6.7 Empirical Formula
Formulas
Percent composition allows you to calculate the
simple ratio of the atoms in a compound.
• Empirical Formula: formula of a compound that
expresses lowest whole number ratios of atoms
• Molecular Formula: Actual formula of a
compound showing the number of atoms present
Formulas
Examples:
Molecular
C8H18

C6H12O6

Empirical
Formulas
Examples:
Molecular
C8H18

C6H12O6

Empirical
C4H9
Formulas
Examples:
Molecular
C8H18

C6H12O6

Empirical
C4H9
CH2O
Formulas
H2O2. Empirical or molecular?
MOLECULAR!!!!
It can be reduced to HO
HO = Empirical formula
Example
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
Al: 4.151g
and
O: 3.692g
Example
2. Convert mass to moles
nAl = m/MM
= 4.151/26.98
= 0.1539 mol Al
nO = m/MM
= 3.692/16.00
= 0.2308 mol O
Example
3. Find ratio by dividing each element by the
smallest amount
0.1539 moles Al = 1.000 mol Al
0.1539
0.2308 moles O = 1.500 mol O
0.1539
4. Multiply by a factor to get whole numbers
O = 1.500 x 2 = 3
Al = 1.000 x 2 = 2
therefore,
Al2O3
Example
1. 4.550 g of Co reacts with 5.475 g Cl to form a
binary compound. Determine the empirical
formula.
2. 2.000 g of iron metal is heated in air. It reacts
with oxygen to achieve a final mass of 2.573
g. Determine the empirical formula.
Example
3. 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.
Exit Ticket
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.
Homework
Read
section 6.7
Questions
p. 292 #1
p. 293 #2-9