[MUSIC]
Lighthouse Scientific
Education presents a lecture
in the Mole series
the topic is Empirical Formula
and it is presented in 2 parts.
A basic part, for those
students who will need it
and, an advanced section.
Material in this lecture
relies on an understanding
of the previous lectures
Atomic Mass and
Molecular Weight
and Percent Composition.
Part 1 of the lecture begins
with a set of definitions.
Specifically, empirical formula
and molecular formula.
A set of rules is offered to
determine empirical formula
and applied with 3
practice problems.
The empirical formula is then
reviewed in the form of a quiz.
For some students this is as
far as they will need to go
with empirical formula.
Part 2 covers a more
advanced topic. It is
making the connection between
molecular formula
and empirical formula.
3 practice problems are
offered to demonstrate
that connection.
Onto part 1 and definitions.
The first definition
is molecular formula
and should be a review.
The molecular formula
is the real ratio of
atoms in a molecule.
Compare that to the
empirical formula.
It is the simplest whole number
ratio of atoms in a molecule.
Sometimes they are the
same thing and sometimes
they are not.
The term empirical,
in this context,
means measured.
That is in contrast to a value
gotten from a theory or logic.
In regards to the empirical
formula it is laboratory
determined mass values
for elements contained in a
known or unknown compound.
At this level of chemistry
students come across
empirical formulas in the form
of homework or test problems
in which the amount or
proportion of each element
in an unknown compound is
given and the simplest whole
number ratio of the elements
is calculated.
For ionic compounds
a formula unit and the
empirical formula are
the same thing since
formula unit is the
simplest whole number ratio
of ions. That is not always
the case with molecules
and molecular formulas.
Consider this table which
compares the molecular
formula of a compound with
its empirical formula as would
be calculated from
empirical data.
Beginning with the
compound hydrogen peroxide.
It has a molecular
formula of H2 O2.
The ratio of atoms in
this compound is 2 to 2.
That ratio can be
reduced to a simpler
whole number ratio
which is 1 to 1.
That is the empirical formula.
If an experiment was
performed on hydrogen
peroxide to determine
the relative masses,
in the form of moles of
the hydrogen and the oxygen,
it would yield a
ratio of 1 to 1.
There is 1 mole of hydrogen
for every 1 mole of oxygen.
We see that 2 to 2 ratio
in the molecular formula
is a whole number
multiple of its
counterpart in the
empirical formula.
In this case the molecular
formula has a ratio that
is two times larger than
the empirical formula.
Oh, and let's not
forget that subscript of 1
is implied, not written.
And then there is ethane, C2H6.
It has a mole ratio of
2 carbons to 6 hydrogens.
That, however, is not
the simplest whole
number ratio. CH3 has the
same mole ratio but uses
smaller whole numbers:
1 carbon to 3 hydrogens.
The ratios in the molecular
and empirical formula
differ by a factor of 2.
An experimental measurement
on an unknown compound
that is actually ethane would
find the masses, in the form
of moles, of carbon and
hydrogen to be in a 1 to 3 ratio.
Without additional information
the molecular formula
of C2H6 could not be gotten.
Measured masses can
only get the lowest whole
number ratio of moles.
That is also the case for
the sugar glucose, C6 H12 O6.
The 6 to 12 to 6 mole ratio
of elements in glucose is not
the simplest
whole number ratio.
That belongs to the 1
to 2 to 1 ratio of CH2O.
The ratio of atoms in glucose
compared to CH2O differ
by a factor of 6.
A closer look at CH2O
shows it to be on a carbon
and a water. CH2O is the
empirical formula for a carbo-
hydrate. Sugars
are carbohydrates.
The next example is diboron
with a molecular formula
of B2 H4.
The 2 to 4 ratio of the
boron to hydrogens is not the
simplest whole number ratio.
That is found with BH2.
The molecular and
empirical formula
differ by a factor of 2.
And to demonstrate that
in some cases the molecular
formula of a compound does
have the simplest whole
number ratio there
is water, H2O.
