1.7: Converting Units
 Page ID
 83044
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Skills to Develop
 To convert a value reported in one unit to a corresponding value in a different unit.
The ability to convert from one unit to another is an important skill. For example, a nurse with 50 mg aspirin tablets who must administer 0.2 g of aspirin to a patient needs to know that 0.2 g equals 200 mg, so 4 tablets are needed. Fortunately, there is a simple way to convert from one unit to another.
Conversion Factors
If you review the SI units described earlier, you can find that one kilogram (kg) is about equal to 2.20 pounds (lbs).
\[1\; \rm{kg} = 2.20\; \rm{lbs}\]
Suppose we divide both sides of the equation by 1 kg (both the number and the unit):
\[\mathrm{\dfrac{1\:kg}{1\:kg}=\dfrac{2.20\:lbs}{1\:kg}}\]
As long as we perform the same operation on both sides of the equals sign, the expression remains an equality. Look at the left side of the equation; it now has the same quantity in the numerator as it has in the denominator. Any fraction that has the same quantity on the top and on the bottom has a value of 1:
\[\mathrm{\dfrac{1\:kg}{1\:kg}= 1 \; \text{ and } \; \dfrac{2.20\:lbs}{1\:kg}}=1\]
We know that one kilogram is equal to 2.20 pounds, so we have the same quantity on the top and the bottom of our fraction, although it is expressed in different units. A fraction that has equivalent quantities in the numerator and the denominator but expressed in different units is called a conversion factor.
Instead of dividing both sides by 1 kg, as above, we could have obtained a conversion factor if we divided both sides by 2.20 kg.
\[1\; \rm{kg} = 2.20\; \rm{lbs}\]
will become
\[\mathrm{\dfrac{1\:kg}{2.20\:lbs}=\dfrac{2.20\:lbs}{2.20\:lbs}}\]
Again, because we performed the same operation on both sides of the equals sign, the expression remains an equality (=1).
\[\mathrm{\dfrac{1\:kg}{2.20\:lbs}= 1 \; \text{ and } \; \dfrac{2.20\:lbs}{2.20\:lbs}}=1\]
This means that every conversion factor will have two forms, both equal to one, but reciprocated (flipped). For 1 kg = 2.20 lbs, the two conversion factors would be:
\[ \dfrac{1\; \rm{kg}}{2.20\; \rm{lbs}} \; \text{or} \; \dfrac{2.20 \; \rm{lbs}}{1\; \rm{kg}}\]
Here is a simple example of how the conversion factor might be used. You are about to return from a European vacation with a suitcase full of gifts. The airline does not allow bags heavier than 55 pounds, but the scale in your hotel only reads in kilograms. How many kg are equal to 55 lbs? To solve the problem with the conversion factor used earlier, we first write the quantity we are given, 55 lbs. Then we multiply this quantity by a conversion factor, which is the same as multiplying it by 1. We can write 1 as either version of the conversion factor, \(\mathrm{\frac{1\:kg}{2.20\:lbs}}\) or \(\mathrm{\frac{2.20\:lbs}{1\:kg}}\). Here we will use \(\mathrm{\frac{1\:kg}{2.20\:lbs}}\) and multiply so that the units we started with (kg) cancel and we end with the units we want (lbs).
\[55 \; \rm{lbs} \times \dfrac{1 \; \rm{kg}}{2.20\; \rm{lbs}}\]
The 55 kg can be thought of as a fraction with a 1 in the denominator (though it is often not shown). Because kg, the abbreviation for kilograms, occurs in both the numerator (top) and the denominator (bottom) of our expression, they cancel out. The 55 on top is divided by 2.20 on bottom and the only units that did not cancel are kg.
\[\dfrac{55 \; \cancel{\rm{lbs}}}{ 1} \times \dfrac{1 \; \rm{kg}}{2.20 \; \cancel{\rm{lbs}}}= 25 \; \rm{kg}\]
In the final answer, we omit the 1 in the denominator. We find that 55 lbs equals 25 kg. A generalized description of this process is as follows:
\[\text{quantity (in old units)} \times \text{conversion factor} = \text{quantity (in new units)} \nonumber\]
This involved conversion only involved one step. We will see that the same basic principle applies for conversions with multiple steps. If you can master the technique of applying conversion factors, you will be able to solve a large variety of problems.
