# 1.1: Unit Conversions

## Writing Conversion Factors

Converting between units is an important part of any science. Below is a table with measurements that are common in chemistry. In the table, useful equalities are given.

Measure Base Unit Abbreviation Conversion to Know
Length meter m 1 inch = 2.54 centimeter
Mass gram g 1 pound = 453.6 g
Volume liter L

1.057 quart = 1 liter

1 cubic centimeter = 1 milliliter

Temperature oCelsius oC

oCelsius = (oFahrenheit - 32)/1.8

Kelvin = oCelsius + 273.15

Energy joule J 1 calorie = 4.184 joule

To convert between different systems of measure, conversion factors can be applied. A conversion factor is written based on the equality between the two units. For example, a conversion factor for inches and centimeters can be written two ways:

Which of these two conversion factors is used depends on the desired conversion.

Example $$\PageIndex{1}$$

Write two conversion factors for the unit equality 12 inches = 1 foot.

Solution

The equality can be written as either $$\dfrac {12 \ in} {1 \ ft}$$ or $$\dfrac {1 \ ft} {12 \ in}$$.

Exercise $$\PageIndex{1}$$

Write two conversion factors for the following unit equalities:

a. 1 mile = 1.61 kilometer

b. 4 cups = 1 quart

a. The equality can be written as either $$\dfrac {1 \ mi} {1.61 \ km}$$ or $$\dfrac {1.61 \ km} {1 \ mi}$$.

b. The equality can be written as either $$\dfrac {4 \ c} {1 \ qt}$$ or $$\dfrac {1 \ qt} {4 \ c}$$.

Another set of conversion factors that you’ll need to know are the metric prefixes. The metric system includes a set of prefixes that are based on factors of 10. These prefixes are extremely useful because they are applied to many different types of measurements (e.g. length measurements, mass measurements, etc.) The following table includes a shortened list of unit prefixes and equalities that Introductory Chemistry students should know.

Prefix Abbreviation Equality in General Notation* Equality in Scientific Notation*
nano- n- 1 m = 1,000,000,000 nm 1 m = 109 nm
micro-

μ-

1 m = 1,000,000 μm

1 m = 106 μm

milli- m- 1 m = 1,000 mm 1 m = 103 mm
centi- c- 1 m = 100 cm 1 m = 102 cm
(base units: g, m, L, J, M, etc.)
kilo- k- 1,000 m = 1 km 103 m = 1 km
mega- M- 1,000,000 m = 1 Mm 106 m = 1 Mm
giga- G- 1,000,000,000 m = 1 Gm 109 m = 1 Gm

*These equalities use meters as an example. The same equalities can be written with an one of the base units.

Example $$\PageIndex{2}$$

Write two conversion factors that can be used to convert between meters and kilometers.

Solution

From the table above, 1000 m = 1 km. This can be written as $$\dfrac {1000 \ m} {1 \ km}$$ or $$\dfrac {1 \ km} {1000 \ m}$$.

Exercise $$\PageIndex{2}$$

Write two conversion factors that can be used to convert between the following units:

1. microliters and liters
2. milligrams and kilograms

a. Add texts here. Do not delete this text first. The conversion can be written as $$\dfrac {10^6 \ µL} {1 \ L}$$ or $$\dfrac {1 \ L} {10^6 \ µL}$$.

b. For this equality, we must use two relationships from the table: 1000 mg = 1 g and 1000 g = 1 kg. We can divide the latter by 1000 to get 1 g = 0.001 kg. This means that $1000 \ mg = 1 \ g = 0.001 \ kg$ The conversion can be written as $$\dfrac {1000 \ mg} { 0.001 \ kg}$$ or $$\dfrac {0.001 \ kg} {1000 \ mg}$$.

Often, conversion factors are not unit conversion factors per say, but are equivalencies that can be derived from physical or chemical properties of substances or systems. For example, the density of a substance is often used to relate its volume to its mass. Other common examples are given below.

Property Common Units
density g/mL, g/cm3, lb/ft3
speed m/s, mi/hr
concentration mol/L, g/mL
percent composition g species/g substance

Example $$\PageIndex{3}$$

Write a conversion factor for the density of gold: 19.32 g/cm3.

Solution

The density can be written as $$\dfrac {19.32 \ g} {1 \ cm ^ {3}}$$ or as $$\dfrac {1 \ cm ^ {3}} {19.32 \ g}$$.

Exercise $$\PageIndex{3}$$

Write a conversion factor for each of the following relationships.

1. 1 tablet contains 250 milligrams of acetaminophen.
2. A water molecule has two hydrogen atoms.
3. Sodium chloride is 39.33% sodium.

Add texts here. Do not delete this text first. a. The relationship can be written as $$\dfrac {1 \ tablet} {250 \ mg}$$ or $$\dfrac {250 \ mg} {1 \ tablet}$$.

b. The relationship can be written as $$\dfrac {1 \ H_2 O \ molecule} {2 \ H \ atoms}$$ or $$\dfrac {2 \ H \ atoms} {1 \ H_2 O \ molecule}$$.

c. The relationship can be written as $$\dfrac {39.33 \ g \ Na} {100 \ g \ NaCl}$$ or $$\dfrac {100 \ g \ NaCl} {39.33 \ g \ Na}$$.

