Chapter 2. Making Measurements

2.4 Mathematical Treatment of Measurement Results – Unit Conversions

Learning Objectives

By the end of this section, you will be able to:

  • Explain the dimensional analysis (factor label) approach to mathematical calculations involving quantities
  • Use dimensional analysis to carry out unit conversions for a given property and computations involving two or more properties
  • Learn about the various temperature scales that are commonly used in chemistry and how to convert from one scale to another.

It is often the case that a quantity of interest may not be easy (or even possible) to measure directly but instead must be calculated from other directly measured properties and appropriate mathematical relationships. For example, consider measuring the average speed of an athlete running sprints. This is typically accomplished by measuring the time required for the athlete to run from the starting line to the finish line, and the distance between these two lines, and then computing speed from the equation that relates these three properties:

[latex]\text{speed}= \frac{\text{distance}}{\text{time}}[/latex]

An Olympic-quality sprinter can run 100 m in approximately 10 s, corresponding to an average speed of

[latex]\frac{\text{100 m}}{\text{10 s}} = \text{10 m/s}[/latex]

Note that this simple arithmetic involves dividing the numbers of each measured quantity to yield the number of the computed quantity (100/10 = 10) and likewise dividing the units of each measured quantity to yield the unit of the computed quantity (m/s = m/s). Now, consider using this same relation to predict the time required for a person running at this speed to travel a distance of 25 m. The same relation between the three properties is used, but in this case, the two quantities provided are a speed (10 m/s) and a distance (25 m). To yield the sought property, time, the equation must be rearranged appropriately:

[latex]\text{time} = \frac{\text{distance}}{\text{speed}}[/latex]

The time can then be computed as:

[latex]\frac{\text{25 m}}{\text{10 m/s}} = \text{2.5 s}[/latex]

Again, arithmetic on the numbers (25/10 = 2.5) was accompanied by the same arithmetic on the units (m/m/s = s) to yield the number and unit of the result, 2.5 s. Note that, just as for numbers, when a unit is divided by an identical unit (in this case, m/m), the result is “1”—or, as commonly phrased, the units “cancel.”

These calculations are examples of a versatile mathematical approach known as dimensional analysis (or the factor-label method). Dimensional analysis is based on this premise: the units of quantities must be subjected to the same mathematical operations as their associated numbers. This method can be applied to computations ranging from simple unit conversions to more complex, multi-step calculations involving several different quantities.

Conversion Factors and Dimensional Analysis

A ratio of two equivalent quantities expressed with different measurement units can be used as a unit conversion factor. For example, the lengths of 2.54 cm and 1 in. are equivalent (by definition), and so a unit conversion factor may be derived from the ratio,

[latex]\frac{\text{2.54 cm}}{\text{1 in.}}\;\text{(2.54 cm = 1 in.) or 2.54}\frac{\text{cm}}{\text{in}}[/latex]

Several other commonly used conversion factors are given in Table 1.

Length Volume Mass
1 m = 1.0936 yd 1 L = 1.0567 qt 1 kg = 2.2046 lb
1 in. = 2.54 cm (exact) 1 qt = 0.94635 L 1 lb = 453.59 g
1 km = 0.62137 mi 1 ft3 = 28.317 L 1 (avoirdupois) oz = 28.349 g
1 mi = 1609.3 m 1 tbsp = 14.787 mL 1 (troy) oz = 31.103 g
Table 1. Common Conversion Factors

When we multiply a quantity (such as distance given in inches) by an appropriate unit conversion factor, we convert the quantity to an equivalent value with different units (such as distance in centimeters). For example, a basketball player’s vertical jump of 34 inches can be converted to centimeters by:

[latex]\text{34 in.} \times \frac{\text{2.54 cm}}{\text{1}\;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{in}} = \text{86 cm}[/latex]

Since this simple arithmetic involves quantities, the premise of dimensional analysis requires that we multiply both numbers and units. The numbers of these two quantities are multiplied to yield the number of the product quantity, 86, whereas the units are multiplied to yield [latex]\frac{\text{in.} \times \text{cm}}{\text{in.}}[/latex]. Just as for numbers, a ratio of identical units is also numerically equal to one, [latex]\frac{\text{in.}}{\text{in.}}=\text{1}[/latex], and the unit product thus simplifies to cm. (When identical units divide to yield a factor of 1, they are said to “cancel.”) Using dimensional analysis, we can determine that a unit conversion factor has been set up correctly by checking to confirm that the original unit will cancel, and the result will contain the sought (converted) unit.

Example 1

The mass of a competition frisbee is 125 g. Convert its mass to ounces using the unit conversion factor derived from the relationship 1 oz = 28.349 g (Table 1).


If we have the conversion factor, we can determine the mass in kilograms using an equation similar the one used for converting length from inches to centimeters.

[latex]x \ \text{oz} = \text{125 \text{g}} \times \text{unit conversion factor}[/latex]

We write the unit conversion factor in its two forms:

[latex]\frac{1 \text{oz}}{28.349 \text{g}} \;\text{and}\;\frac{28.349 \text{g}}{1 \text{oz}}[/latex]

The correct unit conversion factor is the ratio that cancels the units of grams and leaves ounces.

