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2.0.2: Dimensional Analysis

  • Page ID
    210618
  • Learning Objectives

    • 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
    • Perform dimensional analysis calculations with units raised to a power
    • Use density as a conversion factor

    An Introduction to Dimensional Analysis From Crash Course Chemistry

    Video \(\PageIndex{1}\): Watch this video for an introduction to dimensional analysis

    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:

    \[\mathrm{speed=\dfrac{distance}{time}}\]

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

    \[\mathrm{\dfrac{100\: m}{10\: s}=10\: m/s}\]

    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:

    \[\mathrm{time=\dfrac{distance}{speed}}\]

    The time can then be computed as:

    \[\mathrm{\dfrac{25\: m}{10\: m/s}=2.5\: s}\]

    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,

    \[\mathrm{\dfrac{2.54\: cm}{1\: in.}\:(2.54\: cm=1\: in.)\: or\: 2.54\:\dfrac{cm}{in.}}\]

    Several other commonly used conversion factors are given in Table \(\PageIndex{1}\).

    Table \(\PageIndex{1}\): Common Conversion Factors
    Length Volume Mass
    1 m = 1.0936 yd = 3.28 ft 1 L = 1.0567 qt = 10-3 m3 = 1 dm3 = 103 cm3 1 kg = 2.2046 lb
    1 in. = 2.54 cm (exact) 1 qt = 0.94635 L 1 lb = 453.59 g = 16 oz
    1 cm = 0.39370 in 1 gal = 4 qt = 3.7854 L 28.35 g = 1 oz
    1 km = 0.62137 mi 1 ft3 = 28.317 L
    1 mi = 1609.3 m = 5280 ft 1 tbsp = 14.787 mL
    1 cm3 = 1 mL
    1 in3 = 16.4 cm3
    1 fl. oz. = 29.5735 mL

    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:

    \[\mathrm{34\: \cancel{in.} \times \dfrac{2.54\: cm}{1\:\cancel{in.}}=86\: cm}\]

    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

    \[\mathrm{\dfrac{in.\times cm}{in.}}.\]

    Just as for numbers, a ratio of identical units is also numerically equal to one,

    \[\mathrm{\dfrac{in.}{in.}=1}\]

    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 \(\PageIndex{1}\): Using a Unit Conversion Factor

    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 \(\PageIndex{1}\)).

    Solution

    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.

    \[x\:\mathrm{oz=125\: g\times unit\: conversion\: factor} \nonumber\]

    We write the unit conversion factor in its two forms:

    \[\mathrm{\dfrac{1\: oz}{28.349\: g}\:and\:\dfrac{28.349\: g}{1\: oz}} \nonumber\]

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

    \[\begin{align*}
    x\:\ce{oz}&=\mathrm{125\:\cancel{g}\times \dfrac{1\: oz}{28.349\:\cancel{g}}}\\
    &=\mathrm{\left(\dfrac{125}{28.349}\right)\:oz}\\
    &=\mathrm{4.41\: oz\: (three\: significant\: figures)}
    \end{align*}\]

    Exercise \(\PageIndex{1}\)

    Convert a volume of 9.345 qt to liters.

    Answer

    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 \(\PageIndex{2}\): Computing Quantities from Measurement Results

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

    Solution

    Since \(\mathrm{density=\dfrac{mass}{volume}}\), 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 \(\times\) unit conversion factor. The necessary conversion factors are given in Table 1.7.1: 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:

    \[\mathrm{9.26\:\cancel{lb}\times \dfrac{453.59\: g}{1\:\cancel{lb}}=4.20\times 10^3\:g} \nonumber \]

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

    1. Convert quarts to liters.

    \[\mathrm{4.00\:\cancel{qt}\times\dfrac{1\: L}{1.0567\:\cancel{qt}}=3.78\: L} \nonumber\]

    1. Convert liters to milliliters.

    \[\mathrm{3.78\:\cancel{L}\times\dfrac{1000\: mL}{1\:\cancel{L}}=3.78\times10^3\:mL} \nonumber\]

    Then,

    \[\mathrm{density=\dfrac{4.20\times10^3\:g}{3.78\times10^3\:mL}=1.11\: g/mL} \nonumber\]

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

    \[\mathrm{\dfrac{9.26\:\cancel{lb}}{4.00\:\cancel{qt}}\times\dfrac{453.59\: g}{1\:\cancel{lb}}\times\dfrac{1.0567\:\cancel{qt}}{1\:\cancel{L}}\times\dfrac{1\:\cancel{L}}{1000\: mL}=1.11\: g/mL} \nonumber\]

    Exercise \(\PageIndex{2}\)

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

    Answer

    \(\mathrm{2.956\times10^{-2}\:L}\)

    Example \(\PageIndex{3}\): Computing Quantities from Measurement Results

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

    1. What (average) fuel economy, in miles per gallon, did the Roadster get during this trip?
    2. If gasoline costs $3.80 per gallon, what was the fuel cost for this trip?

