Leah here from leah4sci.com and in this MCAT General Chemistry video we're going to look
at percent by mass composition calculations.
Percent by mass as the name implies is the percent of the mass of something.
Before we go nuts with the detailed definition, instead of memorizing all the different percent
formulas let's look at one rule for them all , Percent by Mass, percent composition, percent
yield, percent dissociation, percent of anything can be measured as part over whole times a
hundred percent.
The specific percent calculations will be the part of that specific thing divided by
the whole of that specific thing.
The part gives you piece or the percent that you're calculating and the whole is what you're
comparing it to, what you're taking that percentage from.
The problem is part of a whole is giving you a fraction.
You can turn that fraction into a decimal and you can turn that decimal into a percent.
For example, say you ate two slices of a pizza pie, what percent of that pie did you consume?
Let's set it up as part over whole.
The part that you ate is 2 slices, the whole is the entire pie of 8 slices.
Slices cancel, giving me 2 over 8 which we can simplify as 1 over 4.
The fraction of pizza that you consumed is a fourth.
To turn this into a decimal, divide 1 over 4 which is .25.
If you didn't know this make sure you memorized the mcat math cheat sheet linked below.
And finally if we want to turn that .25 decimal into a percent, we simply multiply by a hundred
or move the decimal over two spaces to the right giving me 25 percent per 100.
Going back to our original question, you ate 25 percent of that pizza pie.
But what does this have to do with composition by percent mass in the context of MCAT Chemistry?
If percent by mass composition is part over whole then we're looking for the percent times
a hundred and mass composition is the composition of the part out of the whole in a molecule,
in an atom, in a big solution, whatever it is we're looking at we're looking for that
mass percent part over whole.
For example say you're asked to find the percent composition of chlorine in 35 grams of NaCl.
First recognize that this is a tricky question.
The percent of something is always a ratio.
It's a ratio of the part over whole so it doesn't what we have as the actual mass.
What we care about is the molar mass of NaCl something we learned how to calculate in the
last video.
How do we set this up?
In this case, part is equal to grams of Cl and whole is equal to grams of NaCl.
Sodium has a mass of 23, Chlorine has a mass of 35 which gives us 23 + 35 is equal to 20
and 30 is 50, 3 and 5 is 8, 58 grams per mol.
In this case we only need grams not grams per mol because they're going to cancel.
If we have grams per mol or grams per gram they cancel.
Our final answer for percent of anything is unitless because it's a ratio of the same
thing, they cancel each other out.
What we do know is chlorine has a mass of 35, sodium chloride has a mass of 58 and then
we just divide 35 over 58.
Question us how do you divide this without a calculator?
This is where you can try a lot of different things, rounding, changing the numbers, just
make sure you don't go too crazy so your numbers aren't too far off.
To round here might give us too much of a rounding error but let's try it.
We can round 35 down to 30 and 58 up to 60 which gives me approximately 30 over 60 which
is 0.5.
Knowing that we rounded 35 down way more than we rounded 58 up so let's assume we're just
looking at a rounding error of a numerator being too small making our answer too small
so that it's something greater than 0.5.
Another trick I like to do when we have a number like 35 and the other number is very
close to a solid or even unit of 10 is to do a go between.
If we look at 35 over 60 which is a little tricky because 30 over 60 is equal to .5 and
then 40 over 60 which 4 over 6, 2 over 3 or .67.
Our answer would be somewhere in between .5 and .67 so let's split the difference and
go with .58 and again, it's close enough.
On the MCAT, close enough is good enough.
On the calculator, 35 over 58 is equal to .6 which as we said is greater than .5 or
pretty close to .58.
This gives us a decimal not a percent.
So our answer would be around 50 to 60 percent or using this version we can say 58 percent.
So what does this tell us?
The percent composition of chlorine in NaCl is 58 percent.
The fact that it's 35 grams doesn't tell us anything unless we specifically wanted the
mass to the mass of Chlorine in 35 grams of NaCl.
To do that you go a different route, you find the number of mols in 35 grams of NaCl and
the number of mols NaCl equals the mols of Cl completely different calculation.
We already hinted it this type of Math in the empirical formula video earlier in this
series.
This is very important, if you're trying to find the percent composition of something
when the empirical formula is much smaller than the molecular formula.
Say you're asked to find the percent composition of each atom in glucose.
We know that glucose is C6H12O6 and by now you should also know that that is equal to
180 grams per mol.
To find the percent composition of each we have to set it up as part of a whole times
a hundred percent` where the part is the specific atom in question and the whole would be glucose.
To find the percent of carbon, we need the grams of carbon over the grams of glucose.
For hydrogen we need the grams of hydrogen divided by the grams of glucose.
For oxygen we need grams of oxygen divided by grams of glucose and multiply each one
by a hundred percent and this is where it gets tedious.
Grams of carbon would be the molar mass of carbon atoms which would be times 12 or 72
grams over 180.
12 Hydrogen would be 12 grams over 180.
For Oxygen we have 6 for a mass of 96 over 180.
Do you see how this math can get very tedious, very fast?
But what if you remember that all we're looking for is a ratio, it's simply the ratio of carbon
to glucose.
Hydrogen to glucose, and oxygen to glucose.
And who said we had to use the molecular formula?
Isn't the empirical formula the same exact ratio as the molecular formula but in a smaller
value?
absolutely!
So let's find the empirical formula.
Greatest common multiple is 6 so we divide everything by 6 giving me C1H2O, CH2O.
Now we simply set it up using the empirical formula mass.
That means we have one carbon at 12 grams plus 2 hydrogen at 2 grams plus 1 oxygen at
16 grams for a total of 12 and 2 is 14 and 16 as 30.
30 grams per mol is the empirical formula weight.
So let's set this up.
Carbon is equal to 12 over 30, hydrogen is equal to 2 over 30, and Oxygen is equal to
16 over 30.
Let's divide, 12 over 30 without a calculator is tricky.
But what if we take 6 out of the numerator and denominator?
12 divided by 6 is 2, 30 divided by 6 is 5 and this is a number you may not recognize
but you do know.
You need to know the value of the fraction 1 over 5 which is .2 and 2 times 1 over 5
is .4.
2 of 30 is simplified to 1 over 15.
This is a fraction you should have memorized from the mcat math cheat sheet linked below
which is equal to 0.07. 16 over 30 is another tricky one.
There aren't any good numbers to pull out except for 2 but what if we estimate 16 over
30 at about 15 over 30 which is equal to a half or .5.
The calculator gives is 0.53 telling us that we're certainly close enough.
But we're not done coz our answers are in decimals.
To find the mass percent we multiply everything by a hundred percent or move the decimals
2 spaces giving me carbon at 40 percent, hydrogen at 7 percent, and oxygen at 50 percent.
When you add up the mass composition, they should be equal to about a hundred percent.
I say about because we estimate and rounded it but 40 and 7 is 47 and 50 is 97, 97 is
close enough to a hundred percent.
Now that we've seen the empirical formula, do you want to try this math without a calculator?
I didn't think so.
So I'm gonna grab the numbers from the calculator real quick just to show you that they're pretty
much the same.
Doing this with the calculator we have 40 percent same as what we got, 6.7 percent close
enough to 7, 53 percent close enough to 50, and we're good to go!
For even more practice and questions like this, first give the video a thumbs up, then
go to the mcat reactions and stoichiometry quiz on my website which you can find along
with this entire video series leah4sci.com/reactions.