Chemistry!!!?

Question

Menthol ( = 156.3 g/mol), a strong-smelling substance used in cough drops, is a compound of carbon, hydrogen, and oxygen. When 0.1595 g menthol was subjected to combustion analysis, it produced 0.449 g CO2 and 0.184 g H2O. What is its molecular formula?

Answer 1

look it up on wikipedia

http://en.wikipedia.org/wiki/Menthol

and could easily be solved by simply comparing the mole ratios of carbon dioxide and water

Answer 2

These problems have a lot of steps and take a while to explain and to work.

Start by taking the mass of CO2 and water and calculating the moles of C and moles of H produced when the compound is burned. So,

0.449 g CO2 (1 mol/44g) = 0.0102 mol C

0.184 g H2O (1 mol/18 g) (2 mol H/mol H2O) = 0.0204 mol H

Now, you have to figure out how much O was present in the original compound.

Convert the mol of C and the mol of H into grams, add them and then subtract that from the mass of the compound burned. The difference should be the mass of O in the sample:

0.0102 mol C (12 g/mol) = 0.1224 g
0.0204 mol H (1 g/mol) = 0.0204 g

Sum of mass C + mass H = 0.1428 g

0.1595 g - 0.1428 g = 0.0167 g O

Now convert grams of O into mol O:
0.0167 g (1 mol/16 g) = 0.00104 mol

Now, divide the moles of C H and O by the smallest number of moles of the three, which is moles of O. That gives:

0.0102 mol C / 0.00104 = 10 ( pretty near)
0.0204 mol H/ 0.00104 = 20 (pretty near)
0.00104 mol O/ 0.00104 = 1.

That says the empirical formula for the compound is C10H20O. The molar mass of that empirical formula would be about 156, so that IS the molecular formula.