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Mathematics in Trades and Life

Megan Ossiander-Gobeille, Mark Fitch

Section 2.5 Medical Ratios

A common application of ratios in medicine is creating drugs of a desired strength. For example some drugs need to be administered based on the body weight of he patient. This requires the medical personnel to mix the drug they have on hand to the needed strength.
We will work the following three types of medical problems.
  • Measure drug concentration
  • Dilute a drug to a lower concentration
  • Determine how much drug to use

Subsection 2.5.1 Terminology

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.
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 total amount into aliquots for each test to be run.
The most important concept is measuring how concentrated a solution is. There are three ways this information can be written.
  • Ratio of concentrated drug to diluent, e.g., 1:4
  • Ratio of concentrated drug to total, e.g., 1:5
  • Dilution Factor, e.g., 5
  • Percent concentration, e.g., 20%

Example 2.5.1.

Suppose we add 500 mL of water to 100 mL of a concentrated solution of hydrochloric acid.
  • The ratio of solution to diluent is \(\frac{100}{500}\)
  • The ratio of solution to total is \(\frac{100}{600}\)
  • The dilution factor is \(\frac{600}{100} = 6\)
  • The percent concentration is \(\frac{100}{600} \cdot 100 = 16.7\%\)
Note this is about diluting the solution, not about determining the concentration of acid in diluent. We need more information for that.

Subsection 2.5.2 Making Solutions

A solution of a desired concentration can be produced to have a desired percent concentration. Note be careful in each problem to determine if this is by weight or by volume. The result can be significantly different.

Example 2.5.2.

A standard concentration of HCl (hydrochloric acid) is 37% by weight. The solvent is water. How much water in volume do we need to add to 37 g of HCl to make this solution?
Solution.
Because the concentration is given as a percent we can calculate this like Example 2.1.5.
\begin{align*} 37 \; \text{g} \amp = 0.37 S \; \text{g}.\\ \frac{37 \; \text{g}}{0.37} \amp = \frac{0.37 S \; \text{g}}{0.37}.\\ 100 \; \text{g} \amp = S. \end{align*}
We have a total of 100 g and 37 g is HCl, thus the water is
\begin{equation*} 100 \; \text{g} - 37 \; \text{g} = 63 \; \text{g} \end{equation*}
of water. 1 g of water at standard conditions has a volume of 1 mL. This is now a unit conversion problem.
\begin{equation*} 63 g \cdot \frac{1 \; \text{mL}}{1 \; \text{g}} = 63 \; \text{mL}\text{.} \end{equation*}
Solutions are sometimes diluted by a specified amount to produce a less concentrated version. First, we will practice calculating dilution factors.

Example 2.5.3.

A solution is produced from 3 mL of concentrated chloroform and 37 mL of water. What is the dilution factor?
Solution.
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 \(\frac{4\bar{0}}{3} \approx 13\text{.}\)

Example 2.5.4.

To produce a solution 3.0 mL of concentrated chloroform that will have a dilution factor of \(5\bar{0}\text{,}\) how much diluent do we add?
Solution.
The dilution factor is the ratio of the total to the substance. The total is substance plus diluent. In this case we have
\begin{align*} \frac{3.0+D}{3.0} \amp = \frac{5\bar{0}}{1.0}.\\ 3.0 \cdot \frac{3.0+D}{3.0} \amp = 3.0 \cdot \frac{5\bar{0}}{1.0}.\\ 3.0+D \amp = 150.\\ -3.0+3.0+D \amp = -3.0+150.\\ D \amp = 147. \end{align*}
So we need 147 mL of diluent.

Subsection 2.5.3 Dilution

A dilution ratio tells us how much an existing solution has been diluted. Dilution ratios are used like percents or applying ratios.

Example 2.5.5.

If we have a solution of HCl and water that has a ratio of solution to total of 37/100, and we apply a dilution ratio of 1/2, what is the new ratio of solution to total?
\begin{equation*} \frac{37}{100} \cdot \frac{1}{2} = \frac{37}{200} \end{equation*}
Note this means there are 37 g of HCl and \(200-37=163\) grams of water.