In the previous example, if we had used the other possible reciprocated conversion factor, \(\mathrm{\frac{2.20\:lbs}{1\:kg}}\), the original unit would not have canceled, and the result would have been meaningless. Here is the incorrect we would have gotten:
\[55 \; \rm{lbs} \times \dfrac{2.20\; \rm{lbs}}{1 \; \rm{kg}} = 0.0355 \dfrac{\rm{lbs}^2}{\rm{kg}}\]
For the answer to be meaningful, we have to construct the conversion factor in a form that causes the original unit to cancel out.
Conversion Factors With Prefix Multipliers (Exponents)
If you learned the prefixes described earlier in this chapter, then you know that the prefix centi (c) is 10^{−2}. The prefixes come in front of some other base unit, like meters (m). The base units must be included on both sides of the equality, so if c = 10^{−2}, then cm = 10^{−2 }m. As before, this can be made into these two different reciprocal conversion factors:
\[ \dfrac{1\; \rm{cm}}{10^{2} \; \rm{m}} \; \text{or} \; \dfrac{10^{2} \; \rm{m}}{1\; \rm{cm}}\]
If you wanted to find out how many centimeters in 3.55 meters, we again follow the process:
\[\text{quantity (in old units)} \times \text{conversion factor} = \text{quantity (in new units)} \nonumber\]
\[\dfrac{3.55 \; \cancel{\rm{m}}}{ 1} \times \dfrac{1 \; \rm{cm}}{10^{2} \; \cancel{\rm{m}}}= 355 \; \rm{cm}\]
Pull out your calculator and ensure that you get 355 as your answer when you divide 3.55 by 10^{−2}. If not, get help with how to properly use your calculator. (For example, in the Casio fx260, 10^{−2 } might be entered as [1], then [EXP], [+/], and [2].)
To avoid using exponents, some instead use that 1 cm is 1/100th of a meter (because centi = 10^{−2} = 1/10^{2 }= 1/100).
\[ 1\; \rm{cm} = \dfrac{1}{100} \; \rm{m}\]
Since fractions of a unit can make conversions difficult later, both sides can be multiplied by 100 to cancel the 1/100 and get this version.
\[100\; \rm{cm} = 1\; \rm{m}\]
The two versions of the conversion factor are therefore:
\[ \dfrac{100 \; \rm{cm}}{1 \; \rm{m}} \; \text{or} \; \dfrac{1 \; \rm{m}}{100\; \rm{cm}}\]
To convert 3.55 meters into centimeters this way, we first write the quantity we are given, 3.55 m. Then we multiply this quantity by a conversion factor, which is the same as multiplying it by 1. We can write 1 as \(\mathrm{\frac{100\:cm}{1\:m}}\) and multiply:
\[\dfrac{3.55 \; \cancel{\rm{m}}}{ 1} \times \dfrac{100 \; \rm{cm}}{1 \; \cancel{\rm{m}}}= 355 \; \rm{cm}\]
Notice that we get the same answer, 355 cm, either way. Using the exponent is most direct and makes things easier with large exponents, like micro = 10^{−6} or giga = 10^{9}. But for those who have trouble with exponential notation, using micro = 1/1,000,000 or giga = 1,000,000,000 is also an option.
