Chapter 2: Measurements

Uncertainty in Measurements • A measurement is any number with a unit attached. • It is not possible to make exact measurements, and all measurements have uncertainty. • We will generally use metric system units, these include. – the meter, m, for length measurements – the gram, g, for mass measurements – the liter, L, for volume measurements Chapter 2

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Scientific Notation • Numbers in science are often very large or very small. To avoid confusion, we use scientific notation. • Scientific notation utilizes the numeric digits in a measurement followed by a power of ten. – The numeric digits are expressed as a number between 1 and 10.

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Applying Scientific Notation Re-write these numbers in Scientific Notation:

345,000

0.00157

Step 1: Identify the first nonzero digit in the number

345,000

0.00157

Step 2: Place a decimal after the first nonzero digit in the number followed by the remaining numeric digits and “x 10”

3.45 x 10

1.57 x 10

345,000 5 spots

0.00157 3 spots

3.45 x 105

1.57 x 10-3

Step 3: Indicate how many places the decimal has moved by the power of 10. • A positive power of 10 indicates that the decimal moves to the right (Large Number). • A negative power of 10 indicates that the decimal moves to the left (Small Number).

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Examples of Scientific Notation • There are 26,800,000,000,000,000,000,000 helium atoms in 1.00 L of helium gas. Express the number in scientific notation. – Step 1: Identify the first nonzero digit: Here it’s a 2 – Step 2: Place the decimal after the 2, followed by the other numerical digits and “x 10”: 2.68 x 10 – Step 3: Count the number of places the decimal has moved and in which direction: 22 places to the right (so a positive power) • Add the positive power of 10 to complete the scientific notation.

Chapter 2

2.68 × 1022 atoms

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Another Example • The typical length between two carbon atoms in a molecule of benzene is 0.000000140 m. What is the length expressed in scientific notation? – Step 1: Identify the first nonzero digit: Here it’s a 1 – Step 2: Place the decimal after the 1, followed by the other numerical digits and “x 10”: 1.40 x 10 (We keep the last zero here!) – Step 3: Count the number of places the decimal has moved and in which direction: 7 places to the left (so a negative power) • Add the negative power of 10 to complete the scientific notation.

1.40 × 10-7 m Chapter 2

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Scientific Notation and Your Calculator • During this class, you will have to be able to add, subtract, multiply and divide numbers that are listed in scientific notation.

The “EE” Button

• That means you need to be able to input these numbers into your calculators. • You DO NOT do this by typing the number then multiplying it by 10 to a power. • Rather, you use a special button on your calculator, such as the “EE” button (shown here) or the “EXP” button, to type in the entire number plus the 10 to the power.

For Example: 1.25 x 1012 1.25

x

^

12

NO

1.25

EE or EXP

12

YES

10

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Dimensional Analysis: Unit Equations • Dimensional analysis (DA) uses the canceling of units to set up calculations. • A unit equation is a simple statement of two equivalent quantities. • For example: 1 hour = 60 minutes 1 minute = 60 seconds Chapter 2

Do you know some others?

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Dimensional Analysis: Conversion Factors • A conversion factor takes the unit equation and converts it into a ratio. • For each unit equation, you can write two conversion factors! • For the unit equation 1 hour = 60 minutes, we can write two unit factors: 1 hour 60 minutes

or

60 minutes 1 hour

Numerator Denominator Chapter 2

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Dimensional Analysis (DA) Problem Solving •

To use dimensional analysis for problem solving, prepare a “plan” by following these steps: Step 1: Write down the given value and units (What you have) Step 2: Write down the unit asked for in the answer (What you want) Step 3: Write out a plan that will let you convert from what you have to what you want. Step 4: Determine which conversion factor(s) is needed to convert the unit in the given value to the unit in the answer.

