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 going to tell you it's 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 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 say let's say 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 6 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 just be 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 6 moles of carbon 12 moles of hydrogen and 6 moles of oxygen so it's going to be what we just had up here it's going to be 6 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 6 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 going to end up with grams in the numerator grams and denominator the units will cancel out and we'll get a pure percentage at the end so let's see in the numerator 6 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 going to worry about significant figures so the denominator we have 72.06 plus see 12 times 1.008 is 12.096 and then we have plus 6 times 16 is 96.00 zero 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 had 72 to 12 i get 84 plus 96 i get 180.156 did i do that right if i were to just 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 hundreds 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 would 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 is equal to and if you round to four significant figures this will be point four zero zero zero so this will be i'll say approximately equal to zero point four zero or we could say 40 or 40.00 percent carbon by mass when we round to four significant figures and we are done