Download Reporting Category 3: Bonding and Chemical Reactions

Alief ISD CHEMISTRY STAAR Review
Reporting Category 3: Bonding and Chemical Reactions
C.7.A
Name ionic
compounds main
group or
transition metals,
covalent
compounds,
acids and bases,
using
International
Union of Pure
and Applied
Chemistry
(IUPAC)
nomenclature
rules.
Naming Compounds
How are ionic compounds named using IUPAC rules?
Background information: An ion forms when an atom or a group of atoms loses or gains electrons. If the
ion has a positive charge, it is a cation. If it has a negative charge, it is an anion. When ionic bonds from
between cations and anions, an ionic compound is formed.
•
•
+
Metals tend to lose electrons making them a cation… ex. Na
Nonmetals tend to gain electrons making them an anion….ex. Cl
Rules for naming ionic compounds:
1.
Binary ionic compound (metal-nonmetal):
• First, name cation using its element name; for a cation with more than one possible charge
2+
3+
(like transition metal Fe or Fe ), add a Roman numeral for the cation’s charge in
parentheses
• Second, name anion with –ide ending
• EXAMPLES:
Formula
CaCl2
BaO
Fe2S3
FeS
2.
Names
calcium chloride (cation only has one charge)
barium oxide
(cation only has one charge)
iron(III) sulfide (cation has more than one charge)
iron(II) sulfide
(cation has more than one charge)
Polyatomic* ionic compound (metal-polyatomic ion):
• First, name cation using same rules as binary ionic compound; remember only the cations
with more than one charge needs a Roman numeral in parentheses.
• Second, name the polyatomic anion (NEVER change their names!)
• EXAMPLES:
Formula
Na3PO4
Al(NO2)3
Pb(CO3)2
sodium phosphate
aluminum nitrate
lead(IV) carbonate
Names
(cation only has one charge)
(cation only has one charge)
(cation has more than one charge)
*See STAAR formula chart for polyatomic ion names
How are covalent compounds named using the IUPAC rules?
Background information: Atoms in covalent compound are held together by covalent bond—bonds formed by
the sharing of electrons between atoms. A covalent bond forms between two atoms with high
electronegativity, usually nonmetals. Because their attractions for a given electron are about equally strong,
they share a pair of electrons. This sharing pulls the atoms into close contact with each other.
Rules for naming covalent compounds:
3. Binary covalent compound (nonmetal-nonmetal):
• First, name the element with the lower electronegativity (which is the first element written in
the formula) by given it its original name.
• Second, name element with higher electronegativity (which is the second element written in
the formula) with –ide ending.
• For both elements in the formula, add the correct Greek prefix(es) for the number of atoms
(mono-is only used for the higher electronegativity element which is the second element in
the formula).
• GREEK PREFIXES: (1) mono-; (2) di-; (3) tri-; (4) tetra-; (5) penta-; ( 6) hexa-; (7) hepta-; (8)
octa-; (9) nona-; (10) deca• EXAMPLES:
Covalent
compound
CO
CO2
NO2
N2O
N2O3
BF3
Names
carbon monoxide
carbon dioxide
nitrogen dioxide
dinitrogen oxide
dinitrogen trioxide
boron trifluoride
How are acids and bases named using the IUPAC rules?
+
Background information: Acids are compounds in which H is bonded to an anion (nonmetal) in aqueous
solution. Acids are named based on the suffix of the anion.
1-
Bases are compounds that produce hydroxide ions [OH ] when dissolved in water. Bases are named in the
same way as other ionic compounds.
Rules for naming acids:
4. Naming acids that do not contain oxygen (most are binary acids)
• Use the prefix hydro- followed by the nonmetal’s stem name with –ic as the suffix (ending) and
then the word “acid”.
• So, it would look like this: hydro-stem name of nonmetal-ic acid.
• EXAMPLES:
Acid Formulas
without oxygen
HF
HCl
H2S
HCN
Names
hydrofluoric acid
hydrochloric acid
hydrosulfuric acid
hydrocyanic acid
*Notice that the anion does NOT contain oxygen.
