[MUSIC]
Lighthouse Scientific Education
Presents a lecture
in The Mole Series
The topic; Percent Composition
Material in this lecture
relies on an understanding
of the previous lectures;
Atomic Mass and
Molecular Weight and
Math for Chemistry Part 2;
specifically the
section on percent.
The lecture begins
with a definition of
percent composition.
A brief review of percent
is provided then a discussion
on a set of rules for solving
percent composition problems.
The rules are then applied
with 4 practice problems
that covered different aspects
of finding percent composition.
Percent composition is
defined as the mass percentage
of each type of
element in a compound.
Basically, it is found by
dividing the mass of an element
in the compound by the
total mass of the compound
and converting that
ratio into percent by
multiplying it by 100.
A useful check is that the sum
of the percent compositions of
all of the elements in a
compound must equal 100.
The sum of the parts
must equal the whole.
To get some perspective,
consider a molecule of water,
H2O. Water's molecular
weight is 18.02 grams per mole
when rounded to the
hundredths place.
In finding the percent
composition of the oxygen
in water, the mass
of the compound in the
percent equation will be
the water's molecular weight
of 18.02 g. Since that is
the mass of a mole of water
the value for the mass of
the element, the numerator,
will be the atomic molar
mass of oxygen, which the
Periodic Table shows
to be 16.00 g per mole.
Substitute that mass of 16 g
in for the mass of the element.
Percent does not have
units, so the unit grams
should cancel out and they do.
The solution to the
percent composition
of oxygen is found by
dividing 16.00 by 18.02
and multiplying
that value by 100.
It is 88.80 percent.
88.80 percent of the mass
of a water molecule
comes from the oxygen.
We could repeat the process
and substitute 2 hydrogens
atomic molar mass for
the mass of the element
and get the percent
composition of hydrogen.
Or, seeing that there are
only two elements in water and
we calculated the percent
composition of one of them
we can deduce the percent
composition of the other.
Early it was remarked that
in a compound the sum of the
percent composition of all
of the elements equals 100.
Mathematically, that is
expressed as 100% equals
the percent composition
of oxygen plus the percent
composition of the hydrogens.
Inputting the 88.80 % for
oxygen leaves an equation
with a single unknown
(% hydrogen) which solves
to 11.20 for the percent
composition of hydrogen.
That mathematical trick
relies on the understanding
of percent.
So, before proceeding
into the step-by-step rules
for getting percent composition
a brief overview of
percent will be presented.
Students who feel comfortable
with percent should
skip the review and
go into the rules.
Percent is simply a ratio,
or fraction, taken 100 times.
The word percent means per 100.
The purpose of percent
is to present a ratio
or fraction value as a
larger number, one people
feel more comfortable with.
It has no other role.
In mathematical terms it
is a ratio or fraction,
numerator over
denominator, times 100.
Written in this form we see that
value of the ratio is converted
in to a larger number
by multiplying it by 100.
Adding some context to
the terms numerator and
denominator we see that
the denominator represents
the size of a whole thing, or
how many parts make a whole.
In percent composition it is
the whole mass of the compound.
The numerator represents the
portion or part of the whole.
In percent composition it
is the mass contribution
of a single element.
More commonly the percent
equation is presented with the
numerator as the PART and
denominator as the WHOLE.
When solving percent
problems a second form
is often used. Part over
whole equals percent over 100.
This form allows cross
multiplying which makes
the solving of percent
problems much easier.
For a more detailed
explanation of percent see the
'Math for Chemistry
Part 2' lecture.
Onto rules for determining
percent composition.
A distinction needs to be
made here because there are 2
forms that percent composition
problems come in. The first
type gives the percent of
each elements in a compound
and asks for a determination
of the empirical formula.
These types of problems
are covered in the
'Empirical Formal' lecture.
Percent composition and
empirical formula are
generally taught together.
It is helpful to
pair these lectures.
The second type has us
given compound information
and asks for the mass
percent of each element.
These are types of problems
covered in this lecture.
To further break down these
problems there will be ones
that just give a
chemical formula
and those that give
no formula but do give
masses for the
elements of a compound.
A set of rules is offered
that solves both types of
percent composition problems.
As these problems are often
in the form of work problems
the general process of
solving word problems
should also be followed.
That is take time
with the problem.
Search the problem for the
known and unknown values.
Write them down.
Then, identify a relationship
or set of relationships
between known and
unknown values. When the
relationship is found to be
'getting percent composition'
then apply these
rules starting with
getting the mass
of the compound.
This will be the denominator
of the percent fraction.
