In science, units are extremely important. The number 12 means absolutely nothing unless we know the units, whether it is time, length or pressure. Consequently, being able to manipulate these units is a key skill, especially for chemistry when certain equations require very specific units. For example, the ideal gas law, PV=nRT, requires pressure to be in atm, volume to be in liters and temperature to be in Kelvin so that everything will properly cancel out.

The process of manipulating our units is called dimensional analysis. This uses the principle that we can multiply a number by fractions that are equivalent to 1 to change the units without changing the actual value of the number. An easy way to think of this is to imagine a ruler that has inches on one side and centimeters on the other. If we measure a piece of string with either side, we get two different numbers with different units, but they represent the same real world length. That is the goal of dimensional analysis: to get the same real world value represented with different units.

To do this, we need to either memorize or reference a table of conversion factors. These are readily available in any chemistry textbook, but some of the most common conversion factors are listed below.

Performing dimensional analysis is a pretty easy process. All you have to do is set up a series of fractions where the units end up canceling out. Remember, when you have fractions and want something to cancel out, you have to make sure it is present in both the numerator and denominator. Here’s a basic example of converting from feet to inches that uses the conversion factor of 1 foot = 12 inches.

1.5\enspace feet\times \dfrac { 12\enspace inches }{ 1\enspace foot } =18\enspace inches

Notice that the same unit, feet, canceled out in our fractions and we were left with just inches.

When we want to convert inches to feet, notice that we have to flip the conversion factor over to make sure we cancel out the correct unit.

18\enspace inches\times \dfrac { 1\enspace foot }{ 1\enspace inches } =1.5\enspace feet

We can also do more challenging problems that require more than one conversion factor. Like the simple problems, the most important thing to remember is that your units need to all cancel out, leaving you with just the units you need for the final answer. Let’s try a multistep problem, converting miles per hour to meters per second. This problem is particularly tricky because we already have something in the denominator of our initial units. So, to cancel them out, we need to make sure we put the corresponding unit in the numerator of our conversion factors.

\dfrac { 60\enspace miles }{ 1\enspace hour } \times \dfrac { 1\enspace hour }{ 3600\enspace seconds } \times \dfrac { 5280\enspace feet }{ 1\enspace mile } \times \dfrac { 1\enspace meter }{ 3.28\enspace feet } =26.82\dfrac { meters }{ second }

Practice Problems

Here are some examples to try. Detailed solutions for each are given below, but make sure you take your time and attempt them all before consulting the solutions. Practice makes perfect! If you get stuck, just remember your ultimate goal, to cancel out all units that are not part of your desired answer.

1. 13,588\enspace g\enspace to\enspace kg

2. { 15 }^{ 0 }F\enspace to\enspace { C }^{ 0 }

3. 18\enspace miles\enspace per\enspace hour\enspace to\enspace kilometers\enspace per\enspace minute

4. 120\enspace cubic\enspace inches\enspace to\enspace cubic\enspace meters

5. 4.58\dfrac { g }{ mL } to\enspace \dfrac { kg }{ L }

6. 5.15\enspace yd\enspace to\enspace cm

7. The density of lead is 11.3\dfrac { g }{ mL }. What is the mass of 45\enspace mL of lead?

8. A particle moves at 15\dfrac { km }{ s }. How far will it move in 7.5\enspace seconds?

9. Seawater contains 0.000245\enspace g of sodium chloride per mL. How much sodium chloride is in 100\enspace mL of seawater?

10. A particle moving through the air at a speed of 48.5\dfrac { m }{ s } hits the wall, bounces, and hits an object 25 \enspace cm away. How long did it take to travel from the wall to the object?

Solutions

1. 13,588\enspace g\enspace to\enspace kg

Necessary conversion factors: 1000 grams = 1 kilogram

13,588\enspace g\times \dfrac { 1\enspace kg }{ 1000\enspace g } =13.588\enspace kg

2. { 15 }^{ 0 }F\enspace to\enspace { C }^{ 0 }

Necessary conversion factors: 1 degree Celsius = 33.8 degrees Fahrenheit

{ 15 }^{ 0 }F\times \dfrac { { 1 }^{ 0 }C }{ { 33.8 }^{ 0 }F } =-{ 9.44 }^{ 0 }C

3. 18\enspace miles\enspace per\enspace hour\enspace to\enspace kilometers\enspace per\enspace minute

Necessary conversion factors: 1 mile = 1.61 kilometers, 1 hour = 60 minutes

18\dfrac { miles }{ hour } \times \dfrac { 1.61\enspace km }{ 1\enspace mile } \times \dfrac { 1\enspace hour }{ 60\enspace minutes } =0.483\dfrac { km }{ min }

4. 120\enspace cubic\enspace inches\enspace to\enspace cubic\enspace meters

Necessary conversion factors: 12 inches = 1 foot, 1 meter = 3.28 feet

Pay careful attention to your conversion factors here. While we know that 12 inches is the same as 1 foot, notice that we are dealing with cubic inches. So, we have to cube all our numbers and units to get units that will actually cancel out.

