You perform a combustion analysis on a 255 mg sample of a substance that contains only C, H, and O, and you find that 561 mg of CO2 is produced, along with 306 mm of H2O.

if the substance contains only C, H, and O, what is the empirical formula

if the molar mass of the compound is 180 g/mol what is the molecular formula of the compound

The combustion of an organic compound is mostly written as,

CaHbOc + O2 - - > CO2 + H2O

where a, b, and c are supposed to be the subscripts of the elements C, H, and O in the compound. Determining the number of moles of C and H in the product which is the same as that in the compound,

(Carbon, C) : (561 mg) x (12/44) = 153 mg x (1 mmole/12 mg) = 12.75

(Hydrogen, H) : (306 mg) x (2/18) = 34 mg x (1 mmole/1 mg) = 34

Calculating for amount of O in the sample,

(oxygen, O) = 255 - 153 mg - 34 mg = 68 mg x (1mmole/16 mg) = 4.25

The empirical formula is therefore,

C (51/4) H34O17/4

C3H8O1

The molar mass of the empirical formula is 60. Therefore, the molecular formula of the compound is,

C9H24O3