1. Chapter 2 Measurements 2.8 Density
Copyright © by Pearson Education, Inc.
Publishing as Benjamin Cummings

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

3. Densities of Common Substances
Copyright © by Pearson Education, Inc. Publishing as Benjamin Cummings

4. Sink or Float
Copyright © 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.

5. Learning Check K W V V W K W V K
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) K W V V W K V W K

6. Solution vegetable oil 0.91 g/mL V W K 1) water 1.0 g/mL
Karo syrup 1.4 g/mL V W K

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?

8. Solution Given: mass = 50.0 g volume = 2.22 cm3
Plan: Write the density expression.
D = mass
volume
Express mass in grams and volume in cm3 mass = 50.0 g volume = 2.22 cm3
Set up problem using mass and volume.
D = 50.0 g = g/cm3
2.22 cm3
= g/cm3 (3 SF)

9. Solve for any of the Three
If d=m/v then…
What does m equal? What does v equal?
Solve it with algebra.
m = d x v
v = m / d

10. 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 mL
9.5 mL
Copyright © by Pearson Education, Inc.
Publishing as Benjamin Cummings

11. Density Using Volume Displacement
The density of the object is
calculated from its mass and
volume.
mass = g = 7.2 g/cm3
volume cm3
Copyright © by Pearson Education, Inc.
Publishing as Benjamin Cummings

12. 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?
25.0 mL mL
object

13. Solution Given: 48.0 g Volume of water = 25.0 mL
Volume of water + metal = 33.0 mL
Need: Density (g/ cm3)
Plan: Calculate the volume difference. Change to cm3,
and place in density expression.
33.0 mL mL = mL = 8.0 cm3
Set up Problem:
Density = g = g/cm3
8.0 cm3

14. Density as a Conversion Factor
Density can be written as an equality.
It converts mass to volume, or volume to mass.
The density of a material is 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 and mL
factors 1 mL g

15. Learning Check
The density of octane, a component of gasoline, is g/mL. What is the mass, in kg, of 875 mL of octane?

16. Solution Given: D = 0.702 g/mL V= 875 mL Need: mass in kg of octane
M = D x V
Unit plan: mL  g  kg
Equalities: density g = 1 mL
and kg = 1000 g
Setup: 875 mL x g x 1 kg = kg
1 mL g
density metric
factor factor

17. 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?

18. Solution Given: D = 0.92 g/mL mass = 285 g Need: volume in liters
V = M / D
Plan: g mL L
Equalities: 1 mL = 0.92 g L = 1000 mL
Set Up Problem:
285 g x 1 mL x L = L
0.92 g mL
density metric
factor factor

19. 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?

20. Solution 125 cans x 1.0 lb x 454 g x 1 cm3 x 1 mL x 1 L
21 cans lb g 1 cm mL
= 1.0 L

21. Learning Check 25 g of aluminum 2.70 g/mL 45 g of gold 19.3 g/mL
Which of the following samples of metals will displace the greatest volume of water?
25 g of aluminum
2.70 g/mL
45 g of gold
19.3 g/mL
75 g of lead
11.3 g/mL

22. Solution 25 g of aluminum 2.70 g/mL 1)
Plan: Calculate the volume for each metal and select
the metal sample with the greatest volume.
1) 25g x 1 mL = mL aluminum g
2) 45 g x 1 mL = mL gold g
3) 75 g x 1 mL = mL lead g 1)
Plan: Calculate the volume for each metal and select
the metal sample with the greatest volume.
1) 25g x 1 mL = mL aluminum g
2) 45 g x 1 mL = mL gold g
3) 75 g x 1 mL = mL lead g 1)
Plan: Calculate the volume for each metal and select
the metal sample with the greatest volume.
1) 25g x 1 mL = mL aluminum g
2) 45 g x 1 mL = mL gold g
3) 75 g x 1 mL = mL lead g 1)
Plan: Calculate the volume for each metal and select
the metal sample with the greatest volume.
1) 25g x 1 mL = mL aluminum g
2) 45 g x 1 mL = mL gold g
3) 75 g x 1 mL = mL lead g