+
5. Naming acids that contain oxygen (tertiary acids; H bonded with a polyatomic anion) NEVER
use hydro- to name these!
• For acids that contain oxygen, look at the polyatomic anion’s name first.
• If the polyatomic anion’s name ends in –ite then the acid becomes an –ous acid
• If the polyatomic anion’s name ends in –ate then the acid becomes an –ic acid.
• EXAMPLES:
Acid formulas
with oxygen
HNO2
HNO3
H2SO3
H2SO4
Names
1-
nitrous acid [(NO2) nitrite…so it
becomes an -ous acid]
1nitric acid
[(NO3) nitrate…so it
becomes a –ic acid]
sulfurous acid
sulfuric acid
*Notice that the polyatomic anion contains oxygen and hydro- is NEVER used!
Rules for naming bases:
1-
6. Naming compounds that contain the hydroxide ion [OH ]
• Use the same rules as naming polyatomic ionic compounds. That is, name the cation first
and then name the polyatomic anion.
• EXAMPLES:
Bases
NaOH
LiOH
Ca(OH)2
KOH
NH3
C.7.B
Write the
chemical
formula of
common
polyatomic
ions, ionic
compounds
containing main
group or
transition
metals,
covalent
compounds,
acid, and
bases.
Name
sodium hydroxide
lithium hydroxide
calcium hydroxide
potassium hydroxide
ammonia
Writing Chemical Formulas
How do you write the chemical formulas of common polyatomic ions?
Many ionic compounds are made of polyatomic ions. A polyatomic ion is a tightly-bound group of atoms
that has a positive or negative charge and behaves as a unit. There is no simple set of rules for naming most
polyatomic ions, so you should memorize the chemical formulas for some of the most common ones, see you
STAAR chart!
Common Chemical Formulas for Some
Common Polyatomic Ions
Name
Formula
1+
Ammonium
(NH4)
1Hydroxide
(OH)
1Nitrite
(NO2)
1Nitrate
(NO3)
2Carbonate
(CO3)
3Phosphate
(PO4)
2Sulfite
(SO3)
2Sulfate
(SO4)
How do you write chemical formulas of ionic compounds containing main group or
transition metals?
•
•
•
•
•
First you must write the symbol of the cation and then the anion.
Next, look up the oxidation number (charges) for the cation and anion.
If the charges DO NOT equal zero, then criss-cross the numbers to make them subscripts (do not
bring down the charges; just the number). This will balance the positive and negative charges to make
the compound neutral (zero charge).
If the charges EQUAL zero, then the formula is already correct.
EXAMPLES:
Name of
compound
Magnesium oxide
Lithium sulfide
Barium nitride
Iron(III) oxide
Write symbols
and check
charges
2+
2Mg O
1+
2-
Li
S
2+
Ba
3+
Fe
N
3-
O
2-
(Roman numeral-charge
of iron)
Lead(IV) sulfate
4+
(SO4)
2+
(OH)
Pb
Criss-cross if
charges DO NOT
equal zero
Charges = 0
So formula correct
Charges ≠ 0
So criss-cross #’s
Charges ≠ 0
So criss-cross #’s
Charges ≠ 0
So criss-cross #’s
2-
Charges ≠ 0
So criss-cross #’s
(Roman numeral-charge
of lead)
Calcium
hydroxide *
Ammonium
phosphate
Ammonium nitrite
Ca
1+
1-
(NH4) (PO4)
(NH4)
1+
3-
(NO2)
1-
Charges ≠ 0
So criss-cross #’s
Charges ≠ 0
So criss-cross #’s
Charges = 0
So formula correct
Write the
correct
formula
MgO
Li2S
Ba3N2
Fe2O3
Pb2(SO4)4
reduce it!
Pb(SO4)2
*Ca(OH)2
*(NH4)3PO4
NH4NO2
*Keep parenthesis around the polyatomic ions only if a number bigger than ONE gets criss-crossed for
a subscript.
