rate in volume over time, but we have a rate in dosage over time: 0.375 mcg/kg/min, and we know that we can get from mcg to mL by using the concentration of the drug: 5 mg of Primacor in 500 mL of D5W.
Because this rate is in terms of mcg/kg/min, we want to go ahead and convert it to mcg/min by mulitplying by 92 kg:
Min: _____

mcg

min

=

0.375 mcg

kg

min

×

92 kg

1

Because we can put kg over 1 without changing it, this becomes:
Min: _____

mcg

min

=

0.375 mcg

kg

min

×

92 kg

1

1

And cancelling kg, which appear in both the top and the bottom of the numerator yields:
Min: _____

mcg

min

=

0.375 mcg

kg

min

×

92 kg

1

1

=

0.375 mcg

min

×

92

1

Multiplying yields:
Min: _____

mcg

min

=

0.375 mcg

min

×

92

1

=

34.5 mcg

min

= 34.5

mcg

min

So this gives us:
Min: _____

mL

h

=

34.5 mcg

1 min

Now we want to get from mcg to mL. So, to cancel out mcg in the bottom of the fraction, we need to multiply by a fraction that has mcg in the top. Because we know that 1 mg = 1000 mcg, we have such a fraction:

1 mg

1000 mcg

. So, multiplying by this yields:
Min: _____

mL

h

=

34.5 mcg

1 min

×

1 mg

1000 mcg

Now we need to cancel out the mg, which appears in the top of the fraction, so we need to multiply by a fraction that has mg in the bottom. The concentration gives us such a fraction:

500 mL

5 mg

. Multiplying by this yields:
Min: _____

mL

h

=

34.5 mcg

1 min

×

1 mg

1000 mcg

×

500 mL

5 mg

Now we notice that we want our answer to be in mL/h, but the only units of time we can see in the bottom of our fractions on the right side of the equation is in min. So we need to multiply by a fraction which is equal to one and has min in the top so that we can cancel out the min in the bottom of the fraction on the right side of the equation. (Remember that it doesn't matter what order we multiply things in; because to multiply we will just multiply straight across, so for two units to cancel we don't need them to be right next to each other - we only need one of them to be somewhere in the top of a fraction and the other to be somewhere in the bottom.) We can convert this min to the unit h, which we want, because we know that 1 h = 60 min. So we can use the fraction:

60 min

1 h

, which is equal to one, and which has min in the top, to cancel out the min in the bottom.

So this yields:
Min: _____

mL

h

=

34.5 mcg

1 min

×

1 mg

1000 mcg

×

500 mL

5 mg

×

60 min

1 h

Now we need to cancel out all units that appear in both the top and the bottom of the equation:
Min: _____

mL

h

=

34.5 ~~mcg~~

1 ~~min~~

×

1 mg

1000 ~~mcg~~

×

500 mL

5 mg

×

60 ~~min~~

1 h

So this simplifies to:
Min: _____

mL

h

=

34.5

1

×

1

1000

×

500

5

×

60

1

mL

h

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 5 and 60 by 5. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Min: _____

mL

h

=

34.5

1

×

1

1000

×

500

5₁

×

60¹²

1

mL

h

Min: _____

mL

h

=

34.5

1

×

1

1000

×

500

1

×

12

1

mL

h

We can divide both 1000 and 12 by 4. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Min: _____

mL

h

=

34.5

1

×

1

~~1000~~₂₅₀

×

500

1

×

1

1

×

12³

1

mL

h

Min: _____

mL

h

=

34.5

1

×

1

250

×

500

1

×

1

1

×

3

1

mL

h

We can divide both 250 and 500 by 250. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Min: _____

mL

h

=

34.5

1

×

1

~~250~~₁

×

~~500~~²

1

×

1

1

×

3

1

mL

h

Min: _____

mL

h

=

34.5

1

×

1

1

×

2

1

×

1

1

×

3

1

mL

h

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:
Min: _____

mL

h

=

207

1

mL

h

= 207

mL

h

= 207

mL

h

So our minimum is 207

mL

h

.
We want our answer to be in

mL

h

, because all the problems in this chapter involve an electronic infusion device to control the flow rate of the IV, so the speed must be in mL/h, because this is the unit of measure accepted by electronic infusion devices.

_____

mL

h

=
Now that we have found the minimum, we will find the maximum.
Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. First we will consider the minimum. We don't have a rate in volume over time, but we have a rate in dosage over time: 0.75 mcg/kg/min, and we know that we can get from mcg to mL by using the concentration of the drug: 5 mg of Primacor in 500 mL of D5W.
Because this rate is in terms of mcg/kg/min, we want to go ahead and convert it to mcg/min by mulitplying by 92 kg:
Max: _____

mcg

min

=

0.75 mcg

kg

min

×

92 kg

1

Because we can put kg over 1 without changing it, this becomes:
Max: _____

mcg

min

=

0.75 mcg

kg

min

×

92 kg

1

1

And cancelling kg, which appear in both the top and the bottom of the numerator yields:
Max: _____

mcg

min

=

0.75 mcg

kg

min

×

92 kg

1

1

=

0.75 mcg

min

×

92

1

Multiplying yields:
Max: _____

mcg

min

=

0.75 mcg

min

×

92

1

=

69 mcg

min

= 69

mcg

min

So this gives us:
Max: _____

mL

h

=

69 mcg

1 min

Now we want to get from mcg to mL. So, to cancel out mcg in the bottom of the fraction, we need to multiply by a fraction that has mcg in the top. Because we know that 1 mg = 1000 mcg, we have such a fraction:

1 mg

1000 mcg

. So, multiplying by this yields:
Max: _____

mL

h

=

69 mcg

1 min

×

1 mg

1000 mcg

Now we need to cancel out the mg, which appears in the top of the fraction, so we need to multiply by a fraction that has mg in the bottom. The concentration gives us such a fraction:

500 mL

5 mg

. Multiplying by this yields:
Max: _____

mL

h

=

69 mcg

1 min

×

1 mg

1000 mcg

×

500 mL

5 mg

Now we notice that we want our answer to be in mL/h, but the only units of time we can see in the bottom of our fractions on the right side of the equation is in min. So we need to multiply by a fraction which is equal to one and has min in the top so that we can cancel out the min in the bottom of the fraction on the right side of the equation. (Remember that it doesn't matter what order we multiply things in; because to multiply we will just multiply straight across, so for two units to cancel we don't need them to be right next to each other - we only need one of them to be somewhere in the top of a fraction and the other to be somewhere in the bottom.) We can convert this min to the unit h, which we want, because we know that 1 h = 60 min. So we can use the fraction:

60 min

1 h

, which is equal to one, and which has min in the top, to cancel out the min in the bottom.

So this yields:
Max: _____

mL

h

=

69 mcg

1 min

×

1 mg

1000 mcg

×

500 mL

5 mg

×

60 min

1 h

Now we need to cancel out all units that appear in both the top and the bottom of the equation:
Max: _____

mL

h

=

69 ~~mcg~~

1 ~~min~~

×

1 mg

1000 ~~mcg~~

×

500 mL

5 mg

×

60 ~~min~~

1 h

So this simplifies to:
Max: _____

mL

h

=

69

1

×

1

1000

×

500

5

×

60

1

mL

h

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 5 and 60 by 5. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

69

1

×

1

1000

×

500

5₁

×

60¹²

1

mL

h

Max: _____

mL

h

=

69

1

×

1

1000

×

500

1

×

12

1

mL

h

We can divide both 1000 and 12 by 4. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

69

1

×

1

~~1000~~₂₅₀

×

500

1

×

1

1

×

12³

1

mL

h

Max: _____

mL

h

=

69

1

×

1

250

×

500

1

×

1

1

×

3

1

mL

h

We can divide both 250 and 500 by 250. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

69

1

×

1

~~250~~₁

×

~~500~~²

1

×

1

1

×

3

1

mL

h

Max: _____

mL

h

=

69

1

×

1

1

×

2

1

×

1

1

×

3

1

mL

h

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:
Max: _____

mL

h

=

414

1

mL

h

= 414

mL

h

= 414

mL

h

So our maximum is 414

mL

h

.
So our answer is 207-414

mL

h

.
Now that we've calculated the ordered flow rate range, we want to calculate the dosage rate for the stabilizing dosage. Since the stabilizing dosage flow rate given in the problem was 250 mL/h, what we actually need to do is to solve the following problem:
The order is for 5 mg of Primacor in 500 mL of D5W and the patient stabilized at 250 mL/h.
Calculate the stabilizing dosage rate in mcg/min to the nearest tenth.
We want our answer to be in

mcg

min

.

