Subsection 2.5.2 Terminology
This section defines terminology used in medicine and sciences about solution concentrations that we need for the ratio examples in this section.
The active ingredient in a drug is often added to a inactive ingredient (often liquid) to administer it. This liquid is known as a diluent. The diluent might be water, saline solution, or other substances.
The substance (active ingredient for medicine) which may be a powder or another liquid to which the diluent is added is called the solute. The solute is disolved in the diluent. For example salt (solute) is disolved in water (diluent) to make saline solution.
Even if the drug can be administered directly (e.g., is already liquid) we sometimes need to dilute the stock solution (undiluted drug) for ease of use.
In some problems the drug mixture will be divided into parts. These parts are sometimes called aliquots. For example when testing substances (like blood samples) we may divide the sample drawn into multiple aliquots, one for each test to be run.
The most important concept is measuring how concentrated a solution is. This enables providing sufficient and safe amounts of drugs. There are three common ways concentration is written. These three are examples of how ratios can present the relationship between quantities in different ways. Being able to change between the different presentations of concentration will demonstrate your ability to understand and use ratios accurately.
Definition 2.5.1. Dilution Ratio.
The Dilution Ratio is the ratio of solute (drug) to diluent.
If the solute is a liquid, then this is in units of volume per volume (e.g., mL per mL). For example a dilution ratio of 1:4 means 1 mL of drug to 4 mL of diluent giving 5 mL of solution.
This expression of concentration is unlikely to be used for dry solutes.
Definition 2.5.2. Dilution Factor.
The Dilution Factor is the ratio of solute (drug) to the resulting solution.
If the solute is liquid, then this is in units of volume per volume. For example a dilution factor of 5 means 1 unit of the drug in every 5 units of solution implying 4 units of diluent (1/4 dilution ratio).
If the solute is solid (e.g, powder) then this is in units of mass per volume. For example, 5 g of drug in a total of 100 mL of solution. Note we do not care how much diluent was added (hence we cannot calculate dilution ratio). This dilution factor can be achieved by putting in the dry ingredient, then adding part of the diluent to disolve the dry ingredient, then pouring in enough additional diluent to reach the desired volume.
Definition 2.5.3. Percent Concentration.
The Percent Concentration is the ratio of mass of solute (drug) to 100 mL of diluent.
If the solute is liquid, then this is in units of volume per volume. For example if the are 2 mL of drug per 100 mL of solution, then the percenet concentration is 2/100 or 2%.
If the solute is solid (e.g, powder) then this is in units of mass per volume. For example, if there are 2 mg of drug per 100 mL of solution, then the percent concentration is 2/100 or 2%. Note this is neither percent by volume nor percent by mass as would be used in science.
These examples illustrate the meaning of these terms.
Example 2.5.4.
A solution is produced from 3 mL of concentrated chloroform and 37 mL of water.
(a)
What is the dilution ratio?
The dilution ratio is the ratio of the solute to the diluent. We are given both. The dilution ratio is \(3/37\text{.}\)
(b)
What is the dilution factor?
The dilution factor is the ratio of the total to the substance. The total is substance plus diluent. Here that is \(3+37=4\bar{0}\text{.}\) The ratio then is \(\frac{4\bar{0}}{3} \approx 13\text{.}\)
(c)
Calculate the percent concentration.
The percent concentration is the ratio of the solute to the total solution by volume written as a percent. We are given the volume of the solute (3 mL) and have calculated the total volume in the previous task (\(4\bar{0}\) mL). Thus the percent concentration is \(\frac{3}{4\bar{0}} = 0.075\) which is 7.5%.
Example 2.5.5.
Saline solution consists of the solute salt (sodium chloride) dissolved in the diluent pure water. The saline solution most commonly used in medical applications is 9 g of the solute salt, which is a solid, dissolved in enough water to make 1 liter of solution. We can express the concentration in the folowing ways.
Because the solution is produced by adding an unspecified amount of water (“enough”), it is easiest to express the concentration as a dilution factor. For medicine these are most commonly expressed in terms of milliliters, so we will convert units (
metric terminology). The dilution factor is
\begin{equation*}
\frac{9 \text{ g}}{1 \text{ L}} \cdot \frac{1 \text{ L}}{1000 \text{ mL}}=\frac{9 \text{ g}}{1000 \text{ mL}}.