Its 2 to 1 ratio of
hydrogen to oxygen
is the not reducible.
The empirical formula
and the molecular formula
are the same thing for water.
So we see the simplest
whole number ratio of the
empirical formula may or may
not be a molecular formula
of a compound.
The exact relationship
between empirical formula and
molecular formula is probed
in part 2 of this lecture.
We are ready to discuss
the process of getting
an empirical formula from
empirical data or given data.
Let's keep the definition
of empirical in mind.
Problems involving
empirical formulas will have
measured values for the
elements in the compound.
Since many of the problems
are word problem it is a good
idea to employ basic
problem solving techniques in
addressing them.
That includes setting
the problem up by determining
known and unknown quantities
from the problem and
identifying the type of problem
and finding a relationship
between known and
unknown values.
That being said, the process
or relationship that is needed
for determining empirical
formula has the student being
given the mass or percent
of each element in an
unknown compound.
If the experimental data
is provided as percent,
a total mass of 100 grams
will be assumed.
That will make more sense
when we actually employ
this assumption in a problem.
Using these masses,
determine the
moles for each element.
This will involve the
atom's atomic molar mass.
If we think about it the
ratio of atoms in the compound
are not mass ratios but
count ratios, mole ratios.
Step three, set up
a chemical formula.
Begin by writing the element
symbol of all elements
in the compound and add the
calculated moles as subscripts
to the corresponding element.
These mole subscript values
here will almost never be
whole numbers which
is required for
a chemical formula.
To get the mole subscript
values to be whole numbers
divide each mole value by the
smallest of the mole values.
This will automatically
make the smallest mole value
subscript a 1 since it is
being divided by itself.
The starting masses
are experimentally
determined values.
This division step frequently
produces values that are
not exactly whole numbers or
not exactly common fractions.
Rounding is generally
required and that falls to
the student's judgment.
For example: the result of
the division may be a bit
less than or a bit more
than the whole number 3.
Rounding gets the value to 3
Or the division value is near
the decimal value of a fraction.
In situations where
the division value
does not round
to whole a number
it will be necessary to
multiply the subscripts
by a common factor.
The logic is to find a common
factor that turns the
fractions into whole numbers.
Generally, this will be the
denominator of the fraction.
These steps may initially
seem confusing but a couple
practice problems will show
how they logically produce
empirical formulas that have
the simplest whole number
ratio of atoms.
Practice is imperative here.
Example 1. A lab experiment
on an unknown substance
shows it to contain 80%
copper and 20% sulfur.
What is its empirical formula?
Begin the process of solving
the problem by identifying the
known and unknown values.
Given is that 80% of
the unknown substance
is copper and 20% is sulfur.
The unknown in the problem
is the empirical formula.
Step one of the process for
getting an empirical formula
is determining whether
mass of the element
or percent of the
element will be used.
We've already
established that it is percent
Since it is percent
we will assume
that there is 100 grams
of the unknown sample.
Why is that?
Well, 100 grams relates
directly to 100%.
80% of 100 g is 80 g.
There are 80 g of copper
in the 100 g sample.
Likewise, 20% of 100 g is 20 g.
There are 20 g of
sulfur in a 100 g sample.
By assuming 100 g of
the unknown we can simply
exchange the percent
sign for grams.
80 g plus 20 g is 100 g.
If we were to assume any other
mass for the unknown sample
then we would have
to calculate what
80% of the mass is and
what 20% of that mass is.
100 grams is just the
easiest value to deal with.
But why get grams at all?
Well, the next step
is to convert those
gram values into moles.
This cannot be
done with percent.
Atomic molar mass is in grams
and atomic molar mass is what
is needed to convert
grams into moles.
The 80 grams of copper
need to be converted
into moles of copper.
The conversion factor
must have grams of copper in
the denominator to cancel out
with grams of copper
in the starting units.
We will get to the
actual atomic weight of
copper in a moment.