The mass will either be found
by calculating the
formula weight of the
compound if the formula
is given. Remember,
formula weight is a broad
term that encompasses
the molar mass or molecular
weight of a covalent compound
and the formula unit
weight of an ionic compound.
Another way that the mass
of a compound is found is by
adding masses of the
elements given in the problem.
This applies when no formula
is given but mass data is.
Adding all of the given
masses will be the mass of the
compound used in
the denominator
of the percent equation.
It is not a formula weight
but that's ok because the
masses of the elements are
not atomic molar masses.
These masses are generally
values determined in
experiment in the lab.
Step 2; set up the percent
equation or equations
(depending on how many
elements are being solved for)
with the element's mass
contribution in numerator
and the compound
mass in denominator.
There should be full
canceling of units.
Finally, step 3, multiply the
ratio by 100 to get percent.
While percent composition
problems look very busy they
are straight forward to solve.
We will solve 4 of them
using these
step-by-step rules.
The first example: an
experiment on an unknown
is found to contain 3.461 g
of carbon, 0.615 g of hydrogen
and 1.635 g of oxygen.
What are the percent
composition of the elements?
Being a word problem
we will take some time
and review the problem before
trying to solve it. Specifically,
what are the known and
unknowns in this problem?
Given are 3.461
grams of carbon,
0.615 grams of hydrogen
and 1.635 grams of oxygen.
What is not known are
the percent composition
of these elements.
From this set up we identify
the problem as one using the
percent composition rules.
This conclusion may seem a
bit obvious but, the practice
of evaluating the
problem before solving it
will generally
serve the student.
The first step is to get
the mass of the compound.
Can it be done by
calculating the formula weight
of the compound?
Well, the actual formula
of the compound is not given.
The problem claims that
the compound is an unknown
and only gives mass values
of its constituent elements.
The compound mass is
therefore gotten by adding the
masses of the elements.
Will this produce a formula
or molecular weight?
No it will not. If the
experiment was performed with
twice as much of the
unknown compound then each of
the element masses
would be doubled.
The element's contribution
to the compound mass
is in relative proportion.
So it doesn't matter
if the compound mass is
determined by adding weights,
like this problem, or by
calculating a formula weight.
Both ways have the same
proportion of compound mass
to element masses.
The ratio is the same.
Ok, getting back
to the compound mass.
To restate the definition;
the mass of the compound
equals the sum of the
masses of the elements.
That is 3.461 g of carbon
plus 0.615 g of hydrogen
plus 1.635 g of oxygen.
The calculator gives
the mass of 5.711 g
for the entire compound.
We will keep
those values handy.
There are 3 elements and
the percent composition
for each has to be
determined individually.
Starting with carbon. Since
we have the mass of a compound
it is time to move into step
2; set up a percent equation.
That begins with the ratio
of 'mass of the element' to
'mass of the compound'.
The 'mass of the
element' carbon is 3.461 grams.
The 'mass of a compound'
was just calculated
and it is 5.711 grams.
The math can be done here
or at the end of problem.
We will do it here.
The first thing to note is that
the units of grams cancel out.
That is important because
the final answer percent
does not have units.
3.461 divided by 5.711
is the decimal 0.606.
The last step is to multiply
by 100 to get percent.
0.606 times 100 equals 60.6%.
The unknown compound is
composed of 60.6% carbon.
We'll keep track of that value.
One down.
What about hydrogen? What
is its percent composition?
For that we return to the
setup equation. This time the
'mass of the element' is the
mass of hydrogen; 0.615 g.
The compound mass
is still the 5.711 g.
Canceling units and dividing
0.615 by 5.711 yields 0.108.
To get percent multiply
the value by 100.
Hydrogen's percent
composition is 10.8%.
Add that value to
the panel with carbon.
We will return to these
percents after getting
the percent composition
of all 3 elements.
We only have oxygen to deal
with. Like with hydrogen,
solving the percent
composition has us returning
to the percent equation.
This time the element mass
belongs to oxygen with
a value of 1.635 g.
The mass of a compound
is unchanged at 5.711 g.
Canceling units and
dividing 1.635 by 5.711
yields the decimal 0.286
which is converted into percent
by multiplying it by 100.
The percent composition
of oxygen, in the compound,
is the 28.6 percent.
Add that to the panel
and the question is solved.
Are we done?
The short answer is yes
the long answer is no.
We should double
check our results.
How do we do that?
By recognizing that the
whole is the sum of the parts.
That is the sum of the
percent from each element
must equal 100.
The addition of these 3
percent values should be 100
or something that
rounds to 100.
Again, these are experimental
values and will have
some error so exactly 100
is not always obtainable.
In this case 60.6 +10.8
+28.6 does give 100%.
