How can we represent matter?
Particle Model
Copyright © 2008 by Pearson Education, Inc.
Publishing as Benjamin Cummings
1
Matter
Matter
• Is the material that
makes up all things.
• Has mass and
occupies space.
Copyright © 2008 by Pearson Education, Inc.
Publishing as Benjamin Cummings
2
Matter is made of particles, which we can think of
as small round spheres or BBs.
3
A few particles
far apart
Many particles
crowded together
Increasing Density
lighter particles
less massive
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heavier particles
more massive
How can we Quantify Matter?
Density
Copyright © 2008 by Pearson Education, Inc.
Publishing as Benjamin Cummings
5
Density
Density
• Compares the mass of an object to its
volume.
• Is the mass of a substance divided by its
volume.
• Note: 1 mL = 1 cm3
• Is expressed as
D = mass = g or g = g/cm3
volume
mL
cm3
6
Densities of Common
Substances
Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings
7
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: Write the density expression.
D =
mass
volume
Express mass in grams and volume in cm3
mass = 50.0 g
volume = 22.2 cm3
Set up problem using mass and volume.
D = 50.0 g
= 22.522522 g/cm3
2.22 cm3
= 22.5 g/cm3 (3 SF)
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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 © 2008 by Pearson Education, Inc.
Publishing as Benjamin Cummings
10
Density Using Volume
Displacement
The density of the object is
calculated from its mass and
volume.
mass = 68.60 g = 7.2 g/cm3
volume 9.5 cm3
Copyright © 2008 by Pearson Education, Inc.
Publishing as Benjamin Cummings
11
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
2) 6.0 g/cm3
3) 380 g/cm3
25.0 mL
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33.0 mL
object
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
13
=
8.0 cm3
6.0 g = 6.0 g/cm3
1 cm3
Sink or Float
Copyright © 2008 by Pearson Education, Inc.
Publishing as Benjamin Cummings
• 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.
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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)
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1
2
3
V
W
K
W
K
V
K
V
W
Solution
1)
V
W
K
16
vegetable oil 0.91 g/mL
water 1.0 g/mL
Karo syrup 1.4 g/mL
Density as a Conversion
Factor
Density can be written as an equality.
• For a density of 3.8 g/mL, the equality is:
3.8 g = 1 mL
• From this equality, two conversion factors can be written:
Conversion 3.8 g
factors
1 mL
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and
1 mL
3.8 g
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
Need: mass in kg of octane
Unit plan: mL  g  kg
Equalities: density 0.702 g
and
= 1 mL
1 kg = 1000 g
Setup: 875 mL x 0.702 g x 1 kg = 0.614 kg
1 mL
1000 g
density
metric
factor
factor
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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)
2)
3)
20
0.26 L
0.31 L
310 L
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
1 L = 1000 mL
Set Up Problem:
285 g x 1 mL x
0.92 g
density
factor
21
1L
=
1000 mL
metric
factor
0.31 L
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
22
2) 2.0 L
3) 4.0 L
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
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2
45 g of gold
19.3 g/mL
3
75 g of lead
11.3 g/mL
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
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