Chapter 6 Notes
Chemical Composition
Section 6.1: Counting By Weighing
• We can weigh a large number of the
objects and find the average mass.
• Once we know the average mass we can
equate that to any number of the objects.
• EXAMPLE:
– The average mass of a book is 40.0 grams.
– How many books are present in a sample with
a mass of 2000.0 grams?
– 2000.0g/40.0g = 50.0 books
6-2
Counting By Weighing
• When we know the average mass of the
atoms of an element, as that element
occurs in nature, we can calculate the
number of atoms in any given sample of
that element by weighing the sample.
• The atomic mass of an element, as found
on the periodic table, allows us to count
by weighing.
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Section 6.2: Atomic Masses –
• Atoms are very small, a single carbon
atom has a mass of 1.99 x 10-23 grams
(0000000000000000000000199g). So,
scientists had to come up with a smaller
unit to measure their mass: the atomic
mass unit (amu).
1 amu = 1.66 x 10-24 grams
• According to the periodic table, the
average mass of a carbon atom is 12.01
amu.
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Section 6.3: The Mole
• Not this type of
mole!
• The mole (mol) is known as
the “chemists dozen” and
represents
• 6.022 x 1023 things (atoms,
particles, molecules, etc).
• Just as we could have a
dozen donuts, we could also
have a mole of donuts;
that’d be A LOT of donuts!
6-5
The Mole: Interesting Mole Facts
• 6.022 X 1023 watermelon seeds: would be found
inside a melon slightly larger than the moon.
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• 6.02 X 1023 grains of sand: would be
more than all of the sand on Miami
Beach.
• 6.022 X 1023 pennies: would make at
least 7 stacks that would reach the moon.
• 6.02 X 1023 blood cells: would be more
than the total number of blood cells found
in every human on earth.
• A mol is A LOT of particles:
• 602,200,000,000,000,000,000,000
The Mole
• It is VERY important to understand that
the value listed as the mass number on
the periodic table really tells us TWO
things:
– The mass of one atom in units of amu’s.
– The mass of one mole of atoms in grams.
–
–
–
–
One mol of Al = 26.98 g & one atom = 26.98 amu
One mol of Au = 196.97 g & one atom = 196.97 amu
One mol of B = 10.81 g
& one atom = 10.81 amu
Remember one mol = 6.022 X 1023 atoms (or particles
or molecules)
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The Mole
• To summarize: a sample of any element
that weighs a number of grams equal to
the average atomic mass (from the
periodic table) contains 6.022 x 1023
atoms (1 mol) of that element.
• Therefore, atomic weight of an element =
#g in one mol of that element
• We write that as g/mol.
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Using the mol in calculations
• How many mols are in 7.0 g of N?
7.0 g
1
1 mol
= 0.50 mol
14 g
How many atoms are in 7.0 g of N?
23 atoms
6.022
x
10
(0.5mol)(
/1 mol)=
OR
(7.0
3.011 x 1023 atoms.
23 atoms
1
mol
6.022
x
10
g)(
/14.0 g)(
/1 mol)=3.011 x 1023atoms
The Mole: Practice
1) Calculate the number of moles and the
number of atoms in a 25.0 g sample of
calcium.
2) Calculate the number of moles and the
number of atoms in a 57.7 g sample of
sulfur.
3) Calculate the number of atoms in a 23.6
mg sample of zinc.
4) Calculate the number of atoms in a 128.3
mg sample of silver.
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•
•
•
•
1.
2.
3.
4.
6.24 mol and 3.76 x 1023 atoms
1.80 mol and 1.08 x 1024 atoms
3.61 x 10-4 mol and 2.17 x 1019 atoms
1.189 x 10-3 mol and 7.160 x 1020 atoms
The Mole: More Interesting Mole Facts
• A one-liter bottle of water contains 55.5 moles
H20 molecules.
• A five-pound bag of sugar contains 6.6 moles of
C12H22O11 (sucrose).