That should give us
confidence in our results.
Okay, let's have
another problem.
What are the percent
composition of the
elements in Epsom
salt? Epsom salt
usually comes as a hydrate.
That is, it has attached waters.
Example 4 also has a
hydrate for a compound.
In that example we will
consider the attached waters,
here we will not.
Word problems should
be met with patients
and carefully considered.
We should ask
"What are the
knowns and unknowns?"
The chemical formula
of magnesium sulfate
is given and the unknowns
are the percent composition
of the elements in
the formula MgSO4.
What is the relationship
between the known
and unknown values?
It is the process of
determining percent
composition which
begins by getting the
mass of the compound.
There are two options for
getting the compound mass,
calculating the formula
weight or adding the
masses of the elements.
Which one should we
use in this problem? Well,
there are no given masses
so the choice clearly falls to
calculating the formula weight
from the compound formula.
At this point in their studies,
the student should feel
reasonably comfortable
getting formula weights.
Still, it can't hurt to
run through the process again.
In one mole of Mg S O4
there is 1 mole of
magnesium, 1 mole sulfur and
4 moles of oxygen.
The '1 to 1 to 4'
comes straight from the
subscripts of the elements
in the compound with
an absence of subscript
implying a 1. A second
point needs to be made
considering the Mg S O4.
It is an ionic compound
with a polyatomic anion.
If we were completely correct
the magnesium would not
be 1 moles of magnesium
but rather 1 moles of
the magnesium ion Mg+2.
While this doesn't
change the formula weight
it does keep us on our toes.
To get a formula weight
of the compound we need to
convert the moles of
these particles into grams.
The conversion factor for
moles to grams of an element
is the atomic molar mass.
The actual masses will
be gotten momentarily.
As written the unit moles
will cancel out leaving
the unit grams.
To get the actual masses
we will consult
a Periodic Table.
We find the atomic molar mass
of magnesium to be 24.31 g.
Sulfur has an atomic
molar mass of 32.07 g and
the atomic molar mass
of oxygen is 16.00 g.
Returning to the formula
weight calculation and
inserting 24.31 g
for magnesium ion,
32.07 g for sulfur and
16.00 g for oxygen we are
ready to get the masses.
For the magnesium ion
cancel out units and
doing the math gives
24.31 g of magnesium
ion. For sulfur,
cancel moles and do the
math for 32.07 g of sulfur.
And finally, 4 times
16;64.00 g of oxygen.
All that is left is to add
up the masses of the elements:
24.31 plus 32.07
plus 64.00 is 120.38.
In one mole of magnesium
sulfate there is
120.38 grams of
magnesium sulfate.
We will need these values
as we proceed in determining
percent composition
of the three elements.
Starting with
the magnesium ion.
Since step one is
complete, move to step 2
and write out the
percent equation. For the
mass of element inset the
24.31 g of the magnesium ion.
For the mass of
the compound weight
insert the 120.38
g formula weight
of magnesium sulfate.
The unit grams will cancel out
and dividing 24.31 by 120.38
gives a value of 0.202.
Step 3 is turning that
decimal value into a percent
by multiplying it by 100.
0.202 times 100 is 20.2
percent. 20.2 percent of
the mass of magnesium sulfate
is magnesium ion.
We'll keep track
of that value for the
final double check.
Magnesium is done,
how about the sulfur?
That requires us
to return to step 2.
This time for the mass
of the element is the
32.07 g of sulfur.
The compound mass is still
the formula weight of 120.38 g.
Gram units cancels and
the math of 32.07 divided
by 120.38 is a value of 0.266.
This value is converted
into percent by multiplying
it by 100.
That is 26.6 percent.
26.6% of magnesium
sulfate is sulfur.
We will keep that value too.
That just leaves the oxygen
and again we will
return to step 2
and insert 64.00 g of
oxygen into the mass
of the element and that same
120.38 g of magnesium sulfate
for the mass of compound.
The division yields
a value of 0.532.
The last step is to
convert that value
into percent by
multiplying it by 100.
This will yield a final value
of 53.2 percent.
Oxygen comprises 53.2%
of the mass of
magnesium sulfate.
And the problem is solved.
As a double check
we note that the sum of the
percent composition of the
elements must equal 100%.
Therefore, the value of
20.2 added to 26.6 and 53.2
should give a sum of 100%.
All the mass in magnesium
sulfate has been accounted for.
For the third example
is a word problem with an
extra dimension.
When heated, calcium
metal can react
with the oxygen in the air
to produce a metal oxide.
5.00 grams of Ca is
heated and forms 7.70 grams
of calcium oxide. What are
the percent composition of the
elements calcium and oxygen?