The 20 grams of sulfur
will need to be converted
into moles of sulfur with
the atomic molar mass as
the conversion factor.
Again, grams of sulfur
will need to be in the
denominator. This lecture
assumes that the student has
practiced conversions with
atomic molar mass
and can follow along.
To get the atomic weights
of copper and sulfur we
bring up the Periodic Table.
Sulfur, atomic number
of 16, has an atomic
molar mass of 32.07 grams.
Copper, atomic number
29 has an atomic molar
mass of 63.55 grams.
Returning to our conversion,
there are 63.55 grams of
copper in one mole of copper.
And, there are 32.07 grams
sulfur in one mole sulfur.
The conversion for copper
is completed by canceling out
units of grams of
copper and taking 80 and
dividing it by 63.55
80 grams of copper is
1.26 moles of copper
Units are also canceled out
with the sulfur and moles of
sulfur is found by dividing 20
by 32.07. 20 grams of sulfur
is 0.624 moles of sulfur.
Conversion from mass into
moles is an essential skill
that should not be neglected.
At this point the percent of
copper and sulfur have been
converted into relative
moles of copper and sulfur.
We are now ready to use those
mole values as a subscript in a
chemical formula.
The chemical formula has an
element symbol for all
elements in the compound.
Cu for copper, S for sulfur.
To the copper symbol
add the subscript 1.26 from
the value of moles of copper.
To the sulfur symbol add
the subscript 0.624 from
the moles of sulfur.
At this point,
there is no problem with
the subscripts not being
whole numbers.
There is no reason
for 80% to yield a
whole number of moles.
Moles will vary in
size dependeng on
the amount of
starting material.
If we assumed 50 grams of
starting material, instead
of 100 grams, these mole
values would be half as large.
To get subscripts
to be whole numbers
each subscript needs to be
divided by the smallest of
the mole subscripts.
Sulfur's 0.624 is smaller
than copper's 1.26.
Divide both of those
subscripts by 0.624.
The mole subscript values
are now in fractional form.
To clear the faction do
the division. For sulfur,
0.624 divided by 0.624
yields a value of 1.
And doing the math for copper,
1.26 divided by 0.624
yields a value of 2.02.
As mentioned earlier,
since the starting percents
are laboratory measurements
the values are not exact.
The numbers will
contain a certain amount of
error. 2.02 is very close to 2.
This level of rounding falls
to the student to conclude.
Now this looks like a
chemical formula. Cu2 S1.
Actually, that 1 is
implied and has to go. Cu2S.
The 2 to 1 mole ratio is the
lowest simple whole number
ratio and is the
empirical formula.
Is this also the
molecular formula?
Well, that can't be
determined from the data.
In part 2 of the lecture
that deals with the
molecular formula we will
explore the process of taking
the empirical formula with a
bit of additional information
and converting it into
a molecular formula.
For now we can lay the ground
work for that exploration by
comparing the formula weight
from the empirical formula
with molecular weight of
potential molecular formulas.
We need to start by getting
the formula weight of Cu2S
That is, we will need
to convert the 2 moles of
copper in Cu2S into
grams of copper and the
1 mole sulfur in Cu2S
into grams of sulfur.
The conversion factor from
moles of copper to grams of
copper is the atomic molar
mass with moles of copper in
the denominator so that moles
cancel out leaving grams.
2 times 63.55 equals
127.10 grams of copper.
For sulfur, the conversion
factor from moles into grams
is the atomic molar mass
with moles of sulfur in the
denominator so that moles
of sulfur cancel out.
The math is trivial,
1 times 32.07 is 32.07.
The formula weight of Cu2S is
127.10 grams plus 32.07 g
for a value of 159.17 g.
The significant figures for this
value is correct since atomic
masses are given
to the hundredths.
The empirical formula is the
simplest whole number ratio
of the atoms in a molecule.
A molecular formula will
retain that ratio but can
have a whole number multiple
of the subscripts.
For instance, doubling the
subscripts would yield a
molecular formula of Cu4 S2.