Significant Figures in Conversions
How do conversion factors affect the determination of significant figures? Numbers in conversion factors based on prefix changes, such as kilograms to grams, are not considered in the determination of significant figures in a calculation because the numbers in such conversion factors are exact. Exact numbers are defined or counted numbers, not measured numbers, and can be considered as having an infinite number of significant figures. (In other words, 1 kg is exactly 1,000 g, by the definition of kilo.) Counted numbers are also exact. If there are 16 students in a classroom, the number 16 is exact. In contrast, conversions that do not involve just a prefix change, like those between different systems of units, are usually not exact. For example, kilograms are S.I. units, but pounds come from an older system. If given 1 kg = 2.20 lbs, you sould assume three significant figures (but can be found to six or more figures, 1 kg = 2.20462 lbs). Conversion factors that come from measurements (such as density, as we will see shortly) or are approximations have a limited number of significant figures and should be considered in determining the significant figures of the final answer.
Example \(\PageIndex{1}\)
 The average volume of blood in an adult male is 4.7 L. What is this volume in milliliters?
 A hummingbird can flap its wings once in 18 ms. How many seconds are in 18 ms?
SOLUTION

We start with what we are given, 4.7 L. We want to change the unit from liters to milliliters. The prefix milli is 10^{−3}. If m = 10^{−3}, adding the liters to both sides, makes it 1 mL = 10^{−3 }L. From this relationship, we can construct two conversion factors:
\[ \dfrac{10^{3} \; \rm{L}}{1 \; \rm{mL}} \; \text{or} \; \dfrac{1 \; \rm{mL}}{10^{3} \; \rm{L}}\]
We use the conversion factor that will cancel out the original unit, liters, and introduce the unit we are converting to, which is milliliters. The conversion factor that does this is the one on the right.
\[ 4.7 \cancel{\rm{L}} \times \dfrac{1 \; \rm{mL}}{10^{3} \; \cancel{\rm{L}}} = 4,700\; \rm{mL}\]
Because the numbers in the conversion factor are exact, we do not consider them when determining the number of significant figures in the final answer. Thus, we report two significant figures in the final answer.

We can construct two conversion factors from the relationships between milliseconds and seconds, 1 ms = 10^{−3 }s:
\[ \dfrac{1 \; \rm{ms}}{10^{3} \; \rm{s}} \; \text{or} \; \dfrac{10^{3} \; \rm{s}}{1 \; \rm{ms}}\]
To convert 18 ms to seconds, we choose the conversion factor that will cancel out milliseconds and introduce seconds. The conversion factor on the right is the appropriate one. We set up the conversion as follows:
\[ 18 \; \cancel{\rm{ms}} \times \dfrac{10^{3} \; \rm{s}}{1 \; \cancel{\rm{ms}}} = 0.018\; \rm{s}\]
The conversion factor’s numerical values do not affect our determination of the number of significant figures in the final answer.
(If you prefer the method without exponents, you could instead use that milli is 1/1,000. If you do it that way, ensure that you get the same answers.)
Example \(\PageIndex{2}\)
Perform each conversion.
 101,000 ns to seconds
 32.08 kg to grams
SOLUTION

The prefix nano is 10^{−9}. If n = 10^{−9}, adding the seconds to both sides, makes it 1 ns = 10^{−9 }s. From this relationship, we can construct two conversion factors:
\[ \dfrac{10^{9} \; \rm{s}}{1 \; \rm{ns}} \; \text{or} \; \dfrac{1 \; \rm{ns}}{10^{9} \; \rm{s}}\]
We use the conversion factor that will cancel out the original unit, nanoseconds, and introduce the unit we are converting to, which is seconds. The conversion factor that does this is the one on the left.
\[ 101,000 \cancel{\rm{ns}} \times \dfrac{10^{9} \; \rm{s}}{1 \; \cancel{\rm{ns}}} = 0.000101 \; \rm{s}\]

We can construct two conversion factors from the relationships between kilograms and grams, 1 kg = 10^{3 }g:
\[ \dfrac{1 \; \rm{kg}}{10^3 \; \rm{g}} \; \text{or} \; \dfrac{10^3 \; \rm{g}}{1 \; \rm{kg}}\]
To convert 32.08 kg to grams, we choose the conversion factor that will cancel out kilograms and introduce grams. The conversion factor on the left is the appropriate one. We set up the conversion as follows:
\[ 32.08 \cancel{\rm{kg}} \times \dfrac{10^{3} \; \rm{g}}{1 \; \cancel{\rm{kg}}} = 32080 \; \rm{g}\]
To more clearly show that the answer has four significant figures, it can be written in scientific notation as 3.208x10^{4 }g.
