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

Lesson 3: Elemental composition of pure substances

# 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?
• You apply sig figs to the end of a calculation not any intermediate steps.
• 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``
• 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.
• 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?
• 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.
• 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?
• 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.
• I have a question. How does significant figures work, like how do I know to round to 2 or 3? Thanks!
• dont u have to have to account for the sig figs during every step
(1 vote)
• You should be accounting for sig figs at the end of the calculation otherwise you’re introducing rounding errors.
• 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?
• 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.
• Why is 12.01 u = 12.01 g/mole of c?
• 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.