Question

What is the empirical formula of a compound that contains 40 percent carbon, 6.7 percent hydrogen, and 53.3 percent oxygen

Answer 1

40g C 6.7g H 53.3g O -------------- 100g total Divide each element by its molar mass 40g C x 1mol / 12.01g = 3.33mol C 6.7g H x 1mol / 1.01g = 6.09mol H 53.3g O x 1mol/ 16.00g = 3.33mol O Divide each mole number by the smallest # of moles. In this case, 3.33 is the smallest. 3.33mol C / 3.33mol = 1 C 6.09mol H / 3.33mol = 1.83 H 3.33mol O / 3.33mol = 1 O Empirical Formula = CH2O

Answer 2

 The  empirical  formula  is  CH2O

calculation

 step 1:  calculate the   moles  of each  element   present  in  the compound.

moles =    %   composition  /  molar  mass

• from periodic  table  the  molar mass  for  carbon =12 g/mol  ,

                                         for hydrogen = 1 g/mol  

                                          for  oxygen  = 16 g/mol

• moles  is therefore

                                   for Carbon (C) =  40 /12  = 3.33 moles

                                    for    hydrogen (H) =  6.7 /1= 6.70   moles

                                    for   oxygen(O)  =  53.3 / 16=3.33  moles

Step 2 :  find the  mole ratio  of  C:H:O  by  diving each  mole by smallest  mole (  3.33 moles)

 that is   carbon (C) =  3.33/3.33 =1

                Hydrogen(H)  =   6.70 / 3.33=2

                 Oxygen(O)  =  3.33/3.33 =1

Therefore  the  empirical formula = CH2O