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## Molecular composition

Current time:0:00Total duration:5:38

# Worked example: Calculating mass percent

AP Chem: SPQ‑2 (EU), SPQ‑2.A (LO), SPQ‑2.A.2 (EK)

## 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.