_____

mcg

min

=
Now since we are looking for a rate, which is dosage over time, we need to find something to put on the right side of the equation that will be a rate which expresses dosage over time. Well, we don't have a rate in dosage over time, but we have a rate in volume over time: 250 gtt/min. We know that all problems in this section use microdrip tubing, so we know that our flow rate in gtt/min is equal to our flow rate in ml/h (since the 60 gtt/mL will cancel out the 60 min/h in our equation). So we can rewrite 250 gtt/min as:250 mL/h, and we know that we can get from mL to mcg by using the concentration of the drug: 5 mg of Primacor in 500 mL of D5W.
So this gives us:
_____

mcg

min

=

250 mL

1 h

Now we want to cancel out the mL units which appear in the top of the fraction at right and convert it to some kind of dosage units, because we are trying to get to mcg units. We can use the concentration to do this: 5 mg of Primacor in 500 mL of D5W. So now we multiply by

5 mg

500 mL

:
_____

mcg

min

=

250 mL

1 h

×

5 mg

500 mL

Now we want to get from mg to mcg. So, to cancel out mg in the top of the fraction, we need to multiply by a fraction that has mg in the bottom. Because we know that 1000 mcg = 1 mg, we have such a fraction:

1000 mcg

1 mg

. So, multiplying by this yields:
_____

mcg

min

=

250 mL

1 h

×

5 mg

500 mL

×

1000 mcg

1 mg

Now we notice that we want our answer to be in mcg/min, but the only units of time we can see in the bottom of our fractions on the right side of the equation is in h. So we need to multiply by a fraction which is equal to one and has h in the top so that we can cancel out the h in the bottom of the fraction on the right side of the equation. (Remember that it doesn't matter what order we multiply things in; because to multiply we will just multiply straight across, so for two units to cancel we don't need them to be right next to each other - we only need one of them to be somewhere in the top of a fraction and the other to be somewhere in the bottom.) We can convert this h to the unit min, which we want, because we know that 1 h = 60 min. So we can use the fraction:

1 h

60 min

, which is equal to one, and which has h in the top, to cancel out the h in the bottom.

So this yields:
_____

mcg

min

=

250 mL

1 h

×

5 mg

500 mL

×

1000 mcg

1 mg

×

1 h

60 min

Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____

mcg

min

=

250 mL

1 h

×

5 mg

500 mL

×

1000 mcg

1 mg

×

1 h

60 min

So this simplifies to:
_____

mcg

min

=

250

1

×

5

500

×

1000

1

×

1

60

mcg

min

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 5 and 500 by 5. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mcg

min

=

250

1

×

1

100

×

1000

1

×

1

1

mcg

min

We can divide both 1 and 60 by 1. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mcg

min

=

250

1

×

1

100

×

1000

1

×

1

60

mcg

min

We can divide both 250 and 60 by 10. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mcg

min

=

25

1

×

1

100

×

1000

1

×

1

6

mcg

min

We can divide both 1000 and 6 by 2. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mcg

min

=

25

1

×

1

1

×

1

100

×

500

1

×

1

3

mcg

min

We can divide both 500 and 100 by 100. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mcg

min

=

25

1

×

1

1

×

1

1

×

5

1

×

1

3

mcg

min

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:
_____

mcg

min

=

125

3

mcg

min

= 41.666666666667

mcg

min

= 41.7

mcg

min

So our stabilizing dosage rate is 41.7

mcg

min

.
Example:
The order is for 50 mg of Nipride in 250 mL of D5W to infuse at 3-6 mcg/kg/min. The patient weighs 70 kg.
Calculate the flow rate range in mL/h to the nearest whole mL. The patient stabilizes at 108 mL/h. Calculate the dosage infusion rate at which the patient stabilized in mg/h.
We want our answer to be in

mL

h

, because all the problems in this chapter involve an electronic infusion device to control the flow rate of the IV, so the speed must be in mL/h, because this is the unit of measure accepted by electronic infusion devices.

_____

mL

h

=
We have a range for the dosage infusion rate: 3-6 mcg/kg/min So this means that we need our answer to be a range as well; in other words, we need a minimum and a maximum flow rate for our answer.
We will find the minimum first, and then we will use the same method to find the maximum.
Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. First we will consider the minimum. We don't have a rate in volume over time, but we have a rate in dosage over time: 3 mcg/kg/min, and we know that we can get from mcg to mL by using the concentration of the drug: 50 mg of Nipride in 250 mL of D5W.
Because this rate is in terms of mcg/kg/min, we want to go ahead and convert it to mcg/min by mulitplying by 70 kg:
Min: _____

mcg

min

=

3 mcg

kg

min

×

70 kg

1

Because we can put kg over 1 without changing it, this becomes:
Min: _____

mcg

min

=

3 mcg

kg

min

×

70 kg

1

1

And cancelling kg, which appear in both the top and the bottom of the numerator yields:
Min: _____

mcg

min

=

3 mcg

kg

min

×

70 kg

1

1

=

3 mcg

min

×

70

1

Multiplying yields:
Min: _____

mcg

min

=

3 mcg

min

×

70

1

=

210 mcg

min

= 210

mcg

min

So this gives us:
Min: _____

mL

h

=

210 mcg

1 min

Now we want to get from mcg to mL. So, to cancel out mcg in the bottom of the fraction, we need to multiply by a fraction that has mcg in the top. Because we know that 1 mg = 1000 mcg, we have such a fraction:

1 mg

1000 mcg

. So, multiplying by this yields:
Min: _____

mL

h

=

210 mcg

1 min

×

1 mg

1000 mcg

Now we need to cancel out the mg, which appears in the top of the fraction, so we need to multiply by a fraction that has mg in the bottom. The concentration gives us such a fraction:

250 mL

50 mg

. Multiplying by this yields:
Min: _____

mL

h

=

210 mcg

1 min

×

1 mg

1000 mcg

×

250 mL

50 mg

Now we notice that we want our answer to be in mL/h, but the only units of time we can see in the bottom of our fractions on the right side of the equation is in min. So we need to multiply by a fraction which is equal to one and has min in the top so that we can cancel out the min in the bottom of the fraction on the right side of the equation. (Remember that it doesn't matter what order we multiply things in; because to multiply we will just multiply straight across, so for two units to cancel we don't need them to be right next to each other - we only need one of them to be somewhere in the top of a fraction and the other to be somewhere in the bottom.) We can convert this min to the unit h, which we want, because we know that 1 h = 60 min. So we can use the fraction:

60 min

1 h

, which is equal to one, and which has min in the top, to cancel out the min in the bottom.

