Introduction to Matter and its Properties: A Comprehensive Guide for High School Chemistry, Schemes and Mind Maps of Chemistry

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GENERAL CHEMISTRY 1 REVIEWER

(FIRST QUARTER)

Prepared by: Engr. Niui Pye T. Nagar

TOPIC #1: MATTER: AN

INTRODUCTION AND ITS PROPERTIES

Matter is anything that occupies space and

has mass and volume. Matter occurs

commonly in three physical forms called

states : SOLID, LIQUID, and GAS.

A solid has a fixed shape that does not conform

to the container shape. Solids are not defined by

rigidity or hardness: solid iron is rigid, but

solid lead is flexible and solid wax is soft.

A liquid conforms to the container shape but fills

the container only to the extent of the liquid’s

volume; thus, a liquid forms a surface.

A gas conforms to the container shape also, but

it fills the entire container, and thus, does not

form a surface.

The particles in the solid lie next to each

other in a regular, three-dimensional array

with a definite pattern. Particles in the liquid

also lie close together but move randomly

around one another. Particles in the gas have

great distances between them as they move

randomly throughout the container.

Properties of Matter

A substance is a type of matter that has a

defined, fixed composition. The properties of

matter can be categorized as physical or

chemical.

A substance shows physical properties

when it shows by itself, without changing

into or interacting with another substance.

Some physical properties are the following:

Chemical properties are present when

substances change into or interact with

another substance/s. Some of chemical

properties are the following:

A substance undergoes physical change if a

substance changes its physical appearance

but not its composition. A physical change

results in different physical properties.

For example, when ice melts, several

physical properties change, such as

hardness, density, and ability to flow. But the

composition of the sample does not change,

it is still water.

Other examples of physical change include:

• Crushing a can
• Melting an ice cube
• Chopping wood
• Sublimation of dry ice

A substance undergoes chemical change ,

also called a chemical reaction , occurs when

a substance is converted into a different

substance.

The picture shows the chemical change

(reaction) that occurs when you pass an

electric current through water: the water

decomposes (breaks down) into two other

substances, hydrogen, and oxygen, each

with physical and chemical properties

different from each other and from water.

Other examples of chemical change include:

• Evolving bubbles or releasing gas
• Absorbing or releasing heat
• Changing color
  • Formation of a new chemical species

All measurable properties of matter fall into

two categories: extensive properties and

intensive properties.

Extensive property depends on how much

matter is being considered. Some extensive

properties are:

• Mass
• Length
• Volume

Intensive property does not depend on the

amount of matter being considered. Some

intensive properties are:

• Temperature
• Melting point
• Boiling point

TOPIC #2: FORMS OF MATTER

Pure Substance

A pure substance is matter that has distinct

properties and a composition that does not

vary from sample to sample.

Elements are substances that cannot be

decomposed into simpler substances, while

compounds are composed of two or more

elements, and they contain two or more

atoms.

From the left: A, Most elements consist of a

large collection of identical atoms. B, Some

elements occur as molecules. C, A molecule

of a compound consists of characteristic

numbers of atoms of two or more elements

chemically bound together. A molecule is an

independent structure consisting of two or

more atoms chemically bound together.

Mixture

A mixture is a combination of two or more

substances in which the substances retain

their distinct identities.

Homogenous mixtures are those with

uniform appearance, while heterogenous

mixtures are those without uniform

appearance.

(a) Many common materials, including

rocks, are heterogeneous mixtures.

This photograph of granite shows a

heterogeneous mixture of silicon

dioxide and other metal oxides.

(b) Homogeneous mixtures are called

solutions. Many substances,

including the blue solid shown here

[copper(II) sulfate], dissolve in water

to form solutions.

Other examples of mixtures are:

• Soft drinks
• Milk
• Air
• Cement

This figure depicts a mixture that are

physically intermingled. In contrast to a

compound, the components of a mixture can

vary in their parts by mass. Because its

composition is not fixed, a mixture is not a

substance. A mixture of the two compounds

sodium chloride and water, for example, can

have many different parts by mass of salt to

water.

This figure is a representation of the forms of

matter and all the properties that you need to

consider before categorizing the matter as a

pure substance or mixture.

TOPIC #3: SIMPLE SEPARATION

TECHNIQUES

Filtration separates the components of a

mixture on the basis of differences in

particle size. It is used most often to separate

a liquid (smaller particles) from a solid

(larger particles). Filtration is a key step in

the purification of the tap water you drink.

The GIF shows simple filtration of a solid

reaction product. In vacuum filtration,

reduced pressure within the flask speeds the

flow of the liquid through the filter.

Filtration is a key step in the purification of

the tap water you drink.

Crystallization is based on differences in

solubility. The solubility of a substance is the

amount that dissolves in a fixed volume of

solvent at a given temperature. The result

shown in this GIF applies the fact that many

substances are more soluble in hot solvent

than in cold. The purified compound

crystallized as the solution was cooled. Key

substances in computer chips and other

electronic devices are purified by a type of

crystallization.

Distillation separates components through

differences in volatility, the tendency of a

substance to become a gas. Ether, for

example, is more volatile than water, which

is much more volatile than sodium chloride.