Twice as many atoms will give
a formula weight that is twice
as big; 318.34 g per mole.
Tripling the subscripts in
the empirical formula would
yield a molecular
formula of Cu6 S3.
A subsequent tripling of the
empirical formula weight would
yield 477.51 grams per mole
The molecular formula has
a whole number multiple
of the subscripts of
the empirical formula.
The molecular weight has
that same multiple of the
empirical formula weight.
Let's try another problem
and see if the pieces fit together
a little smoother this time.
Example 2: You and your
lab partner measure the
composition of an unknown
substance as containing
22.83 grams of carbon,
2.66 grams of hydrogen
and 7.61 grams of oxygen.
What is the empirical
formula of the substance?
Before heading into the
steps for determining
empirical formula we should
first consider the problem
in more detail. What are
the known and unknowns?
We are given 22.83 g of carbon.
Write that down. 2.66
grams of hydrogen.
Write that down.
And 7.61 grams of oxygen.
Write that down.
What is not known
is the empirical formula.
The relationship between
the knowns and unknown
are the steps to
getting the empirical formula.
Begin with getting the
mass of each element.
We are given mass in grams
for the three elements.
Since the given value are in
grams and not percent there
is no need to assume 100 grams.
Assuming 100 grams
here will only
complicate matters.
Now, these know gram
values have to be
converted into moles.
That requires a
conversion factor.
What conversion factor
converts grams of an element
into moles of that element?
It is the atomic molar
mass of that element.
To get the given unit of
grams to cancel out the atomic
molar mass has to have an
orientation that puts grams
in the denominator.
Moles of carbon are found by
canceling units and
dividing 22.83 by 12 .01.
That gives 1.901 moles of
carbon. For the hydrogen,
canceling units and doing
the division produces
2.88 moles of hydrogen.
The 7.61 grams of oxygen
is equivalent to
0.4756 moles of oxygen.
With that, the given values
of grams have been converted
into moles which is
necessary for the next step.
Set up a chemical formula
that uses the element symbols
and the calculated
moles as subscripts.
The element symbols C, H
and O are arranged in the
formula with 1.901 as the
mole subscript for carbon
2.88 the mole subscript
for hydrogen and
0.4756 as the mole
subscript for oxygen.
None of the subscripts
are whole numbers.
And that is expected
here since whole numbers
are produced in the next step.
Divide the subscripts by
the smallest of the mole
subscript values. That starts
to produce whole numbers
Which subscript
is the smallest?
That of course is the
0.4756 of the oxygen.
Divide 0.4756 by 0.4756.
Divide the 2.88 by 0.4756
and the 1.901 by 0.4756.
This dividing by the smallest
value is called normalizing
and it results in the
smallest value becoming a 1.
Any number divided
by itself is 1.
The 1 subscript is also
the smallest possible
value for a subscript
in a chemical formula.
There is no way to simplify
the value any further.
In the end the 1 may not
be the part of the simplest
whole number ratio of the
empirical formula but that
subscript value is certainly
not going to be smaller than 1.
The other two divisions
yield 6.06 for hydrogen and
3.99 for carbon. With the
error inherent in empirical
data the student will often
need to make judgment calls
as to how to round the
mole subscript results.
The rounding does not need
to be to the nearest whole
number as will be the case
here. Rounding can also be to
decimal values that
correspond to common fraction
like .25 for 1/4, .33
for 1/3 or .5 for 1/2.
The confidence to make
correct rounding will only
come with practice.
Rounding the 6.06 subscript
for hydrogen should
not be too challenging.
It is near the whole number
6. And the 3.99 for carbon
is practically 4.
Rounding can be up or down.
Nearest is the best advice.
Since our ratio of subscripts,
4 to 6 to 1, is of all whole
numbers there is no need
to multiple the subscripts
by a common factor.
The next example will
have that requirement.
The 1 on the oxygen should
be removed since it is implied
and the empirical formula
of the unknown substance is
C 4 H 6 O. As a nod to the
second part of the lecture
we will relate empirical
and molecular formulas.
