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:
_____
= 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:
. So, multiplying by this yields:_____
= ×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:
. Multiplying by this yields:_____
= ××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:
, which is equal to one, and which has min in the top, to cancel out the min in the bottom.
So this yields:
_____
= ×××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ×××So this simplifies to:
_____
= ××× 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:
_____
= ××× _____
= ××× 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:
_____
= ×××× _____
= ×××× 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:
_____
= ×××× _____
= ×××× 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:
_____
= ×××× _____
= ×××× 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 = 6So 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
, 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: 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:
_____
= 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:
. So, multiplying by this yields:_____
= ×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: . So, multiplying by this yields:_____
= ××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:
. Multiplying by this yields:_____
= ×××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
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:_____
= ××××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:
, which is equal to one, and which has min in the top, to cancel out the min in the bottom.
So this yields:
_____
= ×××××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ×××××So this simplifies to:
_____
= ××××× 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:
_____
= ××××× _____
= ××××× 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:
_____
= ××××× _____
= ××××× 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:
_____
= ××××× _____
= ××××× 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:
_____
= ××××× _____
= ××××× Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:
_____
= = 37.2 = 37So 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
, 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: 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:
_____
= 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:
. So, multiplying by this yields:_____
= ×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:
. Multiplying by this yields:_____
= ××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:
, which is equal to one, and which has min in the top, to cancel out the min in the bottom.
So this yields:
_____
= ×××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ×××So this simplifies to:
_____
= ××× 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:
_____
= ××× _____
= ××× 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:
_____
= ×××× _____
= ×××× 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:
_____
= ×××× _____
= ×××× 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:
_____
= ×××× _____
= ×××× 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 = 30So 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
, 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: 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:
_____
=×Because we can put kg over 1 without changing it, this becomes:
_____
=×And cancelling kg, which appear in both the top and the bottom of the numerator yields:
_____
=×=×Multiplying yields:
_____
=×== 3344.2So this gives us:
_____
= 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:
. So, multiplying by this yields:_____
= ×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:
. Multiplying by this yields:_____
= ××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:
, which is equal to one, and which has min in the top, to cancel out the min in the bottom.
So this yields:
_____
= ×××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ×××So this simplifies to:
_____
= ××× 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:
_____
= ××× _____
= ××× 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:
_____
= ×××× _____
= ×××× Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:
_____
= = 50.163 = 50So 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
, 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: 2.5 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:
_____
=×Because we can put kg over 1 without changing it, this becomes:
_____
=×And cancelling kg, which appear in both the top and the bottom of the numerator yields:
_____
=×=×Multiplying yields:
_____
=×== 179.5So this gives us:
_____
= 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:
. So, multiplying by this yields:_____
= ×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:
. Multiplying by this yields:_____
= ××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:
, which is equal to one, and which has min in the top, to cancel out the min in the bottom.
So this yields:
_____
= ×××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ×××So this simplifies to:
_____
= ××× 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:
_____
= ××× _____
= ××× 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:
_____
= ×××× _____
= ×××× 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:
_____
= ×××× _____
= ×××× Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:
_____
= = 35.9 = 36So 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
, 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: 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:
_____
=×Because we can put kg over 1 without changing it, this becomes:
_____
=×And cancelling kg, which appear in both the top and the bottom of the numerator yields:
_____
=×=×Multiplying yields:
_____
=×== 260So this gives us:
_____
= 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:
. So, multiplying by this yields:_____
= ×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:
. Multiplying by this yields:_____
= ××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:
, which is equal to one, and which has min in the top, to cancel out the min in the bottom.
So this yields:
_____
= ×××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ×××So this simplifies to:
_____
= ××× 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:
_____
= ××× _____
= ××× 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:
_____
= ×××× _____
= ×××× 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:
_____
= ×××× _____
= ×××× 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:
_____
= ××× _____
= ××× 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 = 10So 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
.
