South Pasadena • Honors Chemistry
Name
2 • Elements and Compounds
Period
2.5
NOTES
Chemical Formulas
 Empirical Formula
Mole ratio of the elements in a compound, in
simplest form. We have the empirical formula
because experimentally, we cannot precisely
determine the composition of a compound.

Molecular Formula
The actual composition of a compound.
Example: Glucose is C6H12O6; its empirical
formula is CH2O.
– CHEMICAL
Date
ANALYSIS
Example: A compound containing only N and O is
69.6% O by mass. Find the empirical formula of this
compound. If its molar mass is 92.011 g/mol, find its
molecular formula.
Assume we have 100 g of the compound.
1 mol N 
30.4 g N 
= 2.17 mol N
14.01 g N
1 mol O
69.6 g O16.00 g O = 4.35 mol O


N2.17/2.17 O4.35/2.17  NO2
The empirical formula has a mass of:
1(14.01) + 2(16.00) = 46.01 g/mol
Molecular Formula = 92.011 g = 2
46.01 g
 Empirical Formula 
Molecular Formula = 2(NO2) = N2O4
Equations of Formation and Decomposition for Binary Compounds
Equation of Formation (also “Combination” or “Synthesis”) Write the barebones compound and elements
Element + Element  Binary Compound
first. Remember:
 The diatomic elements.
Equation of Decomposition
 Liquids: Hg and Br2
Binary Compound  Element + Element
 Gases: Noble gases, H2, N2, O2, F2, Cl2
 Solids: All other elements.
Examples: Write the equations of formation for and decomposition of the following compounds. Include states.
Compound
Equation of Formation
Equation of Decomposition

N2O (g)
2 N2 (g) + O2 (g)  2 N2O (g)
2 N2O (g) 2 N2 (g) + O2 (g)

Al2S3 (s)
2 Al (s) + 3 S (s)  Al2S3 (s)
Al2S3 (s)  2 Al (s) + 3 S (s)

CaO (s)
2 Ca (s) + O2 (g)  2 CaO (s)
2 CaO (s)  2 Ca (s) + O2 (g)

NH3 (g)
N2 (g) + 3 H2 (g)  2 NH3 (g)
2 NH3 (g)  N2 (g) + 3 H2 (g)

HCl (g)
H2 (g) + Cl2 (g)  2 HCl (g)
2 HCl (g)  H2 (g) + Cl2 (g)
Chemical Analysis
Example 2: The decomposition of a compound containing mercury and oxygen produced 130. g Hg (ℓ) and 10.3 g
O2 (g). What is the empirical formula for the compound?
1 mol Hg
130. g Hg 200.6 g Hg = 0.648 mol Hg


1 mol O2
2 mol O
10.3 g O2 32.00 g O  1 mol O  = 0.644 mol O

2 
2
Hg0.648/0.644O0.644/0.644 = HgO
Example 3: Ethane is a gaseous compound containing carbon and hydrogen. 1.69 g of H2 (g) was reacted with
excess carbon to form 8.40 g ethane.
 Find the empirical formula for the compound.
 Calculate the percent composition of each element
1
mol
H
2
mol
H
in ethane.
2 
 = 1.68 mol H
1.69 g H2 
2.016 g H2 1 mol H2
8.40 − 1.69 g
%C =
× 100% = 79.9% C
1
mol
C
8.40 g
(8.40 g – 1.69 g) C = 6.71 g C 12.01 g C


1.69 g
%H =
× 100% = 20.1% H
= 0.559 mol C
8.40 g
C0.559/0.559H1.68/0.559 = CH3 (Or use % composition)

If the molar mass of ethane is 30.07 g/mol, find
the molecular formula of the compound.
MF
30.07
2
EF = 12.01 + 3(1.008) = 1
Molecular Formula = 2(CH3) = C2H6

What is the volume of 8.40 g ethane at STP?
1 mol 22.4 L
8.40 g 30.07 g  1 mol  = 6.26 L




Write the equation of formation for ethane.
2 C (s) + 3 H2 (g)  C2H6 (g)