Story of a curious mole
Overview
In this lesson, you will learn about
• Empirical and Molecular Formula
and
Little Mole
Professor Mole
How do I know that
water is H2O and
not…er..H2O2 for
example.
Because scientists do
experiments in the lab in
Phew..I
am
getting the
order to
determine
thirsty.
I need
H2O.
empirical
formula
and
then derive the
molecular formula. See..
There are different kinds of chemical formulae…
Structural Formula
•shows how the atoms are joined in the molecule.
In Ethane, each Carbon atom is linked to 3 Hydrogen atoms and a Carbon atom.
Molecular Formula
•shows the actual number and kinds of atoms present.
In Ethane, there are 2 Carbon atoms and 6 Hydrogen atoms.
Hence, the molecular formula of Ethane is C2H6.
Empirical Formula
• shows the simplest whole number ratio of the atoms present.
The empirical formula of Ethane is CH3.
Empirical Formula
• The empirical formula of a compound shows the simplest
whole number ratio of the atoms present.
• It is the first step to writing the molecular formula and is used to
find the mass of a given compound in the sample.
• The empirical formula is determined by experimental results.
Little Mole has just learnt that
magnesium oxide, an ionic compound,
is used as refractory material.
It’s not easy dude.
You’ve gotta do an
experiment in the lab
first. Come let me show
you how.
Prof! Prof! Prof! Teach me
how to find the empirical
Awesome!
You are way
formula of magnesium
too
cool!
oxide!
I wanna go school
and show off to my
friends.
Experiment: Determining formula of Magnesium oxide
• To find the formula of magnesium oxide,
there are two information we need.
– Mass of magnesium in magnesium oxide
– Mass of oxygen in magnesium oxide
Procedure:
• Burn 1.20g of Mg coil in a clean, dry crucible until the
ribbon become a whitish ash.
This is Wicked!=D
Hehe..
Experimental Results
• The following masses were recorded.
– Mass of crucible + lid = 26.52g
– Mass of crucible + lid + magnesium = 27.72g
– Mass of crucible + lid + magnesium oxide = 28.52g
Experimental Results
• The following masses were recorded.
– Mass of crucible + lid = 26.52g
– Mass of crucible + lid + magnesium = 27.72g
– Mass of crucible + lid + magnesium oxide = 28.52g
I know! I know! So that
I know the mass of
Magnesium that I used.
So 27.72 – 26.52
….means I used 1.20g
of Magnesium.
Think about it, Little Mole.
Why did I have to calculate
the first two masses?
Experimental Results
• The following masses were recorded.
– Mass of crucible + lid = 26.52g
– Mass of crucible + lid + magnesium = 27.72g
– Mass of crucible + lid + magnesium oxide = 28.52g
Hmm.. You got me
there. I dunno.
Good! Now, how do I
calculate the mass of
oxygen used?
Experimental Results
• The following masses were recorded.
– Mass of crucible + lid = 26.52g
– Mass of crucible + lid + magnesium = 27.72g
– Mass of crucible + lid + magnesium oxide = 28.52g
Ah…I see…
The mass of oxygen is the
difference of the last two
masses! So the amount of
oxygen reacted is 28.52 – 27.72g
which is 0.80g.
That is……………
– Mass of crucible + lid = 26.52g
– Mass of crucible + lid + magnesium = 27.72g
Mass of Magnesium
= 27.72 – 26.52
= 1.20g
Cool! Now
how+ do
– Mass of crucible
+ lid
magnesium oxide = 28.52g
I find the empirical
formula?
Mass of
Oxygen reacted
= 28.52 – 27.72
= 0.80g
Calculating Empirical Formula
Step 1: Write down the mass of each element.
Step 2: Write down the molar mass of each element.
Step 3: Calculate the number of mole. (Recall: no. of mol. = mass / molar mass)
Step 4: Divide by the smallest number to obtain a whole number ratio.
Mass (g)
Molar mass (g/mol)
No. of mol
Ratio
Mg
O
1.20
0.80
24
16
1.20/24
= 0.05
0.80/16
= 0.05
0.05/0.05
=1
0.05/0.05
=1
Since the ratio of Mg: O is 1:1, the empirical formula is MgO.
0.05 mole of Magnesium combines with 0.05 mole of Oxygen, right?
That means 1 atom of Magnesium combines with 1 atom of Oxygen.
Therefore, the empirical formula is MgO.
IThis
don’t
get
How I’m
is all
soit.
cheem.
comeI have
you could
go
glad
you around
to
teach
me=Dof moles
from
number
to ratio directly?
Another example
Find the empirical formula of a compound consisting 2.8 g of iron
combined with 1.2 g of oxygen.
Step 1: Write down the mass of each element.
Step 2: Write down the molar mass of each element.
Step 3: Calculate the number of moles. ( Recall: no. of mol. = mass / molar mass)
Step 4: Divide by the smallest number to obtain a whole number ratio.