The simple distillation apparatus shown in

this GIF is used to separate components with

large differences in volatility, such as water

from dissolved ionic compounds. As the

mixture boils, the vapor is richer in the more

volatile component, which is condensed and

collected separately. Separating components

with small volatility differences requires

many vaporization-condensation steps.

Extraction is also based on differences in

solubility. In a typical procedure, a natural

(often plant or animal) material is ground in

a blender with a solvent that extracts

(dissolves) soluble compound(s) embedded

in insoluble material. This extract is

separated further by the addition of a second

solvent that does not dissolve in the first.

After shaking in a separatory funnel, some

components are extracted into the new

solvent.

Chromatography is a third technique based

on differences in solubility. The mixture is

dissolved in a gas or liquid called the mobile

phase, and the components are separated as

this phase moves over a solid (or viscous

liquid) surface called the stationary phase. A

component with low solubility in the

stationary phase spends less time there, thus

moving faster than a component that is

highly soluble in that phase.

TOPIC #4: STRUCTURE OF ATOM

An atom is an electrically neutral, spherical

entity composed of a positively charged

central nucleus surrounded by one or more

negatively charged electrons. An atomic

nucleus consists of protons and neutrons.

The proton has a positive charge , and the

neutron has no charge ; thus, the positive

charge of the nucleus results from its

protons. The magnitude of charge possessed

by a proton is equal to that of an electron, but

the signs of the charges are opposite. An

atom is neutral because the number of

protons in the nucleus equals the number of

electrons surrounding the nucleus. An

atom’s diameter (~ 10

− 10

m) is about 100,

times the diameter of its nucleus (~ 10

− 15

m).

TOPIC #5: ATOMIC NUMBER, MASS

NUMBER, AND ATOMIC SYMBOL

The atomic number (Z) of an element equals

the number of protons in the nucleus of each

of its atoms. All atoms of a particular element

have the same atomic number, and each

element has a different atomic number from

that of any other element.

Name

(Symbol)

Relative

Charge

Absolute Mass

(g)

Location

Proton

(𝑝

)

1+ 1. 67262 𝑥 10

− 24

Nucleus

Neutron

(𝑛

0

)

0 1. 67493 𝑥 10

− 24

Nucleus

Electron

(𝑒

−

)

1 - 9. 10939 𝑥 10

− 28

Outside

nucleus

The total number of protons and neutrons in

the nucleus of an atom is its mass number

(A).

The nuclear mass number and charge are

often written with the atomic symbol (or

element symbol).

Since the mass number is the sum of protons

and neutrons, the number of neutrons (N)

equals the mass number minus the atomic

number:

Number of neutrons

= mass number − atomic number or N

= A − Z

For example, a chlorine atom, which is

symbolized as Cl 17

35

has A = 35, Z = 17, so

Number of neutrons = A − Z = 35 − 17 = 18

All atoms of an element are identical in

atomic number but not in mass number.

Isotopes of an element are atoms that have

different numbers of neutrons and

therefore have different mass numbers.

For example, there are three isotopes of

hydrogen. One, simple known as hydrogen ,

has one proton and no neutrons. The

deuterium isotope has one proton and one

neutron, and tritium has one proton and two

neutrons.

Example:

Silicon (Si) is essential to the computer

industry as a major component of

semiconductor chips. It has three naturally

occurring isotopes: Si

⬚

28

, Si

⬚

29

, and Si

⬚

30

.

Determine the numbers of protons,

neutrons, and electrons in each silicon

isotope.

TOPIC #6: NAMING OF COMPOUNDS

Ionic bonding generally results from the

interaction of metals on the left side of the

periodic table with nonmetals on the right

side (excluding the noble gases or Group

8A).

For example, the formation of Na

from Na

and Cl

−

from Cl

2

indicates that an electron

has been lost by a sodium atom and gained

by a chlorine atom—we say there has been

an electron transfer from the Na atom to the

Cl atom.

Each metal atom loses a certain number of its

electrons and becomes a cation , a positively

charged ion. The nonmetal atoms gain the

electrons lost by the metal atoms and become

anions , negatively charged ions. A cation or

anion derived from a single atom is called a

monatomic ion.

Ionic compounds have zero net charge, so

the positive charges of the cations must

balance the negative charges of the anions.

For example, calcium bromide is composed

of Ca

2 +

ions and Br

−

ions; therefore, two Br

−

balance each Ca

2 +

. The formula is CaBr

2

, not

Ca 2

Br. In this and all other formulas,

• The subscript refers to the element

preceding it.

• The subscript 1 is understood from

the presence of the element symbol

alone (that is, we do not write

Ca

1

Br

2

).

• The charge (without the sign) of one

ion becomes the subscript of the

other.

Reduce the subscripts to the smallest whole

numbers that retain the ratio of ions or

reduce the chemical formula to its empirical

form.

For example, the chemical formula Ca 2

O

2

formed from the ions Ca

2 +

and O

2 −

should be

reduced to CaO.

Covalent bonding is a chemical bonding

formed by sharing electrons.

