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}
- Example:
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}
- Examples:
Metric and SI Prefixes That Increase the Size of the Unit
| Prefix | Symbol | Numerical Value | Scientific Notation | Equality |
|---|---|---|---|---|
| tera | T | 1,000,000,000,000 | 10^{12} | 1 \text{ T s} = 1 \times 10^{12} \text{ s}, 1 \text{ s} = 1 \times 10^{-12} \text{ T s} |
| giga | G | 1,000,000,000 | 10^9 | 1 \text{ G m} = 1 \times 10^9 \text{ m}, 1 \text{ m} = 1 \times 10^{-9} \text{ G m} |
| mega | M | 1,000,000 | 10^6 | 1 \text{ M g} = 1 \times 10^6 \text{ g}, 1 \text{ g} = 1 \times 10^{-6} \text{ M g} |
| kilo | k | 1,000 | 10^3 | 1 \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
| Prefix | Symbol | Numerical Value | Scientific Notation | Equality |
|---|---|---|---|---|
| deci | d | 0.1 | 10^{-1} | 1 \text{ d L} = 1 \times 10^{-1} \text{ L}, 1 \text{ L} = 1 \times 10^{1} \text{ d L} |
| centi | c | 0.01 | 10^{-2} | 1 \text{ c m} = 1 \times 10^{-2} \text{ m}, 1 \text{ m} = 100 \text{ c m} |
| milli | m | 0.001 | 10^{-3} | 1 \text{ m s} = 1 \times 10^{-3} \text{ s}, 1 \text{ s} = 1 \times 10^{3} \text{ m s} |
| micro | μ | 0.000 001 | 10^{-6} | 1 \text{ μ g} = 1 \times 10^{-6} \text{ g}, 1 \text{ g} = 1 \times 10^{6} \text{ μ g} |
| nano | n | 0.000 000 001 | 10^{-9} | 1 \text{ n m} = 1 \times 10^{-9} \text{ m}, 1 \text{ m} = 1 \times 10^{9} \text{ n m} |
| pico | p | 0.000 000 000 001 | 10^{-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.
| Nutrient | Amount Recommended |
|---|---|
| Calcium | 1.0 \text{ grams} |
| Copper | 2 \text{ milligrams} |
| Iodine | 150 \text{ μgrams} (150 \text{ mcg}) |
| Iron | 18 \text{ milligrams} |
| Magnesium | 400 \text{ milligrams} |
| Niacin | 20 \text{ milligrams} |
| Phosphorus | 800 \text{ milligrams} |
| Potassium | 3.5 \text{ grams} |
| Selenium | 70 \text{ μgrams} (70 \text{ mcg}) |
| Sodium | 2.4 \text{ grams} |
| Zinc | 15 \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}
- 10 grams
- 100 grams
- 1000 grams
- B. 1 \text{ centiLiter} = \text{ ___ Liters}
- 0.001 Liters
- 0.01 Liters
- 100 Liters
- A. 1 \text{ kilogram} = \text{ ___ grams}
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
| Quantity | Metric (SI) | U.S. | Metric–U.S. |
|---|---|---|---|
| Length | 1 \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} | |
| Volume | 1 \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)}* | |||
| Mass | 1 \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}* | |||
| Time | 1 \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.
- The car travels at 85 kilometers/hour.
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.
- For example, a person has 18% body fat by mass.
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.
- Example: Keflex (cephalexin), an antibiotic used for respiratory and ear infections, is available in 250-milligrams capsules.
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) |
|---|---|---|---|---|---|
| Cork | 0.26 | Gasoline | 0.74 | Hydrogen | 0.090 |
| Body fat | 0.909 | Ethanol | 0.79 | Helium | 0.179 |
| Ice (at 0 °Celsius) | 0.92 | Olive oil | 0.92 | Methane | 0.714 |
| Muscle | 1.06 | Water (at 4 °Celsius) | 1.00 | Neon | 0.902 |
| Sugar | 1.59 | Urine | 1.003–1.030 | Nitrogen | 1.25 |
| Bone | 1.80 | Plasma (blood) | 1.03 | Air (dry) | 1.29 |
| Salt (NaCl) | 2.16 | Milk | 1.04 | Oxygen | 1.43 |
| Aluminum | 2.70 | Blood | 1.06 | Carbon dioxide | 1.96 |
| Iron | 7.86 | Mercury | 13.6 | ||
| Copper | 8.92 | ||||
| Silver | 10.5 | ||||
| Lead | 11.3 | ||||
| Gold | 19.3 |
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}
- STEP 1: State the given and needed quantities:
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.