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.

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.

you need to stop taking that class