• We have 3 types of moles that live underground
in North America: Eastern Mole, Hairy-Tailed
Mole and Star-Nosed Mole
• The "Mexican" Mole is a chocolate sauce or
turkey stew. It comes from the Aztec word
"molli."
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Section 6.4: Molar Mass
• A chemical compound is a collection of
atoms.
• One methane (CH4) molecule contains one C
atom and four H atoms.
• It follows then that one mole of methane
molecules contains one mole of C atoms
and four moles of H atoms.
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Figure 6.3:
Various
numbers
of methane
molecules.
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Molar Mass (g/mol)
• The molar mass of any substance is the
mass (in grams) of one mole of the
substance.
• The molar mass of a compound is obtained
by summing the masses of ALL component
atoms.
• The term “formula weight” may be used
for ionic compounds and is analogous to
molecular weight; both are molar masses.
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Molar Mass
• If we know how many atoms and how many
moles are present, we can calculate the
mass of one mole of a compound.
• This is called the “molar mass” or
sometimes you may also see it referred to
as “molecular weight”.
• Since 1 mol C = 12.01 g and 4 mol H =
4(1.008) or 4.032 g, 1 mol CH4 = 12.01 +
4.032 or 16.04 g (sig figs).
• Remember to use mass numbers from the
periodic table.
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Molar Mass
• Remember: we can talk about one mole of
atoms or one mole of molecules.
• One mole of oxygen atoms (O) weighs
16.00 g.
• One mole of oxygen molecules (O2) weighs
32.00 g.
• Two moles of O atoms weigh 32.00 g.
• Two moles of O2 molecules weigh 64.00 g.
• And so on . . .
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Molar Mass: Practice
1) Calculate the following molar masses:
a)
b)
c)
d)
Water – H2O
Ammonia – NH3
Propane – C3H8
Glucose – C6H12O6
2) Calculate the mass of 1.48 mol C3H8.
3) Calculate the mass of 4.85 mol HC2H3O2.
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Molar Mass: Practice
1. Calculate the number of moles of H2CO3
present in a 7.55 g sample.
2. Calculate the number of moles of
tetraphosphorous decoxide present in a
250.0 g sample.
3. How many water molecules are present in
10.0 g of water? (Hint: find moles first)
4. How many molecules of sucrose
(C12H22O11) are present in a five pound
bag of sugar?
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Representative
Particles=
atoms, molecules,
Items, etc
Section 6.5: Percent Composition
of Compounds
• Sometimes it is not enough to know a
compound’s composition in terms of
numbers of atoms; it may also be useful to
know its composition in terms of the
masses of its elements.
• We can calculate the mass fraction by
dividing the mass of a given element in one
mole of a compound by the mass of one
mole of the compound.
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Percent Composition of Compounds
• Once we know the mass fraction we can
multiply by 100 to get the percent.
• Remember: percent = part/whole X 100.
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ex) In one mole of methane (CH4), there is
one mole of C and four moles of H:
1 mol C = 12.01 g
4 mol H = 4(1.008) = 4.032 g
1 mol CH4 = 12.01 g + 4.032 g = 16.042 g
%C= mass of 1mol C/ mass of 1 mol CH4 x 100
%C = 12.01/16.042 X 100 = 74.87% C
%H= mass of 4 mol H/mass of 1 mol CH4 x 100
%H = 4.032/16.042 X 100 = 25.13% H
Percent Composition Practice
• Determine the mass percent of each
element in the following:
– H2SO4 (sulfuric acid)
– C3H7OH (isopropyl alcohol)
– C6H12O6 (glucose)
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Section 6.6: Formulas of
Compounds
• The object of this section is to do the
opposite of the previous section. Instead
of getting the mass from the formula, we
will determine the formula from the mass.
• To do this, the mass must be converted to
moles using each element’s mass number.
How can we convert mass to moles?
