Download Ch 3 - Humble ISD

Matter - Properties and Changes
Chapter 3
What is matter

Chemistry is the study of matter

Matter is anything that takes up space and has mass

Mass is the measurement of how much matter is in
something
Mass vs Weight

Mass is the same regardless of gravity


Mass is measured using a balance
Weight measures how gravity affects an object

Weight is measured using a scale
Mass vs Weight

Say you travel to the Moon, where gravity is 1/6th
that of Earth?

Which would be affected:
 Your Mass,
 Your Weight,
 Neither,
 Both?
Matter

Matter is subdivided into categories:
Matter
Mixtures
Heterogeneous
Homogeneous
Pure Substances
Element
Compound
Types of Matter

Matter = has mass and takes up space

Pure substance = uniform, unchanging composition


Example: Distilled water always has the formula H2O
Mixture = combo of 2 or more pure substances in
which each ingredient retains its individual
chemical properties

Example: Sea Water contains water, salt, magnesium,
iron, phosphates, all mixed together.
Types of Mixtures

Heterogeneous mix = individual substances remain
distinct
 Examples:
 dirty water
 tacos
 oil & vinegar
 chocolate chip cookies
Types of Mixtures

Homogeneous mix = has the same composition
throughout

Also called a Solution
Examples of solutions:
 Air,
 salt water,
 milk,
 steel
Types of Mixtures


Homogeneous mix = has the same composition
throughout
Alloys = homogeneous mix of metals
Examples:
 steel
 brass
 jewelry
Types of Pure Substances

Pure Substances are either elements or compounds

Element = cannot be further separated or broken
down by physical or chemical means
Examples:
 Na,
 S,
 Fe,
 C
Types of Pure Substances

Pure Substances are either elements or compounds

Compound = 2 or more elements chemically
combined
Examples:
 Water = H2O,
 Salt = NaCl,
 sugar = C12H22O11
Types of Pure Substances

Compounds can be broken down only by chemical
means


not by physical means
Compounds have different properties than the
elements they are made of.

Example:
 H2
 O2
 H2O
Properties of Matter

Physical property has
characteristics that can be
observed or measured without
changing the sample’s
composition.

Examples: density, color, odor,
taste, hardness, melting point,
and boiling point
Properties of Matter

Intensive property –
independent of the
amount of the substance
sample

Examples: density,
melting point, boiling
point
Properties of Matter

Extensive property –
dependent of the amount
of the substance sample

Examples: mass, length,
width, volume
Properties of Matter

Chemical property – the
ability of a substance to
combine with or change
into one or more other
substances

Examples: Iron reacts
with oxygen to form rust,
iron does not react with
nitrogen
Separation of Mixtures

Mixtures are physically (not chemically) combined so
separation uses physical processes based on physical
properties.

Separation Techniques:

The physical properties of a substance allow you to devise a
method of separating the parts of a mixture.
Separation of Mixtures

Separation Techniques:

Filtration


Chromatography
Distillation
Separation of Mixtures

Separation Techniques:
 Manual separation


Settle and Decant
Magnetic

Evaporation or crystallization
Changes in Matter

Physical change



Only physical properties are altered
No new substance formed
Structure of molecules not changed

Examples: water to ice, Al foil crumpled, wood cut
Changes in Matter

Chemical change





New substance formed w/ new
properties
Different molecules
Clues – formation of a
precipitate, new gas, etc.
a.k.a chemical reaction (chem
rxn)
Examples: Na + Cl  NaCl, H2
burned as rocket fuel,
iron nail rust
Changes in Matter

Law of Conservation of Mass
 Mass is neither created nor destroyed
 Mass reactants = Mass products
Problem
 A 10.0 g sample of Mg reacts with O2 to form 16.6g
of magnesium oxide (MgO). How many grams of O2
reacted?
Changes in Matter

Given:




Find:


Mg = 10.0g Reactant
MgO = 16.6 g Product
Mg + O2  MgO
O2 = ? g Reactant
Solve:


10.0 + x = 16.6
X = 6.6 g O2