The hydrogen molecule, H 2

, provides the

simplest example of a covalent bond. When

two hydrogen atoms are close to each other,

the two positively charged nuclei repel each

other, the two negatively charged electrons

repel each other, and the nuclei and electrons

attract each other, as shown in. Because the

molecule is stable, we know that the

attractive forces must overcome the

repulsive ones.

Many ionic compounds contain polyatomic

ions , which consist of two or more atoms

bonded covalently and have a net positive

or negative charge.

For example, the ionic compound calcium

carbonate is an array of polyatomic

carbonate anions and monatomic calcium

cations attracted to each other. The carbonate

ion consists of a carbon atom covalently

bonded to three oxygen atoms, and two

additional electrons give the ion its - 2 charge.

In many reactions, a polyatomic ion stays

together as a unit.

Rules in Naming Compounds

Ionic Compound

All ionic compound names give the positive

ion (cation) first and negative ion (anion)

second.

Compounds Formed from Monoatomic

Ions

• The name of the cation is the same as

the name of the metal. Many metal

names end in – ium.

• The name of the anion takes the root

of the nonmetal name and adds the

suffix – ide.

Compounds with Metals That Can Form

More Than One Ion

Many metals, particularly the transition

elements (B groups), can form more than one

ion, each with its own particular charge. For

example, iron can form Fe

2 +

and Fe

3 +

ions.

In common names, the Latin root of the

metal is followed by either of two suffixes:

• The suffix - ous for the ion with the

lower charge

• The suffix - ic for the ion with the

higher charge

• Names of compounds containing

these elements include a Roman

numeral within parentheses

immediately after the metal ion’s

name to indicate its ionic charge.

Compounds Formed from Polyatomic Ions

• Remember that the polyatomic ion

stays together as a charged unit. For

example, the formula for potassium

nitrate is KNO

3

.

• When two or more of the same

polyatomic ions are present in the

formula unit, that ion appears in

parentheses with the subscript

written outside. For example,

calcium nitrate, its formula is

Ca

(

NO

3

)

2

.

Families of Oxoanions

Most polyatomic ions are oxoanions , those

in which an element, usually a nonmetal, is

bonded to one or more oxygen atoms.

With two oxoanions in the family:

• The ion with more O atoms takes the

nonmetal root and the suffix – ate.

• The ion with fewer O atoms takes the

nonmetal root and the suffix - ite.

Example: SO

4

2 −

is the sulf ate ion, and SO

3

2 −

is the sulf ite ion.

With four oxoanions in the family (usually a

halogen bonded to O):

• The ion with most O atoms has the

prefix per - , the nonmetal root, and

the suffix - ate.

• The ion with one fewer O atom has

just the root and the suffix - ate.

• The ion with two fewer O atoms has

just the root and the suffix - ite.

• The ion with least (three fewer) O

atoms has the prefix hypo - , the root,

and the suffix - ite.

Example: For the four chlorine oxoanions,

ClO

4

−

is the per chlor ate ion, ClO

3

−

is the

chlor ate ion, ClO

2

−

is the chlor ite ion, and

ClO

−

is the hypo chlor ite ion.

Acid Names from Anion Names

The two common types of acids are binary

acids and oxoacids:

• Binary acid solutions form when

certain gaseous compounds dissolve

in water. For example, hydrochloric

acid:

• Oxoacid names are similar to those of

the oxoanions, except for two suffix

changes:

• ate in the anion becomes – ic in the acid.
• ite in the anion becomes – ous in the acid.

The oxoanion prefixes hypo - and per - are

kept. For example,

Covalent Compounds

Binary covalent compounds are formed by

the combination of two elements, usually

nonmetals. Some are so familiar, such as

ammonia (NH 3

), methane (CH

4

), and water

(H

2

O), we use their common names, but most

are named in a systematic way:

1. The element with the lower group

number in the periodic table is the

first word in the name; the element

with the higher group number is the

second word. (Exception: When the

compound contains oxygen and any

of the halogens: chlorine, bromine,

and iodine, the halogen is named

first.)

2. If both elements are in the same

group, the one with the higher period

number is named first.

3. The second element is named with its

root and the suffix - ide.

4. Covalent compounds have Greek

numerical prefixes to indicate the

number of atoms of each element in

the compound. The first word has a

prefix only when more than one atom

of the element is present; the second

word usually has a numerical prefix.

TOPIC #7: CALCULATING MOLECULAR

MASSES FROM CHEMICAL

COMPOUNDS

Using the periodic table and the formula of a

compound to see the number of atoms of

each element, we calculate the molecular

mass (also called molecular weight ) of a

formula unit of the compound as the sum of

the atomic masses:

Example 1:

The molecular mass of a water molecule

(using atomic masses to four significant

figures from the periodic table) is:

Molecular mass of H

2

O

= ( 2 x atomic mass of H)

• ( 1 x atomic mass of O)

Molecular mass of H

2

O

= ( 2 x 1. 008 amu)

• ( 1 x 16 amu)

Molecular mass of H

2

O = 18. 02 amu

Example 2:

For barium nitrate, Ba(NO 3

)

2

, the number of

atoms of each element inside the parentheses

is multiplied by the subscript outside the

parentheses.

