Chapter 2

Metric System

2 Chapter 2

Metric System Basic Units

3 Chapter 2

Original Metric Unit Definitions •  A meter was defined as 1/10,000,000 of the

distance from the North Pole to the equator.

•  A kilogram (1000 grams) was equal to the mass of a cube of water measuring 0.1 m on each side.

•  A liter was set equal to the volume of one kilogram of water at 4 °C.

4 Chapter 2

Metric System Advantage •  Another advantage of the metric system is that it is

a decimal system.

•  It uses prefixes to enlarge or reduce the basic units.

•  For example: – A kilometer is 1000 meters.

– A millimeter is 1/1000 of a meter.

5 Chapter 2

Metric System Prefixes •  The following table lists the common prefixes

used in the metric system:

6 Chapter 2

Metric Prefixes, Continued •  For example, the prefix kilo- increases a base unit

by 1000:

– 1 kilogram is 1000 grams.

•  The prefix milli- decreases a base unit by a factor of 1000:

– There are 1000 millimeters in 1 meter.

7 Chapter 2

Metric Symbols •  The names of metric units are abbreviated using

symbols. Use the prefix symbol followed by the symbol for the base unit, so:

– Nanometer is abbreviated nm.

– Microgram is abbreviated µg.

– Deciliter is abbreviated dL.

– Gigasecond is abbreviated Gs.

8 Chapter 2

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.

•  The unit equation is 1 km = 1000 m.

•  Similarly, a millimeter is 1/1000 of a meter, so the unit equation is 1000 mm = 1 m.

9 Chapter 2

Metric Unit Factors •  Since 1000 m = 1 km, we can write the following

unit factors for converting between meters and kilometers:

1 km or 1000 m 1000 m 1 km

•  Since 1000 mm = 1 m, we can write the following unit factors:

1000 mm or 1 m . 1 m 1000 mm

10 Chapter 2

Metric–Metric Conversions •  We will use the unit analysis method we learned

in Chapter 2 to do metric–metric conversion problems.

•  Remember, there are three steps: 1.  Write down the unit asked for in the answer.

2.  Write down the given value related to the answer.

3.  Apply unit factor(s) to convert the given unit to the units desired in the answer.

11 Chapter 2

Metric–Metric Conversion Problem What is the mass in grams of a 325 mg aspirin

tablet?

•  Step 1: We want grams.

•  Step 2: We write down the given: 325 mg.

•  Step 3: We apply a unit factor (1000 mg = 1 g) and round to three significant figures.

325 mg x = 0.325 g 1000 mg 1 g

12 Chapter 2

Two Metric–Metric Conversions A hospital has 125 deciliters of blood plasma.

What is the volume in milliliters?

•  Step 1: We want the answer in mL.

•  Step 2: We have 125 dL.

•  Step 3: We need to first convert dL to L and then convert L to mL:

1 L and 1000 mL 10 dL 1 L

13 Chapter 2

Two Metric–Metric Conversions

•  Apply both unit factors, and round the answer to three significant digits.

•  Notice that both dL and L units cancel, leaving us with units of mL.

125 dL x = 12,500 mL x 10 dL 1 L 1000 mL

1 L

14 Chapter 2

Another Example The mass of the Earth’s moon is 7.35 × 1022 kg. What is the mass expressed in megagrams, Mg?

•  We want Mg; we have 7.35 x 1022 kg.

•  Convert kilograms to grams, and then grams to megagrams.

7.35 x 1022 kg × = 5.98 x 1019 Mg x 1 kg

1000 g 1 Mg 1000000 g

15 Chapter 2

Metric and English Units •  The English system is still very common in the

United States.

•  We often have to convert between English and metric units.

16 Chapter 2

Metric–English Conversion The length of an American football field,

including the end zones, is 120 yards. What is the length in meters?

•  Convert 120 yd to meters (given that 1 yd = 0.914 m).

120 yd x = 110 m 1 yd 0.914 m

17 Chapter 2

English–Metric Conversion A half-gallon carton contains 64.0 fl oz of milk.