[latex]\begin{array}{r @{{}={}} l} x\;\text{oz} & 125\; \rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{g}\;\times\;\frac{\text{1 oz}}{\text{28.349 \rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}g}} \\[1em] & (\frac{125}{\text{28.349}})\text{oz} \\[1em] & \text{4.41 oz (three significant figures)} \end{array}[/latex]


Test Yourself
Convert a volume of 9.345 qt to liters.



8.844 L

Beyond simple unit conversions, the factor-label method can be used to solve more complex problems involving computations. Regardless of the details, the basic approach is the same—all the factors involved in the calculation must be appropriately oriented to insure that their labels (units) will appropriately cancel and/or combine to yield the desired unit in the result. This is why it is referred to as the factor-label method. As your study of chemistry continues, you will encounter many opportunities to apply this approach.

Example 2

What is the density of common antifreeze in units of g/mL? A 4.00-qt sample of the antifreeze weighs 9.26 lb.


Since density = [latex]\frac{\text{mass}}{\text{volume}}[/latex], we need to divide the mass in grams by the volume in milliliters. In general: the number of units of B = the number of units of A × unit conversion factor. The necessary conversion factors are given in Table 1: 1 lb = 453.59 g; 1 L = 1.0567 qt; 1 L = 1,000 mL. We can convert mass from pounds to grams in one step:

[latex]9.26\;\rule[0.75ex]{1.0em}{0.1ex}\hspace{-1.0em}\text{lb} \times \frac{453.59\;\text{g}}{1\;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{lb}} = 4.20 \times 10^3\;\text{g}[/latex]

We need to use two steps to convert volume from quarts to milliliters.

  1. Convert quarts to liters.
    [latex]4.00\;\rule[0.75ex]{1.0em}{0.1ex}\hspace{-1.0em}\text{qt} \times \frac{1 \text{L}}{1.0567 \rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{qt}} = 3.78 \text{L}[/latex]
  2. Convert liters to milliliters.
    [latex]3.78 \;\rule[0.75ex]{0.75em}{0.1ex}\hspace{-0.75em}\text{L} \times \frac{1000 \;\text{mL}}{\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{L}} = 3.78 \;\text{L} \times 10^3 \;\text{mL}[/latex]


[latex]\text{density} = \frac{4.20 \times 10^3 \;\text{g}}{3.78 \times 10^3 \;\text{mL}} = 1.11 \;\text{g/mL}[/latex]

Alternatively, the calculation could be set up in a way that uses three unit conversion factors sequentially as follows:

[latex]\frac{9.26 \rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{lb}}{4.00 \rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{qt}} \times \frac{453.59 \text{g}}{1 \rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{L}} \times \frac{1.0567 \rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{qt}}{1 \rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{L}} \times \frac{1 \rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{L}}{1000 \text{mL}} = 1.11 \text{g/mL}[/latex]


Test Yourself
What is the volume in liters of 1.000 oz, given that 1 L = 1.0567 qt and 1 qt = 32 oz (exactly)?



[latex]2.956 \times 10^{-2} \text{L}[/latex]

Example 3

While being driven from Philadelphia to Atlanta, a distance of about 1250 km, a 2014 Lamborghini Aventador Roadster uses 213 L gasoline.

a) What (average) fuel economy, in miles per gallon, did the Roadster get during this trip?

b) If gasoline costs $3.80 per gallon, what was the fuel cost for this trip?


a) We first convert distance from kilometers to miles:

[latex]1250\;\text{km} \times \frac{0.62137\;\text{mi}}{1\;\text{km}} = 777\;\text{mi}[/latex]

and then convert volume from liters to gallons:

[latex]213\;\rule[0.75ex]{0.65em}{0.1ex}\hspace{-0.65em}\text{L} \times \frac{1.0567\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{qt}}{1\;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{L}} \times \frac{1\;\text{gal}}{4\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{qt}} = 56.3\;\text{gal}[/latex]


[latex]\text{(average) mileage} = \frac{777 \;\text{mi}}{56.3\;\text{gal}} = 13.8\;\text{miles/gallon} = 13.8\;\text{mpg}[/latex]

Alternatively, the calculation could be set up in a way that uses all the conversion factors sequentially, as follows:

[latex]\frac{1250\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{km}}{213\;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{L}} \times \frac{0.62137\;\text{mi}}{1\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{km}} \times \frac{1\;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{L}}{1.0567\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{qt}} \times \frac{4\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{qt}}{1\;\text{gal}} = 13.8\;\text{mpg}[/latex]

b) Using the previously calculated volume in gallons, we find:

[latex]\text{56.3 gal} \times \frac{\$3.80}{1\;\text{gal}} = \$214[/latex]


Test Yourself
A Toyota Prius Hybrid uses 59.7 L gasoline to drive from San Francisco to Seattle, a distance of 1300 km (two significant digits).

a) What (average) fuel economy, in miles per gallon, did the Prius get during this trip?

b) If gasoline costs $3.90 per gallon, what was the fuel cost for this trip?



a) 51 mpg          b) $62

Example 4

a) Convert 35.9 kL to liters.

b)Convert 555 nm to meters.



a) We will use the fact that 1 kL = 1,000 L. Of the two conversion factors that can be defined, the one that will work is 1,000 L/1 kL. Applying this conversion factor, we get

35.9 kL x (1000 L/1 kL) = 35900 L

b) We will use the fact that 1 nm = 1/1,000,000,000 m, which we will rewrite as 1,000,000,000 nm = 1 m, or 109 nm = 1 m. Of the two possible conversion factors, the appropriate one has the nm unit in the denominator: 1 m/109 nm. Applying this conversion factor, we get

555 nm x (1 m/ 10^9 nm) = 5.55 x 10^-7 m

In the final step, we expressed the answer in scientific notation.