Molecular mass of Ba(NO

3

)

2

= ( 1 x atomic mass of Ba)

• ( 2 x atomic mass of N)

• ( 6 x atomic mass of O)

Molecular mass of Ba(NO

3

)

2

=

(

1 x 137. 3 amu

)

+

(

2 x 14. 01 amu

)

• ( 6 x 16 amu)

Molecular mass of Ba

(

NO

3

)

2

= 261. 3 amu

TOPIC #8: THE MOLE

The mole (abbreviated mol) is the SI unit for

amount of substance. It is defined as the

amount of a substance that contains the

same number of entities as there are atoms

in exactly 12 g of carbon- 12. This number is

called Avogadro’s number, in honor of the

19th-century Italian physicist Amedeo

Avogadro:

One mole (1 mol) contains 𝟔. 𝟎𝟐𝟐 𝐱 𝟏𝟎

𝟐𝟑

entities.

Thus,

The central relationship between the mass of

one atom and the mass of 1 mole of those

atoms is that the atomic mass of an element

expressed in amu is numerically the same as

the mass of 1 mole of atoms of the element

expressed in grams. Thus,

A similar relationship holds for compounds:

the molecular mass of a compound

expressed in amu is numerically the same as

the mass of 1 mole of the compound

expressed in grams. Thus, for example,

The molar mass ( M ) of a substance is the

mass per mole of its entities (atoms,

molecules, or formula units). Thus, molar

mass has units of grams per mole (g/mol).

Interconverting Moles, Mass, and Chemical

Entities

The molar mass, which expresses the

equivalent relationship between 1 mole of a

substance and its mass in grams, can be used

as a conversion factor. We multiply by the

molar mass of an element or compound (M,

in g/mol) to convert a given amount (in

moles) to mass (in grams):

𝐌𝐚𝐬𝐬

(

𝐠

)

= 𝐧𝐨. 𝐨𝐟 𝐦𝐨𝐥𝐞𝐬 𝐱

𝐧𝐨. 𝐨𝐟 𝐠𝐫𝐚𝐦𝐬

𝟏 𝐦𝐨𝐥

Or, we divide by the molar mass (multiply

by 1/M) to convert a given mass (in grams) to

amount (in moles):

𝐍𝐨. 𝐨𝐟 𝐦𝐨𝐥𝐞𝐬 = 𝐦𝐚𝐬𝐬 (𝐠) 𝐱

𝟏 𝐦𝐨𝐥

𝐧𝐨. 𝐨𝐟 𝐠𝐫𝐚𝐦𝐬

In a similar way, we use Avogadro’s

number, which expresses the equivalent

relationship between 1 mole of a substance

and the number of entities it contains, as a

conversion factor. We multiply by

Avogadro’s number to convert amount of

substance (in moles) to the number of entities

(atoms, molecules, or formula units):

𝐍𝐨. 𝐨𝐟 𝐞𝐧𝐭𝐢𝐭𝐢𝐞𝐬

= 𝐧𝐨. 𝐨𝐟 𝐦𝐨𝐥𝐞𝐬 𝐱

𝟔. 𝟎𝟐𝟐 𝐱 𝟏𝟎

𝟐𝟑

𝐞𝐧𝐭𝐢𝐭𝐢𝐞𝐬

𝟏 𝐦𝐨𝐥

Or, we divide by Avogadro’s number to do

the reverse:

𝐍𝐨. 𝐨𝐟 𝐦𝐨𝐥𝐞𝐬

= 𝐧𝐨. 𝐨𝐟 𝐞𝐧𝐭𝐢𝐭𝐢𝐞𝐬 𝐱

𝟏 𝐦𝐨𝐥

𝟔. 𝟎𝟐𝟐 𝐱 𝟏𝟎

𝟐𝟑

𝐞𝐧𝐭𝐢𝐭𝐢𝐞𝐬

Calculating the Mass in a Given Number of

Moles of an Element

Example 1: Silver (Ag) is used in jewelry and

tableware but no longer in U.S. coins. How

many grams of Ag are in 0.0342 mol of Ag?

Mass (g) = no. of moles x

no. of grams

1 mol

Mass of Ag = 0. 0342 mol Ag x

107. 9 g Ag

1 mol Ag

Mass of Ag = 𝟑. 𝟔𝟗 𝐠

Calculating the Number of Atoms in a

Given Mass of an Element

Example 2: Iron (Fe), the main component of

steel, is the most important metal in

industrial society. How many Fe atoms are in

95.8 g of Fe?

Step 1: Converting from grams of Fe to

moles:

Moles of Fe = 95. 8 g Fe x

1 mol Fe

55. 85 g Fe

Moles of Fe = 1. 72 mol

Step 2: Converting from moles of Fe to

number of atoms:

No. of Fe atoms

= 1. 72 mol Fe x

6. 022 x 10

23

atoms Fe

1 mol Fe

No. of Fe atoms = 𝟏. 𝟎𝟒 𝐱 𝟏𝟎

𝟐𝟒

𝐚𝐭𝐨𝐦𝐬 𝐅𝐞

Calculating the Moles and Number of

Formula Units in a Given Mass of a

Compound

Example 3: Ammonium carbonate

((NH

4

)

2

CO

3

) is a white solid that decomposes

with warming. Among its many uses, it is a

component of baking powder, fire

extinguishers, and smelling salts. How many

formula units are in 41.6 g of ammonium

carbonate?

