Dimensional Analysis

Converting units accurately is essential in chemistry. Dimensional analysis is a systematic method that treats units as algebraic quantities that can cancel. By using conversion factors (equalities between units written as fractions), we build a clear path from a starting quantity to a desired one, with unwanted units canceling out along the way.

For all calculations, refer to the rules for Significant Figures to ensure your final answers reflect the correct precision.

The Factor-Label Method

The factor-label method is a problem-solving technique that uses conversion factors to change the units of a quantity. A conversion factor is a fraction built from an equality, where the numerator and denominator represent the same physical quantity expressed in different units. Because the numerator and denominator are equivalent, the fraction equals one. Multiplying a measurement by a conversion factor therefore changes its units, but not the physical quantity.

Building Conversion Factors

A relationship between two units can be stated as an equality. For example, the inch is defined as exactly 2.54 centimeters:

\[ 1~\text{in} = 2.54~\text{cm} \]

From this equality, we can create two conversion factors. We choose the one that allows us to cancel the unwanted unit:

\[ \frac{1~\text{in}}{2.54~\text{cm}} \quad \text{or} \quad \frac{2.54~\text{cm}}{1~\text{in}} \]

NoteExact vs. Inexact Numbers

A critical skill in dimensional analysis is knowing which numbers limit your significant figures.

Exact Numbers have infinite significant figures and never limit your calculation. These include:

  • Numbers by definition: 1 m = 100 cm, 1 dozen = 12 items.
  • Counted quantities: “29 bottles,” “3 measurements.”

Inexact (Measured) Numbers have a limited number of significant figures. These are any numbers that come from a measurement.

  • Examples: A mass of 2.50 g, a volume of 2.0 L, a speed of 70.0 mph.

The Rule: Your final answer is always limited by the precision of the least precise inexact number used in the calculation.

Setting Up a Conversion

Quick Reference

Conversion Factors and Metric Prefixes

Follow these steps for any unit conversion:

  1. Identify the starting quantity and its units
  2. Identify the target units
  3. Find conversion factors that connect them
  4. Arrange factors so unwanted units cancel
  5. Calculate and round to the correct significant figures

Example: Convert 5.2 inches to centimeters.

The conversion factor 1 in = 2.54 cm is exact, so the answer is limited by the starting value (2 sig figs). We arrange the factor with cm on top so that inches cancel:

\[ 5.\bar{2}~\text{in} \left( {\color{green}\frac{2.54~\text{cm}}{1~\text{in}}} \right) = 1\bar{3}.208~\text{cm} = 13~\text{cm} \]

Multiple Conversion Factors

We can string together multiple conversion factors to convert one quantity to another.

Example: Convert 3.2 km to mm.

We need two conversion factors to bridge km → m → mm. Both are exact (metric prefixes), so the answer is limited by the starting value (2 sig figs).

\[ 3.\bar{2}~\text{km} \left( {\color{green}\frac{10^3~\text{m}}{1~\text{km}}} \right) \left( {\color{red} \frac{10^3~\mathrm{mm}}{1~\mathrm{m}}} \right) = 3~\bar{2}00~000~\text{mm} = 3.2\times 10^{6}~\mathrm{mm} \]

Although we string multiple conversion factors together, each factor applies sequentially: km cancels first, then m cancels, leaving mm.

When converting compound units like speed, we handle the numerator and denominator separately:

Example: Convert 65 mph to m s−1.

We need to convert miles → meters and hours → seconds. The conversion 1 mi = 1.609344 km is exact; metric prefixes are exact; 1 h = 3600 s is exact. The answer is limited by the starting value (2 sig figs).

\[ 6\bar{5}~\frac{\text{mi}}{\text{h}} \left( \frac{1.609~344~\text{km}}{1~\text{mi}} \right) \left( \frac{10^3~\text{m}}{1~\text{km}} \right) \left( \frac{1~\text{h}}{3600~\text{s}} \right) = 2\bar{9}.057~\text{m}~\text{s}^{-1} = 29~\text{m}~\text{s}^{-1} \]

Practice Problems

Practice

Assume a lifetime is exactly 80 years. Determine this time frame in days, hours, and minutes. Repeat the calculation assuming the lifetime is an inexact measurement of 80.0 years. Treat years as non-leap years.

Solution

Required Conversion Factors: All of these are exact by definition.

  • 1 y = 365 d
  • 1 d = 24 h
  • 1 h = 60 min

Part A: Using an Exact Starting Value (80 years)

When all numbers in a calculation are exact, the result is also exact and has infinite significant figures.

