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# 3.2: Unit Conversions

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

• Identify and use conversion factors.
• Use the method of dimensional analysis to convert between units.
• Understand density as a physical property of matter.
• Perform calculations with derived units, including density.

## Lesson Vocabulary

• conversion factor
• density
• derived unit
• dimensional analysis

### Recalling Prior Knowledge

• Why are units required when reporting the results of a measured quantity?
• When a quantity with a large unit (such as km) is changed into a quantity with a smaller unit (such as cm), will the numerical value of the quantity increase or decrease?

When traveling in another country, you may be faced with a unit problem. For example, if you are driving, you may encounter a sign saying that the next town is 30 km away. If your car’s odometer measures distances in miles, how far will you need to go to get to that town? In this lesson, you will learn to solve this and other unit-conversion problems with a technique called dimensional analysis.

## Conversion Factors

Many quantities can be expressed in several different ways. For example, the English system measurement of 4 cups is also equal to 2 pints, 1 quart, and 1/4 of a gallon.

\begin{align*}4 \ \text{cups} = 2 \ \text{pints} = 1 \ \text{quart} = 0.25 \ \text{gallon}\end{align*}

Notice that the numerical component of each quantity is different, while the actual amount of material that it represents is the same. That is because the units are different. We can establish the same set of equalities for the metric system:

\begin{align*}1 \ \text{meter} = 10 \ \text{decimeters} = 100 \ \text{centimeters} = 1000 \ \text{millimeters}\end{align*}

The metric system’s use of powers of 10 for all conversions makes this quite simple.

Whenever two quantities are equal, a ratio can be written that is numerically equal to 1. Using the metric examples above:

\begin{align*}\dfrac{1 \ \text{m}}{100 \ \text{cm}}=\dfrac{100 \ \text{cm}}{100 \ \text{cm}}=\dfrac{1 \ \text{m}}{1 \ \text{m}}=1\end{align*}

The fraction 1 m/100 cm is called a conversion factor. A conversion factor is a ratio of equivalent measurements. Because both 1 m and 100 cm represent the exact same length, the value of the conversion factor is 1. The conversion factor is read as “1 meter per 100 centimeters.” Other conversion factors from the cup measurement example can be:

\begin{align*}\dfrac{4 \ \text{cups}}{2 \ \text{pints}}=\dfrac{2 \ \text{pints}}{1 \ \text{quart}}=\dfrac{1 \ \text{quart}}{0.25 \ \text{gallon}}=1\end{align*}

Since the numerator and denominator represent equal quantities in each case, all are valid conversion factors.

## Dimensional Analysis

Conversion factors are used in solving problems in which a certain measurement must be expressed with different units. When a given measurement is multiplied by an appropriate conversion factor, the numerical value changes, but the actual size of the quantity measured remains the same. Dimensional analysis is a technique that uses the units (dimensions) of the measurement in order to correctly solve problems. Dimensional analysis is best illustrated with an example.

Sample Problem 3.1: Dimensional Analysis

How many seconds are in a day?

Step 1: List the known quantities and plan the problem.

Known

• 1 day = 24 hours
• 1 hour = 60 minutes
• 1 minute = 60 seconds

Unknown

• 1 day = ? seconds

The known quantities above represent the conversion factors that we will use. The first conversion factor will have day in the denominator so that the “day” unit will cancel. The second conversion factor will then have hours in the denominator, while the third conversion factor will have minutes in the denominator. As a result, the unit of the last numerator will be seconds and that will be the units for the answer.

Step 2: Calculate

\begin{align*}1 \ \text{d} \times \frac{24 \ \text{h}}{1 \ \text{d}} \times \frac{60 \ \text{min}}{1 \ \text{h}} \times \frac{60 \ \text{s}}{1 \ \text{min}}=86,400 \ \text{s}\end{align*}

Applying the first conversion factor, the “d” unit cancels and 1 × 24 = 24. Applying the second conversion factor, the “h” unit cancels and 24 × 60 = 1440. Applying the third conversion factor, the “min” unit cancels and 1440 × 60 = 86400. The unit that remains is “s” for seconds.

