7.321 mg of an organic compound containing carbon, hydrogen, and oxygen was analyzed by combustion. the amount of carbon dioxide produced was 17.873 mg and the amount of water produced was 7.316 mg. determine the empirical formula of the compound.

The molecular formula is C₄H₈O.

We must calculate the masses of C, H, and O from the masses given.

Mass of C = 17.873 mg CO₂ × (12.01 mg C/44.01 mg CO₂) = 4.8774 mg C

Mass of H = 7.316 mg H₂O × (2.016 mg H/18.02 mg H₂O) = 0.818 48 mg H

Mass of O = Mass of compound - Mass of C - Mass of H

= (7.321 – 4.8774 – 0.818 48) mg = 1.6251 mg

Now, we must convert these masses to moles and find their ratios.

From here on, I like to summarize the calculations in a table.

Element     m/mg      n/mmol     Ratio    Integers  

     C          4.877 4    0.406 11   3.9984         4

     H         0.818 48   0.811 99   7.9944         8

     O         1.625 1      0.101 57   1                    1

The empirical formula is C₄H₈O.

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.