Percentage Composition
(by mass...not atoms)
24.305
35.453
Mg
Cl
12
17
magnesium
chlorine
partg
24
% Mg
% = whole x 100
95 g
25.52% Mg
Mg2+
Cl174.48% Cl
MgCl2
It is not 33% Mg and 66% Cl
1 Mg @ 24.305 amu = 24.305 amu
2 Cl @ 35.453 amu = 70.906 amu
95.211 amu
Empirical and Molecular Formulas
A pure compound always consists of the same
elements combined in the same proportions by
weight.
Therefore, we can express molecular
composition as PERCENT BY WEIGHT.
Ethanol, C2H6O
52.13% C
13.15% H
34.72% O
Empirical Formula
Quantitative analysis shows that a compound contains 32.38% sodium,
22.65% sulfur, and 44.99% oxygen.
sodium sulfate
Find the empirical formula of this compound.
 1mol Na 
 =
 23 g Na 
1.408 mol Na / 0.708 mol = 2 Na
32.38% Na
32.38 g Na 
22.65% S
22.65 g S
 1mol S 


32
g
S


= 0.708 mol S / 0.708 mol
=1S
44.99% O
44.99 g O
 1 mol O 


16
g
O


= 2.812 mol O / 0.708 mol
=4O
Step 1) %  g
Step 2) g  mol
Step 3) mol
mol
Na2SO4
Empirical Formula
A sample weighing 250.0 g is analyzed and found to contain the following:
27.38 g
27.38%
1.19%
1.19
g
14.29%
14.29
g
57.14%
57.14
g
Na
sodium
H
hydrogen
C
carbon
O
oxygen
Assume sample is 100 g.
Determine the empirical formula of this compound.
Step 1) convert %  gram
Step 2) gram  moles
  1.1904 mol Na
x mol Na  27.38 g Na1mol Na
/ 1.19 mol = 1 Na
23 g Na 

  1.19 mol H / 1.19 mol = 1 H
x mol H  1.19 g H1mol H

1
g
H


  1.1908 mol C / 1.19 mol = 1 C
x mol C  14.29 g C1mol C

12
g
C



  3.5712 mol O / 1.19 mol = 3 O
x mol O  57.14 g O1mol O

16
g
O


Step 3) mol / mol
NaHCO3
Empirical & Molecular Formula
(contains only hydrogen + carbon)
(~17% hydrogen)
A 175 g hydrocarbon sample is analyzed and found to contain ~83% carbon.
The molar mass of the sample is determined to be 58 g/mol.
Determine the empirical and molecular formula for this sample.
Determine the empirical formula of this compound.
Step 1) convert %  gram
Assume sample is 100 g.
Then, 83 g carbon and 17 g hydrogen.
Step 2) gram  moles
  6.917 mol C
x mol C  83 g C1mol C

/ 6.917 mol = 1 C
12
g
C


  17 mol H / 6.917 mol = 2.5 H
x mol H  17 g H1mol H

1
g
H


(2.4577 H)
2 C @ 12 g = 24 g
5H@ 1g = 5g
29 g
MMempirical = 29 g/mol
CH2.5
C2H5
MMmolecular = 58 g/mol
Step 3) mol / mol
58/29 = 2
Therefore 2(C2H5) = C4H10
butane
Common Mistakes when Calculating
Empirical Formula
Given: Compound consists of 36.3 g Zn and 17.8 g S.
Find: empirical formula
36.3 g Zn
17.8
= 2 Zn
Zn2S
17.8 g S
17.8
= 1S
36.3 g Zn 1 mol Zn
65.4 g Zn
17.8 g S
1 mol S
32.1 g S
1
1
= 0.555 mol Zn
0.555 mol
= 0.555 mol S
0.555 mol
Chemical formula
indicates MOLE ratio,
not GRAM ratio
Zn
ZnS
S
zinc sulfide
Empirical Formula of a Hydrocarbon
burn
in O2
x 1 mol CO2
44.01 g
g CO2
mol CO2
x
2 mol C
1 mol CO2
mol C
mol H
CxHy
g H2 O
x 1 mol H2O
18.02 g
Kotz & Treichel, Chemistry & Chemical Reactivity, 3rd Edition , 1996, page 224
mol H2O
x
2 mol H
1 mol H2O
Empirical
formula
Find the molar mass and percentage composition of zinc acetate
Zn2+ CH3COO1acetate = CH3COO1-
Zn(CH3COO)2
1 Zn @ 65.4 g/mol = 65.4 g / 183.4 g x 100% = 35.6 % Zn
4 C @ 12 g/mol
6 H @ 1 g/mol
= 48 g
= 6g
4 O @ 16 g/mol
= 64 g
Zn(CH3COO)2
183.4 g
/ 183.4 g x 100% = 26.2 % C
/ 183.4 g x 100% = 3.3 % H
/ 183.4 g x 100% = 34.9 % O
A compound is found to be 45.5% Y and 54.5% Cl.
Its molar mass (molecular mass) is 590 g.
Assume a 100 g sample size
a) Find its empirical formula
45.5 g Y
1 mol Y
88.9 g Y
= 0.5118 mol Y / 0.5118 mol = 1 Y
YCl3
54.5 g Cl
1 mol Cl
35.5 g Cl
= 1.535 mol Cl / 0.5118 mol = 3 Cl
1 Y @ 88.9 g/mol = 88.9g
3 Cl @ 35.5 g/mol = 106.5 g
b) Find its molecular formula
590 / 195.4 = 3
3 (YCl3)
YCl3
Y3Cl9
195.4 g
6.02x1023
Molar Mass
Atomic Mass
vs.
2g
H2 = _____
H2 = _______
2 amu
18 g
H2O = _____
H2O = ________
18 amu
120 g
MgSO4 = _____
MgSO4 = ________
120 amu
g
(NH4)3PO4 = 149
_____
(NH4)3PO4 = ________
149 amu
Percentage Composition (by mass)
% =
part
x 100 %
whole
Empirical vs.
(lowest ratio)
Molecular Formula
Empirical Formula



% g
g  mol
mol
mol