Test Yourself

a) Convert 67.08 μL to liters.          b) Convert 56.8 m to kilometers.



a) 6.708 × 10−5 L          b) 5.68 × 10−2 km

Example 5

Complete the following conversions
      (use table 1.1 for any metric conversions;  Note: 1 L = 1.0567 qt )

a) 125 m = ? mm          b) 2.3 x 10-6 Mg = ? g              c) 2.5 L = ? qt



a)   [latex]125\;\rule[0.75ex]{1.0em}{0.1ex}\hspace{-1.0em}\text{m} \times \frac{1\;\text{mm}}{0.001\;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{m}} = 1.25 \times 10^5\;\text{mm}[/latex]

Note that this equivalence statement comes from the literal “meaning” of milli in Table 1.1. Many students prefer to use the equivalence statement 1000 mm = 1 m. That is fine too, as long as you always remember to put the appropriate unit and number.
b)  [latex]2.3\times 10^-6\;\rule[0.75ex]{1.0em}{0.1ex}\hspace{-1.0em}\text{Mg} \times \frac{1\times 10^6\;\text{g}}{1\;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{Mg}} = 2.3\;\text{g}[/latex]
c)  [latex]2.5\;\rule[0.75ex]{1.0em}{0.1ex}\hspace{-1.0em}\text{L} \times \frac{1.0567\;\text{qt}}{1\;\rule[0.25ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{L}} = 2.6\;\text{qt}[/latex]
Test Yourself
Complete the following conversions             (note: 1 in = 2.54 cm exactly)
a) 124 mL = ? L           b) 256 days = ? hrs                   c) 63.2 cm = ? in
a) 0.124 L          b) 6.14 x 103 hrs          c) 24.9 in

Example 6

How many nm are in 26.5 feet? (12 in = 1 ft ,  and 2.54 cm = 1 in exactly)



Likely, you do not have a direct conversion between feet and nm. So, instead, ask yourself “where can I go from feet”. The only possibility is inches. Then, where can you go from inches? …and so on. The overall path becomes:

 feet [latex]\longrightarrow[/latex] inches [latex]\longrightarrow[/latex] cm [latex]\longrightarrow[/latex] m [latex]\longrightarrow[/latex] nm


[latex]26.5\;\rule[0.75ex]{0.65em}{0.1ex}\hspace{-0.65em}\text{feet} \times \frac{12\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{in}}{1\;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{foot}} \times \frac{2.54\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{cm}}{1\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{in}} \times \frac{0.01\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{m}}{1\;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{cm}} \times \frac{1\;\text{nm}}{10^-9\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{m}} = 8.08 \times 10^9\;\text{nm}[/latex]


Test Yourself

A marathon is 26.4 miles. If 1 mile = 1760 yards, and 1 m = 1.094 yards, how many km are in a marathon?



42.5 km

Example 7

a)   A car is moving at 35.2 km/h. How many cm/s is this speed?
b)   How many cm3are in 5 x 102m3?



a)  We must convert km [latex]\longrightarrow[/latex] m [latex]\longrightarrow[/latex] cm AND convert h [latex]\longrightarrow[/latex] min [latex]\longrightarrow[/latex] s. It does not matter which order we do this in. Note that if a unit is on the bottom of a fraction, we cancel it by putting that undesired unit on the top of a conversion factor.


[latex]\frac{35.2\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{km}}{1\;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{h}} \times \frac{1000\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{m}}{1\;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{km}} \times \frac{1\;\text{cm}}{0.01\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{m}} \times \frac{1\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{h}}{60\;\rule[0.5ex]{0.5em}{0.2ex}\hspace{-0.5em}\text{min}} \times \frac{1\;\rule[0.5ex]{0.6em}{0.1ex}\hspace{-0.6em}\text{min}}{60\;\text{sec}} = 978\;\text{cm/s}[/latex]


1 m3= 1 m x 1 m x 1 m, so  1 m3= 100 cm x 100 cm x 100 cm, NOT 100 cm3


[latex]5\times 10^2\;\rule[0.75ex]{1.0em}{0.1ex}\hspace{-1.0em}\text{m}^{3} \times \frac{100\;\text{cm}}{1\;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{m}}\times \frac{100\;\text{cm}}{1\;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{m}}\times \frac{100\;\text{cm}}{1\;\rule[0.5ex]{0.5em}{0.1ex}\hspace{-0.5em}\text{m}} = 5 \times 10^8\;\text{cm}^{3}[/latex]


Test Yourself

Complete the following conversions            

a)  75 mi/h = ? m/s  (1760 yd = 1 mi and 1 m = 1.094 yd)
b)  4.1 g/cm3= ? kg/m3



a) 34 m/s          b) 4.1 x 103 kg/m3

Example 8

How many cubic centimeters are in 0.883 m3?