\end{equation*}
We can also calculate the percent concentration. To do this we need to express volume in milliliters (percent concentration is ratio of solute to 100 mL). We can use unit conversion from above. The percent concentration is
\begin{align*}
\frac{9 \text{ g}}{1000 \text{ mL}} & =\\
\frac{9 \text{ g}}{1000 \text{ mL}} \cdot \frac{1/10}{1/10} & =\\
\frac{0.9 \text{ g}}{100 \text{ mL}} & = 0.009
\end{align*}
or 0.9%.
Note how comparing the information provided to the definition showed us we needed to perform a unit conversion. The definitions also stated which number is numerator and denominator and what form to use for a final expression (e.g., percent or fraction).
Example 2.5.6.
Clorox® Disinfecting Bleach contains 7.0% sodium hypochlorite which is a liquid. This means the percent concentration is 7.0%. One commercially available size contains 11 oz. What are the dilution ratio and dilution factor?
Looking at the defintions of both dilution ratio and dilution factor we see that we need to know the amount of solute (sodium hypochlorite) in the 11 oz. Because we are given a percent concentration, we can calculate this amount using the technique of
Example 2.1.5 There is
\(11 \text{ oz} \cdot 0.070 = 0.77\) oz of sodium hypochlorite.
We know the total volume and the amount of solute which is all we need to calculate the dilution factor. It is \(\frac{0.77}{11}\text{.}\) This ratio would be easier to read if we express it with a denominator of one. We start by converting the ratio to a decimal. \(\frac{0.77}{11}=0.7\text{.}\) Thus the dilution factor is \(\frac{0.7}{1}\) expressed in ounces of bleach to ounces of water.
For the dilution ratio we need the amount of water added. Because the total solution is 11 oz and 0.77 oz is bleach, the amount of water is \(11-0.77=10.23\) oz. Thus the dilution ratio is \(\frac{0.77}{10.23}\text{.}\)
This ratio is hard to interpret. We could express it with the numerator as one to indicate how much diluent is added to one part solute. We can calculate this multiple ways. The first way is to scale the ratio so the numerator is one.
\begin{align*}
\frac{0.77}{10.23} \amp = \frac{0.77}{10.23} \cdot \frac{1/0.77}{1/0.77} \\
\amp = \frac{1}{10.23/0.77}\\
\amp \approx \frac{1}{13.29}.
\end{align*}
Another option is to write this as a proportion.
\begin{align*}
\frac{0.77}{10.23} \amp = \frac{1}{R}.\\
\frac{10.23}{0.77} \amp = \frac{R}{1}.\\
\frac{10.23}{0.77} \amp = R.\\
13.28571429 \amp \approx R.\\
13.29 \amp \approx R.
\end{align*}
The ratio \(\frac{1}{13.29}\) means there is one ounce of bleach for every 13.29 ounces of water.
Note a pure substance has dilution factor 1/1 (the total volume of the solution is just the volume of the solute). The percent concentration for a pure substance is 100%.
Use these Checkpoints to test your ability to calculate these ratios.
Subsection 2.5.3 Dilution
This section shows how to use knowledge of proportions to perform calculations required in medicine. Dilution ratios or factors tell us a desired ratio, and we know the initial ratio. This pair allows us to setup a proportion.
This first example shows how to produce a solution with a desired dilution factor.
Example 2.5.7.
How much diluent do we need to add to produce a solution containing 3.0 mL of concentrated chloroform that will have a dilution factor of \(50\text{?}\) Because this is in a medical context, we will round to one milliliter which can be reasonably achieved with their tools.
We start by reviewing the definition. Dilution factor is the ratio of the total to the substance. The goal is for that to equal \(50\text{,}\) so we can write the proportion
\begin{equation*}
\frac{\text{total}}{\text{solute}}=\frac{50}{1}\text{.}
\end{equation*}
We are not given the total volume, but we do know the volume of solute (3.0 mL). Thus far we have
\begin{equation*}
\frac{\text{total}}{3.0 mL}=\frac{50}{1}.
\end{equation*}
The total is (always) the volume of the solute plus the volume of the diluent. The volume of diluent is what we want to calculate. We can call the volume of diluent \(D\text{.}\) The volume of the solution then is \(3.0+D\) where \(D\) is the volume of diluent to add.
\begin{align*}
\frac{3.0+D}{3.0} \amp = \frac{50}{1.0}.\\
3.0 \cdot \frac{3.0+D}{3.0} \amp = 3.0 \cdot \frac{50}{1.0}. \text{ Multiply to isolate } D\\
3.0+D \amp = 150.\\
-3.0+3.0+D \amp = -3.0+150. \text{ Subtract to isolate } D\\
D \amp = 147.