Conversion factors can also be constructed for converting between different kinds of units. For example, density can be used to convert between the mass and the volume of a substance. Consider mercury, which is a liquid at room temperature and has a density of 13.6 g/mL. The density tells us that 13.6 g of mercury have a volume of 1 mL. We can write that relationship as follows:
13.6 g mercury = 1 mL mercury
This relationship can be used to construct two conversion factors:
\[\mathrm{\dfrac{13.6\:g}{1\:mL}\:and\:\dfrac{1\:mL}{13.6\:g}}\]
Which one do we use? It depends, as usual, on the units we need to cancel and introduce. For example, suppose we want to know the mass of 16 mL of mercury. We would use the conversion factor that has milliliters on the bottom (so that the milliliter unit cancels) and grams on top so that our final answer has a unit of mass:
\[\mathrm{16\:\cancel{mL}\times\dfrac{13.6\:g}{1\:\cancel{mL}}=217.6\:g=220\:g}\]
In the last step, we limit our final answer to two significant figures because the volume quantity has only two significant figures; the 1 in the volume unit is considered an exact number, so it does not affect the number of significant figures. The other conversion factor would be useful if we were given a mass and asked to find volume, as the following example illustrates.
Density can be used as a conversion factor between mass and volume.
Example \(\PageIndex{3}\): Mercury Thermometer
A mercury thermometer for measuring a patient’s temperature contains 0.750 g of mercury. What is the volume of this mass of mercury?
SOLUTION
Because we are starting with grams, we want to use the conversion factor that has grams in the denominator (bottom). The gram unit will cancel algebraically, and milliliters will be introduced in the numerator.
\[ 0.750 \; \cancel{\rm{g}} \times \dfrac{1\; \rm{mL}}{13.6 \; \cancel{\rm{g}}} = 0.055147 ... \; \rm{mL} \approx 0.0551\; \rm{mL}\]
We have limited the final answer to three significant figures.
You could instead rearrange the density formula, d = m/V, to solve for volume, V = m/d = (0.750g) / (13.6 g/mL) = 0.0551 mL, to arrive at the same answer.
Looking Closer: Density and the Body
The densities of many components and products of the body have a bearing on our health.
Bones. Bone density is important because bone tissue of lowerthannormal density is mechanically weaker and susceptible to breaking. The density of bone is, in part, related to the amount of calcium in one’s diet; people who have a diet deficient in calcium, which is an important component of bones, tend to have weaker bones. Dietary supplements or adding dairy products to the diet seems to help strengthen bones. As a group, women experience a decrease in bone density as they age. It has been estimated that fully half of women over age 50 suffer from excessive bone loss, a condition known as osteoporosis. Exact bone densities vary within the body, but for a healthy 30yearold female, it is about 0.95–1.05 g/cm^{3}. Osteoporosis is diagnosed if the bone density is below 0.6–0.7 g/cm^{3}.
Urine. The density of urine can be affected by a variety of medical conditions. Sufferers of diabetes produce an abnormally large volume of urine with a relatively low density. In another form of diabetes, called diabetes mellitus, there is excess glucose dissolved in the urine, so that the density of urine is abnormally high. The density of urine may also be abnormally high because of excess protein in the urine, which can be caused by congestive heart failure or certain renal (kidney) problems. Thus, a urine density test can provide clues to various kinds of health problems. The density of urine is commonly expressed as a specific gravity, which is a unitless quantity defined as
\[ \dfrac{\text{density of some material}}{\text{density of water}}\]
Normal values for the specific gravity of urine range from 1.002 to 1.028.