Example 2.5.6.

If we know the original ratio of solution to diluent and the resulting ratio of solution to diluent, then we can calculate the dilution ratio.
Suppose the original solution of HCl and water had a ratio of solution to total of 37/100 and the resulting solution had a ratio of 74/300. What was the dilution ratio? We setup the same calculation as in Example 2.5.5 but instead of 1/2 we have the unknown ratio.
\begin{align*} \frac{37}{100} r \amp = \frac{74}{300}\\ \frac{100}{37} \cdot \frac{37}{100} r \amp = \frac{100}{37} \cdot \frac{74}{300}\\ r \amp = \frac{2}{3}. \end{align*}
The dilution ratio was 2/3.
One usage of dilution is to reduce the concentration so that instruments can accurately measure it. Consider trying to measure an acid without disolving the tools used to measure it.

Example 2.5.7.

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 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.
Solution.
Before we jump into an equation, let’s try an experiment. 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. 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 of the diluted serum would be
\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}} \amp \; \text{clearing the denominators}\\ (C \; \text{mg})(\text{dL}) \amp = (70 \; \text{mg})(10 \; \text{dL})\\ \frac{(C \; \text{mg})(\text{dL})}{\text{dL}} \amp = \frac{(70 \; \text{mg})(10 \; \text{dL})}{\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 2.3.5
\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}} \amp \; \text{clearing the denominators}\\ (2C \; \text{mg})(\text{dL}) \amp = (70 \; \text{mg})(20 \; \text{dL})\\ \frac{(2C \; \text{mg})(\text{dL})}{2 \text{dL}} \amp = \frac{(70 \; \text{mg})(20 \; \text{dL})}{2 \text{dL}}\\ C \amp = 700 \; \text{mg} \end{align*}
Notice the result is the same. This makes sense, because we are setting up a proportion, and ratios do not depend on the amount.
Sometimes we dilute more than one time. Here we experiment to determine what the effect of serial dilution is upon the dilution factor.

Example 2.5.8.

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 2.4.3.
\begin{align*} \frac{1}{2} \amp = \frac{100 \; \text{mL}}{T \; \text{mL}}\\ 1 \cdot (T \; \text{mL}) \amp = 2 \cdot (100 \; \text{mL})\\ T \amp = 200 \; \text{mL}. \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). We set this up like Example 2.5.6
\begin{align*} \frac{1}{10} \cdot F \amp = \frac{1}{60} \amp \; \text{clear the denominator on the left}\\ 10 \cdot \frac{1}{10} \cdot F \amp = 10 \cdot \frac{1}{60}.\\ F \amp = \frac{1}{6}. \end{align*}
Notice that \(\frac{1}{6} = \frac{1}{2} \cdot \frac{1}{3}\text{.}\) 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.
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 2.5.9.

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?
Solution.
This is a unit conversion problem. We need to convert from IU to mL. We know the ratio is 100 IU/mL. Because we want to end up with mL, we will write the ratio as
\begin{align*} \frac{\text{mL}}{100 \; \text{IU}} \amp = \frac{v \; \text{mL}}{2 \; \text{IU}}\\ (1 \; \text{mL})(2 \; \text{IU}) \amp = (100 \; \text{IU})(v \; \text{mL}) \amp \text{clear denominators}\\ \frac{(1 \; \text{mL})(2 \; \text{IU})}{(100 \; \text{IU})} \amp = \frac{(100 \; \text{IU})(v \; \text{mL})}{(100 \; \text{IU})} \amp \; \text{divide to isolate the variable} \\ \frac{1}{50} \; \text{mL} \amp = v.\\ 0.02 \; \text{mL} \amp = v. \end{align*}

Example 2.5.10.