How do you write the chemical formulas of covalent compounds?
Because the names of covalent compounds usually indicate the amounts of each atom as a prefix, writing the
formula is often a matter of taking the prefixes and translating them into subscripts after each element
symbol. No need to check the elements charges!
EXAMPLES:
Name of Covalent
Compounds
Dinitrogen difluoride
Carbon dioxide
Carbon tetrachloride
Boron trifluoride
Formula
Ni2F2
CO2
CCl4
BF3
How do you write the chemical formulas for acids and bases?
If you know the name of an acid, you can write its formula by using the rules for writing the name of an
acid, in reverse. You still have to balance the ionic charges just as you would for any ionic compound. That is,
check the charges to see if they equal zero. If not, crisscross the numbers.
EXAMPLE:
First, you need to look at the acid’s name. If you see that the acid’s name is a hydro-base name-ic acid, you
should recognize that that acid has no oxygen in its formula. Almost all of these acids are binary acid (made
up of two elements only; one element being hydrogen and the other element coming from the base name
(anion w/-ide ending).
•
Hydrochloric acid: This would be H and Cl…now check charges to write formula.
1+
1H
Cl …formula: HCl
•
Hydronitric acid:
This would be H and N…now check charges to write formula.
1+
3H
N ..formula: H3N
If you see that the acid’s name has no hydro- prefix, then the acid has oxygen in its formula. The anion is a
polyatomic. The -ous ending comes for an -ite polyatomic anion ending, and -ic comes from an -ate
polyatomic anion ending.
C.7.C
Construct
electron dot
formulas to
illustrate ionic
and covalent
bonds.
•
Sulfurous acid: This would be H and (SO3)…now check charges to write formula.
1+
2(SO3) ..formula: H2SO3
H
•
Sulfuric acid:
This would be H and (SO4)….now check charges to write formula.
1+
2(SO4) …formula H2SO4
H
Constructing Electron Dot Formulas
How to construct electron dot formulas:
Different kinds of bonds often form between different atoms. A simple way to show how electrons are
transferred or shared during bond formation is with an electron dot formula.
Learning how to draw and interpret these diagrams is an important skill. An electron dot formula (or
diagram) consists of a chemical symbol surrounded by one to eight dots representing valence electrons.
Valence electrons are the electrons in the outermost energy level of an element’s atom. All of the elements
within a given group (with the exception of helium) have the same number of electron dots in their structures.
Hydrogen has 1 valence electron, so you would write the electron dot formula for hydrogen as:
H·
Fluorine has 7 valence electrons, so you would write the electron dot formula for fluorine as:
··
:F:
·
Ions can be represented in electron dot formulas by depicting the charge and the corresponding number of
1valence electrons. For example, the ion fluorine (F ), which has gained one electron, has eight valence
electrons, shown below.
··
: F:··
2+
Magnesium ion (Mg ), which has lost both of its valence electrons, is illustrated below without any dots.
Mg2+
How to construct electron dot formulas to illustrate ionic bonds:
The electrons involved in an ionic bond are removed or transferred from the cation and found on the anion.
Electron dot formulas illustrate this by indication oppositely charged ions side by side.
For example, you would write the electron dot formula for sodium chloride (NaCl) as shown.
Na+
The electron dot formula for aluminum oxide (Al2O3) is written as:
3+
Al
3+
Al
How to construct electron dot formulas to illustrate covalent bonds:
Covalent bonds are formed from two electrons that are shared between two atoms. The shared pair can be
illustrated in electron dot formulas as two dots, or one short straight line, such as the bond between two
hydrogen atoms.
H:H
H
H
Single bonds are represented by single lines. Double bonds are represented by double lines. Triple bonds are
represented by triple lines.
How to construct electron dot formulas for complex covalent molecules or ions:
Use these steps to help you construct the electron dot formulas for more complex covalent molecules or ions.
1.
Add the total number of valence electrons for every atom in the molecule. If it is an ion, add or
subtract electrons to produce the correct charge.