Step 1: Calculating the molar mass of

ammonium carbonate,

Molecular mass of (NH

4

)

2

CO

3

= ( 2 x M of N) + ( 8 x M of H)

+

(

1 x M of C

)

• ( 3 x M of O)

Molecular mass of

(

NH

4

)

2

CO

3

=

(

2 x 14. 01 g/mol of N

)

+

(

8 x 1. 008 g/mol of H

)

+

(

1 x 12. 01 g/mol of C

)

• ( 3 x 16 g/mol of O)

Molecular mass of (NH

4

)

2

CO

3

= 96. 09 g ⁄mol

Step 2: Converting from grams to moles:

Moles of (NH

4

)

2

CO

3

= mass of (NH

4

)

2

CO

3

x

1 mol

(

NH

4

)

2

CO

3

96. 09 g

(

NH

4

)

2

CO

3

Moles of (NH

4

)

2

CO

3

= 41. 6 g (NH

4

)

2

CO

3

x

1 mol

(

NH

4

)

2

CO

3

96. 09 g (NH

4

)

2

CO

3

Moles of (NH

4

)

2

CO

3

= 0. 433 mol

Step 3: Converting from moles to formula

units:

Formula units of (NH 4

)

2

CO

3

= 0. 433 g

( NH

4

)

2

CO

3

x

6. 022 x 10

23

formula units (NH

4

)

2

CO

3

1 g

( NH

4

)

2

CO

3

= 𝟐. 𝟔𝟏 𝐱 𝟏𝟎

𝟐𝟑

𝐟𝐨𝐫𝐦𝐮𝐥𝐚 𝐮𝐧𝐢𝐭𝐬

(

𝐍𝐇

𝟒

)

𝟐

𝐂𝐎

𝟑

Calculating Mass Percents and Masses of

Elements in a Sample of a Compound

Example 4: In mammals, lactose (milk sugar)

is metabolized to glucose (C 6

H

12

O

6

), the key

nutrient for generating chemical potential

energy. What is the mass percent of each

element in glucose?

Step 1: Calculate the molar mass of C 6

H

12

O

6

:

Molecular mass of C

6

H

12

O

6

=

(

6 x M of C

)

+

(

12 x M of H

)

• ( 6 x M of O)

Molecular mass of C

6

H

12

O

6

= ( 6 x 12. 01 g/mol of C)

• ( 12 x 1. 008 g/mol of H)

• ( 6 x 16 g/mol of O)

Molecular mass of C

6

H

12

O

6

= 180. 16 g ⁄mol

Step 2: Converting moles of each element to

grams:

Mass of C = 6 mol C x

12. 01 g C

1 mol C

= 72. 06 g C

Mass of H = 12 mol H x

1. 008 g H

1 mol H

= 12. 096 g H

Mass of O = 6 mol O x

16 g O

1 mol O

= 96 g O

Step 3: Determining the mass fraction and

mass percentage of each element:

Mass fraction of C =

total mass of C

mass of 1 mol of glucose

=

72. 06 g C

73. 16 g C

6

H

12

O

6

Mass fraction of C = 0. 4000

Mass % of C = mass fraction of C x 100

= 0. 4000 x 100

= 𝟒𝟎. 𝟎𝟎% 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐂

Mass fraction of H =

total mass of H

mass of 1 mol of glucose

=

12. 096 g H

13. 16 g C

6

H

12

O

6

Mass fraction of H = 0. 06714

Mass % of H = mass fraction of H x 100

= 0. 06714 x 100

= 𝟔. 𝟕𝟏𝟒% 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐇

Mass fraction of O =

total mass of O

mass of 1 mol of glucose

=

96 g O

180. 16 g C

6

H

12

O

6

Mass fraction of C = 0. 5329

Mass % of O = mass fraction of O x 100

= 0. 5329 x 100

= 𝟓𝟑. 𝟐𝟗% 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐎

TOPIC # 9 : DETERMINING THE

FORMULA OF AN UNKNOWN

COMPOUND

An analytical chemist investigating a

compound decomposes it into simpler

substances, finds the mass of each

component element, converts these masses

to numbers of moles, and then arithmetically

converts the moles to whole-number

(integer) subscripts. This procedure yields

the empirical formula , the simplest whole-

number ratio of moles of each element in the

compound.

For example,

Analysis of an unknown compound shows

that the sample contains 0.21 mol of zinc, 0.

mol of phosphorus, and 0.56 mol of oxygen.

a) Write the preliminary formula that

contains the fractional subscripts.

Zn

1. 21

P

1. 14

O

1. 56

b) Divide each subscript by the smallest

subscript:

Zn

1. 21

2. 14

P

1. 14

2. 14

O

1. 56

2. 14

= Zn

1. 5

P

1. 0

O

4. 0

c) If any of the subscripts is still not an

integer, multiply through by the

smallest integer that will turn all

subscripts into integers. Here, we

multiply by 2, the smallest integer

that will make 1.5 (the subscript for

Zn) into an integer:

Zn

( 1 .5x2)

P

( 1 .0x2)

O

( 4 .0x2)

→ 𝐙𝐧

𝟑

𝐏

𝟐

𝐎

𝟖

Determining an Empirical Formula from

Masses of Elements

Example 5 : Elemental analysis of a sample of

an ionic compound showed 2.82 g of Na, 4.

g of Cl, and 7.83 g of O. What is the empirical

formula and name of the compound?