A label reads “2.5 mL of solution for injection contains 1000 mg of streptomycin sulfate.” How many millilitres are needed to contain 800 mg of streptomycin?
Solution.
The label gives us a ratio of 1000 mg per 2.5 mL. We can find the desired volume by setting up the following proportion.
\begin{align*} \frac{1000 \; \text{mg}}{2.5 \; \text{mL}} & = \frac{800 \; \text{mg}}{v \; \text{mL}}. \amp \; \text{clear the denominators}\\ 1000 \; \text{mg} \cdot v \; \text{mL} & = 800 \; \text{mg} \cdot 2.5 \; \text{mL}. \amp \; \text{divide to isolate the variable}\\ v \; \text{mL} & = \frac{800 \; \text{mg} \cdot 2.5 \; \text{mL}}{1000 \; \text{mg}}.\\ v & = 2 \; \text{mL}. \end{align*}

Example 2.5.11.

Give 1500 mL of saline solution IV with a drop factor of 10 drops per mL as a rate of 50 drops per minute to an adult patient. Determine how long in hours the IV should be administered.
Solution.
This is also a unit conversion problem like Example 2.3.12. We need to convert from drops per minute to drops per mL.
\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, we can determine how long it will take to give the whole IV solution. Notice how we invert the rate to make the mL units match.
\begin{equation*} 1500 \; \text{mL} \cdot \frac{\text{minute}}{5 \; \text{mL}} = 300 \; \text{minutes} \end{equation*}
The final step is to convert minutes to hours. This is another unit conversion problem using a conversion from Table 1.1.2
\begin{equation*} 300 \; \text{minutes} \cdot \frac{1 \; \text{hr}}{60 \; \text{min}} = 5 \; \text{hours} \end{equation*}

Example 2.5.12.

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 \(15\bar{0}\) 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 \(15\bar{0}\) 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{15\bar{0} \; \text{mg}}{A \; \text{mL}} \amp \; \text{clear the denominators}\\ (125 \; \text{mg})(A \; \text{mL}) \amp = (5 \; \text{mL})(15\bar{0} \; \text{mg}) \amp \; \text{divide to isolate the variable}\\ \frac{(125 \; \text{mg})(A \; \text{mL})}{125 \; \text{mg}} \amp = \frac{(5 \; \text{mL})(15\bar{0} \; \text{mg})}{(125 \; \text{mg})}\\ A \; \text{mL} \amp = \frac{(5 \; \text{mL})(15\bar{0} \; \text{mg})}{(125 \; \text{mg})}\\ A \; \text{mL} \amp = 6. \end{align*}

Checkpoint 2.5.13.

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?
Solution.
Once again we need to scale the amount (from 100 mL to 500 mL). This is also 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{5 \; \text{g}}{100 \; \text{mL}} \amp = \frac{D \; \text{g}}{500 \; \text{mL}} \amp \; \text{clear the denominators}\\ (5 \; \text{g})(500 \; \text{mL}) \amp = (D \; \text{g})(100 \; \text{mL}) \amp \; \text{divide to isolate the variable}\\ \frac{(5 \; \text{g})(500 \; \text{mL})}{(100 \; \text{mL})} \amp = \frac{(D \; \text{g})(100 \; \text{mL})}{(100 \; \text{mL})}\\ 25 \; \text{g} \amp = D \end{align*}

Example 2.5.14.

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}} \amp \; \text{clear the denominators}\\ (1 \; \text{mL})(T \; \text{mL}) \amp = (2 \; \text{mL})(40 \; \text{mL})\\ \frac{(1 \; \text{mL})(T \; \text{mL})}{1 \; \text{mL}} \amp = \frac{(2 \; \text{mL})(40 \; \text{mL})}{1 \; \text{mL}}\\ T \amp = 80 \; \text{mL}. \end{align*}

Exercises 2.5.5 Exercises

1. Medical Ratio.

2. Medical Ratio.

3. Medical Ratio.

4. Medical Proportion.

5. Medicine to Solution.

6. Medical Ratio with Rounding.

7. Concentration.

8. Concentration.

9. Concentration.

10. Dilution Ratio.

11. Dilution Ratio.

12. Dilution Ratio.

13. Serial Dilution.

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