How many milliliters of milk are in a carton? •  We want mL; we have 64.0 fl oz. •  Use 1 qt = 32 fl oz, and 1 qt = 946 mL.

64.0 fl oz x = 1,890 mL x 32 fl oz

1 qt 946 mL 1 qt

18 Chapter 2

Compound Units •  Some measurements have a ratio of units.

•  For example, the speed limit on many highways is 55 miles per hour. How would you convert this to meters per second?

•  Convert one unit at a time using unit factors. 1.  First, miles → meters

2.  Next, hours → seconds

19 Chapter 2

Compound Unit Problem A motorcycle is traveling at 75 km/hour.

What is the speed in meters per second?

•  We have km/h; we want m/s.

•  Use 1 km = 1000 m and 1 h = 3600 s.

= 21 m/s x 1 km

1000 m 1 hr 3600 s

75 km hr

x

20 Chapter 2

Volume by Calculation •  The volume of an object is calculated by

multiplying the length (l) by the width (w) by the thickness (t).

volume = l x w x t

•  All three measurements must be in the same units.

•  If an object measures 3 cm by 2 cm by 1 cm, the volume is 6 cm3 (cm3 is cubic centimeters).

21 Chapter 2

Cubic Volume and Liquid Volume •  The liter (L) is the basic unit of volume in the

metric system. •  One liter is

defined as the volume occupied by a cube that is 10 cm on each side.

22 Chapter 2

Cubic and Liquid Volume Units •  1 liter is equal to 1000 cubic centimeters.

– 10 cm x 10 cm x 10 cm = 1000 cm3

•  1000 cm3 = 1 L = 1000 mL.

•  Therefore, 1 cm3 = 1 mL.

23 Chapter 2

Cubic–Liquid Volume Conversion An automobile engine displaces a volume of

498 cm3 in each cylinder. What is the displacement of a cylinder in cubic inches,

in3?

•  We want in3; we have 498 cm3.

•  Use 1 in = 2.54 cm three times.

= 30.4 in3 x 1 in 2.54 cm

x 498 cm3 x 1 in 2.54 cm

1 in 2.54 cm

24 Chapter 2

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.

•  Volume by displacement can also be used to determine the volume of a gas.

25 Chapter 2

Solid Volume by Displacement You want to measure the volume of an irregularly

shaped piece of jade. •  Partially fill a volumetric flask with water and

measure the volume of the water.

•  Add the jade, and measure the difference in volume.

•  The volume of the jade is 10.5 mL.

26 Chapter 2

Gas Volume by Displacement You want to measure the volume of gas given off

in a chemical reaction. •  The gas produced displaces the water in the flask

into the beaker. The volume of water displaced is equal to the volume of gas.

27 Chapter 2

The Density Concept •  The density of an object is a measure of its

concentration of mass.

•  Density is defined as the mass of an object divided by the volume of the object.

= density volume mass

28 Chapter 2

Density •  Density is expressed in different units. It is

usually grams per milliliter (g/mL) for liquids, grams per cubic centimeter (g/cm3) for solids, and grams per liter (g/L) for gases.

29 Chapter 2

Densities of Common Substances

30 Chapter 2

Estimating Density •  We can estimate the density of a

substance by comparing it to another object.

•  A solid object will float on top of a liquid with a higher density.

•  Object S1 has a density less than that of water, but larger than that of L1.

•  Object S2 has a density less than that of L2, but larger than that of water.

Critical Thinking Question •  A glass cylinder contains four liquid layers:

mercury (d = 13.6 g/mL), chloroform (d = 1.49 g/mL), water (d = 1.00 g/mL), and ether (d = 0.708 g/mL). If an ice cube (d = 0.92 g/mL) is dropped into the cylinder, where does it come to rest?

A) on top of the ether layer B) on top of the water layer C) on top of the chloroform layer D) on top of the mercury layer E) on the bottom of the cylinder

31 Chapter 2

32 Chapter 2

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 ? Recall, density is mass/volume.