A second is a much smaller unit of time than a day, so it makes sense that there are a very large number of seconds in one day.

Practice Problems
1. How many minutes are in a year?
2. How many days are equal to one million seconds?

### Dimensional Analysis and the Metric System

The metric system’s many prefixes allow quantities to be expressed in many different units. Dimensional analysis is useful to convert from one metric system unit to another.

Sample Problem 3.2: Metric Unit Conversions

A particular experiment requires 120 mL of a solution. The teacher knows that he will need to make enough solution for 40 experiments to be performed throughout the day. How many liters of solution should he prepare?

Step 1: List the known quantities and plan the problem.

Known

• 1 experiment requires 120 mL of solution
• 1 L = 1000 mL

Unknown

• 40 experiments require ? L of solution

Since each experiment requires 120 mL of solution and the teacher needs to prepare enough for 40 experiments, multiply 120 by 40 to get 4800 mL of solution needed. Now you must convert mL to L by using a conversion factor.

Step 2: Calculate

\begin{align*}4800 \ \text{mL} \times \frac{1 \ \text{L}}{1000 \ \text{mL}}=4.8 \ \text{L}\end{align*}

Note that conversion factor is arranged so that the mL unit is in the denominator and thus cancels out, leaving L as the remaining unit in the answer.

A liter is much larger than a milliliter, so it makes sense that the number of liters required is less than the number of milliliters.

Practice Problems
1. Perform the following conversions.
1. 0.074 km to m
2. 24,600 μg to g
3. 1300 ms to s
4. 3.8 × 10-5 L to μL

Some metric conversion problems are most easily solved by breaking them down into more than one step. When both the given unit and the desired unit have prefixes, one can first convert to the simple (unprefixed) unit, followed by a conversion to the desired unit. An example will illustrate this method.

Sample Problem 3.3: Two-Step Metric Conversion

Convert 4.3 cm to μm.

Step 1: List the known quantities and plan the problem.

Known

• 1 m = 100 cm
• 1 m = 106 μm

Unknown

• 4.3 cm = ? μm

You may need to consult the table in the “The International System of Units” lesson for the multiplication factor represented by each metric prefix. First convert cm to m, then convert m to μm.

Step 2: Calculate

\begin{align*}4.3 \ \text{cm} \times \frac{1 \ \text{m}}{100 \ \text{cm}} \times \frac{10^6 \ \mu \text{m}}{1 \ \text{m}}=43,000 \ \mu \text{m}\end{align*}

Each conversion factor is written so that unit of the denominator cancels with the unit of the numerator of the previous factor.

A micrometer is a smaller unit of length than a centimeter, so the answer in micrometers is larger than the number of centimeters given.

Practice Problems
1. Perform the following conversions.
1. 4.9 × 107 μg to kg
2. 84 dm to mm
3. 355 nm to cm
4. 70.5 ML to mL

### Dimensional Analysis and Derived Units

Some units are combinations of SI base units. A derived unit is a unit that results from a mathematical combination of SI base units. We have already discussed volume and energy as two examples of derived units. Some others are listed in Table below.

Derived SI Units
Quantity Symbol Unit Unit Abbreviation Derivation
Area A square meter m2 length × width
Volume V cubic meter m3 length × width × height
Density D kilograms per cubic meter kg/m3 mass / volume
Concentration c moles per liter mol/L amount / volume
Speed (velocity) v meters per second m/s length / time
Acceleration a meters per second per second m/s2 speed / time
Force F newton N mass × acceleration
Energy E joule J force × length

Using dimensional analysis with derived units requires special care. When units are squared or cubed as with area or volume, the conversion factors themselves must also be squared. Shown below is the conversion factor for cubic centimeters and cubic meters.

\begin{align*}\left ( \frac{1 \ \text{m}}{100 \ \text{cm}} \right )^3=\frac{1 \ \text{m}^3}{10^6 \ \text{cm}^3}=1\end{align*}

Because a cube has 3 sides, each side is subject to the conversion of 1 m to 100 cm. Since 100 cubed is equal to 1 million (106), there are 106 cm3 in 1 m3. Two convenient volume units are the liter, which is equal to a cubic decimeter, and the milliliter, which is equal to a cubic centimeter. The conversion factor would be:

\begin{align*}\left ( \frac{1 \ \text{dm}}{10 \ \text{cm}} \right)^3=\frac{1 \ \text{dm}^3}{1000 \ \text{cm}^3}=1\end{align*}

There are thus 1000 cm3 in 1 dm3, which is the same thing as saying there are 1000 mL in 1 L (Figure below).

