#### Transcript Empirical & Molecular Formulas

```Empirical & Molecular Formulas
Chapter 3
Mathematical Methods
STEP 1
Obtain in the laboratory, the number of grams
OR the weight percentage of each element in
the compound. This can be found by:
• breaking down the existing compound into
its elements
• building the compound from the elements
STEP 2
• Determine the number of moles of each
element in the compound.
• Use dimensional analysis with the atomic
mass of the element.
STEP 3
• Find the simplest whole number ratio of the
moles of each element. Since each mole
contains the same number of atoms, the
simplest whole number ratio of the moles is
also the simplest whole number ratio of the
atoms of each element.
• To do this, take all the mole values and divide
them by the SMALLEST one
• The answers are the subscripts in the
empirical formula
.05 Rule
• After step #3, if the value is within .05 of a
whole number (+ 0.05 or - 0.05), then the value
may be rounded to that whole number.
• The values used in these problems are obtained
by experimentation. The 0.05 rule allows for
experimental error.
IF the application of the .05 Rule does not
produce whole numbers, then ALL of the
results of step 3 must be multiplied by the
same smallest integer that WILL produce
values that can be rounded to whole numbers
by the .05 Rule.
EXAMPLE: A compound is found to contain
72.3% Fe and 27.7% O by weight. Calculate
the empirical formula.
STEP 1 In 100 g of compound there would be
72.3 g Fe and 27.7 g O
1 mole Fe
72.3g Fe X —————— = 1.296 mole Fe
55.8 g Fe
1.296 mole
= 1.000
1.296 mole
1 mole O
27.7g O
X ——————
16.0 g O
X 3 = 3.00 = 3
Fe3O4
= 1.731 mole O
1.731 mole =1.336 X 3 = 4.01 = 4
1.296 mole
Example of a hydrated compound
CaSO4  7 H2O
• Compounds with molecules of water held in
their crystal structure
• Very common in nature
• Water can be removed by heating, leaving
behind what is called the anhydrous
compound (CaSO4 for the above example)
• Naming -- the following is tacked on the name obtained
from the ions
H2O monohydrate
2 H2O
dihydrate
3 H2O
trihydrate
4 H2O
tetrahydrate
5 H2O
pentahydrate
6 H2O
hexahydrate
7 H2O
heptahydrate
8 H2O
octahydrate
9 H2O
nonahydrate
10 H2O
dekahydrate
CaSO4  7 H2O -- named as
calcium sulfate heptahydrate
Finding the empirical formula of a
hydrate
• Find the empirical formula of a hydrate of CaSO4
hydrate that is 28.5% H2O
• To solve this problem, find the simplest mole
ratio between the anhydrous part of the
compound (CaSO4) and the water (H2O)
• H2O =28.5 %
CaSO4 =71.5 %
(100% - 28.5%)
1 mole CaSO4
71.5 g CaSO4 X ------------------- = .5250 mole CaSO4
136.2 g CaSO4
.
(molar mass of CaSO4)
.5250
------- = 1.00 = 1
.5250
1 mole H2O
28.5 g H2O X ----------------- = 1.583 mole H2O
18.0 g H2O
•When you are finding formulas of
1.583
hydrates they ALWAYS come out
------- = 3.01 = 3
even!
.5250
CaSO  3 H O
4
2
Molecular Formula
• A molecular formula tells the actual number
of atoms of each element in a molecule.
• It is a multiple of the empirical formula.
Molecular Formula Problem:
The compound with a molar mass of 171.0
g/mole that contains 14.0% carbon, 41.5%
chlorine, and 44.4% fluorine is a
chlorofluorocarbon (CFC) that was once used
as a refrigerant, but is now on the list of
chemicals known to be ozone depleting.
What is the molecular formula of this
compound?
1 𝑚𝑜𝑙 𝐶
14.0 𝑔 𝐶 ∗
= 1.167
12.0 𝑔 𝐶
1.167
= 1.000
1.167
1 𝑚𝑜𝑙 𝐶𝑙
41.5 𝑔 𝐶𝑙 ∗
= 1.169
35.5 𝑔 𝐶𝑙
1.169
= 1.002
1.167
1 𝑚𝑜𝑙 𝐹
44.4 𝑔 𝐹 ∗
= 2.337
19.0 𝑔 𝐹
2.337
= 2.003
1.167
Empirical formula =CClF2
Molar Mass of Empirical Formula:
C: 1(12.0) = 12.0
Cl: 1(35.5) = 35.5
F: 2(19.0) = 38.0
85.5
𝑀𝑜𝑙𝑎𝑟 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑒 (𝑖𝑛 𝑝𝑟𝑜𝑏) 171.0
=
=2
𝑀𝑜𝑙𝑎𝑟 𝑀𝑎𝑠𝑠 𝑜𝑓 𝐸𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
85.5
(CClF2)2=C2Cl2F4
Combustion Analysis
A 3.489g sample of a compound containing C, H,
and O yields 7.832 g of CO2, and 1.922g of water
upon combustion. What is the simplest formula
of the compound?
Since it’s combustion, you know
CxHyOz + O2  CO2 + H2O
All the C goes into the CO2
All the H goes into the H2O
(our job is to find x, y, and z!)
• First find the amounts of C, H, and O
Using the MM of CO2:
12.0 𝑔 𝐶
7.832 𝑔 𝐶𝑂2 ∗
= 2.136 𝑔 𝐶
44.0 𝑔 𝐶𝑂2
Using the MM of H2O:
1.922 𝑔 𝐻2 𝑂 ∗
2.02 𝑔 𝐻
18.0 𝑔 𝐻2 𝑂
= 0.2157 𝑔 𝐻
2.136 𝑔 𝐶 + 0.2157𝑔 𝐻 + 𝑥 𝑔 𝑂 = 3.489 𝑔 𝑡𝑜𝑡𝑎𝑙
x=1.137 g O
1 𝑚𝑜𝑙 𝐶
2.136 𝑔 𝐶 ∗
= 0.1780 𝑚𝑜𝑙 𝐶
12.0 𝑔 𝐶
0.1780
=2.520
0.07063
1 𝑚𝑜𝑙 𝐻
0.2157 𝑔 𝐻 ∗
= 0.2136 𝑚𝑜𝑙 𝐻
1.01 𝑔 𝐻
0.2136
0.07063
X2≈5
=3.024 X 2 ≈ 6
1 𝑚𝑜𝑙 𝑂
1.137 𝑔 𝑂 ∗
= 0.07063 𝑚𝑜𝑙 𝑂0.07063
16.0 𝑔 𝑂
=1.000 X 2 = 2
0.07063
C5H6O2
```