## Performing Simple Conversions

To perform single step conversions, a conversion factor that relates the given value to the desired value is identified. The given value is multiplied by the conversion factor so that the given units are divided from the quantity. $given \ units * \dfrac {desired \ units} {given \ units} = desired \ units \nonumber$ When using this approach, it is important to include all units in calculations, treating them as algebraic quantities.

Example $$\PageIndex{4}$$

Perform the following conversion: 36 in. = ____ cm

Solution

We know that 1 in. = 2.54 cm. To convert 36 in. to centimeters, $36 \ \cancel{in.} \ * \dfrac {2.54 \ cm} {1 \ \cancel{in.}} = 91.4 \ cm \nonumber$

Notice that the units of inches 'cancel'. The answer is in the remaining unit, centimeters.

Exercise $$\PageIndex{4}$$

Perform the following conversions.

1. 50.0 cm = _____ in.
2. 4.57 x 104 nm = _____ m

a. $50.0 \ \cancel{cm} * \dfrac {1 \ in.} {2.54 \ \cancel{cm}} = 19.7 \ cm \nonumber$

b. $4.57*10^4 \ \cancel{nm} * \dfrac {1 \ m} {10^9 \ \cancel{nm} } = 4.57*10^{-5} \ m \nonumber$

Sometimes the conversion factors are based on chemical or physical properties.

Example $$\PageIndex{5}$$

The density of iron is 7.87 g/cm3. How much volume does 24.5 g of iron occupy?

Solution

$24.5 \ \cancel{g} * \dfrac {1 \ cm^3 } {7.87 \ \cancel{g}} = 3.11 \ cm^3 \nonumber$

The iron will occupy 3.11 cm3.

Exercise $$\PageIndex{5}$$

Magnesium chloride is 25.5% magnesium. How many grams of magnesium are present in 5.24 x104 g of magnesium chloride (MgCl2)?

The sample contains 1.34 x 104 g magnesium. $5.24x10^4 \ \cancel {g \ MgCl_2} * \dfrac {25.5 \ g \ Mg} {100 \ \cancel {g \ MgCl_2}} = 1.34x10^4 \ g \ Mg \nonumber$

## Special Cases in Unit Conversions – Temperature and Units with Powers

Temperature conversions between the Celsius scale and the Kelvin scale require only subtraction, since the size of a degree in the Celsius and Kelvin scales is identical. When converting to and from the Fahrenheit scale, both the size of the degree and the occurrence of 0° must be accounted for. The conversion equations are given in the first table.

Example $$\PageIndex{6}$$

What is 350 °F in the Kelvin scale?

Solution

The first step is to convert the temperature to Celsius. $^oC = \dfrac { 350 \ ^oF - 32 } {1.8} = 177 \ ^oC \nonumber$ Then the temperature in Celsius can be converted to the Kelvin scale. $K = 177 \ ^oC + 273.15 = 450. \ K \nonumber$

Finally, to convert between units with powers, we must account for the power in the unit conversion. To do so, the entire conversion equality is raised to the desired power. For example, the conversion factor between cubic inches (in.3) and cubic centimeters (cm3) can be determined from the inches to centimeters conversion factor. We know $$1 \ in. = 2.54 \ cm$$. To find the relationship between in.3 and cm3, we can cube the equality. ${(1 \ in. = 2.54 \ cm )}^3 \nonumber$

$1^3 \ in.^3 = 2.54^3 \ cm^3 \nonumber$

$1 \ in.^3 = 16.39 \ cm^3 \nonumber$

Exercise $$\PageIndex{7}$$

How many square meters does an 80. ft2 rug occupy? (1 ft = 0.305 m)

First, we must determine the relationship between ft2 and m2.

${(1 \ ft = 0.305 \ m )}^2 \nonumber$

$1 \ ft^2 = 0.305^2 \ m^2 = 0.0930 \ m^2 \nonumber$

Then we can use this conversion factor to determine the area of the rug.

$80. \ \cancel {ft^2} * \dfrac {0.0930 \ m^2} {1 \ \cancel {ft^2}} = 7.4 \ m^2 \nonumber$

The rug occupies 7.4 square meters.

## Extra Practice

1. Perform the following unit conversions.

a. 12 mg = ______________ lb. h. 12 L = ______________ cm3

b. 67.3 km = ______________ mm i. 0.00245 m = ______________ in.

c. 39.3 in. = ______________ cm j. 1.00 g = ______________ oz. (1 lb. = 16 oz.)

d. 60.0 cal = ______________ J k. 170 g/cm2 = ______________ lb/in2

e. 1.2 x 10-9 mg = ______________ μg l. 2.5 x 107 km = ______________ in.

f. 5.2 m3 = ______________ ft3 m. 45 °F = ______________ °C

g. 62 °C = ______________ K n. -264 °F = ______________ K

2. How many liters of air are in a room that is 1200 m3?

3. What is the volume of a 59.5 g silver spoon? (densityAg = 10.5 g/cm3).

4. The combustion of one gallon of gasoline will produce approximately 8.39 kg of carbon dioxide (CO­2). What volume will the CO2 occupy? (density of CO2 = 1.96 g/L).

5. The density of silver is 10.5 g/cm3. Express this density in terms of lb/ft3.

6. The USDA recommends that a person’s sodium intake be limited to 2,400 mg. Table salt is 39.33% sodium. How many grams of table salt can Jenny consume without surpassing this limit? (Assume there are no other sources of sodium in her diet.)

7. If one cup of coffee contains 95 milligrams of caffeine, how many cups of coffee will contain 5.00 grams of caffeine?