combustion analysis of 63.8 mg of C,H, and O containing compound produced 145.0 mg of CO2 and 59.38 mg of H2O. What is the empirical formula for the compound?

Answer 1

1. Convert 145.0 mg CO2 to grams C, then divide by 63.8 mg to obtain %C.
2. Convert 59.38 mg H2O to grams H, then divide by 63.8 mg to obtain
%H.
3. %O = 100% - %C - %H
4. Take a 100 g sample which will give you the percentages in #3 as grams.
5. Convert grams to mols. mols = g/atomic mass.
6. Find the ratio of the elements to each other with the smallest number being 1.00. The easy way to do that is to divide the smallest number by itself (thereby getting 1.000 for that one), then divide the other numbers by the same small number.
This is the empirical formula..
Post your work if you get stuck.

Answer 2

CHO

Answer 3

C3H6O