6-27
Formulas of Compounds
• ex) An unknown compound with a mass of
0.2015 g is is found to contain:
– 0.0806 g C
– 0.01353 g H
– 0.1074 g O
• It must contain:
0.0806g (1 mol C/12.01 g C) = 0.00671 mol C
0.01353g(1 mol H/1.008 g H) = 0.01342 mol H
0.1074g(1 mol/16.00 g O) = 0.00671 mol O
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• The numbers from the previous slide allow
us to determine the C:H:O ratio.
• 0.00671 (C) : 0.01342 (H) : 0.00671 (O)
• If we divide each number by the smallest
number we get 1:2:1 for the C:H:O ratio.
• This leads us to a formula of C1H2O1 or
CH2O.
• This is not necessarily the TRUE formula
of the compound, but represents the
RELATIVE numbers of atoms.
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• This represents the lowest whole number
ratio of the compound.
• The actual formula could be CH2O,
C2H4O2, C3H6O3, C4H8O4, C5H10O5,
C6H12O6, . . .
• Any formula with a C:H:O ratio of 1:2:1
is possible (in theory, an infinite number).
• C1H2O1 represents the simplest possible
formula or the EMPIRICAL FORMULA.
• The multiples represent possible
MOLECULAR FORMULAS.
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Formulas of Compounds: Practice
• Determine the empirical formula from each
of the following molecular formulas:
– H2O2 (hydrogen peroxide)
– C4H10 (butane)
– CCl4 (name?)
– HC2H3O2 (acetic acid)
– C6H12O6 (glucose)
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•
1.
2.
3.
4.
Section 6.7: Calculation of
Empirical Formulas
There are four steps to determine the
empirical formula of a compound:
Obtain the mass of each element present (in
grams).
Determine the number of moles of each
type of atom present (use atomic mass).
Divide each number by the smallest number.
Multiply all numbers by the smallest integer
that will make them all integers
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Empirical Formulas: Practice
1) A 1.500 g sample of a compound containing
only carbon and hydrogen is found to contain
1.198 g of carbon. Determine the empirical
formula for this compound.
2) A 3.450 g sample of nitrogen reacts with
1.970 g of oxygen. Determine the empirical
formula for this compound.
3) When a 2.000 g sample of iron metal is heated
in air, it reacts with oxygen to achieve a final
mass of 2.573g. Determine the empirical
formula for this compound.
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• If the relative amounts of elements are
presented as percentages, assume we are
starting with a 100 g sample (100%).
Then each percentage simply becomes a
mass (in grams).
• For example if 15% of a compound is
carbon, we just assume it is 15 g of a 100
g sample; from there we convert to moles.
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Empirical Formulas: More Practice
1) The simplest amino acid, glycine, has the
following mass percents: 32.00% carbon,
6.714% hydrogen, 42.63% oxygen, and
18.66% nitrogen. Determine the
empirical formula for glycine.
2) A compound has been analyzed and found
to have the following mass percent
composition: 66.75% copper, 10.84%
phosphorous, and 22.41% oxygen.
Determine the empirical formula for this
compound.
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Section 6.8: Calculation of
Molecular Formulas
• If we know the empirical formula AND the
molar mass, we can calculate a compound’s
molecular formula.
• Note: without the molar mass, the best
you can find is the empirical formula.
• Once the molar mass is known, one must
ALWAYS find the empirical formula before
one can calculate the molecular formula.
It is impossible to do the reverse.
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Calculation of Molecular Formulas
• It is also important to note that the
molecular formula is ALWAYS an integer
multiple of the empirical formula. We can
represent the molecular formula in terms
of the empirical formula:
(Empirical Formula)n = molecular formula
• It should also be noted when n = 1, the
empirical and molecular formulas are
identical to each other.
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Molecular Formulas: Practice
1) A compound containing carbon, hydrogen,
and oxygen is found to be 40.00% carbon
and 6.700% hydrogen by mass. The
molar mass of this compound is between
115 and 125 g/mol. Determine the
molecular formula for this compound.
2) Caffeine is composed of 49.47% C,
5.191% H, 28.86% N, and 16.48% O.
The molar mass is about 194 g/mol.
Determine the molecular formula for
caffeine.
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