Lecture 3 Conversion Factors and Density

Prefixes

  • In metric and SI systems, prefixes modify unit sizes by factors of 10.
    • Example:
      • 1 \text{ kilometer (km)} = 1000 \text{ meters}
      • 1 \text{ millimeter (mm)} = 0.001 \text{ meters}

Prefixes and Equalities

  • Prefixes can be replaced with their numerical values to show their relationship to a unit.
    • Examples:
      • 1 \text{ kilometer} = 1000 \text{ meters}
      • 1 \text{ kiloliter} = 1000 \text{ liters}
      • 1 \text{ kilogram} = 1000 \text{ grams}

Metric and SI Prefixes That Increase the Size of the Unit

PrefixSymbolNumerical ValueScientific NotationEquality
teraT1,000,000,000,00010^{12}1 \text{ T s} = 1 \times 10^{12} \text{ s}, 1 \text{ s} = 1 \times 10^{-12} \text{ T s}
gigaG1,000,000,00010^91 \text{ G m} = 1 \times 10^9 \text{ m}, 1 \text{ m} = 1 \times 10^{-9} \text{ G m}
megaM1,000,00010^61 \text{ M g} = 1 \times 10^6 \text{ g}, 1 \text{ g} = 1 \times 10^{-6} \text{ M g}
kilok1,00010^31 \text{ km} = 1 \times 10^3 \text{ m}, 1 \text{ m} = 1 \times 10^{-3} \text{ km}

Metric and SI Prefixes That Decrease the Size of the Unit

PrefixSymbolNumerical ValueScientific NotationEquality
decid0.110^{-1}1 \text{ d L} = 1 \times 10^{-1} \text{ L}, 1 \text{ L} = 1 \times 10^{1} \text{ d L}
centic0.0110^{-2}1 \text{ c m} = 1 \times 10^{-2} \text{ m}, 1 \text{ m} = 100 \text{ c m}
millim0.00110^{-3}1 \text{ m s} = 1 \times 10^{-3} \text{ s}, 1 \text{ s} = 1 \times 10^{3} \text{ m s}
microμ0.000 00110^{-6}1 \text{ μ g} = 1 \times 10^{-6} \text{ g}, 1 \text{ g} = 1 \times 10^{6} \text{ μ g}
nanon0.000 000 00110^{-9}1 \text{ n m} = 1 \times 10^{-9} \text{ m}, 1 \text{ m} = 1 \times 10^{9} \text{ n m}
picop0.000 000 000 00110^{-12}1 \text{ p s} = 1 \times 10^{-12} \text{ s}, 1 \text{ s} = 1 \times 10^{12} \text{ p s}
  • Note:
    • In medicine, the abbreviation "mc" is used for micro to avoid misreading the symbol μ; thus, 1 \text{ μg} is written as 1 \text{ mcg}.

Daily Values for Selected Nutrients

  • The U.S. Food and Drug Administration uses metric prefixes to express daily nutrient requirements.
NutrientAmount Recommended
Calcium1.0 \text{ grams}
Copper2 \text{ milligrams}
Iodine150 \text{ μgrams} (150 \text{ mcg})
Iron18 \text{ milligrams}
Magnesium400 \text{ milligrams}
Niacin20 \text{ milligrams}
Phosphorus800 \text{ milligrams}
Potassium3.5 \text{ grams}
Selenium70 \text{ μgrams} (70 \text{ mcg})
Sodium2.4 \text{ grams}
Zinc15 \text{ milligrams}

Sample Exercise 13

  • Fill in the blanks with the correct prefix:
    • A. 1000 \text{ meters} = 1 \text{ ___ m}
    • B. \text{ ___ g} = 0.000001 \text{ g}
    • C. 0.1 \text{ L} = 1 \text{ ___ L}

Solution 13

  • A. The prefix for 1000 is kilo; 1000 \text{ meters} = 1 \text{ kilometer}.
  • B. The prefix for 0.000001 is micro; 1 \text{ μg} = 0.000001 \text{ g}.
  • C. The prefix for 0.1 is deci; 0.1 \text{ Liters} = 1 \text{ deciLiter}.

