# 5.1: Unit conversions

- Page ID
- 435655

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### 13

## Harper College Chemistry Department

## Summary

By the end of this section, you will be able to:

- 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

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:

$\text{speed} = \dfrac{\text{distance}}{\text{time}}$

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

$\text{speed} = \dfrac{\text{distance}}{\text{time}} = \dfrac{\text{100 m}}{\text{10 s}} = \dfrac{\text{10 m}}{\text{s}} = 10 m/s$

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:

$\text{time} = \dfrac{\text{distance}}{\text{speed}} = \dfrac{\text{25 m}}{\text{10 m/s}} = \text{2.5 s}$

or

^{-1}s/m) = s

The time can then be computed as:

$\text{time s} = \text{distance m} \times \dfrac{\text{time s}}{\text{distance m}}$

$\text{time s} = \text{25 m} \times \dfrac{\text{1 s}}{\text{10 m}} = 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. So from

2.54 cm = 1 in

we get

$\frac{\text{2.54 cm}}{\text{1 in}}$ and $\frac{\text{1 in}}{\text{2.54 cm}}$ which can be used as conversion factors.

Several other commonly used conversion factors are given in Table 1.

Table 1: Common Conversion Factors | ||
---|---|---|

Length | Volume | Mass |

1 m = 1.0936 yd | 1 L = 1.0567 qt | 1 kg = 2.2046 lb |

1 in. = 2.54 cm (exact) | 1 qt = 0.94635 L | 1 lb = 453.59 g |

1 km = 0.62137 mi | 1 ft^{3} = 28.317 L |
1 (avoirdupois) oz = 28.349 g |

1 mi = 1609.3 m | 1 tbsp = 14.787 mL | 1 (troy) oz = 31.103 g |

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:

$\mathrm{34\ \cancel{in.} \times \dfrac{2.54\ cm}{\cancel{in.}} = 86\ cm}$

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 $\mathrm{\frac{in. \times cm}{in.}}$. Just as for numbers, a ratio of identical units is also numerically equal to one, $\mathrm{\frac{in.}{in.} = 1}$, 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.

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 1).

Solution: If we have the conversion factor, we can determine the mass in ounces using an equation similar the one used for converting length from inches to centimeters.

We write the unit conversion factor in its two forms:

$\mathrm{\dfrac{1\ oz.}{28.349\ g}\ and\ \dfrac{28.349\ g}{1\ oz.}}$

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

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

Check Your Learning: Convert a volume of 9.345 qt to liters.

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.

**Computing Quantities from Measurement Results and Known Mathematical Relations: **

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 $\mathrm{density = \frac{mass}{volume}}$, 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 × unit conversion factor. The necessary conversion factors are given in Table 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:

$9.26\text{ \cancel{lb}} \times \frac{453.59\text{ g}}{1\text{ \cancel{lb}}} = \SI{4.20d3}{g}$ (need 3 significant figures in the answer)

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

*Convert quarts to liters.*$4.00\text{ \cancel{qt}} \times \dfrac{\SI{1}{L}}{1.0567 \text{ \cancel{qt}}} = \SI{3.78}{L}$*Convert liters to milliliters.*

$\mathrm{3.78\ \cancel{L} \times \dfrac{1000\ mL}{1\ \cancel{L}}} = \SI{3.78d3}{mL} = \SI{3780}{mL} $

Then,

$\mathrm{density} = \dfrac{\SI{4.20d3}{g}}{\SI{3.78d3}{mL}} = \SI{1.11}{g/mL}$

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

$\dfrac{9.26\text{ \cancel{lb}}}{4.00\text{ \cancel{qt}}} \times \dfrac{453.59\text{ g}}{1\text{ \cancel{lb}}} \times \dfrac{1.0567\text{ \cancel{qt}}}{1\text{ \cancel{L}}} \times \dfrac{1\text{ \cancel{L}}}{1000\text{ mL}} = \SI{1.11}{g/mL}$

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

2.956×10^{−2}L

**Computing Quantities from Measurement Results and Known Mathematical Relations:**

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

(a) What (average) fuel economy, in miles per gallon, did the Roadster get during this trip?