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Dimensional Analysis Problem • How many days are in 2.5 years? – Step 1: Determine what you have: 2.5 years – Step 2: Determine what you want: ??? days – Step 3: Write a plan to convert from what you have to what you want. – Step 4: Select conversion factor(s) that allows you to perform your plan: 1 year = 365 days

2.5 years ×

365 days = 912.5 days = 910 days 1 year

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Another DA Problem • A can of Coca-Cola contains 12 fluid ounces. What is the volume in quarts? (1 qt = 32 fl oz) – Step 1: Determine what you have: 12 fl oz – Step 2: Determine what you want: ??? quarts. – Step 3: Write out your plan to convert from fl oz to qts – Step 4: Select conversion factor(s): 1 qt = 32 fl oz

1 2 fl oz. × Chapter 2

1 qt = 0 .375 qt = 0 .38 qt 3 2 fl oz. 12

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Another DA Problem • A marathon is 26.2 miles. What is the distance in yards (1 mi = 1760 yards)? – Step 1: Determine what is given: 26.2 miles – Step 2:. Determine what you want: ??? yards. – Step 3: Write out your plan to convert from miles to yds – Step 4: Select conversion factor(s): 1 mi = 1760 yards

26 .2 mi ×

1760 yd = 46 ,112 yd = 46 ,100 yd 1 mi

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The Metric System • The metric system is used by scientists and health professionals throughout the world for scientific measurements. • In 1960, a modification of the metric system, called the International System of Units (SI), was adopted by scientists to provide uniformity of units throughout the world. • The metric system offers simplicity over the English system by having a single base unit for each measurement. Chapter 2

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Mass Measurements • The mass of an object is a measure of the amount of matter it posses. – Mass is measured with a balance and is not affected by gravity.

• Weight is the force exerted by gravity on an object – Mass and weight are not interchangeable

• The SI unit for mass is the kilogram (kg) 1 kg = 2.20 lb Chapter 2

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Length Measurements • The metric and SI unit for length is the meter (m) • This unit is slightly longer than a yard (the value used in common measurements in the US)

Chapter 2

1 m = 39.4 in 1 in = 2.54 cm

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Volume Measurements • Volume is the amount of space occupied by matter. • The SI unit for volume is the cubic meter (m3) • The metric unit (and the more often used unit) is the liter (L) • There are several instruments for measuring volume 1 mL = 1 cm3 Chapter 2

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Temperature and Time • Temperature is a measure of how hot or cold an object is. • There are three temperature scales: – Celsius °C (Metric) – Fahrenheit °F (English) – Kelvin K (SI)

• The SI unit for time is the second (s) Chapter 2

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Common Units of Measurement Physical Quantity

Metric

SI

Length

Meter (m)

Meter (m)

Mass

Gram (g)

Kilogram(kg)

Volume

Liter (L)

Cubic meter (m3)

Time

Second (s)

Second (s)

Temperature

Celsius (°C)

Kelvin (K)

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Metric System Advantage • An advantage of the metric system is that it is a decimal system. • It uses prefixes to enlarge or reduce the basic units by factors of 10. • For example, the base unit for length is the meter (m): – A kilometer is 1000 meters. – A centimeter is 1/100 of a meter. Chapter 2

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Metric System Prefixes

You will need to be able to convert between the units boxed in Chapter 2

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Metric Equivalents • We can write unit equations for the conversion between different metric units. • The prefix kilo- means 1000 basic units, so 1 kilometer is 1000 meters. • What is the unit equation? 1 km = 1000 m • What are the two conversion factors for this unit equation?

Chapter 2

1 km 1000 m

or

1000 m 1 km

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Metric-Metric Conversion Problem • What is the mass in grams of a 325 mg aspirin tablet? – Step 1: Determine what is given: 325 mg. – Step 2: Determine want you want: ??? grams. – Step 3: Write out your plan to convert from mg to grams. – Step 4: Select conversion factor(s): 1 mg = 1 x 10-3 g

1 x 10-3 g 325 mg × = 0.325 g 1 mg Chapter 2

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Two Metric-Metric Conversions • A hospital has 125 deciliter bags of blood plasma. What is the volume in milliliters? – Step 1: Determine what you have: 125 dL – Step 2: Determine what you want: ??? mL. – Step 3: Write out your plan to convert from dL to mL. – Step 4: Select conversion factor(s): 1 dL = 0.1 L AND 1 mL = 1 x 10-3 L

125 dL × Chapter 2

0.1 L 1 dL

×

1 mL = 12,500 mL 1 x 10-3 L 24

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Another Example • The mass of the Earth is 5.98 × 1024 kg. What is the mass expressed in megagrams, Mg? – Step 1: Determine what is given: 5.98 × 1024 kg – Step 2: Determine what you want: ??? Mg – Step 3: Write out your plan to convert from kg to Mg. – Step 4: Select conversion factor(s):

1 kg = 1000 g AND 1 Mg = 1,000,000 g

5.98 × 1024 kg ×

1000 g 1 Mg × = 5.98 × 1021 Mg 1000000 g 1 kg

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Measured Numbers • Measured numbers are the numbers you obtain when you measure a quantity such as your height or weight. – A measurement is a number with a unit attached.

