Chapter 7.3
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How do we use these?
These indicate which of the elements make
up a substance.
These also indicate the number of ions or
atoms that make up a given substance.
C6H12O6
 Atomic
mass is the weighted
average of the masses of the
isotopes of that element.
 This
reflects the mass and the
relative abundance of the isotopes
as they occur in nature.
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A mole is a unit of measurement of number
of atoms of an element
◦ 602,000,000,000,000,000,000,000 atoms are in
one mole of an element.
◦ 1 mole = 6.02 x 1023 particles
◦ This is called Avogadro’s number
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The atomic mass of an element is the mass of
one mole of that element in grams g/mol.
Using moles allows us to work with formulas
in a measurable way – using grams.
“Atomic masses and chemical formulas”
 Take out your periodic tables
 Complete this assignment
We will review this assignment in five minutes
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One mole of an element is equal to its atomic
mass in grams
1 mole of
Nitrogen
Equals
Equals
14.007 g of
Nitrogen
Equals
6.02 x 1023 atoms
of Nitrogen
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Moles are a way to make working with teeny
tiny particles manageable.
Have you ever purchased a dozen eggs?
Moles are not that much different. These are
both ways of counting a number of items
The mole is how we can measure something …
…very small (atoms of an element)
…in a manner that we can see (mass in grams).
For Bromine, the mass
of one mole is
79.9 grams.
This is also the mass of
6.02 x 1023 particles of
Bromine …
AND one mole of
Bromine is 6.02 x 1023
particles of Bromine
This activity will help you to see the similarities
between the mole (which is unfamiliar) and the
dozen (which is quite familiar).
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Now that we know how to work with single
elements’ atomic masses, how can we apply
this to more complicated substances?
If we can determine the mass of mole of a
single element in grams, we can determine
the mass of a mole of a compound in grams
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As you know, a mole reflects 6.02 x 1023
representative particles.
What is a representative particle?
A representative particle is one piece of the
substance …
Depending on what you are talking about, it
might be
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An atom of a single element
A molecule of a substance
An ion
Anything!
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The molar mass of a substance is the mass of
one mole of a substance in grams.
To determine it, you must know the chemical
formula for the substance, and the atomic
masses for each of the elements in it.
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MgCl2
Each particle of this compound contains:
◦ 1 atom of Magnesium
◦ 2 atoms of chlorine
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The molar mass of MgCl2 =
◦ 1 x atomic mass of Mg
◦ 2 x atomic mass of Cl
plus
=
◦ 1x 24.305 g Mg/mole
◦ +2x35.453 g Cl/mole = 95.211 g/mol MgCl2
Remember, the sum of the atomic masses times the
number of atoms of each kind of element is equal to
the mass of one mole of the substance.
Examples:
Na = 22.990 g/mol +
Cl = 35.453 g/mol
Therefore, NaCl has a molar mass of 58.4743 g/mol
Nitric acid is HNO3. Its molar mass is
H = 1.0079 x 1 = 1.0079 g/mol
N = 14.007 x 1 = 14.007 g/mol
O = 15.9997 x 3 = 47.9991 g/mol
Total
= 63.014 g/mol HNO3
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The subscripts in the formula tell you how
many of each atom to include in your
calculations
Ca3N2 will have three Ca atoms and 2 N atoms
If there are parenthesis in a formula,
remember that the number outside the
parenthesis acts as a multiplier for everything
within the parenthesis.
Mg(OH)2 particles contain 1 Mg, 2 O and 2 H
atoms per particle
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There are certain compounds called hydrates
which are salts that bind water molecules in
their crystal structure.
For example, consider ZnSO4 * 7 H2O
Zinc sulfate heptahydrate
When determining the molar mass of a
hydrate, you add the molar masses of that
number of water molecules to the salt portion
of the formula
ZnSO4 * 7 H2O
Molar mass of zinc sulfate + 7 x molar mass of water
ZnSO4 molar mass =
1 Zn (65.39 g/mol) + 1 S (32.06 g/mol) + 4 O (15.999 g/mol) =
161.4476 g/mol zinc sulfate
2H + 1 O = 18.0148 g/mol water x 7 (hepta) = 126.1036 g/mol H2O
Total molar mass is
 161.4476 g/mol zinc sulfate plus 126.1036 g/ 7 mol H2O
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287.5512 g/mol ZnSO4 * 7 H2O
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Now that we know the amount of mass
contributed by each individual element in a
formula, we can determine the percentage of
that element by mass
Mass of element in one mole x 100 = % element in compound
molar mass of compound
The percentage composition of each element in a
compound can be determined using only the correct
formula and the atomic masses.