_____
=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:
_____
= 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
:_____
= ×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:
. So, multiplying by this yields:_____
= ××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:
, which is equal to one, and which has h in the top, to cancel out the h in the bottom.
So this yields:
_____
= ×××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ×××So this simplifies to:
_____
= ×××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:
_____
= ×××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:
_____
= ×××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:
_____
= ×××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:
_____
= ××××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:
_____
= ××××Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:
_____
= = 2 = 2So our answer is 2
. 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
.
_____
=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:
_____
= 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
:_____
= ×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:
. So, multiplying by this yields:_____
= ××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:
, which is equal to one, and which has h in the top, to cancel out the h in the bottom.
So this yields:
_____
= ×××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ×××So this simplifies to:
_____
= ×××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:
_____
= ×××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:
_____
= ×××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:
_____
= ×××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:
_____
= ××××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:
_____
= ××××Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:
_____
= = 616 = 616So our answer is 616
. 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:
_____
= 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
:_____
= ×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:
. So, multiplying by this yields:_____
= ××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ××So this simplifies to:
_____
= ××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:
_____
= ××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:
_____
= ×××Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:
_____
= = 257.14285714286 = 257.1So 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 24 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
, 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.
_____
=We have a range for the dosage infusion rate: 24 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: _____
= 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:
. So, multiplying by this yields:Min: _____
= ×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:
. Multiplying by this yields:Min: _____
= ××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:
, which is equal to one, and which has min in the top, to cancel out the min in the bottom.
So this yields:
Min: _____
= ×××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
Min: _____
= ×××So this simplifies to:
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 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: _____
= ××× Min: _____
= ××× 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: _____
= ×××× Min: _____
= ×××× 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: _____
= ×××× Min: _____
= ×××× 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: _____
= ×××× 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 = 30So our minimum is 30
. 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.
_____
=We have a range for the dosage infusion rate: 24 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: _____
= 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:
. So, multiplying by this yields:Max: _____
= ×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:
. Multiplying by this yields:Max: _____
= ××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:
, which is equal to one, and which has min in the top, to cancel out the min in the bottom.
So this yields:
Max: _____
= ×××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
Max: _____
= ×××So this simplifies to:
Max: _____
= ××× 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: _____
= ××× Max: _____
= ××× 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: _____
= ×××× Max: _____
= ×××× 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: _____
= ×××× Max: _____
= ×××× 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: _____
= ×××× Max: _____
= ×××× 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: _____
= = 60 = 60So our maximum is 60
. So our answer is 3060
. 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
.
_____
=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:
_____
= 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
:_____
= ×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:
. So, multiplying by this yields:_____
= ××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:
, which is equal to one, and which has h in the top, to cancel out the h in the bottom.
So this yields:
_____
= ×××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ×××So this simplifies to:
_____
= ×××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:
_____
= ×××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:
_____
= ×××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:
_____
= ×××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:
_____
= ××××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:
_____
= ××××Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:
_____
= = 3 = 3So our stabilizing dosage rate is 3
. Example:
The order is for 75 mg of a medication in 100 mL of NS to infuse at 130641 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
, 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.
_____
=We have a range for the dosage infusion rate: 130641 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: _____
= 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:
. So, multiplying by this yields:Min: _____
= ×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:
. Multiplying by this yields:Min: _____
= ××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:
, which is equal to one, and which has min in the top, to cancel out the min in the bottom.
So this yields:
Min: _____
= ×××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
Min: _____
= ×××So this simplifies to:
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 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: _____
= ××× 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:
Min: _____
= ×××× Min: _____
= ×××× 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: _____
= ×××× Min: _____
= ×××× 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: _____
= ×××× 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: _____
= = 10.4 = 10So our minimum is 10
. 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.
_____
=We have a range for the dosage infusion rate: 130641 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: _____
= 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:
. So, multiplying by this yields:Max: _____
= ×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:
. Multiplying by this yields:Max: _____
= ××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:
, which is equal to one, and which has min in the top, to cancel out the min in the bottom.