So this yields:
Min: _____

mL

h

=

210 mcg

1 min

×

1 mg

1000 mcg

×

250 mL

50 mg

×

60 min

1 h

Now we need to cancel out all units that appear in both the top and the bottom of the equation:
Min: _____

mL

h

=

210 ~~mcg~~

1 ~~min~~

×

1 mg

1000 ~~mcg~~

×

250 mL

50 mg

×

60 ~~min~~

1 h

So this simplifies to:
Min: _____

mL

h

=

210

1

×

1

1000

×

250

50

×

60

1

mL

h

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 50 and 60 by 10. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Min: _____

mL

h

=

210

1

×

1

1000

×

250

50₅

×

60⁶

1

mL

h

Min: _____

mL

h

=

210

1

×

1

1000

×

250

5

×

6

1

mL

h

We can divide both 1000 and 6 by 2. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Min: _____

mL

h

=

210

1

×

1

~~1000~~₅₀₀

×

250

5

×

1

1

×

6³

1

mL

h

Min: _____

mL

h

=

210

1

×

1

500

×

250

5

×

1

1

×

3

1

mL

h

We can divide both 500 and by 10. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Min: _____

mL

h

=

~~210~~²¹

1

×

1

~~500~~₅₀

×

250

5

×

1

1

×

3

1

mL

h

Min: _____

mL

h

=

21

1

×

1

50

×

250

5

×

1

1

×

3

1

mL

h

We can divide both 50 and 250 by 50. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Min: _____

mL

h

=

21

1

×

1

50₁

×

~~250~~⁵

5

×

1

1

×

3

1

mL

h

Min: _____

mL

h

=

21

1

×

1

1

×

5

5

×

1

1

×

3

1

mL

h

We can divide both 5 and 5 by 5. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Min: _____

mL

h

=

21

1

×

1

1

×

5¹

5₁

×

3

1

mL

h

Min: _____

mL

h

=

21

1

×

1

1

×

1

1

×

3

1

mL

h

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:
Min: _____

mL

h

=

63

1

mL

h

= 63

mL

h

= 63

mL

h

So our minimum is 63

mL

h

.
We want our answer to be in

mL

h

, because all the problems in this chapter involve an electronic infusion device to control the flow rate of the IV, so the speed must be in mL/h, because this is the unit of measure accepted by electronic infusion devices.

_____

mL

h

=
Now that we have found the minimum, we will find the maximum.
Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. First we will consider the minimum. We don't have a rate in volume over time, but we have a rate in dosage over time: 6 mcg/kg/min, and we know that we can get from mcg to mL by using the concentration of the drug: 50 mg of Nipride in 250 mL of D5W.
Because this rate is in terms of mcg/kg/min, we want to go ahead and convert it to mcg/min by mulitplying by 70 kg:
Max: _____

mcg

min

=

6 mcg

kg

min

×

70 kg

1

Because we can put kg over 1 without changing it, this becomes:
Max: _____

mcg

min

=

6 mcg

kg

min

×

70 kg

1

1

And cancelling kg, which appear in both the top and the bottom of the numerator yields:
Max: _____

mcg

min

=

6 mcg

kg

min

×

70 kg

1

1

=

6 mcg

min

×

70

1

Multiplying yields:
Max: _____

mcg

min

=

6 mcg

min

×

70

1

=

420 mcg

min

= 420

mcg

min

So this gives us:
Max: _____

mL

h

=

420 mcg

1 min

Now we want to get from mcg to mL. So, to cancel out mcg in the bottom of the fraction, we need to multiply by a fraction that has mcg in the top. Because we know that 1 mg = 1000 mcg, we have such a fraction:

1 mg

1000 mcg

. So, multiplying by this yields:
Max: _____

mL

h

=

420 mcg

1 min

×

1 mg

1000 mcg

Now we need to cancel out the mg, which appears in the top of the fraction, so we need to multiply by a fraction that has mg in the bottom. The concentration gives us such a fraction:

250 mL

50 mg

. Multiplying by this yields:
Max: _____

mL

h

=

420 mcg

1 min

×

1 mg

1000 mcg

×

250 mL

50 mg

Now we notice that we want our answer to be in mL/h, but the only units of time we can see in the bottom of our fractions on the right side of the equation is in min. So we need to multiply by a fraction which is equal to one and has min in the top so that we can cancel out the min in the bottom of the fraction on the right side of the equation. (Remember that it doesn't matter what order we multiply things in; because to multiply we will just multiply straight across, so for two units to cancel we don't need them to be right next to each other - we only need one of them to be somewhere in the top of a fraction and the other to be somewhere in the bottom.) We can convert this min to the unit h, which we want, because we know that 1 h = 60 min. So we can use the fraction:

60 min

1 h

, which is equal to one, and which has min in the top, to cancel out the min in the bottom.

So this yields:
Max: _____

mL

h

=

420 mcg

1 min

×

1 mg

1000 mcg

×

250 mL

50 mg

×

60 min

1 h

Now we need to cancel out all units that appear in both the top and the bottom of the equation:
Max: _____

mL

h

=

420 ~~mcg~~

1 ~~min~~

×

1 mg

1000 ~~mcg~~

×

250 mL

50 mg

×

60 ~~min~~

1 h

So this simplifies to:
Max: _____

mL

h

=

420

1

×

1

1000

×

250

50

×

60

1

mL

h

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 50 and 60 by 10. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

420

1

×

1

1000

×

250

50₅

×

60⁶

1

mL

h

Max: _____

mL

h

=

420

1

×

1

1000

×

250

5

×

6

1

mL

h

We can divide both 1000 and 6 by 2. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

420

1

×

1

~~1000~~₅₀₀

×

250

5

×

1

1

×

6³

1

mL

h

Max: _____

mL

h

=

420

1

×

1

500

×

250

5

×

1

1

×

3

1

mL

h

We can divide both 500 and by 20. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

~~420~~²¹

1

×

1

~~500~~₂₅

×

250

5

×

1

1

×

3

1

mL

h

Max: _____

mL

h

=

21

1

×

1

25

×

250

5

×

1

1

×

3

1

mL

h

We can divide both 25 and 250 by 25. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

21

1

×

1

25₁

×

~~250~~¹⁰

5

×

1

1

×

3

1

mL

h

Max: _____

mL

h

=

21

1

×

1

1

×

10

5

×

1

1

×

3

1

mL

h

We can divide both 5 and 10 by 5. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

21

1

×

1

1

×

10²

5₁

×

3

1

mL

h

Max: _____

mL

h

=

21

1

×

1

1

×

2

1

×

3

1

mL

h

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:
Max: _____

mL

h

=

126

1

mL

h

= 126

mL

h

= 126

mL

h

So our maximum is 126

mL

h

.
So our answer is 63-126

mL

h

.
Now that we've calculated the ordered flow rate range, we want to calculate the dosage rate for the stabilizing dosage. Since the stabilizing dosage flow rate given in the problem was 108 mL/h, what we actually need to do is to solve the following problem:
The order is for 50 mg of Nipride in 250 mL of D5W and the patient stabilized at 108 mL/h.
Calculate the stabilizing dosage rate in mg/h to the nearest tenth.
We want our answer to be in

mg

h

.

