Chapter 1 Measurements 1.7 Density Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings
Density Density Compares the mass of an object to its volume. Is the mass of a substance divided by its volume. Density expression Density = mass = g or g = g/cm3 volume mL cm3 Note: 1 mL = 1 cm3
Densities of Common Substances
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
Solution Given: mass = 50.0 g volume = 22.2 cm3 Plan: Place the mass and volume of the osmium metal in the density expression. D = mass = 50.0 g volume 2.22 cm3 calculator = 22.522522 g/cm3 final answer (2) = 22.5 g/cm3
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
Density Using Volume Displacement The density of the zinc object is then calculated from its mass and volume. mass = 68.60 g = 7.2 g/cm3 volume 9.5 cm3 Copyright © 2005 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? 1) 0.17 g/cm3 2) 6.0 g/cm3 3) 380 g/cm3 33.0 mL 25.0 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 = 8.0 cm3 1 mL Set up Problem: Density = 48.0 g = 6.0 g = 6.0 g/cm3 8.0 cm3 1 cm3
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
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) 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
Density as a Conversion Factor 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
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
Solution 1) 0.614 kg Given: D = 0.702 g/mL V= 875 mL 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? 1) 0.26 L 2) 0.31 L 3) 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 and 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? 1) 1.0 L 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 1 L 21 cans 1 lb 2.70 g 1 cm3 1000 mL = 1.0 L
Learning Check 1 2 3 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 1) 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