So this yields:
Max: _____
= ×××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
Max: _____
= ×××So this simplifies to:
Max: _____
= ××× 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: _____
= ××× Max: _____
= ××× 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: _____
= ×××× Max: _____
= ×××× 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: _____
= ×××× Max: _____
= ×××× 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: _____
= = 51.28 = 51So our maximum is 51
. So our answer is 1051
. 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
.
_____
=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:
_____
= 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
:_____
= ×Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ×So this simplifies to:
_____
= ×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:
_____
= ×Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:
_____
= = 18 = 18So our stabilizing dosage rate is 18
. Example:
The order is for 50 mg of nitroglycerin in 250 mL of D5W to infuse at 1080 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
, 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.
_____
=We have a range for the dosage infusion rate: 1080 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: _____
= 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:
. So, multiplying by this yields:Min: _____
= ×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:
. Multiplying by this yields:Min: _____
= ××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:
, which is equal to one, and which has min in the top, to cancel out the min in the bottom.
So this yields:
Min: _____
= ×××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
Min: _____
= ×××So this simplifies to:
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 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: _____
= ××× 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:
Min: _____
= ×××× Min: _____
= ×××× 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: _____
= ×××× 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:
Min: _____
= ×××× Min: _____
= ×××× 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: _____
= ××× 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: _____
= = 3 = 3So our minimum is 3
. 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.
_____
=We have a range for the dosage infusion rate: 1080 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: _____
= 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:
. So, multiplying by this yields:Max: _____
= ×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:
. Multiplying by this yields:Max: _____
= ××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:
, which is equal to one, and which has min in the top, to cancel out the min in the bottom.
So this yields:
Max: _____
= ×××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
Max: _____
= ×××So this simplifies to:
Max: _____
= ××× 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: _____
= ××× Max: _____
= ××× 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: _____
= ×××× Max: _____
= ×××× 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: _____
= ×××× Max: _____
= ×××× 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: _____
= ×××× Max: _____
= ×××× 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: _____
= ××× Max: _____
= ××× 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: _____
= = 24 = 24So our maximum is 24
. So our answer is 324
. 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
.
_____
=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:
_____
= 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
:_____
= ×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:
. So, multiplying by this yields:_____
= ××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:
, which is equal to one, and which has h in the top, to cancel out the h in the bottom.
So this yields:
_____
= ×××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ×××So this simplifies to:
_____
= ×××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:
_____
= ×××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:
_____
= ×××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:
_____
= ×××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:
_____
= ××××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:
_____
= ××××Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:
_____
= = 30 = 30So our stabilizing dosage rate is 30
. Example:
The order is for 5 mg of Primacor in 500 mL of D5W to infuse at 0.3750.75 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
, 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.
_____
=We have a range for the dosage infusion rate: 0.3750.75 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: _____
=×Because we can put kg over 1 without changing it, this becomes:
Min: _____
=×And cancelling kg, which appear in both the top and the bottom of the numerator yields:
Min: _____
=×=×Multiplying yields:
Min: _____
=×== 34.5So this gives us:
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:
. So, multiplying by this yields:Min: _____
= ×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:
. Multiplying by this yields:Min: _____
= ××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:
, which is equal to one, and which has min in the top, to cancel out the min in the bottom.
So this yields:
Min: _____
= ×××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
Min: _____
= ×××So this simplifies to:
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 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: _____
= ××× Min: _____
= ××× 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: _____
= ×××× Min: _____
= ×××× 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: _____
= ×××× 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: _____
= = 207 = 207So our minimum is 207
. 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 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: _____
=×Because we can put kg over 1 without changing it, this becomes:
Max: _____
=×And cancelling kg, which appear in both the top and the bottom of the numerator yields:
Max: _____
=×=×Multiplying yields:
Max: _____
=×== 69So this gives us:
Max: _____
= 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:
. So, multiplying by this yields:Max: _____
= ×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:
. Multiplying by this yields:Max: _____
= ××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:
, which is equal to one, and which has min in the top, to cancel out the min in the bottom.