\end{align*}
So we need 147 mL of diluent. Notice once we had the proportion set up we needed only algebra.
This example shows us how to apply a dilution ratio (dilute our solution). We can calculate the resulting dilution factor afterward.
Example 2.5.8.
A doctor orders 120 mL of 50% solution of Ensure every two hours. How much Ensure (liquid) and water is needed?
50% is a percent concentration. This means the Ensure should be 50% of the total volume (120 mL). \(120 \cdot 0.50 = 60\) mL of Ensure. This leaves \(120-60=60\) mL of water (diluent).
In the example above the dilution ratio is 1/1, because there is the same volume of solute (Ensure) and diluent (water). The dilution factor is 1/2, because we have 60 mL of Ensure in 120 mL of solution (\(60/120=1/2\)).
Working on dilutions is a proportion problem. This next example presents a scenario where we work a dilution problem backwards. Notice that the setup is still a proportion, and the solving is still just the algebra steps needed.
Example 2.5.9.
One usage of dilution is to reduce the concentration so that instruments can accurately measure it. Consider trying to measure an acid without dissolving the tools used to measure it.
A sample of a suspected high blood glucose value was obtained. According to the manufacturer of the instrument used to read blood glucose values, the highest glucose result which can be obtained on this particular instrument is 500 mg/dL. When the sample was run, the machine gave an error message (concentration too high).
The serum was then diluted to 1/10 and retested. The machine gave a result of 70 mg/dL. What was the initial concentration?
Note that the ratio is milligrams to decilitres (weight to volume). In these types of problems the amount of substance is so small that it does not affect the volume.
Before we jump into an equation, let’s try an experiment. That’s right, in math we do not have to know what to do when we start. We will try something, learn from it, and maybe revise our approach based on what we learned.
This is a dilution problem which means we can setup the proportion
\begin{equation*}
\frac{\text{solute mg}}{\text{diluent dL}}=\frac{70 \text{ mg}}{\text{dL}}.
\end{equation*}
We are trying to determine the amount of blood sugar in the sample, so the solute portion is unknown. We also do not have the size of sample taken. We will experiment to see how this affects the problem.
Suppose we take 1 dL of the original serum. Because the blood sample is so small, we can calculate as if all the volume is the diluent. That is we started with 1 dL and added more to dilute. To dilute to a ratio of 1/10 we need to add \(10-1=9\) dL of diluent. No blood glucose was added thus the concentration is changed only by the diluent. Thus the concentration proportion is now
\begin{align*}
\frac{C+0 \text{ mg}}{1+9 \text{ dL}} \amp = \frac{70 \text{ mg}}{\text{dL}}\\
\frac{C \text{ mg}}{10 \text{ dL}} \amp = \frac{70 \text{ mg}}{\text{dL}}\\
C \text{ mg} \amp = \frac{70 \text{ mg}}{\text{dL}}(10 \text{ dL})\\
C \amp = 700 \text{ mg}.
\end{align*}
Did this result depend on our selecting 1 dL of the original serum? If we are uncertain we can try the problem again and select 2 dL of the original serum. To figure out the total amount of which 2 is 1/10, we can treat this like
Example 2.3.7
\begin{align*}
\frac{1}{10} \amp = \frac{1}{10} \frac{2}{2}\\
\amp = \frac{2}{20}.
\end{align*}
This means we need \(20-2=18\) dL of diluent to have the desired dilution ratio. Also we will have twice as much of the blood glucose.
\begin{align*}
\frac{2C+0 \text{ mg}}{2+18 \text{ dL}} \amp = \frac{70 \text{ mg}}{\text{dL}}\\
\frac{2C \text{ mg}}{20 \text{ dL}} \amp = \frac{70 \text{ mg}}{\text{dL}}\\
2C \text{ mg} \amp = \frac{70 \text{ mg}}{\text{dL}}(20 \text{ dL})\\
2C \text{ mg} \amp = 1400 \text{ mg}\\
\frac{2C \text{ mg}}{2} \amp = \frac{1400 \text{ mg}}{2}\\
C \amp = 700 \text{ mg}
\end{align*}
The result is the same. This makes sense, because we are setting up a proportion, and ratios do not depend on the amount.
We can be confident that the original serum sample had a blood glucose level of 700 mg/dL.
Sometimes we dilute more than one time. Here we experiment to determine what the effect of serial dilution is upon the resulting dilution factor.
Example 2.5.10.
