What is the empirical formula for fructose given its percent composition: 40.00% #C#, 6.72% #H#, and 53.29% #O#?

1 Answer
May 26, 2016

Answer:

#CH_2O#

Explanation:

The empirical formula is the simplest whole number ratio defining constituent atoms in a species.

In all of these problems it is useful to assume #100# #g# of compund, and calculate the ATOMIC composition.

So, in #100*g# fructose, there are:

#(40.00*g)/(12.01*g*mol^-1)# #=# #3.33*mol*C#,

#(6.72*g)/(1.00794*g*mol^-1)# #=# #6.66*mol*H#,

#(53.29*g)/(15.999*g*mol^-1)# #=# #3.33*mol*O#,

If we divide thru by the smallest molar quantity, we get a formula of #CH_2O#.

The molecuar mass of fructose is #180.16*g*mol^-1#. How do we use these data to provide a molecular formula?