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9/30/2011
Chapter 7
Chemical Formula Relationships
Elements == > Formula
In each chemical formula, one can determine
the number of atoms of each element by its
corresponding subscripts in each component
or formula unit.
Exception: If the subscript is 1, it is usually
omitted for that component in the formula.
Let’s look at some examples!!!
EXAMPLES
Chemical Formulas
Mg(NO3)2
In the formula for
magnesium nitrate, this
formula is comprised of the
following:
1 Mg atom
2 N atoms
6 O atoms
HC2H3O2 (aq)
In the formula for acetic
acid, this formula is
comprised of the following:
4 H atoms
2 O atoms
2 C atoms
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ATOMIC Mass
It is the average mass of
the element as they exist
in nature.
The atomic mass is
determined in
comparison to the mass
of an atom of carbon-12,
which by definition is
exactly 12 atomic mass
units (12u).
Calculate
ATOMIC Mass
CO2
Figure 7.1-Molecular (or formula)
mass for carbon dioxide.
To calculate the formula mass of a compound, it equals:
Formula Mass = Σ atomic masses in the formula unit
Rounding
Molar Mass
When determining the final molar mass of a
compound, one must follow these rules:
1. Calculate the formula mass using atomic
masses in as many decimal places as they are
shown.
2. Round off the answer to the same number of
decimal places as the smallest number of
decimal places in any atomic mass used in the
calculation.
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Sample Question
Calculate the formula mass of barium nitrate.
Solution:
FORMULA: Ba(NO3)2
1 Ba
137.3 u
137.3u
2N
2(14.01u)
28.02u
6O
6(16.00u)
96.00u
----------------------------------------------------Molecular Mass = 261.32u
The final molar mass of Ba(NO3)2 is 261.3u.
Reminders
Molar Mass
In our textbook, molar mass can be as accurate as
the number of significant figures of the atomic
mass.
In our class, each element has been written to 4
significant figures. For determining the molar
mass of a compound, write the molar mass
rounded to the element which has the least
number of significant figures right of the decimal
point.
Mole Concept
By definition, the “mole” is the amount of any
substance that contains the same number of units as
the number of atoms in exactly 12 grams of a carbon12 isotope.
The definition of a mole refers to a number of particles,
but it doesn’t say what that number is.
By experiment, it has been found that to 3 significant
figures:
1 mol of any substance = 6.02 x 1023 of that substance
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Avogadro’s NUMBER (Na)
Avogadro’s number (by experiment):
6.02214179 × 1023/mol
Avogadro’s number can be written as a conversion factor:
1 mole
6.02 × 1023 items
6.02 × 1023 items
1 mole
It is similar as to if you were determining the number of eggs of 1
dozen eggs by analogy as the following manner:
12 eggs
dozen
6.02 × 1023 items
1 mole
Sample Problem
How many moles of NaCl are in 5.66 x 1024
molecules of NaCl?
ANSWER:
5.66 ×10 24 molecules NaCl ×
1 mole NaCl
6.02 × 10 23 molecules NaCl
= 9.40 mole NaCl
Molar Mass
Molar mass is defined as the mass of one mole of a
substance.
Molar mass is a bridge between a macroscopic quantity of
matter, grams, and a particulate-level quantity of matter,
moles.
The units of molar mass by definition is in units of grams
per mole (g/mol).
molar mass =
mass
g
=
mole mol
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Molar Mass
The definitions of atomic mass, the mole, and molar
mass are all directly or indirectly related to carbon-12.
This leads to two important facts:
1. The mass of one atom of carbon-12—the atomic mass
of carbon-12—is exactly 12 atomic mass units.
2. The mass of one mole of carbon-12 atoms is exactly
12 grams; its molar mass is 12 grams per mole.
This leads to the conclusion:
The molar mass of any substance in grams per mole is
numerically equal to the atomic, molecular, or formula
mass of that substance in atomic mass units.
Illustration
Molar Mass
Figure 7.2-(a) One mole of elements and (b) One mole of molecular and ionic
compounds
Molar Mass
What is the molar mass of CH4? Express your
answer in grams/mole.
1C
12.01 g
12.01g
4H
4(1.008g)
4.032g
----------------------------------------------------Molecular Mass = 16.042g
The final molar mass of CH4 is 16.04g.
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Moles, Grams, and Particles
With knowing that
molar mass can be
determined by using
the periodic table, and
Avogadro’s number, one
can make the
relationship between
each unit for an atom or
molecule.
Figure 7.3-Conversion among grams,
moles, and number of particles.
Sample Problem
Determine the number of water molecules in
1.5kg of water.