Suppose you have a solution consisting of 10 mL of acyl chloride and 90 mL of water. If this is diluted to a dilution ratio of 1/2 and then diluted again to a dilution ratio of 1/3, what is the final dilution ratio?
Solution.
We can do the calculations one at a time. First we calculate the original concentration.
\begin{align*}
\frac{10 \text{ mL}}{10+90 \text{ mL}} \amp = \frac{10}{100}\\
\amp = \frac{1}{10}.
\end{align*}
To dilute to a ratio of 1/2 we can calculate the amount of diluent to add as a proportion problem like in
Example 2.4.1.
\begin{align*}
\frac{1}{2} \amp = \frac{100 \text{ mL}}{T \text{ mL}}\\
\frac{2}{1} \amp = \frac{T \text{ mL}}{100 \text{ mL}}\\
200 \text{ mL} \amp = T.
\end{align*}
The total will be 200 mL so we need to add \(200 \text{ mL}- 100 \text{ mL} = 100 \text{ mL}\) of additional diluent. Note at this point the concentration is
\begin{equation*}
\frac{10 \text{ mL acyl chloride}}{200 \text{ mL diluent}} = \frac{1}{20}.
\end{equation*}
To dilute again to a ratio of 1/3 we can calculate the amount of diluent to add
\begin{align*}
\frac{1}{3} \amp = \frac{200 \text{ mL}}{T \text{ mL}}\\
1 \cdot (T \text{ mL}) \amp = 3 \cdot (200 \text{ mL})\\
T \amp = 600 \text{ mL}.
\end{align*}
The total will be 600 mL so we need to add \(600 \text{ mL} - 200 \text{ mL} = 400 \text{ mL}\) of additional diluent. Note at this point the concentration is
\begin{equation*}
\frac{10 \text{ mL acyl chloride}}{600 \text{ mL diluent}} = \frac{1}{60}.
\end{equation*}
Now we can determine what the resulting dilution ratio after diluting twice (1/2 and then 1/3).
\begin{align*}
\frac{1}{10} \cdot F \amp = \frac{1}{60}\\
10 \cdot \frac{1}{10} \cdot F \amp = 10 \cdot \frac{1}{60}.\\
F \amp = \frac{1}{6}.
\end{align*}
Notice that \(\frac{1}{6} = \frac{1}{2} \cdot \frac{1}{3}\text{,}\) that is, the resulting dilution factor is the product of the serial dilutions. This relationship is always true for serial dilution.
Subsection 2.5.4 Dosage
If we know the concentration of a drug, we can determine how much is needed for a given dose. These are proportion problems that require change of units.
In medicine some substances are measured in International Unit or IU. For each substance this is defined by the effect of that amount of the drug.
Example 2.5.11.
One IU of insulin is 0.0347 mg. A common concentration of insulin is U-100 which is 100 IU/mL. This is produced by combining 100 units of insulin in one mL of diluent.
If a person needs 2 units of insulin, how many mL of solution will that be?
This can be solved as a proportion because we are asked for an amount that matches a ratio (concentration). The concentration is 100 IU/mL. Because we are solving for a number of mL, we write the proportion as
\begin{align*}
\frac{\text{mL}}{100 \text{ IU}} \amp = \frac{v \text{ mL}}{2 \text{ IU}}\\
\frac{\text{mL}}{100 \text{ IU}} \cdot (2 \text{ IU}) \amp = \frac{v \text{ mL}}{2 \text{ IU}} \cdot (2 \text{ IU})\\
\frac{2 \text{ mL}}{100} \amp = v \text{ mL}\\
\frac{1}{50} \text{ mL} \amp = v.\\
0.02 \text{ mL} \amp = v.
\end{align*}
Because we have a ratio of desired amount to provided amount we can also solve this problem as a percent.
The ratio is desired amount per provided amount. In this case \(\frac{2 \text{ IU}}{100 \text{ IU}}=\frac{1}{50}=0.02\) which is 2%. We therefore want 2% of the 1 mL (from 100 IU/1 mL) or 0.02 mL.
Example 2.5.12.
A label reads “2.5 mL of solution for injection contains 1000 mg of streptomycin sulfate.” How many milliliters are needed to contain 800 mg of streptomycin?
Solution 1.
Because a ratio is given (1000 mg/2.5 mL) and we want to scale this down (to 800 mg), we can set this up as a proportion.
We want to solve for volume (mL), so we set up the proportion as follows.
\begin{align*}
\frac{2.5 \text{ mL}}{1000 \text{ mg}} & = \frac{v \text{ mL}}{800 \text{ mg}}.\\
\frac{2.5 \text{ mL}}{1000 \text{ mg}} \cdot (800 \text{ mg}) & = \frac{v \text{ mL}}{800 \text{ mg}} \cdot (800 \text{ mg}).\\
2 \text{ mL} & = v.