2. Write the structure of the skeleton. The more electronegative atom usually belongs in the center.
Connect atoms with lines (or pairs of dots).
3. Distribute electrons to each of the outer atoms to satisfy the octet rule.
4. Distribute the remaining electrons to the central atom. Is the octet rule satisfied for the central atom? If
not, then there is probably at least one multiple bond in the molecule or ion. Move pairs of electrons
from neighboring atoms to form double bonds with the central atom until the octet rule is satisfied for
all of the atoms in the molecule.
For example, you would write the electron dot formula for the following as follows:
Sulfur dioxide
Oxygen
carbon dioxide
nitrogen
C.7.D
Describe the
nature of
metallic
bonding and
apply the
theory to
explain metallic
properties such
as thermal and
electrical
conductivity,
malleability,
and ductility.
The Nature of Metallic Bonding
How can you describe the nature of metallic bonding?
Metallic atoms have few valence electrons and low ionization energies. The bonds holding metallic atoms
together in the solid and liquid phases, however, are apparently strong, as metals have fairly high melting and
boiling points. A metallic atom may be considered to have a central portion, or core, made up of its nucleus
and its nonvalence electrons. The atom’s valence electrons surround the core. The cores of the metallic
atoms making up a metallic solid are arranged in the fixed positions of a crystalline lattice.
The valence electrons of metal atoms can be modeled as a “sea of electrons.” That is, the valence electrons
are mobile and can drift freely from one part of the metal to another. Metallic bonds consist of the attraction
of the free-floating valance electrons for the positively charged metal ions. These bonds are the forces of
attraction that hold metals together. A representative of this model in shown in below.
Sea of electrons
Metal cation
How can you apply metallic bonding theory to explain metallic properties?
The nature of metallic bonding explains many physical properties of metals. For example, most metals are
excellent conductors of thermal energy. When a difference in thermal energy is applied across a metal, it is
quickly and evenly transmitted throughout. This is because the crystal lattice of cations making up a metal is
held together very tightly. Thermal energy makes cations move about more rapidly. Since the lattice is held
together so tightly, the energy of motion is quickly distributed throughout the solid.
Most metals are also excellent conductors of electrical energy. When a difference in electrical energy is
applied across a metal, it is quickly and evenly distributed throughout. This happen because the electrons flow
freely. Electrons move in response to the difference in electrical energy, moving from high electric potential
energy to low electric potential energy.
Metals are also malleable, which means that they can be shaped and hammered into thin sheets. A force,
such as the strike of a hammer, applied to the solid reshapes the lattice of cations because the cations can
move through the “sea” of electrons without breaking the metallic bonds. For this same reason, metals are
also ductile, which means that they can be drawn into long wires without breaking.
Another property of metals is their high boiling points. It takes a great deal of thermal energy to break the
bonds holding the ions of a metal together.
STUDY TIP: If you remember that metals have a tendency to give up electrons, then it will be easy to
remember the nature of metallic bonds as positively charged ions in a “sea” of electrons.
What are alloys?
Metallic bonds bind elemental metals, such as pure sodium. However, metallic bonds also bind alloys of two
or more elements. Alloys are mixtures composed of two or more elements, at least one of which is a metal.
Brass, for example, is an alloy of copper and zinc. Alloys are important because their properties are often
superior to those of their component elements. The most important alloys today are steels. The table below
lists some common alloys and the different elements that compose them.
Composition
Copper and zinc
Common Alloys
Common Name
Common Uses
of Alloy
Locks, gears, musical instruments
Brass
Sculptures, bearings, springs
Copper and tin
Bronze
Cookware, appliances
Iron and chromium
Stainless steel
Statuettes, figurines
Tin and copper
Pewter
Fine tableware, jewelry
Silver and copper or other
metal such as zinc
Sterling silver
C.7.E
Predict
molecular
structure for
molecules with
linear, trigonal
planar, or
tetrahedral
electron pair
geometries
using Valence
Shell Electron
Pair Repulsion
(VSEPR) theory.