• It is not possible to make exact measurements, and all measurements have uncertainty. Chapter 2

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Uncertainty: Accuracy and Precision • Uncertainty is expressed in two ways: Accuracy and Precision • Accuracy is how close to the true value a given measurement is. • Precision is how well a number of independent measurements agree with one another. Chapter 2

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Significant Figures • One way to indicate uncertainty in a measurement is to use significant figures (or significant digits). • The number of significant figures in a measurement is the number of digits that are known accurately plus one uncertain digit. • The uncertain digit is always the last digit that is written!

Counting Significant Figures: Rule 1: Count the number of digits in a measurement from the left to right: • Start with the first non-zero digit (ignore the decimal place!) • Do not count place-holder zeros

Rule 2: The rules for significant digits apply only to measurements and not to exact numbers. Chapter 2

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Place-holder Zeros • Zeros found at the beginning of a number ARE never significant. – Therefore, 0.5 cm, 0.05 cm, and 0.005 cm all have one significant digit.

• Zeros found at the end of a number with no decimal point ARE NOT significant. – Therefore, 50 cm, 500 cm, and 5000 cm all have one significant digit.

• All other zeros are significant – Therefore, 50.0 cm, 0.0500 cm, and 501cm all have three significant digits. Chapter 2

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Exact Numbers • When we count (not measure!) something, it is an exact number. • Exact numbers are also obtained when you compare two units in a measurement system – Example: 1 minute = 60 seconds • Significant digit rules do not apply to exact numbers. Other Examples?? Chapter 2

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Rounding Numbers • All numbers from a measurement are significant. However, we often generate nonsignificant digits when performing calculations, which are removed by rounding off numbers. Rules for Rounding Numbers: Rule 1: If the first nonsignificant digit is less than 5, drop all nonsignificant digits. Rule 2: If the first nonsignificant digit is greater than or equal to 5, increase the last significant digit by 1 and drop all nonsignificant digits. Chapter 2

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Adding & Subtracting Measurements • When adding or subtracting measurements, always look at the number of decimal places in each number – the number with the least number of decimal places has the most uncertainty and determines the number of significant digits! • Lets add three mass measurements. • The measurement 5 g has the least number of decimal places and thus the greatest uncertainty (± 1 g) • The answer has to have the same number of decimal places as the number with the least: Chapter 2

15 g

5 5.0 + 5.00 15.00

g g g g

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Multiplying & Dividing Measurements • When multiplying or dividing measurements, you look at the total number of significant digits (don’t look at decimal places here!) – Here, the number of sig figs in the answer is limited by the value with the fewest total significant figures.

• Lets multiply two length measurements.

5.15 cm

• The measurement 2.3 cm has the fewest significant digits, two.

X

2.3 cm

11.845 cm2

• The correct answer is 12 cm2. Chapter 2

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Metric and English Units • The English system is still very common in the United States and we often have to convert between these units. Mass

Volume

1 lb

=

454 g

1 qt

=

1 oz

=

28.3 g

1 qt

=

2 pts

1 kg

=

2.20 lb

1 qt

=

32 fl. oz.

1 gal

=

4 qts

1 ft

=

12 in

1 mL

=

1 cm3

Length

4 cups (cu)

1 yd

=

3 ft

1L

=

1.06 qt

1 mile

=

5280 ft

1 gal

=

3.785 L

1 in

=

2.54 cm

1m

=

39.4 in

1 calorie (cal)

=

Other 4.184 Joules (J)

1 ft

=

30.5 cm

1 atm

=

760 mm Hg or torr

1 km

=

0.6214 mi

1 atm

=

101.3 kPa

You will need to bring these equalities to class for example problems! You don’t need to memorize these equalities. They will be given to you on the quizzes and exams. Chapter 2

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English-Metric Conversion • A half gallon carton contains 64.0 fl oz of milk. How many milliliters of milk are in a carton? (1 qt = 32 fl oz and 1 qt = 946 mL ) – Step 1: Determine what is given: 64.0 fl oz. – Step 2: Determine what you want: ??? mL – Step 3: Write out your plan to convert fl oz. to mL – Step 4: Select conversion factor(s): 1 qt = 32 fl oz; 1 L = 1.06 qt; 1 mL = 0.001 L – Step 5: Round to three significant figures (WHY??)