This problem seems
similar to the problem in
the first example. It is
however a bit more complicated
and it will show why it is
always a good idea to consider
the problem before
pursuing the answer.
What are the known and the
unknowns in this problem?
The starting quantity of 5.00
g of calcium is given as is
The mass of the final
product calcium oxide.
The 7.70 g is the mass of
the compound and will be
the denominator in
the percent equation.
What is unknown, aside for
the percent composition,
is the mass of the oxygen
that combines with the
calcium to form calcium oxide.
The mass is not given but
the question specifically asks
for the percent composition
of oxygen. Add
oxygen as an unknown.
We've seen a couple examples
of percent composition so far
and in each case the
mass of the element was
either given or its
atomic molar mass was used.
As we are not using formula
weight for the compound mass
the atomic molar mass is
not an appropriate mass
for the element.
We need another way of getting
the mass of oxygen.
We need a relationship
between what we know
and what we don't know.
In this case we can
use the relationship
that says the whole is
the sum of its parts.
Put into a mathematical form
it says the mass of the
whole of the compound, 7.70 g,
is equal to the mass of
its parts which are the
5.00 g of calcium and the
unknown amount of oxygen.
When shown in this perspective
the amount of 2.70 g for
oxygen's contribution
becomes evident, or at
least can be calculated.
Add that value to set up and
we are ready to move on to
the rest the problem.
Find the percent composition
of the calcium and oxygen.
That begins with step 1;
get the mass of the compound.
A look at the information we
have already gathered shows
the mass of the calcium oxide
was given in the problem.
There will be no need to
calculate the formula weight
or add the masses. With
that in hand, get the percent
composition of the calcium.
Step 2 is to write out the
percent equation that has
mass of element over mass
of compound and substitute in
the 5.00 g of calcium
for the element mass and
7.70 g of calcium oxide
for the mass of a compound.
Note that grams cancel
out and doing the math,
5.00 divided by 7.70 is 0.649.
To get percent composition
multiply 0.649 by 100
for a final value
of 64.9 percent.
64.9% of the mass of calcium
oxide is due to the calcium.
As for the oxygen, return
to step 2 and insert into
mass of element 2.70
g of oxygen and into
mass of compound, 7.70
g of calcium oxide.
That math produces a value
of 0.351 which we can convert
into percent by
multiplying it by 100.
The final value is
35.1%. 35.1% of that mass
of calcium oxide is
contributed by the oxygen.
That should complete
the problem. However,
were going to look at a
short cut in solving for the
percent composition of oxygen.
It involves the expression
that 'the sum of all elements
percent must equal 100'
which is what we
find if we look at
64.9+35.1 It does give 100%.
Consider for a moment, that
we never calculated the 2.70 g
of oxygen therefore we
were unable to come up with
35.1 % oxygen.
We can still get that percent
in a manner very similar to
how we came up with the 2.70 g.
If follows from the
equation that says the
'sum of the parts
is equal to 100'.
We calculated the percent
due to the calcium as 64.9
The part due to the oxygen
is, we'll say, unknown.
We have another
example of an equation
with a single unknown.
A bit of math has oxygen
needing to be 35.1%,
that is, to sum to 100.
Same basic concept as when
we calculated it is a mass.
A lot of time and effort can
be saved in recognizing an
unknown value can be solved
if it can be put into equation
where it is the only unknown.
The last practice
problem involves a hydrate
What is the percent
composition of water
in the hydrate
BaCl2 with 2 waters?
Barium chloride is considered
a hydrate because it is a
compound that includes
water molecules.
They are physically attached.
Usually it is a very specific
number of water molecules.
Here it is 2 waters and the dot
between the barium chloride
and the water is what
signifies that it is a hydrate.
For Epson salt the
dot says that there are
7 attached water molecules.
For both of these compounds
those waters can be driven off
with heat but given the
opportunity the compound will
attach waters molecules.
The question should
be reviewed for what
is known and not known.
The molecular formula of the
compound is known and the
percent composition of water
is unknown.
Note that the question
asks for the percent
composition of a
molecule, water.
The previous examples have
asked for the percent
composition of elements.
The difference, however,
is only one of terminology.
The water in this problem is
a part of the whole compound
just as an element is a
part of the whole compound.
As such this problem can be
solved in the same manner.
Solving the problem requires
an application of the steps for
finding percent composition.
Moving into the first step,
getting the mass
of the compound and
we consider our options. Is
the mass found by calculating
the formula weight or
by adding the masses of
the individual elements?
By being given the
compound formula and, not
any masses, the choice is
made for us. There are
several ways to approach
getting the mass
of the hydrate.
