Then if the moles of CO2 and H2O are the same, then there are twice as many moles of H as C in the methanol.
CnH2nOx we have so far. where n is some number, and x is another number.
Now consider the law of mass conservation (sum masses reactants=sum mass products). From that , you know how much O2, and the moles of O2 were required.
Now you are approaching then end. You need to know the moles of O coming from the menthol.
from the masses and moles of the products, you can determine the total moles of O in the products, subtract twice the moles of O2 as a reactant: what is left? the moles of O in methol.
Now compare that to the number n. You have that already (number C moles in CO2 = .102=n notice x then is some number times n.
Now you have the formula ratio n, 2n, x
divide all of those subscripts by n
CH2Ox/n
now, probably x/n is a fraction. find some number which multiplied by x/n is a whole number. That is the common ratio to make the subscripts whole number ratios. Look up the structure of menthol, is is something like C10H20O
It has several isomers.
Combustion of a 0.1595 g of menthol
(156 g/mol) produces .449 g of Carbon Dioxide and 0.184 g of Water. What is the formula of menthol?
I am really confused as to where to start. I converted the Carbon Dioxide and Water to moles and got .102 mol for both of them but now I don't know what to do with those values and figure out how oxygen factors into it.
2 answers
Thank you!