2.2: Dimensional Analysis

Learning Objective

• Convert metric and English based units.
• Report converted values in correct scientific notation with the corresponding units.
• Master using your calculator.

Conversion Factors

We have an expression,

\[\dfrac{3\,ft}{1\,yd} =1 \label{1} \]

Equation \ref{1} is a strange way to write 1, but it makes sense: 3 ft equal 1 yd, so the quantities in the numerator and denominator are the same quantity, just expressed with different units. The expression in Equation \ref{1} is called a conversion factor and it is used to formally change the unit of a quantity into another unit. (The process of converting units in such a formal fashion is sometimes called dimensional analysis or the factor label method.)

To see how this happens, let us start with the original quantity: \(4\, yd\) and let us multiply this quantity by 1. Rather than multiplying by just 1, let us write 1 as

\[ 1= \dfrac{3 \, ft}{1 \, yd} \nonumber \]

When you multiply anything by 1, you don’t change the value of the quantity.

\[\begin{align*} 4 \, yd &= 4 \, yd \times 1 \\[4pt] &= 4 \, yd \times \dfrac{3 \, ft}{1 \, yd} \end{align*} \]

The 4 yd term can be thought of as yd/1; that is, it can be thought of as a fraction with 1 in the denominator. We are essentially multiplying fractions. If the same thing appears in the numerator and denominator of a fraction, they cancel. In this case, what cancels is the unit yard:

\[4 \, \cancel{yd}\times \dfrac{3 \, ft}{1 \, \cancel{yd}} \nonumber \]

That is all that we can cancel. Now, multiply and divide all the numbers to get the final answer:

\[\dfrac{4\times 3 \, ft}{1}= \dfrac{12 \, ft}{1}= 12 \, ft \nonumber \]

Again, we get an answer of 12 ft, just as we did originally. But in this case, we used a more formal procedure that is applicable to a variety of problems.

How many millimeters are in 14.66 m? To answer this, we need to construct a conversion factor between millimeters and meters and apply it correctly to the original quantity. One method would entail using the conversion factor of \(1\, mm = 1 \times 10^{-3}\,m\)

\[\begin{align} 14.66\,\cancel{m}\times \dfrac{1\,mm}{10^{-3}\,\cancel{m}} &= 14660\,mm \\[5pt] &= 1.46\times 10^{4}\,mm \end{align} \nonumber \]

Which conversion factor do we use? The answer is based on what unit you want to get rid of in your initial quantity. The original unit of our quantity is meters, which we want to convert to millimeters. Because the original unit is assumed to be in the numerator, to get rid of it, we want the meter unit in the denominator; then they will cancel. Using the positive metric exponents, we have an alternative method to calculate the same answer as shown previously. Canceling units and performing the mathematics, we get:

\[\begin{align} 14.66\, \cancel{m} \times \dfrac{1000\,mm}{1\,\cancel{m}} &= 14660\,mm \\[5pt] &= 1.46\times 10^{4}\,mm \end{align} \nonumber \]

Note how \(m\) cancels, leaving \(mm\), which is the unit of interest.

The ability to construct and apply proper conversion factors is a very powerful mathematical technique in chemistry. You need to master this technique if you are going to be successful in this and future science courses.

 

Example \(\PageIndex{1}\)

1. Convert 35.9 kL to liters.
2. Convert 555 nm to meters.

Solution

1. 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 1000L/ 1kL. Applying this conversion factor, we get:

\[35.9\, \cancel{kL} \times \dfrac{1000\,L}{1\, \cancel{kL}}= 35,900\,L \nonumber \]

2. 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:

\[\dfrac{1\,m}{10^{9}\,m} \nonumber \]

Applying this conversion factor, we get

\[\begin{align*}555\,\cancel{nm} \times \dfrac{1\,m}{10^{9}\,\cancel{nm}} &= 0.000000555\,m \\[5pt] &= 5.55\times 10^{-7}\,m \end{align*} \]

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

Exercise \(\PageIndex{1}\)

1. Convert 67.08 μL to liters.
2. Convert 56.8 m to kilometers.

Answer a

6.708 × 10⁻⁵ L

Answer b

5.68 × 10⁻² km

Exercise \(\PageIndex{2}\)

How many milliliters are in 607.8 kL?

Answer

This problem involves multiple steps that require going through the metric base of liters. Think of a pathway of a conversion before setting up these types of problems. The suggested pathway would be mL ⇒ L ⇒ kL

\[ \begin{align*} 607.8 \,kL \times \dfrac{1000\,L}{1\,kL} &= 6.08\times 10^{5}\,L \nonumber \\[5pt] &= 6.08\times 10^5 \, L \dfrac{1000\,mL}{1\,L} \\[5pt] &= 608000000\,mL \\[5pt] &= 6.08\times 10^{8}\,mL \nonumber \end{align*} \]

Exercise \(\PageIndex{3}\)

Kathleen Hsu Barron (Furman University, Class of 2011) is exploring Switzerland with her husband Patrick Barron (Furman University, Class of 2011). The sign she is pointing to is a typical speed in km/hr. Convert this distance to miles/hr and feet/hr.

                                           Figure \(\PageIndex{1}\):  Kathleen Hsu Barron standing by speed limit sign. (CC-BY-NC-ND, Patrick Barron)

Answer

Again, this problem is a multi-step problem. There are multiple ways that one could proceed to solve this problem. With the conversion factors in this course, the suggested round would be miles ⇒ kilometers ⇒ meters ⇒ centimeters ⇒ inches ⇒ feet. Another method could utilize miles ⇒ feet.

18.6 miles/hr, 9.8 x 10⁴ ft/hr

In this course, we will not use significant figures. You will need to read the directions on your assignment, quiz, or test and round accordingly.

Applications of Conversions and the Metric System

                                   

                                   Figure \(\PageIndex{2}\):  Why is folic acid important in pregnancy. (CC-BY-NC-ND, Andy Brunning).

Question:  The FDA recommends taking .400 mg of folic acid a day before pregnancy.  How many micrograms is this dosage?  Hint- think of the pathway mg⇒? ⇒μg

 

Need More Practice?

• Turn to Section 2.E of this OER and work problems #1, #3, #5, #6, and #8.

Contributors and Attributions

• Elizabeth R. Gordon (Furman University)

• Hayden Cox (Furman University)
• Makena Song (Furman University)