Empirical and
Molecular
Formulas
How to find out what an unknown
compound is
A chemist obtains a new product

What is the formula for the compound?

First step – determine constituent elements
and their amounts

This info can be used to determine chemical
formula
Formula of a compound
represents relative numbers
of atoms present

E.g. “CO2” tells us that in a molecule of this
compound there is 1 carbon atom to every 2
oxygen atoms

To determine the formula of a substance we
need to count the atoms – we can do this by
weighing
An example

You have a compound that you know
contains only Carbon, Hydrogen and
Oxygen.

You have a 0.2015 g sample

Analysis shows you have 0.0806 g C,
0.01353 g H, and 0.1074 g O

We can convert these masses to moles, and
then moles to atoms using dimensional
analysis…..
Determining # of moles of
elements in unknown
compound
Determining number of atoms
of unknown compound
To summarize up to this point

We have .00671 moles of Carbon

We have .01342 moles of Hydrogen

We have .006713 moles of Oxygen

Amount of Carbon = Amount of Oxygen

.01342/.006713 = 2, so we have twice as
much Hydrogen as we have of Oxygen and
as we have of Carbon

We have a ratio of 1:2:1
We know the ratio of
elements

1 Carbon : 2 Hydrogens : 1 Oxygen

We can write this as CH2O

Is this the molecular formula? Maybe…but the
molecule might also have 2 Carbons, 4
Hydrogens and 2 Oxygens, or it might have 16
Carbons, 32 Hydrogens and 16 Oxygens

We have found the EMPIRICAL FORMULA – a
formula that represents the ratio of elements in
a compound.

This is also called “simplest formula” since it is
smallest whole-number ratio of elements in the
compound
Empirical Formula Vs
Molecular Formula

Empirical formula gives relative numbers of
atoms e.g. CH2O

Molecular formula gives the actual numbers
of atoms e.g. C6H12O6

C6H12O6 = (CH2O)6
Find the Empirical Formula
Empirical Formula of Benzene = CH
Empirical Formula of Dioxin = C6H2Cl2O
How to calculate empirical
formula

We have 0.2636 grams of nickel. We heat it
in the presence of oxygen to produce 0.3354
grams of a nickel oxide. What is the formula
of the compound we made?

First - what is the mass of oxygen that
reacted with the copper?

Mass of copper oxide – Mass of copper =
mass of oxygen

0.3354 g copper oxide – 0.2636 g copper =
0.0718 g oxygen

Next – Find number atoms involved…
Find number of atoms in the
compound
•Mole quantities represent number of atoms
•We have same number of moles of oxygen as of nickel
•Empirical formula will be NiO
Another example

You have a metal oxide made by reacting 4.151g Al with 3.692 g O. What
is the empirical formula?

Atomic mass Al = 26.98 g/mol

Atomic mass O = 16.00 g/mol

Need to know relative numbers of atoms, so need to convert grams to
moles and then find the whole number ratio of atoms
Find whole number ratios
Do this by dividing both numbers by
the smallest of the two. This converts
the smallest number to 1
This is not a whole number ratio – to get to a
whole number, all we have to do is multiply by 2
Finally we have found our
empirical formula
Al2O3
Summary
If you have percent composition your
can also find the empirical formula

You have a compound that is 27% Carbon
and 73% Oxygen by mass. What is the
empirical formula?

Assume you have 100 grams. That
means you would have 27 g C and 73 g O

Convert grams to moles:

27 g C (1 mol/12 g C) = 2.25 mol C

73 g O (1 mol/16 g O) = 4.6 mol O

Divide by smallest # of moles:

4.6 /2.25 = 2.04 (O)

2.25/2.25 = 1 (C)

Empirical formula = CO2
A poem to help
Percent to mass
Mass to mole
Divide by small
Multiply ‘til whole
Find empirical formula from
percent composition

Nylon-6 is a compound that is 63.68 C,
12.38% N, 9.80% H and 14.4% O. Find the
empirical formula
C6H11NO
One more step to find
molecular formula
 To find the molecular
formula of your unknown
compound, you need
another piece of information
– the molar mass of the
compound, in addition to the
percent composition
You have the empirical
formula – what is the
molecular formula?

Empirical formula = P2O5

Molar mass of compound (determined
experimentally) is 238.88 g/mol

P = 30.97 g/mol ; O = 16.00 g/mol

Find empirical formula mass - the molar
mass that the compound would have
based on the empirical formula:

2(30.97g/mol) + 5(16.00 g/mol) =
141.94 g/mol
Finding molecular formula

The molar mass of the compound will be
some multiple of the empirical mass – find
the multiple by dividing molar mass by
empirical mass:

(238.88 g/mol)/(141.94 g/mol) = 1.7

1.7 ≈ 2

Molar mass of unknown is about twice
empirical mass. We will multiply empirical
formula subscripts by 2

Molecular formula = P4O10
One last problem
Homework

Read Ch 6, section 3 (pp. 196 – 208)

Do empirical formula homework sheets

Do molecular formula homework sheets

Monday – we will go over homework and
talk about hydrates

Test Review

Test on Wednesday!!!!

We have 12 class meetings to go until
semester final