Example: Sodium chloride or NaCl
Na = 22.99
% Na = 22.99 x 100 = 39.3%Na
Cl = 35.453
58.443
58.443g/mol
% Cl = 35.453 x 100 = 60.7% Cl
58.443
Notice that the total of the percentages is always
equal or very close to 100%.
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Percent of phosphorus in phosphate (PO4)
In each phosphate, there are 1 P and 4 O
atoms
1 P x 30.97376 g/mol = 30.97376 g/mol
4 O x 15.9994 g/mol = 63.9976 g/mol
94.97136 g/mol PO4
30.97376 g P *100 = 32.6% P
94.97136 g PO4
What is the percent of water in ZnSO4 * 7 H2O?
From our previous calculations, the total molar
mass is
 161.4476 g/mol zinc sulfate plus +126.1036
g/ 7 mol H2O
 287.5512 g/mol ZnSO4 * 7 H2O
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126.1036 g/ 7 mol H2O x 100 = 43.9% water
287.5512 g/mol ZnSO4 * 7 H2O
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We use dimensional analysis to work with
moles, grams and particles.
Atomic masses, one mole, and Avogadro’s
number can be used as conversion factors to
convert between moles, grams and particles
of an element
1 mole of
Nitrogen
Equals
Equals
14.007 g of
Nitrogen
Equals
6.02 x 1023 atoms
of Nitrogen
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How many grams are there in 5.40 moles of
Nitrogen?
Converting from MOLES to GRAMS
The conversion factor you will use is:
1 mole N = 14.007 g N
5.40 moles N x 14.007 g N = 75.6 g N
1 mole
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How many atoms are there in 40.6 g Nitrogen?
Converting from GRAMS to ATOMS
The conversion factor that you will use is 14.007
g Nitrogen = 6.02 x 1023 atoms
1023
40.6 g N x 6.02 x
atoms =
14.007 g N
1.74 x 1024
atoms of N
How many moles are there in 2.3 x 1023
atoms of Nitrogen?
 Converting from ATOMS to MOLES
 The conversion factor that you will use is:
1 mole Nitrogen = 6.02 x 1023 atoms
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2.3 x 1023 atoms x
1 mole =
0.38 moles N
6.02 x 1023 atoms
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What is an empirical formula?
The empirical formula represents the
lowest whole number ratio between
elements in a compound.
For example, peroxide has 1 H for
every 2 O in the compound
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BUT … the empirical formula may not match
the way that the actual molecules form.
For that, we have the molecular formula,
which represents the whole number ratio of
atoms in a molecule.
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Sometimes these do match!
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Ex: For water, H2O, the empirical and
molecular formulas are the same.
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Let’s work through a problem in which you
need to determine the empirical and
molecular formulas for a substance.
Peroxide is a substance made from hydrogen
and oxygen atoms.
We will determine the empirical and
molecular formulas for peroxide:
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A 6.8 g sample of a compound contains 4.0 g
H and 6.4 g O. It has a molar mass of
34.0146 g/mol.
Step 1. Determine the number of moles of
each of the elements in the formula.
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From our previous calculation, we determine
that there is a one-to-one ratio between H
and O in this substance.
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0.4 moles H = 0.4 moles O = 1:1 ratio
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But this does not match the molar mass!
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HO = 1.0079 g/mol H + 15.9994 g/mol O =
17.0073 g/mol HO
But …..
17.0073 g/mol X 2 does equal 34.0146
g/mol.
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So, the correct molecular formula for
peroxide is H2O2.
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We will work through some problems on this
topic.
And, this is the last topic in this chapter, so
we will move into chemical reactions next!