Book Chapter 8 - Unit Conversions

Chapter 8: Unit Conversions

8.1 Unit Analysis

• Unit analysis is a technique for doing unit conversions, applicable in chemistry and everyday life.

• It involves a stepwise thought process that provides confidence in the correctness of answers.

• Alternative names for unit analysis include the factor-label method, the conversion factor method, and dimensional analysis.

General Procedure Overview

• The general procedure involves identifying the desired unit, the given value, and multiplying by conversion factors to cancel unwanted units and generate desired ones.

• A conversion factor is a ratio that describes the relationship between two units.

• For example, to convert 2 teaspoons (tsp) to milliliters (mL), the relationship 1 tsp = 5 mL is used.

• When multiplying by a conversion factor, the value (number and unit) changes, but the actual amount does not.

• Units in a unit analysis setup cancel like variables in an algebraic equation.

Metric-Metric Conversions

• Relationships between metric units can be derived from metric prefixes.

• For example, milli- means $10^{-3}$$$10^{-3}$$, so 1 mL = $10^{-3}$$$10^{-3}$$ L, meaning there are 1000 mL in a liter.

Example 8.1: Conversion Factors

• To relate nanometers (nm) and meters (m) with positive exponents: 1 nm = $10^{-9}$$$10^{-9}$$ m, so $10^9$$$10^9$$ nm = 1 m.

Example 8.2: Unit Conversions

• To convert 365 nm to kilometers (km), one can change nanometers to meters and then meters to kilometers.

• nm → m → km

• $? \text{ km} = 365 \text{ nm} \times \frac{1 \text{ m}}{10^9 \text{ nm}} \times \frac{1 \text{ km}}{10^3 \text{ m}} = 3.65 \times 10^{-10} \text{ km}$$$? \text{ km} = 365 \text{ nm} \times \frac{1 \text{ m}}{10^9 \text{ nm}} \times \frac{1 \text{ km}}{10^3 \text{ m}} = 3.65 \times 10^{-10} \text{ km}$$

English-Metric Conversions

• Unit analysis is useful for converting between English and metric units.

• The English inch is defined as exactly 2.54 cm.

Example 8.3: Unit Conversions

• To convert the mass of a hydrogen atom ($1.67 × 10^{-18}$$$1.67 × 10^{-18}$$ μg) to pounds (lb):

• μg → g → lb

• $? \text{ lb} = 1.67 × 10^{-18} \text{ μg} \times \frac{1 \text{ g}}{10^6 \text{ μg}} \times \frac{1 \text{ lb}}{453.6 \text{ g}}$$$? \text{ lb} = 1.67 × 10^{-18} \text{ μg} \times \frac{1 \text{ g}}{10^6 \text{ μg}} \times \frac{1 \text{ lb}}{453.6 \text{ g}}$$

8.2 Rounding Off and Significant Figures

• Calculators often provide more digits than justified by scientific data, so rounding is necessary.

Measurements, Calculations, and Uncertainty

• Measurements have uncertainty, expressed as ±1 in the last decimal place reported.

• For example, 5.0 g implies a mass between 4.9 g and 5.1 g.

• Calculations using inexact values yield uncertain answers.

Rounding Off Answers Derived from Multiplication and Division

• Three steps:

• Determine which numbers affect uncertainty (inexact values).

• Count significant figures to determine relative uncertainties.

• Round off the answer to the same number of significant figures as the inexact value with the fewest significant figures.

Sample Study Sheet 8.1: Rounding Off Numbers Calculated Using Multiplication and Division

• Exact values (from definitions or counting) do not affect uncertainty.

• Non-zero digits are always significant.

• Zeros between nonzero digits are significant.

• Zeros to the left of nonzero digits are not significant.

• Zeros to the right of nonzero digits in numbers with decimal points are significant.

• If the digit to the right of the final digit is less than 5, round down; if 5 or greater, round up.

Example 8.4: Rounding Off Answers Derived from Multiplication and Division

• Convert 5.2 L of blood to quarts.

• $5.2 \text{ L} \times \frac{1 \text{ gal}}{3.785 \text{ L}} \times \frac{4 \text{ qt}}{1 \text{ gal}} = 5.5 \text{ qt}$$$5.2 \text{ L} \times \frac{1 \text{ gal}}{3.785 \text{ L}} \times \frac{4 \text{ qt}}{1 \text{ gal}} = 5.5 \text{ qt}$$

• 5.2 L has two significant figures, 3.785 L has four significant figures. The 4 qt/gal is an exact value.