= 22.5 g/cm3 10.0 cm3 224.50 g

33 Chapter 2

Density as a Unit Factor •  We can use density as a unit factor for conversions

between mass and volume.

•  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?

– We have 1275 mL; we want grams:

1275 mL x = 2350 g mL 1.84 g

34 Chapter 2

Critical Thinking: Gasoline The density of gasoline is 730 g/L at 0 ºC (32 ºF)

and 713 g/L at 25 ºC (77 ºF). What is the mass difference of 1.00 gallon of gasoline at these two

temperatures (1 gal = 3.784L)?

35 Chapter 2

Critical Thinking: Gasoline The density of gasoline is 730 g/L at 0 ºC (32 ºF)

and 713 g/L at 25 ºC (77 ºF). What is the mass difference of 1.00 gallon of gasoline at these two

temperatures (1 gal = 3.784L)?

•  The difference is about 60 grams (about 2 %).

= 2760 g x At 0 ºC: 1.00 gal x 730 g L

3.784 L 1 gal

= 2700 g x At 25 ºC: 1.00 gal x 713 g L

3.784 L 1 gal

36 Chapter 2

Temperature •  Temperature is a measure of the average kinetic

energy of the individual particles in a sample.

•  There are three temperature scales: 1.  Celsius

2.  Fahrenheit

3.  Kelvin

•  Kelvin is the absolute temperature scale.

37 Chapter 2

Temperature Scales •  On the Fahrenheit scale, water freezes at 32 °F and

boils at 212 °F.

•  On the Celsius scale, water freezes at 0 °C and boils at 100 °C. These are the reference points for the Celsius scale.

•  Water freezes at 273 K and boils at 373 K on the Kelvin scale.

38 Chapter 2

Temperature Conversions •  This is the equation for converting °C to °F.

(°C * 9/5) + 32 = °F

•  This is the equation for converting °F to °C.

(°F – 32) * 5/9 = °C

•  To convert from °C to K, add 273. °C + 273 = K

39 Chapter 2

Fahrenheit–Celsius Conversions •  Body temperature is 98.6 °F. What is body

temperature in degrees Celsius? In Kelvin?

K = °C + 273 = 37.0 °C + 273 = 310 K

( ) 180°F 100°C

= 37.0°C (98.6°F - 32°F) x

40 Chapter 2

Heat •  Heat is the flow of energy from an object at a

higher temperature to an object at a lower temperature.

•  Heat measures the total energy of a system.

•  Temperature measures the average energy of particles in a system.

•  Heat is often expressed in terms of joules (J) or calories (cal).

41 Chapter 3

Heat

Same temperature but Beaker (b) has more heat than beaker (a)

42 Chapter 2

The Percent Concept •  A percent, %, expresses the amount of a single

quantity compared to an entire sample.

•  A percent is a ratio of parts per 100 parts.

•  The formula for calculating percent is shown below:

100% sample total

interest of quantity % x=

43 Chapter 2

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 of silver in sterling silver?

silver 92.5% % 100g 1.5) (18.5

silver g 18.5=

+x

44 Chapter 2

Percent Unit Factors •  A percent can be expressed as parts per 100 parts.

•  25% can be expressed as 25/100 and 10% can be expressed as 10/100.

•  We can use a percent expressed as a ratio as a unit factor.

– A rock is 4.70% iron, so sample of g 001

iron g 4.70

45 Chapter 2

Percent Unit Factor Calculation The Earth and Moon have a similar composition;

each contains 4.70% iron. What is the mass of iron in a lunar sample that weighs 235 g?

•  Step 1: We want g iron. •  Step 2: We write down the given: 235 g sample. •  Step 3: We apply a unit factor (4.70 g iron = 100 g

sample) and round to three significant figures.

iron g 0.11sample g100

iron g 4.70sample g235 =x

46 Chapter 2

Chemistry Connection: Coins •  A nickel coin contains 75.0 % copper metal and

25.0 % nickel metal, and has a mass of 5.00 grams. •  What is the mass of nickel metal in a nickel coin?

nickel g 5.12coin g100nickel g 25.0

coin g00.5 =x