The actual molecular formula of Vitamin C is C6H8O6
To begin we can start by subtracting the final mass of the CO2 absorber after combustion from the initial would give 1.500g of CO2 that was formed from the reaction.
We can use this mass of CO2 to determine how much carbon was inside Vitamin C by doing the following:
1.500 grams of CO2*(12.01 g/mol of C/44.01 g/mol of CO2)
to get the mass of carbon. THe units should cancel out and leave us with .4093 grams of carbon.
Now we do the same with water with the following equation which is basically the same method except we use the mass of water and hydrogen instead.
.410 grams of H2O absorbed*(2.016 g/mol H/18.016 g/mol H2O) = .0459 grams of hydrogen
Don't forget that there are TWO hydrogen atoms in 1 water molecule. This is crucial so you don't mess up the calculation later on.
Now we can use this to find the mass of oxygen that was in the compound before combustion. We find oxygen last because in combustion analysis, oxygen is also a reactant in the form of O2 as well as from vitamin C.
To find the mass of oxygen we do 1.000g - (.0459g + .4093g) = .5449g of oxygen
Now we have the mass of all three elements in the compound vitamin C
.0459 g of hydrogen, .5449 g of oxygen, and .4093 g of carbon
We convert each to moles now and get .0455 moles of hydrogen, .03406 moles of oxygen, and .03408 moles of carbon
Using this we can write a formula of C.03408 H.0459 O.03406
Divide each of the number by the lowest mole number which is .03408...approximately C1H1.35O which is like an empirical formula of vitamin C.
Find the molar mass of CH1.35O and you would get 29.3708 g/mol
Now we can divide vitamin C molar mass which is 176.12 by the 29.3708 to get 6 (this number is unitless).
Multiply each of the subscripts in CH1.35O by 6 and you will get C6H8O6
P.S. your teacher is crazy as hell. This is a type of question in a college level introductory chem course -_- even at the top colleges. I aced chem and currently taking organic chem as part of my prerequisites for my major and medical school. At least this is good practice and review for me for MCAT