We will get the formula weight
of C 4 H 6 O and compare
it to molecular weights
of compounds that share the
4 to 6 to 1 ratio but do
so with larger numbers.
Since this lecture is part of the
Mole Series of lectures we will
continue to demonstrate
the process of getting
formula weights.
A mole of a compound with
a molecular formula
of C 4 H 6 O will have
4 moles of carbon, 6
moles of hydrogen and
1 mole of oxygen.
These mole values will need
to be converted into gram
values that can be totaled to
a final molecular weight.
The conversion factor
that converts moles to grams
is the atomic molar mass
of the element. The
orientation of the atomic
mass is such that the current
units of moles cancel out.
4 times 12.01 is 48.04.
The atomic molar mass of
hydrogen is also oriented
so that moles cancel out
leaving grams and
giving a mass of 6.048 g.
Finally, oxygen's atomic molar
mass is used to gets oxygen's
contribution to the
formula weight; 16.00 g
And that final value is found
by adding 48.04 g to 6.048 g
and 16.00 grams.
The formula weight of C4 H6 O
is 70.088 grams per mole.
Being that the final step is
addition, getting the
appropriate number of
sig. figs. requires us to note
the starting number with the
least precision, and that is
both that for the carbon and the
oxygen which only
go to the hundredths.
As such the formula weight
can only go to the hundredths
and the weight is
appropriately determined.
With this formula weight we
can consider some possible
molecules that the unknown
substance might actually be.
And those will have whole
number multiples of the
subscripts in the
empirical formula.
For instance multiplying
all subscripts by 2
will get a molecule: C8 H12 O2.
Doubling all the atoms
will produce a doubling
of the molecular weight.
Doubling 70 gives 140.
Tripling the subscripts
in the empirical formula
gets the molecule: C12 H18 O3.
It has a molecular weight
that is 3 times that
of the empirical formal.
Any one of these molecules
could have been the source of
the measured data
listed in the problem.
Masses alone are not sufficient
to specifically identify
the molecule.
A third example
explores a final step
in the determination of
empirical formulas that
we have yet to come across.
Phosphorus and oxygen
can combine in many ways.
A compound was found to
contain 56.5% phosphorus
and 43.5% oxygen.
What is the empirical
formula of this substance?
Before jumping into the
problem it is a good idea to
stop and consider the
problem as a whole.
What are the
known and unknowns?
What type of problem is it?
Well, we are given that
a compound is composed
of 56.5 percent phosphorus
and 43.5 percent oxygen.
What is the unknown
in the problem?
Yes, it is the empirical formula.
Now that we are comfortable
with the problem we can
call up the first step and
it is to get the mass of
each element in grams or as %.
A look at the problem
shows mass in percent.
The second part of the first
step has us assuming 100 g
of the compound or substance
when percent is given.
In practice any starting amount
can be assumed but a 100 g
sample of a substance that
is 56.5 percent phosphorus
will be composed of
56.5 grams of phosphorus.
The 43.5 % oxygen means
that 100 g of the substance
is composed of
43.5 g of oxygen.
When the given values are
in percent, directly convert
percent into grams by
assuming 100 g of the sample
Assuming 100 grams is
an invaluable technique
when dealing with percent.
The next step is to convert
these masses into moles.
The 56.5 g of phosphorus
needs to be converted
into moles of phosphorus.
While the 43.5 g of oxygen
will need to be converted
into moles of oxygen.
The conversion factor is the
atomic molar mass
for the element.
For phosphorus, a check of
the Periodic Table tells us
that its atomic molar
mass is 30.97 g/mole.
The correct orientation
of the conversion factor
has grams in the denominator.
56.5 g of phosphorus is
1.82 moles of phosphorus.
Oxygen's atomic molar mass
is also oriented in the
same manner. 43.5 grams
of oxygen is 2.72
moles of oxygen.
The given percents are
now in relative moles.
