This is "Unit 2", section 2.1 from the book General Chemistry (v. 1.0).

Has this book helped you? Consider passing it on:
Help Creative Commons
Creative Commons supports free culture from music to education. Their licenses helped make this book available to you.
Help a Public School
DonorsChoose.org helps people like you help teachers fund their classroom projects, from art supplies to books to calculators.

2.1 Metric-Metric Conversions

Learning Objective

  1. To be able to convert measurements within the metric system.

A powerful approach to solving problems in chemistry goes by a variety of names. The conversion factor approach to solving problems is also called the factor-label method. Using conversion factors is a form of dimensional analysis where the units cross out. In this section we will use conversion factors to solve problems that require changing units within the metric system. As we move through the material we will see more and more applications for conversion factors with increasing level of complexity. It's best to start using the conversion factor approach now while the material is simple. If you master the conversion factor approach now, while the problems are easy, when the complexity increases, it's just a matter of plugging in new conversion factors. So, yes, thanks to the mathematical ease associated with multiplication and division by powers of 10, along with society's ever-increasing familiarity with the metric system, you can probably already do these conversions in your head without using a formal problem-solving strategy; however, this is a wonderful opportunity to get acquainted with the way we will be doing most of the math problems in this course.

Metric-Metric Conversions

In Section 1.4, Measurement , the metric system base units and prefixes were detailed. We can use the information in Table 1.4(2) as conversion factors to change units within the metric system. Mathematically speaking we are simply multiplying a quantity by the number 1. But we are writing the number 1 as a factor such that the undesired units cancel leaving the desired unit. For example, if we wanted to cook 10 quarter pound hamburgers and we wanted to know how many pounds of ground beef required, we would set up the following-

( 10 burgers ) ( 1 lb ground beef 4 burgers ) = 2.5 lbs ground beef

Since 1 pound is equal to 4 burgers, the quantity 1 pound 4 burgers is the number 1. Likewise the quantity 4 burgers 1 pound is the number 1; however, had we used 4 burgers 1 pound as the factor, the units would not have canceled. This problem illustrates the importance of including units in our conversion factors. Had we just written 1 4 or 4 1 we would not be multiplying by one. In the former case we would have got the correct number, but the units would have been incorrect. In the latter case we would have calculated both the wrong value AND the wrong units. To use the conversion factor method of problem solving you must always include the units in your calculations.

For doing the calculations within the metric system we often face several possibilities for conversion factors. Table 1.4(2) says 1 μm = 1 ×10-6 m. It is often more convenient however to follow the general rule to avoid (-) powers in conversion factors. Mathematically, it doesn't matter but in terms of the workings of the human mind, rather than saying 1 μm = 1 × 10-6 m, instead, say 1 m = 1×106 μm. These are equivalent statements. Mentally, it makes the calculation simpler if you avoid the negative exponents.

All measurements must be expressed in the correct units to have any meaning. Never write down a measurement with just the number. Does it really take that long to scribble some units? When making unit conversions, use arithmetic steps accompanied by unit cancellation.

Suppose you have measured a mass in milligrams but need to report the measurement in kilograms. In problems that involve SI units, you can use the definitions of the prefixes given in Table 1.3(2) "Approximate Elemental Composition of a Typical 70 kg Human" to get the necessary conversion factors. For example, you can convert milligrams to grams and then convert grams to kilograms:

milligrams → grams → kilograms 1000 mg → 1 g 1000 g → 1 kilogram

If you have measured 928 mg of a substance, you can convert that value to kilograms as follows:

928 mg × 1 g 1000 mg = 0 .928 g 0 .928 g × 1 kg 1000 g = 0 .000928 kg = 9 .28 × 10 − 4 kg

In each arithmetic step, the units cancel as if they were algebraic variables, leaving us with an answer in kilograms. In the conversion to grams, we begin with milligrams in the numerator. Milligrams must therefore appear in the denominator of the conversion factor to produce an answer in grams. The individual steps may be connected as follows:

928 mg × 1 g 1000 mg × 1 kg 1000 g = 928 kg 10 6 = 928 × 10 − 6 kg = 9 .28 × 10 − 4 kg

The following examples provide practice converting between units.

