Percent Composition
From masses (Example)
A sample of steel wool is placed in an aluminum pan and ignited. The following
data is collected:
A
B
C
Description
Aluminum pan
Al pan with steel wool
Al pan and product
Particle
Fe
O
Total=
Mass
(g)
1.324
0.353
1.677
Mass (g)
1.238
2.562
2.915
%
%
(6 significant digits)
(rounded)
78.950 5
78.95
21.049 4
21.0
Check Sum= √= 99.95
% Composition, From masses (Practice)
A 7.885 g sample of aluminum is reacted with sulfuric acid. Aluminum sulfate is
separated from the resulting solution and dried. The mass of the aluminum sulfate
is found to be 50.000 g. Determine the percent aluminum and percent sulfate in
this compound.
% by mass of
Mass of
%
Particle
particle
Particle (g)
(Rounded)
(6 significant digits)
Aluminum ions
Sulfate ions
Total=
7.855
42.115
50.000
15.710 0
84.230 0
15.71
84.230
99.94
√= 100.0
% Composition, From a chemical name or formula (Example)
Determine the percent composition of manganese(III) oxide.
Formula: Mn2O3
Number
%
Atomic/Molar Mass of
(6
significant
Particle
of
digits)
mass
particle
particles
Mn
2
54.94
109.88
69.5971 %
O
3
16.00
48.00
30.4028 %
Total=
157.88
99.9999 %
%
(Rounded)
69.597 %
30.40 %
99.997 %
√=100.00%
% Composition, From a chemical name or formula (Practice)
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Determine the percent composition of manganese(IV) phosphate.
Formula: Mn3(PO4)4
%
Number
Atomic/Molar Mass of
%
Particle
(6 significant
of particles
mass
particle
(Rounded)
digits)
Mn
3
54.94
164.82
30.258 8
30.259
P
4
30.97
123.88
22.742 7
22.743
O
16
16.00
256.00
46.998 3
46.998
Total=
544.70
99.999 8 √=100.000
Determining Empirical Formulas
From masses (Example)
A sample of steel wool is placed in an aluminum pan and ignited. The following
data is collected:
Description
Mass (g)
A Aluminum pan
1.238
B Al pan with steel wool
2.562
C Al pan and product
2.915
Fe 1.324
55.85
O 0.353
16.00
Formula:
Smaller=
Fe1O1 
mass
1.324

 0.023 71
at.mass 55.85
0.353
 0.0221
16.00
0.022 1
FeO
Name:
moles of Fe 0.023 71

 1.073
smaller
0.0221
moles of O 0.0221

 1.00
smaller
0.0221
1
1
iron(II) oxide
Empirical Formulas, From masses (Practice)
A 7.885 g sample of aluminum is reacted with sulfuric acid. Aluminum sulfate is
separated from the resulting solution and dried. The mass of the aluminum sulfate
is found to be 50.000 g. Determine the percent of aluminum and percent sulfate in
this compound.
Whole
Atomic/Molar
Part
Mass
Moles
Quotient Number
Mass
Ratio
Al
7.885
26.98
0.292 3
1.000
2
-2
SO4
42.155
96.07
0.438 8
1.501
3
Smaller =
0.292 3
Formula: Al2(SO4)3
Name:
aluminum sulfate
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Determining Molecular Formulas (Type III [N-N] and Organic [CxHy])
Example:
1.50 g of an unknown white powder is analyzed and found to contain phosphorus
and oxygen. Through further testing the mass of oxygen in the compound is
determined to be 0.85 g and the molar mass of the compound is found to be 284
g/mol. Determine the molecular formula for this compound.
Whole
Atomic/Molar
Part
Mass
Moles
Quotient Number
Mass
Ratio
P
0.65
30.97
0.021
1.0
2
O
0.85
16.00
0.053
2.5
5
Empirical Formula:
P2O5
Empirical Mass:
2(30.97) + 5(16.00) = 141.94 g/mol

R
284
MolarMass

= 2.000 84 = 2
141.94
EmpiricalM ass
Molecular Formula
= (P2O5)R = (P2O5)2 = P2*2O5*2 = P4O10
Can you name this Type III compound?
tetraphosphorus decoxide
Practice:
A compound containing chlorine, carbon, and hydrogen is used as an additive for
gasoline to help prevent engine knock. This compound shows the following
percentage composition:
71.65 % Cl
24.27 % C
?
%H
The molar mass is known to be 98.96 g. Determine the molecular formula for this
compound.
Part
Mass
Atomic/Molar
Mass
Cl
C
H
71.65
24.27
4.08
35.45
12.01
1.008
Moles
Quotient
2.021
2.021
4.05
1.000
1.000
2.00
Whole
Number
Ratio
1
1
2
Empirical Formula:
ClCH2
Empirical Mass: 35.45 + 12.01 + 2(1.008) = 49.476 = 49.48 g
R
= Molar Mass / Empirical Mass = 98.96 / 49.48 = 2.000 = 2
Molecular Formula
= (ClCH2)R = (ClCH2)2 = Cl1*2C1*2H2*2 = Cl2C2H4
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Hydrates
% Water and Molecular Formula From masses: (Example)
A 12.73 g sample of hydrated cobalt(II) chloride is heated to remove the water.
Once cool, the anhydrous crystal is found to have a mass of 6.95 g. Determine the
percent water and the formula of the hydrate.
CoCl2·xH2O
Mass
%
(6
significant
digits)
(g)
CoCl2
6.95
54.595 4
H2O
5.78
45.404 5
Total=
12.73
Formula of a Hydrate (from masses):
Atomic or
Part
Mass
Molar
Mass
CoCl2
6.95
129.83
H2O
5.78
18.016
12.73
%
Particle
(rounded)
54.6
45.4
100.0
Moles
Quotient
0.053 5
0.321
1.00
6.00
Whole
Number
Ratio
1
6
Formula: CoCl2·6H2O
Name: cobalt(II) chloride hexahydrate
% Water in a Hydrate (from formula):
Number
Atomic/Molar
Particle
of
mass
particles
CoCl2
1
129.83
H2O
6
18.016
Total=
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Mass of
particle
129.83
108.096
237.93
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%
(6 significant
digits)
54.566 4
45.431 8
%
(Rounded)
54.566
45.432
√= 99.998
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