Percent By Mass Composition MCAT General Chemistry
Articles Blog

Percent By Mass Composition MCAT General Chemistry

September 19, 2019


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.

Leave a Reply

Your email address will not be published. Required fields are marked *