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Dimensional Analysis

dimensional analysis examples

Core Concepts

In this tutorial, you will learn what dimensional analysis is in the field of chemistry, how to use it, see examples, and learn how it can be applied to chemistry. 

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What is Dimensional Analysis?

What is the definition of dimensional analysis? Dimensional analysis is an essential skill used widely in the field of chemistry. Using this technique can answer questions like: “How much of this chemical do I need in my reaction?” and “What is the concentration of my solution?” At its simplest form, dimensional analysis is the methodical canceling-out of units. Take the example below:

    \begin{gather*} {\text{If } 2\text{Red} = 1\text{Blue,}} \\ {4\text{Red is ....}} \\ {\frac{4\text{Red}}{1}\cdot\frac{1\text{Blue}}{2\text{Red}}=2\text{Blue}} \end{gather*}

In more real-world applications, dimensional analysis is used to convert between different units of measurement, and find unknown characteristics from those that we do know. 

Unit Conversions

It is often necessary to convert between units of measurement. Units of measurement are used to define the qualities of something. 

  • A gold block weighs 12 kilograms. 
  • There are 150 milliliters of water in the container. 

You will encounter Imperial and SI (International System) Units. Imperial units are measurements like feet, inches, and pounds. SI units are measurements like meters, centimeters, and kilograms. SI units are most common in chemistry; furthermore, one of the most important units of measurement is the mole (mol).

Below are some of the relationships between these units you might see throughout various sources, such as textbooks, or online:

Imperial Units
Quantity Unit of Measurement Relationship
Length Mile, mi
Yard, yd
Feet, ft
Inches, in		1 \text{mi} = 1760 \text{yd}
1 \text{yd} = 3\text{ft}
1 \text{ft} = 12 \text{in}
WeightPound, lb
Ounce, oz		1 \text{lb} = 12 {oz}
VolumeGallon, gal
Quart, qt
Pint, pt
Cup, c		1 \text{gal} = 4 \text{qt}
1 \text{qt} = 2 \text{pt}
1 \text{pt} = 2 \text{c}
SI Units 
QuantityUnit of MeasurementRelationship
LengthMeters, m
Centimeters, cm
Millimeters, mm
Nanometers, nm		1 \text{m} = 100 \text{cm} = 10^{2} \text{cm}
1 \text{cm} = 10 \text{mm}
1 \text{mm} = 10^{6} \text{nm}
WeightKilograms, kg
Grams, g
Milligram, mg
Micrograms, ug		1 \text{kg} = 1000 \text{g}=10^{3}\text{g}
1 \text{g} = 10^{3}\text{mg}
1 \text{mg} = 10^{3}\text{mg}
VolumeLiters, L
Milliliters, ml		1 \text{L} = 1000 \text{ml} = 10^{3}\text{ml}

Additionally, it is also possible to convert between the two systems of measurement. 

what is dimensional analysis

If the gold block weighs 12 kilograms, how many pounds does it weigh?

1.0 pound is about 0.45 kilograms, so we find …

    \begin{gather*} {\frac{12\text{kg}}{1}\cdot\frac{1.0\text{lb}}{0.45\text{kg}}=26\text{lbs}} \end{gather*}

The relationship 1.0 \text{lbs} = 0.45 \text{kg} is first rewritten as a ratio.

When writing the ratio, place \text{kg} in the denominator and \text{lbs} in the numerator so that \text{kgs} later cancels out and only \text{lbs} is left in our answer. 

    \begin{gather*} {1.0\text{lb} = 0.45\text{kg} \rightarrow \frac{1.0\text{lb}}{0.45\text{kg}}} \end{gather*}

Then, we multiply our known value, 12\text{kg}, by the ratio \frac{1.0\text{lb}}{0.45\text{kg}} 

    \begin{gather*} {\frac{12\text{kg}}{1}\cdot\frac{1.0\text{lb}}{0.45\text{kg}}=26\text{lbs}} \end{gather*}

We are left with the solution: For every 12 kilograms, there are 26 pounds. 

Finding Unknowns

Dimensional analysis is not only useful for converting between one unit to another, but can help in solving for a number of different properties. It becomes important to be aware of how quantities like mass, volume, and density are related:

QuantityRelationship
Densitymass/volume
Energy force✕distance
Volume area✕length
Pressureforce/area
Adapted from WebAssign

It is also increasingly important to pay attention to units of measurement. 

If there are 150\text{mL} of water in a container, and the water in the container weighs 150\text{g}, what is the density of the water?

    \begin{gather*} {\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{\text{grams}}{\text{milliliter}}} \end{gather*}

In this case, mass is given in grams, and volume is given in milliliters. We define density as mass/volume, so for this case density is grams/milliliter.

    \begin{gather*} {\text{Density}_{H_{2}O} = \frac{150\text{g}}{150\text{mL}} = 1.0\text{g/mL}} \end{gather*}

Water has the density of 1\text{g/ml}.

Dimensional Analysis Examples in Chemistry

In chemistry, dimensional analysis can be used in the ways described above, but it can also be used to relate the quantities of chemicals in a reaction. Ratios can be created using stoichiometric relationships.

    \begin{gather*} {CaI_{2}(s) \rightarrow Ca^{2+}(aq) + 2I^{-}{aq}} \end{gather*}

Since there are two mols of Iodide (I– ) for every one mol of Calcium Iodide (CaI2), the following ratio can be written: 

    \begin{gather*} {\frac{2\text{mol}I^{-}}{1\text{mol}CaI_{2}}} \end{gather*}

How many moles of iodide ions are produced when 4 mols of CaI2 are fully dissolved?

    \begin{gather*} {\frac{4\text{mol}CaI_{2}}{1}\cdot\frac{2\text{mol}I^{-}}{1\text{mol}CaI_{2}}=8\text{mol}I^{-}} \end{gather*}

For a more in depth tutorial, see Solving Stoichiometry Problems.

Dimensional Analysis Practice Problems

Problem 1

You mix 2.00\text{mol} of MgCl2 into a solution with excess AgNO3, hence MgCl2 is your limiting reactant. Assuming a 100% yield, how much AgCl would you expect to produce, in grams? (AgCl has a molecular weight of 143.32\text{g/mol}).

MgCl2 and AgNO3 react according to the following equation:

    \begin{gather*} {MgCl_{2} + 2AgNO_{3} \rightarrow Mg\left(NO_{3}\right)_{2} + 2AgCl} \end{gather*}

Problem 2

Under standard temperature and pressure (STP), an ideal gas occupies 22.4\text{L} per mole of gas particles (\left(\frac{22.4\text{L}}{1\text{mol}\right)}). If you sublimate 30.0\text{g} of solid CO2 into gas, what would the resulting volume of the CO2 under STP? (CO2 has a molecular weight of 44.01\text{g/mol}).

For More Help, Watch Our Interactive Video Explaining Dimensional Analysis!

Dimensional Analysis Practice Problems

1: 573\text{g}AgCl

2: 15.3\text{L}CO_{2}

Further Reading