Adapted conversion factors: 1728 cubic inches = 1 cubic foot, 1 cubic meter = 35.32 cubic feet

123{ in }^{ 3 }\times \dfrac { { 1ft }^{ 3 } }{ { 1728in }^{ 3 } } \times \dfrac { { 1m }^{ 3 } }{ { 35.32ft }^{ 3 } } =0.00196{ m }^{ 3 }

5. 4.58\dfrac { g }{ mL } to\enspace \dfrac { kg }{ L }

Necessary conversion factors: 1000 grams = 1 kilogram, 1000 milliliters = 1 liter

4.58\dfrac { g }{ mL } \times \dfrac { 1\enspace kg }{ 1000\enspace g } \times \dfrac { 1000\enspace mL }{ 1\enspace L } =4.58\dfrac { kg }{ L }

6. 5.15\enspace yd\enspace to\enspace cm

Necessary conversion factors: 1 yard = 3 feet, 1 foot = 12 inches, 1 inch = 2.54 cm

5.15\enspace yd\times \dfrac { 3\enspace ft }{ 1\enspace yd } \times \dfrac { 12\enspace in }{ 1\enspace ft } \times \dfrac { 2.54\enspace cm }{ 1\enspace in } =470.9\enspace cm

7. The density of lead is 11.3\dfrac { g }{ mL }. What is the mass of 45\enspace mL of lead?

For word problems, a necessary conversion factor is almost always given in the problem itself. In this case, 11.3 g/mL can help us find the mass of 45 mL of lead. We also need the fact that 1000 grams is equivalent to 1 kilogram.

45\enspace mL\times \dfrac { 11.3\enspace g }{ 1\enspace mL } \times \dfrac { 1\enspace kg }{ 1000\enspace g } =0.509\enspace kg

8. A particle moves at 15\dfrac { km }{ s }. How far will it move in 7.5\enspace seconds?

Necessary conversion factors: 15 km/s, 1000 meters = 1 kilometer

7.5\enspace s\times \dfrac { 15\enspace km }{ 1\enspace s } \times \dfrac { 1000m }{ 1\enspace km } =112500\enspace m

9. Seawater contains 0.000245\enspace g of sodium chloride per mL. How much sodium chloride is in 100\enspace mL of seawater?

Necessary conversion factors: 0.000245g/mL, 1000 milliliters = 1 liter

1\enspace L\times \dfrac { 1000\enspace mL }{ 1\enspace L } \times \dfrac { 0.000245\enspace g }{ 1\enspace mL } =0.245\enspace g

10. A particle moving through the air at a speed of 48.5\dfrac { m }{ s } hits the wall, bounces, and hits an object 25 \enspace cm away. How long did it take to travel from the wall to the object?

Necessary conversion factors: 100 centimeters = 1 meter, 48.5 m/s

25\enspace cm\times \dfrac { 1\enspace m }{ 100\enspace cm } \times \dfrac { 1\enspace s }{ 48.5\enspace m } =0.005\enspace s

Bonus Concept

In chemistry, a lot of your dimensional analysis will involve the number of molecules and atoms involved in a reaction (these calculations are called stoichiometry). Using the given information, use the skills of dimensional analysis to answer the following question.

A sample of calcium nitrate, Ca(NO₃)₂, has a formula weight of 164 g/mole and one mole of any molecule has 6.022 x 10²³ molecules. How many molecules are present in 0.25 pounds of calcium nitrate?

Necessary conversion factors: 1 mole = 6.022 x 10²³ molecules, 164 g/mole, 1 kilogram = 2.2 pounds, 1000 grams = 1 kilogram

0.25\enspace lbs\times \dfrac { 1\enspace kg }{ 2.2\enspace lbs } \times \dfrac { 1000\enspace g }{ 1\enspace kg } \times \dfrac { 1\enspace mole }{ 164\enspace g } \times \dfrac { { 6.022\times 10 }^{ 23 }\enspace molecules }{ 1\enspace mole } ={ 4.17\times 10 }^{ 23 }\enspace molecules

If you were able to understand that problem, you are doing a great job! Otherwise, keep practicing. Dimensional analysis, especially with chemical concepts, can be difficult to grasp, but as with most things, it gets easier with practice. Use the skills covered here and work hard. Soon, dimensional analysis will be simple for you.

Let us know if you have any tips for doing dimensional analysis easily and efficiently by leaving a comment!

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