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1.8Density
Chapter 1 Chemical Foundations
Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings
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Density• Compares the mass of an object to its volume.• Is the mass of a substance divided by its
volume.
Density expressionDensity = mass = g or g = g/cm3
volume mL cm3
Note: 1 mL = 1 cm3
Density
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Densities of Common Substances
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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
Learning Check
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Given: mass = 50.0 g volume = 22.2 cm3
Plan: Place the mass and volume of the osmium metalin the density expression.
D = mass = 50.0 g volume 2.22 cm3
calculator = 22.522522 g/cm3
final answer (2) = 22.5 g/cm3
Solution
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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 isthen calculated from its massand 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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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 2) 6.0 g/cm3 3) 380 g/cm3
Learning Check
object
33.0 mL25.0 mL
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Solution
Given: 48.0 g Volume of water = 25.0 mLVolume 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 mL8.0 mL x 1 cm3 = 8.0 cm3
1 mLSet up Problem:
Density = 48.0 g = 6.0 g = 6.0 g/cm3
8.0 cm3 1 cm3
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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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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
K
K
W
W
W
V
V
VK
Learning Check
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1)
vegetable oil 0.91 g/mL
water 1.0 g/mL
Karo syrup 1.4 g/mLKW
V
Solution
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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 3.8 g and 1 mL factors 1 mL 3.8 g
Density as a Conversion Factor
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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
Learning Check
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Solution
1) 0.614 kgGiven: D = 0.702 g/mL V= 875 mLUnit plan: mL → g → kgEqualities: density 0.702 g = 1 mL
and 1 kg = 1000 g
Setup: 875 mL x 0.702 g x 1 kg = 0.614 kg1 mL 1000 g
density metric factor factor
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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 L2) 0.31 L3) 310 L
Learning Check
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Solution
2) 0.31 LGiven: D = 0.92 g/mL mass = 285 gNeed: volume in litersPlan: g → mL → LEqualities: 1 mL = 0.92 g and 1 L = 1000 mLSet Up Problem:
285 g x 1 mL x 1 L = 0.31 L0.92 g 1000 mL density metricfactor factor
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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
Learning Check
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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 1 L 21 cans 1 lb 2.70 g 1 cm3 1000 mL
= 1.0 L
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Which of the following samples of metals will displace the greatest volume of water?
1 2 3
25 g of aluminum2.70 g/mL
45 g of gold19.3 g/mL
75 g of lead11.3 g/mL
Learning Check
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Solution
25 g of aluminum2.70 g/mL
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Plan: Calculate the volume for each metal and selectthe metal sample with the greatest volume.
1) 25g x 1 mL = 9.3 mL aluminum2.70 g
2) 45 g x 1 mL = 2.3 mL gold19.3 g
3) 75 g x 1 mL = 6.6 mL lead11.3 g
Ref: Timberlake, “Chemistry”, Pearson/Benjamin Cummings, 2006, 9th Ed.