_____

mg

h

=
Now since we are looking for a rate, which is dosage over time, we need to find something to put on the right side of the equation that will be a rate which expresses dosage over time. Well, we don't have a rate in dosage over time, but we have a rate in volume over time: 108 gtt/min. We know that all problems in this section use microdrip tubing, so we know that our flow rate in gtt/min is equal to our flow rate in ml/h (since the 60 gtt/mL will cancel out the 60 min/h in our equation). So we can rewrite 108 gtt/min as:108 mL/h, and we know that we can get from mL to mg by using the concentration of the drug: 50 mg of Nipride in 250 mL of D5W.
So this gives us:
_____

mg

h

=

108 mL

1 h

Now we want to cancel out the mL units which appear in the top of the fraction at right and convert it to some kind of dosage units, because we are trying to get to mg units. We can use the concentration to do this: 50 mg of Nipride in 250 mL of D5W. So now we multiply by

50 mg

250 mL

:
_____

mg

h

=

108 mL

1 h

×

50 mg

250 mL

Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____

mg

h

=

108 mL

1 h

×

50 mg

250 mL

So this simplifies to:
_____

mg

h

=

108

1

×

50

250

mg

h

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 50 and 250 by 50. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mg

h

=

108

1

×

1

5

mg

h

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:
_____

mg

h

=

108

5

mg

h

= 21.6

mg

h

= 21.6

mg

h

So our stabilizing dosage rate is 21.6

mg

h

.
Example:
The order is for 1 g of dobutamine in 250 mL of D5W to infuse at 1-5 mcg/kg/min. The patient weighs 109 kg.
Calculate the flow rate range in mL/h to the nearest whole mL. The patient stabilizes at 7.5 mL/h. Calculate the dosage infusion rate at which the patient stabilized in mg/min.
We want our answer to be in

mL

h

, because all the problems in this chapter involve an electronic infusion device to control the flow rate of the IV, so the speed must be in mL/h, because this is the unit of measure accepted by electronic infusion devices.

_____

mL

h

=
We have a range for the dosage infusion rate: 1-5 mcg/kg/min So this means that we need our answer to be a range as well; in other words, we need a minimum and a maximum flow rate for our answer.
We will find the minimum first, and then we will use the same method to find the maximum.
Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. First we will consider the minimum. We don't have a rate in volume over time, but we have a rate in dosage over time: 1 mcg/kg/min, and we know that we can get from mcg to mL by using the concentration of the drug: 1 g of dobutamine in 250 mL of D5W.
Because this rate is in terms of mcg/kg/min, we want to go ahead and convert it to mcg/min by mulitplying by 109 kg:
Min: _____

mcg

min

=

1 mcg

kg

min

×

109 kg

1

Because we can put kg over 1 without changing it, this becomes:
Min: _____

mcg

min

=

1 mcg

kg

min

×

109 kg

1

1

And cancelling kg, which appear in both the top and the bottom of the numerator yields:
Min: _____

mcg

min

=

1 mcg

kg

min

×

109 kg

1

1

=

1 mcg

min

×

109

1

Multiplying yields:
Min: _____

mcg

min

=

1 mcg

min

×

109

1

=

109 mcg

min

= 109

mcg

min

So this gives us:
Min: _____

mL

h

=

109 mcg

1 min

Now we want to get from mcg to mL. So, to cancel out mcg in the bottom of the fraction, we need to multiply by a fraction that has mcg in the top. Because we know that 1 mg = 1000 mcg, we have such a fraction:

1 mg

1000 mcg

. So, multiplying by this yields:
Min: _____

mL

h

=

109 mcg

1 min

×

1 mg

1000 mcg

But we notice that our concentration is 1 g of dobutamine in 250 mL of D5W, which contains g, but our set of fractions still contains mg instead of g. We can fix this; all we need to do is multiply by a fraction that has mg on the bottom to cancel out the mg which appears on the top. Because we know that 1 g = 1000 mg, we have such a fraction:

1 g

1000 mg

. So, multiplying by this yields:
Min: _____

mL

h

=

109 mcg

1 min

×

1 mg

1000 mcg

×

1 g

1000 mg

Now we need to cancel out the g, which appears in the top of the fraction, so we need to multiply by a fraction that has g in the bottom. The concentration gives us such a fraction:

250 mL

1 g

. Multiplying by this yields:
Min: _____

mL

h

=

109 mcg

1 min

×

1 mg

1000 mcg

×

1 g

1000 mg

×

250 mL

1 g

Now we notice that we want our answer to be in mL/h, but the only units of time we can see in the bottom of our fractions on the right side of the equation is in min. So we need to multiply by a fraction which is equal to one and has min in the top so that we can cancel out the min in the bottom of the fraction on the right side of the equation. (Remember that it doesn't matter what order we multiply things in; because to multiply we will just multiply straight across, so for two units to cancel we don't need them to be right next to each other - we only need one of them to be somewhere in the top of a fraction and the other to be somewhere in the bottom.) We can convert this min to the unit h, which we want, because we know that 1 h = 60 min. So we can use the fraction:

60 min

1 h

, which is equal to one, and which has min in the top, to cancel out the min in the bottom.

So this yields:
Min: _____

mL

h

=

109 mcg

1 min

×

1 mg

1000 mcg

×

1 g

1000 mg

×

250 mL

1 g

×

60 min

1 h

Now we need to cancel out all units that appear in both the top and the bottom of the equation:
Min: _____

mL

h

=

109 ~~mcg~~

1 ~~min~~

×

1 mg

1000 ~~mcg~~

×

1 g

1000 mg

×

250 mL

1 g

×

60 ~~min~~

1 h

So this simplifies to:
Min: _____

mL

h

=

109

1

×

1

1000

×

1

1000

×

250

1

×

60

1

mL

h

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 1 and 60 by 1. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Min: _____

mL

h

=

109

1

×

1

1000

×

1

1000

×

250

1₁

×

60⁶⁰

1

mL

h

Min: _____

mL

h

=

109

1

×

1

1000

×

1

1000

×

250

1

×

60

1

mL

h

We can divide both 1000 and 60 by 20. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Min: _____

mL

h

=

109

1

×

1

~~1000~~₅₀

×

1

1000

×

250

1

×

1

1

×

60³

1

mL

h

Min: _____

mL

h

=

109

1

×

1

50

×

1

1000

×

250

1

×

1

1

×

3

1

mL

h

We can divide both 50 and 250 by 50. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Min: _____

mL

h

=

109

1

×

1

50₁

×

1

1000

×

~~250~~⁵

1

×

1

1

×

3

1

mL

h

Min: _____

mL

h

=

109

1

×

1

1

×

1

1000

×

5

1

×

1

1

×

3

1

mL

h

We can divide both 1000 and 5 by 5. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Min: _____

mL

h

=

109

1

×

1

1

×

1

~~1000~~₂₀₀

×

5¹

1

×

1

1

×

3

1

mL

h

Min: _____

mL

h

=

109

1

×

1

1

×

1

200

×

1

1

×

1

1

×

3

1

mL

h

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:
Min: _____

mL

h

=

327

200

mL

h

= 1.635

mL

h

= 2

mL

h

So our minimum is 2

mL

h

.
We want our answer to be in

mL

h

, because all the problems in this chapter involve an electronic infusion device to control the flow rate of the IV, so the speed must be in mL/h, because this is the unit of measure accepted by electronic infusion devices.

