Transcript bond energies, empirical, & molecular formulas

PERCENT COMPOSITION,
EMPIRICAL &
MOLECULAR FORMULAS
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Percent Composition
oNew food labels are required to
describe the ingredients using
percents of the daily reccommended allowance
• These numbers tell
what part of the total #
of calories can be obtained from a product
• AKA percent composition
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Percent Composition
oTo get the information found on
food labels the chemists had to
know what fraction of the
whole was each component
• Component/total and then
multiply by 100
• There are a couple of
procedures used to calculate
percent compositions
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Calculating PC given formula
What percentage of
Hydrogen and Oxygen is
in Water (H2O)?
Assume you have 1 mole of
water, and calculate its molar
mass
(2•1.008g) + (1•15.994g) =
18.01g
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Calculating PC given formula
oThere are 2 mols of H atoms for
every 1 mol of Water molecules
oHow much do 2 mols of H atoms
weigh?
H: (2•1.008g)= 2.016g H
oPercent of H in Water?
2.016g H
X 100%= 11.2%
18.01 g H2O
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Calculating PC given formula
oThere is 1 mol of O atoms for
every 1 mol of Water molecules
oHow much does 1 mol of O
atoms weigh?
O: (1•15.994g)= 15.994g O
oPercent of O in Water?
15.994 O
X 100%= 88.8%
18.01 g H2O
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Percent Composition
o Another method of calculating
the percent composition is by
experimental analysis.
• the overall mass of the sample
is measured.
• then the sample is decomposed or separated into its
component elements
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Percent Composition
o The masses of the component
elements are then determined
and the percent composition is
calculated as before
• by dividing the mass of each
element by the total mass of
the sample
• then multiplying by 100
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Calculating PC given sample
Find the percent composition
of a compnd that contains
1.94g of carbon, 0.48g of
Hydrogen, and 2.58g of
Sulfur in a 5.0g sample of the
compnd.
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Calculating PC given sample
o Calculate the percents for
each element much like you
would calculate the percents
for anything.
C: 1.94g/5.0g X 100% = 38.8%
H: 0.48g/5.0g X 100% = 9.6%
S: 2.58g/5.0g X 100% = 51.6%
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Empirical Formulas
o Once the percent compositions
are determined then they can
be used to calculate a simple
chem formula for the compnd
• key is to convert the percents
by mass into amounts in moles
• Then, compare the moles using
ratios to determine coefficients
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Calculating Empirical Formulas
What is the empirical formula
of a compound that is 80%C
and 20%H by mass
oSince we have been given percents rather than masses we
need to make an assumption.
• Let’s suppose we have a total
sample that weighs 100 g.
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Calculating Empirical Formulas
o This allows us to say that if we
had a 100 grams of sample,
• 80 g is Carbon
• 20 g is Hydrogen
o Now that we have a set of
masses we need to convert
them to moles
• Divide by the molar masses
from the Periodic Table
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Calculating Empirical Formulas
80g C
20g H
1 mole C
12 g C
1 mole H
1gH
= 6.7mol C
= 20 mol H
• Now calculate the simplest ratio
of each by dividing both values
by the smallest value
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Calculating Empirical Formulas
Divide each mole value by the
smaller of the two values:
C: 6.7/6.7=1
H: 20/6.7 = 2.98  3
Ratio is 1 C’s for every 3 H’s;
so the formula is = CH
3
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Calculating Empirical Formulas
Determine the empirical
formula of a compound
containing 25.9g of N and
74.1g of O.
Notice we have masses
this time not percents,
we can convert masses
directly to moles
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Calculating Empirical Formulas
25.9g N
74.1g O
1 mol N
14 g N
= 1.85 mol N
1.85 mol
1 mol O
= 4.63 mol O
16 g O
1.85 mol
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Calculating Empirical Formulas
Is the final answer N1O2.5?
Of course not!
We need a whole number
ratio…
Each part of the ratio is multiplied
by a number that converts the
fraction to a whole number
N2(1)O2(2.5)= N2O5
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Molecular Formulas
o The empirical formula indicates
the simplest ratio of the atoms
in the compnd
• However, it does not tell you the
actual numbers of atoms in
each molecule of the compnd
•For instance, glucose has the
molecular formula of C6H12O6
• Empirical form would be CH2O
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Molecular Formulas
o The empirical formula of CH2O,
could be several compnds.
• C2H4O2 or C3H6O3 or
C100H200O100
o It’s more important to know the
exact numbers of atoms
involved
The numbers of atoms define
the properties of the compnd
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Molecular Formulas
o The molecular formula is
always a whole-number multiple
of the emp. formula
o In order to calculate the
molecular formula you must
have 2 pieces of information
• Empirical formula
• Molar mass of the unknown
compound (must be given)
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Calculating Molecular Formulas
Find the molecular formula of a
compound that contains
56.36 g of O and 54.6 g of P.
If the molar mass of the
compound is 189.5 g/mol.
1) Find the Empirical Formula
2) Find the MM of the Emp. Form.
3) Find the ratio of the 2 molar
masses (Mol MM/Emp MM)
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1)Find the Empirical Formula
1
mol
O
56.36g O
= 3.5 mol O
16 g O
1.8 mol
54.6g P
1 mol P
= 1.8 mol P
31g P
1.8 mol
Empirical formula: P1O2
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2) Find the MM of the Emp
Form.
MM of PO2: (1•31g P) + (2•16g O)
= 63g/mol
3)Find the ratio of the 2 molar
masses (mol MM/emp MM)
GIVEN
189.5 g/mol
= 3.00
CALCULATED
63 g/mol
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Calculating Molecular Formulas
o So the Molecular formula is
3 times heavier than the
Empirical formula
• Therefore, the molecular
formula has 3 times more
atoms than the emp. formula
P3(1)O2(3)= P3O6
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