Example 8.5: Rounding Off Answers Derived from Multiplication and Division

• Calculate time for an ant to travel 6.0 feet at 0.01 m/s.

• $6.0 \text{ ft} \times \frac{12 \text{ in}}{1 \text{ ft}} \times \frac{2.54 \text{ cm}}{1 \text{ in}} \times \frac{1 \text{ m}}{100 \text{ cm}} \times \frac{1 \text{ s}}{0.01 \text{ m}} \times \frac{1 \text{ min}}{60 \text{ s}} = 3 \text{ min}$$$6.0 \text{ ft} \times \frac{12 \text{ in}}{1 \text{ ft}} \times \frac{2.54 \text{ cm}}{1 \text{ in}} \times \frac{1 \text{ m}}{100 \text{ cm}} \times \frac{1 \text{ s}}{0.01 \text{ m}} \times \frac{1 \text{ min}}{60 \text{ s}} = 3 \text{ min}$$

• 6.0 has two significant figures, 0.01 has one significant figure.

Rounding Off Answers Derived from Addition and Subtraction

Sample Study Sheet 8.2: Rounding Off Numbers Calculated Using Addition and Subtraction

• Round off to the same number of decimal places as the inexact value with the fewest decimal places.

Example 8.6: Rounding Off Answers Derived from Addition and Subtraction

• Calculate the mass of a liquid added to a beaker.

• 60.2 g (beaker with liquid) - 52.3812 g (beaker) = 7.8 g (liquid).

• 60.2 has one decimal place; 52.3812 has four decimal places.

8.3 Density and Density Calculations

• Density (mass density) is mass divided by volume.

• The densities of liquids and solids generally decrease with increasing temperature.

• Densities of liquids and solids are usually described in grams per milliliter (g/mL) or grams per cubic centimeter (g/cm³).

• Gases are described in grams per liter (g/L).

Using Density as a Conversion Factor

• Density converts between mass and volume.

Example 8.7: Density Conversions

• Calculate the mass of 75.0 mL of water at 20 °C, given density ≈ 0.9982 g/mL.

Example 8.8: Density Conversions

• Calculate the volume of 25.00 kg of water at 20 °C.

Determination of Mass Density

• Density is calculated from measured mass and volume.

Example 8.9: Density Calculations

• Calculate density of methanol given mass and volume.

• $\text{Density} = \frac{\text{Mass}}{\text{Volume}}$$$\text{Density} = \frac{\text{Mass}}{\text{Volume}}$$

Example 8.10: Density Calculations

• Determine if a bracelet is silver or platinum by comparing its density to known densities.

8.4 Percentage and Percentage Calculations

• Percentage provides ratios used as conversion factors.

• Percentage by mass is mass units of the part for each 100 mass units of the whole.

• Volume percentages are designated as such (% by volume).

Example 8.11: Unit Conversions

• Calculate the mass of blood in kilograms for a 145-pound person (8.0% blood).

8.5 A Summary of the Unit Analysis Process

• Unit analysis is a procedure for navigating unit conversion problems.

Sample Study Sheet 8.3: Calculations Using Unit Analysis

• Steps:

• State the question.

• Multiply by conversion factors.

• Check units.

• Calculate and round.

Figure 8.3: Types of Unit Conversions

• Summary of common conversions and conversion factors.

Example 8.12: Metric-Metric Unit Conversions

• Convert micrograms to kilograms.

• Convert to base unit, then to the desired unit.

Example 8.13: English-Metric Unit Conversions

• Convert miles to kilometers.

• Use 2.54 cm/in or 1.609 km = 1 mi.

Example 8.14: Unit Conversions Using Density

• Calculate volume using density.

Example 8.15: Unit Conversions Using Percentage

• Calculate grams of calcium in cat food using percentages.

Example 8.16: Converting a Ratio of Two Units

• Calculate heat of combustion in J/g.

8.6 Temperature Conversions

• Use equations to convert between Celsius, Fahrenheit, and Kelvin scales.

• $$T{\text{°C}} = \frac{T{\text{°F}} - 32}{1.8}$$

• $$T{\text{°F}} = 1.8 \times T{\text{°C}} + 32$$

• $$T{\text{K}} = T{\text{°C}} + 273.15$$

• The numbers 1.8, 32, and 273.15 are exact.

Example 8.17: Temperature Conversions

• Convert 38.9 °F to °C.

Example 8.18: Temperature Conversions

• Convert 46.6 °C to °F.

Example 8.19: Temperature Conversions

• Convert 961 °C to K.

Example 8.20: Temperature Conversions

• Convert 1155 K to °C.