Molecular Structures
How to predict molecular structures:
Linear
Trigonal planar
Tetrahedral
The simplest way to predict the molecular structures involves the Valence Shell Electron Pair Repulsion
theory. The Valence Shell Electron Pair Repulsion (VSEPR) theory states that the repulsion between
electron pairs causes molecular shapes to adjust so that the valence-electron pairs stay as far apart as
possible. Valence electrons, like all electrons, repel each other. The repulsions of valence electrons affect the
angles of the covalent bonds, giving the overall shape to each molecule.
How to use the VSEPR theory to predict geometric shapes:
Once you have drawn the electron dot structure of a molecule and identified any lone pairs of electrons,
you can use the VSEPR theory to help determine the geometric shape of the molecule. Since they repel each
other, the electron pairs are as far away as possible from each other. The shape of a molecule depends on
the arrangement of electron pairs.
Linear Arrangement
There are two electron pairs around the central atom in a molecule with linear molecular geometry, 2 bonding
electron pairs and 0 lone pairs. The ideal bond angle is 180°.
Trigonal Planar Arrangement
If there are three electron pairs on an atom in a molecule, they tend to spread out from each other on the same
0.
plane. The ideal bond angle is 120
Tetrahedral Arrangement
If there are four electron pairs on an atom in a molecule, they tend to spread out from each other into four
0
points that form a pyramid. The ideal bond angle is 109.5 .
To understand any molecule, one must first complete a Lewis dot structure. It is then possible to predict the
molecular shape using TWO BASIC PRINCIPLES:
1. The shapes of molecules are determined by the repulsion between electron pairs in the outer shell of the
central atom. Both bond pairs (electron pairs shared by two atoms) and lone pairs (those located on a
central atom but not shared) must be considered.
2. Lone pairs repel more than bond pairs.
The application of these principles is best seen by referring to specific examples. We shall start looking only at
molecules with single bonds (for simplicity).
A. Central atoms with less than octet configurations:
The BeCl2 molecule has a Lewis dot structure as shown. The central Be atom has two
bond pairs in its outer shell. Repulsion between these two pairs (first principle above)
causes the molecule to be linear (see above). If the molecule were bent in any direction,
the two bond pairs would be brought closer together, increasing the repulsion.
The molecule BF3 has a dot symbol as:
Here the B atom has three bond pairs in its outer shell.
Minimizing the repulsion causes this molecule to have a trigonal
planar shape, with the F atoms forming an equilateral triangle
about the B atom. The F-B-F bond angles are all 120°, and all the
atoms are in the same plane.
BF3
B. Central atoms with octet (noble gas) configurations:
The CH4 molecule has a dot structure as
shown. The shape of this molecule, however,
is not planar, as is suggested by the way we
draw this dot structure.
Carbon has 4 bond pairs. The four H atoms are arranged about the C
atom in a tetrahedral shape . This shape minimizes the repulsion
between the bond pairs. The 109.5° angle is the same for all H-C-H
bond angles and is called the tetrahedral angle.
CH4
There are many molecules that have four bond pairs and this regular tetrahedral shape; CCl4, SiF4, and SnCl4
are just a few examples.
C.8.A
Define and use
the concept of a
mole
Defining and Using the Mole
How do chemists define a mole?
Atoms and molecules are so small that even a small amount of a substance consists of an extremely large
number of them. A drop of water, for example, contains trillions of water molecules. In chemistry, the number
of particles of a substance is expressed by a unit called a mole, which is the SI unit for the amount of a
substance. It is abbreviated as mol.
23
A mole of a substance is 6.02 x 10 representative particles of that substance. The number of
23
representative particles in a mole, 6.02 x 10 , is called Avogadro’s number.
