2 8 density
TRANSCRIPT
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2.8
Density
Chapter 2 Measurements
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
Objects that sink in water are more dense than water; objects that float in water are less dense.
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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 expression:
D = mass = g or g = g/cm3 volume mL cm3
Note: 1 mL = 1 cm3
Density
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Guide to Calculating Density
3 Basic Chemistry Copyright © 2011 Pearson Education, Inc.
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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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STEP 1 Given 50.0 g; 22.2 cm3
Need density, in g/cm3
STEP 2 Plan Write the density expression. D = mass
volumeSTEP 3 Express mass in grams and volume in
cm3
mass = 50.0 g volume = 22.2 cm3
STEP 4 Set up problem D = 50.0 g = 22.522522 g/cm3
2.22 cm3
= 22.5 g/cm3 (3 SF)
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
The density of a solid is determined from its mass and the volume it displaces when submerged in water.
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Density Using Volume Displacement
The density of the object iscalculated from its mass andvolume. mass = 68.60 g = 7.2 g/cm3
volume 9.5 cm3
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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
25.0 mL 33.0 mL
object
Learning Check
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STEP 1 Given 48.0 g volume of water = 25.0 mL volume of water + metal = 33.0
mL Need Density (g/mL)
STEP 2 Plan Calculate the volume difference, change unit to cm3, and place in density expression.
STEP 3 Express mass in grams and volume in cm3
volume of solid = 33.0 mL - 25.0 mL = 8.0 mL
8.0 mL x 1 cm3 = 8.0 cm3
1 mL
STEP 4 Set Up Problem
Density = 48.0 g = 6.0 g = 6.0 g/cm3
8.0 cm3 1 cm3
Solution
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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.
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
Objects that sink in water are more dense than water.
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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
W
W
K
K
K
V
V
V
W
Learning Check
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1)
vegetable oil 0.91 g/mL
water 1.0 g/mL
Karo syrup 1.4 g/mLK
W
V
Solution
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For a density of 3.8 g/mL, • an equality is written as
3.8 g = 1 mL
• and two conversion factors are written as
Conversion 3.8 g and 1 mL factors 1 mL 3.8 g
Density as a Conversion Factor
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
Guide to Using Density
15 Basic Chemistry Copyright © 2011 Pearson Education, Inc.
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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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1) 0.614 kg
STEP 1 Given D = 0.702 g/mL; V= 875 mL
Need mass in kg of octane
STEP 2 Plan mL g kg
STEP 3 Equalities
0.702 g = 1 mL 1 kg = 1000 g
STEP 4 Set Up Problem 875 mL x 0.702 g x 1 kg = 0.614 kg (1)
1 mL 1000 g density metric factor factor
Solution
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
18
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
Learning Check
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2) 0.31 L
STEP 1 Given D = 0.92 g/mL mass = 285 g
Need volume in liters
STEP 2 Plan g mL L
STEP 3 Equalities 1 mL = 0.92 g 1 L = 1000 mL
STEP 4 Set Up Problem
285 g x 1 mL x 1 L = 0.31 L 0.92 g 1000 mL density metric
factor factor
Solution
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
20
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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1)
Plan: Calculate the volume for each metal and select the metal sample with the greatest volume.
1) 25 g 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
Solution
25 g of aluminum2.70 g/mL
Basic Chemistry Copyright © 2011 Pearson Education, Inc.