    Solution

    (a) We first convert distance from kilometers to miles:

    \[\mathrm{1250\: km\times\dfrac{0.62137\: mi}{1\: km}=777\: mi} \nonumber\]

    and then convert volume from liters to gallons:

    \[\mathrm{213\:\cancel{L}\times\dfrac{1.0567\:\cancel{qt}}{1\:\cancel{L}}\times\dfrac{1\: gal}{4\:\cancel{qt}}=56.3\: gal} \nonumber\]

    Then,

    \[\mathrm{(average)\: mileage=\dfrac{777\: mi}{56.3\: gal}=13.8\: miles/gallon=13.8\: mpg} \nonumber\]

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

    \[\mathrm{\dfrac{1250\:\cancel{km}}{213\:\cancel{L}}\times\dfrac{0.62137\: mi}{1\:\cancel{km}}\times\dfrac{1\:\cancel{L}}{1.0567\:\cancel{qt}}\times\dfrac{4\:\cancel{qt}}{1\: gal}=13.8\: mpg} \nonumber \]

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

    \[\mathrm{56.3\: gal\times\dfrac{$3.80}{1\: gal}=$214} \nonumber \]

    Exercise \(\PageIndex{3}\)

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

    1. What (average) fuel economy, in miles per gallon, did the Prius get during this trip?
    2. If gasoline costs $3.90 per gallon, what was the fuel cost for this trip?
    Answer a

    51 mpg

    Answer b

    $62

    Units Raised to a Power

    When calculating area or volume, you multiply together lengths, widths, and heights. Just like in our dimensional analysis above, our units and our numbers both undergo the mathematical operation, meaning that multiplying the quantity of length by the quantity of width also multiplies the units.

    $$42cm*33cm=1386 cm^{^{2}}\]

    This complicates the conversion of units, however, since our GIVEN conversion factors often only account for one dimension, not two or three. When building a conversion factor for units raised to a power, we simply raise the conversion factor to the power we want our final units in. The trick is to remember to raise the QUANTITY and UNITS by the power.

    Example \(\PageIndex{4}\): Computing Quantities from Measurement Results

    Convert 5.70 L to in3.

    Solution

    We must first convert L to mL, which as we saw in Section 1.1, is equivalent to cm3.

    $$5.70 L*\frac{1000 mL}{1 L}*\frac{1 cm^{3}}{1 mL}=5700cm^{3}\]

    Now, if we examine the table of conversion factors (Table \(\PageIndex{1}\)), we find that there is 16.4 cm3 in 1 in3.

    $$5700cm^{3}*\frac{1in^{^{3}}}{16.4cm^{3}}=347.6cm^{3}\]

    We could have just as easily have done this if we hadn't been given the direct conversion factor between cm3 and in3. We simply would have had to raise the conversion factor between cm and in to the third power.

    $$5700cm^{3}*\left ( \frac{1in}{2.54cm} \right )^{3}=347.6in^{3}\]

    The trick with this way of doing the calculation is you have to remember to apply the power to EVERYTHING:

    $$\left ( \frac{1in}{2.54cm} \right )^{3}=\frac{\left ( 1^{3}in^{3} \right )}{2.54^{3}cm^{3}}\]

    Note: We are ignoring a concept known as "significant figures" in this example. You will cover the rules for significant figures in next week's lab.

    Exercise \(\PageIndex{4}\)

    Convert 274 m2 to in2.

    Answer

    424701 in2

    Using Density as your Conversion Factor

    Density can also be used as a conversion factor. This is useful for determining how much the volume of something weighs without having to mass the object - which is particularly useful if your object is too heavy to actually weigh, like a jet or rocket.

    Exercise \(\PageIndex{5}\)

    A commercial jet is fueled with 156,874 L of jet fuel.

    a) If the density of the fuel is 0.768 g/cm3, what is the mass of the fuel in kilograms?

    b) If the jet weights 443.613 Mg without passengers or fuel, what is the mass when the fuel is added?

    Answer a

    120,479 kg

    Answer b

    564,092 kg or 564.092 Mg

    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.

    Glossary

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

    Contributors and Attributions

    • Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193-2bd...a7ac8df6@9.110).

    • Adelaide Clark, Oregon Institute of Technology
    • Crash Course Chemistry, Crash Course is a division of Complexly and videos are free to stream for educational purposes.

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