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.