Percent Composition, Empirical Formulas, Molecular Formulas

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Percent Composition, Empirical Formulas,
Molecular Formulas
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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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Hydrates
Hydrated salt salt that has water molecules
trapped within the crystal lattice
Examples CuSO45H2O , CuCl22H2O
Anhydrous salt salt without water molecules
Examples CuCl2
Can calculate the percentage of water in a
hydrated salt.
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Percent Composition
Calculate the percentage of water in sodium
carbonate decahydrate, Na2CO310H2O.
1. Molar Mass of Na2CO310H2O
Na 2(22.99) 45.98
C 1(12.01) 12.01
H 20(1.01) 20.2
3.
O 13(16.00) 208.00
180.2 g
MM 286.2
_______
67.97
x 100
286.2 g
Water
2.
H 20(1.01) 20.2
O 10(16.00) 160.00
MM 180.2
H 2(1.01) 2.02
or
So 10 H2O 10(18.02) 180.2
O 1(16.00) 16.00
MM H2O 18.02
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Percent Composition
Calculate the percentage of water in Aluminum
bromide hexahydrate, AlBr36H2O.
1. Molar Mass of AlBr36H2O
Al 1(26.98) 26.98
Br 3(79.90) 239.7
H 12(1.01) 12.12
O 6(16.00) 96.00
MM 374.8
3.
108.1 g
_______
Water
2.
28.85
x 100
374.8 g
H 12(1.01) 12.1
O 6(16.00) 96.00
MM 108.1
MM 18.02 For 6 H2O 6(18.02) 108.2
or
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Percent Composition
If 125 grams of magnesium sulfate heptahydrate is
completely dehydrated, how many grams of
anhydrous magnesium sulfate will remain?
MgSO4 . 7 H2O
1. Molar Mass
2. MgSO4
Mg 1 x 24.31 24.31 g S 1 x 32.06 32.06
g O 4 x 16.00 64.00 g MM 120.37 g
120.4 g
X 100
48.84
246.5 g
3. Grams anhydrous MgSO4
H 2 x 1.01 2.02 g O 1 x 16.00 16.00 g
MM 18.02 g
0.4884 x 125
61.1 g
MM H2O 7 x 18.02 g 126.1 g
Total MM 120.4 g 126.1 g 246.5 g
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Percent Composition
If 145 grams of copper (II) sulfate pentahydrate
is completely dehydrated, how many grams of
anhydrous copper sulfate will remain?
CuSO4 . 5 H2O
1. Molar Mass
2. CuSO4
Cu 1 x 63.55 63.55 g S 1 x 32.06 32.06
g O 4 x 16.00 64.00 g MM 159.61 g
159.6 g
X 100
63.92
249.7 g
3. Grams anhydrous CuSO4
H 2 x 1.01 2.02 g O 1 x 16.00 16.00 g
MM 18.02 g
0.6392 x 145
92.7 g
MM H2O 5 x 18.02 g 90.1 g
Total MM 159.6 g 90.1 g 249.7 g
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Percent Composition
A 5.0 gram sample of a hydrate of BaCl2 was
heated, and only 4.3 grams of the anhydrous salt
remained. What percentage of water was in the
hydrate?
2. Percent of water
1. Amount water lost
0.7 g water

• 5.0 g hydrate
• 4.3 g anhydrous salt
• 0.7 g water

x 100
14
5.0 g hydrate
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Percent Composition
A 7.5 gram sample of a hydrate of CuCl2 was
heated, and only 5.3 grams of the anhydrous salt
remained. What percentage of water was in the
hydrate?
2. Percent of water
1. Amount water lost
2.2 g water

• 7.5 g hydrate
• 5.3 g anhydrous salt
• 2.2 g water

x 100
29
7.5 g hydrate
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Percent Composition
A 5.0 gram sample of Cu(NO3)2nH2O is heated, and
3.9 g of the anhydrous salt remains. What is the
value of n?
1. Amount water lost

• 5.0 g hydrate
• 3.9 g anhydrous salt
• 1.1 g water

3. Amount of water
0.22 x 18.02
4.0
2. Percent of water
1.1 g water
x 100
22
5.0 g hydrate
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Percent Composition
A 7.5 gram sample of CuSO4nH2O is heated, and
5.4 g of the anhydrous salt remains. What is the
value of n?
1. Amount water lost

• 7.5 g hydrate
• 5.4 g anhydrous salt
• 2.1 g water

3. Amount of water
0.28 x 18.02
5.0
2. Percent of water
2.1 g water
x 100
28
7.5 g hydrate
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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