Transcript Chapter 1
Chapter 3 The Properties of Matter and Energy
Setting the Stage – Matter and Energy
Understanding the properties of matter allows us to draw conclusions about compounds and elements found on other planets
Energy, which has no mass, is a second component of the universe, in addition to matter
We can quantify energy and how it interacts with matter
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Setting a Goal - Part A The Properties of Matter
You will learn how a sample of matter can be described by its properties and how they can be quantitatively expressed
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Objective for Section 3-1
List and define several properties of matter and distinguish them as physical or chemical
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3-1 The Physical and Chemical Properties of Matter
Properties describe the particular characteristics of a substance Pure substances have definite composition and definite, unchanging properties Physical properties - can be observed without changing the substance Chemical properties - require that the substance changes into another
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The Physical States of Matter
The three fundamental physical states are solid, liquid and gas
solids have a definite shape and volume
liquids have a definite volume but not a definite shape Gases have neither a definite volume nor shape A substance exists in a particular physical state under defined conditions
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The Physical States of Matter
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Changes in Physical State
Melting point (reverse: freezing point)
temperature at which a substance changes from solid to liquid (reverse: liquid to solid)
Boiling point (reverse: condensation point
temperature at which a substance changes from liquid to gas (reverse: gas to liquid)
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Types of Physical Properties
Intensive properties – those properties that depend on the type or identity (but not the amount) of material present
Examples: color, density, melting point
Extensive properties – those properties that depend on the amount of material present
Examples: mass, volume
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Chemical Changes and Properties
Chemical properties – tendency of a pure substance to undergo chemical changes Sometimes quite difficult to determine Some examples of chemical changes are burning (as opposed to boiling) and color changes Law of the Conservation of Mass - matter is neither created nor destroyed in chemical reactions
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Chemical Changes
Fast reaction: “Thermite Reaction” Medium fast reaction: Zn and HCl
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Slow reaction: rusting of iron
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Objective for Section 3-2
Perform calculations involving the density of liquids and solids
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3-2 Density – A Physical Property
Density is the ratio of the mass of a
substance to the volume of that mass
– it is usually measured in g/mL for solids and liquids; g/L for gases
Specific gravity is the ratio of the mass of a substance to the mass of an equal volume of water
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Density as a Conversion Factor
Density can also be used to convert between mass and volume Typically g → mL or mL → g The density of table salt is 2.16 g/mL. What is the volume in mL occupied by 485 g of table salt?
Unit map g mL Conversion factor is SOLUTION 485 g x 1mL 2.16 g = 225 mL 1 mL 2.16 g
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Determination of Density
This example shows just one way of determining the density of a solid substance
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Objective for Section 3-3
Describe the differences in properties between a pure substance and a mixture
Perform calculations involving percent as applied to mixtures
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3-3 The Properties of Mixtures
A mixture is an aggregation of two or more pure substances Can be separated by physical means (filtration, distillation, crystallization, chromatography) Have chemical and physical properties that are different from the substances that make them up The percentages by mass of the components of a mixture can be varied continuously
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Types of Mixtures
Heterogeneous mixture - nonuniform mixture containing two or more phases with definite boundaries between the phases (e.g. ice and water; sand and water) Homogeneous mixture - same throughout and contains only one phase (substances are mixed at the atomic or molecular level) (e.g. air; aqueous solution of glucose) A phase is one physical state with distinct boundaries and uniform properties
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Separation of Mixtures
A heterogeneous mixture of a solid and liquid can be separated by filtration A heterogeneous mixture of two liquids can be separated using a separating funnel Several substances in solution can be separated by chromatography A solution of a solid in a liquid or of two liquids can be separated by distillation
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Basic Distillation Kit
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Distillation is Important in Industry
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Solutions
A type of homogeneous mixture
Usually involves a liquid phase, but can be solid-solid, gas-gas, solid-liquid, etc.
The pure substances can be in different phases but form a homogeneous mixture (table salt or glucose and water, for example)
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Alloys
Alloys are homogeneous mixtures of metallic elements existing in one phase
Important solid solutions of two or more metals include:
brass (copper and zinc)
dental fillings (silver and mercury)
stainless steel (iron, chromium and nickel)
solder (tin and lead)
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Analysis
Solder is an alloy made from 60.0 % tin and 40.0 % lead. What mass of lead is present in 72 g of solder?