(b) 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:

$1250\text{ \cancel{km}} \times \dfrac{\SI{0.62137}{mi}}{1\text{ \cancel{km}}} = \SI{777}{mi}$

and then convert volume from liters to gallons:

$213\text{ \cancel{L}} \times \dfrac{1.0567\text{ \cancel{qt}}}{\text{\cancel{L}}} \times \dfrac{1\text{ gal}}{4\text{ \cancel{qt}}} = \SI{56.3}{gal}$

Then,

$\text{(average) mileage} = \dfrac{777\text{ mi}}{56.3\text{ gal}} = 13.8 \text{ miles/gallon} = 13.8\text{ mpg(average)}$

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

$\dfrac{1250\text {\cancel{km}}}{213\text{ \cancel{L}}} \times \dfrac{0.62137\text{ mi}}{1\text{ \cancel{km}}} \times \dfrac{1\text{ \cancel{L}}}{1.0567\text{ \cancel{qt}}} \times \dfrac{4\text{ \cancel{qt}}}{1\text{ gal}}=13.8\text{ mpg}$

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

\[\SI{56.3}{gal} \times \dfrac{\$3.80}{\SI{1}{gal}} = \$214\]

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

(a) What (average) fuel economy, in miles per gallon, did the Prius get during this trip?

(b) If gasoline costs \$ 3.90 per gallon, what was the fuel cost for this trip?

(a) 51 mpg; (b) \$ 62

### Key Concepts and 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.

### Key Equations

- T°C=(5/9×T°F)−32 T°C=(5/9×T°F)−32
- T°F=(9/5×T°C)+32 T°F=(9/5×T°C)+32
- TK=°C+273.15 TK=°C+273.15
- T°C=K−273.15 T°C=K−273.15

### Chemistry End of Chapter Exercises

1. Write conversion factors (as ratios) for the number of:

(a) yards in 1 meter

(b) liters in 1 liquid quart

(c) pounds in 1 kilogram

2. Write conversion factors (as ratios) for the number of:

(a) kilometers in 1 mile

(b) liters in 1 cubic foot

(c) grams in 1 ounce

3. The label on a soft drink bottle gives the volume in two units: 2.0 L and 67.6 fl oz. Use this information to derive a conversion factor between the English and metric units. How many significant figures can you justify in your conversion factor?

4. The label on a box of cereal gives the mass of cereal in two units: 978 grams and 34.5 oz. Use this information to find a conversion factor between the English and metric units. How many significant figures can you justify in your conversion factor?

5. Soccer is played with a round ball having a circumference between 27 and 28 in. and a weight between 14 and 16 oz. What are these specifications in units of centimeters and grams?

6. A woman’s basketball has a circumference between 28.5 and 29.0 inches and a maximum weight of 20 ounces (two significant figures). What are these specifications in units of centimeters and grams?

7. How many milliliters of a soft drink are contained in a 12.0-oz can?

8. A barrel of oil is exactly 42 gal. How many liters of oil are in a barrel?

9. The diameter of a red blood cell is about 3 × 10^{−4} in. What is its diameter in centimeters?

10. The distance between the centers of the two oxygen atoms in an oxygen molecule is 1.21 × 10^{−8} cm. What is this distance in inches?

11. Is a 197-lb weight lifter light enough to compete in a class limited to those weighing 90 kg or less?

12. A very good 197-lb weight lifter lifted 192 kg in a move called the clean and jerk. What was the mass of the weight lifted in pounds?

13. Many medical laboratory tests are run using 5.0 μL blood serum. What is this volume in milliliters?

14. If an aspirin tablet contains 325 mg aspirin, how many grams of aspirin does it contain?

15. Use scientific (exponential) notation to express the following quantities in terms of the SI base units in [link]:

(a) 0.13 g

(b) 232 Gg

(c) 5.23 pm

(d) 86.3 mg

(e) 37.6 cm

(f) 54 μm

(g) 1 Ts

(h) 27 ps

(i) 0.15 mK

16. Complete the following conversions between SI units.

(a) 612 g = ________ mg

(b) 8.160 m = ________ cm

(c) 3779 μg = ________ g

(d) 781 mL = ________ L

(e) 4.18 kg = ________ g

(f) 27.8 m = ________ km

(g) 0.13 mL = ________ L

(h) 1738 km = ________ m

(i) 1.9 Gg = ________ g

17. Gasoline is sold by the liter in many countries. How many liters are required to fill a 12.0-gal gas tank?

18. Milk is sold by the liter in many countries. What is the volume of exactly 1/2 gal of milk in liters?

19. A long ton is defined as exactly 2240 lb. What is this mass in kilograms?

20. Make the conversion indicated in each of the following:

(a) the men’s world record long jump, 29 ft 4¼ in., to meters

(b) the greatest depth of the ocean, about 6.5 mi, to kilometers

(c) the area of the state of Oregon, 96,981 mi^{2}, to square kilometers

(d) the volume of 1 gill (exactly 4 oz) to milliliters

(e) the estimated volume of the oceans, 330,000,000 mi^{3}, to cubic kilometers.