There are 1000 cm3 in 1 dm3. Since 1 cm3 is equal to 1 mL and 1 dm3 is equal to 1 L, we can say that there are 1000 mL in 1 L.

You can participate in an interactive version of this cube at www.dlt.ncssm.edu/core/Chapter1-Introduction/Chapter1-Animations/M3_DM3_CM3.html.

Sample Problem 3.4: Derived Unit Conversion

Convert 3.6 × 108 mm3 to mL.

Step 1: List the known quantities and plan the problem.

Known

• 1 m = 1000 mm
• 1 mL = 1 cm3
• 1 m = 100 cm

Unknown

• 3.6 mm3 = ? mL

This problem requires multiple steps and the technique for converting with derived units. Simply proceed one step at a time: mm3 to m3 to cm3 = mL.

Step 2: Calculate

\begin{align*}3.6 \ \text{mm}^3 \times \left ( \frac{1 \ \text{m}}{1000 \ \text{mm}} \right )^3 \times \left ( \frac{100 \ \text{cm}}{1 \ \text{m}} \right )^3 \times \frac{1 \ \text{mL}}{1 \ \text{cm}^3}=0.0036 \ \text{mL}\end{align*}

Numerically, the steps are to divide 3.6 by 109, followed by multiplying by 106. Alternatively, you can shorten the calculation by one step if you first determine the conversion factor between mm and cm. Using the fact that there are 10 mm in 1 cm, you can avoid the intermediate step of converting to meters.

\begin{align*}3.6 \ \text{mm}^3 \times \left ( \frac{1 \ \text{cm}}{10 \ \text{mm}} \right )^3 \times \frac{1 \ \text{mL}}{1 \ \text{cm}^3} = 0.0036 \ \text{mL}\end{align*}

Both conversion methods give the same result of 0.0036 mL.

Cubic millimeters are much smaller than cubic centimeters, so the final answer is much less than the original number of mm3.

Practice Problems
1. Perform the following conversions.
1. 0.00722 km2 to m2
2. 129 cm3 to L
3. 4.9 × 105 μm3 to mm3

You can find more help with dimensional analysis by watching this video at www.chemcollective.org/stoich/dimensionalanalysis.php.

You can download an instructional packet that explains dimensional analysis step by step and provides practice problems at https://docs.google.com/open?id=0B_ZuEGrhVEfMS09WeUtOMFNTaFk.

## Density

A golf ball and a table tennis ball are about the same size. However, the golf ball is much heavier than the table tennis ball. Now imagine a similar size ball made out of lead. That would be very heavy indeed! What are we comparing? By comparing the mass of an object relative to its size, we are studying a property called density. Density is the ratio of the mass of an object to its volume.

\begin{align*}\text{Density} = \dfrac{\text{mass}}{\text{volume}} \ \ \ \ \text{or} \ \ \ \ \text{D}=\dfrac{\text{m}}{\text{V}}\end{align*}

Density is an intensive property, meaning that it does not depend on the amount of material present in the sample. For example, water has a density of 1.0 g/mL. That density is the same whether you have a small glass of water or a swimming pool full of water. Density is a property that is constant for a specific type of matter at a given temperature.

The SI units of density are kilograms per cubic meter (kg/m3), because the kg and the m are the SI units for mass and length, respectively. Unfortunately, this unit is awkwardly large for everyday usage in the laboratory. Most solids and liquids have densities that are conveniently expressed in grams per cubic centimeter (g/cm3). Since a cubic centimeter is equal to a milliliter, density units can also be expressed as g/mL. Gases are much less dense than solids and liquids, so their densities are often reported in g/L. Densities of some common substances at 20°C are listed in Table below. Since most materials expand as temperature increases, the density of a substance is temperature-dependent and usually decreases as temperature increases.

