What is the empirical formula of a compound composed of 3.25% hydrogen (H), 19.36% carbon (C), and 77.39% oxygen (O) by mass?

2 Answers
May 7, 2016

#CH_2O_3#

Explanation:

We assume #100*g# of unknown compound, and work out the elemental proportions in terms of moles:

#H:# #(3.25*g)/(1.00794*g*mol^-1)# #=# #3.22*mol;#

#C:# #(19.36*g)/(12.01*g*mol^-1)# #=# #1.61*mol;#

#O:# #(77.39*g)/(15.99*g*mol^-1)# #=# #4.84*mol;#

We divide thru by the SMALLEST molar quantity, that of carbon, to give an empirical formula:

#CH_2O_3#

May 7, 2016

#CH_2O_3#

Explanation:

Assume that there is 100g of the substance, so there would be

#3.25g\ H# #19.36g\ C# and #77.39g\ O#

Divide each mass by the molar mass of their respective element.

#(3.25g)/1 H# #(19.36g)/12 C# #(77.39g) /16 O#

#=3.25 H# #=1.61 C# #=4.84 O#

Divide by the smallest number. In this case, #1.61#

#=2 H# #=1 C# #=3O#

So the empirical formula is #CH_2O_3#