_____

mL

h

=
Now that we have found the minimum, we will find the maximum.
Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. First we will consider the minimum. We don't have a rate in volume over time, but we have a rate in dosage over time: 5 mcg/kg/min, and we know that we can get from mcg to mL by using the concentration of the drug: 1 g of dobutamine in 250 mL of D5W.
Because this rate is in terms of mcg/kg/min, we want to go ahead and convert it to mcg/min by mulitplying by 109 kg:
Max: _____

mcg

min

=

5 mcg

kg

min

×

109 kg

1

Because we can put kg over 1 without changing it, this becomes:
Max: _____

mcg

min

=

5 mcg

kg

min

×

109 kg

1

1

And cancelling kg, which appear in both the top and the bottom of the numerator yields:
Max: _____

mcg

min

=

5 mcg

kg

min

×

109 kg

1

1

=

5 mcg

min

×

109

1

Multiplying yields:
Max: _____

mcg

min

=

5 mcg

min

×

109

1

=

545 mcg

min

= 545

mcg

min

So this gives us:
Max: _____

mL

h

=

545 mcg

1 min

Now we want to get from mcg to mL. So, to cancel out mcg in the bottom of the fraction, we need to multiply by a fraction that has mcg in the top. Because we know that 1 mg = 1000 mcg, we have such a fraction:

1 mg

1000 mcg

. So, multiplying by this yields:
Max: _____

mL

h

=

545 mcg

1 min

×

1 mg

1000 mcg

But we notice that our concentration is 1 g of dobutamine in 250 mL of D5W, which contains g, but our set of fractions still contains mg instead of g. We can fix this; all we need to do is multiply by a fraction that has mg on the bottom to cancel out the mg which appears on the top. Because we know that 1 g = 1000 mg, we have such a fraction:

1 g

1000 mg

. So, multiplying by this yields:
Max: _____

mL

h

=

545 mcg

1 min

×

1 mg

1000 mcg

×

1 g

1000 mg

Now we need to cancel out the g, which appears in the top of the fraction, so we need to multiply by a fraction that has g in the bottom. The concentration gives us such a fraction:

250 mL

1 g

. Multiplying by this yields:
Max: _____

mL

h

=

545 mcg

1 min

×

1 mg

1000 mcg

×

1 g

1000 mg

×

250 mL

1 g

Now we notice that we want our answer to be in mL/h, but the only units of time we can see in the bottom of our fractions on the right side of the equation is in min. So we need to multiply by a fraction which is equal to one and has min in the top so that we can cancel out the min in the bottom of the fraction on the right side of the equation. (Remember that it doesn't matter what order we multiply things in; because to multiply we will just multiply straight across, so for two units to cancel we don't need them to be right next to each other - we only need one of them to be somewhere in the top of a fraction and the other to be somewhere in the bottom.) We can convert this min to the unit h, which we want, because we know that 1 h = 60 min. So we can use the fraction:

60 min

1 h

, which is equal to one, and which has min in the top, to cancel out the min in the bottom.

So this yields:
Max: _____

mL

h

=

545 mcg

1 min

×

1 mg

1000 mcg

×

1 g

1000 mg

×

250 mL

1 g

×

60 min

1 h

Now we need to cancel out all units that appear in both the top and the bottom of the equation:
Max: _____

mL

h

=

545 ~~mcg~~

1 ~~min~~

×

1 mg

1000 ~~mcg~~

×

1 g

1000 mg

×

250 mL

1 g

×

60 ~~min~~

1 h

So this simplifies to:
Max: _____

mL

h

=

545

1

×

1

1000

×

1

1000

×

250

1

×

60

1

mL

h

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 1 and 60 by 1. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

545

1

×

1

1000

×

1

1000

×

250

1₁

×

60⁶⁰

1

mL

h

Max: _____

mL

h

=

545

1

×

1

1000

×

1

1000

×

250

1

×

60

1

mL

h

We can divide both 1000 and 60 by 20. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

545

1

×

1

~~1000~~₅₀

×

1

1000

×

250

1

×

1

1

×

60³

1

mL

h

Max: _____

mL

h

=

545

1

×

1

50

×

1

1000

×

250

1

×

1

1

×

3

1

mL

h

We can divide both 50 and by 5. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

~~545~~¹⁰⁹

1

×

1

50₁₀

×

1

1000

×

250

1

×

1

1

×

3

1

mL

h

Max: _____

mL

h

=

109

1

×

1

10

×

1

1000

×

250

1

×

1

1

×

3

1

mL

h

We can divide both 10 and 250 by 10. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

109

1

×

1

10₁

×

1

1000

×

~~250~~²⁵

1

×

1

1

×

3

1

mL

h

Max: _____

mL

h

=

109

1

×

1

1

×

1

1000

×

25

1

×

1

1

×

3

1

mL

h

We can divide both 1000 and 25 by 25. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
Max: _____

mL

h

=

109

1

×

1

1

×

1

~~1000~~₄₀

×

25¹

1

×

1

1

×

3

1

mL

h

Max: _____

mL

h

=

109

1

×

1

1

×

1

40

×

1

1

×

1

1

×

3

1

mL

h

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:
Max: _____

mL

h

=

327

40

mL

h

= 8.175

mL

h

= 8

mL

h

So our maximum is 8

mL

h

.
So our answer is 2-8

mL

h

.
Now that we've calculated the ordered flow rate range, we want to calculate the dosage rate for the stabilizing dosage. Since the stabilizing dosage flow rate given in the problem was 7.5 mL/h, what we actually need to do is to solve the following problem:
The order is for 1 g of dobutamine in 250 mL of D5W and the patient stabilized at 7.5 mL/h.
Calculate the stabilizing dosage rate in mg/min to the nearest tenth.
We want our answer to be in

mg

min

.

_____

mg

min

=
Now since we are looking for a rate, which is dosage over time, we need to find something to put on the right side of the equation that will be a rate which expresses dosage over time. Well, we don't have a rate in dosage over time, but we have a rate in volume over time: 7.5 gtt/min. We know that all problems in this section use microdrip tubing, so we know that our flow rate in gtt/min is equal to our flow rate in ml/h (since the 60 gtt/mL will cancel out the 60 min/h in our equation). So we can rewrite 7.5 gtt/min as:7.5 mL/h, and we know that we can get from mL to mg by using the concentration of the drug: 1 g of dobutamine in 250 mL of D5W.
So this gives us:
_____

mg

min

=

7.5 mL

1 h

Now we want to cancel out the mL units which appear in the top of the fraction at right and convert it to some kind of dosage units, because we are trying to get to mg units. We can use the concentration to do this: 1 g of dobutamine in 250 mL of D5W. So now we multiply by

1 g

250 mL

:
_____

mg

min

=

7.5 mL

1 h

×

1 g

250 mL

Now we want to get from g to mg. So, to cancel out g in the top of the fraction, we need to multiply by a fraction that has g in the bottom. Because we know that 1000 mg = 1 g, we have such a fraction:

1000 mg

1 g

. So, multiplying by this yields:
_____

mg

min

=

7.5 mL

1 h

×

1 g

250 mL

×

1000 mg

1 g

Now we notice that we want our answer to be in mg/min, but the only units of time we can see in the bottom of our fractions on the right side of the equation is in h. So we need to multiply by a fraction which is equal to one and has h in the top so that we can cancel out the h in the bottom of the fraction on the right side of the equation. (Remember that it doesn't matter what order we multiply things in; because to multiply we will just multiply straight across, so for two units to cancel we don't need them to be right next to each other - we only need one of them to be somewhere in the top of a fraction and the other to be somewhere in the bottom.) We can convert this h to the unit min, which we want, because we know that 1 h = 60 min. So we can use the fraction:

1 h

60 min

, which is equal to one, and which has h in the top, to cancel out the h in the bottom.

So this yields:
_____

mg

min

=

7.5 mL

1 h

×

1 g

250 mL

×

1000 mg

1 g

×

1 h

60 min

Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____

mg

min

=

7.5 mL

1 h

×

1 g

250 mL

×

1000 mg

1 g

×

1 h

60 min

So this simplifies to:
_____

mg

min

=

7.5

1

×

1

250

×

1000

1

×

1

60

mg

min

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 1 and 250 by 1. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mg

min

=

7.5

1

×

1

250

×

1000

1

×

1

1

mg

min

We can divide both 1 and 60 by 1. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mg

min

=

7.5

1

×

1

250

×

1000

1

×

1

60

mg

min

We can divide both 7.5 and 60 by 1. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mg

min

=

7.5

1

×

1

250

×

1000

1

×

1

60

mg

min

We can divide both 1000 and 60 by 20. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mg

min

=

7.5

1

×

1

1

×

1

250

×

50

1

×

1

3

mg

min

We can divide both 50 and 250 by 50. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mg

min

=

7.5

1

×

1

1

×

1

5

×

1

1

×

1

3

mg

min

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:
_____

mg

min

=

7.5

15

mg

min

= 0.5

mg

min

= 0.5

mg

min

So our stabilizing dosage rate is 0.5

mg

min

.
One Particular Critical IV Medication: Heparin

Heparin

What is it?