Measuring Length: Equalities

  • An equality shows the relationship between two units that measure the same quantity.
  • Each of the following equalities describes the same length in a different unit.

Measuring Length: Equalities (Continued)

  • The metric length of 1 meter is the same length as 10 decimeters, 100 centimeters, and 1000 millimeters.

Measuring Volume: Equalities

  • Volumes of 1 Liter or smaller are common in the health sciences.
  • When a liter is divided into 10 equal portions, each portion is called a deciliter (dL).

Measuring Volume

  • The cubic centimeter (\text{cm}^3 or cc) is the volume of a cube with the dimensions 1 \text{ centimeter} \times 1 \text{ centimeter} \times 1 \text{ centimeter}.
  • A cube measuring 10 centimeters on each side has a volume of 1 \text{ Liter}; a cube measuring 1 centimeter on each side has a volume of 1 \text{ cubic centimeter}, or 1 \text{ milliLiter}.

Measuring Volume (Continued)

  • A cubic centimeter (cc) has the same volume as a milliliter, and the units are often used interchangeably.

Measuring Mass: Equalities

  • When you visit the doctor for a physical examination, he or she records your mass in kilograms (kg) and laboratory results in micrograms (\text{μg} or mcg).

Sample Exercise 14

  • Identify the larger unit in each of the following:
    • A. millimeter or centimeter
    • B. kilogram or centigram
    • C. deciLiter or μLiter
    • D. m g or μicrograms

Solution 14

  • A. Centimeter is larger. A millimeter is 0.001 \text{ meter}, smaller than a centimeter, 0.01 \text{ meter}.
  • B. Kilogram is larger. A kilogram is 1000 \text{ grams}, larger than a centigram, 0.01 \text{ grams}.
  • C. Deciliter is larger. A deciliter is 0.1 \text{ Liter}, larger than a microliter, 0.000 001 \text{ Liter}.
  • D. Milligram is larger. A milligram is 0.001 \text{ grams}, larger than a microgram, 0.000 001 \text{ grams}.

Sample Exercise 15

  • Complete each of the following equalities:
    • A. 1 \text{ kilogram} = \text{ ___ grams}
      1. 10 grams
      2. 100 grams
      3. 1000 grams
    • B. 1 \text{ centiLiter} = \text{ ___ Liters}
      1. 0.001 Liters
      2. 0.01 Liters
      3. 100 Liters

Solution 15

  • A. 1 \text{ kilogram} = 1000 \text{ grams}
  • B. 1 \text{ centiLiter} = 0.01 \text{ Liter}

Equalities

  • Equalities:
    • use two different units to describe the same quantity
    • can be between units of the metric system, or between U.S. units, or between metric and U.S. units
      • 1 \text{ meter} = 1000 \text{ millimeter}
      • 1 \text{ lb} = 16 \text{ oz}
      • 2.20 \text{ lb} = 1 \text{ kilogram}

Equalities on Food Labels

  • The contents of many packaged foods:
    • are listed in both metric and U.S. units.
    • indicate the same amount of a substance in two different units.
  • In the United States, the contents of many packaged foods are listed in both U.S. and metric units.

Equalities and Significant Figures

  • The numbers in any equality between two metric units or between two U.S. system units are definitions.
  • Because numbers in a definition are exact, they are not used to determine significant figures (SFs).
    • 1 \text{ gram} = 1000 \text{ milligrams}
    • 1 \text{ foot} = 12 \text{ inches}
    • 1 \text{ minutes} = 60 \text{ seconds}

Equalities and Significant Figures (Continued)

  • When an equality consists of a metric unit and a U.S. unit, one of the numbers in the equality is obtained by measurement and counts toward the significant figures (SFs) in the answer.
    • 454 \text{ g} = 1 \text{ lb}
    • 946 \text{ milliLiters} = 1 \text{ quart}
    • 39.4 \text{ inches} = 1 \text{ meter}
  • Exception:
    • The equality 1 \text{ inch} = 2.54 \text{ centimeters} has been defined as an exact relationship, and therefore 2.54 is an exact number.