\end{align*}
Solution 2.
Because we have a ratio of desired amount to provided amount we can also solve this problem as a percent.
The ratio is desired amount/provided amount. In this case \(\frac{800 \text{ IU}}{1000 \text{ IU}}=\frac{4}{5}=0.8\) which is 80%. We therefore want 80% of the 2.5 mL dose or \(0.8 \cdot 2.5 = 2\) mL.
Checkpoint 2.5.13.
A physician may prescribe a medicine and specify a total amount and a speed at which it should be delivered. For IV’s this is called drop factor and is specified as a number of drops per minute. Medical personnel calculate how long to operate the IV so that the total amount of drug prescribed is delivered in the specified time.
Example 2.5.14.
Give 1500 mL of saline solution IV with a drop factor of 10 drops per mL at a rate of 50 drops per minute to an adult patient. Determine how long in hours the IV should be administered.
The rate is specified in drops and the amount is specified in mL which means we need to convert units. This will be done like
Example 1.1.20. The units suggest we should multiply the conversion ratios as follows.
\begin{equation*}
\frac{50 \text{ drops}}{\text{minute}} \cdot \frac{\text{mL}}{10 \text{ drops}} = \frac{5 \text{ mL}}{\text{minute}}
\end{equation*}
Now that we know the rate in mL, we can set up a proportion so that time calculated per total grams of medication matches the specified rate. Notice how we invert the rate to make the algebra easier.
\begin{align*}
\frac{T \text{ min}}{1500 \text{ mL}} & = \frac{1 \text{ min}}{5 \text{ mL}}.\\
\frac{T \text{ min}}{1500 \text{ mL}} \cdot (1500 \text{ mL}) & = \frac{1 \text{ min}}{5 \text{ mL}} \cdot (1500 \text{ mL}).\\
T & = 300 \text{ min}.
\end{align*}
The final step is to convert minutes to hours. This is another unit conversion problem using a conversion from
Table 1.1.2. The units suggest we can multiply the 300 minutes by the conversion ratio.
\begin{equation*}
300 \text{ minutes} \cdot \frac{1 \text{ hour}}{60 \text{ minutes}} = 5 \text{ hours}
\end{equation*}
Example 2.5.15.
Amoxicillin is an antibiotic obtainable in a liquid suspension form, part medication and part water, and is frequently used to treat infections in infants. One formulation of the drug contains 125 mg of amoxicillin per 5 mL of liquid. A pediatrician orders \(150\) mg per day for a 4-month-old child with an ear infection. How much of the amoxicillin suspension would the parent need to administer to the infant in order to achieve the recommended daily dose?
Solution.
Here we need to scale the amount (from 125 mg to \(150\) mg). This is a proportion problem, that is, the ratio of medicine to volume is the same so we can setup an equation based on the drug concentration.
\begin{align*}
\frac{125 \text{ mg}}{5 \text{ mL}} \amp = \frac{150 \text{ mg}}{A \text{ mL}}\\
\frac{5 \text{ mL}}{125 \text{ mg}} \amp = \frac{A \text{ mL}}{150 \text{ mg}}\\
\frac{5 \text{ mL}}{125 \text{ mg}} \cdot (150 \text{ mg}) \amp = A \text{ mL}\\
6 \text{ mL} \amp = A.
\end{align*}
Checkpoint 2.5.16.
A 5% dextrose solution (D5W) contains 5 g of pure dextrose per 100 mL of solution. A doctor orders 500 mL of D5W IV for a patient. How much dextrose does the patient receive from that IV?
Example 2.5.17.
A sample of chloroform water has a dilution factor of 40. If 2 mL of chloroform are needed how many milliliters total are needed?
Solution.
A dilution factor of 40 indicates that 1 mL of chloroform is in 40 mL total of solution. We can setup a proportion to answer this.
\begin{align*}
\frac{1 \text{ mL}}{40 \text{ mL}} \amp = \frac{2 \text{ mL}}{T \text{ mL}}\\
\frac{40 \text{ mL}}{1 \text{ mL}} \amp = \frac{T \text{ mL}}{2 \text{ mL}}\\
\frac{40 \text{ mL}}{1 \text{ mL}} \cdot 2 \text{ mL} \amp = T \text{ mL}\\
80 \text{ mL}.
\end{align*}