Heparin is a drug that works as an anticoagulant, which means that it prevents blood clots from forming and prevents existing blood clots from expanding. It was orginally extracted from the liver, but now it is mostly manufactured synthetically. Heparin, however does not actually dissolve existing blood clots; it only prevents them from forming or expanding.

What is it used for?

Heparin is usually given to patients who are suffering from a stroke, a heart attack, or a blood clot; it is used to prevent the blood clots from expanding to allow the body's natural mechanism, or other blood clot dissolving drugs, to break down the clot. It may also be used to prevent blood clots from forming when a patient undergoes open-heart or bypass surgery or dialysis or when a patient has a condition that puts them at risk for heart attact, stroke or other blood clots (for example, certain kinds of irregular heartbeat rhythms).

Heparin is also often used as a flush, which is a small amout of fluid run through an IV port or lock to dissolve any blood clots the patient might have formed that could block the IV fluid from entering the vein, particularly when the IV has been connected but no fluid has been administered to the patient in several hours. However, the concentration of heparin used to give a flush is significantly lower than that used for medical treatment.

How is it given?

Heparin must be given by injection; when given orally, the body digests it before it can be made of any use.

How is it measured?

Heparin is measured in Units, but these units are not the same volume as the Units used to measure insulin or penicillin. Common concentrations of heparin are 1000 units per mL, 5000 units per mL, and 10,000 units per mL; these can be used for IM or s.q. dosing, or diluted in a diluent liquid for IV use. For heparin administration as an IV flush, common concentrations are 2 units per mL, 10 units per mL, 50 units per mL, or 100 units per mL.

Often Heparin comes in premixed IV bags containing heparin in normal saline or heparin in D5W. Such bags are marked with red writing indicating that they contain heparin; this is to prevent accidental confusion of a premixed heparin bag with bags containing other IV solutions, which is very dangerous, since accidental dosage of heparin can have such severe consequences. Below we can see the label for such a premixed bag:

If a premixed solution of heparin is not available at the concentration that you need, you can draw the appropriate dosage out of a vial of heparin and add it to a compatible IV fluid (compatible IV fluids include dextrose in water, saline, dextrose in saline, and lactated ringers/ringers lactate). Below we can see the labels of several heparin vials of varying strengths which can be used to make a heparin IV solution by withdrawing the volume necessary to obtain the desired number of units and then injecting the medication into a prepared bag of IV fluid:

The above heparin vials all have high concentrations that would be used for drug orders of heparin intended to treat specific conditions; however, below we can see two vials of heparin in a concentration that would be used for an IV flush:

What is the correct dosage?

The correct dosing of heparin is critical; the safe range for an adult is 20,000 units to 40,000 units per day. If an order is written that would cause the patient to receive less than 20,000 units or more than 40,000 units in a single day, the dosage should not be administered, and the nurse should check with the doctor who wrote the order. Any patient dosed with heparin will have regular blood tests done to check their coagulation times so that the dosage can be adjusted to the patient's response to the drug.

If too little heparin is given, new blood clots could form or existing blood clots could expand causing or worsening stroke, heart attack, damage to limbs cut off from the blood supply by the clot, or damage to the lungs by clots that become dislodged and travel to the lungs.

If too much heparin is given, the blood may not clot correctly, causing internal bleeding, which if severe enough, can result in death.

Heparin Drug Orders: Calculating Flow Rates and Dosage Infusion Rates

Heparin can be ordered either by a mL/h flow rate, or by a units/h dosage infusion rate. (Our flow rate is always in mL/h instead of gtt/min because heparin will always be infused using an electronic infusion device.)

Heparin Flow Rates

If a heparing dose is ordered by a units/h dosage infusion rate, then in order to administer the medication, we must calculate the flow rate in mL/h so that we know what rate to enter into the electronic infusion device that is being used to administer the IV. This will be very similar to the problems we have already done for other medications.

Heparin Dosage Infusion Rate

If heparin is ordered by a flow rate in mL/h, then the physician has already done the calculations needed to set the flow rate and we are ready to set up the electronic infusion device to administer the medication using these flow rates; however, it is our responsibility to use the given flow rate to find the units/h dosage infusion rate so that we can verify that the order is within the 20,000-40,000 units per day of heparin which is the safe range for an adult. We must do this before we actually administer the IV.

Let's look at some examples:

First we will look at few examples where heparin is ordered by dosage infustion rate, and we will check this for safety and then calculate the corresponding flow rate:

First we will look at some simple examples. In these problems, the IV medication has been ordered by dosage infusion rate, and we must convert this to flow rate in mL/h so that we can set up the IV.

Example:
The order is for 25000 units of heparin in 1 L of D5W to infuse at 1200 units/h.
Calculate the flow rate in mL/h to the nearest whole mL.(You should check to make sure the order is safe before you do your calculations.)
First we should check to make sure that this order is safe. The order is for heparin to infuse at a rate of 1200 units/h, and we know that 20,000-40,000 units of heparin per day is safe. Since there are 24 hours in a day, we multiply 1200 units/h by 24 to get the total number of units this patient will get per day: 28800 units.
Because 28800 units falls between 20,000 and 40,000 units, this order is safe, so we can procede to calculate the flow rate:
We want our answer to be in

mL

h

, because all the problems in this chapter involve an electronic infusion device to control the flow rate of the IV, so the speed must be in mL/h, because this is the unit of measure accepted by electronic infusion devices.

_____

mL

h

=
Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. Well, we don't have a rate in volume over time, but we have a rate in dosage over time: 1200 units/h, and we know that we can get from units to mL by using the concentration of the drug: 25000 units of heparin in 1 L of D5W.
So this gives us:
_____

mL

h

=

1200 units

1 h

Now we need to cancel out the units, which appears in the top of the fraction, so we need to multiply by a fraction that has units in the bottom. The concentration gives us such a fraction:

1 L

25000 units

. Multiplying by this yields:
_____

mL

h

=

1200 units

1 h

×

1 L

25000 units

We need to multiply by a fraction which is equal to one and has L in the bottom so that we can cancel out the L in the top of the fraction on the right of the equation.
So, we first need to multiply by a fraction that is equal to one and which cancels out L and gets us to mL. Since we know that 1L=1000mL, the fraction

1000 mL

1 L

will be equal to one, and since it has L in the bottom, will also cancel out the L in the top of the fraction in the right side of the equation we already have:
_____

mL

h

=

1200 units

1 h

×

1 L

25000 units

×

1000 mL

1 L

Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____

mL

h

=

1200 ~~units~~

1 h

×

1 L

25000 ~~units~~

×

1000 mL

1 L

So this simplifies to:
_____

mL

h

=

1200

1

×

1

25000

×

1000

1

mL

h

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 25000 and 1 by 1000. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mL

h

=

1200

1

×

1

~~25000~~₂₅

×

~~$new1000~~¹

1

×

1

1

mL

h

_____

mL

h

=

1200

1

×

1

25

×

1

1

×

1

1

mL

h

We can divide both 25 and 1200 by 25. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mL

h

=

~~1200~~⁴⁸

1

×

1

25₁

×

1

1

mL

h

_____

mL

h

=

48

1

×

1

1

×

1

1

mL

h

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:
_____

mL

h

=

48

1

mL

h

= 48

mL

h

= 48

mL

h

So our answer is 48

mL

h

.
Example:
The order is for 20000 units of heparin in 1 L of D51/2S to infuse at 1000 units/h.
Calculate the flow rate in mL/h to the nearest whole mL.(You should check to make sure the order is safe before you do your calculations.)
First we should check to make sure that this order is safe. The order is for heparin to infuse at a rate of 1000 units/h, and we know that 20,000-40,000 units of heparin per day is safe. Since there are 24 hours in a day, we multiply 1000 units/h by 24 to get the total number of units this patient will get per day: 24000 units.
Because 24000 units falls between 20,000 and 40,000 units, this order is safe, so we can procede to calculate the flow rate:
We want our answer to be in

mL

h

, because all the problems in this chapter involve an electronic infusion device to control the flow rate of the IV, so the speed must be in mL/h, because this is the unit of measure accepted by electronic infusion devices.

