Chapter 2 Measurements 2.8 Density Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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

Densities of Common Substances Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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.

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) 1 2 3 K W V V W K V W K

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

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?

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 = 22.522522 g/cm3 2.22 cm3 = 22.5 g/cm3 (3 SF)

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

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 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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

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 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/ cm3) Plan: Calculate the volume difference. Change to cm3, and place in density expression. 33.0 mL - 25.0 mL = 8.0 mL = 8.0 cm3 Set up Problem: Density = 48.0 g = 6.0 g/cm3 8.0 cm3

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 1 mL factors 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?

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 0.702 g = 1 mL and 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

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?

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 1 L = 1000 mL Set Up Problem: 285 g x 1 mL x 1 L = 0.31 L 0.92 g 1000 mL density metric factor factor

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?

Solution 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

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? 1 2 3 25 g of aluminum 2.70 g/mL 45 g of gold 19.3 g/mL 75 g of lead 11.3 g/mL

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 = 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 1) 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 1) 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 1) 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