Percent has been converted
into count. From here
a chemical formula is set
up using elements from the
problem and the calculated
moles as subscripts.
The two elements in the
problem are phosphorus and
oxygen. The subscript
for phosphorus is 1.82
and the subscript
for oxygen is 2.72.
To begin the process of
getting the subscripts into all
whole numbers start by
dividing subscripts by
the lowest subscript value.
Phosphorous has a subscript
of 1.82 which is lower
than oxygen's 2.72.
Divide both subscripts by 1.82.
The math for phosphorus is
of course 1; anything
divided by itself is 1.
The math for the oxygen
comes out to 1.49.
In the previous examples we
rounded the subscript values
to the nearest whole
number and we did so because
the values were very
close to the whole number.
1.49 does not fit
that description.
Rounding it to a whole number
is not the correct option here.
If there is rounding to be done
it will be to a decimal value
of a common fraction. 1.49
is very close to the fraction
1 and 1/2;
1.49 should therefore be
converted to the decimal 1.5.
That still leaves us with an
issue of getting the decimal
value into a whole number.
Returning to the fourth rule
in solving empirical formulas;
if all the subscripts are
not rounded to whole numbers
multiply each subscript
by a common factor.
The question now becomes
what common factor should
the subscripts be multiply by.
The answer is based on the
subscript, or
subscripts, that are not
close to whole numbers.
In this case it is the 1.50.
What number can 1.50
be multiplied by to
make it a whole number?
2 seems a good value because
the '.5' represents the
fraction, 1/2, where
the denominator is a 2.
We can try it out and,
if it doesn't work out will
simply choose a
different number.
Notice that we are also
going to multiple the
subscript of the
phosphorus by that
same value of 2.
If we did not, the ratio of
1 to 1.5 would be different
from what we calculated.
Staring with the oxygen;
2 times 1.5 is 3.
That is a whole number. That
is good. To prevent the ratio
of 1 to 1.5 from becoming 1
to 3 we multiple the subscript
of phosphorus by
2 making it a 2.
2 to 3 is the same
ratio as 1 to 1.5
but is all in whole
numbers and can be
considered the empirical
formula of the compound.
Selection of the
appropriate value to multiple
the subscripts
by is usually the
most confusing
part of the process.
A couple of additional
examples of fractions in
decimal form might
help clarify the matter.
To get a subscript value
that ends in .25 (or something
near that) to a whole number
multiple all the
subscripts by 4.
0.25 is the decimal 1/4.
Multiplying this subscript
by the denominator of the
fraction (4) will give
convert the fraction to 1.
To get a subscript value
that ends in .33, which is the
decimal form of 1/3, into
a whole number a common
multiple of 3 can be applied.
Multiplying 1/3 by 3 gives
the whole number of 1.
If the subscript value ends
in 67, which is the decimal
form of 2/3, multiplied by
the denominator 3 to clear
the fraction and give a whole
number value of
the numerator, 2.
And finally, if the subscript
value ends in .75, which is
the decimal form of 3/4,
turn that into a whole number
by multiply by 4.
And remember, all of the
subscripts in the formula need
to be multiplied by the
common factor to keep the
ratio that was
determined by the masses.
This is a good time to review
what has been covered so
far and we will do so
in a quiz. 5 questions.
Question 1. The simplest
whole-number ratio of the
atoms in a molecule is the
empirical formula or
the molecular formula?
The clue here is simplest
whole number ratio.
The molecular formula gives
the type and number of atoms
in a molecule. It has no
restriction as to simplest
whole number ratio. It
is the exact number ratio.
The empirical formula is the
simplest whole number ratio.
That is the ratio they can be
determined from the relative
masses of the elements
in the compound. If the
question also includes
the option of picking 'formula
unit' that too would be correct.
Question 2, which of
the following could be
an empirical formula?
A, H2O B, C2H2 or C, P1 O1.5.
An inspection of these ratios
shows that answer C does
not contain all whole numbers.