_____

mL

h

=
Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. Well, we don't have a rate in volume over time, but we have a rate in dosage over time: 1000 units/h, and we know that we can get from units to mL by using the concentration of the drug: 20000 units of heparin in 1 L of D51/2S.
So this gives us:
_____

mL

h

=

1000 units

1 h

Now we need to cancel out the units, which appears in the top of the fraction, so we need to multiply by a fraction that has units in the bottom. The concentration gives us such a fraction:

1 L

20000 units

. Multiplying by this yields:
_____

mL

h

=

1000 units

1 h

×

1 L

20000 units

We need to multiply by a fraction which is equal to one and has L in the bottom so that we can cancel out the L in the top of the fraction on the right of the equation.
So, we first need to multiply by a fraction that is equal to one and which cancels out L and gets us to mL. Since we know that 1L=1000mL, the fraction

1000 mL

1 L

will be equal to one, and since it has L in the bottom, will also cancel out the L in the top of the fraction in the right side of the equation we already have:
_____

mL

h

=

1000 units

1 h

×

1 L

20000 units

×

1000 mL

1 L

Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____

mL

h

=

1000 ~~units~~

1 h

×

1 L

20000 ~~units~~

×

1000 mL

1 L

So this simplifies to:
_____

mL

h

=

1000

1

×

1

20000

×

1000

1

mL

h

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 20000 and 1 by 1000. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mL

h

=

1000

1

×

1

~~20000~~₂₀

×

~~$new1000~~¹

1

×

1

1

mL

h

_____

mL

h

=

1000

1

×

1

20

×

1

1

×

1

1

mL

h

We can divide both 20 and 1000 by 20. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mL

h

=

~~1000~~⁵⁰

1

×

1

20₁

×

1

1

mL

h

_____

mL

h

=

50

1

×

1

1

×

1

1

mL

h

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:
_____

mL

h

=

50

1

mL

h

= 50

mL

h

= 50

mL

h

So our answer is 50

mL

h

.
Example:
The order is for 30000 units of heparin in 500 mL of D5NS to infuse at 1500 units/h.
Calculate the flow rate in mL/h to the nearest whole mL.(You should check to make sure the order is safe before you do your calculations.)
First we should check to make sure that this order is safe. The order is for heparin to infuse at a rate of 1500 units/h, and we know that 20,000-40,000 units of heparin per day is safe. Since there are 24 hours in a day, we multiply 1500 units/h by 24 to get the total number of units this patient will get per day: 36000 units.
Because 36000 units falls between 20,000 and 40,000 units, this order is safe, so we can procede to calculate the flow rate:
We want our answer to be in

mL

h

, because all the problems in this chapter involve an electronic infusion device to control the flow rate of the IV, so the speed must be in mL/h, because this is the unit of measure accepted by electronic infusion devices.

_____

mL

h

=
Now since we are looking for a rate, which is volume over time, we need to find something to put on the right side of the equation that will be a rate which expresses volume over time. Well, we don't have a rate in volume over time, but we have a rate in dosage over time: 1500 units/h, and we know that we can get from units to mL by using the concentration of the drug: 30000 units of heparin in 500 mL of D5NS.
So this gives us:
_____

mL

h

=

1500 units

1 h

Now we need to cancel out the units, which appears in the top of the fraction, so we need to multiply by a fraction that has units in the bottom. The concentration gives us such a fraction:

500 mL

30000 units

. Multiplying by this yields:
_____

mL

h

=

1500 units

1 h

×

500 mL

30000 units

Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____

mL

h

=

1500 ~~units~~

1 h

×

500 mL

30000 ~~units~~

So this simplifies to:
_____

mL

h

=

1500

1

×

500

30000

mL

h

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 30000 and 1500 by 1500. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mL

h

=

~~1500~~¹

1

×

500

~~30000~~₂₀

mL

h

_____

mL

h

=

1

1

×

500

20

mL

h

We can divide both 20 and 500 by 20. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

mL

h

=

1

1

×

~~500~~²⁵

20₁

mL

h

_____

mL

h

=

1

1

×

25

1

mL

h

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:
_____

mL

h

=

25

1

mL

h

= 25

mL

h

= 25

mL

h

So our answer is 25

mL

h

.
Example:
The order is for 35000 units of heparin in 1 L of NS to infuse at 2000 units/h.
Calculate the flow rate in mL/h to the nearest whole mL.(You should check to make sure the order is safe before you do your calculations.)
First we should check to make sure that this order is safe. The order is for heparin to infuse at a rate of 2000 units/h, and we know that 20,000-40,000 units of heparin per day is safe. Since there are 24 hours in a day, we multiply 2000 units/h by 24 to get the total number of units this patient will get per day: 48000 units.
Because 48000 units dose not fall between 20,000 and 40,000 units, this order is not safe, so we should not procede to calculate the flow rate. Instead, we should check this order with the physician.
Now we will look at some examples where heparin is ordered based on the flow rate, and we will convert this flow rate to a dosage infusion rate and check this dosage rate for safety.

Example:
The order is for 40000 units of heparin in 1 L of D51/2NS to infuse at 40 mL/h.
Calculate the dosage rate in units/h to the nearest tenth. Is this order safe?
We want our answer to be in

units

h

.

_____

units

h

=
Now since we are looking for a rate, which is dosage over time, we need to find something to put on the right side of the equation that will be a rate which expresses dosage over time. Well, we don't have a rate in dosage over time, but we have a rate in volume over time: 40 mL/h, and we know that we can get from mL to units by using the concentration of the drug: 40000 units of heparin in 1 L of D51/2NS.
So this gives us:
_____

units

h

=

40 mL

1 h

Now we need to cancel out the mL which appear in the bottom of the fraction, and we want to be able to use the concentration of the drug: 40000 units of heparin in 1 L of D51/2NS to get from mL to a dosage unit. However, we notice that the units used in the concentration are L, and not mL. So, before we can use the concentration to get from volume to dosage measurements, we must first convert mL to L. We can do this because we know that 1000 mL = 1 L, so muliplying by the fraction

1 L

1000 mL

will cancel out the mL in the top, which we want to get rid of without changing the value of the right side of our equation:
_____

units

h

=

40 mL

1 h

×

1 L

1000 mL

Now we want to cancel out the L units which appear in the top of the fraction at right and convert it to some kind of dosage units, because we are trying to get to units units. We can use the concentration to do this: 40000 units of heparin in 1 L of D51/2NS. So now we multiply by

40000 units

1 L

:
_____

units

h

=

40 mL

1 h

×

1 L

1000 mL

×

40000 units

1 L

Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____

units

h

=

40 mL

1 h

×

1 L

1000 mL

×

40000 units

1 L

So this simplifies to:
_____

units

h

=

40

1

×

1

1000

×

40000

1

units

h

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 40000 and 1 by 1. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

units

h

=

40

1

×

1

1

×

40000

1

units

h

We can divide both 40000 and 1 by 1000. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

units

h

=

40

1

×

1

1

×

40

1

units

h

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:
_____

units

h

=

1600

1

units

h

= 1600

units

h

= 1600

units

h

So our answer is 1600

units

h

.
Now we should check to make sure that this order is safe. The order is for heparin to infuse at a rate of units/h, and we know that 20,000-40,000 units of heparin per day is safe. Since there are 24 hours in a day, we multiply 1600 units/h by 24 to get the total number of units this patient will get per day: 38400 units.
Because 38400 units falls between 20,000 and 40,000 units, this order is safe, so we can procede to give this order to the patient.
Example:
The order is for 45000 units of heparin in 1 L of D51/4S to infuse at 40 mL/h.
Calculate the dosage rate in units/h to the nearest tenth. Is this order safe?
We want our answer to be in

units

h

.

