Simplest (Empirical) and
Molecular Formulas
Molecular Formula
- shows the actual number of atoms
Example: C6H12O6
Simplest Formula
- shows the ratio between atoms
Example: CH2O
Given that a compound contains 12.7%
C, 2.1% H and 85.2% Br, calculate its
simplest (empirical) formula
Step 1:
Assume you have 100 g of the substance therefore,
mass of C = 12.7 g
mH = 2.1 g
mBr = 85.2g
MolarMass=12.01g/mol MH = 1.01 g/mol
MBr = 79.90
Step 2:
Calculate the number of moles of each using n= m/M
nC = 12.7g
nH = 2.1g
nBr = 85.2g
12.01
= 1.06 mol
1.01
= 2.1 mol
79.90
= 1.07 mol
Step 3: Ratio Calculation
Divide by the smallest number of moles to
figure out the ratio between the atoms.
2.1 mol
1.07 mol
1.06 mol
C=
H = 1.06 mol
Br = 1.06 mol
1.06 mol
C=1
H = 1.98
Br = 1.01
Therefore the simplest formula is CH2Br
Calculate the molecular formula for the
compound in the previous example if its
molar mass is 190 g/mol
Step 1. Calculate the molar mass for
the empirical formula, CH2Br.
M
CH2Br
= 12.01 + 2(1.01) + 79.90
= 93.93 g/mol
Step 2. Divide the molar mass by the
simplest (empirical) formula molar mass.
Molar mass
Simplest formula molar mass
190 g/mol
93.93 g/mol
=
=2
Step 3. Multiply this number by the empirical
formula.
2 x CH2Br
Therefore, the molecular formula is C2H4Br2
Example 2: What is the empirical
formula of a compound that contains
69.9 g Fe, and 30.1 g O?
Step 1: Assume the mass is 100 g.
Step 2: Calculate the # of moles of each
element using n=m/M
mFe = 69.9g
mO = 30.1 g
Mfe = 55.85 g/mol
MO = 16.00 g/mol
nFe = 1.25 mol
nO = 1.88 mol
Step 3: Find the ratio by dividing by the
smallest # of moles.
Fe = 1.25 mol
1.25 mol
O=
1.88 mol
1.25 mol
Fe = 1 x 2
O = 1.5 x 2
=2
=3
In this case, multiply by a factor (2)to get a
whole number ratio.
Step 4: Use the simplest whole number ratio:
Fe2O3
Try these:
p. 209, 211, 218:
#11, 14, 15, 16, 18, 19, 20
p. 214
# 2, 6
p. 230
#12, 17, 18a
Answers: page 231