Empirical & Molecular Formulas

Chapter 3

Mathematical Methods

STEP 1Obtain 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

72.3g Fe 1 mole FeX —————— 55.8 g Fe

= 1.296 mole Fe

27.7g O 1 mole O X —————— 16.0 g O

= 1.731 mole O

1.296 mole 1.296 mole

1.731 mole 1.296 mole

= 1.000

=1.336

X 3 = 3.00 = 3

X 3 = 4.01 = 4

Fe3O4

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 ionsH2O monohydrate 2 H2O dihydrate

3 H2O trihydrate 4 H2O tetrahydrate5 H2O pentahydrate 6 H2O hexahydrate7 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)

1 mole H2O 28.5 g H2O X ----------------- = 1.583 mole H2O 18.0 g H2O

.5250 ------- = 1.00 = 1 .5250

1.583------- = 3.01 = 3 .5250CaSO4 3 H2O

•When you are finding formulas of hydrates they ALWAYS come out even!

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?

14.0𝑔𝐶∗ 1𝑚𝑜𝑙𝐶12.0𝑔𝐶=1.167

41.5𝑔𝐶𝑙∗ 1𝑚𝑜𝑙𝐶𝑙35.5𝑔𝐶𝑙=1.169

44.4𝑔𝐹∗ 1𝑚𝑜𝑙𝐹19.0 𝑔𝐹=2.337

1.1671.167=1.000

1.1691.167=1.002

2.3371.167=2.003

Empirical formula =CClF2

Molar Mass of Empirical Formula:C: 1(12.0) = 12.0

Cl: 1(35.5) = 35.5F: 2(19.0) = 38.0

85.5

𝑀𝑜𝑙𝑎𝑟 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑒( 𝑖𝑛𝑝𝑟𝑜𝑏)𝑀𝑜𝑙𝑎𝑟 𝑀𝑎𝑠𝑠 𝑜𝑓 𝐸𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎=171.0

85.5=2

(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 knowCxHyOz + O2 CO2 + H2OAll 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 OUsing the MM of CO2:

Using the MM of H2O:

7.832𝑔𝐶𝑂2∗12.0𝑔𝐶

44.0𝑔𝐶𝑂2=2.136𝑔𝐶

x=1.137 g O

1.137𝑔𝑂∗ 1𝑚𝑜𝑙𝑂16.0𝑔𝑂=0.07063𝑚𝑜𝑙𝑂

2 .136𝑔𝐶∗ 1𝑚𝑜𝑙𝐶12.0𝑔𝐶=0.1780𝑚𝑜𝑙𝐶

0 .2157𝑔𝐻∗ 1𝑚𝑜𝑙𝐻1.01𝑔𝐻 =0.2136𝑚𝑜𝑙𝐻

=2.520

=3.024

=1.000

X 2 ≈ 5

X 2 ≈ 6

X 2 = 2

C5H6O2