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## AP®︎/College Chemistry

### Course: AP®︎/College Chemistry > Unit 1

Lesson 3: Elemental composition of pure substances- Empirical, molecular, and structural formulas
- Worked example: Calculating mass percent
- Worked example: Determining an empirical formula from percent composition data
- Worked example: Determining an empirical formula from combustion data
- Elemental composition of pure substances

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# Worked example: Calculating mass percent

To calculate the mass percent of an element in a compound, we divide the mass of the element in 1 mole of the compound by the compound's molar mass and multiply the result by 100. Created by Sal Khan.

## Want to join the conversation?

- why does sal wait until the end of his calculations to worry about sig figs? shouldn't you have to account for sig figs at each step since we are multiplying, dividing, and adding measured numbers?(3 votes)
- You apply sig figs to the end of a calculation not any intermediate steps.(32 votes)

- Why did Sal use 1 mole of glucose rather than 1 molecule(if it's right way to say) of glucose for his calculation? Is it just for a practice of dimensional analysis or any other reason?

I did it following way using 1 molecule of glucose:`The ratio of mass of Carbon to 1 (molecule of) glucose`

= total mass of Carbon in 1 glucose / total mass of 1 glucose

= 6*C / 6*C+12*H+6*O

= 6(12.01u) / 6(12.01u)+12(1.008u)+6(16.00u)

= 72.06u / 180.156u

= about 0.4(6 votes)- You could really use any amount of glucose for calculating this percentage because you'd get the same result anyway. He probably used a mole because then the atomic mass units go to g/mol easily. Also, mass for a mole of glucose is way easier to measure or imagine than with just one molecule.(9 votes)

- I still don't understand where Sal got the molar mass from and why you multiply it by the number of moles of a element; is it to convert the moles to grams?(2 votes)
- moles = mass / molar mass

-> mass = moles x molar mass

You can get the molar mass of an element from the periodic table. For a molecule or compound you just add up the molar masses of all the elements that make it up.(12 votes)

- I have a question. How does significant figures work, like how do I know to round to 2 or 3? Thanks!(3 votes)
- You generally round your final answer to the significant figures of your given with the fewest sig figs. If your givens are 9.5, 5.07, 7.993, and 3.32, you'd round to 2 significant figures because the fewest in your givens is two in 9.5.

https://www.khanacademy.org/math/arithmetic-home/arith-review-decimals#arithmetic-significant-figures-tutorial(6 votes)

- instead of having a complicated fraction that needs to be simplified and puzzle more if quantity doesn't matter why not use a molecule. OR instead of making a complicated fraction and taking out grams per mole and so many complicated calculations why not do this instead:

C6 H12 O6

6*12+12*1+6*16 grams or AMU(if you're using a single molecule

=180 grams

c=6*12=72 grams

%c=72/180 * 100 %

=40%

now no mole or grams or any dimensional anylisis required. just simple calculations.

Why not use this less confusing way?(3 votes)- I mean you basically did the same calculation as Sal, so really you used grams and moles like he did. But you omitted all the decimal digits. Doing so can affect the accuracy of an answer.(5 votes)

- Is the mass percent really independent of given mass? I mean what if I take 2 moles of glucose? Would it affect the mass percent?(3 votes)
- If we had two moles of glucose, then the moles of carbon, hydrogen, and oxygen would be twice that for the one mole of glucose calculation (since we have twice as many glucose molecules). So yeah, the mass percent would be consistent regardless of how many glucose molecules we had.

Hope that helps.(3 votes)

- Why is 12.01 u = 12.01 g/mole of c?(2 votes)
- Chemists defined the molar masses of elements such that a mole of atoms of a single element would be equal to its atomic mass, but in grams instead of u. So carbon's relative atomic mass is 12.01 u, but a mole of carbon atoms is 12.01 gram simply because we defined it to be so.

Hope that helps.(4 votes)

- dont u have to have to account for the sig figs during every step(0 votes)
- You should be accounting for sig figs at the end of the calculation otherwise you’re introducing rounding errors.(8 votes)

- Couldn't you just say

6(12.01u)/6(12.01u) + 12 (1.008u) + 6 (16.00u) = mass %

72.06u/72.06u + 12.096u + 96.00u = mass %

72.06u/180.156u = mass %

72.06/180.156 = mass %

0.4000 = mass & (Rounded to 4 significant figures)

Why does Sal convert unified atomic mass (u) into g/mol instead of solving for the mass % directly using unified atomic mass?(2 votes)- "72.06u/72.06u + 12.096u + 96.00u = mass %

72.06u/180.156u = mass %"

Not sure how you arrived at that.