With an exponent of 3, we have three length units, so by extension we need to use three conversion factors between meters and centimeters. Thus, we have

0.883m^3 x (100cm/1m) x (100cm/1m) x (100cm/1m) = 883000 cm^3

You should demonstrate to yourself that the three meter units do indeed cancel.


Test Yourself

How many cubic millimeters are present in 0.0923 m3?



9.23 × 107 mm3

Example 9

Convert 88.4 m/min to meters/second.



We want to change the unit in the denominator from minutes to seconds. Because there are 60 seconds in 1 minute (60 s = 1 min), we construct a conversion factor so that the unit we want to remove, minutes, is in the numerator: 1 min/60 s. Apply and perform the math:

88.4m/m x 1min/60s = 1.47 m/s

Notice how the 88.4 automatically goes in the numerator. That’s because any number can be thought of as being in the numerator of a fraction divided by 1.


Test Yourself

Convert 0.203 m/min to meters/second.



0.00338 m/s or 3.38 × 10−3 m/s

Figure 1. How fast is fast? A common garden snail moves at a rate of about 0.2 m/min, which is about 0.003 m/s, which is 3 mm/s!  Source: “Grapevine snail”by Jürgen Schoneris licensed under the Creative Commons Attribution-Share Alike 3.0 Unported license.

Example 10

How many nanoseconds are in 368.09 μs?



You can either do this as a one-step conversion from microseconds to nanoseconds or convert to the base unit first and then to the final desired unit. We will use the second method here, showing the two steps in a single line. Using the definitions of the prefixes micro- and nano-,

368.09 us x 1s/10^6us x 10^9ns /1s = 368090 ns = 3.608 x 10^5 ns


Test Yourself

How many milliliters are in 607.8 kL?



6.078 × 108 mL

Example 11

A rectangular plot in a garden has the dimensions 36.7 cm by 128.8 cm. What is the area of the garden plot in square meters? Express your answer in the proper number of significant figures.



Area is defined as the product of the two dimensions, which we then have to convert to square meters and express our final answer to the correct number of significant figures, which in this case will be three.

36.7 cm x 128.8 cm x 1 m/100cm x 1 m/100 cm = 0.472696 m^2 = 0.473 m^2

The 1 and 100 in the conversion factors do not affect the determination of significant figures because they are exact numbers, defined by the centi- prefix.


Test Yourself

What is the volume of a block in cubic meters whose dimensions are 2.1 cm × 34.0 cm × 118 cm?



0.0084 m3

Chemistry Is Everywhere: The Gimli Glider

On July 23, 1983, an Air Canada Boeing 767 jet had to glide to an emergency landing at Gimli Industrial Park Airport in Gimli, Manitoba, because it unexpectedly ran out of fuel during flight. There was no loss of life in the course of the emergency landing, only some minor injuries associated in part with the evacuation of the craft after landing. For the remainder of its operational life (the plane was retired in 2008), the aircraft was nicknamed “the Gimli Glider.”

The 767 took off from Montreal on its way to Ottawa, ultimately heading for Edmonton, Canada. About halfway through the flight, all the engines on the plane began to shut down because of a lack of fuel. When the final engine cut off, all electricity (which was generated by the engines) was lost; the plane became, essentially, a powerless glider. Captain Robert Pearson was an experienced glider pilot, although he had never flown a glider the size of a 767. First Officer Maurice Quintal quickly determined that the aircraft would not be able make it to Winnipeg, the next large airport. He suggested his old Royal Air Force base at Gimli Station, one of whose runways was still being used as a community airport. Between the efforts of the pilots and the flight crew, they managed to get the airplane safely on the ground (although with buckled landing gear) and all passengers off safely.

What happened? At the time, Canada was transitioning from the older English system to the metric system. The Boeing 767s were the first aircraft whose gauges were calibrated in the metric system of units (liters and kilograms) rather than the English system of units (gallons and pounds). Thus, when the fuel gauge read 22,300, the gauge meant kilograms, but the ground crew mistakenly fueled the plane with 22,300 pounds of fuel. This ended up being just less than half of the fuel needed to make the trip, causing the engines to quit about halfway to Ottawa. Quick thinking and extraordinary skill saved the lives of 61 passengers and 8 crew members—an incident that would not have occurred if people were watching their units.

Conversion of Temperature Units

We use the word temperature to refer to the hotness or coldness of a substance. One way we measure a change in temperature is to use the fact that most substances expand when their temperature increases and contract when their temperature decreases. The mercury or alcohol in a common glass thermometer changes its volume as the temperature changes. Because the volume of the liquid changes more than the volume of the glass, we can see the liquid expand when it gets warmer and contract when it gets cooler.