Conversion Factors

  • Any equality can be written as fractions called conversion factors.
  • Be sure to include units when you write conversion factors.
  • Conversion Factors for the Equality 1 \text{ h} = 60 \text{ min}
    • These conversion factors are read as “60 minutes per hour” and “1 hour per 60 minutes.”
    • The per means “divide.”

Some Common Equalities

QuantityMetric (SI)U.S.Metric–U.S.
Length1 \text{ kilometer} = 1000 \text{ meters}1 \text{ foot} = 12 \text{ inches}2.54 \text{ centimeters} = 1 \text{ inch} (exact)
1 \text{ meter} = 1000 \text{ millimeters}1 \text{ yard} = 3 \text{ foot}1 \text{ meter} = 39.4 \text{ inches}
1 \text{ centimeter} = 10 \text{ millimeters}1 \text{ mile} = 5280 \text{ foot}1 \text{ kilometer} = 0.621 \text{ miles}
Volume1 \text{ Liter} = 1000 \text{ milliLiters}1 \text{ quart} = 4 \text{ cups}946 \text{ milliLiter} = 1 \text{ quart}
1 \text{ deciLiter} = 100 \text{ milliLiters}1 \text{ quart} = 2 \text{ pints}1 \text{ Liter} = 1.06 \text{ quart}
1 \text{ milliliter} = 1 \text{ cubic centimeter}1 \text{ gallon} = 4 \text{ quart}473 \text{ milliLiters} = 1 \text{ pints}
1 \text{ milliLiter} = 1 \text{ cubic centimeter}*5 \text{ milliLiters} = 1 \text{ t (tsp)}*
15 \text{ milliLiters} = 1 \text{ T (tbsp)}*
Mass1 \text{ kilogram} = 1000 \text{ grams}1 \text{ lb} = 16 \text{ oz}1 \text{ kilogram} = 2.20 \text{ lb}
1 \text{ gram} = 1000 \text{ milligrams}454 \text{ grams} = 1 \text{ lb}
1 \text{ milligram} = 1000 \text{ micrograms}*
Time1 \text{ hour} = 60 \text{ minute}1 \text{ hour} = 60 \text{ minute}
1 \text{ minute} = 60 \text{ seconds}1 \text{ minute} = 60 \text{ seconds}
  • *Used in medicine.

Writing Conversion Factors

  • We can write metric conversion factors:
    • \frac{100 \text{ cm}}{1 \text{ m}}
    • \frac{1 \text{ m}}{100 \text{ cm}}
  • 1 \text{m} = 100 \text{ cm}
  • Metric-U.S. system conversion factors:
    • \frac{2.20 \text{ lb}}{1 \text{ kg}}
    • \frac{1 \text{ kg}}{2.20 \text{ lb}}

Sample Exercise 16

  • Write the equality and two conversion factors for each of the following pairs of units:
    • A. liters and milliliters
    • B. inches and meters
    • C. milligrams and micrograms

Solution 16

  • (Solution not provided in transcript)

Conversion Factors Within a Problem

  • An equality may also be stated within a problem that only applies to that problem.
    • The car travels at 85 kilometers/hour.
      • Equality: 85 \text{ kilometers} = 1 \text{ hour}
      • Conversion Factors: \frac{85 \text{ kilometers}}{1 \text{ hour}}, \frac{1 \text{ hour}}{85 \text{ kilometers}}
      • Significant Figures: The 85 \text{ kilometer} is measured (two significant figures), and the 1 \text{ h} is exact.

