Classification of Matter

MaterialMaterial is used to refer to a specific kind of matter.

Examples

A mixturemixture is matter that contains more than one material.

Examples

Heterogeneous Materials

Mixtures in which the materials are not uniform are called

heterogeneous.

Granite

• Granite is heterogeneous because it is composed of several different minerals. If it were crushed, you could pick out sand-sized particles of quartz, biotite, and feldspar.

Milk

• Appears uniform, but it can be separated into visible parts.

Phase

• Each separate part of a material is called a phase.

• OR, it is any region with a uniform set of properties.

Interfaces

• The different phases in a heterogeneous mixture are separated from each other by definite boundaries called interfaces.

Homogeneous Materials

Materials that consist of only one phase are called homogeneous.

Examples

Solution

• Only one kind of homogeneous material can be classified as a mixture, a solution.

• Solutions are composed of more than one material:– Solute – dissolved material– Solvent – dissolving material

• Solute particles are dispersed among the solvent uniformly.

Example

Sugar Water

Solute?Solvent?

Not all solutions are liquids!

Made of nitrogen, oxygen, and other gases…

Molarity• Solutions can be different concentrations.• The letter “M” is used to represent the

term molarity.• Molarity is the amount of solute in a given

amount of solvent.– A 6M (6 molar) solution contains 6 times as

much solute as 1M (1 molar) solution of the same volume.

• Concentrated solutions have a higher ratio of solute to solvent than dilute solutions.

Substances

Homogeneous materials that always have the same

composition.

Examples

Substances can be divided into two categories:

• Elements – substances composed of only one kind of atom– Examples – sulfur, oxygen, hydrogen, copper,

and gold

• Compounds – substances composed of more than one kind of atom– Example – Water, H2O (Atoms in a compound

are always in definite proportion, like 2 hydrogen to 1 oxygen in water.)

Organic vs. Inorganic

• Organic compounds mean that carbon is contained.

• Inorganic compounds mean that no carbon is contained

There are a few exceptions…

Physical and Chemical Changes

A quick review…

Physical Changes

• A physical change occurs when a substance is subjected to some condition, and the substance remains.

• Examples:– Pounding copper sheets– Cutting wood– Tearing Paper– Dissolving sugar in water

Chemical Changes• Whenever a substance undergoes a

change so that one or more new substance with different characteristics is formed, a chemical change (or chemical reaction) has taken place.– *A hint… If a precipitate, gas, color change,

or energy change occurs, a chemical change has taken place. (There are some exceptions.)

– A precipitate is a solid substance that forms from a solution.

Physical and Chemical Properties

A quick review…

Physical Properties

• A physical property is a description of the behavior of a substance undergoing a physical change.– Extensive properties – depend on the amount

of matter present• Mass, length, and volume

– Intensive properties – do not depend on the amount of matter present

• Density, malleability, ductility, conductivity, color, melting point, and boiling point

Chemical Properties• A chemical property describes the reaction

of a substance with other materials such as air, water, acid, or a reaction within the substances itself.– Example: Iron and water rust

Energy Transfer

• The most common form of energy change involves heat.

• Heat is the energy transferred as a result of a temperature difference and is represented by the letter, q.

• Two ways that heat can be transferred:1. Contact

- Energy will transfer from matter with a higher temp to an object with a lower temp until the objects are equal in temp

2. Work- Surroundings can do work on a system

Quantitative measurements of energy changes are expressed in

joules, J.Calories are used to measure

energy changes, too.

1 calorie = 4.184 joules

Energy and Chemical Changes

• Chemical changes are always accompanied by a change in energy.

• Two types of reactions:1. Endothermic – energy is absorbed

These reactions get cold because they release no heat.

2. Exothermic – energy is releasedThese reactions get hot because they are giving off

energy.

Activation Energy

• Both of these reactions require a certain amount of energy to get started called activation energy.

• Example

Striking a match – friction is the activation energy, causing an exothermic reaction.

Measuring Energy Changes

• A calorimeter is a device used to measure the energy given off or absorbed during a chemical/physical change.

• To change the temp of a substance, heat must be added or removed.

• Some substances require little heat, while others require a lot for the same temp change

• Example– 1 gram of liquid water needs 4.184 J of heat

to raise its temp 1˚C– 1 gram of aluminum needs only 0.902 J to

raise its temp 1˚C

Specific Heat

• The heat needed to raise 1 gram of a substance by one degree Celsius is called its specific heat (Cp).

• Every substance has its own Cp

– Example

The heat required to raise the temp of 1g of water 1˚C is 4.184 J. The Cp of water is 4.184 J/g·C˚ (joule per gram Celsius degree).

Specific Heats (con’t)

• Specific heats can be used to find the change in temp of a specific mass of a substance.

• The Law of Conservation of Energy states that energy is always conserved.

• So, heat lost by one quantity of matter is gained by another through a energy transfer.

q = m(ΔT)(Cp)

heat gained/lost = mass in grams x change in temp x specific heat

ΔT = change in temperature

- Tfinal – Tinitial when heat is gained

- Tinitial – Tfinal when heat is lost

Problem

How much heat is lost when a solid aluminum ingot with a mass of 4110 g cools from 660.0˚C to 25˚C?

Given:

m = 4110 g

ΔT = Tinitial – Tfinal = 660.0˚C - 25˚C = 635˚C

Cp = 0.903 J/g·C˚

Unknown – q = ?

Equation

q = m(ΔT)(Cp)

q = (4110 g)(635˚C)(0.903 J)

g·C˚

q = 2,400,000J

Suppose a piece of iron with a mass of 21.5g at a temp of 100.0˚C is dropped into an insulated container of water. The mass of the water is 132g and its temp before adding the iron is 20.0˚C. What will be the final temp of the system?

Given:

miron = 21.5g Tinitial = 100.0 ˚C

mwater = 132g Tfinal = 20.0˚C

Unknown: Tfinal = ?

Equation: q = m(ΔT)(Cp)

Step 1

• Heat lost by the iron

q = m(ΔT)(Cp)

q = (21.5g)(100.0˚C - 20.0˚C)(0.449J)

g·C˚

Step 2

• Heat gained by water

q = m(ΔT)(Cp)

q = (132g)(Tf - 20.0˚C)(4.184J)

g·C˚

Step 3

• Heat gained must equal heat lost

(132g)(Tf - 20.0˚C)(4.184J) = (21.5g)(100.0˚C - 20.0˚C)(0.449J)

g·C g·C˚