Step 1: Determine the moles of each element:

Moles of Na = 2. 82 g Na x

1 mol Na

22. 99 g Na

= 0. 123 mol Na

Moles of Cl = 4. 35 g Cl x

1 mol Cl

35. 45 g Cl

= 0. 123 mol Na

Moles of O = 7. 83 g O x

1 mol O

16 g O

= 0. 489 mol O

Step 2: Construct the preliminary formula:

Na

1. 123

Cl

1. 123

O

1. 489

Step 3: Divide each subscript by the smallest

subscript to get the empirical formula:

Na

1. 123

2. 123

Cl

1. 123

2. 123

O

1. 489

2. 123

= 𝐍𝐚

𝟏

𝐂𝐥

𝟏

𝐎

𝟒

𝐨𝐫 𝐍𝐚𝐂𝐥𝐎

If we know the molar mass of a compound,

we can use the empirical formula to obtain

the molecular formula , the actual number of

moles of each element in 1 mol of

compound.

Whole − number multiple

=

molar mass (g mol

⁄

)

empirical formula mass (g ⁄mol )

Determining the Molecular Formula from

Masses of Elements

Example 6 : During excessive physical

activity, lactic acid ( M = 90.08 g/mol) forms

in muscle tissue and is responsible for

muscle soreness. Elemental analysis shows

that this compound contains 40.0 mass % C,

6.71 mass % H, and 53.3 mass % O.

a) Determine the empirical formula of

lactic acid.

b) Determine the molecular formula.

Solution:

a) Determining the empirical formula:

Step 1: Express the mass % as grams, so we

assume 100 g of lactic acid and converting

from grams to moles of each element:

Moles of C = 40. 00 g C x

1 mol Na

12. 01 g C

= 3. 33 mol C

Moles of H = 6. 71 g H x

1 mol H

1. 008 g H

= 6. 66 mol H

Moles of O = 53. 3 g O x

1 mol O

16 g O

= 3. 33 mol O

Step 2: Construct the preliminary formula:

C

3. 33

H

6. 66

O

3. 33

Step 3: Divide each subscript by the smallest

subscript to get the empirical formula:

C 3. 33

3. 33

H 6. 66

3. 33

O 3. 33

3. 33

= 𝐂

𝟏

𝐇

𝟐

𝐎

𝟏

𝐨𝐫 𝐂𝐇

𝟐

𝐎

b) Determining the molecular formula:

Step 1: Determine the molecular weight of

the empirical formula:

Molecular mass of CH

2

O

=

(

1 x M of C

)

+

(

2 x M of H

)

+

(

1 x M of O

)

Molecular mass of CH

2

O

= ( 1 x 12. 01 g/mol of C)

• ( 2 x 1. 008 g/mol of H)

• ( 1 x 16 g/mol of O)

Molecular mass of CH

2

O = 30. 03 g ⁄mol

Step 2: Get the whole-number multiple and

multiply it to the subscripts of the empirical

formula to determine the molecular formula:

Whole − number multiple

=

molar mass (g ⁄mol )

empirical formula mass (g ⁄mol )

Whole − number multiple =

90. 08 g mol

⁄

30. 03 g mol

⁄

Whole − number multiple = 3

C

(1x3)

H

(2x3)

O

(1x3)

→ 𝐂

𝟑

𝐇

𝟔

𝐎

𝟑

TOPIC #10: CHEMICAL REACTIONS

AND CHEMICAL EQUATIONS

A chemical equation uses formulas to

express the identities and quantities of

substances involved in a chemical change.

For an equation to depict these amounts

accurately, it must be BALANCED ; that is,

the same number of each atom must appear

on both sides of the equation.

Example 7: Consider the chemical change

that occurs in an old-fashioned photographic

flashbulb: a magnesium strip burns in

oxygen gas to yield powdery magnesium

oxide.

The following are the steps in constructing a

chemical equation:

Step 1: TRANSLATING THE

STATEMENT. We first translate the

chemical statement into a “skeleton”

equation: chemical formulas arranged in an

equation format.

Step 2: BALANCING THE ATOMS. The

next step involved shifting our attention

back and forth from right to left to match the

number of each type of atom on each side. At

the end of this step, each blank will contain a

BALANCING (STOICHIOMETRIC)

COEFFICIENT , a numerical multiplier of all

the atoms in the formula that follows it.

Step 3: ADJUSTING THE COEFFICIENTS.

There are several conventions about the final

form of the coefficients:

• In most cases, the smallest whole-

number coefficients are preferred.

• We use the coefficient 1 to remind us

to balance each substance. In the final

form, a coefficient of 1 is implied just

by the presence of formula of the

substance.

Step 4: CHECKING. After balancing and

adjusting the coefficients, always check that

the equation is balanced.