To mark a scale on a thermometer, we need a set of reference values: Two of the most commonly used are the freezing and boiling temperatures of water at a specified atmospheric pressure. On the Celsius scale, 0 °C is defined as the freezing temperature of water and 100 °C as the boiling temperature of water. The space between the two temperatures is divided into 100 equal intervals, which we call degrees. On the Fahrenheit scale, the freezing point of water is defined as 32 °F and the boiling temperature as 212 °F. The space between these two points on a Fahrenheit thermometer is divided into 180 equal parts (degrees).

Defining the Celsius and Fahrenheit temperature scales as described in the previous paragraph results in a slightly more complex relationship between temperature values on these two scales than for different units of measure for other properties. Most measurement units for a given property are directly proportional to one another (y = mx). Using familiar length units as one example:

[latex]\text{length in feet} = \left( \frac {1\;\text{ft}}{12\;\text{in.}}\right) \times \text{length in inches}[/latex]

where y = length in feet, x = length in inches, and the proportionality constant, m, is the conversion factor. The Celsius and Fahrenheit temperature scales, however, do not share a common zero point, and so the relationship between these two scales is a linear one rather than a proportional one (y = mx + b). Consequently, converting a temperature from one of these scales into the other requires more than simple multiplication by a conversion factor, m, it also must take into account differences in the scales’ zero points (b).

The linear equation relating Celsius and Fahrenheit temperatures is easily derived from the two temperatures used to define each scale. Representing the Celsius temperature as x and the Fahrenheit temperature as y, the slope, m, is computed to be:

[latex]m = \frac{\Delta y}{\Delta x} = \frac{212 \;^{\circ}\text{F} - 32 \;^{\circ}\text{F}}{100 \;^{\circ}\text{C} - 0 \;^{\circ}\text{C}} = \frac{180 \;^{\circ}\text{F}}{100 \;^{\circ}\text{C}} = \frac{9 \;^{\circ}\text{F}}{5 \;^{\circ}\text{C}}[/latex]

The y-intercept of the equation, b, is then calculated using either of the equivalent temperature pairs, (100 °C, 212 °F) or (0 °C, 32 °F), as:

[latex]b = y - mx = 32\;^{\circ}\text{F} - \frac{9\;^{\circ}\text{F}}{5\;^{\circ}\text{C}} \times 0\;^{\circ}\text{C} = 32\;^{\circ}\text{F}[/latex]

The equation relating the temperature scales is then:

[latex]T_{^\circ\text{F}} = (\frac{9 \;^\circ\text{F}}{5 \;^\circ\text{C}} \times T_{^\circ\text{C}}) + 32\;^\circ\text{C}[/latex]

An abbreviated form of this equation that omits the measurement units is:

[latex]T_{^\circ\text{F}} = \frac{9}{5} \times T_{^\circ\text{C}} + 32[/latex]

Rearrangement of this equation yields the form useful for converting from Fahrenheit to Celsius:

[latex]T_{^\circ\text{C}} = \frac{5}{9} (T_{^\circ\text{F}} - 32)[/latex]

As mentioned earlier in this chapter, the SI unit of temperature is the kelvin (K). Unlike the Celsius and Fahrenheit scales, the kelvin scale is an absolute temperature scale in which 0 (zero) K corresponds to the lowest temperature that can theoretically be achieved. The early 19th-century discovery of the relationship between a gas’s volume and temperature suggested that the volume of a gas would be zero at −273.15 °C. In 1848, British physicist William Thompson, who later adopted the title of Lord Kelvin, proposed an absolute temperature scale based on this concept (further treatment of this topic is provided in this text’s chapter on gases).

The freezing temperature of water on this scale is 273.15 K and its boiling temperature 373.15 K. Notice the numerical difference in these two reference temperatures is 100, the same as for the Celsius scale, and so the linear relation between these two temperature scales will exhibit a slope of [latex]1\;\frac{\text{K}}{^{\circ}\text{C}}[/latex]. Following the same approach, the equations for converting between the kelvin and Celsius temperature scales are derived to be:

[latex]T_{\text{K}} = T_{^\circ\text{C}} + 273.15[/latex]
[latex]T_{^\circ\text{C}} = T_{\text{K}} - 273.15[/latex]

The 273.15 in these equations has been determined experimentally, so it is not exact. Figure 3 shows the relationship among the three temperature scales. Recall that we do not use the degree sign with temperatures on the kelvin scale.

A thermometer is shown for the Fahrenheit, Celsius and Kelvin scales. Under the Fahrenheit scale, the boiling point of water is 212 degrees while the freezing point of water is 32 degrees. Therefore, there are 180 Fahrenheit degrees between the boiling point of water and the freezing point of water. Under the Celsius scale, the boiling point of water is 100 degrees while the freezing point of water is 0 degrees. Therefore, there are 100 Celsius degrees between the boiling point and freezing point of water. Under the kelvin scale, the boiling point of water is 373.15 K, while the freezing point of water is 273.15 K. 233.15 K is equal to negative 40 degrees Celsius, which is also equal to negative 40 degrees Fahrenheit.
Figure 3. The Fahrenheit, Celsius, and kelvin temperature scales are compared.

