>> narrator: In this video,
we'll take a closer look
at percent composition
and empirical formulas.
Percent composition is
the percentage, by mass,
of each element in a compound.
That's why it is also
called "mass percent."
You calculate it by finding
the mass of the element
in the compound,
and dividing by the total
molar mass of the compound,
times 100.
You do this for each
element in the compound,
and then you should have
mass percents for each element.
The first problem asks to
find the percent composition
of sodium chloride.
There are two elements
in the compound,
sodium and chlorine.
You need to figure out
the percent, by mass,
of each element.
Start with the sodium
in sodium chloride.
Take the mass of the sodium
in the compound
from the periodic table,
22.99 grams.
and there's only one sodium
per formula unit
of sodium chloride.
Put that number over the
molar mass of the compound,
which is 58.44,
and then multiply it times 100
to convert it to a percentage.
And you find that the percent
of sodium in this compound
is 39.34.
You do the same thing
for chlorine to figure out
its mass percent.
Put the mass of chlorine,
which is 35.45,
over 58.44, which, again,
is the molar mass
of the compound.
Then multiply that times 100
to get the mass percent
of chlorine, which is 60.66%.
When you do percent
composition problems,
all of the percentages should
add up to 100 at the end.
So let's check.
39.34 plus 60.66 would,
indeed, add up to 100.
Here's another problem
in which you're asked to
calculate percent composition.
This time,
it's potassium nitride.
From the periodic table,
you get the molar mass
of potassium at 39.098.
In this compound,
there are three potassiums
in the formula.
So multiply 39.098 times three.
That is 117.29, and that is
the mass contribution of
potassium in this compound.
If you're calculating the
percentage of the compound
that is potassium,
take the mass contributed
from potassium, 117.29,
and put that over the total
mass of the compound,
which is 131.0.
Now, divide those two
and multiply the result
times 100,
to get the percentage of
potassium in this compound,
which is 89.33%.
To solve for mass percent
for the nitrogen,
use the same approach
as for the potassium.
Only, this time,
put the mass of nitrogen
that is found in the compound
in the numerator,
and divide it by the total
mass of the compound.
Plug in 14.01,
the mass of the nitrogen,
found from the periodic table.
Use one nitrogen, because there
was only one in the formula,
and divide 14.01 by 131.30,
which is the molar mass
from the compound.
After dividing 14.01 by 131.30,
multiply that times 100
and you find that
the mass percent of nitrogen
is 10.67.
Again, double-check
the calculation by looking
at the two percentages
and making sure that they add up
to be 100%, and they do.
So this example is correct.
Let's move on from
percent composition
to empirical formulas.
An empirical formula is the
smallest whole number ratio of
atoms that are in a compound.
Let's review how you
can solve to determine
the empirical formula
of a compound.
First, find the mass percent
of each element.
Then figure out how many moles
of each element there are,
if you were given
percent composition
in a problem,
which you commonly are
for these types of problems.
You can assume that
you have a 100 gram sample
of the compound, and that way,
you can convert the percentages
directly into masses and grams.
Once you have moles
of each element,
identify the number of moles
calculated that is
the smallest value.
Then, divide every molar amount
by the smallest number of moles
to figure out the value of
the subscript for each element
in the empirical formula.
When necessary,
multiply those subscripts
by two, three or four
to get whole numbers,
because you must have
whole numbers as subscripts
in an empirical formula.
Here is an empirical formula
practice problem.
It asks you to find
the empirical formula
for a compound
containing 63.5% silver,
8.2% nitrogen,
and 28.3% oxygen.
The first step is to change
the percentages to grams.
And remember,
you can do this by assuming
that you have 100 grams
of the compound.
The 63.5% silver would be
63.5 grams of silver
and 100 grams of the compound.
Do the same thing
with the other elements.
Think in grams.
So it's not 8.2% of nitrogen.
It's 8.2 grams of nitrogen.
And instead of thinking of it
as 28.3% oxygen, write it
as 28.3 grams of oxygen.
At this point,
you can start converting
each of the masses to moles,
which we need to know
for each element.
Start making conversion charts,
and when you're converting
from grams to moles,
you bring down
the units first.
So bring down grams of silver,
and then use the molar mass
of silver to convert
from grams of silver
to moles of silver.
Use 107.87 grams
from the periodic table
for the molar mass for silver.
You can cancel
the units for grams,
and you find that there
are .589 moles of silver
if there are
63.5 grams of silver.
This is found by
taking 63.5 times one,
and then dividing it
by 107.87.
Now do the same for
the other two elements,
nitrogen and oxygen.
Use the molar mass
of 14.01 grams of nitrogen,
to convert from
grams of nitrogen
to moles of nitrogen.
And then take 8.2 times one,
divided by 14.01,
and you find that
8.2 grams of nitrogen
is .585 moles of nitrogen.
For oxygen, use 16.00 grams
for the molar mass,
and that will allow you
to cancel grams of oxygen.
And then take 28.3 times one,
divided by 16,
and you find that there
are 1.77 moles of oxygen.
So you now have three
different molar amounts
for the three elements
in the compound.
Next, divide all
the molar amounts by
the smallest molar amount
to get the subscripts
for the empirical formula.
There are two amounts
that are very similar.
The .589 and the .585.
But .585 is smaller,
so use that.
Divide .585 into
all three amounts
to determine the subscripts.
So, .589 divided by .585
equals 1.01.
.585 divided by .585 is one.
And 1.77 divided by .585
is equal to 3.03.
These numbers, 1.01 and 3.03,
are close enough to round
and use as subscripts
for the empirical formula.
Write the metal first,
which is silver, Ag.
Since you just calculated that
there was a subscript of one,
leave it as just Ag.
And since you also just
calculated that there is a
subscript of one for nitrogen,
just write one N.
Last is oxygen, and since
there are three oxygens,
based on the calculation,
the subscript for oxygen
is three,
in the empirical formula,
And the empirical formula
for this compound is AgNO3,
and that's silver nitrate.
Here's one for you to try.
Find the empirical formula
for a compound that
contains 68.4% chromium
and 31.6% oxygen.
When you're ready,
press pause and give it a try.
Then press play
and see if you got it right.
If your final answer was Cr2O3,
that is correct.
Let's walk through
how you got there.
We took the percentages
that were given
and converted those to grams.
So the 68.4% chromium
becomes 68.4 grams of chromium.
Then, we used the molar mass
from the periodic table
to get from grams to moles,
and found that there are
1.315 moles of chromium.
Then we followed the same
steps with the percentage
that was given for oxygen.
We changed it into grams,
and then brought the units down
and used the molar mass
from the periodic table to
convert from grams of oxygen
to moles of oxygen.
And we found that we have
1.975 moles of oxygen.
That was the first part,
determining the number of moles
of each element in the compound.
In the second part
of the calculation,
we took the molar amounts
that were given to us,
and divided each molar amount
by the lowest molar amount,
which, in this case,
was 1.315.
Then, we divided 1.315
into 1.315, and of course,
that's one mole of chromium.
Next, 1.975 was the molar
amount for the oxygen,
and we divided 1.315 into it
and found 1.50 moles of oxygen.
This is one of the situations
in which we have to multiply
by two, three or four,
because we cannot have an
empirical formula of CRO 1.5.
We can't have 1.5
as a subscript.
So we have to multiply
both the molar amounts that
we just calculated by two,
because 1.5 times two
equals three.
For chromium, one times two
equals two moles of chromium.
And for the oxygen,
1.5 times two gives
three moles of oxygen.
And, we found that
the empirical formula
for the compound is Cr2O3.
That's a Closer Look
at percent composition
and empirical formulas.
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