Sal begins with the element's relative atomic masses (in u though) and converts them into molar masses. He can do this because the magnitude of an element's relative atomic mass on the periodic table is defined to be the same as the magnitude of the mass of a mole of the element, but in grams. Then he assumes he has a mole of glucose and calculates the mass percent of carbon in grams technically. If you keep the mass in unified atomic mass units and assume you're dealing with a single molecule of glucose then the math is identical.

Hope that helps.(2 votes)

- Why does Sal write 16.00 as oxygen? Can you just put 16, since those zeros don't count as anything?(1 vote)
- Mathematically yes, 16 is the same as 16.00, but not when it comes to scientific calculations. In science we’re concerned with significant figures. A number like 16.00 has more significant figures than 16. By using 16 instead of 16.00, we’re essentially omitting significant digits which can affect the accuracy of the answer.(3 votes)

## Video transcript

- [Instructor] So right over here, I have the molecular formula for glucose. And so let's just say that I had a sample of pure glucose right over here, this is my little pile of glucose. I'm not even gonna tell you its mass, but based on the molecular formula, can you figure out the
percentage of carbon by mass of my sample? Pause this video and think about it. And as a hint, I've given
you the average atomic masses of carbon, hydrogen, and oxygen. All right, now let's work
through this together. Now, the reason why the amount
of glucose doesn't matter is because the percent carbon by mass should be the same
regardless of the amount. But to help us think this through, we can imagine amount. Let's just assume that this is a mole, this is a mole of glucose. So one way we could think about it is, we say okay, for every mole of glucose, we have six moles of carbon. Because every glucose
molecule has six carbon atoms. So we could say, what
is going to be the mass of six moles of carbon divided by the mass of one mole of glucose? And once again, the reason
why it's six moles of carbon divided by one mole of
glucose is because this, if we assume this is a mole of glucose, every molecule of glucose has six carbons. So it's going to be six
times as many carbon atoms or six moles of carbon. Now, what is this going to be? Well, this is going to be equal to, it's going to be in our numerator, we're going to have six moles of carbon times the molar mass of carbon. Well, what's that going to be? Well, we can get that from the average atomic mass of carbon. If the average atomic mass is 12.01 universal atomic mass units, the molar mass is going to be 12.01 grams per mole of carbon. So times 12.01 grams per mole of carbon. And notice the numerator
will be just left with grams. And then in the denominator, what are we going to have? Well, the mass of one mole of glucose, for every glucose molecule, we have six carbons, 12
hydrogens, and six oxygens. So it's going to be the
mass of six moles of carbon, 12 moles of hydrogen,
and six moles of oxygen. So it's going to be what
we just had up here, it's going to be six moles of carbon times the molar mass of carbon, 12.01 grams per mole of carbon. To that, we are going to add the mass of 12 moles of hydrogen. So 12 moles of hydrogen times the molar mass of hydrogen, which is going to be 1.008
grams per mole of hydrogen. Plus six moles of oxygen, times the molar mass of oxygen, which is going to be 16.00
grams per mole of oxygen. And the good thing is, down
here, the units cancel out, so we're left with just
grams in the denominator. And that makes sense. We're gonna end up with
grams in the numerator, grams in the denominator,
the units will cancel out, and we'll get a pure
percentage at the end. So let's see, in the numerator, six times 12.01 is 72.06. And then in the denominator, I'm just going to do the
pure calculation first, and then I'm gonna worry
about significant figures. So in the denominator, we have 72.06 plus, let's see, 12 times 1.008 is 12.096, and then we have plus
six times 16 is 96.00, and this will be equal to 72, if we're just thinking
about the pure calculation, before we think about significant figures, 72.06 divided by, let's see, if I add 72 to 12, I get 84, plus 96, I get 180.156. Did I do that right? If I were just to add up everything, not even think about significant figures. So we can type this into a calculator but we should remind ourselves that our final answer should have no more than four significant figures. Because even down here, if we were just doing this
blue calculation here, that should only have
four significant figures, it would have gotten us
to the hundredths place, and so when we add things together, we should get no more
than the hundredths place, but even if we rounded over there for significant figures purposes, we would still have at least four, we'd actually have five
significant figures. So this four significant figures is our significant
figures limiting factor. So we just have to calculate this and round to four significant figures. 72.06 divided by 180.156 is equal to, and if we round to four
significant figures, this will be .4000. So this will be, I'll say, approximately equal to 0.4000. Or we could say 40% or 40.00% carbon by mass when we round to four significant figures. And we are done.