Although the kelvin (absolute) temperature scale is the official SI temperature scale, Celsius is commonly used in many scientific contexts and is the scale of choice for nonscience contexts in almost all areas of the world. Very few countries (the U.S. and its territories, the Bahamas, Belize, Cayman Islands, and Palau) still use Fahrenheit for weather, medicine, and cooking.

Example 12

Normal body temperature has been commonly accepted as 37.0 °C (although it varies depending on time of day and method of measurement, as well as among individuals). What is this temperature on the kelvin scale and on the Fahrenheit scale?



[latex]\text{K} = ^\circ\text{C} + \text{273.15} = \text{37.0} + \text{273.2} = \text{310.2 K}[/latex]
[latex]^\circ\text{F} = \frac{9}{5} ^\circ\text{C} + \text{32.0} = (\frac{9}{5} \times \text{37.0}) + \text{32.0} = \text{66.6} + \text{32.0} = \text{98.6}^\circ\text{F}[/latex]


Test Yourself
Convert 80.92 °C to K and °F.



354.07 K, 177.7 °F

Example 13

Baking a ready-made pizza calls for an oven temperature of 450 °F. If you are in Europe, and your oven thermometer uses the Celsius scale, what is the setting? What is the kelvin temperature?



[latex]^\circ\text{C} = \frac{5}{9}(^\circ\text{F} - \text{32}) = \frac{5}{9} \text{(450 - 32)} = \frac{5}{9} \times \text{418} = \text{232} \;^\circ\text{C} \longrightarrow \text{set oven to 230} \;^\circ\text{C} \text{(two significant figures)}[/latex]
[latex]\text{K} = ^\circ\text{C} + \text{273.15} = \text{230} + \text{273} = \text{503 K} \longrightarrow \text{5.0} \times \text{10}^2\text{K} \text{(two significant figures)}[/latex]


Test Yourself
Convert 50 °F to °C and K.



10 °C, 280 K

Example 14

a) What is 98.6 °F in degrees Celsius?

b) What is 25.0 °C in degrees Fahrenheit?



a) Using the first formula from above, we have

[latex]^\circ\text{C} = \frac{5}{9}(^\circ\text{F} - \text{32}) = \frac{5}{9} \text{(98.6 - 32)} = \frac{5}{9} \times \text{66.6} = \text{37.0} \;^\circ\text{C}[/latex]


b) Using the second formula from above, we have

[latex]^\circ\text{F} = \frac{9}{5} ^\circ\text{C} + \text{32.0} = (\frac{9}{5} \times \text{25.0}) + \text{32.0} = \text{45.0} + \text{32.0} = \text{77.0}^\circ\text{F}[/latex]


Test Yourself

a) Convert 0 °F to degrees Celsius.

b) Convert 212 °C to degrees Fahrenheit.



a) −17.8 °C          b) 414 °F

Example 15

If normal room temperature is 72.0 °F, what is room temperature in degrees Celsius and kelvins?



First, we use the formula to determine the temperature in degrees Celsius:

[latex]^\circ\text{C} = \frac{5}{9}(^\circ\text{F} - \text{32}) = \frac{5}{9} \text{(72.0 - 32)} = \frac{5}{9} \times \text{40.0} = \text{22.2} \;^\circ\text{C}[/latex]

Then we use the appropriate formula above to determine the temperature in the Kelvin scale:

K = 22.2 °C + 273.15 = 295.4 K

So, room temperature is about 295 K.


Test Yourself

What is 98.6 °F on the Kelvin scale?



310.2 K

Food and Drink App: Cooking Temperatures

Because degrees Fahrenheit is the common temperature scale in the United States, kitchen appliances, such as ovens, are calibrated in that scale. A cool oven may be only 150°F, while a cake may be baked at 350°F and a chicken roasted at 400°F. The broil setting on many ovens is 500°F, which is typically the highest temperature setting on a household oven.

People who live at high altitudes, typically 2,000 ft above sea level or higher, are sometimes urged to use slightly different cooking instructions on some products, such as cakes and bread, because water boils at a lower temperature the higher in altitude you go, meaning that foods cook slower. For example, in Cleveland water typically boils at 212°F (100°C), but in Denver, the Mile-High City, water boils at about 200°F (93.3°C), which can significantly lengthen cooking times. Good cooks need to be aware of this.

At the other end is pressure cooking. A pressure cooker is a closed vessel that allows steam to build up additional pressure, which increases the temperature at which water boils. A good pressure cooker can get to temperatures as high as 252°F (122°C); at these temperatures, food cooks much faster than it normally would. Great care must be used with pressure cookers because of the high pressure and high temperature. (When a pressure cooker is used to sterilize medical instruments, it is called an autoclave.)

Other countries use the Celsius scale for everyday purposes. Therefore, oven dials in their kitchens are marked in degrees Celsius. It can be confusing for US cooks to use ovens abroad—a 425°F oven in the United States is equivalent to a 220°C oven in other countries. These days, many oven thermometers are marked with both temperature scales.

Need a refresher or more practice with unit conversion? Visit this site ( to go over the basics of unit conversions.