(f) the mass of a 3525-lb car to kilograms

(g) the mass of a 2.3-oz egg to grams

21. Make the conversion indicated in each of the following:

(a) the length of a soccer field, 120 m (three significant figures), to feet

(b) the height of Mt. Kilimanjaro, at 19,565 ft the highest mountain in Africa, to kilometers

(c) the area of an 8.5 × 11-inch sheet of paper in cm^{2}

(d) the displacement volume of an automobile engine, 161 in.^{3}, to liters

(e) the estimated mass of the atmosphere, 5.6 × 10^{15} tons, to kilograms

(f) the mass of a bushel of rye, 32.0 lb, to kilograms

(g) the mass of a 5.00-grain aspirin tablet to milligrams (1 grain = 0.00229 oz)

22. Many chemistry conferences have held a 50-Trillion Angstrom Run (two significant figures). How long is this run in kilometers and in miles? (1 Å = 1 × 10^{−10} m)

23. A chemist’s 50-Trillion Angstrom Run (see Exercise) would be an archeologist’s 10,900 cubit run. How long is one cubit in meters and in feet? (1 Å = 1 × 10^{−8} cm)

24. The gas tank of a certain luxury automobile holds 22.3 gallons according to the owner’s manual. If the density of gasoline is 0.8206 g/mL, determine the mass in kilograms and pounds of the fuel in a full tank.

25. As an instructor is preparing for an experiment, he requires 225 g phosphoric acid. The only container readily available is a 150-mL Erlenmeyer flask. Is it large enough to contain the acid, whose density is 1.83 g/mL?

26. To prepare for a laboratory period, a student lab assistant needs 125 g of a compound. A bottle containing 1/4 lb is available. Did the student have enough of the compound?

27. A chemistry student is 159 cm tall and weighs 45.8 kg. What is her height in inches and weight in pounds?

28. In a recent Grand Prix, the winner completed the race with an average speed of 229.8 km/h. What was his speed in miles per hour, meters per second, and feet per second?

29. Solve these problems about lumber dimensions.

(a) To describe to a European how houses are constructed in the US, the dimensions of “two-by-four” lumber must be converted into metric units. The thickness × width × length dimensions are 1.50 in. × 3.50 in. × 8.00 ft in the US. What are the dimensions in cm × cm × m?

(b) This lumber can be used as vertical studs, which are typically placed 16.0 in. apart. What is that distance in centimeters?

30. The mercury content of a stream was believed to be above the minimum considered safe—1 part per billion (ppb) by weight. An analysis indicated that the concentration was 0.68 parts per billion. What quantity of mercury in grams was present in 15.0 L of the water, the density of which is 0.998 g/ml? (1 ppb Hg=1 ng Hg/1 g water)

31. Calculate the density of aluminum if 27.6 cm^{3} has a mass of 74.6 g.

32. Osmium is one of the densest elements known. What is its density if 2.72 g has a volume of 0.121 cm^{3}?

33. Calculate these masses.

(a) What is the mass of 6.00 cm^{3} of mercury, density = 13.5939 g/cm^{3}?

(b) What is the mass of 25.0 mL octane, density = 0.702 g/cm^{3}?

34. Calculate these masses.

(a) What is the mass of 4.00 cm^{3} of sodium, density = 0.97 g/cm^{3} ?

(b) What is the mass of 125 mL gaseous chlorine, density = 3.16 g/L?

Calculate these volumes.

(a) What is the volume of 25 g iodine, density = 4.93 g/cm^{3}?

(b) What is the volume of 3.28 g gaseous hydrogen, density = 0.089 g/L?

35. Calculate these volumes.

(a) What is the volume of 11.3 g graphite, density = 2.25 g/cm^{3}?

(b) What is the volume of 39.657 g bromine, density = 2.928 g/cm^{3}?

36. Convert the boiling temperature of gold, 2966 °C, into degrees Fahrenheit and kelvin.

37. Convert the temperature of scalding water, 54 °C, into degrees Fahrenheit and kelvin.

38. Convert the temperature of the coldest area in a freezer, −10 °F, to degrees Celsius and kelvin.

39. Convert the temperature of dry ice, −77 °C, into degrees Fahrenheit and kelvin.

40. Convert the boiling temperature of liquid ammonia, −28.1 °F, into degrees Celsius and kelvin.

41. The label on a pressurized can of spray disinfectant warns against heating the can above 130 °F. What are the corresponding temperatures on the Celsius and kelvin temperature scales?

42. The weather in Europe was unusually warm during the summer of 1995. The TV news reported temperatures as high as 45 °C. What was the temperature on the Fahrenheit scale?

### 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

- Fahrenheit
- unit of temperature; water freezes at 32 °F and boils at 212 °F on this scale

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