Step 5: SPECIFYING THE STATES OF

MATTER. The final equation also indicates

the physical state of each substance or

whether it is dissolved in water. Symbols

indicating the physical state of each reactant

and product are often shown in chemical

equations.

The final chemical equation is:

2 Mg

(s)

+ O

2 (g)

→ 2 MgO

(s)

Keep in mind these other key points about

the balancing process:

• A balanced equation remains

balanced even if you multiply all the

coefficients by the same number.

Always simplify the final chemical

equation.

Four Types of Chemical Reactions

• Synthesis/Combination
• Decomposition
• Single Replacement/Displacement
• Double Replacement/Displacement

TOPIC #1 1 : CALCULATING THE

AMOUNTS OF REACTANT AND

PRODUCT

Steps:

1. Write a balanced equation for the

reaction.

2. Convert the given mass of the first

substance to moles.

3. Use the appropriate molar ratio from

the balanced equation to calculate

the amount (mol) of the second

substance.

4. Convert the amount of the second

substance to the desired mass.

State Symbol

Solid ( s )

Liquid ( l )

Gas ( g )

Dissolved in water; in

aqueous solution;

solution

( aq )

Example 8: In a lifetime, the average

American uses 1750 lb (794 kg) of copper in

coins, plumbing, and wiring. Copper is

obtained from sulfide ores, such as

chalcocite, or copper(I) sulfide, by a

multistep process. After an initial grinding,

the first step is to “roast” the ore (heat it

strongly with oxygen gas) to form

powdered copper(I) oxide and gaseous

sulfur dioxide.

Balanced Chemical Equation

2 Cu

2

S

(s)

+ 3 O

2 (g)

→ 2 Cu

2

O

(s)

+ 2 SO

2 (g)

a. How many moles of oxygen are

required to roast 10.0 mol of

copper(I) sulfide?

Plan : We are given the moles of Cu

2

S

and need to find the moles of O

2

. The

balanced equation shows that 3

moles of O

2

is needed for every 2

moles of Cu

2

S consumed, sot the

conversion factor is “3 moles O

2

/

moles Cu

2

S”.

Moles of O

2

= 10. 0 moles Cu

2

S x

3 moles O

2

2 moles Cu

2

S

= 𝟏𝟓. 𝟎 𝐦𝐨𝐥𝐞𝐬 𝐎

𝟐

b. How many grams of sulfur dioxide

are formed when 10.0 mol of

copper(I) sulfide is roasted?

Plan : We need the grams of product

(SO

2

) that form from the given moles

of reactant (Cu

2

S). We first find the

moles of SO

2

using the molar ratio

from the balanced equation (2 moles

SO

2

/2 moles Cu

2

S) and then multiply

by its molar mass of SO

2

to find

grams of SO

2

.

Mass

(

g

)

of SO

2

= 10. 0 moles Cu

2

S x

2 moles SO

2

2 moles Cu

2

S

x

64. 07 g SO

2

1 mole SO

2

= 𝟔𝟒𝟏 𝐠 𝐒𝐎

𝟐

Practice Problems

1. Potassium nitrate decomposes on

heating, producing potassium oxide

and gaseous nitrogen and oxygen. To

produce 56.6 g of oxygen, how many

(a) moles of potassium nitrate must

be heated? (b) grams of potassium

nitrate must be heated?

Ans.: (a) 1.42 moles potassium nitrate ; (b)

143.56 g potassium nitrate

2. Phosphine, PH

3 (g)

, combusts in

oxygen gas to form water vapor and

solid tetraphosphorus decaoxide.

How many (a) moles of water vapor

will be produced from the

combustion of 36.7 moles phosphine?

(b) grams of tetraphosphorus

decaoxide produced from 42.3 g of

oxygen gas?

Ans.: (a) 55.05 moles water vapor ; (b) 46.91 g

tetraphosphorus decaoxide

Limiting and Excess Reactant

An analogous situation occurs in chemical

reactions when one reactant is used up

before the others. The reaction stops as soon

as any reactant is totally consumed, leaving

the excess reactants as leftovers.

The reactant that is completely consumed in

a reaction is called the limiting reactant

because it determines, or limits, the amount

of product formed. The other reactant is

called the excess reactant.

Example 9: A fuel mixture used in the early

days of rocketry is composed of two liquids,

hydrazine (N 2

H

4

) and dinitrogen tetraoxide,

which ignite on contact to form nitrogen gas

and water vapor. How many grams of

nitrogen gas form when 1. 00 x 10

2

g of N

2

H

4

and 2. 00 x 10

2

g of dinitrogen tetraoxide are

mixed?

