Howdy! Mr. Causey here coming to you from
beautiful Atascocita, Texas in my virtual studios
and today we're going to talk about percent compositions
and percentages. This is a very important,
empirical lesson that chemists use to determine
the composition of unknown compounds.
Get out your calculators and get out your
periodic tables and let's get started. In this lesson,
you're going to talk about percent by mass, percent composition,
empirical formulas, molecular formulas,
and much, much more. You'll need your periodic table,
you'll need your calculator and you must know
the periodic table. You must know how to find
molar mass and you need to be able to write formulas
and of course all of this can be found
in some of my videos on my video channel. And you need to know how to do percent.
So are you ready?
Let's get started!
One of the first things we want to talk about is compounds
and some things that we need to know about compounds is that they have a
definite composition
and a definite ratio and this has been determined by the law of definite
composition and the law
of multiple proportions. These two laws, determined
in the eighteenth-century, help us to understand
that compounds are always going to be the same.
So that means that water, H2O, is always
H2O. It's always going to be 2 moles of Hydrogen
and 1 mole of Oxygen and with
that knowledge we can use it to
help us write chemical formulas and figure out the percent
composition and the percentage composition
of different compounds. A percentage composition
is the composition of a compound given
in percentages, of course that's using percent,
and then we can determine our empirical formulas
and we can even determine molecular formulas using this information
and you're going to want to listen to the whole thing because I do two examples
at the end dealing with finding the empirical formula
and then finding a molecular formula but before we do that
let's look at percent and percentages.
Percent by mass, we're going to take the mass of an element
divided by the mass of the compound
and then multiply it by 100 and that's how you find a percent.
So if I take grams of Hydrogen
and divide it by the grams
of Ammonia and then multiply it by 100, that gives me the percent
of Hydrogen. On a percentage,
it's just taking the mass of the compound multiplied by the percent.
Grams Ammonia multiplied by the
percent Hydrogen and don't forget to
change the percent to a decimal and that will give us the grams Hydrogen.
Let's do a practice problem. What percent of HNO3 is Hydrogen?
Well, first we need to determine the molar masses and then
we need to realize that HNO3 is the whole, it's the
main part, and
the Hydrogen is the part of the whole.
And if you know about fractions, we can take
the part divided by the whole,
multiply by 100 and that is our percent.
so let's look at that problem, let's go to the magic blackboard.
HNO3's molar mass, well we have 1 mole of Hydrogen,
we have 1 mole of Nitrogen and 3 moles of Oxygen
and we can get all these values off of the periodic table. That adds up to
63.02 grams
of Nitric acid and of course there's one
1 mole of Hydrogen and that's 1.01 grams. So there's our two values,
let's take them
divide them. We take the part, right here the part,
divide it by the whole, times 100
and it equals a percent.
Now finding percentage,
how much Iron can be obtained from 125.0 grams
of Fe2O3 which is
Iron 3 Oxide,
commonly known as rust? So we need to determine the molar masses,
we then need to find the percent Iron
and then calculate the percentage. Three steps.
Let's go to the magic blackboard. First thing is taking
Iron 3 Oxide and determining the mole amounts.
There are 2 moles of
iron and 3 moles of oxygen. Add those together
and there we go, there's our molar mass
of Iron 3 Oxide and we want to compare it to the
moles Iron and there's 2 moles of Iron
in there so we want to make sure that we use that number.
And we take the part, divide it by the whole, times 100
and it's 69.94 percent Iron.
So take the total amount of Iron 3 Oxide that we had, 125 grams,
multiply it by the
percent and now notice we have to change it to a decimal.
That gives us 87.43 grams of Iron
and so in 125 grams of
Iron 3 Oxide, there are 87.43
grams of Iron. And I think our sig figs are right there but if they aren't
you can check them.
I believe we have four sig figs. Let's recap!
I've told you or answered the question, what is percentage composition?
I've shown you how to find percent by mass and I've shown you how to find a
percentage.
Alright, now let's go ahead and use this
information to find empirical formulas
and molecular formulas. Find the empirical formula of a compound that has the following
following composition:
40.0% carbon, 6.70% hydrogen,
and 53.3% oxygen.
Alright, one of the first things you want to do is step one, assume that we have
100.0 grams
of the compound. The reason I've chosen 100 grams is because it makes
the math easier, it's easier to work with 100 grams
or multiples of 10 than is
anything else. So that would give us
40 grams of carbon,
6.70 grams of hydrogen, and 53.3 grams of oxygen.
Now, let's convert the masses to moles using the periodic table.
We can go and use the molar mass to convert these 2 moles.
So to the blackboard!
First thing we do is multiply 40 grams of carbon by its molar
mass ratio and we get 3.33
moles of carbon. We do the same thing to hydrogen
and we get 6.63 moles of hydrogen. And then we
take oxygen, do the same thing,
and we get 3.33 moles of oxygen.
So now we have our molar values and
we can use this to do step number three. In step 3, we want to write a ratio
using the smallest value
as the denominator and if you remember, on the blackboard,
3.33 is the smallest of the values.
So let's write our ratios. Again,
we knew that there were 3.33 moles of carbon,
we divide that by 3.33 and we get 1
mole of carbon. Do the same thing with hydrogen and we get 2 moles
of hydrogen
and we get one mole
of oxygen. Well if we look at that, that means we get
1 carbon, 2 hydrogen and
1 oxygen and our empirical formula is CH2O
but that's the empirical formula and not necessarily the molecular formula.
Let's take a look at the molecular formula. If the sample
in the preceding problem has a molar mass of
60.06 grams, find the molecular formula.
We have three steps in this
calculation. Step 1: Determine the mass
of the empirical formula, the empirical formula of course being CH2O.
To the blackboard!
We get CH2O,
1 mole of carbon, 2 moles of hydrogen,
and 1 mole of Oxygen and it all adds up to
30.03 grams of that compound or at least the empirical formula.
Step 2: Divide the actual molar mass by the empirical formula.
Now remember the actual molar mass is 60.06 grams.
60.06 grams divided by
30.03 grams gives us ratio of 2
I really don't mean a ratio but a value of 2 which then tells us that
everything in the empirical formula needs to be multiplied by two
and that should give us C2H4O2
and we see
that is our molecular formula. aAlright, let's recap.
We looked at how to find the empirical formula, we looked at how to find the molecular formula,
and if you have any questions,
send an email to mrcausey
@mrcausey.com and you can check it out
by going to mrcausey.com or mrcauseysworld.com
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