Conversion Factors Within a Problem (Continued)

  • The tablet contains 500 \text{ milligrams} of vitamin C.
    • Equality: 1 \text{ tablet} = 500 \text{ milligrams of vitamin C}
    • Conversion Factors: \frac{500 \text{ milligrams vitamin C}}{1 \text{ tablet}}, \frac{1 \text{ tablet}}{500 \text{ milligrams vitamin C}}
    • Significant Figures: The 500 \text{ milligrams} is measured (one significant figure). The 1 \text{ tablet} is exact.

Conversion Factors from a Percentage

  • A percent (%) is written as a conversion factor by choosing a unit and expressing the numerical relationship of the parts of this unit to 100 parts of the whole.
    • For example, a person has 18% body fat by mass.
      • Equality: 18 \text{ kilograms of body fat} = 100 \text{ kilograms of body mass}
      • Conversion Factors: \frac{18 \text{ kilograms of body fat}}{100 \text{ kilograms of body mass}}, \frac{100 \text{ kilograms of body mass}}{18 \text{ kilograms of body fat}}
      • The 18 \text{ kilograms} is measured: It has two significant figures.
      • The 100 \text{ kilograms} is exact.

Conversion Factors from Dosage Problems

  • Equalities stated within dosage problems for medications can also be written as conversion factors.
    • Example: Keflex (cephalexin), an antibiotic used for respiratory and ear infections, is available in 250-milligrams capsules.
      • Equality: 1 \text{ capsule} = 250 \text{ milligrams of Keflex}
      • Conversion Factors: \frac{250 \text{ milligrams of Keflex}}{1 \text{ capsule}}, \frac{1 \text{ capsule}}{250 \text{ milligrams of Keflex}}
      • The 250 \text{ milligrams} is measured: It has two significant figures.
      • The 1 \text{ capsule} is exact.

Sample Exercise 17

  • Write the equality and two conversion factors for each of the following:
    • A. meters and decimeters
    • B. jewelry that contains 18% gold
    • C. One gallon of gas is $2.40.

Solution 17

  • (Solution not provided in transcript)

Problem Solving Using Conversion Factors

  • The problem-solving process begins by analyzing the problem in order to:
    • identify the given unit and needed unit
    • write a plan that converts the given unit to the needed unit
    • identify one or more conversion factors that cancel units and provide the needed unit
    • set up the calculation

Solving Problems Using Conversion Factors

  • Example: If a person weighs 178 lb, what is the body mass in kilograms?

    • STEP 1: State the given and needed quantities:

      • Given: 178 \text{ lb}
      • Need: kilograms
      • Connect: conversion factor (kilograms/lb)
    • STEP 2: Write a plan to convert the given unit to the needed unit.

    • STEP 3: State the equalities and conversion factors.

    • STEP 4: Set up the problem to cancel units and calculate the answer.

      178 \text{ lb} \times \frac{1 \text{ kg}}{2.20 \text{ lb}} = 80.9 \text{ kg}

Sample Exercise 18

  • A rattlesnake is 2.44 \text{ m} long. How many centimeters long is the snake?

Solution 18

  • STEP 1: State the given and needed quantities:
    • Given: 2.44 \text{ m}
    • Need: centimeters
    • Connect: conversion factor (cm/m)
  • STEP 2: Write a plan to convert the given unit to the needed unit.
  • STEP 3: State the equalities and conversion factors.
    • 1 \text{ m} = 100 \text{ cm}
    • \frac{100 \text{ cm}}{1 \text{ m}}
  • STEP 4: Set up the problem to cancel units and calculate the answer.
    • 2.44 \text{ m} \times \frac{100 \text{ cm}}{1 \text{ m}} = 244 \text{ cm}

Using Two or More Conversion Factors

  • In problem solving:
    • two or more conversion factors are often needed to complete the change of units
    • \text{Unit 1} \rightarrow \text{Unit 2} \rightarrow \text{Unit 3}
    • to set up these problems, one factor follows the other
    • each factor is arranged to cancel the preceding unit until the needed unit is obtained
    • \text{Given unit} \times \text{Factor 1} \times \text{Factor 2} = \text{Needed unit}

Using Two or More Conversion Factors (Continued)

  • Example: A doctor’s order prescribed a dosage of 0.150 \text{ milligrams} of Synthroid. If you only have 50 \text{.-micrograms} tablets in stock, how many tablets are required to provide the prescribed dosage?

