Percent Composition
Percent Composition
• Percent Composition – the
percentage by mass of
each element in a
compound
Part
_______
Percent =
x 100%
Whole
Percent composition Mass of element in 1 mol
of a compound or = ____________________ x 100%
Mass of 1 mol
molecule
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
Percent Composition
Example: What is the percent composition of
Potassium Permanganate (KMnO4)?
Molar Mass of KMnO4
% K
39.1 g K
158 g
= 158 g
x 100 = 24.7 %
54.9 g Mn x 100 = 34.8 %
% Mn 158 g
K=
1(39.10) = 39.1
Mn = 1(54.94) = 54.9
O = 4(16.00) = 64.0
MM = 158
64.0 g O x 100 = 40.5 %
% O
158 g
Percent Composition
Determine the percentage composition of sodium carbonate
(Na2CO3)?
Molar Mass
Na = 2(23.00) = 46.0
C = 1(12.01) = 12.0
O = 3(16.00) = 48.0
MM= 106 g
Percent Composition
46.0 g
% Na =106 g
12.0 g
% C = 106 g
48.0 g
% O = 106 g
x 100% = 43.4 %
x 100% = 11.3 %
x 100% = 45.3 %
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%
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.
3.
79.90 g
___________
= 0.6714
119.0 g
0.6714 x 50.0g = 33.6 g Br
Hydrates
Hydrated salt – salt that has water molecules trapped
within the crystal lattice
Examples: CuSO4•5H2O , CuCl2•2H2O
Anhydrous salt – salt without water molecules
Examples: CuCl2
Can calculate the percentage of water in a hydrated
salt.
Percent Composition
Calculate the percentage of water in sodium carbonate
decahydrate, Na2CO3•10H2O.
1. Molar Mass of Na2CO3•10H2O
Na
C
H
O
2.
= 2(22.99) = 45.98
= 1(12.01) = 12.01
= 20(1.01) = 20.2
= 13(16.00)= 208.00
3.
MM = 286.2
Water
180.2 g
_______
x 100%= 62.96 %
286.2 g
H = 20(1.01) = 20.2
O = 10(16.00)= 160.00
MM = 180.2
or H = 2(1.01) = 2.02
O = 1(16.00) = 16.00
MM H2O = 18.02
So…
10 H2O = 10(18.02) = 180.2
Percent Composition
Calculate the percentage of water in Aluminum bromide
hexahydrate, AlBr3•6H2O.
1. Molar Mass of AlBr3•6H2O
Al
Br
H
O
2.
= 1(26.98) = 26.98
= 3(79.90) = 239.7
= 12(1.01) = 12.12
= 6(16.00) = 96.00
MM = 374.8
3.
Water
H = 12(1.01) = 12.1
O = 6(16.00)= 96.00
MM = 108.1
or
MM = 18.02
For 6 H2O = 6(18.02) = 108.2
108.1 g
_______
x 100%= 28.85 %
374.8 g
EMPIRICAL AND MOLECULAR
FORMULAS
Formulas
Percent composition allows you to calculate the simplest
ratio among the atoms found in a 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:
C4H10 - molecular
C2H5 - empirical
C6H12O6 - molecular
CH2O
- empirical
Formulas
Is H2O2 an empirical or molecular formula?
Molecular, it can be reduced to HO
HO = empirical formula
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.
4.151 g Al 1 mol Al
= 0.1539 mol Al
26.98 g Al
3.692 g O
1 mol O
16.00 g O
= 0.2308 mol O
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
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.
4.550 g Co 1 mol Co
58.93 g Co
5.475 g Cl 1 mol Cl
35.45 g Cl
0.07721 mol Co
=1
0.07721
CoCl2
= 0.07721 mol Co
= 0.1544 mol Cl
0.1544 mol Cl
0.07721
=2
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
2.000 g Fe 1 mol Fe
55.85 g Fe
0.573 g O
1 mol O
16.00 g
= 0.03581 mol Fe
= 0.03581 mol O
1:1
FeO
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.
1.3813 g Pb 1 mol Pb
= 0.006667 mol Pb
207.2 g Pb
0.00672 gH 1 mol H
1.008 g H
= 0.00667 mol H
0.4995 g As 1 mol As
= 0.006667 mol As
74.92 g As
0.4267g Fe
1 mol O
16.00 g O
= 0.02667 mol O
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
0.006667
= 1.00 mol H
0.006667 mol As
= 1.000 mol As
0.006667
0.02667 mol O
0.006667
= 4.000 mol O
PbHAsO4
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
12.01 g C
12.38 g N 1 mol N
14.01 g N
= 5.302 mol C
9.80 g H
1 mol H
1.01 g H
= 0.8837 mol N
14.14 g O
1 mol O
16.00 g O
= 9.72 mol H
= 0.8832 mol O
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
0.8837
= 6.000 mol C
0.8837 mol N
= 1.000 mol N
0.8837
9.72 mol H
0.8837
= 11.0 mol H
0.8837 mol O
= 1.000 mol O
0.8837
6:1:11:1
C6NH11O
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 compound’s 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
Step 2: Divide MM by
Empirical Formula Mass
238.88 g
=2
141.94g
(P2O5)2 =
P4O10
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
78 g/mol
13.01 g/mol
C6H6
=6