It cannot be an
empirical formula.
A look at answer B shows the
subscript ratio to be 2 to 2.
Well, that is a whole number
ratio but it is not simplest.
2 to 2 can be expressed as
1 to 1 which is simpler.
The answer is A because the
ratio of hydrogen and
oxygen in water is 2 to 1.
Those are whole numbers
and they cannot be reduced
to a simpler ratio.
Question 3, True or false. Even
if gram values are given for
the elements assume a total
of 100 grams when solving for
the empirical formula.
The reason that the amount
of each element in the unknown
has to be given in, or
converted to, mass is that
the masses are themselves
converted into moles.
If the given values are
already in grams then move on
to the next step of
converting those into moles.
It is only with element values
that are given in percent that
an assumption of 100 grams is
made so that there is a direct
conversion from
percent into grams.
So, False.
Question 4, the purpose of
dividing each mole value by
the smallest mole value is to:
get the mole values
to be whole numbers
or is it to get the
smallest mole value to be 1?
An illustration of this
process can be taken
from the copper
and sulfur example.
The division here
produces a 2 to 1 ratio.
It seems that both answers
A and B are correct.
However, if the illustration
is replaced by what we found
in example 3, where the
division produced a 1 to 1.5
ratio then answer A
is not always correct.
Answer B is always correct.
Regardless of the need to
multiply the subscripts by
a common value, the division
step puts the smallest
mole value as the
simplest non-zero,
positive whole
number which is 1.
Question 5, what molecular
formula can come from the
empirical formula CH2O?
is it CH2O? C2H4O2?
C4H8O4? C6H12O6?
or is it all of the above?
This question is really a
lead-in to Part 2 of the
lecture; getting a molecular
formula from an empirical
formula. We've seen that the
molecular formula is a
whole number multiple of
the empirical formula.
With that we see that
answer is A is the empirical
formula. The empirical formula
can be a molecular formula so
A is correct. Answer B has the
molecular formula that would
come from multiplying the
subscripts in the empirical
formula by 2. Answer C is the
molecular formula when
multiplying subscripts by 3.
Answer D when multiplying by 4.
Therefore, answer E is
correct because all of
these are possible
molecular formulas.
But which one is the
correct molecular formula?
For those who only need what
we have covered so far go to
the Recap and then visit the
Concept and Definition page.
Part 2 of the lecture is
concerned with molecular
formulas. Specifically
it is concern with how to
get the molecular formula
from the empirical formula.
The molecular formula
can be gotten from
the empirical formula if:
1, the empirical formula is
obtained (such as was done in
the three examples of Part 1).
2, the empirical formula
weight is obtained.
This was also done for
2 of the 3 compounds in
our examples. There is also
one new piece of information
needed and it has
to be provided.
It is the compounds
molecular weight.
Without this piece of
information the problem stops
at empirical formula.
Of course, if the two weight
values are the same then
the empirical formula is the
molecular formula. With these
three pieces of information
in hand the molecular formula
can be found by following a
couple of steps.
Begin by dividing weights.
Divide molecular weight by
empirical weight (that is the
big number by small
number): this should give a
whole number or a
value that is readily
reduced to a whole number.
Once this whole number
value is determined
multiply subscripts;
multiple each
subscript value in the
empirical formula by
the whole number value
obtained in step 1.
That's it for getting
the molecular formula.
It is actually easier than
getting the empirical formula.
We can demonstrate the
process using the empirical
formulas determined in the
3 example problems in Part 1.
Starting with the
copper 2 sulfide.
The empirical formula was
found to be Cu2S and the
calculated formula
weight was 159.17 g/mol.
What we still needed is to
be given the molecular weight
of the compound.
Say it was provided
as 318.31 grams per mole.
The first step is to divide
the molecular weight by the
empirical weight.
The molecular weight is
318.31 grams per mole.
The empirical weight is 159.17.
If the weights are equivalent
then the empirical formula is
the molecular formula.