    • STEP 1: State the given and needed quantities:

      • Given: 0.150 \text{ milligrams of Synthroid}.
      • Need: number of tablets 1 tablet = 50 micrograms of Synthroid
      • Connect: 1 tablet = 50 \text{ micrograms} of Synthroid
    • STEP 2: Write a plan to convert the given unit to the needed unit.

    • STEP 3: State the equalities and conversion factors.

    • STEP 4: Set up the problem to cancel units and calculate the answer.

      0.150 \text{ mg Synthroid} \times \frac{1000 \text{ μg}}{1 \text{ mg}} \times \frac{1 \text{ tablet}}{50 \text{ μg}} = 3.0 \text{ tablets}

Sample Exercise 19

  • How many minutes are in 1.6 days?

Solution 19

  • STEP 1: State the given and needed quantities:
    • Given: 1.6 days
    • Need: minutes
    • Connect: conversion factors (hours/day, minutes/hour)
  • STEP 2: Write a plan to convert the given unit to the needed unit.
  • STEP 3: State the equalities and conversion factors.
    • 1 \text{ day} = 24 \text{ hr}
    • 1 \text{ hr} = 60 \text{ min}
    • \frac{24 \text{ hr}}{1 \text{ day}}
    • \frac{60 \text{ min}}{1 \text{ hr}}
  • STEP 4: Set up the problem to cancel units and calculate the answer.
    • 1.6 \text{ days} \times \frac{24 \text{ hr}}{1 \text{ day}} \times \frac{60 \text{ min}}{1 \text{ hr}} = 2300 \text{ min}

Sample Exercise 20

  • If your pace on a treadmill is 65 meters/minute, how many minutes will it take for you to walk a distance of 7.5 kilometers?

Solution 20

  • STEP 1: State the given and needed quantities:
    • Given: 7.5 kilometers
    • Need: minutes
    • Connect: conversion factors (meters/kilometer, 65 meters/minute)
  • STEP 2: Write a plan to convert the given unit to the needed unit.
  • STEP 3: State the equalities and conversion factors.
    • 1 \text{ km} = 1000 \text{ m}
    • \frac{65 \text{ m}}{1 \text{ min}}
  • STEP 4: Set up the problem to cancel units and calculate the answer.
    • 7.5 \text{ km} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ min}}{65 \text{ m}} = 120 \text{ min}

Density

  • Density:
    • compares the mass of an object to its volume.
    • is the mass of a substance divided by its volume.
    • is defined as Density = \frac{\text{mass}}{\text{volume}}

Density (Continued)

  • Substances that have:
    • higher densities contain particles that are closely packed together.
    • lower densities contain particles that are farther apart.
  • Metals such as gold and lead have higher densities because their atoms are packed closely together.

Density: Units

  • In the metric system, densities of solids, liquids, and gases are expressed with different units.
    • The density of a solid or liquid is usually given in grams per cubic centimeter (\text{g/cm}^3) or grams per milliliter (g/mL).
    • The density of a gas is usually given in grams per liter (g/L).

Densities of Common Substances

Solids (at 25 °Celsius)Density (grams/milliLiters)Liquids (at 25 °Celsius)Density (grams/milliLiters)Gases (at 0 °Celsiuselsius)Density (grams/Liters)
Cork0.26Gasoline0.74Hydrogen0.090
Body fat0.909Ethanol0.79Helium0.179
Ice (at 0 °Celsius)0.92Olive oil0.92Methane0.714
Muscle1.06Water (at 4 °Celsius)1.00Neon0.902
Sugar1.59Urine1.003–1.030Nitrogen1.25
Bone1.80Plasma (blood)1.03Air (dry)1.29
Salt (NaCl)2.16Milk1.04Oxygen1.43
Aluminum2.70Blood1.06Carbon dioxide1.96
Iron7.86Mercury13.6
Copper8.92
Silver10.5
Lead11.3
Gold19.3

Chemistry Link to Health: Bone Density

  • A condition of severe thinning of bone known as osteoporosis may occur.
  • Scanning electron micrographs (SEMs) show (a) normal bone and (b) bone with osteoporosis due to loss of bone minerals.

