Percent Composition, Empirical Formulas, Molecular Formulas

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Percent Composition, Empirical Formulas,
Molecular Formulas
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Why do we need to know composition?

• How much sugar is in a soft drink?
• How many calories are in a candy bar?
• How much of your paycheck goes to pay Uncle Sam?
• These are examples of percent composition that we
  use every day.

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Percent Composition

• Percent Composition the percentage by mass of
  each element in a compound

Part
_______
Percent
x 100
Whole
So
Percent composition of a compound or molecule
Mass of element in 1 mol
____________________
x 100
Mass of 1 mol
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Percent Composition
Example What is the percent composition of
Potassium Permanganate (KMnO4)?
Molar Mass of KMnO4
K 1(39.1) 39.1
Mn 1(54.9) 54.9
O 4(16.0) 64.0
MM 158 g
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Percent Composition
Example What is the percent composition of
Potassium Permanganate (KMnO4)?
158 g
Molar Mass of KMnO4
39.1 g K
K
x 100
24.7
158 g
54.9 g Mn
34.8
x 100
Mn
158 g
K 1(39.10) 39.1
64.0 g O
x 100
40.5
O
Mn 1(54.94) 54.9
158 g
O 4(16.00) 64.0
MM 158
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Percent Composition
Determine the percentage composition of sodium
carbonate (Na2CO3)?
Molar Mass
Percent Composition
46.0 g
x 100
43.4
Na
Na 2(23.00) 46.0 C 1(12.01) 12.0 O
3(16.00) 48.0 MM 106 g
106 g
12.0 g
x 100
11.3
C
106 g
48.0 g
x 100
45.3
O
106 g
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Percent Composition
Determine the percentage composition of ethanol
(C2H5OH)?
C 52.13, H 13.15, O 34.72
_______________________________________________
Determine the percentage composition of sodium
oxalate (Na2C2O4)?
Na 34.31, C 17.93, O 47.76
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Percent Composition
Calculate the mass of bromine in 50.0 g of
Potassium bromide.
1. Molar Mass of KBr
K 1(39.10) 39.10
Br 1(79.90) 79.90
MM 119.0
2.
79.90 g
___________
0.6714
119.0 g
3. 0.6714 x 50.0g 33.6 g Br
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Percent Composition
Calculate the mass of nitrogen in 85.0 mg of the
amino acid lysine, C6H14N2O2.
1. Molar Mass of C6H14N2O2
C 6(12.01) 72.06
H 14(1.01) 14.14
N 2(14.01) 28.02
O 2(16.00) 32.00
MM 146.2
28.02 g
2.
___________
0.192
146.2 g
3. 0.192 x 85.0 mg 16.3 mg N
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Formulas
Percent composition allow you to calculate the
simplest ratio among the atoms found in compound.
Empirical Formula formula of a compound that
expresses lowest whole number ratio of
atoms. Molecular Formula actual formula of a
compound showing the number of atoms present
Examples
C6H12O6
- molecular
C4H10
- molecular
- empirical
- empirical
C2H5
CH2O
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Formulas
Is H2O2 an empirical or molecular formula?
Molecular, it can be reduced to HO HO
empirical formula
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Calculating Empirical Formula
An oxide of aluminum is formed by the reaction of
4.151 g of aluminum with 3.692 g of oxygen.
Calculate the empirical formula.
1. Determine the number of grams of each element
in the compound.
4.151 g Al and 3.692 g O
2. Convert masses to moles.
1 mol Al
4.151 g Al

