1.8 Density Chapter 1 Chemical Foundations

Is the mass of a substance divided by its volume. Density expression. Density = mass = g or g = g/cm3 volume mL cm3. Note: 1 mL = 1 cm3. Density ... 4...

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Chapter 1

Chemical Foundations 1.8 Density

Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

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Density Density

• Compares the mass of an object to its volume. • Is the mass of a substance divided by its volume. Density expression Density = mass = g or g = g/cm3 volume mL cm3 Note: 1 mL = 1 cm3

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Densities of Common Substances

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Learning Check Osmium is a very dense metal. What is its density in g/cm3 if 50.0 g of osmium has a volume of 2.22 cm3? 1) 2.25 g/cm3 2) 22.5 g/cm3 3) 111 g/cm3

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Solution Given: mass = 50.0 g volume = 22.2 cm3 Plan: Place the mass and volume of the osmium metal in the density expression. D =

mass volume

calculator final answer (2)

=

50.0 g 2.22 cm3

= 22.522522 g/cm3 = 22.5 g/cm3

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Volume by Displacement • A solid completely



submerged in water displaces its own volume of water. The volume of the solid is calculated from the volume difference. 45.0 mL - 35.5 mL = 9.5 mL = 9.5 cm3 Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

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Density Using Volume Displacement The density of the zinc object is then calculated from its mass and volume. mass = 68.60 g = 7.2 g/cm3 volume 9.5 cm3

Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

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Learning Check What is the density (g/cm3) of 48.0 g of a metal if the level of water in a graduated cylinder rises from 25.0 mL to 33.0 mL after the metal is added? 1) 0.17 g/cm3

25.0 mL

2) 6.0 g/cm3

3) 380 g/cm3

33.0 mL object 8

Solution Given: 48.0 g

Volume of water = 25.0 mL Volume of water + metal = 33.0 mL Need: Density (g/mL) Plan: Calculate the volume difference. Change to cm3, and place in density expression. 33.0 mL - 25.0 mL = 8.0 mL 8.0 mL x

1 cm3 1 mL

Set up Problem: Density = 48.0 g = 8.0 cm3

=

8.0 cm3

6.0 g = 6.0 g/cm3 1 cm3 9

Sink or Float • Ice floats in water because the density of ice is less than the density of water.

• Aluminum sinks because its density is greater than the density of water.

Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

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Learning Check Which diagram correctly represents the liquid layers in the cylinder? Karo (K) syrup (1.4 g/mL), vegetable (V) oil (0.91 g/mL,) water (W) (1.0 g/mL) 1

2

3

V

W

K

W

K

V

K

V

W 11

Solution 1)

V W K

vegetable oil 0.91 g/mL water 1.0 g/mL Karo syrup 1.4 g/mL

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Density as a Conversion Factor Density can be written as an equality. • For a substance with a density of 3.8 g/mL, the equality is 3.8 g = 1 mL

• From this equality, two conversion factors can be written for density. Conversion factors

3.8 g 1 mL

and

1 mL 3.8 g

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Learning Check The density of octane, a component of gasoline, is 0.702 g/mL. What is the mass, in kg, of 875 mL of octane? 1) 0.614 kg 2) 614 kg 3) 1.25 kg

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Solution 1) 0.614 kg Given: D = 0.702 g/mL

V= 875 mL

Unit plan: mL → g → kg Equalities: density and

0.702 g

= 1 mL

1 kg = 1000 g

Setup: 875 mL x 0.702 g x 1 kg = 0.614 kg 1 mL 1000 g density factor

metric factor 15

Learning Check If olive oil has a density of 0.92 g/mL, how many liters of olive oil are in 285 g of olive oil? 1) 0.26 L 2) 0.31 L 3) 310 L

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Solution 2) 0.31 L Given: D = 0.92 g/mL mass = 285 g Need: volume in liters Plan: g → mL → L Equalities: 1 mL = 0.92 g and 1 L = 1000 mL Set Up Problem: 285 g x 1 mL x 0.92 g density factor

1L = 1000 mL

0.31 L

metric factor

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Learning Check A group of students collected 125 empty aluminum cans to take to the recycling center. If 21 cans make 1.0 lb aluminum, how many liters of aluminum (D=2.70 g/cm3) are obtained from the cans? 1) 1.0 L

2) 2.0 L

3) 4.0 L

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Solution 1) 1.0 L 125 cans x 1.0 lb x 454 g x 1 cm3 x 1 mL x 1L 21 cans 1 lb 2.70 g 1 cm3 1000 mL = 1.0 L

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Learning Check Which of the following samples of metals will displace the greatest volume of water? 1 25 g of aluminum 2.70 g/mL

2 45 g of gold 19.3 g/mL

3 75 g of lead 11.3 g/mL

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Solution 1)

25 g of aluminum 2.70 g/mL

Plan: Calculate the volume for each metal and select the metal sample with the greatest volume. 1) 25g x 1 mL = 9.3 mL aluminum 2.70 g 2) 45 g x 1 mL = 2.3 mL gold 19.3 g 3) 75 g x 1 mL = 6.6 mL lead 11.3 g Ref: Timberlake, “Chemistry”, Pearson/Benjamin Cummings, 2006, 9th Ed. 21