Example 2.1-1

Convert 75.2 nm to μm

Given: a metric measurement of length

Asked for: different metric unit

Strategy:

Apply the prefixes from Table 1.4(2) and arrange units such that the given unit cancels leaving the desired unit.

Solution:

( 75.2 nm ) ( 1 m 10 9 nm ) ( 10 6 μm 1 m ) = 0.0752 μm

Exercise

Convert 0.00572 km to cm.

Answer: 572 cm

Example 2.1-2

Use the information in Table 1.4(2) to convert each measurement. Be sure that your answers contain the correct number of significant figures and are expressed in scientific notation where appropriate.

  1. 59.2 cm to decimeters
  2. 3.7 × 105 mg to kilograms
  3. 270 mL to cubic decimeters
  4. 9.024 × 1010 s to years

Solution

  1. 59 .2 cm × 1 m 100 cm × 10 dm 1 m = 5 .92 dm
  2. 3 .7 × 10 5 mg × 1 g 1000 mg × 1 kg 1000 g = 3 .7 × 10 − 1 kg
  3. 270 mL × 1 L 1000 mL × 1 dm 3 1 L = 270 × 10 − 3 dm 3 = 2 .70 × 10 − 1 dm 3
  4. 9 .024 × 10 10 s × 1 min 60 s × 1 h 60 min × 1 d 24 h × 1 yr 365 d = 2 .86 × 10 3 yr

Example 2.1-3

Courtesy of Fred Redmore, Highland Community College

Carry out the following conversions.

  1. 42 cm to m
  2. 250 mL to L
  3. 45 μg to g
  4. 0.12 m to km
  5. 0.015 cm3 to L
  6. 1.2 m to cm
  7. 0.15 L to mL
  8. 6.2 kg to g
  9. 1.2 m to μm
  10. 4.5 L to cm3

Solution

  1. (42 cm) ( m 100 cm ) = 0.42 m
  2. (250 mL) ( L 1000 mL ) = 0.25 L
  3. (45 μg) ( g 10 6 μg ) = 45 × 10-6 g = 4.5 × 10-5 g
  4. (0.12 m) ( km 10 3 m ) = 0.12 ×10-3 km = 1.2 × 10-4 km
  5. (0.015 mL) ( L 10 3 mL ) = 0.015 ×10-3 L = 1.5 × 10-5 L
  6. (1.20 m) ( 100 cm m ) = 120 cm
  7. (0.15 L) ( 1000 mL L ) = 150 mL
  8. (6.2 kg) ( 1000 g kg ) = 6200 g
  9. (1.2 m) ( 10 6 μm m ) = 1.2 × 106 μm
  10. (4.5 L) ( 1000 cm 3 L ) = 4.5 × 103 cm3

Example 2.1-4

Courtesy of Fred Redmore, Highland Community College

Carry out the following conversions.

  1. 1.5 cm to mm
  2. 12.8 mg to kg
  3. 95 mL to μL
  4. 0.23 km to cm
  5. 86 mm to km
  6. 47.2 μL to mL
  7. 6.3 kg to mg
  8. 0.825 cm to mm

Solution

  1. (1.5 cm) ( m 100 cm ) ( 10 3 mm m ) = 15 mm

  2. (12. mg) ( g 10 3 mg ) ( kg 10 3 g ) = 12.8 × 10-6 kg = 1.28 × 10-5 kg

  3. (95 mL) ( L 10 3 mL ) ( 10 6 μL L ) = 95 × 103 μL = 9.5 × 104 μL
  4. (0.23 km) ( 10 3 m km ) ( 10 2 cm m ) = 0.23 × 105 cm = 2.3 × 104 cm
  5. (86 mm) ( m 10 3 mm ) ( km 10 3 m ) = 86 × 10-6 km = 8.6 × 10-5 km
  6. (47.2 μL) ( L 10 6 μL ) ( 10 3 mL L ) = 47.2 × 10-3 mL = 0.0472 mL

  7. (6.3 kg) ( 10 3 g kg ) ( 10 3 mg g ) = 6.3 × 106 mg

  8. (0.825 cm) ( m 10 2 cm ) ( 10 3 mm m ) = 8.25 mm

Summary

Within the metric system, units are related by powers of 10. Conversion factors equivalent to the number 1 are applied as a power of 10 such that the units cancel to give the desired unit.

Key Takeaway

  • Conversion factors are applied to change units.