Video source: Unit conversion by keyj

Key Concepts and Summary

Measurements are made using a variety of units. It is often useful or necessary to convert a measured quantity from one unit into another. These conversions are accomplished using unit conversion factors, which are derived by simple applications of a mathematical approach called the factor-label method or dimensional analysis. This strategy is also employed to calculate sought quantities using measured quantities and appropriate mathematical relations.

Key Equations

  • [latex]T_{^\circ\text{C}} = \frac{5}{9} \times T_{^\circ\text{F}} - 32[/latex]
  • [latex]T_{^\circ\text{F}} = \frac{9}{5} \times T_{^\circ\text{C}} + 32[/latex]
  • [latex]T_\text{K} = {^\circ\text{C}} + 273.15[/latex]
  • [latex]T_{^\circ\text{C}} = \text{K} - 273.15[/latex]


1.  Perform the following conversions.

a)  255°F to degrees Celsius                 b)  −255°F to degrees Celsius

c)  50.0°C to degrees Fahrenheit          d)  −50.0°C to degrees Fahrenheit

2.  Perform the following conversions.

a)  100.0°C to kelvins                           b)  −100.0°C to kelvins

c)  100 K to degrees Celsius                 d)  300 K to degrees Celsius

3.  Convert 0 K to degrees Celsius. What is the significance of the temperature in degrees Celsius?

4.  The hottest temperature ever recorded on the surface of the earth was 136°F in Libya in 1922. What is the temperature in degrees Celsius and in kelvins?

5.  Write the two conversion factors that exist between the two given units.

a)  milliliters and laters      b)  microseconds and seconds       c)  kilometers and meters

6.  Perform the following conversions.

a)  5.4 km to meters      b)  0.665 m to millimeters      c)  0.665 m to kilometers

7.  Perform the following conversions.

a)  17.8 μg to grams      b)  7.22 × 102 kg to grams      c)  0.00118 g to nanograms

8.  Perform the following conversions.

a)  9.44 m2 to square centimetres      b)  3.44 × 108 mm3 to cubic meters

9.  Why would it be inappropriate to convert square centimeters to cubic meters?

10.  Perform the following conversions.

a)  45.0 m/min to meters/second       b)  0.000444 m/s to micrometers/second

c)  60.0 km/h to kilometers/second

11.  Perform the following conversions.

a)  0.674 kL to milliliters      b)  2.81 × 1012 mm to kilometers      c)  94.5 kg to milligrams

12.  Perform the following conversions.

a)  6.77 × 1014 ms to kilo seconds       b)  34,550,000 cm to kilometers

13.  Perform the following conversions. Note that you will have to convert units in both the numerator and the denominator.

a)  88 ft/s to miles/hour (Hint: use 5,280 ft = 1 mi.)      b)  0.00667 km/h to meters/second

14.  What is the area in square millimeters of a rectangle whose sides are 2.44 cm × 6.077 cm? Express the answer to the proper number of significant figures.
15.  The formula for the area of a triangle is 1/2 × base × height. What is the area of a triangle in square centimeters if its base is 1.007 m and its height is 0.665 m? Express the answer to the proper number of significant figures.

16.  Write conversion factors (as ratios) for the number of:

a) yards in 1 meter      b) liters in 1 liquid quart      c) pounds in 1 kilogram

17.  The label on a soft drink bottle gives the volume in two units: 2.0 L and 67.6 fl oz. Use this information to derive a conversion factor between the English and metric units. How many significant figures can you justify in your conversion factor?

18.  Soccer is played with a round ball having a circumference between 27 and 28 in. and a weight between 14 and 16 oz. What are these specifications in units of centimeters and grams?

19.  How many milliliters of a soft drink are contained in a 12.0-oz can?

20.  The diameter of a red blood cell is about 3 × 10−4 in. What is its diameter in centimeters?

21.  Is a 197-lb weight lifter light enough to compete in a class limited to those weighing 90 kg or less?

22.  Many medical laboratory tests are run using 5.0 μL blood serum. What is this volume in milliliters?

23.  Use scientific notation to express the following quantities in terms of the SI base units:

a) 0.13 g      b) 232 Gg      c) 5.23 pm      d) 86.3 mg      e) 37.6 cm      f) 54 μm      g) 1 Ts      h) 27 ps               i) 0.15 mK

24.  Gasoline is sold by the liter in many countries. How many liters are required to fill a 12.0-gal gas tank?

25.  A long ton is defined as exactly 2240 lb. What is this mass in kilograms?

26.  Make the conversion indicated in each of the following:

a) the length of a soccer field, 120 m (three significant figures), to feet

b) the height of Mt. Kilimanjaro, at 19,565 ft the highest mountain in Africa, to kilometers

c) the area of an 8.5 t 11-inch sheet of paper in cm2

d) the displacement volume of an automobile engine, 161 in.3, to liters

e) the estimated mass of the atmosphere, 5.6 t 1015 tons, to kilograms

f) the mass of a bushel of rye, 32.0 lb, to kilograms

g) the mass of a 5.00-grain aspirin tablet to milligrams (1 grain = 0.00229 oz)

27.  A chemist’s 50-Trillion Angstrom Run would be an archeologist’s 10,900 cubit run. How long is one cubit in meters and in feet? (1 Å = 1 × 10−8 cm)

28.  As an instructor is preparing for an experiment, he requires 225 g phosphoric acid. The only container readily available is a 150-mL Erlenmeyer flask. Is it large enough to contain the acid, whose density is 1.83 g/mL?