So this yields:
Max: _____
= ×××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
Max: _____
= ×××So this simplifies to:
Max: _____
= ××× 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: _____
= ××× Max: _____
= ××× 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: _____
= ×××× Max: _____
= ×××× 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: _____
= ×××× Max: _____
= ×××× 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: _____
= = 414 = 414So our maximum is 414
. So our answer is 207414
. 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
.
_____
=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:
_____
= 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
:_____
= ×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:
. So, multiplying by this yields:_____
= ××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:
, which is equal to one, and which has h in the top, to cancel out the h in the bottom.
So this yields:
_____
= ×××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ×××So this simplifies to:
_____
= ×××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:
_____
= ×××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:
_____
= ×××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:
_____
= ×××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:
_____
= ××××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:
_____
= ××××Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:
_____
= = 41.666666666667 = 41.7So our stabilizing dosage rate is 41.7
. Example:
The order is for 50 mg of Nipride in 250 mL of D5W to infuse at 36 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
, 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.
_____
=We have a range for the dosage infusion rate: 36 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: _____
=×Because we can put kg over 1 without changing it, this becomes:
Min: _____
=×And cancelling kg, which appear in both the top and the bottom of the numerator yields:
Min: _____
=×=×Multiplying yields:
Min: _____
=×== 210So this gives us:
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:
. So, multiplying by this yields:Min: _____
= ×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:
. Multiplying by this yields:Min: _____
= ××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:
, which is equal to one, and which has min in the top, to cancel out the min in the bottom.
So this yields:
Min: _____
= ×××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
Min: _____
= ×××So this simplifies to:
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 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: _____
= ××× 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:
Min: _____
= ×××× Min: _____
= ×××× 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: _____
= ×××× 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:
Min: _____
= ×××× Min: _____
= ×××× 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: _____
= ××× 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: _____
= = 63 = 63So our minimum is 63
. 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 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: _____
=×Because we can put kg over 1 without changing it, this becomes:
Max: _____
=×And cancelling kg, which appear in both the top and the bottom of the numerator yields:
Max: _____
=×=×Multiplying yields:
Max: _____
=×== 420So this gives us:
Max: _____
= 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:
. So, multiplying by this yields:Max: _____
= ×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:
. Multiplying by this yields:Max: _____
= ××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:
, which is equal to one, and which has min in the top, to cancel out the min in the bottom.
So this yields:
Max: _____
= ×××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
Max: _____
= ×××So this simplifies to:
Max: _____
= ××× 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: _____
= ××× Max: _____
= ××× 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: _____
= ×××× Max: _____
= ×××× 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: _____
= ×××× Max: _____
= ×××× 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: _____
= ×××× Max: _____
= ×××× 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: _____
= ××× Max: _____
= ××× 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: _____
= = 126 = 126So our maximum is 126
. So our answer is 63126
. 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
.
_____
=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:
_____
= 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
:_____
= ×Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ×So this simplifies to:
_____
= ×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:
_____
= ×Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:
_____
= = 21.6 = 21.6So our stabilizing dosage rate is 21.6
. Example:
The order is for 1 g of dobutamine in 250 mL of D5W to infuse at 15 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
, 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.
_____
=We have a range for the dosage infusion rate: 15 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: _____
=×Because we can put kg over 1 without changing it, this becomes:
Min: _____
=×And cancelling kg, which appear in both the top and the bottom of the numerator yields:
Min: _____
=×=×Multiplying yields:
Min: _____
=×== 109So this gives us:
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:
. So, multiplying by this yields:Min: _____
= ×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: . So, multiplying by this yields:Min: _____
= ××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:
. Multiplying by this yields:Min: _____
= ×××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:
, which is equal to one, and which has min in the top, to cancel out the min in the bottom.