0.1539 mol Al
26.98 g Al
1 mol O
3.692 g O

0.2308 mol O
16.00 g O
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Calculating Empirical Formula
An oxide of aluminum is formed by the reaction of
4.151 g of aluminum with 3.692 g of oxygen.
Calculate the empirical formula.
3. Find ratio by dividing each element by
smallest amount of moles.
0.1539 moles Al
1.000 mol Al
0.1539
0.2308 moles O
1.500 mol O
0.1539
4. Multiply by common factor to get whole
number. (cannot have fractions of atoms in
compounds)
O 1.500 x 2 3 Al 1.000 x 2 2
therefore,
Al2O3
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Calculating Empirical Formula
A 4.550 g sample of cobalt reacts with 5.475 g
chlorine to form a binary compound. Determine
the empirical formula for this compound.
1 mol Co
4.550 g Co
0.07721 mol Co
58.93 g Co
1 mol Cl
5.475 g Cl
0.1544 mol Cl
35.45 g Cl
0.07721 mol Co
0.1544 mol Cl
2
1
0.07721
0.07721
CoCl2
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Calculating Empirical Formula
When a 2.000 g sample of iron metal is heated in
air, it reacts with oxygen to achieve a final
mass of 2.573 g. Determine the empirical formula.
Fe 2.000 g
O 2.573 g 2.000 g 0.5730 g
1 mol Fe
2.000 g Fe
0.03581 mol Fe
55.85 g Fe
1 mol O
0.573 g O
0.03581 mol Fe
16.00 g
1 1
FeO
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Calculating Empirical Formula
A sample of lead arsenate, an insecticide used
against the potato beetle, contains 1.3813 g
lead, 0.00672g of hydrogen, 0.4995 g of arsenic,
and 0.4267 g of oxygen. Calculate the empirical
formula for lead arsenate.
1 mol Pb
1.3813 g Pb
0.006667 mol Pb
207.2 g Pb
0.00672 gH
1 mol H
0.00667 mol H
1.008 g H
1 mol As
0.4995 g As
0.006667 mol As
74.92 g As
1 mol O
0.4267g Fe
0.02667 mol O
16.00 g O
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Calculating Empirical Formula
A sample of lead arsenate, an insecticide used
against the potato beetle, contains 1.3813 g
lead, 0.00672g of hydrogen, 0.4995 g of arsenic,
and 0.4267 g of oxygen. Calculate the empirical
formula for lead arsenate.
0.006667 mol Pb
1.000 mol Pb
0.006667
0.00667 mol H
1.00 mol H
PbHAsO4
0.006667
0.006667 mol As
1.000 mol As
0.006667
0.02667 mol O
4.000 mol O
0.006667
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Calculating Empirical Formula
The most common form of nylon (Nylon-6) is 63.38
carbon, 12.38 nitrogen, 9.80 hydrogen and
14.14 oxygen. Calculate the empirical formula
for Nylon-6.
Step 1 In 100.00g of Nylon-6 the masses of
elements present are 63.38 g C, 12.38 g n, 9.80 g
H, and 14.14 g O.
Step 2
63.38 g C
1 mol C
9.80 g H
1 mol H
5.302 mol C
9.72 mol H
12.01 g C
1.01 g H
12.38 g N
1 mol N
0.8837 mol N
14.01 g N
14.14 g O
1 mol O
0.8832 mol O
16.00 g O
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Calculating Empirical Formula
The most common form of nylon (Nylon-6) is 63.38
carbon, 12.38 nitrogen, 9.80 hydrogen and
14.14 oxygen. Calculate the empirical formula
for Nylon-6.
Step 3
5.302 mol C
6.000 mol C
61111
0.8837
0.8837 mol N
1.000 mol N
0.8837
C6NH11O
9.72 mol H
11.0 mol H
0.8837
0.8837 mol O
1.000 mol O
0.8837
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Calculating Molecular Formula
A white powder is analyzed and found to have an
empirical formula of P2O5. The compound has a
molar mass of 283.88g. What is the compounds
molecular formula?
Step 3 Multiply
Step 1 Molar Mass
P 2 x 30.97 g 61.94g O 5 x 16.00g 80.00
g 141.94 g
(P2O5)2
P4O10
Step 2 Divide MM by Empirical Formula Mass
238.88 g
2
141.94g
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Calculating Molecular Formula
A compound has an experimental molar mass of 78
g/mol. Its empirical formula is CH. What is its
molecular formula?
(CH)6
C 12.01 g H 1.01 g 13.01 g
C6H6
78 g/mol
6
13.01 g/mol