  1. Time in Days: \[ 80~\text{y} \left( \frac{365~\text{d}}{1~\text{y}} \right) = 29~200~\text{d} \]

  2. Time in Hours: \[ 80~\text{y} \left( \frac{365~\text{d}}{1~\text{y}} \right) \left( \frac{24~\text{h}}{1~\text{d}} \right) = 700~800~\text{h} \]

  3. Time in Minutes: \[ 80~\text{y} \left( \frac{365~\text{d}}{1~\text{y}} \right) \left( \frac{24~\text{h}}{1~\text{d}} \right) \left( \frac{60~\text{min}}{1~\text{h}} \right) = 42~048~000~\text{min} \]

Part B: Using an Inexact Starting Value (80.0 years)

The starting value, 80.0 y, has 3 significant figures. This will limit the precision of our final answers.

  1. Time in Days: \[ 80.\bar{0}~\text{y} \left( \frac{365~\text{d}}{1~\text{y}} \right) = 29~\bar{2}00~\text{d} \rightarrow 2.92 \times 10^{4}~\text{d} \]

  2. Time in Hours: \[ 29~\bar{2}00~\text{d} \left( \frac{24~\text{h}}{1~\text{d}} \right) = 70\bar{1}~000~\text{h} \rightarrow 7.01 \times 10^{5}~\text{h} \]

  3. Time in Minutes: \[ 70\bar{1}~000~\text{h} \left( \frac{60~\text{min}}{1~\text{h}} \right) = 4.2\bar{0}6~0 \times 10^{7}~\text{min} \rightarrow 4.21 \times 10^{7}~\text{min} \]

Practice

A water tank contains exactly 15 gal of water. You need to transfer this water into bottles that each hold 2.0 L.

  1. How many bottles are needed?
  2. If each bottle costs $0.74, what is the total cost?
  3. If you earn $8.25 per hour, how many hours must you work to afford the bottles?
Solution

Required Conversion Factors:

  • 1 gal = 3.785412 L (inexact; rounded from exact value)
  • 1 bottle = 2.0 L (inexact)
  • 1 bottle = $0.74 (stipulated, 2 sig figs)
  • 1 h = $8.25 (stipulated, 3 sig figs)

1. Bottles Needed

The calculation is limited by the bottle volume (2.0 L), which has 2 significant figures. \[ 15~\text{gal} \left( \frac{3.78541\bar{2}~\text{L}}{1~\text{gal}} \right) \left( \frac{1~\text{bottle}}{2.\bar{0}~\text{L}} \right) = 2\bar{8}.39059...~\text{bottles} \] The calculated result rounds to 28 bottles. However, 28 bottles would not be enough to hold all the water. In a practical scenario like this, we must round up to the next whole number.

Final Answer: You would need 29 bottles.

2. Cost of Bottles

For this step, the number of bottles becomes an exact, counted quantity (29). The cost per bottle ($0.74) is a stipulated value with 2 significant figures. The precision of the original volume measurement no longer applies. \[ 29~\text{bottles} \left( \frac{\$0.7\bar{4}}{1~\text{bottle}} \right) = \$2\bar{1}.46 = \$21 \]

3. Work Hours

The total cost $21 has 2 significant figures. The wage ($8.25) is a stipulated value with 3 significant figures. The answer is limited to 2 significant figures. \[ \$2\bar{1} \left( \frac{1~\text{h}}{\$8.2\bar{5}} \right) = 2.5\bar{4}54...~\text{h} = 2.5~\text{h} \]

Practice

A road trip from New York, NY to Mexico City, Mexico is reported to be 4,145.67 km long.

  1. How long is the trip in miles?
  2. Assuming an average speed of 70.0 mph the entire trip with no stops, how long (in h) would the trip be?
  3. Suppose the traveler made the trip at a constant 80.0 mph. How much time (in h) would you save if travelling at this higher speed?
Solution

Required Conversion Factor:

  • 1 mi = 1.609344 km (exact by definition)

1. Length of Trip in Miles

The initial distance (4,145.67 km) has 6 significant figures. The conversion factor is exact. The result must have 6 significant figures. \[ 4~145.6\bar{7}~\text{km} \times \left( \frac{1~\text{mi}}{1.609~344~\text{km}} \right) = 2~575.99\bar{9}~9...~\text{mi} \rightarrow \mathbf{2~576.00~\text{mi}} \]

