## Determining an Emperical Formula using Combustion Analysis data

the problem im working with is:
Naphthalene is a carbon-hydrogen compound that finds use as mothballs. A sample of naphthalene is subjected to combustion analysis, producing 1.100g of CO2 and .1802g of H2O. Based on these data, calculate the empirical formula of naphthalene.

So when i set up the demensional analysis and get the moles of C and H i get for C .02499/
.02499 and for H i get .02000/.02499. For H the book multiplies the fraction for H by 5 because .02000/.02499 = .800320128 which is not close to a whole number.

My question is: How do they determine what the fraction is. Which is in this case is 4/5ths.

The Emperical Formula for napthalene is C5H4
 Admin 0.8=8/10=4/5. It is not always that obvious, but here it is pretty simple.
 Ya but how do i know when to multiply is by some fraction? Like .99999 is pretty obvious but what about .4154 or .5900 etc etc. What is the rule of thumb?

## Determining an Emperical Formula using Combustion Analysis data

 Quote by Crookedsky Ya but how do i know when to multiply is by some fraction? Like .99999 is pretty obvious but what about .4154 or .5900 etc etc. What is the rule of thumb?
At a simple level, just multiply by 2, 3, 4, 5 etc until you get reasonably close to whole numbers
 Ok thanks Sjb, Now i have a different problem that i have been working: Ethyl Alcohol, the alcohol present in alcoholic beverages, is a carbon-hydrogen-oxygen compound. Combustion analysis of a 1.000g sample of ethyl alcohol produces 1.913g of CO2 and 1.174g of H2O. Based on these data, calculate the empirical formula of ethyl alcohol. So i set the dimensional analysis and solve for grams of Carbon(.5220) and grams of Hydrogen(.1316). Then i add C and H up and get .6536g mass. Next i solve for O by sutracting .6536 from 1.000 and come up with a .3464g of O. Then I convert H, C, and O to moles and come up with .04346 mol of C, .130 mol of H, and .02165 mol of O. At this point either the book is wrong or i am wrong. I used the moles of C (.04346) as my divisor to calculate the ratio. I came up with an empirical formula of CH3O5 but the book uses the moles of O (.02165) as the divisor and comes up with a empirical formula of C2H6O. Can someone check my calculations because i have done this problem twice. Thanks
 What is .04346 / .02165 ?
 Admin Numbers of moles you calculated so far are OK. Answer sjb's question.