Formula of water: H2O
Molar Mass (H2O) = 18.02g
Avogadro’s Number = 6.02 x 1023 molecules/mol
1 mole H2O 6.02 × 1023 molecules H2O
1.5kg × 1000g ×
×
= 5.01×1025 molecules H 2O
1kg 18.02g H 2O
1 mol H 2O
Percent Composition
For percent composition, one can determine
the percent of one component of a mixture by
the following component using the following
equation:
%A =
parts of A
×100%
total parts
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Sample Problem
What percentage of oxygen is in calcium oxide?
Formula of calcium oxide (CaO)
Molar Mass = 40.08 g/mol Ca + 16.00g/mol O
= 56.08g
%O=
16.00g
×100% = 28.53%
56.08g
Sample Problem
Calculate the percentage composition of sodium
nitrate.
FORMULA: NaNO3
Molar Mass = 22.99g/mol Na
14.01g/mol N
+3(16.00g/mol O)
------------------85.00 g/mol NaNO3
Sample Problem (cont’d)
Now, to find the percentage of each element.
Na:
% Na =
22.99g
×100% = 27.05%
85.00g
N:
%N =
14.01g
× 100% = 16.48%
85.00g
O:
%O =
48.00g
× 100% = 56.47%
85.00g
TOTAL Percentage: 27.05% + 16.48% + 56.47% = 100.00%
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Empirical Formula
An empirical formula is the lowest wholenumber ratio between the atoms in a compound.
EXAMPLES:
For C2H4, the empirical formula is CH2.
For C3H6, the empirical formula is CH2.
Therefore, the empirical formula of CH2 will have
the same empirical formula and same percent
composition.
Practice Problems
Determine the empirical formula (EF) for each
of the following substances.
a)
b)
c)
d)
e)
C3H3
P4O10
B2H6
C6H14
Ag2O2
EF:
EF:
EF:
EF:
EF:
Determining an Empirical Formula
1) Find the masses of different elements in a
sample of the compound.
2) Convert the masses into moles of atoms of
the different elements.
3) Determine the ratio of moles of atoms.
4) Express the moles of atoms as the smallest
possible ratio of integers.
5) Write the empirical formula, using the
number for each atom in the integer ratio as
the subscript in the formula.
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Practice Problem
Hydroquinone, used as a photographic developer is 65.4%
C, 5.5% H, and 29.1% O by mass. What is the empirical
formula of hydroquinone?
Solution:
It is usually helpful in an empirical formula problem to
organize the calculations in a table with the following
headings:
Element
Element
C
H
O
Grams
Moles
Mole Ratio
Grams
Moles
Mole
Ratio
65.4
65.4g
-------------12.01g/mol
5.45
--------1.82
= 5.45 mol C
=2.99
5.5g
-------------1.008g/mol
5.5
--------1.82
= 5.5 mol H
=3.02
29.1g
-------------16.00g/mol
1.82
--------1.82
= 1.82 mol O
=1
5.5
29.1
Formula
Ratio
Formula
Ratio
Empirical
Formula
Empirical
Formula
3
3
C3H3O
1
Molecular Formula
A molecular formula illustrates the total number of
atoms present of each element in the compound. This
formula can be determined by knowing how many
empirical formula units are present in the substance.
For example, if the empirical formula is CH, the
molecule could be the following:
CH, C2H2, C3H3, C4H4, C5H5, C6H6
empirical formula units in 1 molecule =
molar mass of compound
molar mass of empirical formula
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Determining an Molecular Formula
1) Determine the empirical formula of the
compound.
2) Calculate the molar mass of the empirical
formula unit.
3) Determine the molar mass of the compound
(which will be given in this book).
4) Divide the molar mass of the compound by
the molar mass of the empirical formula unit
to get n (the number of empirical formula
units per molecule).
5) Write the molecular formula.
Practice Problem
Sorbic acid is added to food as a mold
inhibitor. Its composition is 64.3% C, 7.2% H,
and 28.5% O, and its molecular mass is 112
g/mol. What is its molecular formula?
Element
Grams
Moles
Mole Ratio
Element
Grams
Moles
Mole
Ratio
C
64.3
64.3g
-------------12.01g/mol
5.35
--------1.78
= 5.35 mol C
=3.01
7.2g
-------------1.008g/mol
7.1
--------1.78
= 7.1 mol H
=3.99
28.5g
-------------16.00g/mol
1.78
--------1.78
= 1.78 mol O
=1
H
O
7.2
28.5
Formula
Ratio
Formula
Ratio
Empirical
Formula
Empirical
Formula
3
4
C3H4O
1
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Solution (cont’d)
Finding the molar mass of C3H4O.
Molar Mass (EF) =
n=
3(12.01 g/mol C)
4(1.008 g/mol H)
+16.00 g/mol O
------------56.06 g/mol C3H4O
112 g/mol
= 2.00 = 2
56.06 g/mol
(C H O) = C H O
3 4 2
6 8 2
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