Transcript Percent Composition

Percent Composition



AKA “Mass Percent”
AKA “Percent by Mass”
Component Mass
X 100%
Total Mass
Example:
C2H4O2
2(12.01 g/mol) = 24.02 g/mol
4( 1.01 g/mol) = 4.04 g/mol
+ 2(16.00 g/mol) = 32.00 g/mol
60.06 g/mol
Mass
Mass
Mass
Mass
C
H
O
C2H4O2
% Composition

%C


24.02 g/mol
Component Mass
Total Mass
100% 
60.06 g/mol

%H

4.04 g/mol

%O

60.06 g/mol
39.99% C
 100% 
6.73% H
 100% 
53.28% O
60.06 g/mol
32.00 g/mol
100%
Total = 100.00%

Practice:
What is the percent composition for each component
in sodium carbonate, Na2CO3?
+
2(22.99 g/mol) = 45.98 g/mol
12.01 g/mol = 12.01 g/mol
3(16.00 g/mol) = 48.00 g/mol
105.99 g/mol
Na2CO3
Na
C
O

Na:
45.98 g/mol
100%  43.38% Na
105.99 g/mol


C:
O:
12.01 g/mol
100% 
11.33% C
105.99 g/mol
48.00 g/mol
100% 
45.29% O
105.99 g/mol
Total = 100.00%
Converting Mass Percent to an
Empirical Formula (Mole Ratio)

Determine the empirical formula of a
compound that contains 58.84% Barium,
13.74% Sulfur and 27.43% Oxygen.
Assume 100 g
58.84% Ba
13.74% S
27.43% O
 58.84 g Ba
 13.74 g S
 27.43 g O
1 mol
Ba : 58.84 g 
 0.4285 mol Ba
137.32 g
1 mol
S : 13.74 g 
 0.4286 mol S
32.06 g
O : 27.43 g 
1 mol
 1.714 mol O
16.00 g
Did somebody ask
for a mole?
Ba : S : O
0.4285 mol : 0.4286 mol : 1.714 mol
0.4285 mol
0.4285 mol
0.4285 mol
1.000 : 1.000 : 4.000
BaSO4
Barium Sulfate