When you refer to the particles in a mole of a substance, you are referring to representative particles of that
substance. A representative particle is the type of particle that makes up the substance. It is the smallest
particle into which the substance can be divided and still be that substance. Representative particles might
be atoms, molecules, formula units, or ions depending on the substance. See chart below:
Representative Particles of Selected Substances
Substance
Lithium
Iron
Argon gas
Bromine gas
Sodium ion
Sodium chloride
Water
Representative Particle
Chemical Formula
Atom
Atom
Atom
Molecule
Ion
Formula unit
Molecule
Li
Fe
Ar
Br2
NaNaCl
H2O
Number of
representative
particles per mole
23
6.02 x 10
23
6.02 x 10
23
6.02 x 10
23
6.02 x 10
23
6.02 x 10
23
6.02 x 10
23
6.02 x 10
How can you determine the representative particle of a given substance?
Substance
Most elements
Diatomic molecules and all
molecular compounds
Ionic compounds
Representative Particle
Atom
Molecule
Examples
Cu, Ag, Mg
H2O, C12H22O11
Formula units
CaCO3, MgO
How can you determine the representative particle of a given substance?
23
The relationship 1 mol = 6.02 x 10 representative particles is the basis for the following conversion factors
that you can use to convert number of representative particles to moles:
1 mol
23
6.02 x 10 representative particles
You can also use the conversion factor below to convert from moles to number of representative particles.
23
6.02 x 10 representative particles
1 mol
If you know the number of moles of a substance in a sample, you can calculate the number of representative
particles. Conversely, you can calculate the number of moles of a substance if you know the number of
representative particles. These conversions can be useful in calculation involving the mass of a substance.
EXAMPLE #1:
20
Suppose a sample consists of 6.85 x 10 atoms of carbon. How can you determine the number of moles?
First list what you know:
20
• number of atoms: 6.85 x 10 atoms C
23
• 1 mol C = 6.02 x 10 atoms C
Next, multiply the number of atoms of C by the conversion factor, and this will give you the answer.
20
6.85 x 10
atoms C
x
1 mol C
23
6.02 x 10 atoms C
-3
=
1.14 x 10 mol C
EXAMPLE #2:
Another sample consists of 2.58 moles of water, H2O. How can you determine the number of water molecules
in the sample?
2.58 mol H20
x 6.02 x 10
23
molecules H2O
1 mol H2O
=
1.55 x 10
24
molecules H2O
QUESTION: Why are the representative particles called atoms in the first example and molecules in the
second example? When do you use formula units for the representative particles?
C.8.B
Use the mole
concept to
calculate the
number of
atoms, ions, or
molecules in a
sample of
material.
Calculating Atoms, Ions, or Molecules Using Moles
You can use the mole concept to calculate how many particles make up the mass of a sample of material. In
order to perform the calculation, you need to know the molar mass.
Molar mass is the mass of one mole of the material. One mole of atoms of an element has a molar mass
in grams that is equal to the atomic mass(given in periodic table in amu) Example: 1 mol C = 12.011 grams C
23
= 6.02 x 10 atoms C
6
C
12.011
atomic mass
So, 1 mol Al = 26.982 grams Al = 6.02 x 10
1 mol Mg = 24.305 grams Mg = 6.02 x 10
23
23
atoms Al;
atoms Mg;
1 mol Mg3N2 = 100.929 grams Mg3N2 = 6.02 x 10
23
formula units Mg3N2
CALCULATIONS:
You will need to use your STAAR periodic table to do these calculations.
Example 1:
How many atoms are in 2 moles of He? (start out with what is given)
2 mol He
x
23
6.02 x 10 atoms He
1 mol He
24
= 1.20 x 10 atoms He
Example 2:
What is the mass in grams of 3 mol N2? (start out with what is given)
3 mol N2
x
28.014 grams N2
1 mol N2
=
84.042 grams N2
Example 3:
How many molecules are in 7.1 grams H2O? (start out with what is given)
7.1 g H2O
x
1 mol H2O
18.015 g H2O
23
x 6.02 x 10 molecules H2O = 2.4 x 10
1 mol H2O
23
molecules H2O
Example 4:
Find the number of chlorine ions (Cl ) in 5 grams of CaCl2. (start our with what is given)
5 g CaCl2
C.8.C
Calculate
percent
composition
and empirical
and molecular
formulas.
x
1 mol CaCl2
111 g CaCl2
x
-
2 mol Cl ions
1 mol CaCl2
x
-
22
-
6.02 x 1023 Cl ions = 5 x 10 Cl ions
1 mol Cl ions
Calculating Percent Composition, Empirical Formulas, and Molecular
Formulas
How to calculate percent composition:
% by mass of element =
mass of element in mol compound
molar mass of compound
Example 1:
Calculate the percent compositions of each element in MgCl2.