Chemistry Link to Health: Bone Density (Continued)

  • Bone density is often determined by passing low-dose X-rays through the narrow part at the top of the femur (hip) and the spine (c).
  • Bones with high density will block more of the X-rays compared to bones that are less dense.

Density Using Volume Displacement

  • The density of the zinc object is calculated from its mass and volume.
  • A solid completely submerged in water displaces its own volume of water (its volume is calculated from the volume increase).

Sample Exercise 21

  • What is the density (grams/milliLiters) of a 48.0-grams sample of a metal if the level of water in a graduated cylinder rises from 25.0 milliLiters to 33.0 milliLiters after the metal is added?

Solution 21

  • V = 33.0 \text{ mL} - 25.0 \text{ mL} = 8.0 \text{ mL}
  • Density = \frac{48.0 \text{ g}}{8.0 \text{ mL}} = 6.0 \text{ g/mL}

Problem Solving Using Density

  • If the volume and the density of a sample are known, the mass in grams of the sample can be calculated by using density as a conversion factor.

Problem Solving Using Density (Continued)

  • Example: An unknown liquid has a density of 1.32 \frac{\text{grams}}{\text{milliLiters}}. What is the volume (milliLiters) of a 14.7 \text{-grams} sample of the liquid?
    • STEP 1: State the given and needed quantities:
      • Given: 14.7 \text{ g}
      • Need: milliliters
      • Connect: density (\frac{\text{grams}}{\text{milliLiters}}
    • STEP 2: Write a plan to calculate the needed quantity.
    • STEP 3: Write the equalities and their conversion factors including density.
      • \frac{1.32 \text{ g}}{1 \text{ mL}}, \frac{1 \text{ mL}}{1.32 \text{ g}}
    • STEP 4: Set up the problem to calculate the needed quantity.
      • 14.7 \text{ g} \times \frac{1 \text{ mL}}{1.32 \text{ g}} = 11.1 \text{ mL}

Sample Exercise 22

  • John took 2.0 teaspoons (tsp) of cough syrup for a cough. If the syrup had a density of 1.20 grams/milliLiters and there is 5.0 milliLiters in 1 teaspoons, what was the mass, in grams, of the cough syrup?

Solution 22

  • STEP 1: State the given and needed quantities:
    • Given: 2.0 teaspoons of syrup
    • Need: grams
    • Connect: conversion factors (milliLiters/teaspoons, density of syrup)
  • STEP 2: Write a plan to calculate the needed quantity.
  • STEP 3: Write the equalities and their conversion factors including density.
    • 1 \text{ tsp} = 5.0 \text{ mL}
    • \frac{1.20 \text{ g}}{1 \text{ mL}}
  • STEP 4: Set up the problem to calculate the needed quantity.
    • 2.0 \text{ tsp} \times \frac{5.0 \text{ mL}}{1 \text{ tsp}} \times \frac{1.20 \text{ g}}{1 \text{ mL}} = 12 \text{ g}

Specific Gravity

  • Specific gravity (sp gr) is a relationship between the density of a substance and the density of water.
  • Specific gravity is calculated by dividing the density of a sample by the density of water, which is 1.00 grams/milliLiters at 4 °Celsius.
  • A substance with a specific gravity of 1.00 has the same density as water (1.00 grams/milliLiters).

Specific Gravity (Continued)

  • A hydrometer is used to measure the specific gravity of urine.
  • The normal range of specific gravity for urine is 1.003 to 1.030.
  • The specific gravity can decrease with type 2 diabetes and kidney disease.
  • Increased specific gravity may occur with dehydration, kidney infection, and liver disease.

Specific Gravity (