Mass (g)
Molar mass (g/mol)
No. of mol
Ratio
Fe
O
2.80
1.20
56
16
2.80/56
= 0.05
1.20/16
= 0.075
0.05/0.05
=1
0.075/0.05
= 1.5
Since the ratio of Fe: O is 2:3, the empirical formula is Fe2O3.
That’s Mole,
right! You
alsoyou
have to
Little
have
include workings
forabout
the final
noticed
anything
step for number of moles of
the
way
we present
each
element
and theirour
answers?
simplest ratio.
Hmm..You always
present it in a
table…..and include
units of g for mass and
g/mol for molar mass.
Find the empirical formula of a
compound consisting of 2.8 g of
When your number’s close to
iron combined with 1.2 g of oxygen. See.
What
happens
a whole
number.
If you when
have
You
multiply
when
you
Ihave
multiply
2 toof
that
ratio
1.9
forratios
example,
then
will
1.5you
and
round
up it
to becomes
2. Likewise,2:3
if
of
1:1.5?
1.25
in 1.001,
order
toyou
getround
a
you get
then
right?
down tonumber.
1. It all comes with
whole
practice, my child.
Wait! I am
When
Ahh..I
do
see..All
I round
this
upisIthe
or
Err...ok..So
how
do
confused..You
said
down
starting
then?
toFe:O
make
know
when
to multiply?
ratio
of
issense.
1:1.5
right? How come it
became Fe2O3?
An oxide of sulfur contains 40%
sulfur and 60% oxygen. Find its
empirical formula.
60g! You
rock How am I
*Gulps*
Huh??!
ProfessortoMole!
What
supposed
start
doing
Err...40g?
would
I do
without
this
question?
you?
And when you have
Simple. Assume that you
60% oxygen, how much
have 100g of the sample.
oxygen
do you have in
So if you have 40%
100g
ofhow
themuch
sample?
sulfur,
sulfur
do you have in 100g?
An oxide of sulfur contains 40% sulfur and 60% oxygen by mass.
its empirical formula.
This is how you present…
Find
Assuming 100g of the compound,
40% x 100g = 40g Sulfur
60% x 100g = 60g oxygen
the compound will contain 40g Sulfur and 60g oxygen.
S
O
Mass (g)
40
60
Molar mass (g/mol)
32
16
40/32 =
1.25
60/16 =
3.75
1.25/1.25
=1
3.75/1.25
=3
No. of mol
Ratio
Since the ratio of S : O is 1:3, the empirical formula is SO3.
Let’s look next at:
Molecular Formula
Molecular Formula of Ethane is C2H6
Ball- and- stick model
H
H
H
C
C
H
Structural Formula
H
H
• The ratio of the numbers of carbon and hydrogen atoms is 1 : 3.
• The empirical formula of ethane is CH3.
• From the ball-and-stick model, we can tell that the molecular
formula is a multiple of the empirical formula.
•The molecular formula of a compound shows
the actual number and kinds of atoms present.
Molecular
Formula
Derived
from
Empirical
Formula
• If Empirical formula = AxBy ,
molecular formula = (AxBy)n where n = 1,2,3 etc
Example…
The empirical formula of ethane is CH3. Given that the
relative molecular mass of ethane is 30, what is the
molecular formula?
Let the molecular formula of ethane be CnH3n
Mr of ethane = (n x 12) + (3n x 1) = 15n
I.e.
15 n
n
= 30
= 30 ÷ 15
=2
Hence, the molecular formula of ethane is
C12H32= C2H6
Another Example…
The empirical formula of a compound is H2CO2. Given its
relative molecular mass is 46, find the molecular formula.
Let the molecular formula of the compound be H2nCnO2n.
Mr of compound = (2n x 1) + (n x 12) + (2n x 16) = 46n
i.e.
46 n = 46
n = 46 ÷ 46
=1
Hence, the molecular formula of the compound is
H21C11O21= H2CO2
Think Time!
Is this true?
“The empirical formula is always simpler than
the molecular formula of a compound”.
NO!
In compounds such as MgO and NaCl, the empirical formula
is the same as the molecular formula.
Empirical formula of an ionic compound is also its molecular
formula.
Think Time!
Is this true?
“The empirical formula for different
compounds are different.”
NO!
The empirical formula of ethane (C2H4) and propane (C3H6)
are the same.
But what’s the point of
Then the
knowing
theempirical
empirical
formula
is worked out?
formula?
To
the formula
of
It
is find
useful
Organic
That’s
right!inMy
mere
the new substance,
a a
Chemistry.
Everyday,
presence must
be making
sample
is analyzed
to
new
compound
is you
you smarter.
Then
obtaindiscovered
the mass or or
either
work out
the Mr and
percentage composition
made
in the
lab.
then the
molecular
of each element in the
formula.
compound
Certificate of Completion
This is to certify that
Little Mole
has successfully completed the e-learning module on
empirical & molecular formula.
Presented by
Professor Mole
Professor Mole
Prof, it looks like I’m
done with learning
empirical formula.
Yup. You’ve successfully
graduated. Don’t forget
to practise though. Do
the questions in
Worksheet 2 and the
handout.