Densities of Some Common Substances
Liquids and Solids Density at 20°C (g/mL) Gases Density at 20°C (g/L)
Ethanol 0.79 Hydrogen 0.084
Ice (0°C) 0.917 Helium 0.166
Corn oil 0.922 Air 1.20
Water 0.998 Oxygen 1.33
Water (4°C) 1.000 Carbon dioxide 1.83
Aluminum 2.70
Copper 8.92
Mercury 13.6
Gold 19.3

You know from experience that ice floats in water. This is consistent with the values in Table above, which show that ice is less dense than liquid water. Alternatively, corn syrup would sink if placed into water, because it has a higher density. Figure below shows a number of substances arranged into a density column.

A density column made of some familiar materials.

With gases, a balloon filled with helium floats because helium is less dense than air. However, a balloon filled with carbon dioxide sinks because carbon dioxide is denser than air.

Sample Problem 3.5: Density Calculations

An 18.2 g sample of zinc metal has a volume of 2.55 cm3. Calculate the density of zinc.

Step 1: List the known quantities and plan the problem.

Known

• mass = 18.2 g
• volume = 2.55 cm3

Unknown

• density = ? g/cm3

Use the equation for density, D = m/V, to solve the problem.

Step 2: Calculate

\begin{align*}\text{D}=\dfrac{\text{m}}{\text{V}}=\dfrac{18.2 \ \text{g}}{2.55 \ \text{cm}^3}=7.14 \ \text{g/cm}^3\end{align*}

If 1 cm3 of zinc has a mass of about 7 grams, then a sample that is approximately 2.5 cm3 will have a mass between 2 and 3 times its density, which is consistent with the values given in this problem. Additionally, metals are expected to have a density that is greater than that of water, and zinc’s density falls within the range of the other metals listed in Table above.

Since density values are known for many substances, density can be used to determine an unknown mass or an unknown volume. Dimensional analysis will be used to ensure that units cancel appropriately.

Sample Problem 3.6: Using Density to Determine Mass and Volume

1. What is the mass of 2.49 cm3 of aluminum?
2. What is the volume of 50.0 g of aluminum?

Step 1: List the known quantities and plan the problem.

Known

• density = 2.70 g/cm3
• 1. volume = 2.49 cm3
• 2. mass = 50.0 g

Unknown

• 1. mass = ? g
• 2. volume = ? cm3

Use the equation for density, D = m/V, and dimensional analysis to solve each problem.

Step 2: Calculate

1. \begin{align*}2.49 \ \text{cm}^3 \times \dfrac{2.70 \ \text{g}}{1 \ \text{cm}^3} = 6.72 \ \text{g}\end{align*}
2. \begin{align*}50.0 \ \text{g} \times \dfrac {1 \ \text{cm}^3}{2.70 \ \text{g}} = 18.5 \ \text{cm}^3\end{align*}

In problem 1, the mass is equal to the density multiplied by the volume. In problem 2, the volume is equal to the mass divided by the density.

Because a 1 cm3 sample of aluminum has a mass of 2.70 g, the mass of a 2.49 cm3 sample should be greater than 2.70 g. The mass of a 50-g block of aluminum is substantially more than the value of its density in g/cm3, so that amount should occupy a volume that is significantly larger than 1 cm3.

Practice Problems
1. A student finds the mass of a “gold” ring to be 41.7 g and its volume to be 3.29 cm3. Calculate the density of the ring. Is it pure gold? (See Table above.)
2. What is the mass of 125 L of oxygen gas?
3. The density of silver is 10.5 g/cm3. What is the volume of a 13.4 g silver coin?

Finally, conversion problems involving density or other derived units like speed or concentration may involve a separate conversion of each unit. To convert a density in g/cm3 to kg/m3, two steps must be used. One step converts g to kg, while the other converts cm3 to m3. They may be performed in either order.