Presented here will
be calculating the mass
of the individual compounds,
the mass of barium chloride,
and the mass of the 2 waters.
These masses will then
be combed into the
mass of the hydrate.
The main reason for
taking this tact is that
we are going to need
the mass of the waters
themselves in the
percent equation.
A secondary reason is that
we want to start to build a
collection of common
molecular weights that come
up so often that a recalculation
is not always in order.
Water fits that description.
Starting with the Ba Cl2.
An interesting question here
is to ask whether this
compound is ionic or covalent.
Well, it contains
an element from the
left hand side of
the Periodic Table and
an element from the right hand
side of the Periodic Table .
That is a good indication
that BaCl2 an ionic compound.
From the formula it can we seen
that 1 mole of Ba Cl2 contains
1 mole of barium ion and 2
moles of the chloride ion.
These mole values need to
be converted into gram values
using the conversion
factor of atomic molar mass.
Remember that we can use
the element's atomic mass
for the ion's
atomic molar mass.
Referring to the Periodic
Table, barium is found to have
an atomic mass of
137.33 grams per mole and
chlorine a value of
35.45 grams per mole.
Insert in as a conversion
factor: 137.33 grams per mole for
barium ion and 35.45 grams
per mole for the chloride.
The mass of the barium is
solved by canceling units
and multiplying to get
137.33 grams of barium ion.
The moles of chloride
cancel and 2 times 35.45
gives a mass of 70.90
grams for chloride.
Summing the masses
of the ions gives a
formula weight of 208.23
grams for barium chloride.
That takes care of
the barium chloride.
Now to the water.
And there are 2
common approaches
for getting water's
mass contribution.
Both begin with the recognition
that for every 1 mole
of barium chloride hydrate
there are 2 moles of H2O
which can be written as
2 times 1 mole of H2O.
The first approach is to
treat water like a molecule
and gets its mass by getting
the masses of the elements
that make up water. And in
one mole of H2O there are
2 mole of hydrogen
and 1 mole of oxygen.
Therefore, in 2 moles of H2O,
there are 4 moles hydrogen
and 2 moles of oxygen.
We can take these values,
along with the
atomic molar masses
of the hydrogen and oxygen,
and calculate the mass
contribution of the waters.
Or we can start to
build a library of common
molecular weights and simplify
the problem of calculating
the grams of 2 mole of H2O
by multiplying the 2 moles
by molecular weight of
H2O. Earlier in lecture
we calculated that
as 18.02 g per mole.
The molecular weight of water
comes up frequently in the
study of chemistry and it is
worth memorizing its value.
This math gives a gram
value of 36.04 grams.
Quite a bit easier.
Both approaches
work; one is easier.
That is the mass contribution
of the two compounds.
Now we need to find the
mass of the whole compound,
the whole hydrate,
the barium chloride
with the waters.
That is the sum of
the compound masses:
208.33 for barium chloride
plus 36.04 g of water.
That gives 244.27 grams for
the barium chloride hydrate.
It's time to calculate the
percent composition of water
in the compound. Step 2
calls up the percent equation.
A slight modification
in terminology is needed
here because the problem
asks for percent composition
of water and not of an element.
Changing the numerator
'mass elements' to 'mass
part' broadens the equation
and allows for the mass
of water, 36.04 g to be
added in the numerator.
The denominator mass
is the hydrate 244.27 g.
Gram units cancel out and
the math of 36.04 divided
by 244.27 gives
a value of 0.148.
The last step is to
convert it into a percent
by multiplying by100
producing a percent of 14.8.
14.8% of the barium
chloride hydrate is water.
And that is the
last of our examples
Recapping the lecture.
In a review of percent
we saw that percent
is a ratio taken 100 times.
Percent composition is the
mass percentage of each type
of element in a compound.
It can be expanded to include
any part of the compound such
as the water in the hydrate.
It is found by dividing the
mass of an element, or part,
by the total mass of the
compound and converting
it into percent.
Importantly,
the sum of all elements'
percent must be 100.
To find the percent composition
of an element or part
first get the mass
of the compound.
Generally that involves
one of the two procedures;
calculate the formula weight
if the formula is given
or add the masses of the
elements if they are given.
From there set up the percent
equation or equations with
element's mass contribution
in the numerator and compound
mass in the denominator.
Finally, multiply
the ratio by 100.
A double check can be done
by adding all of the elements
percent compositions together.
It should sum to 100
and that concludes our lecture.
Be sure follow up this
lecture with a look at the
'Empirical Formula'
lecture that it
also contains percent
composition problems.
These two lectures are
usually studied in tandem.
[MUSIC]