Molar mass of Mg in formula: 24.305 grams x 1 = 24.305 grams
Molar mass of Cl in formula: 34.453 grams x 2 = 70.906 grams
Molar mass of compound (MgCl2): 24.305 + 70.906 = 95.211 grams
% Mg = 24.304 g
95.211 g
x
100 = 26% Mg
% Cl = 70.906 g
95.211 g
x
100
= 74% Cl
x 100 %
How to calculate empirical formula:
An empirical formula (reduced formula) is a formula with the lowest whole-number ratio of elements in a
compound. It is not always the same as the molecular formula, which tells exactly how many atoms of
each element are contained in a molecule or a formula unit of a compound. For example, methane’s
molecular formula is C2H6, and its empirical formula is CH3.
If you know the percent compositions of the elements in a compound, you can calculate the empirical formula.
•
•
•
First, find the number of moles of each element by dividing the percent composition of each element
by its atomic mass.
Next, divide each result by the lowest number of moles of all the elements contained in the compound.
If the results are not integers (whole numbers), multiply by the smallest factor that will make them
integers. Remember, whatever you do to one element in the formula you must do it to all!
EXAMPLE 1:
Find the empirical formula for a compound that is 79.8% C and 20.2% H. To make the calculations
easier, consider a 100 g sample.
First: Change % to grams and find moles for each element
79.8g C
x
1 mol C
12.011g C
= 6.64 mol C
20.2g H
x
1 mol H
1.008g H
= 20.0 mol H
Next: Divide each answer by the smallest mole answer. In this case, the smallest mole answer is
6.64 mol.
6.64 mol C = 1
6.64
20.0 mol H
6.64
= 3
Therefore, the ratio of C to H is 1:3.
Now, write the empirical formula:
CH3
How to calculate a molecular formula:
Subscript multiplier =
actual molecular weight
empirical formula weight
EXAMPLE 1:
The empirical formula for a molecule is NH2. If its molecular weight is 32 amu, what is the compound’s
molecular formula?
•
•
First find the empirical formula mass of NH2 = 14.007 + (2 x 1.008) = 16.023 amu
Next, find the subscript multiplier.
actual molecular weight = 32 amu
= 2
empirical formula weight 16.023 amu
•
Now multiple both subscripts of NH2 by 2 to get the molecular formula: N2H4.
C.8.D
Use the law of
conservation of
mass to write
and balance
chemical
equations
Balancing Chemical Equations
Law of Conservation of Mass: matter is not created or destroyed in an ordinary chemical
reaction (atoms are not created or destroyed)
Total mass of reactants = Total mass of products
Example: If 4 sulfur atoms are in the reactants, then 4 sulfur atoms are in the product.
How to use the law of conservation of mass to write and balance chemical equations:
Use coefficients: number to left of element or compound in a chemical equation; to determine the number of
a particular atom in any term; multiply the term’s coefficient by the atom’s subscript.
Example: 3H2O has 6 H and 3 O
Balanced chemical equation: gives the number and kind of substances in a reaction; reactants are to the
left of the arrow and products are to the right of the arrow.
2H2 + O2
Reactants
2 H2O
Products
The following steps describe how to write and balance a chemical equation:
1. Write a skeleton equation. Write formulas for reactants on the left and formulas for products on the
right. Place an arrow between them.