If not this division will produce
a number larger than 1. If
that is not true then the
fraction needs to be flipped
or the values were incorrectly
put into the calculator.
Note that the grams per mole
in the numerator cancel out
with the grams per
mole in the denominator.
Division gives a
value of 2. No units.
Some minor rounding may be
necessary after the division.
Step 2 is multiplying the
subscript values in the
empirical formula by the
whole number value obtained in
step 1. In this case it is a 2.
The empirical formula is Cu2S.
Multiply copper's subscript
of 2 by the whole number
value of 2 gives a
molecular formula with Cu 4.
Multiplying the implied 1
on the sulfur by the whole
number 2 finishes the
molecular formula with S2.
Cu4S2 is the molecular
formula. Not too bad.
Moving on to the second
example which uses the
solution to the second
problem, the empirical formula
is C4H6O with a formula
weight of 70.09 g/mole.
The new piece of information
needed for this problem is
the molecular weight
of the compound.
That is given, in this
instance, as 205 grams per mole.
Step 1 has us dividing
the molecular weight
of 205 grams per mole
by the empirical weight
of 70.09 grams per mole.
Units cancel and the math
yields a value of 2.92.
Not exactly 3 but it should
not be too big of a leap
to round it to 3. A
whole number is needed
for the next step and if the
math doesn't provide one then
the answer will need to be
rounded. The next step has the
subscripts in the
empirical formula multiplied
by the whole number 3.
That is the 4 on the carbon
in C4H6O is multiplied by 3
to give a molecular formula
that starts with C12.
Multiply the 6 on the
hydrogen by 3 gives H18 and
multiply the implied 1 on
oxygen by 3 finishes the
molecular formula
with O3. C12 H18 O3.
Still the same 4 to 6 to 1
ratio but with larger numbers.
The final example use the
empirical formula of P2O3 and
empirical formula weight
of 109.94 grams per mole.
The molecular weight of
this compound is given here
as 225 grams per mole.
The whole number that we
will need for step 2 is
found by dividing
the 225 grams per mole
of the molecular weight
by the 109.94 grams per
mole of the empirical weight.
Units cancel and the math
provides a value of 2.05.
Not an exact whole number
value but a value that
easily rounds to the number 2
which is what we multiply
the subscripts in the P2O3 by.
Starting with the
phosphorous, the 2 subscript
multiplied by the whole number
2 gives a molecular formula
starting with P4. Multiplying
the 3 subscript on the
oxygen by the whole number
2 completes the molecular
formula with O6. P4 O6.
The molecular formula
from the empirical formula.
Recapping the lecture.
A molecular formula is
the real ratio of the atoms
in a molecule while the
empirical formula is the
simplest whole-number ratio
of the atoms in a compound.
They can be the same thing
but they do not have to be.
The subscripts in the
molecular formula and
empirical formula differ
by a whole number multiple
Importantly, the term
empirical means measured.
Empirical formulas are
going to be associated
with measured values.
In a chemistry
problem these values
are given as mass or
percent of each element.
If values are given as
percent, assume 100 g as the
total mass for the sample.
This allows the direct
conversion into grams.
Next determine moles for
each element (from the mass in
step 1 and the atomic molar
mass from the Periodic Table).
After that set up a chemical
formula, using the element
symbols, with determined
mole as subscripts.
Divide each mole value by the
smallest mole value getting a
subscript of 1 for the
smallest mole value.
If this does not produce
subscripts that are all
whole numbers, or numbers
that readily round to
whole numbers, multiple the
subscripts by a common factor.
In part 2 of the lecture we
saw that the molecular formula
can be gotten from
empirical values if:
The empirical formula and the
formula weight are obtained.
Also required is a
compound formula weight.
To get the compound formula
divide formula weight by
empirical weight and multiple
subscripts in the empirical
formula by that value.
And that completes our lecture.
At this point in our study
of chemistry we find that
many aspects of
our study involve
multiple step processes.
Practice and follow
the process each time.
It is the process
that you are learning
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