Body Fat. The overall density of the body is one indicator of a person’s total body fat. Fat is less dense than muscle and other tissues, so as it accumulates, the overall density of the body decreases. Measurements of a person’s weight and volume provide the overall body density, which can then be correlated to the percentage of body fat. (The body’s volume can be measured by immersion in a large tank of water. The amount of water displaced is equal to the volume of the body.)
Multiple Conversions
Sometimes you will have to perform more than one conversion to obtain the desired unit. For example, suppose you want to convert 54.7 km into millimeters. You can either memorize the relationship between kilometers and millimeters, or you can do the conversion in steps. Most people prefer to convert in steps.
To do a stepwise conversion, we first convert the given amount to the base unit. In this example, the base unit is meters. We know that there are 10^{3} m in 1 km:
\[ 54.7\; \cancel{\rm{km}} \times \dfrac{10^3 \; \rm{m}}{1\; \cancel{\rm{km}}} = 54,700\; \rm{m}\]
Then we take the result (54,700 m) and convert it to millimeters, remembering that there are \(1\; \rm{mm}\) for every \(10^{3}\; \rm{m}\):
\[ 54,700 \; \cancel{\rm{m}} \times \dfrac{1 \; \rm{mm}}{10^{3} \; \cancel{\rm{m}}} = 54,700,000 \; \rm{mm} = 5.47 \times 10^7\; \rm{mm}\]
Expressing the final answer in scientific notation more clearly shows the three significant figures.
As a shortcut, both steps in the conversion can be combined into a single, multistep expression:
\[ 54.7\; \cancel{\rm{km}} \times \dfrac{10^3 \; \cancel{\rm{m}}}{1\; \cancel{\rm{km}}} \times \dfrac{1 \; \rm{mm}}{10^{3} \; \cancel{\rm{m}}} = 54,700,000 \; \rm{mm} = 5.47 \times 10^7\; \rm{mm}\]
NOTE: When using the prefixes and exponents, the prefix should always be on one side and the exponent that it equals should be on the other side. (In the first step here, since kilo is on bottom, the 10^{3} must be on top. Then in the other step, because milli is on the top, the 10^{3 }should be on bottom.)
Either method—one step at a time or all the steps together—is acceptable. If you do all the steps together, the restriction for the proper number of significant figures should be done after the last step. As long as the math is performed correctly, you should get the same answer no matter which method you use.
Example \(\PageIndex{4}\)
Convert 58.2 ms to megaseconds in one multistep calculation.
SOLUTION
The milli prefix abbreviatioin is lower case, m = 10^{3}, so with base units of seconds it would be, ms = 10^{3} s. The mega abbreviation is upper case, M = 10^{6}, so Ms = 10^{6 }s.
First, convert the given unit to the base unit—in this case, seconds—and then convert seconds to the final unit, megaseconds.
\[ 58.2 \; \cancel{\rm{ms}} \times \dfrac{\cancel{10^{3} \rm{s}}}{1 \; \cancel{\rm{ms}}} \times \dfrac{1 \; \rm{Ms}}{10^6 \; \cancel{ \rm{s}}} =0.0000000582\; \rm{Ms} = 5.82 \times 10^{8}\; \rm{Ms}\]
Neither conversion factor affects the number of significant figures in the final answer.
Example \(\PageIndex{5}\)
Convert 43.00 ng to kilograms in one multistep calculation.
SOLUTION
For nano, n = 10^{9}, so with base units of grams it would be, ng = 10^{9} g. For kilo, k = 10^{3}, so kg = 10^{3 }g.