2. Count the number of atoms of each element in the reactants and of each element in the products. For
1-,
22+
simplicity, if a polyatomic ion such as NO3 SO4 , CO3 , or NH4 , is present unchanged on both sides
of the equation, treat it as a single unit.
3. Use coefficients to make the number of each element and polyatomic ion the same on both sides of
the equation.
4. Make sure the coefficients are in the lowest possible ratio that balances.
EXAMPLE:
Balance the chemical equation: ___H2 + ___N2  ___NH3
1. Write a skeleton equation by checking elements to see if they are diatomic molecules and then check
compounds for correct formulas.
2. Count the atoms: There are 2 H on the left and 3 H on the right. There are 2 N on left and 1 N on right.
3. Insert coefficients so that there are 6 H on each side: 3H2 + __N2 2NH3
4. Now check the N. There are 2 N on the left and 2 N on the right. It is balanced with a coefficient of 1 on
N2: balanced equations: 3H2 + N2 2NH3
5. Check again by counting each element per side. Left side: 3H2 = 6 H and N2 = 2 N
Right side: 2NH3 = 2 N and 6 H
(left = right)
C.8.E
Perform
stoichiometric
calculation,
including
determination
of mass
relationships
between
reactants and
products,
calculation of
limiting
reagents, and
percent yield.
Stoichiometric Calculations
How to determine mass relationships between reactants and products:
The calculation of quantities in chemical reactions is a subject of chemistry called stoichiometry.
BALANCED equations are used to perform stoichiometric calculations.
A balanced chemical equation shows molar ratios among substances in a reaction (reactants or products);
use molar mass values to convert from moles to grams or from grams to moles.
EXAMPLE:
Suppose 8.75 g of propane (C3H8) react with oxygen gas (O2) to produce carbon dioxide (CO2) and water
(H2O). How many grams of water are produced?
1. First, write a balanced equation for the reaction:
C3H8
+
5O2
3CO2 + 4H2O
2. Find the two substances that will be compared in the equation. The two substances to compare are
propane and water since the problem gave grams of propane and asks for grams of water.
3. This will be a grams to grams problem: Start with grams of propane and take it to grams of water.
Use the balance equation to get the mole ratio of propane to water.
8.75 g C3H8
x
1 mol C3H8
44.097 g C3H8
x
4 mol H2O
1 mol C3H8
x
18.015 g H2O = 14.3 g H2O
1 mol H2O
How to determine the limiting reagent in a reaction:
The limiting reagent in a chemical reaction is the reactant that determines the amount of each product that
will form. When all of the limiting reagent has been used, the reaction stops. Any other reactants that remain
are excess reagents.
There are several ways to find the limiting reagent. The one that follows is one of the simplest ways to find the
limiting reagent.
EXAMPLE:
What is the limiting reagent when 36 g of CH4 reacts with 98 g O2 to produce CO2 and H2O?
1. Write the balanced equation for the reaction.
CH4 + 2O2  CO2 + 2H2O
2. Change grams of each reactant to moles (If moles are given for each reactant then skip this step)
36 g CH4 x 1 mol CH4
16 g CH4
= 2.25 mol CH4
98 g O2
= 3.06 mol O2
x
1 mol O2
32 g CH4
3. Now, look at the balanced equation and divide each mole answer by the coefficient of each
reactant.
CH4 + 2O2  CO2 + 2H2O
So, take 2.25 mol CH4 and divide it by 1. You will get 2.25 for your answer.
Then take 3.06 mol O2 and divide it by 2. You will get 1.53 for your answer.
The lowest answer will be the limiting reagent. That means O2 is the limiting reagent.
CH4 is the excess reagent.
How to calculate percent yield:
Percent yield: describes the portion of the theoretical yield that is actually recovered
percent yield =
actual yield (from experiment)
x 100 %
theoretical yield (from calculation)
Suppose, for example, that you calculate a theoretical yield of 55.0 g H2O, but the actual yield from your
experimentation was 48.8 g H2O. The percent yield is:
percent yield =
48.8 g
55.0 g
x 100% = 88.7 %