_____

units

h

=
Now since we are looking for a rate, which is dosage over time, we need to find something to put on the right side of the equation that will be a rate which expresses dosage over time. Well, we don't have a rate in dosage over time, but we have a rate in volume over time: 40 mL/h, and we know that we can get from mL to units by using the concentration of the drug: 45000 units of heparin in 1 L of D51/4S.
So this gives us:
_____

units

h

=

40 mL

1 h

Now we need to cancel out the mL which appear in the bottom of the fraction, and we want to be able to use the concentration of the drug: 45000 units of heparin in 1 L of D51/4S to get from mL to a dosage unit. However, we notice that the units used in the concentration are L, and not mL. So, before we can use the concentration to get from volume to dosage measurements, we must first convert mL to L. We can do this because we know that 1000 mL = 1 L, so muliplying by the fraction

1 L

1000 mL

will cancel out the mL in the top, which we want to get rid of without changing the value of the right side of our equation:
_____

units

h

=

40 mL

1 h

×

1 L

1000 mL

Now we want to cancel out the L units which appear in the top of the fraction at right and convert it to some kind of dosage units, because we are trying to get to units units. We can use the concentration to do this: 45000 units of heparin in 1 L of D51/4S. So now we multiply by

45000 units

1 L

:
_____

units

h

=

40 mL

1 h

×

1 L

1000 mL

×

45000 units

1 L

Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____

units

h

=

40 mL

1 h

×

1 L

1000 mL

×

45000 units

1 L

So this simplifies to:
_____

units

h

=

40

1

×

1

1000

×

45000

1

units

h

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 45000 and 1 by 1. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

units

h

=

40

1

×

1

1

×

45000

1

units

h

We can divide both 45000 and 1 by 1000. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

units

h

=

40

1

×

1

1

×

45

1

units

h

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:
_____

units

h

=

1800

1

units

h

= 1800

units

h

= 1800

units

h

So our answer is 1800

units

h

.
Now we should check to make sure that this order is safe. The order is for heparin to infuse at a rate of units/h, and we know that 20,000-40,000 units of heparin per day is safe. Since there are 24 hours in a day, we multiply 1800 units/h by 24 to get the total number of units this patient will get per day: 43200 units.
Because 43200 units dose not fall between 20,000 and 40,000 units, this order is not safe, so we should not procede to administer this order to the patient. Instead, we should check this order with the physician.
Example:
The order is for 25000 units of heparin in 500 mL of D5W to infuse at 30 mL/h.
Calculate the dosage rate in units/h to the nearest tenth. Is this order safe?
We want our answer to be in

units

h

.

_____

units

h

=
Now since we are looking for a rate, which is dosage over time, we need to find something to put on the right side of the equation that will be a rate which expresses dosage over time. Well, we don't have a rate in dosage over time, but we have a rate in volume over time: 30 mL/h, and we know that we can get from mL to units by using the concentration of the drug: 25000 units of heparin in 500 mL of D5W.
So this gives us:
_____

units

h

=

30 mL

1 h

Now we want to cancel out the mL units which appear in the top of the fraction at right and convert it to some kind of dosage units, because we are trying to get to units units. We can use the concentration to do this: 25000 units of heparin in 500 mL of D5W. So now we multiply by

25000 units

500 mL

:
_____

units

h

=

30 mL

1 h

×

25000 units

500 mL

Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____

units

h

=

30 mL

1 h

×

25000 units

500 mL

So this simplifies to:
_____

units

h

=

30

1

×

25000

500

units

h

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 25000 and 500 by 500. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

units

h

=

30

1

×

50

1

units

h

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:
_____

units

h

=

1500

1

units

h

= 1500

units

h

= 1500

units

h

So our answer is 1500

units

h

.
Now we should check to make sure that this order is safe. The order is for heparin to infuse at a rate of units/h, and we know that 20,000-40,000 units of heparin per day is safe. Since there are 24 hours in a day, we multiply 1500 units/h by 24 to get the total number of units this patient will get per day: 36000 units.
Because 36000 units falls between 20,000 and 40,000 units, this order is safe, so we can procede to give this order to the patient.
Example:
The order is for 30000 units of heparin in 1 L of D5NS to infuse at 25 mL/h.
Calculate the dosage rate in units/h to the nearest tenth. Is this order safe?
We want our answer to be in

units

h

.

_____

units

h

=
Now since we are looking for a rate, which is dosage over time, we need to find something to put on the right side of the equation that will be a rate which expresses dosage over time. Well, we don't have a rate in dosage over time, but we have a rate in volume over time: 25 mL/h, and we know that we can get from mL to units by using the concentration of the drug: 30000 units of heparin in 1 L of D5NS.
So this gives us:
_____

units

h

=

25 mL

1 h

Now we need to cancel out the mL which appear in the bottom of the fraction, and we want to be able to use the concentration of the drug: 30000 units of heparin in 1 L of D5NS to get from mL to a dosage unit. However, we notice that the units used in the concentration are L, and not mL. So, before we can use the concentration to get from volume to dosage measurements, we must first convert mL to L. We can do this because we know that 1000 mL = 1 L, so muliplying by the fraction

1 L

1000 mL

will cancel out the mL in the top, which we want to get rid of without changing the value of the right side of our equation:
_____

units

h

=

25 mL

1 h

×

1 L

1000 mL

Now we want to cancel out the L units which appear in the top of the fraction at right and convert it to some kind of dosage units, because we are trying to get to units units. We can use the concentration to do this: 30000 units of heparin in 1 L of D5NS. So now we multiply by

30000 units

1 L

:
_____

units

h

=

25 mL

1 h

×

1 L

1000 mL

×

30000 units

1 L

Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____

units

h

=

25 mL

1 h

×

1 L

1000 mL

×

30000 units

1 L

So this simplifies to:
_____

units

h

=

25

1

×

1

1000

×

30000

1

units

h

Now we need to cancel as much as possible by dividing the top and bottom of the fractions by any factors they have in common:
We can divide both 30000 and 1 by 1. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

units

h

=

25

1

×

1

1

×

30000

1

units

h

We can divide both 30000 and 1 by 1000. So, dividing these numbers, which we can do because one is on the bottom of the fractions and one is on the top, yields:
_____

units

h

=

25

1

×

1

1

×

30

1

units

h

Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:
_____

units

h

=

750

1

units

h

= 750

units

h

= 750

units

h

So our answer is 750

units

h

.
Now we should check to make sure that this order is safe. The order is for heparin to infuse at a rate of units/h, and we know that 20,000-40,000 units of heparin per day is safe. Since there are 24 hours in a day, we multiply 750 units/h by 24 to get the total number of units this patient will get per day: 18000 units.
Because 18000 units dose not fall between 20,000 and 40,000 units, this order is not safe, so we should not procede to administer this order to the patient. Instead, we should check this order with the physician.