First, cancel ng and convert the given unit to the base unit (g), and then convert grams to the final unit, kilograms.
\[ 43.00 \; \cancel{\rm{ng}} \times \dfrac{\cancel{10^{9} \rm{g}}}{1 \; \cancel{\rm{ng}}} \times \dfrac{1 \; \rm{kg}}{10^3 \; \cancel{ \rm{g}}} =0.00000000004300\; \rm{kg} = 4.300 \times 10^{11}\; \rm{kg}\]
Neither conversion factor affects the number of significant figures (four) in the final answer.
Squared or Cubed Units
Area is often found by multiplying two lengths, and thus has squared units. For example length in feet, times width in feet, gives area in square feet, ft^{2}. Volume is often found by multiplying three lengths, and thus has cubed units. For a box, length in centimeters, times width in centimeters, times depth in centimeters, gives volume in cubic centimeters, cm^{3}. A length conversion factor must be used twice for squared area and three times for cubed volume.
Assume we find that the area of notebook page is 93.5 square inches and we want to convert that to square centimeters. Most often, you will only be able to find the relationship between inches and centimeters, 1 in = 2.54 cm. The conversion factor will need to be used twice here (so that both inches in in^{2} cancel and so that the two centimeters on top leave you with cm^{2}).
\[ 93.5 \; \cancel{\rm{in^2}} \times \dfrac{2.54 \;\rm{cm}}{1\; \cancel{\rm{in}}} \times \dfrac{2.54 \; \rm{cm}}{1 \; \cancel{\rm{in}}} = 603 \; \rm{cm^2}\]
Example \(\PageIndex{6}\)
Tire pressure is often measured in pounds per square inch, abbreviated psi or lbs/in^{2}. Convert 32.0 psi to units of kilograms per square centimeter.
SOLUTION
The pressures have weight units (lbs or kg) on top and area units (in^{2} or cm^{2}) on bottom. We will use 1 kg = 2.20 lbs and 1 in = 2.54 cm to create our conversion factors.
\[ \dfrac{\cancel{32.0 \; \rm{lbs}}}{\cancel{\rm{in^2}}} \times \dfrac{1 \rm{kg}}{2.20 \; \cancel{\rm{lbs}}} \times \dfrac{\cancel{1 \; \rm{in}}}{2.54 \; \rm{cm}} \times \dfrac{\cancel{1 \; \rm{in}}}{2.54 \; \rm{cm}} = \dfrac{2.25 \: \rm{kg}}{\rm{cm^2}}\]
Since 32.0 lbs/in^{2} had inches twice on the bottom, \(\mathrm{\frac{1\:in}{2.54\:cm}}\) was used twice. The final answer is rounded to three significant figures.
Career Focus: Pharmacist
A pharmacist dispenses drugs that have been prescribed by a doctor. Although that may sound straightforward, pharmacists in the United States must hold a doctorate in pharmacy and be licensed by the state in which they work. Most pharmacy programs require four years of education in a specialty pharmacy school.
Pharmacists must know a lot of chemistry and biology so they can understand the effects that drugs (which are chemicals, after all) have on the body. Pharmacists can advise physicians on the selection, dosage, interactions, and side effects of drugs. They can also advise patients on the proper use of their medications, including when and how to take specific drugs properly. Pharmacists can be found in drugstores, hospitals, and other medical facilities.
Curiously, an outdated name for pharmacist is chemist, which was used when pharmacists formerly did a lot of drug preparation, or compounding. In modern times, pharmacists rarely compound their own drugs, but their knowledge of the sciences, including chemistry, helps them provide valuable services in support of everyone’s health.
Concept Review Exercises
 How do you determine which quantity in a conversion factor goes in the denominator of the fraction?
 State the guidelines for determining significant figures when using a conversion factor.
Answers
 The unit you want to cancel from the numerator goes in the denominator of the conversion factor.
 Exact numbers that appear in many conversion factors do not affect the number of significant figures; otherwise, the normal rules of multiplication and division for significant figures apply.
Key Takeaway
 A unit can be converted to another unit of the same type with a conversion factor.
There are plenty more conversion exercises to practice in Section 1.E at the end of this chapter.
Contributors
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