The balanced equation is:

2 N

2

H

4 (l)

+ N

2

O

4 (l)

→ 3 N

2 (g)

+ 4 H

2

O

(g)

Here’s a roadmap showing the steps in

solving the problem:

Finding the moles of N 2

from the moles of

N

2

H

4

:

Moles of N

2

H

4

= 1. 00 x 10

2

g N

2

H

4

x

1 mole N

2

H

4

32. 05 g N

2

H

4

= 3. 12 moles N

2

H

4

Moles of N

2

= 3. 12 moles N

2

H

4

x

3 moles N

2

2 moles N

2

H

4

= 4. 68 moles N

2

Finding the moles of N

2

from the moles of

N

2

O

4

:

Moles of N

2

O

4

= 2. 00 x 10

2

g N

2

O

4

x

1 mole N

2

O

4

92. 02 g N

2

O

4

= 2. 17 moles N

2

O

4

Moles of N

2

= 2. 17 moles N

2

O

4

x

3 moles N

2

1 mole N

2

O

4

= 6. 51 moles N

2

Thus, 𝐍

𝟐

𝐇

𝟒

is the limiting reactant because

it yields fewer moles of N

2

.

Converting from moles of N

2

to grams N

2

:

Mass (g) of N

2

= 4. 68 moles N

2

x

28. 02 g N

2

1 mole N

2

= 𝟏𝟑𝟏 𝐠 𝐍

𝟐

To get the amount of reacted N

2

O

4

,

Amount of reacted N

2

O

4

= 4. 68 moles N

2

x

1 mole N

2

O

4

3 mole N

2

x

92. 02 g N

2

O

4

1 mole N

2

O

4

= 𝟏𝟒𝟑. 𝟓𝟓 𝐠 𝐍

𝟐

𝐎

𝟒

Calculating for amount of unreacted N

2

O

4

or

the amount of N

2

O

4

left after the reaction,

Amount of unreacted N

2

O

4

= 2. 00 x 10

2

g − 143. 55 g

= 𝟓𝟔. 𝟒𝟓 𝐠 𝐍

𝟐

𝐎

𝟒

Actual, Theoretical, and Percent Yield

The quantity of product calculated to form

when all of a limiting reactant is consumed is

called the theoretical yield. The amount of

product actually obtained is called the actual

yield. The percent yield relates actual and

theoretical yields:

% yield =

actual yield

theoretical yield

x 100

Example 10: Adipic acid, H 2

C

6

H

8

O

4

, used to

produce nylon, is made commercially by a

reaction between cyclohexane (C 6

H

12

) and

O

2

:

2 C

6

H

12 (l)

+ 5 O

2 (g)

→ 2 H

2

C

6

H

8

O

4 (l)

+ 2 H

2

O

(g)

a. Assume that you carry out this

reaction with 25.0 g of cyclohexane

and that cyclohexane is the limiting

reactant. What is the theoretical yield

of adipic acid?

b. If you obtain 33.5 g of adipic acid,

what is the percent yield for the

reaction?

Solution:

a. Calculating for the theoretical yield

of adipic acid:

Mass (g) of H 2

C

6

H

8

O

4

= 25. 0 g C 6

H

12

x

1 mole C

6

H

12

84. 0 g C

6

H

12

x

2 moles H

2

C

6

H

8

O

4

2 moles C

6

H

12

x

146. 0 g H

2

C

6

H

8

O

4

1 mole H

2

C

6

H

8

O

4

= 𝟒𝟑. 𝟓 𝐠 𝐇

𝟐

𝐂

𝟔

𝐇

𝟖

𝐎

𝟒

b. Calculating the percent yield,

% yield =

actual yield

theoretical yield

x 100

=

33. 5 g

34. 5 g

x 100

= 𝟕𝟕. 𝟎%

Practice Problems

1. How many grams of solid aluminum

sulfide can be prepared by the

reaction of 10.0 g of aluminum and

15.0 g of sulfur? How much of the

nonlimiting reactant is in excess?

Answers: (a) 23.4 g aluminum

sulfide ; (b) 1.6 g

2. Imagine you are working on ways to

improve the process by which iron

ore containing Fe

2

O

3

is converted

into iron:

Fe

2

O

3 (s)

+ 3 CO

(g)

→ 2 Fe

(s)

+ 3 CO

2 (g)

a. If you start with 150 g of Fe

2

O

3

as

the limiting reactant, what is the

theoretical yield of Fe?

b. If your actual yield is 87.9 g, what

is the percent yield?

Answers: (a) 105 g Fe ; (b) 83.7%

Textbook References:

Brown, T.L., LeMay, Jr., H., Bursten, B.E.,

Murphy, C.J., Woodward, P.M., & Stoltzfus,

M.W. (2015). Chemistry: The Central Science

(13th ed.). Pearson Education, Inc.

Silberberg, M.S. (2009). Chemistry: The

Molecular Nature of Matter and Change (5th

ed.). McGraw-Hill.

Image References:

https://www.readersdigest.ca/home-

garden/tips/5-things-do-chalk/

https://www.scienceabc.com/nature/why-is-

there-sand-on-beaches.html

https://www.health.harvard.edu/staying-

healthy/how-much-water-should-you-drink

https://health.clevelandclinic.org/is-salt-

bad-for-you/

https://philusa.com.ph/our-brand/gleam-

muriatic-acid/

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filter-coffee-filtering-gif- 16846839

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crystallization-gifs

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4791679/

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gif

https://www.athenology.com/post/2019/09/

03/the-investigation-of-chlorophyll-using-

thin-layer-chromatography

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content/uploads/2023/03/Screen-Shot- 2021 -

02 - 02 - at-6.23.35-AM-1024x574-1.png