64.0 fl oz ×

1 qt 1L 1 mL = 1,890 mL × × 0.001 L 32 fl oz 1.06 qt

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Compound Units • Some measurements have a ratio of units. Can you think of any?

• For example, the speed limit on many highways is 55 miles per hour. How would you convert to this to meters per second? – Convert one unit at a time using the correct conversion factors. Chapter 2

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Compound Unit Problem • A Corvette is traveling at 95 km/hour. What is the speed in meters per second? – Step 1: Determine what is given: 95 km/h – Step 2: Determine what you want: ??? m/s. – Step 3: Write out your plan to convert km to m AND h to s – Step 4: Select conversion factor(s): 1 km = 1000 m and 1 h = 3600 s. – Step 5: Round to two significant figures. (WHY??)

95 km 1 hr 1000 m × × = 26 m/s 1 km hr 3600 s Chapter 2

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The Percent Concept • A percent, %, expresses the amount of a single portion compared to an entire sample.

%=

portion of interest (" Part" ) × 100% total sample (" Whole")

• The percent composition of a compound lists the mass percent of each element. % = Chapter 2

Mass of element Mass of compound X 100% 38

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Calculating Percentages • Sterling silver contains silver and copper. If a sterling silver chain contains 18.5 g of silver and 1.5 g of copper, what is the percent silver in sterling silver? Step 1: Identify the “Part”. Here it is the silver (18.5 g). Step 2: Identify the “Whole”. Here, the total sample (the sterling silver) is the mass of the silver (18.5 g) plus the mass of the copper (1.5 g). Step 3: Plug these values into the formula for percent composition and multiply by 100%.

18.5 g silver × 100% = 92.5% silver (18.5 + 1.5) g Chapter 2

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The Density Concept • The density of an object is its mass per unit of volume – Density (D) is defined as the mass (m) of an object divided by the volume (V) of the object.

mass m = density (D) = volume V Density is a physical property of a chemical substance Density can vary with the physical state of a substance! What do you think is the most dense? A solid, liquid or gas? Chapter 2

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Estimating Density • We can estimate the density of a substance by comparing it to another object. – A solid object will float on top a liquid with a higher density.

What do you think happens if the solid has a higher density than the liquid? Chapter 2

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Calculating Density • What is the density of a platinum nugget that has a mass of 224.50 g and a volume of 10.0 cm3 ? – Step 1: We have a mass (224.50 g) and a volume (10.0 cm3) – Step 2: We want the density (g/cm3) – Step 3: Use the formula for density with your given values to solve for density – Step 4: Round to three significant digits

224.50 g = 22.5 g/cm3 3 10.0 cm Chapter 2

D=

m V 42

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Volume by Displacement • If a solid has an irregular shape, its volume cannot be determined by measuring its dimensions. • You can determine its volume indirectly by measuring the amount of water it displaces. • This technique is called volume by displacement. The Volume of the Object = Final Volume – Initial Volume

VObject = VF - VI Chapter 2

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Calculating Density • A blob of aluminum has a mass of 62.181 g. When the aluminum is placed in a graduated cylinder with a water level of 11.22 mL the water level rises to 34.25 mL. Find the density of the aluminum blob. – Step 1: We have a mass (62.181 g) and two volume measurements (before Al added (VI): 11.22 mL AND after Al added (VF): 34.25 mL) – Step 2: We want the density (g/mL) – Step 3: You must calculate the volume of the aluminum blob (VF - VI) then use the formula for density with you given values to solve for density – Step 4: Round to four significant digits

VAl = VF – VI = 34.25 mL – 11.22 mL = 23.03 mL Chapter 2

D=

D=

m V

62.181 g = 2.700 g/mL 23.03 mL 44

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How to Solve a Quantitative Chemistry Problem

Chapter 2

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Another Density Problem • An automobile battery contains 1275 mL of acid. If the density of battery acid is 1.84 g/mL, how many grams of acid are in an automobile battery? – Step 1: We have volume (1275 mL) – Step 2: We want mass (grams) – Step 3: These quantities are related to one another by the formula for density so rearrange that formula and use the given value for density and volume to solve for our unknown mass – Step 4: Round to three significant digits

D=

m V

Rearrange

1275 mL × Chapter 2

m=D•V

1.84 g = 2350 g mL 46

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More Density Problems

There are additional practice problems available on eRES and my website! Chapter 2

m D= V

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