So this yields:
Min: _____
= ××××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
Min: _____
= ××××So this simplifies to:
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 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: _____
= ×××× 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:
Min: _____
= ××××× 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:
Min: _____
= ××××× 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:
Min: _____
= ××××× 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: _____
= = 1.635 = 2So our minimum is 2
. 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 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: _____
=×Because we can put kg over 1 without changing it, this becomes:
Max: _____
=×And cancelling kg, which appear in both the top and the bottom of the numerator yields:
Max: _____
=×=×Multiplying yields:
Max: _____
=×== 545So this gives us:
Max: _____
= 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:
. So, multiplying by this yields:Max: _____
= ×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: . So, multiplying by this yields:Max: _____
= ××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:
. Multiplying by this yields:Max: _____
= ×××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:
, which is equal to one, and which has min in the top, to cancel out the min in the bottom.
So this yields:
Max: _____
= ××××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
Max: _____
= ××××So this simplifies to:
Max: _____
= ×××× 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: _____
= ×××× Max: _____
= ×××× 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: _____
= ××××× Max: _____
= ××××× 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: _____
= ××××× Max: _____
= ××××× 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: _____
= ××××× Max: _____
= ××××× 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: _____
= ××××× Max: _____
= ××××× 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: _____
= = 8.175 = 8So our maximum is 8
. So our answer is 28
. 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
.
_____
=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:
_____
= 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
:_____
= ×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:
. So, multiplying by this yields:_____
= ××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:
, which is equal to one, and which has h in the top, to cancel out the h in the bottom.
So this yields:
_____
= ×××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ×××So this simplifies to:
_____
= ×××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:
_____
= ×××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:
_____
= ×××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:
_____
= ×××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:
_____
= ××××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:
_____
= ××××Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:
_____
= = 0.5 = 0.5So our stabilizing dosage rate is 0.5
. 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 openheart 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 20,00040,000 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 20,00040,000 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
, 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: 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:
_____
= 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:
. Multiplying by this yields:_____
= ×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
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:_____
= ××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ××So this simplifies to:
_____
= ×× 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:
_____
= ××× _____
= ××× 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:
_____
= ×× _____
= ×× Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:
_____
= = 48 = 48So our answer is 48
. 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 20,00040,000 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
, 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: 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:
_____
= 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:
. Multiplying by this yields:_____
= ×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
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:_____
= ××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ××So this simplifies to:
_____
= ×× 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:
_____
= ××× _____
= ××× 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:
_____
= ×× _____
= ×× Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:
_____
= = 50 = 50So our answer is 50
. 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 20,00040,000 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
, 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: 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:
_____
= 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:
. Multiplying by this yields:_____
= ×Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ×So this simplifies to:
_____
= × 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:
_____
= × _____
= × 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:
_____
= × _____
= × Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest whole number yields:
_____
= = 25 = 25So our answer is 25
. 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 20,00040,000 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
.
_____
=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:
_____
= 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
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:_____
= ×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
:_____
= ××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ××So this simplifies to:
_____
= ××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:
_____
= ××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:
_____
= ××Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:
_____
= = 1600 = 1600So our answer is 1600
. 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 20,00040,000 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
.
_____
=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:
_____
= 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
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:_____
= ×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
:_____
= ××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ××So this simplifies to:
_____
= ××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:
_____
= ××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:
_____
= ××Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:
_____
= = 1800 = 1800So our answer is 1800
. 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 20,00040,000 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
.
_____
=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:
_____
= 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
:_____
= ×Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ×So this simplifies to:
_____
= ×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:
_____
= ×Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:
_____
= = 1500 = 1500So our answer is 1500
. 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 20,00040,000 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
.
_____
=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:
_____
= 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
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:_____
= ×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
:_____
= ××Now we need to cancel out all units that appear in both the top and the bottom of the equation:
_____
= ××So this simplifies to:
_____
= ××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:
_____
= ××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:
_____
= ××Multiplying across the top and the bottom, then dividing the top by the bottom and rounding our answer to the nearest tenth yields:
_____
= = 750 = 750So our answer is 750
. 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 20,00040,000 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.