2. Trip Duration at 70.0 mph

This is a multi-step calculation. We must use the unrounded value from the previous step. The calculation is limited by the speed (70.0 mph), which has 3 significant figures. \[ 2~575.999~9...~\text{mi} \times \left( \frac{1~\text{h}}{70.0~\text{mi}} \right) = 36.79\bar{9}...~\text{h} \rightarrow \mathbf{36.8~\text{h}} \]

3. Time Saved

First, calculate the trip duration at the new speed. This is also limited to 3 significant figures.

\[ \text{Time at 80.0 mph} = 2~575.999~9...~\text{mi} \times \left( \frac{1~\text{h}}{80.0~\text{mi}} \right) = 32.19\bar{9}...~\text{h} \]

Now, subtract the two times. Per the addition/subtraction rule, the answer is limited by the least precise decimal place. Let’s look at the intermediate values before rounding: \(36.799...\) and \(32.199...\) The least precise of these is the hundredths place.

\[ \text{Time Saved} = 36.799...~\text{h} - 32.199...~\text{h} = 4.60\bar{0}...~\text{h} \]

The result is limited to the hundredths place.

Final Answer: \(4.60~\text{h}\)

Practice

The goal of this problem is to use multiple significant figure rules in a multi-step calculation to determine the final answer to the correct number of significant figures.

What is the heat of vaporization of water (ΔvapH in J mol−1) if the vapor pressure of water is 0.01212 atm (P1) at 10.0 °C (T1) and 0.46794 atm (P2) at 80.0 °C (T2)? The molar ideal gas constant (R) is 8.314 J mol−1 K−1.

Use the following equation:

\[ \ln \left ( \frac{P_2}{P_1} \right ) = \frac{\Delta_{\mathrm{vap}} H}{R} \left ( \frac{1}{T_1} - \frac{1}{T_2} \right ) \]

Solution

1. Convert Temperatures to Kelvin

The least precise value is in the tenths place.

\[ T_1 = 10.\bar{0} + 273.15 = 283.\bar{1}5~\text{K} \]

\[ T_2 = 80.\bar{0} + 273.15 = 353.\bar{1}5~\text{K} \]

2. Rearrange and Solve for ΔvapH

\[ \Delta_{\mathrm{vap}} H = R ~ \frac{\ln \left ( \frac{P_2}{P_1} \right )}{\left ( \frac{1}{T_1} - \frac{1}{T_2}\right )} \]

Let’s calculate the terms inside the fraction first, tracking sig figs. These values will be used in a final calculation so the unrounded numbers will be retained with two guard digits.

  • Pressure Ratio:

    \[ \begin{align*} \frac{P_2}{P_1} &= \frac{0.4679\bar{4}~\text{atm}}{0.0121\bar{2}~\text{atm}} \\[1.5ex] &= 38.6\bar{0}89 \quad (\text{limited to 4 sig figs by } P_1) \end{align*} \]

  • Log of Pressure Ratio:

    Since this calculation uses a result from a previous calculation, use an unrounded result (at least 2 guard digits).

    \[ \begin{align*} \ln\left ( \frac{P_2}{P_1} \right ) &= \ln(38.60\bar{8}91) \\[1.5ex] &= 3.6534\bar{8}30 \end{align*} \]

  • Temperature Terms:

    \[ \begin{align*} \dfrac{1}{T_1} - \dfrac{1}{T_2} &= \frac{1}{283.\bar{1}5~\mathrm{K}} - \frac{1}{353.\bar{1}5~\mathrm{K}} \\[1.5ex] &= 0.00353\bar{1}69~\mathrm{K^{-1}} - 0.00283\bar{1}65~\mathrm{K^{-1}} \\[1.5ex] &= 0.000700\bar{0}39~\mathrm{K^{-1}} \quad (\text{4 sig figs}) \end{align*} \]

Now, plug these back into the main equation.

\[ \begin{align*} \Delta_{\mathrm{vap}} H &= R ~ \dfrac{\ln \left ( \frac{P_2}{P_1} \right )}{\left ( \frac{1}{T_1} - \frac{1}{T_2}\right )} \\[1.5ex] &= \left ( \dfrac{8.31\bar{4}~\mathrm{J}}{\mathrm{mol~K}} \right ) \dfrac{3.6534\bar{8}30}{0.000700\bar{0}39~\mathrm{K^{-1}}} \\[1.5ex] &= 43,3\bar{9}2.6~\mathrm{J~mol^{-1}} \end{align*} \] The calculation is limited by the gas constant, R, which has 4 significant figures. Rounding to 4 significant figures gives: Final Answer: 43,390 J mol−1