Sample Problem 3.7: Density Conversions

The average density of the planet Jupiter is 1.33 g/cm3. What is Jupiter’s density in kg/m3?

Step 1: List the known quantities and plan the problem.

Known

• density = 1.33 g/cm3
• 1 kg = 1000 g
• 1 m = 100 cm

Unknown

• density = ? kg/m3

Use separate conversion factors to convert the mass from g to kg and the volume from cm3 to m3.

Step 2: Calculate

\begin{align*}\dfrac{1.33 \ \text{g}}{\text{cm}^3} \times \dfrac{1 \ \text{kg}}{1000 \ \text{g}} \times \left ( \dfrac{100 \ \text{cm}}{1 \ \text{m}} \right )^3 = 1330 \ \text{kg/m}^3\end{align*}

Since a cubic meter is so much larger (1 million times) than a cubic centimeter, the density of Jupiter is larger in kg/m3 than in g/cm3.

You can perform a density experiment to identify a mystery object online. Find this simulation at http://phet.colorado.edu/en/simulation/density.

## Lesson Summary

• Conversion factors are ratios of equivalent quantities expressed in different units. When multiplying by a conversion factor, the numerical value and the unit changes while the actual size of the quantity remains the same.
• Dimensional analysis employs conversion factors to solve problems in which the units are changing. Dimensional analysis can be used to solve metric system conversion problems.
• Density is a derived unit of mass per unit volume and is a physical property of a substance. Density problems can be solved using dimensional analysis.

## Lesson Review Questions

### Reviewing Concepts

1. What must be true for a ratio of two measurements to be a conversion factor?
2. Which of the following ratios qualify as conversion factors? For the ones that do not, explain why.
1. \begin{align*}\dfrac{10 \ \text{pennies}}{1 \ \text{dime}}\end{align*}
2. \begin{align*}\dfrac{3 \ \text{dogs}}{\text{several}}\end{align*}
3. \begin{align*}\dfrac{1 \ \text{hour}}{60 \ \text{seconds}}\end{align*}
4. \begin{align*}\dfrac{1 \ \text{dozen donuts}}{12 \ \text{donuts}}\end{align*}
3. How do you decide which unit should go in the denominator of a conversion factor?
4. What is a derived unit?
5. Explain what is wrong with this statement: “The density of a heavy bar of pure gold is greater than the density of a small ingot of pure gold.”

### Problems

1. Make the following conversions.
1. 128 mL to L
2. 2.5 × 105 μg to g
3. 0.481 km to m
4. 1890 cm to km
5. 6.2 × 10-5 ms to ns
6. 75,000 pg to cg
2. Make the following conversions.
1. 2800 cm3 to m3
2. 5.8 g/cm3 to g/L
3. A speed of 60.0 miles per hour to m/s (1 mile = 1608 m)
4. A flow rate of 125 mL/min to liters per hour
3. The speed of light is 3.0 × 108 m/s. If the distance from Earth to the Sun is 1.5 × 108 km, how many minutes does it take for light from the Sun to reach the Earth?
4. A regular solid has dimensions of 3.20 cm by 4.90 cm by 5.40 cm. The mass of the solid is 235 g. What is its density in g/cm3?
5. What is the mass of a cube of copper that is 1.80 cm on each side? The density of copper is 8.92 g/cm3.
6. A balloon is filled with 2300 mL of an unknown gas. The mass of the gas is 3.24 g. Will the balloon float or sink in air?
7. A cube of lead (density = 11.35 g/cm3) has a mass of 145.7 g. What is the length of each side of the cube in cm?

## Points to Consider

Measurements all must have a certain amount of uncertainty in them, since no measuring tool is 100% accurate. The uncertainty in a measurement must be considered both when reporting measured values and when doing calculations.

• How is the uncertainty in a given measurement indicated in the reported value?
• When a quantity such as density is calculated from two measurements (mass and volume), is it important to measure both accurately or is just one sufficient?
• What are the meanings of the terms precision and accuracy when dealing with measurements?

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