29.  A chemistry student is 159 cm tall and weighs 45.8 kg. What is her height in inches and weight in pounds?

30.  Solve these problems about lumber dimensions.

a) To describe to a European how houses are constructed in the US, the dimensions of “two-by-four” lumber must be converted into metric units. The thickness × width × length dimensions are 1.50 in. × 3.50 in. × 8.00 ft in the US. What are the dimensions in cm × cm × m?

b) This lumber can be used as vertical studs, which are typically placed 16.0 in. apart. What is that distance in centimeters?

31.  Calculate the density of aluminum if 27.6 cm3 has a mass of 74.6 g.

32.  Calculate these masses.

a) What is the mass of 6.00 cm3 of mercury, density = 13.5939 g/cm3?

b) What is the mass of 25.0 mL octane, density = 0.702 g/cm3?

33.  Calculate these volumes.

a) What is the volume of 25 g iodine, density = 4.93 g/cm3?

b) What is the volume of 3.28 g gaseous hydrogen, density = 0.089 g/L?

34.  Convert the boiling temperature of gold, 2966 °C, into degrees Fahrenheit and kelvin.

35.  Convert the temperature of the coldest area in a freezer, −10 °F, to degrees Celsius and kelvin.

36.  Convert the boiling temperature of liquid ammonia, −28.1 °F, into degrees Celsius and kelvin.

37.  The weather in Europe was unusually warm during the summer of 1995. The TV news reported temperatures as high as 45 °C. What was the temperature on the Fahrenheit scale?



1.  a)  124°C      b)  −159°C      c)  122°F      d)  −58°F

2.  a)  373 K      b)  173 K      c)  −173°C      d)  27°C

3.   −273°C. This is the lowest possible temperature in degrees Celsius.

4.  57.8°C;  331 K

5.  a)  1,000 mL/1 L and 1 L/1,000 mL      b)  1,000,000 μs/1 s and 1 s/1,000,000 μs

c)  1,000 m/1 km and 1 km1,000 m

6.  a)  5,400 m      b)  665 mm      c)  6.65 × 10−4 km

7.  a)  1.78 × 10−5 g      b)  7.22 × 105 g      c) 1.18 × 106 ng

8.  a)  94,400 cm2      b)  0.344 m3

9.  One is a unit of area, and the other is a unit of volume.

10.  a)  0.75 m/s       b)  444 µm/s      c)  1.666 × 10−2 km/s

11.  a)  674,000 mL      b)  2.81 × 106 km      c)  9.45 × 107 mg

12.  a)  6.77 × 108 ks      b)  345.5 km

13.  a)  6.0 × 101 mi/h      b)  0.00185 m/s

14.  1.48 × 103 mm2

15.  3.35 × 103 cm2

16.  a) [latex]\frac{\text{1.0936 yd}}{\text{1 m}}[/latex]   b) [latex]\frac{\text{0.94635 L}}{\text{1 qt}}[/latex]          c) [latex]\frac{\text{2.2046 lb}}{\text{1 kg}}[/latex]

17. [latex]\frac{\text{2.0 L}}{\text{67.6 fl oz}} = \frac{\text{0.030 L}}{\text{1 fl oz}}[/latex]
Only two significant figures are justified.

18. 68–71 cm; 400–450 g

19. 355 mL

20. 8 × 10−4 cm

21. yes; weight = 89.4 kg

22. 5.0 × 10−3 mL

23. a) 1.3 × 10−4 kg      b) 2.32 × 108 kg      c) 5.23 × 10−12 m      d) 8.63 × 10−5 kg      e) 3.76 × 10−1 m          f) 5.4 × 10−5 m      g) 1 × 1012 s      h) 2.7 × 10−11 s      i) 1.5 × 10−4 K

24. 45.4 L

25. 1.0160 × 103 kg

26.  a) 394 ft      b) 5.9634 km      c) 6.0 × 102      d) 2.64 L      e) 5.1 × 1018 kg      f) 14.5 kg      g) 324 mg

27. 0.46 m; 1.5 ft/cubit

28. Yes, the acid’s volume is 123 mL.

29. 62.6 in (about 5 ft 3 in.) and 101 lb

30. (a) 3.81 cm × 8.89 cm × 2.44 m; (b) 40.6 cm

31. 2.70 g/cm3

32. (a) 81.6 g; (b) 17.6 g

33. (a) 5.1 mL; (b) 37 L

34. 5371 °F, 3239 K

35. −23 °C, 250 K

36. −33.4 °C, 239.8 K

37. 113 °F


dimensional analysis: (also, factor-label method) versatile mathematical approach that can be applied to computations ranging from simple unit conversions to more complex, multi-step calculations involving several different quantities

Fahrenheit: unit of temperature; water freezes at 32 °F and boils at 212 °F on this scale

unit conversion factor: ratio of equivalent quantities expressed with different units; used to convert from one unit to a different unit


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