72 g solder x 40.0 g Pb 100 g solder = 29 g Pb
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Setting a Goal - Part B The Properties of Energy
You will be able to qualitatively and quantitatively describe processes in terms of the forms and types of energy associated with them
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Objectives for Section 3-4
Distinguish among the forms and types of energy
Define the terms endothermic and exothermic, providing several examples of each type of process
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3-4 The Forms and Types of Energy
Energy is the capacity to do work
There are many forms of energy
heat
light
chemical (stored energy)
electrical energy
mechanical
nuclear
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Energy
Law of the Conservation of Energy - energy can neither be created nor destroyed, but can only transformed from one form to another The transformation from one type to another may not be efficient (the efficiency of transforming chemical energy to electricity energy (Figure 3-8) is only about 35% efficient). The other 65% is lost as heat but the total amount of energy is constant
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Energy Flow
Exothermic reactions - produce energy (release energy to the surroundings)
Endothermic reactions - require energy input (store energy)
Potential energy is that energy available due to position or composition
Kinetic energy is that energy resulting from motion
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Objective for Section 3-5
Perform calculations involving the specific heat of a substance, and use it to identify a substance
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3-5 Energy Measurement and Specific Heat
Specific heat - the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius (or Kelvin) Reflects how some substances heat up faster than others amount of heat energy Specific heat = mass x
t ( o C)
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Temperature and Specific Heat
Recall that we measure temperature in ° C or K Energy units
calorie (cal) - amount of heat required to raise the temperature of one gram of water from 14.5 ° C to 15.5 ° C
* joule: 1 cal = 4.184 J (exactly) Nutritional ‘Calorie’ is actually 1000 cal (indicated as 1 C)
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Heat Flow
When two substances at different temperatures are put in contact with each other, or mixed, heat flows spontaneously from the substance at higher temperature to the substance at lower temperature
This heat flow continues until the temperatures are the same
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Worked Example 1 (Ch. 3)
Label the following as physical (P) or chemical (C) changes
Brewing beer
Distilling rice wine
Making mist from “dri ice”
Dyeing clothes
Frying chicken
C
P
P
C
C
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Worked Example 2 (Ch. 3)
Classify the following as pure substances, (PS) homogeneous (Ho M) or heterogeneous (He M) mixtures
Pewter
Cloudy apple juice
Sugar (for coffee)
Ho M
He M
PS
Clear apple juice
Distilled water
Tap water
Ho M PS
Ho M
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Worked Example 3 (Ch. 3)
An unknown metal with a mass of 23.6 g is placed in a graduated cylinder that had an initial water volume of 22.0 mL. After the sample is submerged in the water, the level of water in the cylinder was observed to be 24.1 mL. Determine the identity of the metal
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Worked Example 3 (Ch. 3)
Solution 23.6 g occupies 2.1 mL, hence the density of the metal is 23.6 g/2.1 mL = 11.2 g/mL The metal is likely to be Pb (density = 11.3
g/mL)
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Worked Example 4 (Ch. 3)
A gas can (438 g) contains 33.5 mL of kerosene. If the mass of the can plus the kerosene is 465 g, what is the density of kerosene?
Solution: The 33.5 mL of kerosene weighs 465 g – 438 g = 27.0 g, hence its density is 27.0/33.5 = 0.806 g/mL
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Worked Example 5 (Ch. 3)
At room temperature, the specific heat of benzene is 1.060 J/g o C. If a 30.-g sample of benzene releases 450 J of energy, what is the change in temperature?
Solution: Energy (heat) transferred = m x sp ht x Change in temp (
T) 450 (J) = 30 (g) x 1.060 (J/g o C) x
T ( o C) Change in temp = 14 o C
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Worked Example 6 (Ch. 3)
A 440 g piece of metal at 100.0 o C is placed in 258 g of water initially at 25.0 o C. If the final temperature is 36.5 o C, what is the specific heat of the metal?
Solution: Heat lost (metal) = heat gained (water) -m(m) x sp ht (m) x
t (m) = m(w) x sp ht (w) x
t (w)
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Worked Example 6 (Ch. 3)…contd.
Hence, -440 g x sp ht(m) x (36.5 - 100.0 ) o C = 258 g x 4.184 J/g o C x (36.5 – 25.0) o C Therefore, sp ht (m) = 258 g x 4.184 J/g o C x 11.5 o C 440 g x 63.5 o C = 0.444 J /g o C (iron)
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Objective for Special Topic – Units of Energy
Distinguish amongst different units of energy and carry out interconversions between them
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Special Topic – Units of Energy
The SI (m-kg-s or MKS system) unit of energy is the joule (J) (= kg m 2 s -2 )
The cgs (cm-g-s) system unit of energy is the erg (= g cm 2 s -2 ) The conversion factor is 1 J = 10 7 ergs
The joule is a much bigger unit of energy than the erg
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An Energy Unit for Atomic Systems
The potential energy associated with a proton and an electron separated by 1 Å (= 10 -10 m), a typical atomic distance, is only –2.307 x 10 -18 J (or –2.307 x 10 -11 ergs)
Hence a more convenient unit is needed to express energies relating to atoms.
This unit is the electron-volt (eV)
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The Electron-Volt Unit
The kinetic energy (KE) gained by an electron falling through an electrical potential difference (voltage) of 1 volt (V) is defined as 1 eV
1 V is defined as the voltage that gives 1 joule of energy to 1 Coulomb (C) of charge; that is, 1 V = 1 J C -1
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The Electron-Volt Unit
KE = electronic charge x potential difference
Hence, 1 eV = 1.602 x 10 -19 C x 1 V = 1.602 x 10 -19 C x 1 J C -1 = 1.602 x 10 -19 J (1 J = 6.173 x 10 18 eV) Therefore, the potential energy associated with a proton and an electron separated by 1A is _ 2.307 x 10 -18 J x 6.173 x 10 18 1 J eV =
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_ 14.40 eV
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The Electron-Volt Unit
With this value, the Coulomb equation for the potential energy associated with two particles of charge q 1 and q 2 PE = kq r 1 q 2 q 1 and q 2 are in coulombs; k is a proportionality constant; r is the distance between q in meters. PE is in J 1 and q 2 can be reduced to a much more convenient 'engineering formula', PE = 14.40 (eVA) q 1 q 2 r(A) PE is in eV Here, q 1 and q 2 are merely the unit electronic charges on the particles (e.g. -1 for an electron, +1 for a proton, +2 for a He nucleus, etc)
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